LNTwww - User contributions [en]

LNTwww:Glossary

2023-04-30T18:17:43Z

Noah:

{{BlaueBox|TEXT=
Due to the fact,  that our e–learning project LNTwww was first conceived in German and the wish for an English version came much later,  in the English version the assignment between  »Formula signs«   and  »Designation«  is not quite easy.   The following alphabetically ordered entries can help in this case:
::  »Formula sign«   ⇒   »German name«   ⇒   »English name« 

* First select from the list below the category to which the "formula sign" you are looking for belongs.

*A few explanations are given under the last menu item  »Some remarks to the Glossary«.}}

===Upper case letters  A, ... , G ===
 
  $A$   ⇒   $(1)$  Ereignis,  $(2)$  Impulsamplitude   ⇒   $(1)$  event,  $(2)$  pulse amplitude 
:  $\overline{A}$   ⇒   Komplementärmenge des Ereignisses  $A$   ⇒   complementary set of event  $A$ 
:  $A_0$   ⇒   Gleichsignalkoeffizient der Fourierreihe   ⇒   DC coefficient of Fourier series 
:  $A_n$   ⇒   $n$–ter Cosinuskoeffizient der Fourierreihe   ⇒   $n^{\rm th}$  cosine coefficient of Fourier series 

  $B$   ⇒   einseitige Bandbreite   ⇒   one-sided bandwidth 
:  $B_{\rm K}$   ⇒   einseitige Kanalbandbreite  $($des Kanals$)$  ⇒   one-sided channel bandwidth 
:  $B_{x}$   ⇒   einseitige Bandbreite  des Signals  $x(t)$  ⇒   one-sided bandwidth of signal  $x(t)$ 
:  $B_n$   ⇒   $n$–ter Sinuskoeffizient der Fourierreihe   ⇒   $n^{\rm th}$  sine coefficient of Fourier series 

  $C$   ⇒   Kanalkapazität   ⇒   channel capacity 
:  $C_{\rm A}$   ⇒   Kanalkapazität bei Amplitudenbegrenzung   ⇒   channel capacity under peak-value limitation 
:  $C_{\rm L}$   ⇒   Kanalkapazität bei Leistungsbegrenzung   ⇒   channel capacity under power limitation 
:  $C_0=A_0$   ⇒   Gleichsignalkoeffizient der Fourierreihe   ⇒   DC coefficient of Fourier series 
:  $C_n$   ⇒   $n$–ter Betragskoeffizient der Fourierreihe   ⇒   $n^{\rm th}$  magnitude coefficient of Fourier series 
:  $C_x(\Omega)$   ⇒   charakteristische Funktion der Zufallsgröße  $x$   ⇒   characteristic function   of random variable  $x$  $($Fourier retransform of PDF$)$ 

  $D$   ⇒   Dummy-Variable für "Verzögerung"   ⇒   dummy variable for "delay" 
:  $D_{\mu}$   ⇒   Spektralkoeffizienten der DFT   ⇒   spectral coefficients of the DFT 

  $ E$   ⇒   $(1)$  Schwellenwert,  $(2)$  Energie,  $(3)$  Ergebnis eines Zufallsexperiments   ⇒   $(1)$  threshold value,  $(2)$  energy, $(3)$  outcome of a random experiment 

:  $ E_{\rm opt}$   ⇒   optimaler Schwellenwert   ⇒   optimum threshold value 
:  $ E_{\mu}$   ⇒   Schwellenwerte eines mehrstufigen Systems   ⇒   thresholds of a multilevel system  $g(t)$ 

:  $ E_{g}$   ⇒   Energie des Impules  $g(t)$   ⇒   energy of pulse  $g(t)$ 
:  $ E_{\rm B}$   ⇒   Energie pro Bit,  Bitenergie   ⇒   energy per bit 
:  $ E_{\rm S}$   ⇒   Energie pro Symbol,  Symbolenergie   ⇒   energy per symbol 

:  $ {\rm E}(x)$   ⇒   Erwartungswert  der Zufallsgröße  $x$   ⇒   expected value  of random variable  $x$ 
:  ${\rm E}\big[g (x ) \big] = \int_{-\infty}^{+\infty} g(x)\cdot f_{x}(x) \,{\rm d}x$   ⇒   Erwartungswert  $($der mit  $g(x)$  gewichteten Zufallsgröße  $x)$   ⇒   expected value  of random variable  $x$ weighted by  $g(x)$ 

  $F$   ⇒   Rauschzahl   ⇒   noise figure 
:  $F_{x}(r) ={\Pr}(x \le r)$   ⇒   Verteilungsfunktion  $\rm (VTF)$  der Zufallsgröße  $x$   ⇒   cumulative distribution function  $\rm (CDF)$  of random variable  $x$ 

  $G$   ⇒     $(1)$  Störabstandsgewinn in dB,  $(2)$ Grundmenge   ⇒   $(1)$  signal-to-noise ratio gain in dB,  $(2)$  universal set 

:  $G(D)$   ⇒     Generatorpolynom   ⇒   generator polynomial 
:  $G_{\rm R}(D)=D^{L}\cdot G(D^{-1})$   ⇒     reziprokes Polynom des Generatorpolynoms  $G(D)$   ⇒   reciprocal polynomialof the generator polynomial  $G(D)$ 

:  $G(f)$   ⇒   Grundimpulsspektrum   ⇒   spectrum of the basic pulse  $g(t)$ 
:  $G_d(f)$   ⇒   Detektionsgrundimpulsspektrum   ⇒   spectrum of the basic detection pulse  $g_d(t)$ 
:  $G_r(f)$   ⇒   empfangsgrundimpulsspektrum   ⇒   spectrum of the basic receiver pulse  $g_r(t)$ 
:  $G_s(f)$   ⇒   Sendegrundimpulsspektrum   ⇒   spectrum of the basic transmission pulse  $g_s(t)$ 

===Upper case letters  H, ... , O ===
 
  $\rm H$   ⇒   Symbolwert  »High«  einer binären Zufallsgröße $z \in \{ \text{L, H}\}$   ⇒   symbol value  »High«  of a binary random variable  $z \in \{ \text{L, H}\}$ 

:  $H(f)$   ⇒   Frequenzgang, Übertragungsfunktion   ⇒   frequency response, transfer function 
:  $H_{\rm E}(f)$   ⇒   Empfängerfrequenzgang   ⇒   receiver frequency response 
:  $H_{\rm K}(f)$   ⇒   Kanalfrequenzgang   ⇒   channel frequency response 
:  $H_{\rm MF}(f)$   ⇒   Frequenzgang des Matched-Filters  ⇒   frequency response of the Matched Filter 
:  $H_{\rm S}(f)$   ⇒   Senderfrequenzgang   ⇒   transmitter frequency response 
:  $|H(f)|$   ⇒   Betragsfrequenzgang   ⇒   magnitude frequency response 

:  $H(X)$   ⇒   Quellenentropie   ⇒   source entropy 
:  $H(Y)$   ⇒   Sinkenentropie   ⇒   sink entropy 
:  $H(X|Y)$   ⇒   Äquivokation   ⇒   equivocation 
:  $H(Y|X)$   ⇒   Irrelevanz   ⇒   irrelevance 
:  $H(XY)$   ⇒   Verbundentropie   ⇒   joint entropy 

  $I$ 
:  $I(X; Y)$   ⇒   Transinformation   ⇒   mutual information 

  $J$   ⇒   Spreizfaktor   ⇒   spreading factor 

  $K = \mu_4/σ^4$   ⇒   Kurtosis   ⇒   kurtosis 

  $\rm L$   ⇒   Symbolwert  »Low«  einer binären Zufallsgröße $z \in \{ \text{L, H}\}$   ⇒   symbol value  »Low«  of a binary random variable  $z \in \{ \text{L, H}\}$ 

  $ M$   ⇒   $(1)$  Symbolumfang,  $(2)$  Stufenzahl   ⇒   $(1)$  symbol set size,  $(2)$  level number 
:  $ M_c$   ⇒   Stufenzahl des Codersignals   ⇒   level number of the encoded signal 
:  $ M_q$   ⇒   Stufenzahl des Quellensignals   ⇒   level number of the source signal 

  $ N$  ⇒   Dimension des Signal-Vektorraums   ⇒   dimension of the signal vector space 
:  $ N_0$   ⇒   physikalische Rauschleistungsdichte  $($einseitig$)$   ⇒   physical noise power density  $($one-sided$)$ 
:  $ N_0/2$   ⇒   systemtheoretische Rauschleistungsdichte  $($zweiseitig$)$   ⇒   system– theoretical noise power density  $($two-sided$)$ 

  $\mathcal{O}$  ⇒   Anzahl der Operationen eines Algorithmus   ⇒   number of operations of an algorithm 

===Upper case letters  P, ... , Z===
 
  $ P$   ⇒   $(1)$  Leistung,    $(2)$  Periodendauer   ⇒   $(1)$  power,  $(2)$  period duration 

:  $P_{\rm max} = 2^L - 1$   ⇒   maximale Periodendauer eines Schieberegisters der Länge  $L$   ⇒   maximum period of a shift register with length  $L$ 

:  $ P_{x}$   ⇒   Leistung des Signals  $x(t)$   ⇒   power of the signal  $x(t)$ 

:  $P_{\hspace{0.01cm}Y\hspace{0.03cm} \vert \hspace{0.01cm}X}(Y\hspace{0.03cm} \vert \hspace{0.03cm} X)$   ⇒   Matrix bedingter Wahrscheinlichkeiten   ⇒   conditional probability matrix 
:  $P_{XY}(X,\hspace{0.1cm}Y)$   ⇒   Verbundwahrscheinlichkeitsmatrix   ⇒   joint probability matrix 
:  $P_{\hspace{0.01cm}X\hspace{0.03cm} \vert \hspace{0.03cm}Y}(X\hspace{0.03cm} \vert \hspace{0.03cm} Y)$   ⇒   Rückschlusswahrscheinlichkeitsmatrix   ⇒   inference probability matrix 

:  $P_X(X)$   ⇒   Wahrscheinlichkeitsfunktion der Zufallsgröße  $X$   ⇒   probability mass function  $\rm (PMF)$  of random variable  $X$ 
:  $P_{\hspace{0.01cm}Y\hspace{0.03cm} \vert \hspace{0.01cm}X}(Y\hspace{0.03cm} \vert \hspace{0.03cm} X)$   ⇒   Übergangswahrscheinlichkeitsmatrix der Zufallsgröße  $X$   ⇒   transition probabilitiy matrix  of random variable  $X$ 
:  $P_{\hspace{0.01cm}X\hspace{0.03cm} \vert \hspace{0.01cm}Y}(X\hspace{0.03cm} \vert \hspace{0.03cm} Y)$   ⇒   ????? Übergangswahrscheinlichkeitsmatrix der Zufallsgröße  $X$   ⇒   transition probabilitiy matrix  of random variable  $X$ 
:  ${\rm Pr} (A_i)$   ⇒   Wahrscheinlichkeit des Ereignisses  $A_i$   ⇒   probability of event  $A_i$ 
:  ${\rm Pr} (A \cup B)$   ⇒   Wahrscheinlichkeit der Vereinigungsmenge von  $A$  und $B$   ⇒   probability of the union set of  $A$  and $B$. 
:  ${\rm Pr} (A \cap B)$   ⇒   Wahrscheinlichkeit der Schnittmenge von  $A$  und $B$   ⇒   probability of the intersection set of  $A$  and $B$. 
:  ${\rm Pr} (A \hspace{0.03cm} \vert \hspace{0.01cm} B)$   ⇒   bedingte Wahrscheinlichkeit von  $A$  unter der Bedingung  $B$   ⇒   conditional probability of  $A$  under the condition  $B$ 

  $ R$   ⇒   Rate   ⇒   rate 
:  $ R_{\rm C}=k/n$   ⇒   Coderate bei Blockcodes   ⇒   code rate for block codes 

  $S=μ3/σ3$   ⇒   Charliersche Schiefe   ⇒   Charlier's skewness 

  $ T$   ⇒   (1) Symboldauer,  (2) ???   ⇒   (1) symbol duration,  (2) ??? 
:  $ T_{\rm A}$   ⇒   Abtastintervall  $x(t)$   ⇒   sampling interval  $x(t)$ 
:  $ T_{\rm B}$   ⇒   Bitdauer  $x(t)$   ⇒   bit duration  $x(t)$ 
:  $ T_{\rm D}$   ⇒   Detektionszeitpunkt  $x(t)$   ⇒   detection time  $x(t)$ 
:  $ T_{\rm M}$   ⇒   Messdauer  $x(t)$   ⇒   measure duration  $x(t)$ 

===Lower case letters  a, ... , g ===
 
  $a(f)$   ⇒   Dämpfungsfunktion  ⇒   attenuation function 
:  $a_{\rm K}(f)$   ⇒   Dämpfungsfunktion  $($eines Kabels$)$  ⇒   attenuation function  $($of a cable$)$ 
:  $a_0$   ⇒   Gleichsignaldämpfung  ⇒   DC signal attenuation 
:  $a_\star$   ⇒   charakteristische Kabeldämpfung  $($bei halber Bitrate$)$  ⇒   characteristic cable attenuation value  $($at half bitrate$)$ 
:  $ a_\nu $   ⇒   Amplitudenkoeffizient  $($sendeseitig$)$   ⇒   amplitude coefficient  $($transmitter side) 
:  $ a_\nu '$   ⇒   Amplitudenkoeffizient  $($empfängerseitig$)$   ⇒   amplitude coefficient  $($receiver side$)$ 
:  $\langle a_\mu \rangle$   ⇒   zeitliche Folge der Amplitudenkoeffizienten   ⇒   temporal sequence of amplitude coefficients 
:  $ \{ a_\nu \} $   ⇒   Menge der möglichen Amplitudenkoeffizienten   ⇒   set of possible amplitude coefficients

  $b(f)$   ⇒   Phasenfunktion   ⇒   phase function 
:  $b_{\rm K}(f)$   ⇒   Phasenfunktion  $($eines Kabels$)$   ⇒   phase function  $($of a cable$)$ 

  $c= 3 \cdot 10 ^8\ \rm m/s$   ⇒   Lichgeschwindigkeit   ⇒   velocity of light 
:  $c(t)$   ⇒   Codersignal   ⇒   encoded signal 
:  $ \langle c_\nu \rangle$   ⇒   Codesymbolfolge   ⇒   encoded symbol sequence 
:  $ \{ c_\mu \}$   ⇒   Codesymbolvorrat   ⇒   encoded symbol set 

  $d$   ⇒   Leitungsdurchmesser   ⇒   line diameter 
:  $d(t)$   ⇒   Detektionssignal   ⇒   detection signal 
:  $d_\nu$   ⇒   $(1)$  Detektionsabtastwert,  $(2)$  Zeitkoeffizienten der DFT  ⇒   $(1)$  detection sample value,  $(2)$  time coefficients of the DFT 

:  $d_{\rm N}(t)$   ⇒   Detektionsstörsignal   ⇒   detection noise signal 
:  $d_{\rm N\nu}$   ⇒   Detektionsstörabtastwert   ⇒   detection noise sample value 
:  $d_{\rm S}(t)$   ⇒   Detektionsnutzsignal   ⇒   useful detection signal 
:  $d_{\rm S\nu}$   ⇒   Detektionsnutzabtastwert   ⇒   useful detection sample value 
:  $d_{\rm H}(\underline{x}, \ \underline{x}\hspace{0.03cm}')$   ⇒   Hamming–Distanz zwischen den Codeworten  $\underline{x}$  und  $\underline{x}\hspace{0.03cm}'$   ⇒   Hamming–Distance between codewords  $\underline{x}$  and  $\underline{x}'$ 

$e= 2.718281828456$...   ⇒   Eulersche Zahl   ⇒   Eulerian number 

  $f$   ⇒   Frequenz   ⇒   frequency 
:  $f_{\rm G}$   ⇒   Grenzfrequenz   ⇒   cutoff frequency 
:  $f_{\rm Nyq}$   ⇒   Nyquistfrequenz   ⇒   Nyquist frequency 
:  $f_{\rm T}$   ⇒   Trägerfrequenz   ⇒   carrier frequency 
:  $f_{x}(x)$   ⇒   Wahrscheinlichkeitsdichtefunktion  $\rm (WDF)$  der Zufallsgröße  $x$   ⇒   probability density function  $\rm (PDF)$  of the random variable  $x$ 
:  $f_{X}(X=x)$   ⇒   exaktere Schreibweise der WDF mit Zufallsgröße  $X$  und Realisierung  $x$   ⇒   more exact notation of WDF with random variable  $X$  and realization  $x$ 
  $g$ 
:  $g_l$   ⇒   Rückkopplungskoeffizienten eines Schieberegisters   ⇒   feedback coefficients of a shift register 

:  $g(t)$   ⇒   Grundimpuls   ⇒   basic pulse 
:  $g_d(t)$   ⇒   Detektionsgrundimpuls   ⇒   basic detection pulse 
:  $g_r(t)$   ⇒   Empfangssgrundimpuls   ⇒   basic receiver pulse 
:  $g_s(t)$   ⇒   Sendegrundimpuls   ⇒   basic transmission pulse 

===Lower case letters  h, ... , o ===
 
  $h$   ⇒   Modulationsindex bei FSK   ⇒   modulation index at FSK 
:  $h(t)$   ⇒   Impulsantwort   ⇒   impulse response 
:  $h_{\rm MF}(t)$   ⇒   Impulsantwort des Matched-Filters  ⇒   impulse response of the Matched Filter 
  $i$   
:  $i(t)$   ⇒   Stromverlauf   ⇒   current curve 

  $\rm j$   ⇒   imaginäre Einheit   ⇒   imaginary unit 

  $k$   ⇒   Informationsblocklänge bei Blockcodes   ⇒   information block length for block codes 
:  $k_{\rm B}= 1.38 \cdot 10 ^{23}\ \rm Ws/s$   ⇒   Boltzmann–Konstante   ⇒   Boltzmann's constant 

  $l$   ⇒   Leitungslänge   ⇒   line length 
:  $l_{\rm max}$   ⇒   maximale Leitungslänge   ⇒   maximum line length 

  $m$   ⇒   Anzahl der Paritybit bei Blockcodes   ⇒   Number of paritybits for block codes 
:  $m_k = {\rm E}\big[x^k \big]$   ⇒   Moment  $k$–ter Ordnung   ⇒   moment of order  $k$ 
:  $m_1$   ⇒   erstes Moment  $($Mittelwert$)$   ⇒   first moment  $($mean$)$ 
:  $m_2$   ⇒   zweites Moment  $($Leistung$)$   ⇒   second moment  $($power$)$ 
:  $m_i \hspace{0.1cm} \Leftrightarrow \hspace{0.1cm} s_i(t)$   ⇒   Nachrichten,  die den Signalen  $s_i(t)$  zugeordnet sind   ⇒   Messages,  associated with the signals  $s_i(t)$ 

  $n$   ⇒   Codewortlänge bei Blockcodes   ⇒   code word length for block codes 

  $o$   
===Lower case letter  ö ===
 
  ${\ddot{o}(t)}$   ⇒   vertikale Augenöffnung   ⇒   $($vertical$)$  eye opening 
:  ${\ddot{o}(T_{\rm D})}$   ⇒   vertikale Augenöffnung zum Detektionszeitpunkt  $T_{\rm D}$   ⇒   $($vertical$)$  eye opening at detection time  $T_{\rm D}$ 
:  ${\ddot{o}_{\rm norm}(T_{\rm D})}$   ⇒   normierte Augenöffnung zum Detektionszeitpunkt  $T_{\rm D}$   ⇒   normalized eye opening at detection time  $T_{\rm D}$ 

===Lower case letters  p, ... , z ===

:  $p_\mu = {\rm Pr}(z=\mu) $   ⇒   mögliche Wahrscheinlichkeiten einer wertdiskreten Zufallsgröße  $z$   ⇒   possible probabilities of a discrete-value random variable  $z$ 
:  $p_{\rm b\hspace{0.03cm}\vert \hspace{0.03cm}A} = {\rm Pr}(Y\hspace{-0.1cm} = {\rm b}\hspace{0.05cm}\vert X \hspace{-0.1cm}= {\rm A}) $   ⇒   Verfälschungswahrscheinlichkeit   ⇒   falsification probability 
:  $p_{\rm B\hspace{0.03cm}\vert \hspace{0.03cm}a} = {\rm Pr}(X\hspace{-0.1cm} = {\rm B}\hspace{0.05cm}\vert y \hspace{-0.1cm}= {\rm a}) $   ⇒   Verfälschungswahrscheinlichkeit   ⇒   inference probability 

  $q(t)$   ⇒   Quellensignal   ⇒   source signal, data signal 
:  $ \langle q_\nu \rangle$   ⇒   Quellensymbolfolge   ⇒   source symbol sequence 
:  $ \{ q_\mu \}$   ⇒   Quellensymbolvorrat   ⇒   source symbol set 

  $r$   ⇒   $(1)$  relative Redundanz,   $(2)$  Rolloff–Faktor   ⇒   relative redundancy,   $(2)$  rolloff factor   
:  $\rm random()$   ⇒   C-Aufruf eines Zufallsgenerator für Gleichverteilung   ⇒   C-function of a random number generator for uniformly distributed random variables 
:  $r(t)$   ⇒   Empfangssignal   ⇒   received signal 
:  $r_{\rm TP}(t)$   ⇒   äquivalentes Tiefpass–Empfangssignal   ⇒   equivalent low-pass received signal 

  $s(t)$   ⇒   Sendesignal   ⇒   transmitted signal 
:  $s_{\rm TP}(t)$   ⇒   äquivalentes Tiefpass–Sendesignal   ⇒   equivalent low-pass transmitted signal 
:  $s_{\rm +}(t)$   ⇒   analytisches Sendesignal   ⇒   analytic transmitted signal 
:  $\mathbf{s}_i = \big( s_{i1}\hspace{0.05cm}, \hspace{0.3cm}s_{i2}\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm},\hspace{0.05cm} s_{iN} \big )$   ⇒   vektorieller Repräsentant der Musterfunktion  $s_i(t)$   ⇒   vectorial representative of the pattern function  $s_i(t)$ 
:  $\vert \vert s_1(t) \vert \vert$   ⇒   Euklidische Norm der Zeitfunktion  $s_1(t)$   ⇒   Euclidean norm of the time function  $s_1(t)$ 
:  ${\rm sign}(t)$   ⇒   Signumfunktion   ⇒   signum function 
:  ${\rm si}(x)= \sin(x)/x={\rm sinc}(\pi \cdot x)$   ⇒   si–Funktion   ⇒   si function 
:  ${\rm sinc}(x)= \sin(\pi x)/(\pi x)={\rm si}(x/\pi)$   ⇒   sinc–Funktion   ⇒   sinc function 

  $t$   ⇒   Zeit   ⇒   time 
:  $t_\nu$   ⇒   zeitliche Folge der Detektionszeitpunkte   ⇒   time 

  $u(t)$   ⇒   Spannungsverlauf   ⇒   voltage curve 

  $ v(t)$   ⇒   Sinkensignal   ⇒   sink signal 
:  $\langle v_\mu \rangle$   ⇒   Sinkensymbolfolge   ⇒   sink symbol sequence 
:  $ \{ v_\nu \}$   ⇒   Sinkensymbolvorrat   ⇒   sink symbol set 

:  $ w_{\rm H}(\underline{x})$   ⇒   Hamming–Gewicht des Codewortes  $\underline{x}$   ⇒   Hamming weight of code word  $\underline{x}$ 
  $x(t)$   ⇒   Eingangssignal   ⇒   input signal 
:  $x_{\rm g}(t)$   ⇒   gerader Anteil des Signals  $x(t)$   ⇒   even portion of the signal  $x(t)$ 
:  $x_{\rm u}(t)$   ⇒   ungerader Anteil des Signals  $x(t)$   ⇒   odd portion of the signal  $x(t)$ 
:  $<\hspace{-0.01cm}x(t), \hspace{0.05cm}y(t) \hspace{-0.01cm}>$   ⇒   inneres Produkt der Signale  $x(t)$  und  $y(t)$   ⇒   inner product of signals  $x(t)$  and  $y(t)$ 

:  $x_i(t)$   ⇒   $i$-tes Mustersignal eines Zufallsprozesses   ⇒   $i$-th pattern signal of a random process 
:  $\{x_i(t)\}$   ⇒   Zufallsprozess   ⇒   random process 

  $y(t)$   ⇒   Ausgangssignal   ⇒   output signal 

=== Upper case greek letters and special characters $(??? \text{...})$ ===

:  $\Delta f$   ⇒   äquivalente Bandbreite   ⇒   equivalent bandwidth 
:  $\Delta t$   ⇒   äquivalente Zeitdauer der Impulsantwort   ⇒   equivalent duration of the impulse response 
:  $\Delta t_{\rm S}$   ⇒   äquivalente Sendeimpulsdauer   ⇒   equivalent pulse duration 

:  $\nabla f$   ⇒   ???   ⇒   ???? 
:  $\square f$   ⇒   äquivalente Rauschbandbreite   ⇒   equivalent noise bandwidth 

  ${\it \Phi}_s(f)$   ⇒   Leistungsdichtespektrum  $\rm (LDS)$  des Sendesignals  $s(t)$    ⇒   power-spectral density  $\rm (PSD)$  of the transmitted signal  $s(t)$ 
  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{gs}(f) = |G_s(f)|^2
$   ⇒   Energiespektrum  des Sendegrundimpulses  $g_s(t)$    ⇒   Energy spectrum  of the basic transmission pulse  $g_s(t)$ 

=== Lower case greek letters  $(\alpha, \beta, \text{...})$ ===

  $\alpha$   ⇒   Dämpfungsfaktor   ⇒   attenuation factor 
:  $\alpha(f)=a(f)/l$   ⇒   Dämpfungsmaß   ⇒   attenuation function per unit length 

  $\beta$   ⇒   ???   ⇒   ''' 
:  $\beta(f)=b(f)/l$   ⇒   Phasenmaß   ⇒   phase function per unit length 

:  $\gamma(f)=\alpha(f) + {\rm j} \cdot \beta(f)$   ⇒   Übertragungsmaß   ⇒   complex propagation function per unit length 
:  $\gamma(x)$   ⇒   Sprungfunktion   ⇒   step function 
:  $\gamma_0(x)$   ⇒   ähnliche Funktion wie  $\gamma(x)$,  aber nicht identisch   ⇒   similar function as  $\gamma(x)$,  but not identical 

  $\delta(t)$   ⇒   Diracfunktion, Diracimpuls   ⇒   Dirac delta function, Dirac delta impulse 
  $\delta_{jk}$   ⇒   Kronecker–Symbol   ⇒   Kronecker icon 

  $\theta$   ⇒   absolute Temperatur in  "Kelvin"   ⇒   absolute temperature in  "Kelvin" 
  $\theta_k(t)$   ⇒   Hilfsfunktion für die Gram-Schmidt-Methode   ⇒   auxiliary function for the Gram-Schmidt method 

  $\kappa$  

  $\lambda$   ⇒   Rate der Poisson–Verteilung  $($Anteil der Einsen pro Zeitintervall$)$   ⇒   rate of the Poisson distribution  $($proportion of "ones" per time interval$)$ 

:  $\mu_k = {\rm E}\big[(x-m_{\rm 1})^k\big]$   ⇒   Zentralmoment  $k$–ter Ordnung   ⇒   central moment of order  $k$ 

  $\rho$   ⇒   Signal-zu-Rauschverhältnis  $\rm (SNR)$   ⇒   signal-to-noise ratio  $\rm (SNR)$ 
  $10 \cdot \lg\ \rho$   ⇒   Signal-zu-Rauschabstand in dB   ⇒   signal-to-noise ratio in dB 
:  $\rho_d$   ⇒   SNR des Detektionssignals  $d(t)$   ⇒   signal-to-noise ratioof the detection signal  $d(t)$ 
:  $\rho_d(T_{\rm D})$   ⇒   SNR für den Detektionszeitpunkt  $T_{\rm D}$   ⇒   SNR for detection time  $T_{\rm D}$ 

:  $\sigma_x$   ⇒   Streuung der Zufallsgröße  $x$   ⇒   standard deviation of the random variable  $x$ 
:  $\sigma_x^2$   ⇒   Varianz der Zufallsgröße  $x$   ⇒   variance of the random variable  $x$ 

  $\tau$   ⇒   Laufzeit   ⇒   delay time 

  $\phi$   ⇒   leere Menge  $({\rm Pr}(\phi) = 0)$   ⇒   empty set  $({\rm Pr}(\phi) = 0)$ 
:  $\phi(x)$   ⇒   Gaußsches Fehlerintegral   ⇒   Gaussian error integral 

  $\varphi = -\phi$   ⇒   Nullphasenwinkel   ⇒   zero phase angle 
:  $\varphi_x(t_1,t_2)$   ⇒   allgemeine Definition der Autokorrelationsfunktion  $\rm (AKF)$    ⇒   general definition of the auto-correlation function  $\rm (ACF)$ 
:  $\varphi_{x}(\tau)$   ⇒   Autokorrelationsfunktion eines ergodischen Zufallsprozesses  $\{x_i(t)\}$   ⇒   auto-correlation function of an ergodic random process  $\{x_i(t)\}$ 
:  $\varphi_{j}(t)$   ⇒   orthonormale Basisfunktionen zur Signalbeschreibung   ⇒   orthonormal basis functions for signal description 
:  $\varphi^{^{\bullet} }_{gs}(\tau)$   ⇒   Energie–AKF des Sendegrundimpulses  $g_s(t)$   ⇒   energy ACF of the basic transmission pulse  $g_s(t)$ 

 
=== Some remarks to the Glossary ===
{{BlaueBox|TEXT=
$\text{Note:}$  
#The categories are arranged alphabetically,  starting with  »uppercase letters«,  then  »lowercase letters« and finally  »upper and lowercase Greek letters«.
#The German umlaut  »$\rm {\ddot{o} }$«  is assigned its own category;  for example,   ${\ddot{o}(T_{\rm D})}$  stands for  »eye opening at detection time  $T_{\rm D}$«.
#Within a letter  $($e.g. »$\rm A«)$,  the order is no longer alphabetical,  but happens according to thematically related terms.
#If a  »formula sign«  has two different meanings like  »$E$«,  then both are specified;  here:  $(1)$  »threshold value«,  $(2)$  »energy«. }}

LNTwww:About LNTwww

2023-04-30T18:14:47Z

Noah:

==Welcome to the English version of LNTwww==
 
»$\text{https://en.lntwww.de}$«  is an e-learning tutorial for Communications Engineering with nine didactic multimedia textbooks including exercises with solutions,  learning videos,  and interactive applets.  It is offered by the  »[https://www.ce.cit.tum.de/en/lnt/home/ Institute for Communications Engineering]«  of the  »[https://www.tum.de/en/ Technical University of Munich]«. 

:⇒   '''It is freely accessible,  registration is not necessary and no system requirements are needed'''.

The German-language version   »$\text{https://www.lntwww.de}$«   ⇒   »$\rm L$erntutorial für $\rm N$achrichten$\rm T$echnik im $\rm w$orld $\rm w$ide $\rm w$eb«  was created between 2001 – 2021 by members of our Institute.  The toolbar entry  »Deutsch«  takes you to the German original.  In spring 2020 we started the English translation,  and in spring 2023 we finished.

*The current version from 2023 is based on the software  [https://en.wikipedia.org/wiki/MediaWiki »MediaWiki«],  known by the encyclopaedia  »WIKIPEDIA«.   The following is a kind of  »user guide«  to our e–learning project.  Corresponding links to this file  »About LNTwww«  can be found at the bottom of each page between  »Privacy policy« and  »Disclaimer«.

*You can find more information on our e-learning project in the PDF document  [https://www.lntwww.de/downloads/Sonstiges/HER-Beitrag_LNTwww.pdf »LNTwww - Praxisbericht zum E-Learning aus den Ingenieurwissenschaften«],  a German-language contribution by Günter Söder in the special issue  »So gelingt E-Learning«,  Reader zum Higher Education Summit 2019,  Munich:  Pearson Studium.

*We consider the present version as final;  an extension is currently not planned.  But of course we will continue to improve detected errors or inaccuracies promptly.  So if you notice any inadequacies regarding content,  presentation or handling,  then please send a detailed message by mail to  »LNTwww@ice.cit.tum.de«.

*On the  [[LNTwww:Information|»Information«]]  page you will find notes about temporary restrictions  $($e.g. in case of unavailability due to service work$)$  and a list of bugs already detected by us,  but not yet fixed.   We wish  that in this list there are only few entries and only for a short time.

We would be pleased if we could arouse your interest in our e-learning offer.  We wish you a successful learning success.

$\text{Have fun and good luck!}$  

[https://www.ce.cit.tum.de/en/lnt/people/professors/kramer/ $\text{Gerhard Kramer}$'''],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Francisco_Javier_Garc.C3.ADa_G.C3.B3mez_.28at_LNT_from_2016-2021.29| $\text{Javier Garcia Gomez}$]],  [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28at_L.C3.9CT_since_2014.29| $\text{Tasnád Kernetzky}$]],  [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Benedikt_Leible.2C_M.Sc._.28at_L.C3.9CT_since_2017.29| $\text{Benedikt Leible}$]],  [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28at_LNT_since_1974.29 |$\text{Günter Söder}$]]

Munich,  in spring 2023          

 

==Content==

===(A)   The didactic concept of LNTwww===

At the beginning of the work on  $\rm LNTwww$  in 2001,  we gave ourselves the following ten rules.  These still apply today:

'''(1)'''   The teaching area  »Information and Communication Technology«  $\text{(I&K)}$  including associated basic subjects  $($Signal Representation,  Fourier and Laplace Transform,  Stochastic Signal Theory, etc.$)$  is presented in a didactically and multimedia prepared form.

'''(2)'''   Nine subject areas were selected,  each of which is covered by a self-contained book in the scope of a one-semester course with three semester hours per week to five semester hours per week.

'''(3)'''   The target group of our online offer are students of  $\text{I&K}$  technology,  especially of Communications Engineering,  as well as practicing engineers  $($Keywords:  »professional training«,  »lifelong learning«$)$.

'''(4)'''   In particular,  the interrelationships between different subfields of our extensive e-leatning offer should also be shown,  which is promoted by a nomenclature that is largely consistent in all books.

'''(5)'''   $\rm LNTwww$  offers two modes of learning:   Beginners should proceed sequentially  –   for advanced learners,  use it as a tutorial  $($work through exercises first,  jump to the theory part if deficits are identified$)$.

'''(6)'''   The theory is explained as in a traditional engineering textbook through texts,  graphics,  and mathematical derivations.  In addition,  each chapter includes at least one multimedia module.

'''(7)'''   $\rm LNTwww$  shall provide the user with multiple interaction options regarding the selection and presentation of theory chapters,  exercises,  learning videos as well as multimedia and calculation modules.

'''(8)'''   The methodology of hyperlinks typical of the  »world wide web«  is extensively used within
the  $\rm LNTwww$  and externally.  This is also intended to show connections between different teaching areas.

'''(9)'''   In order to prevent a user from getting lost in his learning environment and using  $\rm LNTwww$  only for  »surfing«,  a purposeful path must be recognizable for him at all times despite certain freedoms.

'''(10)'''  For reasons of sustainability of learning success,  there are possibilities for printing the texts and graphics,  ignoring the fact that today's students generation often devalues this as a  »relapse into the analog age«.
 

===(B)   Content and scope of LNTwww===

$\rm LNTwww$  is a virtual course totaling  $\text{36}$  semester hours per week 
*with  $\text{23}$  semester hours per week  (quasi) lectures   ⇒   $\text{23L}$

*and  $\text{13}$  semester hours per week exercises   ⇒   $\text{13E}$. 

It is organized in book form.  Each book contains a one-semester course.  For example,  in the case of the third book,  it is indicated that the book  »Theory of Stochastic Signals«  corresponds to a face-to-face–course with three semester hours per week of  »lecture«  and two semester hours per week of  »exercises«   ⇒   $\text{3L +2E}$.

Textbooks:
# [[Signal_Representation|'''Signal Representation''']]   ⇒   [[LNTwww:General_notes_about_"Signal_Representation"|»Impressum«]],
# [[Linear_and_Time_Invariant_Systems|'''Linear and Time Invariant Systems''']]   ⇒  [[LNTwww:General_notes_about_"Linear_and_Time_Invariant_Systems"|»Impressum«]],
# [[Theory_of_Stochastic_Signals|'''Theory of Stochastic Signals''']]   ⇒  [[LNTwww:General_Notes_about_the_Book_"Stochastic_Signal_Theory"|»Impressum«]],
# [[Information_Theory|'''Information Theory''']]   ⇒  [[LNTwww:General_notes_about_"Information_Theory"|»Impressum«]],
# [[Modulation_Methods|'''Modulation Methods''']]   ⇒  [[LNTwww:General_notes_about_"Modulation_Methods"|»Impressum«]],
# [[Digital_Signal_Transmission|'''Digital Signal Transmission''']]   ⇒  [[LNTwww:General_notes_about_"Digital_Signal_Transmission"|»Impressum«]],
# [[Mobile_Communications|'''Mobile Communications''']]   ⇒  [[LNTwww:General_notes_about_"Mobile_Communications"|»Impressum«]],
# [[Channel_Coding|'''Channel Coding''']]   ⇒  [[LNTwww:General_notes_about_"Channel_Coding"|»Impressum«]],
#[[Examples_of_Communication_Systems|'''Examples of Communication Systems''']]   ⇒  [[LNTwww:General_notes_about_"Examples_of_Communication_Systems"|»Impressum«]].

*The theory pages of all books result in the print version in approx.  $1500$  pages  $($DIN A4$)$  and contain on average one and a half graphics per page. 

*In addition,  LNTwww provides via the link  [[Biographies_and_Bibliographies|»'''Biographies & Bibliography'''«]]  a subject-specific bibliography with approx.  $400$  entries,  plus links to the WIKIPEDIA biographies of important scientists.
 

===(C)   Design and structure of LNTwww===

One can reach the nine reference books and »Biographies & Bibliography«  through the link  [[Book Overview|»'''Book Overview'''«]].  From this interface one can reach the individual books.  
*Each book is divided into several  »'''main chapters'''«, 

*each main chapter into several  »'''chapters'''«,  and

*each chapter includes several  »'''sections'''«.

{{GraueBox|TEXT=
$\text{Example 1:}$ 
We consider the book  [[Signal Representation|»Signal Representation«]].  This contains five  »main chapters«.
*By clicking on the first main chapter  »Basic Terms of Communications Engineering«,  one can get to three  »chapters«.  Each chapter corresponds to a MediaWiki file.

*The exemplary chapter  [[Signal_Representation/Principles_of_Communication|»Principles of Communication«]]  contains ten  »sections»  or  »pages».

*The last two pages are almost the same in all chapters,  namely  »Exercises for the chapter«  and  »References«.}}
 

===(D)   Content overviews for LNTwww===

A brief overview of all books is available on the selection interface  [[Book Overview|»'''Book Overview'''«]].
*More information is provided by the  »first page«  of each book.

*The respective main chapter content can be found in the first subchapter on the first page of each.

{{GraueBox|TEXT=
$\text{Example 2:}$ 
The first page  $($title page$)$  of the book  [[Signal_Representation|»Signal Representation«]]  provides the following information:
# A brief summary of the entire book;
# Scope of learning offer:  $2{\rm L} + 1{\rm E}$   ⇒   lecture with two semester hours per week and one additionalhour exercise. 
# Five main chapters,  19 chapters,  127 sections,  58 exercises;
# Links to the five main chapters of the book;
# Links to the associated exercises,  learning videos,  and interactive applets in the book  »Signal Representation«;
# Bibliography for the book;
# The imprint to the book  $($Authors,  other contributors,  materials as a starting point of the book,  referencces$)$.

The content of the first main chapter  »Principles of Communication« can be found on the first page 
[[Signal_Representation/Principles_of_Communication#OVERVIEW_OF_THE_FIRST_MAIN_CHAPTER|»# OVERVIEW OF THE FIRST MAIN CHAPTER #«
]].}}
 
===(E)   LNTwww exercises===

You can find the  »'''exercise overview'''«  for all books  $($approx.  $640$  exercises, approx.  $3100$  subtasks)  on the home page via the link  [[Aufgaben:Aufgabensammlung|»'''Exercises'''«]].  Please note:
*Each exercise consists of several  »subtasks«.   An exercise is only solved correctly if all subtasks are correct.

* For each exercise there is a detailed  "sample solution",  sometimes with the indication of several ways to the goal.

* The exercise types used are:
# "Single Choice"   ⇒   only one of the  $n$  given answers is correct;      ⇒   Marks of alternative answers:  ${\huge\circ}$
# "Multiple Choice"   ⇒   of the  $n$  given answers, between zero and  $n$  answers can be correct;      ⇒   Marks of alternative answers:  $\square$
# "Arithmetic exercise"   ⇒   numerical value query,  possibly with sign;     small deviations  $($usually  $\pm 3\%)$  are allowed when checking real-valued results.

* We distinguish between  »exercises«  $($e.g.  "Exercise 1.1"$)$  and  »additional exercises«  (e.g.  $($e.g.  "Exercise 1.1Z"$)$.
# If you were able to solve all exercises of a chapter without any problems,  we believe that you are familiar with the content of the entire chapter. 
#If you have solved one exercise incorrectly,  you should also work on the following,  usually somewhat easier additional exercise.

{{GraueBox|TEXT=
$\text{Example 3:}$ 
The  $58$  exercises/additional exercises of the book  "Signal Representation"  can be accessed via the link  [https://en.lntwww.de/Category:Signal_Representation:_Exercises »Signal Representation: Exercises«]. 

*From there,  we move on to the individual exercises,  for example,  to  [https://en.lntwww.de/Aufgaben:Exercise_1.1:_Music_Signals »Exercise 1.1: Music Signals«].  This relatively simple exercise consists of
#  one &bdquo;Single Choice”   ⇒   subtask  '''(1)''',
#  two &bdquo;Multiple Choice”   ⇒   subtasks  '''(2)''',  '''(3)''',  and
#  one computational task with two real-valued computational queries   ⇒   Subtask  '''(4)'''.

*But there are also much more difficult exercises in  $\rm LNTwww$.  Although MediaWiki also calls arithmetic exercises  "quizzes",  answering them is usually much more difficult than than on the numerous quiz shows on TV.   Because: 
::There are no predetermined answers in an arithmetic exercise,  and moreover:  Integrals often have to be solved beforehand,  such as in  [[Aufgaben:Exercise_4.4:_Two-dimensional_Gaussian_PDF|»Exercise 4.4: Two-dimensional Gaussian probabilty density function«.]]

*We recommend:  Print the exercise first   ⇒   »$\text{printable version}$«  and solve the exercise  offline  before checking  online      ⇒   Note:  In the  "printable version":  For links,  the target addresses are always given in brackets.
}}
 

 
===(F)   LNTwww learning videos===

You can access approximately  $30$  learning videos via the link  »Videos«  on the start page.  The realization of a learning video required the following individual steps: 
:Writing the script and texts   ⇒   Creating a set of slides with only slight differences between successive slides   ⇒   Voicing texts and audio editing   ⇒;   Combining texts and images into a coherent video stream.
#Clicking on this link brings up a; list of all learning videos,  grouped by textbook.  Some videos appear for multiple books.
#After selecting the desired learning video,  a wiki description page appears with a short content description and user interface.
#From here you can start the video in  »mp4«  and  »ogv«  format.  The browser will search for the appropriate format.
#The videos can be played by many browsers  $($Firefox, Chrome, Safari, ...$)$  as well as smartphones and tablets.
#The bottom link provides all available learning videos in alphabetical order.

Note:   All learning videos are with German language.  English translations are not planned.

{{GraueBox|TEXT=
$\text{Example 4:}$ 
We'll take a look at   [[Analoge_und_digitale_Signale_(Lernvideo)|»Analog and digital signals«]]  as an example.  This provides a two-part video in mp4 and ogv format.
*Each video part can be started by single click and paused by another click.

*The playback speed of the videos can be changed:
** Firefox offers a submenu after right-clicking on the video.
** For Google Chrome you can install e.g. the plugin  »Video Speed Controller«.
}}
 

===(G)   LNTwww applets===

Applets have a similar function as laboratories in mathematical-scientific courses:  Supplementing lecture/exercise with independent work by the student on the topic covered. 

You can access the provided interactive applets via the link of the same name on the home page.  It should be noted:
#Clicking on the link  »'''Applets'''«  a list of all applets appears,  grouped by reference books. 
#We distinguish between the newer  $\text{HTML 5/JavaScript}$  applets  $($in the respective lists above$)$  and the older  $\text{SWF}$  applets  $($below$)$. 
#The SWF applets unfortunately do not work on smartphones and tablets.
#After selecting an HTML 5/JS applet  a wiki description page appears with introductory theory section,  exercises to be solved and sample solutions. 
#At the beginning and end of this wiki description page there are links to the actual applet in German resp. English Language.

{{GraueBox|TEXT=
$\text{Example 5:}$ 
The didactic importance of applets shall be proved by  [[Applets:Eye_Pattern_and_Worst-Case_Error_Probability|»Eye Pattern and Worst-Case Error Probability«]]:
*The  »eye diagram«  is a proven transmission engineering tool,  to capture the influence of  »line dispersion«   ⇒   »intersysmbol interference«  on the quality characteristic  »error probability«  of a digital transmission system.  

*Such applets serve the clarification of more difficult facts,  in this example  »the step-by-step construction of the eye diagram from the symbol sequence«.  The program offers a lot of setting possibilities.  However, not every setting brings the user a relevant learning success and even fewer lead to a so-called "Aha! moment". 

*This is why we guide the user specifically through the program on the basis of the experiment.  He has to solve various tasks:  Predict and evaluate results,  Optimize parameters,  etc.

*A top 10% student has of course the possibility,  to set himself tasks going beyond the execution of experiments with the help of the applet and thus to penetrate very deeply into the presented subject matter.

}}

In addition to these  $\approx\hspace{-0.1cm} 30$  HTML 5/JS based applets  we still offer some of our  $\approx\hspace{-0.1cm}50$  older German-language applets,  which are based on  »Shock Wave Flash«  $\rm (SWF)$.  These were programmed for  »Adobe Flash«. 
#Since the Flashplayer browser plugin is no longer supported for security reasons,  these applets must be opened with the  »projector version«.
#You do not need to install the projector version and it will not be integrated into your browser.  So there are no security concerns in this regard.
#On the corresponding wiki pages you can find the projector version of the flash player and of course the applet itself.
 

===(H)   Glossary===

Due to the fact,  that our e–learning project LNTwww was first conceived in German and the wish for an English version came much later,  in the English version the assignment between  »Formula signs«   and  »Designation«  is not quite easy.   What do for example

#  $f_{\rm T}$,
#  $s_{\rm TP}(t)$, 
#  $e$,
#  $E$?

Here the link  [[LNTwww:Glossary|»Glossary«]]  on the home page below can help with the following alphabetically ordered entries:
::  »Formula sign«   ⇒   »German name«   ⇒   »English name« 

The file is self-explanatory. A few explanations are given under the last menu item  »Some remarks to the Glossary«.

{{GraueBox|TEXT=
$\text{Example 6:}$  In this file you will find the following entries:
:  $f_{\rm T}$   ⇒   Trägerfrequenz   ⇒   carrier frequency 
:  $s_{\rm TP}(t)$   ⇒   äquivalentes Tiefpass–Sendesignal   ⇒   equivalent low-pass transmitted signal 
:  $e= 2.718281828456$...   ⇒   Eulersche Zahl   ⇒   Eulerian number 
:  $ E$   ⇒   $(1)$ Schwellenwert,  $(2)$ Energie   ⇒   $(1)$ threshold value,  $(2)$ energy 

From the context,  the decision for  $(1)$  or  $(2)$  should be easy. }}

 

===(I)    History of LNTwww===

At the  [https://www.ce.cit.tum.de/en/lnt/home/ »Institute for Communications Engineering«]  $\rm (LNT)$  of the  [https://www.tum.de »Technical University of Munich«]  $\rm (TUM)$  two  teaching software packages  $\text{(LNTsim, LNTwin)}$  were realized by  [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28at_LNT_since_1974.29|»Günter Söder«]] from 1984 to 1996, which were used in our practical courses.  Several other universities have also acquired and used these programs.

At the beginning of the first Internet euphoria,  there were inquiries from students whether we could also provide such simulation and demonstration programs online.  After careful consideration  ("Is the expected big effort worth it?")  Günter Söder began 2001 planning the German-language project  »www.LNTwww.de«.  Co-responsible was his colleague  [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Klaus_Eichin_.28at_LNT_from_1972-2011.29|»Klaus Eichin«]].  The project was to be completed by 2011 at the latest,  since both would be retiring this year.

The content was derived from his own teaching materials as well as those of his colleague  [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Prof._Dr.-Ing._Norbert_Hanik_.28at_LNT_from_1989-1995.2C_at_L.C3.9CT_since_2004.29|»Norbert Hanik«]]  $($Associate Professor of Line Transmission Technology$)$.  Other lecture material was also taken into account,  which was produced at the Institute of Communications Engineering under the last four chair holders:
::*Professor [[Biographies_and_Bibliographies/Chair_holders_of_the_LNT_since_1962#Prof._Dr.-Ing._Dr.-Ing._E.h._Hans_Marko_.281962-1993.29|»Hans Marko«]]  $($Head of the LNT from 1962 to 1993$)$,
::*Professor [[Biographies_and_Bibliographies/Chair_holders_of_the_LNT_since_1962#Prof._Dr.-Ing._Dr.-Ing._E.h._Joachim_Hagenauer_.281993-2006.29|»Joachim Hagenauer«]]  $($Head of the LNT from 1993 to 2006$)$,
::*Professor [[Biographies_and_Bibliographies/Chair_holders_of_the_LNT_since_1962#Prof._Dr._Ralf_K.C3.B6tter_.282007-2009.29|»Ralf Kötter»]]  $($Head of the LNT from 2007 to 2009$)$,  and
::*Professor [https://www.ce.cit.tum.de/en/lnt/people/professors/kramer/ »Gerhard Kramer«]  $($Head of the LNT since 2010$)$.

Just a few dates about progress of the German-language LNTwww project,  eleven years after the planned completion :
* First of all our own platform had to be developed by students  $($Marin Winkler,  Yven Winter$)$.  The authoring system  »LNTwww«  was based on the http server  »Apache«,  the database  »MySQL«,  the script language  »Perl«  and  »Shock Wave Flash«  $\rm (SWF)$  as a basis for multimedia applications   ⇒   version  »LNTwww.v1«  $($2003$)$.

*Work of the following years was online adaptation of the manuscripts,  input into the database with the rather complicated LNTwww syntax,  creation of the graphs as well as conception and realization of multimedia elements.  After completion of all nine textbooks the desired final state was reached   ⇒   version  »LNTwww.v2«  $($2016$)$.

* At the same time,  it became known that  »SWF«  would not longer be supported by relevant manufacturers.  This fact and the criticism heard from some users about the meanwhile too staid design  $($our authoring system was on the level of 2003$)$  were decisive for a new start based on  »MediaWiki«   ⇒   version  »LNTwww.v3«  $($2021$)$.

Finally,  a few sentences about the English LNTwww version. 
*At the beginning of the Corona pandemic and the associated lockdowns,  the call for  »e-learning«  also became louder and louder at the universities,  even from professors who had previously rather rejected this form of teaching. 

* Suddenly,  funds were also made available,  to provide as many e-learning courses as possible in as short a time as possible.  Our chair  »Gerhard Kramer«  therefore already submitted a corresponding application for working student funds in spring 2020,  which was approved within a few weeks.

*In June 2020 we started the English translation with support of the  »DEEPL«  program  $($free version$)$  and finished it in April 2023.  From the LNT staff were involved:  »Javier Garcia Gomez, Tasnád Kernetzky, Benedikt Leible and Günter Söder«.  Of the students involved,  »Noah Nagi«  and  »Jiwoo Hwang«  deserve special mention.
 

===(J)   Acknowledgement===

The Institute for Communications Engineering would like to thank the many people involved in the creation of  $\rm LNTwww$:

*The persons responsible for the German and/or English LNTwww projects:
::[[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28at_LNT_since_1974.29 |$\text{Günter Söder}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Klaus_Eichin_.28at_LNT_from_1972-2011.29|$\text{Klaus Eichin}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28at_L.C3.9CT_since_2014.29| $\text{Tasnád Kernetzky}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Francisco_Javier_Garc.C3.ADa_G.C3.B3mez_.28at_LNT_from_2016-2021.29| $\text{Javier Garcia Gomez}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Benedikt_Leible.2C_M.Sc._.28at_L.C3.9CT_since_2017.29| $\text{Benedikt Leible}$]].

* The  $($former$)$  LNT/LÜT colleagues,  who contributed as co–authors or experts or supervised student work:
::[[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Ronald_B.C3.B6hnke_.28at_LNT_from_2012-2014.29|$\text{Ronald Böhnke}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Joschi_Brauchle_.28at_LNT_from_2007-2015.29|$\text{Joschi Brauchle}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Dr.-Ing._Bernhard_G.C3.B6bel_.28at_L.C3.9CT_from_2004-2010.29|$\text{Bernhard Göbel}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LÜT#Prof._Dr.-Ing._Norbert_Hanik_.28at_LNT_from_1989-1995.2C_at_L.C3.9CT_since_2004.29|$\text{Norbert Hanik}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Thomas_Hindelang_.28at_LNT_from_1994-2000_und_2007-2012.29|$\text{Thomas Hindelang}$]],   [[Biographies_and_Bibliographies/External_Contributors_to_LNTwww#Dr._Gianluigi_Liva|$\text{Gianluigi Liva}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Tobias_Lutz_.28at_LNT_from_2008-2014.29|$\text{Tobias Lutz}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Michael_Mecking_.28at_LNT_from_1997-2012.29|$\text{Michael Mecking}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Markus_Stinner_.28at_LNT_from_2011-2016.29|$\text{Markus Stinner}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Thomas_Stockhammer_.28at_LNT_from_1995-2004.29|$\text{Thomas Stockhammer}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Johannes_Zangl_.28at_LNT_from_2000-2006.29|$\text{Johannes Zangl}$]],   [[Biographies_and_Bibliographies/LNTwww_members_from_LNT#Dr.-Ing._Georg_Zeitler_.28at_LNT_from_2007-2012.29|$\text{Georg Zeitler}$]].

*The more than  »50 students«,  who have worked on subareas,  designed learning videos and applets or implemented the porting to the MediaWiki version within the framework of Engineering practice,  diploma,  bachelor and master theses or within the framework of a working student activity.

*The  [https://www.https://www.ei.tum.de/en/welcome/ »Department of Electrical and Computer Engineering«]  and the  [https://www.tum.de/en/ »Technical University of Munich«]  for funding working students in the years since 2016 within the framework of the  [https://www.ei.tum.de/studium/studienzuschuesse/ »MoliTUM«]  resp.  [https://www.tum.de/en/studies/teaching/awards-and-competitions/ideas-competition »EXIni«]  programs.

[https://www.ce.cit.tum.de/en/lnt/people/professors/kramer/ $\text{Gerhard Kramer}$]

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-04-26T17:10:51Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}

===Signal reconstruction===

[[File:EN_Sig_T_5_1_S2_v2.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==

[[File:Exercise_Abtast_v2.png|right|600px|frame|Screenshot]]
    '''(A)'''     Selection of one of the four given source signals,               Adjustment of amplitudes, frequencies and phases.

    '''(B)'''     Output of all set parameters of the source signal:

    '''(C)'''     Sampling & signal reconstruction parameters:
::*Sampling frequency  $f_{\rm A}$, 
::*Limit frequency of the receiving filter  $f_{\rm G}$,
::*Rolloff factor of the receiving filter  $r$,
:::⇒   Trapezoidal–corner frequencies:  $f_{1,\ 2} = f_{\rm G}\cdot (1\mp r)$

    '''(D)'''     Numerical result output:
::*Signal power  $P_{x}$
::*Distortion power  $P_{\varepsilon}$
::*Signal to distortion ratio  $10 \cdot \lg \ P_{x}/P_{\varepsilon}$

    '''(E)'''     Graphical output range for time domain:
::*Source signal  $x(t)$   ⇒   blue,
::*Sampled signal  $x_{\rm A}(t)$   ⇒   blue,
::*Reconstructed signal  $y(t)$   ⇒   green,
::*Differential signal  $\varepsilon(t)=y(t) - x(t)$   ⇒   purple

    '''(F)'''     Graphical output area for frequency domain:
::*$X(f)$   ⇒   blue,
::*$X_{\rm A}(f)$   ⇒   blue,
::*$Y(f)$   ⇒   green,
::*$E(f)=Y(f) - X(f)$   ⇒   purple

    '''(G)'''     Exercise selection

    '''(H)'''     Questions and solutions
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-04-26T17:10:29Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}

===Signal reconstruction===

[[File:EN_Sig_T_5_1_S2_v2.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==

[[File:Exercise_Abtast_v2.png|right|600px|frame|Screenshot]]
    '''(A)'''     Selection of one of the four given source signals,               Adjustment of amplitudes, frequencies and phases.

    '''(B)'''     Output of all set parameters of the source signal:

    '''(C)'''     Sampling & signal reconstruction parameters:
::*Sampling frequency  $f_{\rm A}$, 
::*Limit frequency of the receiving filter  $f_{\rm G}$,
::*Rolloff factor of the receiving filter  $r$,
:::⇒   Trapezoidal–corner frequencies:  $f_{1,\ 2} = f_{\rm G}\cdot (1\mp r)$

    '''(D)'''     Numerical result output:
::*Signal power  $P_{x}$
::*Distortion power  $P_{\varepsilon}$
::*Signal to distortion ratio  $10 \cdot \lg \ P_{x}/P_{\varepsilon}$

    '''(E)'''     Graphical output range for time domain:
::*Source signal  $x(t)$   ⇒   blue,
::*Sampled signal  $x_{\rm A}(t)$   ⇒   blue,
::*Reconstructed signal  $y(t)$   ⇒   green,
::*Differential signal  $\varepsilon(t)=y(t) - x(t)$   ⇒   purple,

    '''(F)'''     Graphical output area for frequency domain:
::*$X(f)$   ⇒   blue,
::*$X_{\rm A}(f)$   ⇒   blue,
::*$Y(f)$   ⇒   green,
::*$E(f)=Y(f) - X(f)$   ⇒   purple.

    '''(G)'''     Exercise selection

    '''(H)'''     Questions and solutions
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-04-26T17:05:42Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}

===Signal reconstruction===

[[File:EN_Sig_T_5_1_S2_v2.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==

[[File:Exercise_Abtast_v2.png|right|600px|frame|Screenshot]]
    '''(A)'''     Selection of one of the four given source signals,               Adjustment of amplitudes, frequencies and phases.

    '''(B)'''     Output of all set parameters of the source signal:

    '''(C)'''     Sampling & signal reconstruction parameters:
::*Sampling frequency  $f_{\rm A}$, 
::*Limit frequency of the receiving filter  $f_{\rm G}$,
::*Rolloff factor of the receiving filter  $r$,
:::⇒   Trapezoidal–corner frequencies:  $f_{1,\ 2} = f_{\rm G}\cdot (1\mp r)$

    '''(D)'''     Numerical result output:
::*Signal power  $P_{x}$
::*Distortion power  $P_{\varepsilon}$
::*Signal to distortion ratio  $10 \cdot \lg \ P_{x}/P_{\varepsilon}$

    '''(E)'''     Graphical output range for time domain:
::*Source signal  $x(t)$   ⇒   blue,
::*Sampled signal  $x_{\rm A}(t)$   ⇒   blue,
::*Reconstructed signal  $y(t)$   ⇒   green,
::*Differential signal  $\varepsilon(t)=y(t) - x(t)$   ⇒   purple,

    '''(F)'''     Graphical output area for frequency domain:
::*$X(f)$   ⇒   blue,
::*$X_{\rm A}(f)$   ⇒   blue,
::*$Y(f)$   ⇒   green,
::*$E(f)=Y(f) - X(f)$   ⇒   purple.

    '''(G)'''     Exercise selection.

    '''(H)'''     Questions and solutions
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Coherent and Non-Coherent On-Off Keying

2023-04-26T16:56:04Z

Noah:

{{LntAppletLinkEnDe|on-off-keying_en|on-off-keying}}

==Applet Description==
 

Considered is the symbol error probability  $p_{\rm S}$  of   "On–off keying"   $\rm (OOK)$  in the presence of white noise,  characterized by the standard deviation  $\sigma_{\rm AWGN}$,  both in the case of  coherent demodulation  and in the case of  noncoherent demodulation.   Plotted for both cases are the probability density functions  $\rm (PDF)$  of the received signal  $r(t)$  for the possible transmitted symbols  $s_0$  and  $s_1 \equiv 0$. 
*In the coherent case, there are two Gaussian functions around  $s_0$  and  $s_1$.

*In the incoherent case,  there is a Rayleigh PDF for the symbol  $s_1 = 0$  and a Rice PDF for  $s_0 \ne 0$,  whose form also depends on the input parameter  $C_{\rm Rice}$.

The applet returns the joint probabilities  ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})$   ⇒   $($filled blue area in the PDF graph$)$  and  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0})$   ⇒   $($red area$)$  and as a final result: 
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r} \ne \boldsymbol{s})= {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) + {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}). $$
*All these quantities also depend on the decision threshold  $G$  whose optimal value in each case is also determined.

*In addition,  the applet shows which error one makes when approximating the generally more complicated Rice PDF by the best possible Gaussian PDF.

==Theoretical Background==
 
===On–Off–Keying with coherent demodulation===
The simplest digital modulation method is  "On–off keying"  $\rm (OOK)$.  This method – also called  "Amplitude Shift Keying"  $\rm (2–ASK)$  – can be characterized as follows: 

[[File:EN_Dig_T_4_4_S3.png|right|frame|Signal space constellations for on-off keying|class=fit]]

*$\rm OOK$  is a binary and one-dimensional modulation method,  for example with  $s_{1} \equiv 0$  and
:*$\boldsymbol{s}_{0} = \{s_0,\ 0\}$  $($for cosinusoidal carrier,  left graph$)$  resp.

:*$\boldsymbol{s}_{0} = \{0,\ -s_0\}$  $($for sinusoidal carrier,  right graph$)$.

*With coherent demodulation,  the signal space constellation of the received signal is equal to that of the transmitted signal and again consists of the two points  $\boldsymbol{r}_0=\boldsymbol{s}_0$  and  $\boldsymbol{r}_1=\boldsymbol{s}_1$.  

*In this case,  the AWGN noise is one-dimensional with variance  $\sigma_{\rm AWGN}^2$  and one obtains  corresponding to the   [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Error_probability_for_symbols_with_equal_probability|"theory section"]]  for the  "symbol error probability":
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s})= {\rm Q} \left ( \frac{s_0/2}{\sigma_{\rm AWGN}}\right )
= {\rm Q} \left ( \sqrt{ {E_{\rm S}}/{N_0}}\right ) \hspace{0.05cm}. $$

To this it should be noted:
#The function  ${\rm Q}(x)$  is called the  [[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Function"]].
#The above equation applies to equally probable symbols with the decision threshold  $G$  midway between  $\boldsymbol{r}_0$  and  $\boldsymbol{r}_1$. 
#The distance of the two signal points from the decision threshold  $G$  is thus respectively  $\Delta G = s_0/2$  $($counter in the argument of the first  $\rm Q$–function$)$.
#$E_{\rm S}=s_0^2/2 \cdot T$  denotes for this case the  "average energy per symbol"  and  $N_0=2T \cdot \sigma_{\rm AWGN}^2$  is the  $($one-sided$)$  AWGN noise power density.

[[File:Applet_Bild3.png|right|frame| BER calculation for coherent demodulation]]
{{GraueBox|TEXT=
$\text{Example 1:}$  Let be  $\sigma_{\rm AWGN}= 0.8$  and  $s_{0} = 2$,  ⇒   $G=1$  $($these values are normalized to  $1\hspace{0.05cm} {\rm V})$.

The graph shows two  "half Gaussian functions"  around  $s_1=0$  $($blue curve$)$  and  $s_0=2$  $($red curve$)$.  The threshold value  $G$.  The shaded areas mark the symbol error probability.

*According to the first equation,  with  $\Delta G = s_{0} -G= G-s_1 = 1$:  
:$$p_{\rm S} = {\rm Q} ( 1/0.8 )= {\rm Q} ( 1.25 )\approx 10.56 \%.$$
*Similarly,  the second equation provides:  $E_{\rm S}/{N_0} = 1/4 \cdot s_0^2/\sigma_{\rm AWGN}^2 = 1.5615$:
:$$p_{\rm S} = {\rm Q} (\sqrt{1.5615} )\approx 10.56 \%.$$

Due to symmetry,  the threshold  $G=1$  is optimal.  In this case,  the red and blue shaded areas are equal   ⇒   the symbols  $\boldsymbol{s}_{0}$  and  $\boldsymbol{s}_{1}$  are falsified in the same way.

With  $G\ne 1$  there is a larger falsification probability.  For example,  with  $G=0.6$:
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s}) = {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})
+ {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0})= 1/2 \cdot {\rm Q} ( 0.75)+ 1/2 \cdot {\rm Q} ( 1.75)\approx 13.33\% .$$

Here the falsification probability for the symbol  $\boldsymbol{s}_{1}$   ⇒   blue filled area ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 11. 33\%$  is much larger than that of the symbol  $\boldsymbol{s}_{0}$   ⇒   red filled area ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 2\%$. }}

===On–Off–Keying with noncoherent demodulation===

The following diagram shows the structure  $($in the equivalent low-pass range$)$  of the optimal OOK receiver for incoherent demodulation.  See  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation#Non-coherent_demodulation_of_.22on.E2.80.93off_keying.22_.28OOK.29|"Detailed description"]].  According to this graph applies:

[[File:EN_Dig_T_4_5_S2b_neu.png|right|frame|Receiver for incoherent OOK demodulation  $($complex signals are labeled blue$)$|class=fit]]

*The input signal  $\boldsymbol{r}(t) = \boldsymbol{s}(t) \cdot {\rm e}^{\hspace{0.02cm}{\rm j}\hspace{0.03cm}\phi} + \boldsymbol{n}(t)$  at the receiver is generally complex because of the current phase angle  $\phi$  and because of the complex noise term  $\boldsymbol{n}(t)$.

*Now the correlation between the complex received signal  $\boldsymbol{r}(t)$  and a  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#Basis_functions_of_complex_time_signals|"complex basis function"]]  $\boldsymbol{\xi}(t)$  is required. 

*The result is the  $($complex$)$  detected value  $\boldsymbol{r}$,  from which the magnitude  $y = |\boldsymbol{r}(t)|$  is formed as a real decision input. 

*If  $y \gt G$,  then the estimated value  $m_0$  for the symbol  $\boldsymbol{s}_{0}$  is output,  otherwise the estimated value  $m_1$  for the symbol  $\boldsymbol{s}_{1}$.

*Once again,  the mean symbol error probability can be represented as the sum of two joint probabilities:
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s}) = {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})
+ {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}).$$

===Error probability calculation considering Rayleigh and Rice distribution===

To calculate the symbol error probability for incoherent demodulation,  we start from the following graph.  Shown is the received signal in the equivalent low-pass region in the complex plane.

[[File:Applet_Bild1.png|right|frame|Incoherent demodulation of On-Off-Keying|class=fit]]

#The point  $\boldsymbol{s_1}=0$  leads in the received signal again to  $\boldsymbol{r_1}=0$.
#In contrast,  $\boldsymbol{r}_0 = \boldsymbol{s}_0 \cdot {\rm e}^{\hspace{0.02cm}{\rm j}\hspace{0.03cm}\phi}$  can lie on any point of a circle with  radius  $1$  since the phase  $\phi$  is unknown. 
#The decision process taking into account that the AWGN noise is now to be interpreted in two dimensions,  as indicated by the arrows in the graph. 
#The decision region  $I_1$  for symbol  $\boldsymbol{s_1}$  is the blue filled circle with radius  $G$,  where the correct value of  $G$  remains to be determined.
#If the received value  $\boldsymbol{r}$ is outside this circle,  i.e. in the red highlighted area  $I_0$,  the decision is in favor of  $\boldsymbol{s_0}$. 

$\rm Rayleigh\ portion$ 

Considering the AWGN–noise,  $\boldsymbol{r_1}=\boldsymbol{s_1} + \boldsymbol{n_1}$.  The noise component  $\boldsymbol{n_1}$  has a  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF|"Rayleigh distribution"]]  $($amount of the two mean-free Gaussian components for  $I$  and  $Q)$.

*Their conditional PDF is with the rotationally symmetric noise component  $\eta$  with  $\sigma=\sigma_{\rm AWGN}$ :
:$$f_{y\hspace{0.05cm}\vert \hspace{0.05cm}\boldsymbol{s_1}} (\eta \hspace{0.05cm}\vert \hspace{0.05cm} \boldsymbol{s_1})=\frac{\eta}{\sigma^2}\cdot {\rm e}^{-\eta^2 / ( 2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma^2) } = f_{\rm Rayleigh}(\eta) .$$
*Thus one obtains for the conditional probability
:$${\rm Pr}(\boldsymbol{r_0}|\boldsymbol{s_1}) = \int_{G}^{\infty}f_{\rm Rayleigh}(\eta) \,{\rm d} \eta
\hspace{0.05cm},$$
:and with the factor  $1/2$  because of the equally probable transmitted symbols, the joint probability:
:$${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) = 1/2 \cdot {\rm Pr}(\boldsymbol{r_0}|\boldsymbol{s_1})= 1/2 \cdot \int_{G}^{\infty}f_{\rm Rayleigh}(\eta) \,{\rm d} \eta
\hspace{0.05cm}.$$

$\rm Rice\ portion$ 

The noise component  $\boldsymbol{n_0}$  has a  [[Theory_of_Stochastic_Signals/Further_Distributions#Rice_PDF|"Rice distribution"]] 
$($magnitude of Gaussian components with mean values  $m_x$  and  $m_y)$   ⇒   constant  $C=\sqrt{m_x^2 + m_y^2}$ $($Note:   In the applet, the constant  $C$  is denoted by  $C_{\rm Rice}$ $)$.

:$$f_{y\hspace{0.05cm}\vert \hspace{0.05cm}\boldsymbol{s_0}} (\eta \hspace{0.05cm}\vert \hspace{0.05cm} \boldsymbol{s_0})=\frac{\eta}{\sigma^2}\cdot{\rm e}^{-({C^2+\it \eta^{\rm 2} })/ ({\rm 2 \it \sigma^{\rm 2} })}\cdot {\rm I_0}(\frac{\it \eta\cdot C}{\sigma^{\rm 2} }) = f_{\rm Rice}(\eta) \hspace{1.4cm}{\rm with} \hspace{1.4cm} {\rm I_0}(\eta) = \sum_{k=0}^{\infty}\frac{(\eta/2)^{2k} }{k! \cdot {\rm \Gamma ({\it k}+1)} }.$$

This gives the second joint probability:

:$${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) = 1/2 \cdot \int_{0}^{G}f_{\rm Rice}(\eta) \,{\rm d} \eta
\hspace{0.05cm}.$$

[[File:P ID3148 Dig T 4 5 S2c version1.png|right|frame|Density functions for "OOK, non-coherent"]]
{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows the result of this equation for  $\sigma_{\rm AWGN} = 0.5$  and  $C_{\rm Rice} = 2$.  The decision threshold is at  $G \approx 1.25$.  One can see from this plot:

*The symbol error probability  $p_{\rm S}$  is the sum of the two colored areas.  As in Example 1 for the coherent case:
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s}) = {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})
+ {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}).$$
*The area marked in blue gives the joint probability  ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 2.2\%$  This is calculated as the integral over half the Rayleigh PDF in the range from  $G$  to  $\infty$.

*The red highlighted area gives the joint probability  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 2.4\%$  This is calculated as the integral over half the Rice PDF in the range from  $0$  to  $G$.

*Thus obtaining  $p_{\rm S} \approx 4.6\%$.  Note that the red and blue areas are not equal and that the optimal decision boundary  $G_{\rm opt}$  is obtained from the intersection of the two curves.

*The optimal decision threshold  $G_{\rm opt}$  is obtained as the intersection of the blue and red curves.}}

==Exercises==
 

*Select the number  $(1,\ 2$, ... $)$  of the task to be processed.  The number "0" corresponds to a "Reset":  Setting as at the program start.
*A task description is displayed.  Parameter values are adjusted.  Solution after pressing "Sample solution". 
*Always interpret the graphics and the numerical results.  The symbols  $s_0$  (adjustable) and  ${s}_{1}\equiv 0$  are equal probability.
*For space reasons, in some of the following questions and sample solutions we also use  $\sigma = \sigma_{\rm AWGN}$  and  $C = C_{\rm Rice}$. 

{{BlueBox|TEXT=
'''(1)'''   We consider  $\text{coherent}$  demodulation with  $\sigma_{\rm AWGN} = 0.5$  and  $s_0 = 2$.  What is the smallest possible value for the symbol error probability  $p_{\rm S}$? }}

*For coherent demodulation, the PDF of the reception signal is composed of two "half" Gaussian functions around  $s_0 = 2$  $($red$)$ and  $s_1 = 0$  $($blue$)$.
*Here the minimum  $p_{\rm S}$ value results with  $G=1$  and  $\Delta G = s_{0} -G= G-s_1 = 1$  to  $p_{\rm S}= {\rm Q} ( \Delta G/\sigma )={\rm Q} ( 1/0.5 )= {\rm Q} ( 2 )\approx 2.28 \%.$
*With  $G=1$  both symbols are falsified equally.   The blue area ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})$  is equal to the red area  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0})$.  Their sum gives  $p_{\rm S}$.
*With  $G=0.5$  the red area is almost zero.  Nevertheless   $p_{\rm S}\approx 8\%$  (sum of both areas)  is more than twice as large as with  $G_{\rm opt}=1$.

{{BlueBox|TEXT=
'''(2)'''   Now let  $\sigma = 0.75$.  With what  $s_0$  value does optimal $G$ give the same symbol error probability as in $(1)$?  Then what is the quotient  $E_{\rm S}/N_0$?}}

*In general  $p_{\rm S}= {\rm Q}\big ( (s_0/2) / \sigma \big )$.  If one increases  $\sigma$  from  $0. 5$  to   $0.75$, then  $s_0$  must also be increased   ⇒   $s_0 = 3$   ⇒   $p_{\rm S}= {\rm Q} ( 1.5/ 0.75 )= {\rm Q} ( 2 )$.
*Except  $p_{\rm S}= {\rm Q}\big ( (s_0/2) / \sigma \big )$  but also holds:  $p_{\rm S}= {\rm Q} ( \sqrt{E_{\rm S}/N_0} )$.  It follows:  $p_{\rm S}= {\rm Q}(2) ={\rm Q} ( \sqrt{E_{\rm S}/N_0})$   ⇒   $\sqrt{E_{\rm S}/N_0}= 2$   ⇒   $E_{\rm S}/N_0= 4$.
*For control:  $E_{\rm S}=s_0^2/2 \cdot T, \ N_0=2T \cdot \sigma^2$   ⇒   $E_{\rm S}/N_0 =s_0^2/(4 \cdot \sigma^2)= 3^2/(4 \cdot 0. 75^2)=4$.  The same  $E_{\rm S}/N_0 =4$  results for the problem  $(1)$.

{{BlueBox|TEXT=
'''(3)'''   Now consider  $\text{non–coherent}$  demodulation with  $\sigma_{\rm AWGN} = 0.75$,  $C_{\rm Rice} = 2.25$  and  $G=2$.  What is the symbol error probability  $p_{\rm S}$? }}

*For non–coherent demodulation, the PDF of the reception signal is composed of "half" a Rayleigh function $($blue$)$ and "half" a Rice function $($red$)$.
*${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 1.43\%$  gives the proportions of the blue curve above  $G =2$, and ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 15. 18\%$  the proportions of the red curve below  $G =2$.
*With  $G=2$  the sum for the symbol error probability is  $p_{\rm S}\approx 16.61\%$ , and with  $G_{\rm opt}=1.58$  a slightly better value:  $p_{\rm S}\approx 12.25\%$.

{{BlueBox|TEXT=
'''(4)'''   Let  $X$  be a Rayleigh random variable in general and  $Y$  be a Rice random variable, each with above parameters.  How large are  ${\rm Pr}(X\le 2)$  and  ${\rm Pr}(Y\le 2)$ ?}}

* It holds  ${\rm Pr}(Y\le 2) = 2 \cdot {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 30.36\%$,  since in the applet the Rice PDF is represented by the factor  $1/2$.
*In the same way  ${\rm Pr}(X> 2) = 2 \cdot {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 2.86\%$   ⇒   ${\rm Pr}(X \le 2)= 1-0.0286 = 97.14\%$.

{{BlueBox|TEXT=
'''(5)'''   We consider the values  $\sigma_{\rm AWGN} = 0.75$,  $C_{\rm Rice} = 2.25$  and  $G=G_{\rm opt}=1. 58$.  How does  $p_{\rm S}$ change when "Rice" is replaced by "Gauss" as best as possible? }}

*After the exact calculation, using the optimal threshold  $G_{\rm opt}=1.58$:     ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 5. 44\%$,  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 6.81\%$   ⇒   $p_{\rm S}\approx 12.25\%$.
*With the Gaussian approximation, for the same  $G$  the first term is not changed.  The second term increases to  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 9.29\%$   ⇒   $p_{\rm S}\approx 14.73\%$.
*The new optimization of the threshold  $G$  considering the Gaussian approximation leads to  $G_{\rm opt}=1.53$  and  $p_{\rm S}\approx 14.67\%$.
*The parameters of the Gaussian distribution are set as follows:  mean  $m_{\rm Gaussian}= C_{\rm Rice}=2.25$,  standard deviation  $\sigma_{\rm Gaussian}= \sigma_{\rm AWGN}=0.75$.

{{BlueBox|TEXT=
'''(6)'''   How do the results change from  $(5)$  with  $\sigma_{\rm AWGN} = 0. 5$,  $C_{\rm Rice} = 1.5$  and with  $\sigma_{\rm AWGN} = 1$,  $C_{\rm Rice} = 3$  respectively,  each with   $G=G_{\rm opt}$? }}

*With the optimal decision threshold  $G_{\rm opt}$, the probabilities are the same, both for the exact Rice distribution and with the Gaussian approximation.
*For all three parameter sets,  $E_{\rm S}/N_0= 2.25$.  This suggests:  The results with non–coherent demodulation depend on this characteristic value alone.

{{BlueBox|TEXT=
'''(7)'''   Let the setting continue to be  $\text{non–coherent/approximation}$  with  $C_{\rm Rice} = 3$,  $G=G_{\rm opt}$.  Vary the AWGN standard deviation in the range  $0.5 \le \sigma \le 1$.          Interpret the relative error   ⇒   $\rm (False - Correct)/Correct$  as a function of the quotient  $E_{\rm S}/N_0$.}}

*With  $\sigma =0.5$   ⇒   $E_{\rm S}/N_0 = 9$  one obtains  $p_{\rm S}^{\ \rm (exact)}\approx 0. 32\%$  and  $p_{\rm S}^{\ \rm (approximate)}\approx 0.38\%$.  The absolute error is  $0.06\%$  and the relative error  $18.75\%$.
*With  $\sigma =1$   ⇒   $E_{\rm S}/N_0 = 2.25$  one obtains  $p_{\rm S}^{\ \rm (exact)}\approx 12. 25\%$  and  $p_{\rm S}^{\ \rm (approximate)}\approx 14.67\%$.  The absolute error is  $2.42\%$  and the relative error  $19.75\%$.
* ⇒   The Gaussian approximation becomes better with larger  $E_{\rm S}/N_0$.  This statement can be seen more clearly from the absolute than from the relative error.

{{BlueBox|TEXT=
'''(8)'''   Now repeat the last experiment with  $\text{coherent}$  demodulation and  $s_0 = 3$,  $G=G_{\rm opt}$.  What conclusion does the comparison with  $(7)$ allow? }}

*The comparison of  $(7)$  and  $(8)$  shows:     For each  $E_{\rm S}/N_0$  there is a greater (worse) symbol error probability with non–coherent demodulation.
*For  $E_{\rm S}/N_0= 9$:     $p_{\rm S}^{\ \rm (coherent)}\approx 0.13\%$  and  $p_{\rm S}^{\ \rm (non–coherent)}\approx 0.32\%$.   And for  $E_{\rm S}/N_0= 2.25$:     $p_{\rm S}^{\ \rm (coherent)}\approx 6.68\%$  and  $p_{\rm S}^{\ \rm (non–coherent)}\approx 12.25\%$. 
*The simpler realization of the incoherent demodulator (no clock synchronization) causes a loss of quality   ⇒   greater error probability.

==Applet Manual==

[[File:Exercise_OnOff.png |right|frame|Screenshot $($English version,  light background$)$ ]]

    '''(A)'''     Selection:
::*coherent,
::*Incoherent,
::*Incoherent with approximation.

    '''(B)'''     Parameter input: 
::*$\sigma_{\rm AWGN}$, 
::*$s_0$, 
::*$E_{\rm S}/N_0$, 
::*$G_{\rm opt}$

    '''(C)'''     Numerical output area of probabilities.

    '''(D)'''     Graphical output area of PDF proportions.

    '''(E)'''     Exercise selection

    '''(F)'''     Questions and solutions
 
==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2011 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Martin_V.C3.B6lkl_.28Diplomarbeit_LB_2010.29|Martin Völkl]]  as part of his diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]  and  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Dr.-Ing._Klaus_Eichin_.28am_LNT_von_1972-2011.29|Klaus Eichin]]).

*In 2020 the program was redesigned via HTML5/JavaScript by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ).

*Last revision and English version 2021 by Carolin Mirschina. 

*The conversion of this applet was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (TUM Department of Electrical and Computer Engineering).  We thank.

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|on-off-keying_en|on-off-keying}}

Applets:Digital Filters

2023-04-26T16:44:13Z

Noah:

{{LntAppletLinkEnDe|digitalFilters_en|digitalFilters}}

==Applet Description==
 
The applet should clarify the properties of digital filters, whereby we confine ourselves to filters of the order $M=2$. Both non-recursive filters $\rm (FIR$,  ''Finite Impulse Response''$)$  as well as recursive filters $\rm (IIR$,  ''Infinite Impulse Response''$)$.

The input signal $x(t)$ is represented by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$. The output sequence $〈y_ν〉$is calculated, i.e. the discrete-time representation of the output signal $y(t)$.

*$T_{\rm A}$ denotes the time interval between two samples.
*We also limit ourselves to causal signals and systems, which means that $x_ν \equiv 0$ and $y_ν \equiv 0$ for $ν \le 0$.

It should also be noted that we denote the initial sequence $〈y_ν〉$ as

'''(1)''' the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:         $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉,$

'''(2)''' the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:         $〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉,$

'''(3)''' the '''discrete-time rectangle response''' $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangle function” is present at the input:     $〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉;$         In quotation marks are the beginning of the ones $(2)$ and the position of the last ones $(4)$.

==Theoretical background==
 
===General block diagram===

Each signal $x(t)$ can only be represented on a computer by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$.
[[File:P_ID552__Sto_T_5_2_S1_neu.png|right |frame| Block diagram of a digital (IIR–) filter $M$–order]]
*The time interval $T_{\rm A}$ between two samples is limited by the [https://en.lntwww.de/Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|"sampling theorem"].
*We limit ourselves here to causal signals and systems, which means that $x_ν \equiv 0$ for $ν \le 0$.

*In order to determine the influence of a linear filter with frequency response $H(f)$ on the discrete-time input signal $〈x_ν〉$, it is advisable to describe the filter discrete-time. In the time domain, this happens with the discrete-time impulse response $〈h_ν〉$.
*On the right you can see the corresponding block diagram. The following therefore applies to the samples of the output signal $〈y_ν〉$ thus holds:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu } \cdot x_{\nu - \mu } + \sum\limits_{\mu = 1}^M {b_\mu } \cdot y_{\nu - \mu } .$$

The following should be noted here:
*The index $\nu$ refers to sequences, for example at the input $〈x_ν〉$ and output $〈y_ν〉$.
*On the other hand, we use the index $\mu$ to identify the $a$ and $b$ filter coefficients.
*The first sum describes the dependency of the current output $y_ν$ on the current input $x_ν$ and on the $M$ previous input values $x_{ν-1}$, ... , $x_{ν-M}$.
*The second sum indicates the influence of $y_ν$ by the previous values $y_{ν-1}$, ... , $y_{ν-M}$ at the filter output. It specifies the recursive part of the filter.
*The integer parameter $M$ is called the order of the digital filter. In the program, this value is limited to $M\le 2$.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

'''(1)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:
:$$〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉 .$$
'''(2)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:
:$$〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉 .$$
'''(3)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time rectangle response'''  $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangular function” is present at the input:
:$$〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉 .$$
:The beginning of ones $(2)$ and the position of the last ones $(4)$ are given in single quotes.
}}

===Non-recursive filter   ⇒   FIR–filter ===

{{BlaueBox|TEXT=
[[File:P_ID553__Sto_T_5_2_S2_neu.png|right |frame| Non-recursive digital filter  $($FIR filter$)$  $M$ order]]
$\text{Definition:}$ If all feedback coefficients $b_{\mu} = 0$ , one speaks of one '''non-recursive filter'''. In the English language literature, the term '''FIR filter''' (''Finite Impulse Response'') is also used for this.

The following applies to the order $M$ applies:

*The output value $y_ν$ depends only on the current and the previous $M$ input values:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu \cdot x_{\mu - \nu } } .$$
*Discrete-time impulse response with $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉$:
:$$〈h_\mu〉= 〈a_0,\ a_1,\ \text{...},\ a_M〉 .$$}}

{{GraueBox|TEXT=
$\text{Example 1:}$  A two-way channel where
*the signal on the main path arrives undamped compared to the input signal but is delayed by $2\ \rm µ s$ arrives with a delay, and
*at $4\ \rm µ s$ distance – so absolutely at time $t = 6\ \rm µ s$ – follows an echo with half the amplitude,

can be simulated by a non-recursive filter according to the sketch above, whereby the following parameter values must be set:
:$$M = 3,\quad T_{\rm A} = 2\;{\rm{µ s} },\quad a_{\rm 0} = 0,\quad a_{\rm 1} = 1, \quad a_{\rm 2} = 0, \quad a_{\rm 3} = 0.5.$$}}

{{GraueBox|TEXT=
$\text{Example 2:}$ Consider a non-recursive filter with the filter coefficients $a_0 = 1,\hspace{0.5cm} a_1 = 2,\hspace{0.5cm} a_2 = 1.$
[[File:P_ID608__Sto_Z_5_3.png|right|frame|Non-recursive filter]]

'''(1)''' The conventional impulse response is: $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$         ⇒   discrete-time impulse response: $〈h_\mu〉= 〈1,\ 2,\ 1〉 .$

'''(2)'''   The frequency response $H(f)$ is the Fourier transform of $h(t)$. By applying the displacement theorem:
:$$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }\cdot }f \cdot T_{\rm A} } )} \big ] \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }fT_{\rm A} }\hspace{0.5cm}\Rightarrow \hspace{0.5cm}H(f = 0) = 4.$$

'''(3)'''   It follows that the '''discrete-time step response''' $〈\sigma_ν〉$ tends to become $4$ for large $\nu$.

'''(4)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;0,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;2,\;1,\;0,\;1,\;2,\;1,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$

'''(5)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;1,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;3,\;2,\;2,\;1,\;0,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$}}

===Recursive filter   ⇒   IIR filter ===

{{BlaueBox|TEXT=
[[File:P_ID607__Sto_A_5_3.png|right|frame|First order recursive filter]]
$\text{Definition:}$ 
*If at least one of the feedback coefficients is $b_{\mu} \ne 0$, then this is referred to as a '''recursive filter''' (see graphic on the right). The term '''IIR filter'''  (''Infinite Impulse Response'') is also used for this, particularly in the English-language literature. This filter is dealt with in detail in the trial implementation.

*If all forward coefficients are also identical $a_\mu = 0$ with the exception of $a_0$, a '''purely recursive filter''' is available (see graphic on the left).

[[File:P_ID554__Sto_T_5_2_S3_neu.png|left|frame| Purely recursive first order filter]] }}

In the following we restrict ourselves to the special case “purely recursive filter of the first order”. This filter has the following properties:
*The output value $y_ν$ depends (indirectly) on an infinite number of input values:
:$$y_\nu = \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot x_{\nu - \mu } .}$$
*This shows the following calculation:
:$$y_\nu = a_0 \cdot x_\nu + b_1 \cdot y_{\nu - 1} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + {b_1} ^2 \cdot y_{\nu - 2} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + a_0 \cdot {b_1} ^2 \cdot x_{\nu - 2} + {b_1} ^3 \cdot y_{\nu - 3} = \text{...}. $$

*By definition, the discrete-time impulse response is the same as the output sequence if there is a single "one" at $t =0$ at the input.
:$$h(t)= \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot \delta ( {t - \mu \cdot T_{\rm A} } )}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉= 〈\hspace{0.05cm}a_0, \ a_0\cdot {b_1}, \ a_0\cdot {b_1}^2 \ \text{...} \hspace{0.05cm}〉.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  With a recursive filter, the (discrete-time) impulse response extends to infinity with $M = 1$:
*For reasons of stability, $b_1 < 1$ must apply.
*With $b_1 = 1$ the impulse response $h(t)$ would extend to infinity and with $b_1 > 1$ the variable $h(t)$ would even continue to infinity.
*With such a recursive filter of the first order, each individual Dirac delta line is exactly the factor $b_1$ smaller than the previous Dirac delta line:
:$$h_{\mu} = h(\mu \cdot T_{\rm A}) = {b_1} \cdot h_{\mu -1}.$$}}

{{GraueBox|TEXT=
[[File:Sto_T_5_2_S3_version2.png |frame| Discrete-time impulse response]]
$\text{Example 3:}$  The graphic opposite shows the discrete-time impulse response $〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉$ of a recursive filter of the first order with the parameters $a_0 = 1$ and $b_1 = 0.6$.
*The (discrete-time) course is exponentially falling and extends to infinity.
*The ratio of the weights of two successive Dirac deltas is $b_1 = 0.6$.
}}

===Recursive filter as a sine generator===
[[File:EN_Sto_A_5_4_version2.png|right|frame|Proposed filter structure]]

The graphic shows a second-order digital filter that is suitable for generating a discrete-time sine function on a digital signal processor (DSP) if the input sequence $\left\langle \hspace{0.05cm} {x_\nu } \hspace{0.05cm}\right\rangle$  a (discrete-time) Dirac delta function is:
:$$\left\langle \hspace{0.05cm}{y_\nu }\hspace{0.05cm} \right\rangle = \left\langle {\, \sin ( {\nu \cdot T_{\rm A} \cdot \omega _0 } )\, }\right\rangle .$$

The five filter coefficients result from the:
[https://en.wikipedia.org/wiki/Z-transform "$Z$-transform"]:
:$$Z \big \{ {\sin ( {\nu T{\rm A}\cdot \omega _0 } )} \big \} = \frac{{z \cdot \sin \left( {\omega _0 \cdot T_{\rm A}} \right)}}{{z^2 - 2 \cdot z \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right) + 1}}.$$
After implementing this equation using a second-order recursive filter, the following filter coefficients are obtained:
:$$a_0 = 0,\quad a_1 = \sin \left( {\omega _0 \cdot T_{\rm A}} \right),\quad a_2 = 0, \quad b_1 = 2 \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right),\quad b_2 = - 1.$$

*The filter coefficients $a_0$ and $a_2$ can be omitted and $b_2=-1$ has a fixed value.
*The angular frequency $\omega_0$ of the sine wave is therefore only determined by $a_0$ and $a_0$.

{{GraueBox|TEXT=
$\text{Example 3:}$  Let $a_1 = 0.5$, $b_1 = \sqrt 3$, $x_0 = 1$ and $x_{\nu \hspace{0.05cm}\ne\hspace{0.05cm} 0} = 0$.

'''(1)'''  Then the following applies to the initial values $y_\nu$ at times $\nu \ge 0$: 
:*  $y_0 = 0;$
:*  $y_1 = 0.5$                                                                                         ⇒  the "$1$" at the input only has an effect at time $\nu = 1$ because of $a_0= 0$ at the output;
:*  $y_2 = b_1 \cdot y_1 - y_0 = {\sqrt 3 }/{2} \approx 0.866$                             ⇒   with $\nu = 2$ the recursive part of the filter also takes effect;
:*  $y_3 = \sqrt 3 \cdot y_2 - y_1 = \sqrt 3 \cdot {\sqrt 3 }/{2} - {1}/{2} = 1$          ⇒  for  $\nu \ge 2$  the filter is purely recursive:     $y_\nu = b_1 \cdot y_{\nu - 1} - y_{\nu - 2}$;
:*  $y_4 = \sqrt 3 \cdot y_3 - y_2 = \sqrt 3 \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$
:*  $y_5 = \sqrt 3 \cdot y_4 - y_3 = \sqrt 3 \cdot {\sqrt 3 }/{2} - 1 = 0.5;$
:*  $y_6 = \sqrt 3 \cdot y_5 - y_4 = \sqrt 3 \cdot {1}/{2} - {\sqrt 3 }/{2} = 0;$
:*  $y_7 = \sqrt 3 \cdot y_6 - y_5 = \sqrt 3 \cdot 0 - {1}/{2} = - 0.5.$

'''(2)'''  By continuing the recursive algorithm one gets for large $\nu$–values:     $y_\nu = y_{\nu - 12}$   ⇒   $T_0/T_{\rm A}= 12.$ }}

==Exercises==

[[File:Exercises_binomial_fertig.png|right]]
*First select the number '''1''' ... '''10''' of the task to be processed.
*A task description is displayed.  The parameter values are adjusted.
*Solution after pressing  "Sample Solution".
*The number '''0''' corresponds to a "reset":  Same setting as when the program was started.
 

{{BlaueBox|TEXT=
'''(1)'''  The filter coefficients are  $a_0=0.25$,  $a_1=0.5$,  $a_2=0.25$,  $b_1=b_2=0$.  Which filter is it?         Interpret the impulse response  $〈h_ν〉$,  the step response  $〈\sigma_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$, each in a discrete-time representation.}}

:*  Due to the missing  $b$ coefficients, it is a non-recursive digital filter ⇒   '''FIR filter''' (''Finite Impulse Response'').
:*  The impulse response consists of  $M+1=3$  Dirac delta lines according to the  $a$  coefficients:  $〈h_ν〉= 〈a_0, \ a_1,\ a_2〉= 〈0.25, \ 0.5,\ 0.25,\ 0, \ 0, \ 0,\text{...}〉 $.
:*  The step response is:  $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1,\text{...}〉 $.  The final value is equal to the DC signal transfer factor  $H(f=0)=a_0+a_1+a_2 = 1$.
:*  The distortions with rise and fall can also be seen from the rectangular response  $〈\rho_ν^{(2, 8)}〉= 〈0,\ 0, 0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1, \ 1, \ 0.75, \ 0.25, \ \text{...}〉$.

{{BlaueBox|TEXT=
'''(2)'''  How do the results differ with  $a_2=-0.25$? }}

:*  Taking into account  $H(f=0)= 0.5$  there are comparable consequences   ⇒   Step response:    $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 0.5,\ 0.5, \ 0.5, \ 0.5,\text{...}〉 $.

{{BlaueBox|TEXT=
'''(3)'''  Now let the filter coefficients  $a_0=1$,  $b_1=0.9$  and  $a_1=a_2= b_2=0$.  Which filter is it?  Interpret the impulse response  $〈h_ν〉$.}}

:*  It is a recursive digital filter   ⇒   '''IIR filter'''  (''Infinite Impulse Response'')  of the first order.  It is the discrete-time analogon of the RC low-pass.
:*  Starting from  $h_0= 1$ is $h_1= h_0 \cdot b_0= 0.9$,  $h_2= h_1 \cdot b_0= b_0^2=0.81$,  $h_3= h_2 \cdot b_0= b_0^3=0.729$,  and so on   ⇒   $〈h_ν〉$  extends to infinity.
:*  Impulse response  $h(t) = {\rm e}^{-t/T}$  with  $T$:   intersection $($Tangente bei  $t=0$, Abscissa$)$   ⇒   $h_\nu= h(\nu \cdot T_{\rm A}) = {\rm e}^{-\nu/(T/T_{\rm A})}$  with  $T/T_{\rm A} = 1/(h_0-h_1)= 10$.
:*  So:  The values of the continuous time differ from the discrete-time impulse response.  This results in the values  $1.0, \ 0.9048,\ 0.8187$ ...

{{BlaueBox|TEXT=
'''(4)'''  The filter setting is retained.  Interpret the step response  $〈h_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$.  What is the value for  $H(f=0)$?}}

:*  The step response is the integral over the impulse response:   $\sigma(t) = T \cdot (1-{\rm e}^{-t/T}) ]$   ⇒   $\sigma_\nu= 10 \cdot (1-{\rm e}^{-\nu/10})$   ⇒   $\sigma_0=1$,  $\sigma_1=1.9$,  $\sigma_2=2.71$, ...
:*  For large  $\nu$  values, the (discrete-time) step response tends to the DC signal transmission factor  $H(f=0)= 10$:  $\sigma_{40}=9.867$,  $\sigma_{50}=9.954$,  $\sigma_\infty=10$.
:* The rectangular response  $〈\rho_ν^{(2, 8)}〉$  increases with a delay of  $2$  in the same way as  $〈\sigma_ν〉$.  In the area  $\nu \ge 8$  the  $\rho_ν$  values decrease exponentially.

{{BlaueBox|TEXT=
'''(5)'''  We continue to consider the filter with  $a_0=1$,  $b_1=0.9$,  $a_1=a_2=b_2=0$.  What is the output sequence  $〈y_ν〉$  for the input sequence  $〈x_ν〉= 〈1,\ 0,\ -0.5〉$?         ''Note'': The task can also be solved with this program, although the constellation considered here cannot be set directly.}}

:*  You can help yourself by setting the coefficient  $a_2=-0.5$  and reducing the input sequence  to $〈x_ν〉= 〈1,\ 0,\ 0,\ \text{ ...}〉$   ⇒   „Dirac delta function”.
:*  The actual impulse response of this filter  $($with  $a_2=0)$  was determined in task  '''(3)''':   $h_0= 1$,   $h_1= 0.9$,   $h_2= 0.81$,   $h_3= 0.729$,   $h_4= 0.646$.  
:*  The solution to this problem is:  $y_0 = h_0= 1$,   $y_1= h_1= 0.9$,   $y_2 =h_2-h_0/2= 0.31$,   $y_3 =h_3-h_1/2= 0.279$,   $y_4 =h_4-h_2/2= 0.251$.  
:*  Caution:  Step response and rectangular response now refer to the fictitious filter  $($with  $a_2=-0.5)$  and not to the actual filter  $($with  $a_2=0)$.

{{BlaueBox|TEXT=
'''(6)'''  Consider and interpret the impulse response and the step response for the filter coefficients  $a_0=1$,  $b_1=1$,  $a_1=a_2= b_2=0$.  }}

:*  '''The system is unstable''':   A discrete-time Dirac delta function at input $($at time  $t=0)$  causes an infinite number of Dirac deltas of the same height in the output signal.
:*  A discrete-time step function at the input causes an infinite number of Dirac deltas with monotonically increasing weights (to infinity) in the output signal.

{{BlaueBox|TEXT=
'''(7)'''  Consider and interpret the impulse response and step response for the filter coefficients  $a_0=1$,  $b_1=-1$,  $a_1=a_2= b_2=0$.  }}
:*  In contrast to exercise  '''(6)''', the weights of the impulse response  $〈h_ν〉$  are not constantly equal to  $1$, but alternating  $\pm 1$.  The system is unstable too.
:*  With the jump response  $〈\sigma_ν〉$, however, the weights alternate between  $0$  $($with even $\nu)$  and  $1$  $($with odd $\nu)$.

{{BlaueBox|TEXT=
'''(8)'''  We consider the "sine generator":  $a_1=0.5$,  $b_1=\sqrt{3}= 1.732$,  $b_2=-1.$  Compare the impulse response with the calculated values in  $\text{Example 4}$.         How do the parameters $a_1$ and $b_1$ influence the period duration  $T_0/T_{\rm A}$  and the amplitude  $A$  of the sine function?}}
:*  $〈x_ν〉=〈1, 0, 0, \text{...}〉$   ⇒   $〈y_ν〉=〈0, 0.5, 0.866, 1, 0.866, 0.5, 0, -0.5, -0.866, -1, -0.866, -0.5, 0, \text{...}〉$   ⇒   '''sine''',  period  $T_0/T_{\rm A}= 12$,  amplitude  $1$.
:*  The increase/decrease of $b_1$  leads to the larger/smaller period  $T_0/T_{\rm A}$  and the larger/smaller amplitude  $A$.  $b_1 < 2$ must apply.
:*  $a_1$  only affects the amplitude, not the period.  There is no value limit for  $a_1$. If  $a_1$  is negative, the minus sine function results.

{{BlaueBox|TEXT=
'''(9)'''  The basic setting is retained.  Which  $a_1$  and  $b_1$  result in a sine function with period  $T_0/T_{\rm A}=16$  and amplitude  $A=1$?}}
:*  Trying with  $b_1= 1.8478$  actually achieves the period duration  $T_0/T_{\rm A}=16$.  However, this increases the amplitude to  $A=1.307$.
:*  Adjusting the parameter  $a_1= 0.5/1.307=0.3826$  then leads to the desired amplitude  $A=1$.
:*  Or you can calculate this as in the example:  $b_1 = 2 \cdot \cos ( {2{\rm{\pi }}\cdot{T_{\rm A}}/{T_0 }})= 2 \cdot \cos (\pi/8)=1.8478$,     $a_1 = \sin (\pi/8)=0.3827$.

{{BlaueBox|TEXT=
'''(10)'''  We continue with the  "sine generator".  What modifications do you have to make to generate a  "cosine"?}}
:*  With  $a_1=0.3826$,  $b_1=1.8478$,  $b_2=-1$  and  $〈x_ν〉=〈1, 1, 1, \text{...}〉$  is the output sequence  $〈y_ν〉$  the discrete-time analogon of the step response  $\sigma(t)$.
:*  The step response is the integral over   $\sin(\pi\cdot\tau/8)$   within the limits of   $\tau=0$   to   $\tau=t$   ⇒   $\sigma(t)=-8/\pi\cdot\cos(\pi\cdot\tau/8)+1$.
:*  If you change   $a_1=0.3826$   on   $a_1=-0.3826\cdot\pi/8=-0.1502$, then   $\sigma(t)=\cos(\pi\cdot\tau/8)-1$   ⇒   Values between  $0$  and  $-2$.
:*  Would you still in the block diagram   $z_\nu=y_\nu+1$   add, then   $z_\nu$   a discrete-time cosine curve with   $T_0/T_{\rm A}=16$   and   $A=1$.

[[File:EN_DIG_Fil_Mannt.png|right |frame| Screenshot]]

==Applet Manual==

    '''(A)'''     Input signal selection  $($Dirac delta,  unit step or rectangular$)$

    '''(B)'''     Settings for abscissa, ordinate and velocity.

    '''(C)'''     Control panel  $($Start, Single step, Total, Pause, Reset$)$

    '''(D)'''     Block diagram with stepwise adjustment of all values.

    '''(E)'''     Graphic area for output of the output sequence

    '''(F)'''     Exercise selection

    '''(G)'''     Questions and solutions
 
==About the authors==
 
This interactive calculation tool was designed and implemented at the [http://www.lnt.ei.tum.de/startseite chair for communications engineering] at the [https://www.tum.de/ Technische Universität München].
*The first version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Andr.C3.A9_Schulz_.28Bachelorarbeit_LB_2020.29|André Schulz]] (Bachelor thesis LB, Supervisors: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_Übertragungstechnik#Benedikt_Leible.2C_M.Sc._.28bei_L.C3.9CT_seit_2017.29|Benedikt Leible]] and [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".

==Once again: Open Applet in new Tab==

{{LntAppletLink|digitalFilters_en}}

Applets:Principle of 4B3T Coding

2023-04-26T16:14:35Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==

[[File:Exercise_4B3T.png|right|600px|frame|Screenshot]]
 
    '''(A)'''     Selection of source symbol sequence:  $\rm A$,  $\rm B$  or  $\rm C$.

    '''(B)'''     Program Options             $($random sequence,  blockwise RDS calculation,  total view,  Reset$)$

    '''(C)'''     "MS43"  or  "MMS43"

    '''(D)'''     Calculation of  "Running Digital Sum"

    '''(E)'''     Blockwise bit change

    '''(F)'''     Graphic area for the source signal  $q(t)$

    '''(G)'''     Graphic area for the encoder signal  $c(t)$

    '''(H)'''     Total plot of values  ${\it \Sigma}_0,\ {\it \Sigma}_1, \ {\it \Sigma}_2, \ {\it \Sigma}_4$  for  "MS43"  and  "MMS43"

    '''(I)'''     Exercise selection.

    '''(I)'''     Questions and solutions

 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Principle of 4B3T Coding

2023-04-26T16:13:12Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==

[[File:Exercise_4B3T.png|right|600px|frame|Screenshot]]
 
    '''(A)'''     Selection of source symbol sequence:  $\rm A$,  $\rm B$  or  $\rm C$.

    '''(B)'''     Program Options             $($random sequence,  blockwise RDS calculation,  total view,  Reset$)$

    '''(C)'''     "MS43"  or  "MMS43"

    '''(D)'''     Calculation of  "Running Digital Sum"

    '''(E)'''     Blockweise Bitänderung

    '''(F)'''     Graphic area for the source signal  $q(t)$

    '''(G)'''     Graphic area for the encoder signal  $c(t)$

    '''(H)'''     Total plot of values  ${\it \Sigma}_0,\ {\it \Sigma}_1, \ {\it \Sigma}_2, \ {\it \Sigma}_4$  for  "MS43"  and  "MMS43"

    '''(I)'''     Exercise selection.

    '''(I)'''     Questions and solutions

 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Principle of 4B3T Coding

2023-04-26T16:13:01Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==

[[File:Exercise_4B3T.png|right|600px|frame|Screenshot]]
 
    '''(A)'''     Selection of source symbol sequence:  $\rm A$,  $\rm B$  or  $\rm C$.

    '''(B)'''     Program Options             $($random sequence,  blockwise RDS calculation,  total view,  Reset$)$

    '''(C)'''     "MS43"§  or  "MMS43"

    '''(D)'''     Calculation of  "Running Digital Sum"

    '''(E)'''     Blockweise Bitänderung

    '''(F)'''     Graphic area for the source signal  $q(t)$

    '''(G)'''     Graphic area for the encoder signal  $c(t)$

    '''(H)'''     Total plot of values  ${\it \Sigma}_0,\ {\it \Sigma}_1, \ {\it \Sigma}_2, \ {\it \Sigma}_4$  for  "MS43"  and  "MMS43"

    '''(I)'''     Exercise selection.

    '''(I)'''     Questions and solutions

 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Principle of 4B3T Coding

2023-04-26T16:12:47Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==

[[File:Exercise_4B3T.png|right|600px|frame|Screenshot]]
 
    '''(A)'''     Selection of source symbol sequence:  $\rm A$,  $\rm B$  or  $\rm C$.

    '''(B)'''     Program Options             $($random sequence,  blockwise RDS calculation,  total view,  Reset$)$

    ''(C)'''     "MS43"§  or  "MMS43"

    '''(D)'''     Calculation of  "Running Digital Sum"

    '''(E)'''     Blockweise Bitänderung

    '''(F)'''     Graphic area for the source signal  $q(t)$

    '''(G)'''     Graphic area for the encoder signal  $c(t)$

    '''(H)'''     Total plot of values  ${\it \Sigma}_0,\ {\it \Sigma}_1, \ {\it \Sigma}_2, \ {\it \Sigma}_4$  for  "MS43"  and  "MMS43"

    '''(I)'''     Exercise selection.

    '''(I)'''     Questions and solutions

 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Principle of 4B3T Coding

2023-04-26T16:12:23Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==

[[File:Exercise_4B3T.png|right|600px|frame|Screenshot]]
 
    '''(A)'''     Selection of source symbol sequence:  $\rm A$,  $\rm B$  or  $\rm C$.

    '''(B)'''     Program Options             $($random sequence,  blockwise RDS calculation,  total view,  Reset$)$

    ''(C)'''     '''MS43'  or  "MMS43"

    '''(D)'''     Calculation of  "Running Digital Sum"

    '''(E)'''     Blockweise Bitänderung

    '''(F)'''     Graphic area for the source signal  $q(t)$

    '''(G)'''     Graphic area for the encoder signal  $c(t)$

    '''(H)'''     Total plot of values  ${\it \Sigma}_0,\ {\it \Sigma}_1, \ {\it \Sigma}_2, \ {\it \Sigma}_4$  for  "MS43"  and  "MMS43"

    '''(I)'''     Exercise selection.

    '''(I)'''     Questions and solutions

 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Principle of 4B3T Coding

2023-04-26T16:11:43Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==

[[File:Exercise_4B3T.png|right|600px|frame|Screenshot]]
 
    '''(A)'''     Selection of source symbol sequence:  $\rm A$,  $\rm B$  or  $\rm C$.

    '''(B)'''     Program Options             $($random sequence,  blockwise RDS calculation,  total view,  Reset$)$

    ''(C)'''     '''MS43'  or  '''MMS43'''

    '''(D)'''     Calculation of  "Running Digital Sum"

    '''(E)'''     Blockweise Bitänderung

    '''(F)'''     Graphic area for the source signal  $q(t)$

    '''(G)'''     Graphic area for the encoder signal  $c(t)$

    '''(H)'''     Total plot of values  ${\it \Sigma}_0,\ {\it \Sigma}_1, \ {\it \Sigma}_2, \ {\it \Sigma}_4$  for  "MS43"  and  "MMS43"

    '''(I)'''     Exercise selection.

    '''(I)'''     Questions and solutions

 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Coherent and Non-Coherent On-Off Keying

2023-04-16T23:04:33Z

Noah:

{{LntAppletLinkEnDe|on-off-keying_en|on-off-keying}}

==Applet Description==
 

Considered is the symbol error probability  $p_{\rm S}$  of   "On–off keying"   $\rm (OOK)$  in the presence of white noise,  characterized by the standard deviation  $\sigma_{\rm AWGN}$,  both in the case of  coherent demodulation  and in the case of  noncoherent demodulation.   Plotted for both cases are the probability density functions  $\rm (PDF)$  of the received signal  $r(t)$  for the possible transmitted symbols  $s_0$  and  $s_1 \equiv 0$. 
*In the coherent case, there are two Gaussian functions around  $s_0$  and  $s_1$.

*In the incoherent case,  there is a Rayleigh PDF for the symbol  $s_1 = 0$  and a Rice PDF for  $s_0 \ne 0$,  whose form also depends on the input parameter  $C_{\rm Rice}$.

The applet returns the joint probabilities  ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})$   ⇒   $($filled blue area in the PDF graph$)$  and  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0})$   ⇒   $($red area$)$  and as a final result: 
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r} \ne \boldsymbol{s})= {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) + {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}). $$
*All these quantities also depend on the decision threshold  $G$  whose optimal value in each case is also determined.

*In addition,  the applet shows which error one makes when approximating the generally more complicated Rice PDF by the best possible Gaussian PDF.

==Theoretical Background==
 
===On–Off–Keying with coherent demodulation===
The simplest digital modulation method is  "On–off keying"  $\rm (OOK)$.  This method – also called  "Amplitude Shift Keying"  $\rm (2–ASK)$  – can be characterized as follows: 

[[File:EN_Dig_T_4_4_S3.png|right|frame|Signal space constellations for on-off keying|class=fit]]

*$\rm OOK$  is a binary and one-dimensional modulation method,  for example with  $s_{1} \equiv 0$  and
:*$\boldsymbol{s}_{0} = \{s_0,\ 0\}$  $($for cosinusoidal carrier,  left graph$)$  resp.

:*$\boldsymbol{s}_{0} = \{0,\ -s_0\}$  $($for sinusoidal carrier,  right graph$)$.

*With coherent demodulation,  the signal space constellation of the received signal is equal to that of the transmitted signal and again consists of the two points  $\boldsymbol{r}_0=\boldsymbol{s}_0$  and  $\boldsymbol{r}_1=\boldsymbol{s}_1$.  

*In this case,  the AWGN noise is one-dimensional with variance  $\sigma_{\rm AWGN}^2$  and one obtains  corresponding to the   [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Error_probability_for_symbols_with_equal_probability|"theory section"]]  for the  "symbol error probability":
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s})= {\rm Q} \left ( \frac{s_0/2}{\sigma_{\rm AWGN}}\right )
= {\rm Q} \left ( \sqrt{ {E_{\rm S}}/{N_0}}\right ) \hspace{0.05cm}. $$

To this it should be noted:
#The function  ${\rm Q}(x)$  is called the  [[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Function"]].
#The above equation applies to equally probable symbols with the decision threshold  $G$  midway between  $\boldsymbol{r}_0$  and  $\boldsymbol{r}_1$. 
#The distance of the two signal points from the decision threshold  $G$  is thus respectively  $\Delta G = s_0/2$  $($counter in the argument of the first  $\rm Q$–function$)$.
#$E_{\rm S}=s_0^2/2 \cdot T$  denotes for this case the  "average energy per symbol"  and  $N_0=2T \cdot \sigma_{\rm AWGN}^2$  is the  $($one-sided$)$  AWGN noise power density.

[[File:Applet_Bild3.png|right|frame| BER calculation for coherent demodulation]]
{{GraueBox|TEXT=
$\text{Example 1:}$  Let be  $\sigma_{\rm AWGN}= 0.8$  and  $s_{0} = 2$,  ⇒   $G=1$  $($these values are normalized to  $1\hspace{0.05cm} {\rm V})$.

The graph shows two  "half Gaussian functions"  around  $s_1=0$  $($blue curve$)$  and  $s_0=2$  $($red curve$)$.  The threshold value  $G$.  The shaded areas mark the symbol error probability.

*According to the first equation,  with  $\Delta G = s_{0} -G= G-s_1 = 1$:  
:$$p_{\rm S} = {\rm Q} ( 1/0.8 )= {\rm Q} ( 1.25 )\approx 10.56 \%.$$
*Similarly,  the second equation provides:  $E_{\rm S}/{N_0} = 1/4 \cdot s_0^2/\sigma_{\rm AWGN}^2 = 1.5615$:
:$$p_{\rm S} = {\rm Q} (\sqrt{1.5615} )\approx 10.56 \%.$$

Due to symmetry,  the threshold  $G=1$  is optimal.  In this case,  the red and blue shaded areas are equal   ⇒   the symbols  $\boldsymbol{s}_{0}$  and  $\boldsymbol{s}_{1}$  are falsified in the same way.

With  $G\ne 1$  there is a larger falsification probability.  For example,  with  $G=0.6$:
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s}) = {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})
+ {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0})= 1/2 \cdot {\rm Q} ( 0.75)+ 1/2 \cdot {\rm Q} ( 1.75)\approx 13.33\% .$$

Here the falsification probability for the symbol  $\boldsymbol{s}_{1}$   ⇒   blue filled area ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 11. 33\%$  is much larger than that of the symbol  $\boldsymbol{s}_{0}$   ⇒   red filled area ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 2\%$. }}

===On–Off–Keying with noncoherent demodulation===

The following diagram shows the structure  $($in the equivalent low-pass range$)$  of the optimal OOK receiver for incoherent demodulation.  See  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation#Non-coherent_demodulation_of_.22on.E2.80.93off_keying.22_.28OOK.29|"Detailed description"]].  According to this graph applies:

[[File:EN_Dig_T_4_5_S2b_neu.png|right|frame|Receiver for incoherent OOK demodulation  $($complex signals are labeled blue$)$|class=fit]]

*The input signal  $\boldsymbol{r}(t) = \boldsymbol{s}(t) \cdot {\rm e}^{\hspace{0.02cm}{\rm j}\hspace{0.03cm}\phi} + \boldsymbol{n}(t)$  at the receiver is generally complex because of the current phase angle  $\phi$  and because of the complex noise term  $\boldsymbol{n}(t)$.

*Now the correlation between the complex received signal  $\boldsymbol{r}(t)$  and a  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces#Basis_functions_of_complex_time_signals|"complex basis function"]]  $\boldsymbol{\xi}(t)$  is required. 

*The result is the  $($complex$)$  detected value  $\boldsymbol{r}$,  from which the magnitude  $y = |\boldsymbol{r}(t)|$  is formed as a real decision input. 

*If  $y \gt G$,  then the estimated value  $m_0$  for the symbol  $\boldsymbol{s}_{0}$  is output,  otherwise the estimated value  $m_1$  for the symbol  $\boldsymbol{s}_{1}$.

*Once again,  the mean symbol error probability can be represented as the sum of two joint probabilities:
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s}) = {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})
+ {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}).$$

===Error probability calculation considering Rayleigh and Rice distribution===

To calculate the symbol error probability for incoherent demodulation,  we start from the following graph.  Shown is the received signal in the equivalent low-pass region in the complex plane.

[[File:Applet_Bild1.png|right|frame|Incoherent demodulation of On-Off-Keying|class=fit]]

#The point  $\boldsymbol{s_1}=0$  leads in the received signal again to  $\boldsymbol{r_1}=0$.
#In contrast,  $\boldsymbol{r}_0 = \boldsymbol{s}_0 \cdot {\rm e}^{\hspace{0.02cm}{\rm j}\hspace{0.03cm}\phi}$  can lie on any point of a circle with  radius  $1$  since the phase  $\phi$  is unknown. 
#The decision process taking into account that the AWGN noise is now to be interpreted in two dimensions,  as indicated by the arrows in the graph. 
#The decision region  $I_1$  for symbol  $\boldsymbol{s_1}$  is the blue filled circle with radius  $G$,  where the correct value of  $G$  remains to be determined.
#If the received value  $\boldsymbol{r}$ is outside this circle,  i.e. in the red highlighted area  $I_0$,  the decision is in favor of  $\boldsymbol{s_0}$. 

$\rm Rayleigh\ portion$ 

Considering the AWGN–noise,  $\boldsymbol{r_1}=\boldsymbol{s_1} + \boldsymbol{n_1}$.  The noise component  $\boldsymbol{n_1}$  has a  [[Theory_of_Stochastic_Signals/Further_Distributions#Rayleigh_PDF|"Rayleigh distribution"]]  $($amount of the two mean-free Gaussian components for  $I$  and  $Q)$.

*Their conditional PDF is with the rotationally symmetric noise component  $\eta$  with  $\sigma=\sigma_{\rm AWGN}$ :
:$$f_{y\hspace{0.05cm}\vert \hspace{0.05cm}\boldsymbol{s_1}} (\eta \hspace{0.05cm}\vert \hspace{0.05cm} \boldsymbol{s_1})=\frac{\eta}{\sigma^2}\cdot {\rm e}^{-\eta^2 / ( 2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sigma^2) } = f_{\rm Rayleigh}(\eta) .$$
*Thus one obtains for the conditional probability
:$${\rm Pr}(\boldsymbol{r_0}|\boldsymbol{s_1}) = \int_{G}^{\infty}f_{\rm Rayleigh}(\eta) \,{\rm d} \eta
\hspace{0.05cm},$$
:and with the factor  $1/2$  because of the equally probable transmitted symbols, the joint probability:
:$${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) = 1/2 \cdot {\rm Pr}(\boldsymbol{r_0}|\boldsymbol{s_1})= 1/2 \cdot \int_{G}^{\infty}f_{\rm Rayleigh}(\eta) \,{\rm d} \eta
\hspace{0.05cm}.$$

$\rm Rice\ portion$ 

The noise component  $\boldsymbol{n_0}$  has a  [[Theory_of_Stochastic_Signals/Further_Distributions#Rice_PDF|"Rice distribution"]] 
$($magnitude of Gaussian components with mean values  $m_x$  and  $m_y)$   ⇒   constant  $C=\sqrt{m_x^2 + m_y^2}$ $($Note:   In the applet, the constant  $C$  is denoted by  $C_{\rm Rice}$ $)$.

:$$f_{y\hspace{0.05cm}\vert \hspace{0.05cm}\boldsymbol{s_0}} (\eta \hspace{0.05cm}\vert \hspace{0.05cm} \boldsymbol{s_0})=\frac{\eta}{\sigma^2}\cdot{\rm e}^{-({C^2+\it \eta^{\rm 2} })/ ({\rm 2 \it \sigma^{\rm 2} })}\cdot {\rm I_0}(\frac{\it \eta\cdot C}{\sigma^{\rm 2} }) = f_{\rm Rice}(\eta) \hspace{1.4cm}{\rm with} \hspace{1.4cm} {\rm I_0}(\eta) = \sum_{k=0}^{\infty}\frac{(\eta/2)^{2k} }{k! \cdot {\rm \Gamma ({\it k}+1)} }.$$

This gives the second joint probability:

:$${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) = 1/2 \cdot \int_{0}^{G}f_{\rm Rice}(\eta) \,{\rm d} \eta
\hspace{0.05cm}.$$

[[File:P ID3148 Dig T 4 5 S2c version1.png|right|frame|Density functions for "OOK, non-coherent"]]
{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows the result of this equation for  $\sigma_{\rm AWGN} = 0.5$  and  $C_{\rm Rice} = 2$.  The decision threshold is at  $G \approx 1.25$.  One can see from this plot:

*The symbol error probability  $p_{\rm S}$  is the sum of the two colored areas.  As in Example 1 for the coherent case:
:$$p_{\rm S} = {\rm Pr}(\boldsymbol{r}\ne \boldsymbol{s}) = {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})
+ {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}).$$
*The area marked in blue gives the joint probability  ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 2.2\%$  This is calculated as the integral over half the Rayleigh PDF in the range from  $G$  to  $\infty$.

*The red highlighted area gives the joint probability  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 2.4\%$  This is calculated as the integral over half the Rice PDF in the range from  $0$  to  $G$.

*Thus obtaining  $p_{\rm S} \approx 4.6\%$.  Note that the red and blue areas are not equal and that the optimal decision boundary  $G_{\rm opt}$  is obtained from the intersection of the two curves.

*The optimal decision threshold  $G_{\rm opt}$  is obtained as the intersection of the blue and red curves.}}

==Exercises==
 

*Select the number  $(1,\ 2$, ... $)$  of the task to be processed.  The number "0" corresponds to a "Reset":  Setting as at the program start.
*A task description is displayed.  Parameter values are adjusted.  Solution after pressing "Sample solution". 
*Always interpret the graphics and the numerical results.  The symbols  $s_0$  (adjustable) and  ${s}_{1}\equiv 0$  are equal probability.
*For space reasons, in some of the following questions and sample solutions we also use  $\sigma = \sigma_{\rm AWGN}$  and  $C = C_{\rm Rice}$. 

{{BlueBox|TEXT=
'''(1)'''   We consider  $\text{coherent}$  demodulation with  $\sigma_{\rm AWGN} = 0.5$  and  $s_0 = 2$.  What is the smallest possible value for the symbol error probability  $p_{\rm S}$? }}

*For coherent demodulation, the PDF of the reception signal is composed of two "half" Gaussian functions around  $s_0 = 2$  $($red$)$ and  $s_1 = 0$  $($blue$)$.
*Here the minimum  $p_{\rm S}$ value results with  $G=1$  and  $\Delta G = s_{0} -G= G-s_1 = 1$  to  $p_{\rm S}= {\rm Q} ( \Delta G/\sigma )={\rm Q} ( 1/0.5 )= {\rm Q} ( 2 )\approx 2.28 \%.$
*With  $G=1$  both symbols are falsified equally.   The blue area ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1})$  is equal to the red area  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0})$.  Their sum gives  $p_{\rm S}$.
*With  $G=0.5$  the red area is almost zero.  Nevertheless   $p_{\rm S}\approx 8\%$  (sum of both areas)  is more than twice as large as with  $G_{\rm opt}=1$.

{{BlueBox|TEXT=
'''(2)'''   Now let  $\sigma = 0.75$.  With what  $s_0$  value does optimal $G$ give the same symbol error probability as in $(1)$?  Then what is the quotient  $E_{\rm S}/N_0$?}}

*In general  $p_{\rm S}= {\rm Q}\big ( (s_0/2) / \sigma \big )$.  If one increases  $\sigma$  from  $0. 5$  to   $0.75$, then  $s_0$  must also be increased   ⇒   $s_0 = 3$   ⇒   $p_{\rm S}= {\rm Q} ( 1.5/ 0.75 )= {\rm Q} ( 2 )$.
*Except  $p_{\rm S}= {\rm Q}\big ( (s_0/2) / \sigma \big )$  but also holds:  $p_{\rm S}= {\rm Q} ( \sqrt{E_{\rm S}/N_0} )$.  It follows:  $p_{\rm S}= {\rm Q}(2) ={\rm Q} ( \sqrt{E_{\rm S}/N_0})$   ⇒   $\sqrt{E_{\rm S}/N_0}= 2$   ⇒   $E_{\rm S}/N_0= 4$.
*For control:  $E_{\rm S}=s_0^2/2 \cdot T, \ N_0=2T \cdot \sigma^2$   ⇒   $E_{\rm S}/N_0 =s_0^2/(4 \cdot \sigma^2)= 3^2/(4 \cdot 0. 75^2)=4$.  The same  $E_{\rm S}/N_0 =4$  results for the problem  $(1)$.

{{BlueBox|TEXT=
'''(3)'''   Now consider  $\text{non–coherent}$  demodulation with  $\sigma_{\rm AWGN} = 0.75$,  $C_{\rm Rice} = 2.25$  and  $G=2$.  What is the symbol error probability  $p_{\rm S}$? }}

*For non–coherent demodulation, the PDF of the reception signal is composed of "half" a Rayleigh function $($blue$)$ and "half" a Rice function $($red$)$.
*${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 1.43\%$  gives the proportions of the blue curve above  $G =2$, and ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 15. 18\%$  the proportions of the red curve below  $G =2$.
*With  $G=2$  the sum for the symbol error probability is  $p_{\rm S}\approx 16.61\%$ , and with  $G_{\rm opt}=1.58$  a slightly better value:  $p_{\rm S}\approx 12.25\%$.

{{BlueBox|TEXT=
'''(4)'''   Let  $X$  be a Rayleigh random variable in general and  $Y$  be a Rice random variable, each with above parameters.  How large are  ${\rm Pr}(X\le 2)$  and  ${\rm Pr}(Y\le 2)$ ?}}

* It holds  ${\rm Pr}(Y\le 2) = 2 \cdot {\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 30.36\%$,  since in the applet the Rice PDF is represented by the factor  $1/2$.
*In the same way  ${\rm Pr}(X> 2) = 2 \cdot {\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 2.86\%$   ⇒   ${\rm Pr}(X \le 2)= 1-0.0286 = 97.14\%$.

{{BlueBox|TEXT=
'''(5)'''   We consider the values  $\sigma_{\rm AWGN} = 0.75$,  $C_{\rm Rice} = 2.25$  and  $G=G_{\rm opt}=1. 58$.  How does  $p_{\rm S}$ change when "Rice" is replaced by "Gauss" as best as possible? }}

*After the exact calculation, using the optimal threshold  $G_{\rm opt}=1.58$:     ${\rm Pr}(\boldsymbol{r_0} \cap \boldsymbol{s_1}) \approx 5. 44\%$,  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 6.81\%$   ⇒   $p_{\rm S}\approx 12.25\%$.
*With the Gaussian approximation, for the same  $G$  the first term is not changed.  The second term increases to  ${\rm Pr}(\boldsymbol{r_1} \cap \boldsymbol{s_0}) \approx 9.29\%$   ⇒   $p_{\rm S}\approx 14.73\%$.
*The new optimization of the threshold  $G$  considering the Gaussian approximation leads to  $G_{\rm opt}=1.53$  and  $p_{\rm S}\approx 14.67\%$.
*The parameters of the Gaussian distribution are set as follows:  mean  $m_{\rm Gaussian}= C_{\rm Rice}=2.25$,  standard deviation  $\sigma_{\rm Gaussian}= \sigma_{\rm AWGN}=0.75$.

{{BlueBox|TEXT=
'''(6)'''   How do the results change from  $(5)$  with  $\sigma_{\rm AWGN} = 0. 5$,  $C_{\rm Rice} = 1.5$  and with  $\sigma_{\rm AWGN} = 1$,  $C_{\rm Rice} = 3$  respectively,  each with   $G=G_{\rm opt}$? }}

*With the optimal decision threshold  $G_{\rm opt}$, the probabilities are the same, both for the exact Rice distribution and with the Gaussian approximation.
*For all three parameter sets,  $E_{\rm S}/N_0= 2.25$.  This suggests:  The results with non–coherent demodulation depend on this characteristic value alone.

{{BlueBox|TEXT=
'''(7)'''   Let the setting continue to be  $\text{non–coherent/approximation}$  with  $C_{\rm Rice} = 3$,  $G=G_{\rm opt}$.  Vary the AWGN standard deviation in the range  $0.5 \le \sigma \le 1$.          Interpret the relative error   ⇒   $\rm (False - Correct)/Correct$  as a function of the quotient  $E_{\rm S}/N_0$.}}

*With  $\sigma =0.5$   ⇒   $E_{\rm S}/N_0 = 9$  one obtains  $p_{\rm S}^{\ \rm (exact)}\approx 0. 32\%$  and  $p_{\rm S}^{\ \rm (approximate)}\approx 0.38\%$.  The absolute error is  $0.06\%$  and the relative error  $18.75\%$.
*With  $\sigma =1$   ⇒   $E_{\rm S}/N_0 = 2.25$  one obtains  $p_{\rm S}^{\ \rm (exact)}\approx 12. 25\%$  and  $p_{\rm S}^{\ \rm (approximate)}\approx 14.67\%$.  The absolute error is  $2.42\%$  and the relative error  $19.75\%$.
* ⇒   The Gaussian approximation becomes better with larger  $E_{\rm S}/N_0$.  This statement can be seen more clearly from the absolute than from the relative error.

{{BlueBox|TEXT=
'''(8)'''   Now repeat the last experiment with  $\text{coherent}$  demodulation and  $s_0 = 3$,  $G=G_{\rm opt}$.  What conclusion does the comparison with  $(7)$ allow? }}

*The comparison of  $(7)$  and  $(8)$  shows:     For each  $E_{\rm S}/N_0$  there is a greater (worse) symbol error probability with non–coherent demodulation.
*For  $E_{\rm S}/N_0= 9$:     $p_{\rm S}^{\ \rm (coherent)}\approx 0.13\%$  and  $p_{\rm S}^{\ \rm (non–coherent)}\approx 0.32\%$.   And for  $E_{\rm S}/N_0= 2.25$:     $p_{\rm S}^{\ \rm (coherent)}\approx 6.68\%$  and  $p_{\rm S}^{\ \rm (non–coherent)}\approx 12.25\%$. 
*The simpler realization of the incoherent demodulator (no clock synchronization) causes a loss of quality   ⇒   greater error probability.

==Applet Manual==
 
[[File:Exercise_impuls.png |right|frame|Screenshot (english version, light background) '''Korrektur''']]
    ''(A)'''     Theme (changeable graphical interface design).
:* Dark:   black background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green–visual impairment
Protanopia:   for users with pronounced red–visual impairment

    '''(B)'''     Preselection for pulse shape  $x_1(t)$  (red curve).

    '''(C)'''     Parameter setting for  $x_1(t)$ 

    '''(D)'''     Numeric output for  $x_1(t_*)$  and  $X_1(f_*)$

    '''(E)'''     Preselection for the pulse shape  $x_2(t)$  (blue curve).

    '''(F)'''     Parameter setting for  $x_2(t)$ 

    '''(G)'''     Numeric output for  $x_2(t_*)$  and  $X_2(f_*)$

    '''(H)'''     Setting the time  $t_*$  for the numeric output.

    '''(I)'''     Setting of frequency  $f_*$  for numeric output

    '''(J)'''     Range of the graphical representation in the time domain

    '''(K)'''     Range of graphical representation in frequency domain

    '''(L)''''     Selection of the exercise according to the exercise number.

    '''(M)'''     Exercise description and question.

    '''(N)'''     Show and hide sample solution

'''Details of the above points  (J ) and  (K)'''

Zoom–options:        "$+$" (Zoom in),      "$-$" (Zoom out),      "$\rm o$" (Reset)

Moving–options:     "$\leftarrow$"     "$\uparrow$"     "$\downarrow$"     "$\rightarrow$"         "$\leftarrow$"  means:     Image frame to the left, ordinate to the right

Other options:

*With the shift key pressed and scrolling, the coordinate system can be zoomed.
*With the shift key pressed and the left mouse button held down, the coordinate system can be moved.
 

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2011 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Martin_V.C3.B6lkl_.28Diplomarbeit_LB_2010.29|Martin Völkl]]  as part of his diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]  and  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Dr.-Ing._Klaus_Eichin_.28am_LNT_von_1972-2011.29|Klaus Eichin]]).

*In 2020 the program was redesigned via HTML5/JavaScript by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ).

*Last revision and English version 2021 by Carolin Mirschina. 

*The conversion of this applet was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (TUM Department of Electrical and Computer Engineering).  We thank.

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|on-off-keying_en|on-off-keying}}

Applets:Two-dimensional Gaussian Random Variables

2023-04-16T20:20:43Z

Noah:

{{LntAppletLinkEnDe|gauss_en|gauss}}

==Applet Description==
 
The applet illustrates the properties of two-dimensional Gaussian random variables  $XY\hspace{-0.1cm}$, characterized by the standard deviations (rms)  $\sigma_X$  and  $\sigma_Y$  of their two components, and the correlation coefficient  $\rho_{XY}$ between them. The components are assumed to be zero mean:  $m_X = m_Y = 0$.

The applet shows
* the two-dimensional probability density function   ⇒   $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$  $f_{XY}(x, \hspace{0.1cm}y)$  in three-dimensional representation as well as in the form of contour lines,
* the corresponding marginal probability density function  ⇒   $\rm 1D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$  $f_{X}(x)$  of the random variable  $X$  as a blue curve; likewise  $f_{Y}(y)$  for the second random variable,
* the two-dimensional distribution function  ⇒   $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}CDF$  $F_{XY}(x, \hspace{0.1cm}y)$  as a 3D plot,
* the distribution function  ⇒   $\rm 1D\hspace{-0.1cm}-\hspace{-0.1cm}CDF$  $F_{X}(x)$  of the random variable  $X$; also  $F_{Y}(y)$  as a red curve.

The applet uses the framework  [https://en.wikipedia.org/wiki/Plotly "Plot.ly"]

==Theoretical Background==
 

===Joint probability density function   ⇒   2D–PDF===

We consider two continuous value random variables  $X$  and  $Y\hspace{-0.1cm}$, between which statistical dependencies may exist. To describe the interrelationships between these variables, it is convenient to combine the two components into a  '''two-dimensional random variable'''  $XY =(X, Y)$  . Then holds:

{{BlaueBox|TEXT=
$\text{Definition:}$ 
The  '''joint probability density function'''  is the probability density function (PDF) of the two-dimensional random variable  $XY$  at location  $(x, y)$:
:$$f_{XY}(x, \hspace{0.1cm}y) = \lim_{\left.{\delta x\rightarrow 0 \atop {\delta y\rightarrow 0} }\right. }\frac{ {\rm Pr}\big [ (x - {\rm \Delta} x/{\rm 2} \le X \le x + {\rm \Delta} x/{\rm 2}) \cap (y - {\rm \Delta} y/{\rm 2} \le Y \le y +{\rm \Delta}y/{\rm 2}) \big] }{ {\rm \Delta} \ x\cdot{\rm \Delta} y}.$$

*The joint probability density function, or in short  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$  is an extension of the one-dimensional PDF.
*$∩$  denotes the logical AND operation.
*$X$  and  $Y$ denote the two random variables, and  $x \in X$  and   $y \in Y$ indicate realizations thereof.
*The nomenclature used for this applet thus differs slightly from the description in the [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables#Joint_probability_density_function|"Theory section"]].}}

Using this 2D–PDF  $f_{XY}(x, y)$  statistical dependencies within the two-dimensional random variable  $XY$  are also fully captured in contrast to the two one-dimensional density functions   ⇒   '''marginal probability density functions''':
:$$f_{X}(x) = \int _{-\infty}^{+\infty} f_{XY}(x,y) \,\,{\rm d}y ,$$
:$$f_{Y}(y) = \int_{-\infty}^{+\infty} f_{XY}(x,y) \,\,{\rm d}x .$$

These two marginal density functions  $f_X(x)$  and  $f_Y(y)$
*provide only statistical information about the individual components  $X$  and  $Y$, respectively,
*but not about the bindings between them.

As a quantitative measure of the linear statistical bindings  ⇒   '''correlation'''  one uses.
* the  '''covariance'''  $\mu_{XY}$, which is equal to the first-order common linear moment for mean-free components:
:$$\mu_{XY} = {\rm E}\big[X \cdot Y\big] = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} X \cdot Y \cdot f_{XY}(x,y) \,{\rm d}x \, {\rm d}y ,$$
*the  '''correlation coefficient'''  after normalization to the two rms values  $σ_X$  and $σ_Y$  of the two components:
:$$\rho_{XY}=\frac{\mu_{XY} }{\sigma_X \cdot \sigma_Y}.$$

{{BlaueBox|TEXT=
$\text{Properties of correlation coefficient:}$ 
*Because of normalization, $-1 \le ρ_{XY} ≤ +1$ always holds .
*If the two random variables  $X$  and  $Y$ are uncorrelated, then  $ρ_{XY} = 0$.
*For strict linear dependence between  $X$  and  $Y$,  $ρ_{XY}= ±1$   ⇒   complete correlation.
*A positive correlation coefficient means that when  $X$ is larger, on statistical average,  $Y$  is also larger than when  $X$ is smaller.
*In contrast, a negative correlation coefficient expresses that  $Y$  becomes smaller on average as  $X$  increases}}.
 

===2D–PDF for Gaussian random variables===

For the special case  '''Gaussian random variables'''  - the name goes back to the scientist  [https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss "Carl Friedrich Gauss"]  - we can further note:
*The joint PDF of a Gaussian 2D random variable  $XY$  with means  $m_X = 0$  and  $m_Y = 0$  and the correlation coefficient  $ρ = ρ_{XY}$  is:
: $$f_{XY}(x, y)=\frac{\rm 1}{\rm 2\it\pi \cdot \sigma_X \cdot \sigma_Y \cdot \sqrt{\rm 1-\rho^2}}\ \cdot\ \exp\Bigg[-\frac{\rm 1}{\rm 2 \cdot (1- \it\rho^{\rm 2} {\rm)}}\cdot(\frac {\it x^{\rm 2}}{\sigma_X^{\rm 2}}+\frac {\it y^{\rm 2}}{\sigma_Y^{\rm 2}}-\rm 2\it\rho\cdot\frac{x \cdot y}{\sigma_x \cdot \sigma_Y}\rm ) \rm \Bigg]\hspace{0.8cm}{\rm with}\hspace{0.5cm}-1 \le \rho \le +1.$$
*Replacing  $x$  by  $(x - m_X)$  and  $y$  by  $(y- m_Y)$, we obtain the more general PDF of a two-dimensional Gaussian random variable with mean.
*The marginal probability density functions  $f_{X}(x)$  and  $f_{Y}(y)$  of a 2D Gaussian random variable are also Gaussian with the standard deviations  $σ_X$  and  $σ_Y$, respectively.
*For uncorrelated components  $X$  and  $Y$, in the above equation  $ρ = 0$  must be substituted, and then the result is obtained:
:$$f_{XY}(x,y)=\frac{1}{\sqrt{2\pi}\cdot\sigma_{X}} \cdot\rm e^{-\it {x^{\rm 2}}\hspace{-0.08cm}/{\rm (}{\rm 2\hspace{0.05cm}\it\sigma_{X}^{\rm 2}} {\rm )}} \cdot\frac{1}{\sqrt{2\pi}\cdot\sigma_{\it Y}}\cdot e^{-\it {y^{\rm 2}}\hspace{-0.08cm}/{\rm (}{\rm 2\hspace{0.05cm}\it\sigma_{Y}^{\rm 2}} {\rm )}} = \it f_{X} \rm ( \it x \rm ) \cdot \it f_{Y} \rm ( \it y \rm ) .$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  In the special case of a 2D random variable with Gaussian PDF  $f_{XY}(x, y)$  it also follows directly from  ''uncorrelatedness''  the  ''statistical independence:''
:$$f_{XY}(x,y)= f_{X}(x) \cdot f_{Y}(y) . $$

Please note:
*For no other PDF can the  ''uncorrelatedness''  be used to infer  ''statistical independence''  .
*But one can always   ⇒   infer  ''uncorrelatedness'' from  ''statistical independence''  for any 2D-PDF  $f_{XY}(x, y)$  because:
*If two random variables  $X$  and  $Y$  are completely (statistically) independent of each other, then of course there are no ''linear''  dependencies between them   ⇒   they are then also uncorrelated  ⇒   $ρ = 0$. }}
 
===Contour lines for uncorrelated random variables===

[[File:Sto_App_Bild2.png |frame| Contour lines of 2D-PDF with uncorrelated variables | right]]
From the conditional equation  $f_{XY}(x, y) = {\rm const.}$  the contour lines of the PDF can be calculated.

If the components  $X$  and  $Y$ are uncorrelated  $(ρ_{XY} = 0)$, the equation obtained for the contour lines is:

:$$\frac{x^{\rm 2}}{\sigma_{X}^{\rm 2}}+\frac{y^{\rm 2}}{\sigma_{Y}^{\rm 2}} =\rm const.$$
In this case, the contour lines describe the following figures:
*'''Circles'''  (if  $σ_X = σ_Y$,   green curve), or
*'''Ellipses'''  (for  $σ_X ≠ σ_Y$,   blue curve) in alignment of the two axes.
 
===Regression line===

As  '''regression line'''  is called the straight line  $y = K(x)$  in the  $(x, y)$–plane through the "center" $(m_X, m_Y)$. This has the following properties:
[[File:Sto_App_Bild1a.png|frame| Gaussian 2D PDF (approximation with $N$ measurement points) and correlation line  $y = K(x)$]]

*The mean square error from this straight line - viewed in  $y$–direction and averaged over all  $N$  measurement points - is minimal:
:$$\overline{\varepsilon_y^{\rm 2} }=\frac{\rm 1}{N} \cdot \sum_{\nu=\rm 1}^{N}\; \;\big [y_\nu - K(x_{\nu})\big ]^{\rm 2}={\rm minimum}.$$
*The correlation straight line can be interpreted as a kind of "statistical symmetry axis". The equation of the straight line in the general case is:
:$$y=K(x)=\frac{\sigma_Y}{\sigma_X}\cdot\rho_{XY}\cdot(x - m_X)+m_Y.$$

*The angle that the correlation line makes to the  $x$–axis is:
:$$\theta={\rm arctan}(\frac{\sigma_{Y} }{\sigma_{X} }\cdot \rho_{XY}).$$

===Contour lines for correlated random variables===

For correlated components  $(ρ_{XY} ≠ 0)$  the contour lines of the PDF are (almost) always elliptic, so also for the special case  $σ_X = σ_Y$.

Exception:  $ρ_{XY}=\pm 1$   ⇒   "Dirac-wall"; see  [[Aufgaben:Exercise_4.4:_Two-dimensional_Gaussian_PDF|"Exercise 4.4"]]  in the book "Stochastic Signal Theory", subtask  ''(5)''.
[[File:Sto_App_Bild3.png|right|frame|height lines of the two dimensional PDF with correlated quantities]]
Here, the determining equation of the PDF height lines is:

:$$f_{XY}(x, y) = {\rm const.} \hspace{0.5cm} \rightarrow \hspace{0.5cm} \frac{x^{\rm 2} }{\sigma_{X}^{\rm 2}}+\frac{y^{\rm 2} }{\sigma_{Y}^{\rm 2} }-{\rm 2}\cdot\rho_{XY}\cdot\frac{x\cdot y}{\sigma_X\cdot \sigma_Y}={\rm const.}$$
The graph shows a contour line in lighter blue for each of two different sets of parameters.

*The ellipse major axis is dashed in dark blue.
*The  [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables#Regression_line|"regression line"]]  $K(x)$  is drawn in red throughout.

Based on this plot, the following statements are possible:
*The ellipse shape depends not only on the correlation coefficient  $ρ_{XY}$  but also on the ratio of the two standard deviations  $σ_X$  and  $σ_Y$  .
*The angle of inclination  $α$  of the ellipse major axis (dashed straight line) with respect to the  $x$–axis also depends on  $σ_X$,  $σ_Y$  and  $ρ_{XY}$  :
:$$\alpha = {1}/{2} \cdot {\rm arctan } \big ( 2 \cdot \rho_{XY} \cdot \frac {\sigma_X \cdot \sigma_Y}{\sigma_X^2 - \sigma_Y^2} \big ).$$
*The (red) correlation line  $y = K(x)$  of a Gaussian 2D-random variable always lies below the (blue dashed) ellipse major axis.
* $K(x)$  can be geometrically constructed from the intersection of the contour lines and their vertical tangents, as indicated in the sketch in green color.
 
===Two dimensional cumulative distribution function   ⇒   2D–CDF===

{{BlaueBox|TEXT=
$\text{Definition:}$  The  '''2D cumulative distribution function'''  like the 2D-CDF, is merely a useful extension of the  [[Theory_of_Stochastic_Signals/Cumulative_Distribution_Function#CDF_for_continuous-valued_random_variables|"one-dimensional distribution function"]]  (PDF):
:$$F_{XY}(x,y) = {\rm Pr}\big [(X \le x) \cap (Y \le y) \big ] .$$}}

The following similarities and differences between the "1D–CDF" and the" 2D–CDF" emerge:
*The functional relationship between "2D–PDF" and "2D–CDF" is given by the integration as in the one-dimensional case, but now in two dimensions. For continuous random variables, the following holds:
:$$F_{XY}(x,y)=\int_{-\infty}^{y} \int_{-\infty}^{x} f_{XY}(\xi,\eta) \,\,{\rm d}\xi \,\, {\rm d}\eta .$$
*Inversely, the probability density function can be given from the cumulative distribution function by partial differentiation to  $x$  and  $y$  :
:$$f_{XY}(x,y)=\frac{{\rm d}^{\rm 2} F_{XY}(\xi,\eta)}{{\rm d} \xi \,\, {\rm d} \eta}\Bigg|_{\left.{x=\xi \atop {y=\eta}}\right.}.$$
*In terms of the cumulative distribution function  $F_{XY}(x, y)$  the following limits apply:
:$$F_{XY}(-\infty,\ -\infty) = 0,\hspace{0.5cm}F_{XY}(x,\ +\infty)=F_{X}(x ),\hspace{0.5cm}
F_{XY}(+\infty,\ y)=F_{Y}(y ) ,\hspace{0.5cm}F_{XY}(+\infty,\ +\infty) = 1.$$
*In the limiting case $($infinitely large  $x$  and  $y)$  thus the value  $1$ is obtained for the "2D–CDF". From this we obtain the  '''normalization condition'''  for the two-dimensional probability density function:
:$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} f_{XY}(x,y) \,\,{\rm d}x \,\,{\rm d}y=1 . $$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Note the significant difference between one-dimensional and two-dimensional random variables:
*For one-dimensional random variables, the area under the PDF always yields $1$.
*For two-dimensional random variables, the PDF volume always equals $1$.}}
 

==Exercises==
 
*Select the number  $(1,\ 2$, ... $)$  of the task to be processed.  The number "0" corresponds to a "Reset":  Setting as at the program start.
*A task description is displayed.  Parameter values are adjusted.  Solution after pressing "Sample solution". 
*In the task description, we use  $\rho$  instead of  $\rho_{XY}$.
*For the one-dimensional Gaussian PDF holds:  $f_{X}(x) = \sqrt{1/(2\pi \cdot \sigma_X^2)} \cdot {\rm e}^{-x^2/(2 \hspace{0.05cm}\cdot \hspace{0.05cm} \sigma_X^2)}$.

{{BlueBox|TEXT=
'''(1)'''  Get familiar with the program using the default  $(\sigma_X=1, \ \sigma_Y=0.5, \ \rho = 0.7)$.  Interpret the graphs for  $\rm PDF$  and  $\rm CDF$.}}

* $\rm PDF$  is a ridge with the maximum at  $x = 0, \ y = 0$.  The ridge is slightly twisted with respect to the  $x$–axis.
* $\rm CDF$  is obtained from  $\rm PDF$  by continuous integration in both directions.  The maximum $($near  $1)$  occurs at  $x=3, \ y=3$.

{{BlueBox|TEXT=
'''(2)'''  The new setting is  $\sigma_X= \sigma_Y=1, \ \rho = 0$.  What are the values for  $f_{XY}(0,\ 0)$  and  $F_{XY}(0,\ 0)$?  Interpret the results}}

* The PDF maximum is  $f_{XY}(0,\ 0) = 1/(2\pi)= 0.1592$, because of  $\sigma_X= \sigma_Y = 1, \ \rho = 0$.  The contour lines are circles.
* For the CDF value:  $F_{XY}(0,\ 0) = [{\rm Pr}(X \le 0)] \cdot [{\rm Pr}(Y \le 0)] = 0.25$.  Minor deviation due to numerical integration.

{{BlueBox|TEXT=
'''(3)'''  The settings of  $(2)$  continue to apply.  What are the values for  $f_{XY}(0,\ 1)$  and  $F_{XY}(0,\ 1)$?  Interpret the results.}}

* It holds  $f_{XY}(0,\ 1) = f_{X}(0) \cdot f_{Y}(1) = [ \sqrt{1/(2\pi)}] \cdot [\sqrt{1/(2\pi)} \cdot {\rm e}^{-0.5}] = 1/(2\pi) \cdot {\rm e}^{-0.5} = 0.0965$.
* The program returns  $F_{XY}(0,\ 1) = [{\rm Pr}(X \le 0)] \cdot [{\rm Pr}(Y \le 1)] = 0.4187$, i.e. a larger value than in  $(2)$,  since it integrates over a wider range.

{{BlueBox|TEXT=
'''(4)'''  The settings are kept.  What values are obtained for  $f_{XY}(1,\ 0)$  and  $F_{XY}(1,\ 0)$?  Interpret the results}}

* Due to rotational symmetry, same results as in  $(3)$.

{{BlueBox|TEXT=
'''(5)'''  Is the statement true: "Elliptic contour lines exist only for  $\rho \ne 0$".  Interpret the  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$  and  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}CDF$  for  $\sigma_X=1, \ \sigma_Y=0.5$  and  $\rho = 0$.}}

* No!  Also, for  $\ \rho = 0$  the contour lines are elliptical  (not circular)  if  $\sigma_X \ne \sigma_Y$.
* For $\sigma_X \gg \sigma_Y$  the  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$  has the shape of an elongated ridge parallel to  $x$–axis, for $\sigma_X \ll \sigma_Y$  parallel to  $y$–axis.
* For $\sigma_X \gg \sigma_Y$  the slope of  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}CDF$  in the direction of the  $y$–axis is much steeper than in the direction of the  $x$–axis.

{{BlueBox|TEXT=
'''(6)'''  Starting from  $\sigma_X=\sigma_Y=1\ \rho = 0.7$  vary the correlation coefficient  $\rho$.  What is the slope angle  $\alpha$  of the ellipse main axis?}}

* For  $\rho > 0$:   $\alpha = 45^\circ$.     For  $\rho < 0$:   $\alpha = -45^\circ$.  For  $\rho = 0$:  The contour lines are circular and thus there are no ellipses main axis.

{{BlueBox|TEXT=
'''(7)'''  Starting from  $\sigma_X=\sigma_Y=1\ \rho = 0.7$  vary the correlation coefficient  $\rho$.  What is the slope angle  $\theta$  of the correlation line  $K(x)$?}}

* For  $\sigma_X=\sigma_Y$:   $\theta={\rm arctan}\ (\rho)$.  The slope increases with increasing  $\rho > 0$.  In all cases,  $\theta < \alpha = 45^\circ$ holds. For  $\rho = 0.7$  this gives  $\theta = 35^\circ$.

{{BlueBox|TEXT=
'''(8)'''  Starting from  $\sigma_X=\sigma_Y=0.75, \ \rho = 0.7$  vary the parameters  $\sigma_Y$  and  $\rho $.  What statements hold for the angles  $\alpha$  and  $\theta$?}}

* For  $\sigma_Y<\sigma_X$:   $\alpha < 45^\circ$.     For  $\sigma_Y>\sigma_X$:   $\alpha > 45^\circ$.  For all settings:  '''The correlation line is below the ellipse main axis'''.

{{BlueBox|TEXT=
'''(9)'''  Assume  $\sigma_X= 1, \ \sigma_Y=0.75, \ \rho = 0.7$.  Vary  $\rho$.  How to construct the correlation line from the contour lines?}}

* The correlation line intersects all contour lines at that points where the tangent line is perpendicular to the contour line.

{{BlueBox|TEXT=
'''(10)'''  Now let be  $\sigma_X= \sigma_Y=1, \ \rho = 0.95$.  Interpret the  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$.  Which statements are true for the limiting case  $\rho \to 1$ ?}}

* The  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}WDF$  only has components near the ellipse main axis.  The correlation line is just below:  $\alpha = 45^\circ, \ \theta = 43.5^\circ$.
* In the limiting case  $\rho \to 1$  it holds  $\theta = \alpha = 45^\circ$.  Outside the correlation line, the  $\rm 2D\hspace{-0.1cm}-\hspace{-0.1cm}PDF$  would have no shares.  That is:
* Along the correlation line, there would be a  "Dirac wall"   ⇒   All values are infinitely large, nevertheless Gaussian weighted around the mean.

==Applet Manual==
 
[[File:Anleitung_2D-Gauss.png|left|500px|frame|Screen shot from the German version]]
 
    '''(A)'''     Parameter input via slider:  $\sigma_X$,  $\sigma_Y$ and  $\rho$.

    '''(B)'''     Selection:  Representation of PDF or CDF.

    '''(C)'''     Reset:  Setting as at program start.

    '''(D)'''     Display contour lines instead of one-dimensional PDF.

    '''(E)'''     Display range for two-dimensional PDF.

    '''(F)'''     Manipulation of the three-dimensional graph (zoom, rotate, ...)

    '''(G)'''     Display range for  "one-dimensional PDF"  or  "contour lines".

    '''(H)'''     Manipulation of the two-dimensional graphics ("one-dimensional PDF")

    '''( I )'''     Area for exercises: Task selection.

    '''(J)'''     Area for exercises: Task description

    '''(K)'''     Area for exercises: Show/hide solution

    '''( L)'''     Area for exercises: Output of the sample solution

Note:    Value output of the graphics  $($both 2D and 3D$)$  via mouse control.
 

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2003 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]   as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2019 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|gauss_en|gauss}}

Applets:Digital Filters

2023-04-16T19:46:41Z

Noah:

{{LntAppletLinkEnDe|digitalFilters_en|digitalFilters}}

==Applet Description==
 
The applet should clarify the properties of digital filters, whereby we confine ourselves to filters of the order $M=2$. Both non-recursive filters $\rm (FIR$,  ''Finite Impulse Response''$)$  as well as recursive filters $\rm (IIR$,  ''Infinite Impulse Response''$)$.

The input signal $x(t)$ is represented by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$. The output sequence $〈y_ν〉$is calculated, i.e. the discrete-time representation of the output signal $y(t)$.

*$T_{\rm A}$ denotes the time interval between two samples.
*We also limit ourselves to causal signals and systems, which means that $x_ν \equiv 0$ and $y_ν \equiv 0$ for $ν \le 0$.

It should also be noted that we denote the initial sequence $〈y_ν〉$ as

'''(1)''' the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:         $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉,$

'''(2)''' the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:         $〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉,$

'''(3)''' the '''discrete-time rectangle response''' $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangle function” is present at the input:     $〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉;$         In quotation marks are the beginning of the ones $(2)$ and the position of the last ones $(4)$.

==Theoretical background==
 
===General block diagram===

Each signal $x(t)$ can only be represented on a computer by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$.
[[File:P_ID552__Sto_T_5_2_S1_neu.png|right |frame| Block diagram of a digital (IIR–) filter $M$–order]]
*The time interval $T_{\rm A}$ between two samples is limited by the [https://en.lntwww.de/Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|"sampling theorem"].
*We limit ourselves here to causal signals and systems, which means that $x_ν \equiv 0$ for $ν \le 0$.

*In order to determine the influence of a linear filter with frequency response $H(f)$ on the discrete-time input signal $〈x_ν〉$, it is advisable to describe the filter discrete-time. In the time domain, this happens with the discrete-time impulse response $〈h_ν〉$.
*On the right you can see the corresponding block diagram. The following therefore applies to the samples of the output signal $〈y_ν〉$ thus holds:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu } \cdot x_{\nu - \mu } + \sum\limits_{\mu = 1}^M {b_\mu } \cdot y_{\nu - \mu } .$$

The following should be noted here:
*The index $\nu$ refers to sequences, for example at the input $〈x_ν〉$ and output $〈y_ν〉$.
*On the other hand, we use the index $\mu$ to identify the $a$ and $b$ filter coefficients.
*The first sum describes the dependency of the current output $y_ν$ on the current input $x_ν$ and on the $M$ previous input values $x_{ν-1}$, ... , $x_{ν-M}$.
*The second sum indicates the influence of $y_ν$ by the previous values $y_{ν-1}$, ... , $y_{ν-M}$ at the filter output. It specifies the recursive part of the filter.
*The integer parameter $M$ is called the order of the digital filter. In the program, this value is limited to $M\le 2$.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

'''(1)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:
:$$〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉 .$$
'''(2)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:
:$$〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉 .$$
'''(3)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time rectangle response'''  $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangular function” is present at the input:
:$$〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉 .$$
:The beginning of ones $(2)$ and the position of the last ones $(4)$ are given in single quotes.
}}

===Non-recursive filter   ⇒   FIR–filter ===

{{BlaueBox|TEXT=
[[File:P_ID553__Sto_T_5_2_S2_neu.png|right |frame| Non-recursive digital filter  $($FIR filter$)$  $M$ order]]
$\text{Definition:}$ If all feedback coefficients $b_{\mu} = 0$ , one speaks of one '''non-recursive filter'''. In the English language literature, the term '''FIR filter''' (''Finite Impulse Response'') is also used for this.

The following applies to the order $M$ applies:

*The output value $y_ν$ depends only on the current and the previous $M$ input values:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu \cdot x_{\mu - \nu } } .$$
*Discrete-time impulse response with $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉$:
:$$〈h_\mu〉= 〈a_0,\ a_1,\ \text{...},\ a_M〉 .$$}}

{{GraueBox|TEXT=
$\text{Example 1:}$  A two-way channel where
*the signal on the main path arrives undamped compared to the input signal but is delayed by $2\ \rm µ s$ arrives with a delay, and
*at $4\ \rm µ s$ distance – so absolutely at time $t = 6\ \rm µ s$ – follows an echo with half the amplitude,

can be simulated by a non-recursive filter according to the sketch above, whereby the following parameter values must be set:
:$$M = 3,\quad T_{\rm A} = 2\;{\rm{µ s} },\quad a_{\rm 0} = 0,\quad a_{\rm 1} = 1, \quad a_{\rm 2} = 0, \quad a_{\rm 3} = 0.5.$$}}

{{GraueBox|TEXT=
$\text{Example 2:}$ Consider a non-recursive filter with the filter coefficients $a_0 = 1,\hspace{0.5cm} a_1 = 2,\hspace{0.5cm} a_2 = 1.$
[[File:P_ID608__Sto_Z_5_3.png|right|frame|Non-recursive filter]]

'''(1)''' The conventional impulse response is: $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$         ⇒   discrete-time impulse response: $〈h_\mu〉= 〈1,\ 2,\ 1〉 .$

'''(2)'''   The frequency response $H(f)$ is the Fourier transform of $h(t)$. By applying the displacement theorem:
:$$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }\cdot }f \cdot T_{\rm A} } )} \big ] \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }fT_{\rm A} }\hspace{0.5cm}\Rightarrow \hspace{0.5cm}H(f = 0) = 4.$$

'''(3)'''   It follows that the '''discrete-time step response''' $〈\sigma_ν〉$ tends to become $4$ for large $\nu$.

'''(4)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;0,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;2,\;1,\;0,\;1,\;2,\;1,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$

'''(5)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;1,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;3,\;2,\;2,\;1,\;0,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$}}

===Recursive filter   ⇒   IIR filter ===

{{BlaueBox|TEXT=
[[File:P_ID607__Sto_A_5_3.png|right|frame|First order recursive filter]]
$\text{Definition:}$ 
*If at least one of the feedback coefficients is $b_{\mu} \ne 0$, then this is referred to as a '''recursive filter''' (see graphic on the right). The term '''IIR filter'''  (''Infinite Impulse Response'') is also used for this, particularly in the English-language literature. This filter is dealt with in detail in the trial implementation.

*If all forward coefficients are also identical $a_\mu = 0$ with the exception of $a_0$, a '''purely recursive filter''' is available (see graphic on the left).

[[File:P_ID554__Sto_T_5_2_S3_neu.png|left|frame| Purely recursive first order filter]] }}

In the following we restrict ourselves to the special case “purely recursive filter of the first order”. This filter has the following properties:
*The output value $y_ν$ depends (indirectly) on an infinite number of input values:
:$$y_\nu = \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot x_{\nu - \mu } .}$$
*This shows the following calculation:
:$$y_\nu = a_0 \cdot x_\nu + b_1 \cdot y_{\nu - 1} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + {b_1} ^2 \cdot y_{\nu - 2} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + a_0 \cdot {b_1} ^2 \cdot x_{\nu - 2} + {b_1} ^3 \cdot y_{\nu - 3} = \text{...}. $$

*By definition, the discrete-time impulse response is the same as the output sequence if there is a single "one" at $t =0$ at the input.
:$$h(t)= \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot \delta ( {t - \mu \cdot T_{\rm A} } )}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉= 〈\hspace{0.05cm}a_0, \ a_0\cdot {b_1}, \ a_0\cdot {b_1}^2 \ \text{...} \hspace{0.05cm}〉.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  With a recursive filter, the (discrete-time) impulse response extends to infinity with $M = 1$:
*For reasons of stability, $b_1 < 1$ must apply.
*With $b_1 = 1$ the impulse response $h(t)$ would extend to infinity and with $b_1 > 1$ the variable $h(t)$ would even continue to infinity.
*With such a recursive filter of the first order, each individual Dirac delta line is exactly the factor $b_1$ smaller than the previous Dirac delta line:
:$$h_{\mu} = h(\mu \cdot T_{\rm A}) = {b_1} \cdot h_{\mu -1}.$$}}

{{GraueBox|TEXT=
[[File:Sto_T_5_2_S3_version2.png |frame| Discrete-time impulse response]]
$\text{Example 3:}$  The graphic opposite shows the discrete-time impulse response $〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉$ of a recursive filter of the first order with the parameters $a_0 = 1$ and $b_1 = 0.6$.
*The (discrete-time) course is exponentially falling and extends to infinity.
*The ratio of the weights of two successive Dirac deltas is $b_1 = 0.6$.
}}

===Recursive filter as a sine generator===
[[File:EN_Sto_A_5_4_version2.png|right|frame|Proposed filter structure]]

The graphic shows a second-order digital filter that is suitable for generating a discrete-time sine function on a digital signal processor (DSP) if the input sequence $\left\langle \hspace{0.05cm} {x_\nu } \hspace{0.05cm}\right\rangle$  a (discrete-time) Dirac delta function is:
:$$\left\langle \hspace{0.05cm}{y_\nu }\hspace{0.05cm} \right\rangle = \left\langle {\, \sin ( {\nu \cdot T_{\rm A} \cdot \omega _0 } )\, }\right\rangle .$$

The five filter coefficients result from the:
[https://en.wikipedia.org/wiki/Z-transform "$Z$-transform"]:
:$$Z \big \{ {\sin ( {\nu T{\rm A}\cdot \omega _0 } )} \big \} = \frac{{z \cdot \sin \left( {\omega _0 \cdot T_{\rm A}} \right)}}{{z^2 - 2 \cdot z \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right) + 1}}.$$
After implementing this equation using a second-order recursive filter, the following filter coefficients are obtained:
:$$a_0 = 0,\quad a_1 = \sin \left( {\omega _0 \cdot T_{\rm A}} \right),\quad a_2 = 0, \quad b_1 = 2 \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right),\quad b_2 = - 1.$$

*The filter coefficients $a_0$ and $a_2$ can be omitted and $b_2=-1$ has a fixed value.
*The angular frequency $\omega_0$ of the sine wave is therefore only determined by $a_0$ and $a_0$.

{{GraueBox|TEXT=
$\text{Example 3:}$  Let $a_1 = 0.5$, $b_1 = \sqrt 3$, $x_0 = 1$ and $x_{\nu \hspace{0.05cm}\ne\hspace{0.05cm} 0} = 0$.

'''(1)'''  Then the following applies to the initial values $y_\nu$ at times $\nu \ge 0$: 
:*  $y_0 = 0;$
:*  $y_1 = 0.5$                                                                                         ⇒  the "$1$" at the input only has an effect at time $\nu = 1$ because of $a_0= 0$ at the output;
:*  $y_2 = b_1 \cdot y_1 - y_0 = {\sqrt 3 }/{2} \approx 0.866$                             ⇒   with $\nu = 2$ the recursive part of the filter also takes effect;
:*  $y_3 = \sqrt 3 \cdot y_2 - y_1 = \sqrt 3 \cdot {\sqrt 3 }/{2} - {1}/{2} = 1$          ⇒  for  $\nu \ge 2$  the filter is purely recursive:     $y_\nu = b_1 \cdot y_{\nu - 1} - y_{\nu - 2}$;
:*  $y_4 = \sqrt 3 \cdot y_3 - y_2 = \sqrt 3 \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$
:*  $y_5 = \sqrt 3 \cdot y_4 - y_3 = \sqrt 3 \cdot {\sqrt 3 }/{2} - 1 = 0.5;$
:*  $y_6 = \sqrt 3 \cdot y_5 - y_4 = \sqrt 3 \cdot {1}/{2} - {\sqrt 3 }/{2} = 0;$
:*  $y_7 = \sqrt 3 \cdot y_6 - y_5 = \sqrt 3 \cdot 0 - {1}/{2} = - 0.5.$

'''(2)'''  By continuing the recursive algorithm one gets for large $\nu$–values:     $y_\nu = y_{\nu - 12}$   ⇒   $T_0/T_{\rm A}= 12.$ }}

==Exercises==

[[File:Exercises_binomial_fertig.png|right]]
*First select the number '''1''' ... '''10''' of the task to be processed.
*A task description is displayed.  The parameter values are adjusted.
*Solution after pressing  "Sample Solution".
*The number '''0''' corresponds to a "reset":  Same setting as when the program was started.
 

{{BlaueBox|TEXT=
'''(1)'''  The filter coefficients are  $a_0=0.25$,  $a_1=0.5$,  $a_2=0.25$,  $b_1=b_2=0$.  Which filter is it?         Interpret the impulse response  $〈h_ν〉$,  the step response  $〈\sigma_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$, each in a discrete-time representation.}}

:*  Due to the missing  $b$ coefficients, it is a non-recursive digital filter ⇒   '''FIR filter''' (''Finite Impulse Response'').
:*  The impulse response consists of  $M+1=3$  Dirac delta lines according to the  $a$  coefficients:  $〈h_ν〉= 〈a_0, \ a_1,\ a_2〉= 〈0.25, \ 0.5,\ 0.25,\ 0, \ 0, \ 0,\text{...}〉 $.
:*  The step response is:  $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1,\text{...}〉 $.  The final value is equal to the DC signal transfer factor  $H(f=0)=a_0+a_1+a_2 = 1$.
:*  The distortions with rise and fall can also be seen from the rectangular response  $〈\rho_ν^{(2, 8)}〉= 〈0,\ 0, 0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1, \ 1, \ 0.75, \ 0.25, \ \text{...}〉$.

{{BlaueBox|TEXT=
'''(2)'''  How do the results differ with  $a_2=-0.25$? }}

:*  Taking into account  $H(f=0)= 0.5$  there are comparable consequences   ⇒   Step response:    $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 0.5,\ 0.5, \ 0.5, \ 0.5,\text{...}〉 $.

{{BlaueBox|TEXT=
'''(3)'''  Now let the filter coefficients  $a_0=1$,  $b_1=0.9$  and  $a_1=a_2= b_2=0$.  Which filter is it?  Interpret the impulse response  $〈h_ν〉$.}}

:*  It is a recursive digital filter   ⇒   '''IIR filter'''  (''Infinite Impulse Response'')  of the first order.  It is the discrete-time analogon of the RC low-pass.
:*  Starting from  $h_0= 1$ is $h_1= h_0 \cdot b_0= 0.9$,  $h_2= h_1 \cdot b_0= b_0^2=0.81$,  $h_3= h_2 \cdot b_0= b_0^3=0.729$,  and so on   ⇒   $〈h_ν〉$  extends to infinity.
:*  Impulse response  $h(t) = {\rm e}^{-t/T}$  with  $T$:   intersection $($Tangente bei  $t=0$, Abscissa$)$   ⇒   $h_\nu= h(\nu \cdot T_{\rm A}) = {\rm e}^{-\nu/(T/T_{\rm A})}$  with  $T/T_{\rm A} = 1/(h_0-h_1)= 10$.
:*  So:  The values of the continuous time differ from the discrete-time impulse response.  This results in the values  $1.0, \ 0.9048,\ 0.8187$ ...

{{BlaueBox|TEXT=
'''(4)'''  The filter setting is retained.  Interpret the step response  $〈h_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$.  What is the value for  $H(f=0)$?}}

:*  The step response is the integral over the impulse response:   $\sigma(t) = T \cdot (1-{\rm e}^{-t/T}) ]$   ⇒   $\sigma_\nu= 10 \cdot (1-{\rm e}^{-\nu/10})$   ⇒   $\sigma_0=1$,  $\sigma_1=1.9$,  $\sigma_2=2.71$, ...
:*  For large  $\nu$  values, the (discrete-time) step response tends to the DC signal transmission factor  $H(f=0)= 10$:  $\sigma_{40}=9.867$,  $\sigma_{50}=9.954$,  $\sigma_\infty=10$.
:* The rectangular response  $〈\rho_ν^{(2, 8)}〉$  increases with a delay of  $2$  in the same way as  $〈\sigma_ν〉$.  In the area  $\nu \ge 8$  the  $\rho_ν$  values decrease exponentially.

{{BlaueBox|TEXT=
'''(5)'''  We continue to consider the filter with  $a_0=1$,  $b_1=0.9$,  $a_1=a_2=b_2=0$.  What is the output sequence  $〈y_ν〉$  for the input sequence  $〈x_ν〉= 〈1,\ 0,\ -0.5〉$?         ''Note'': The task can also be solved with this program, although the constellation considered here cannot be set directly.}}

:*  You can help yourself by setting the coefficient  $a_2=-0.5$  and reducing the input sequence  to $〈x_ν〉= 〈1,\ 0,\ 0,\ \text{ ...}〉$   ⇒   „Dirac delta function”.
:*  The actual impulse response of this filter  $($with  $a_2=0)$  was determined in task  '''(3)''':   $h_0= 1$,   $h_1= 0.9$,   $h_2= 0.81$,   $h_3= 0.729$,   $h_4= 0.646$.  
:*  The solution to this problem is:  $y_0 = h_0= 1$,   $y_1= h_1= 0.9$,   $y_2 =h_2-h_0/2= 0.31$,   $y_3 =h_3-h_1/2= 0.279$,   $y_4 =h_4-h_2/2= 0.251$.  
:*  Caution:  Step response and rectangular response now refer to the fictitious filter  $($with  $a_2=-0.5)$  and not to the actual filter  $($with  $a_2=0)$.

{{BlaueBox|TEXT=
'''(6)'''  Consider and interpret the impulse response and the step response for the filter coefficients  $a_0=1$,  $b_1=1$,  $a_1=a_2= b_2=0$.  }}

:*  '''The system is unstable''':   A discrete-time Dirac delta function at input $($at time  $t=0)$  causes an infinite number of Dirac deltas of the same height in the output signal.
:*  A discrete-time step function at the input causes an infinite number of Dirac deltas with monotonically increasing weights (to infinity) in the output signal.

{{BlaueBox|TEXT=
'''(7)'''  Consider and interpret the impulse response and step response for the filter coefficients  $a_0=1$,  $b_1=-1$,  $a_1=a_2= b_2=0$.  }}
:*  In contrast to exercise  '''(6)''', the weights of the impulse response  $〈h_ν〉$  are not constantly equal to  $1$, but alternating  $\pm 1$.  The system is unstable too.
:*  With the jump response  $〈\sigma_ν〉$, however, the weights alternate between  $0$  $($with even $\nu)$  and  $1$  $($with odd $\nu)$.

{{BlaueBox|TEXT=
'''(8)'''  We consider the "sine generator":  $a_1=0.5$,  $b_1=\sqrt{3}= 1.732$,  $b_2=-1.$  Compare the impulse response with the calculated values in  $\text{Example 4}$.         How do the parameters $a_1$ and $b_1$ influence the period duration  $T_0/T_{\rm A}$  and the amplitude  $A$  of the sine function?}}
:*  $〈x_ν〉=〈1, 0, 0, \text{...}〉$   ⇒   $〈y_ν〉=〈0, 0.5, 0.866, 1, 0.866, 0.5, 0, -0.5, -0.866, -1, -0.866, -0.5, 0, \text{...}〉$   ⇒   '''sine''',  period  $T_0/T_{\rm A}= 12$,  amplitude  $1$.
:*  The increase/decrease of $b_1$  leads to the larger/smaller period  $T_0/T_{\rm A}$  and the larger/smaller amplitude  $A$.  $b_1 < 2$ must apply.
:*  $a_1$  only affects the amplitude, not the period.  There is no value limit for  $a_1$. If  $a_1$  is negative, the minus sine function results.

{{BlaueBox|TEXT=
'''(9)'''  The basic setting is retained.  Which  $a_1$  and  $b_1$  result in a sine function with period  $T_0/T_{\rm A}=16$  and amplitude  $A=1$?}}
:*  Trying with  $b_1= 1.8478$  actually achieves the period duration  $T_0/T_{\rm A}=16$.  However, this increases the amplitude to  $A=1.307$.
:*  Adjusting the parameter  $a_1= 0.5/1.307=0.3826$  then leads to the desired amplitude  $A=1$.
:*  Or you can calculate this as in the example:  $b_1 = 2 \cdot \cos ( {2{\rm{\pi }}\cdot{T_{\rm A}}/{T_0 }})= 2 \cdot \cos (\pi/8)=1.8478$,     $a_1 = \sin (\pi/8)=0.3827$.

{{BlaueBox|TEXT=
'''(10)'''  We continue with the  "sine generator".  What modifications do you have to make to generate a  "cosine"?}}
:*  With  $a_1=0.3826$,  $b_1=1.8478$,  $b_2=-1$  and  $〈x_ν〉=〈1, 1, 1, \text{...}〉$  is the output sequence  $〈y_ν〉$  the discrete-time analogon of the step response  $\sigma(t)$.
:*  The step response is the integral over   $\sin(\pi\cdot\tau/8)$   within the limits of   $\tau=0$   to   $\tau=t$   ⇒   $\sigma(t)=-8/\pi\cdot\cos(\pi\cdot\tau/8)+1$.
:*  If you change   $a_1=0.3826$   on   $a_1=-0.3826\cdot\pi/8=-0.1502$, then   $\sigma(t)=\cos(\pi\cdot\tau/8)-1$   ⇒   Values between  $0$  and  $-2$.
:*  Would you still in the block diagram   $z_\nu=y_\nu+1$   add, then   $z_\nu$   a discrete-time cosine curve with   $T_0/T_{\rm A}=16$   and   $A=1$.

==Applet Manual==
 
'''Korrektur'''
==About the authors==
 
This interactive calculation tool was designed and implemented at the [http://www.lnt.ei.tum.de/startseite chair for communications engineering] at the [https://www.tum.de/ Technische Universität München].
*The first version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Andr.C3.A9_Schulz_.28Bachelorarbeit_LB_2020.29|André Schulz]] (Bachelor thesis LB, Supervisors: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_Übertragungstechnik#Benedikt_Leible.2C_M.Sc._.28bei_L.C3.9CT_seit_2017.29|Benedikt Leible]] and [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".

==Once again: Open Applet in new Tab==

{{LntAppletLink|digitalFilters_en}}

Applets:Digital Filters

2023-04-16T19:44:11Z

Noah:

{{LntAppletLinkEnDe|digitalFilters_en|digitalFilters}}

==Applet Description==
 
The applet should clarify the properties of digital filters, whereby we confine ourselves to filters of the order $M=2$. Both non-recursive filters $\rm (FIR$,  ''Finite Impulse Response''$)$  as well as recursive filters $\rm (IIR$,  ''Infinite Impulse Response''$)$.

The input signal $x(t)$ is represented by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$. The output sequence $〈y_ν〉$is calculated, i.e. the discrete-time representation of the output signal $y(t)$.

*$T_{\rm A}$ denotes the time interval between two samples.
*We also limit ourselves to causal signals and systems, which means that $x_ν \equiv 0$ and $y_ν \equiv 0$ for $ν \le 0$.

It should also be noted that we denote the initial sequence $〈y_ν〉$ as

'''(1)''' the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:         $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉,$

'''(2)''' the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:         $〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉,$

'''(3)''' the '''discrete-time rectangle response''' $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangle function” is present at the input:     $〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉;$         In quotation marks are the beginning of the ones $(2)$ and the position of the last ones $(4)$.

==Theoretical background==
 
===General block diagram===

Each signal $x(t)$ can only be represented on a computer by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$.
[[File:P_ID552__Sto_T_5_2_S1_neu.png|right |frame| Block diagram of a digital (IIR–) filter $M$–order]]
*The time interval $T_{\rm A}$ between two samples is limited by the [https://en.lntwww.de/Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|"sampling theorem"].
*We limit ourselves here to causal signals and systems, which means that $x_ν \equiv 0$ for $ν \le 0$.

*In order to determine the influence of a linear filter with frequency response $H(f)$ on the discrete-time input signal $〈x_ν〉$, it is advisable to describe the filter discrete-time. In the time domain, this happens with the discrete-time impulse response $〈h_ν〉$.
*On the right you can see the corresponding block diagram. The following therefore applies to the samples of the output signal $〈y_ν〉$ thus holds:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu } \cdot x_{\nu - \mu } + \sum\limits_{\mu = 1}^M {b_\mu } \cdot y_{\nu - \mu } .$$

The following should be noted here:
*The index $\nu$ refers to sequences, for example at the input $〈x_ν〉$ and output $〈y_ν〉$.
*On the other hand, we use the index $\mu$ to identify the $a$ and $b$ filter coefficients.
*The first sum describes the dependency of the current output $y_ν$ on the current input $x_ν$ and on the $M$ previous input values $x_{ν-1}$, ... , $x_{ν-M}$.
*The second sum indicates the influence of $y_ν$ by the previous values $y_{ν-1}$, ... , $y_{ν-M}$ at the filter output. It specifies the recursive part of the filter.
*The integer parameter $M$ is called the order of the digital filter. In the program, this value is limited to $M\le 2$.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

'''(1)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:
:$$〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉 .$$
'''(2)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:
:$$〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉 .$$
'''(3)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time rectangle response'''  $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangular function” is present at the input:
:$$〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉 .$$
:The beginning of ones $(2)$ and the position of the last ones $(4)$ are given in single quotes.
}}

===Non-recursive filter   ⇒   FIR–filter ===

{{BlaueBox|TEXT=
[[File:P_ID553__Sto_T_5_2_S2_neu.png|right |frame| Non-recursive digital filter  $($FIR filter$)$  $M$ order]]
$\text{Definition:}$ If all feedback coefficients $b_{\mu} = 0$ , one speaks of one '''non-recursive filter'''. In the English language literature, the term '''FIR filter''' (''Finite Impulse Response'') is also used for this.

The following applies to the order $M$ applies:

*The output value $y_ν$ depends only on the current and the previous $M$ input values:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu \cdot x_{\mu - \nu } } .$$
*Discrete-time impulse response with $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉$:
:$$〈h_\mu〉= 〈a_0,\ a_1,\ \text{...},\ a_M〉 .$$}}

{{GraueBox|TEXT=
$\text{Example 1:}$  A two-way channel where
*the signal on the main path arrives undamped compared to the input signal but is delayed by $2\ \rm µ s$ arrives with a delay, and
*at $4\ \rm µ s$ distance – so absolutely at time $t = 6\ \rm µ s$ – follows an echo with half the amplitude,

can be simulated by a non-recursive filter according to the sketch above, whereby the following parameter values must be set:
:$$M = 3,\quad T_{\rm A} = 2\;{\rm{µ s} },\quad a_{\rm 0} = 0,\quad a_{\rm 1} = 1, \quad a_{\rm 2} = 0, \quad a_{\rm 3} = 0.5.$$}}

{{GraueBox|TEXT=
$\text{Example 2:}$ Consider a non-recursive filter with the filter coefficients $a_0 = 1,\hspace{0.5cm} a_1 = 2,\hspace{0.5cm} a_2 = 1.$
[[File:P_ID608__Sto_Z_5_3.png|right|frame|Nichtrekursives Filter]]

'''(1)''' The conventional impulse response is: $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$         ⇒   discrete-time impulse response: $〈h_\mu〉= 〈1,\ 2,\ 1〉 .$

'''(2)'''   The frequency response $H(f)$ is the Fourier transform of $h(t)$. By applying the displacement theorem:
:$$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }\cdot }f \cdot T_{\rm A} } )} \big ] \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }fT_{\rm A} }\hspace{0.5cm}\Rightarrow \hspace{0.5cm}H(f = 0) = 4.$$

'''(3)'''   It follows that the '''discrete-time step response''' $〈\sigma_ν〉$ tends to become $4$ for large $\nu$.

'''(4)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;0,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;2,\;1,\;0,\;1,\;2,\;1,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$

'''(5)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;1,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;3,\;2,\;2,\;1,\;0,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$}}

===Recursive filter   ⇒   IIR filter ===

{{BlaueBox|TEXT=
[[File:P_ID607__Sto_A_5_3.png|right|frame|First order recursive filter]]
$\text{Definition:}$ 
*If at least one of the feedback coefficients is $b_{\mu} \ne 0$, then this is referred to as a '''recursive filter''' (see graphic on the right). The term '''IIR filter'''  (''Infinite Impulse Response'') is also used for this, particularly in the English-language literature. This filter is dealt with in detail in the trial implementation.

*If all forward coefficients are also identical $a_\mu = 0$ with the exception of $a_0$, a '''purely recursive filter''' is available (see graphic on the left).

[[File:P_ID554__Sto_T_5_2_S3_neu.png|left|frame| Purely recursive first order filter]] }}

In the following we restrict ourselves to the special case “purely recursive filter of the first order”. This filter has the following properties:
*The output value $y_ν$ depends (indirectly) on an infinite number of input values:
:$$y_\nu = \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot x_{\nu - \mu } .}$$
*This shows the following calculation:
:$$y_\nu = a_0 \cdot x_\nu + b_1 \cdot y_{\nu - 1} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + {b_1} ^2 \cdot y_{\nu - 2} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + a_0 \cdot {b_1} ^2 \cdot x_{\nu - 2} + {b_1} ^3 \cdot y_{\nu - 3} = \text{...}. $$

*By definition, the discrete-time impulse response is the same as the output sequence if there is a single "one" at $t =0$ at the input.
:$$h(t)= \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot \delta ( {t - \mu \cdot T_{\rm A} } )}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉= 〈\hspace{0.05cm}a_0, \ a_0\cdot {b_1}, \ a_0\cdot {b_1}^2 \ \text{...} \hspace{0.05cm}〉.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  With a recursive filter, the (discrete-time) impulse response extends to infinity with $M = 1$:
*For reasons of stability, $b_1 < 1$ must apply.
*With $b_1 = 1$ the impulse response $h(t)$ would extend to infinity and with $b_1 > 1$ the variable $h(t)$ would even continue to infinity.
*With such a recursive filter of the first order, each individual Dirac delta line is exactly the factor $b_1$ smaller than the previous Dirac delta line:
:$$h_{\mu} = h(\mu \cdot T_{\rm A}) = {b_1} \cdot h_{\mu -1}.$$}}

{{GraueBox|TEXT=
[[File:Sto_T_5_2_S3_version2.png |frame| Discrete-time impulse response | rechts]]
$\text{Example 3:}$  The graphic opposite shows the discrete-time impulse response $〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉$ of a recursive filter of the first order with the parameters $a_0 = 1$ and $b_1 = 0.6$.
*The (discrete-time) course is exponentially falling and extends to infinity.
*The ratio of the weights of two successive Diracs is $b_1 = 0.6$.
}}

===Recursive filter as a sine generator===
[[File:EN_Sto_A_5_4_version2.png|right|frame|Proposed filter structure]]

The graphic shows a second-order digital filter that is suitable for generating a discrete-time sine function on a digital signal processor (DSP) if the input sequence $\left\langle \hspace{0.05cm} {x_\nu } \hspace{0.05cm}\right\rangle$  a (discrete-time) Dirac delta function is:
:$$\left\langle \hspace{0.05cm}{y_\nu }\hspace{0.05cm} \right\rangle = \left\langle {\, \sin ( {\nu \cdot T_{\rm A} \cdot \omega _0 } )\, }\right\rangle .$$

The five filter coefficients result from the:
[https://en.wikipedia.org/wiki/Z-transform "$Z$-transform"]:
:$$Z \big \{ {\sin ( {\nu T{\rm A}\cdot \omega _0 } )} \big \} = \frac{{z \cdot \sin \left( {\omega _0 \cdot T_{\rm A}} \right)}}{{z^2 - 2 \cdot z \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right) + 1}}.$$
After implementing this equation using a second-order recursive filter, the following filter coefficients are obtained:
:$$a_0 = 0,\quad a_1 = \sin \left( {\omega _0 \cdot T_{\rm A}} \right),\quad a_2 = 0, \quad b_1 = 2 \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right),\quad b_2 = - 1.$$

*The filter coefficients $a_0$ and $a_2$ can be omitted and $b_2=-1$ has a fixed value.
*The angular frequency $\omega_0$ of the sine wave is therefore only determined by $a_0$ and $a_0$.

{{GraueBox|TEXT=
$\text{Example 3:}$  Let $a_1 = 0.5$, $b_1 = \sqrt 3$, $x_0 = 1$ and $x_{\nu \hspace{0.05cm}\ne\hspace{0.05cm} 0} = 0$.

'''(1)'''  Then the following applies to the initial values $y_\nu$ at times $\nu \ge 0$: 
:*  $y_0 = 0;$
:*  $y_1 = 0.5$                                                                                         ⇒  the "$1$" at the input only has an effect at time $\nu = 1$ because of $a_0= 0$ at the output;
:*  $y_2 = b_1 \cdot y_1 - y_0 = {\sqrt 3 }/{2} \approx 0.866$                             ⇒   with $\nu = 2$ the recursive part of the filter also takes effect;
:*  $y_3 = \sqrt 3 \cdot y_2 - y_1 = \sqrt 3 \cdot {\sqrt 3 }/{2} - {1}/{2} = 1$          ⇒  for  $\nu \ge 2$  the filter is purely recursive:     $y_\nu = b_1 \cdot y_{\nu - 1} - y_{\nu - 2}$;
:*  $y_4 = \sqrt 3 \cdot y_3 - y_2 = \sqrt 3 \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$
:*  $y_5 = \sqrt 3 \cdot y_4 - y_3 = \sqrt 3 \cdot {\sqrt 3 }/{2} - 1 = 0.5;$
:*  $y_6 = \sqrt 3 \cdot y_5 - y_4 = \sqrt 3 \cdot {1}/{2} - {\sqrt 3 }/{2} = 0;$
:*  $y_7 = \sqrt 3 \cdot y_6 - y_5 = \sqrt 3 \cdot 0 - {1}/{2} = - 0.5.$

'''(2)'''  By continuing the recursive algorithm one gets for large $\nu$–values:     $y_\nu = y_{\nu - 12}$   ⇒   $T_0/T_{\rm A}= 12.$ }}

==Exercises==

[[File:Exercises_binomial_fertig.png|right]]
*First select the number '''1''' ... '''10''' of the task to be processed.
*A task description is displayed.  The parameter values are adjusted.
*Solution after pressing  "Sample Solution".
*The number '''0''' corresponds to a "reset":  Same setting as when the program was started.
 

{{BlaueBox|TEXT=
'''(1)'''  The filter coefficients are  $a_0=0.25$,  $a_1=0.5$,  $a_2=0.25$,  $b_1=b_2=0$.  Which filter is it?         Interpret the impulse response  $〈h_ν〉$,  the step response  $〈\sigma_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$, each in a discrete-time representation.}}

:*  Due to the missing  $b$ coefficients, it is a non-recursive digital filter ⇒   '''FIR filter''' (''Finite Impulse Response'').
:*  The impulse response consists of  $M+1=3$  Dirac delta lines according to the  $a$  coefficients:  $〈h_ν〉= 〈a_0, \ a_1,\ a_2〉= 〈0.25, \ 0.5,\ 0.25,\ 0, \ 0, \ 0,\text{...}〉 $.
:*  The step response is:  $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1,\text{...}〉 $.  The final value is equal to the DC signal transfer factor  $H(f=0)=a_0+a_1+a_2 = 1$.
:*  The distortions with rise and fall can also be seen from the rectangular response  $〈\rho_ν^{(2, 8)}〉= 〈0,\ 0, 0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1, \ 1, \ 0.75, \ 0.25, \ \text{...}〉$.

{{BlaueBox|TEXT=
'''(2)'''  How do the results differ with  $a_2=-0.25$? }}

:*  Taking into account  $H(f=0)= 0.5$  there are comparable consequences   ⇒   Step response:    $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 0.5,\ 0.5, \ 0.5, \ 0.5,\text{...}〉 $.

{{BlaueBox|TEXT=
'''(3)'''  Now let the filter coefficients  $a_0=1$,  $b_1=0.9$  and  $a_1=a_2= b_2=0$.  Which filter is it?  Interpret the impulse response  $〈h_ν〉$.}}

:*  It is a recursive digital filter   ⇒   '''IIR filter'''  (''Infinite Impulse Response'')  of the first order.  It is the discrete-time analogon of the RC low-pass.
:*  Starting from  $h_0= 1$ is $h_1= h_0 \cdot b_0= 0.9$,  $h_2= h_1 \cdot b_0= b_0^2=0.81$,  $h_3= h_2 \cdot b_0= b_0^3=0.729$,  and so on   ⇒   $〈h_ν〉$  extends to infinity.
:*  Impulse response  $h(t) = {\rm e}^{-t/T}$  with  $T$:   intersection $($Tangente bei  $t=0$, Abscissa$)$   ⇒   $h_\nu= h(\nu \cdot T_{\rm A}) = {\rm e}^{-\nu/(T/T_{\rm A})}$  with  $T/T_{\rm A} = 1/(h_0-h_1)= 10$.
:*  So:  The values of the continuous time differ from the discrete-time impulse response.  This results in the values  $1.0, \ 0.9048,\ 0.8187$ ...

{{BlaueBox|TEXT=
'''(4)'''  The filter setting is retained.  Interpret the step response  $〈h_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$.  What is the value for  $H(f=0)$?}}

:*  The step response is the integral over the impulse response:   $\sigma(t) = T \cdot (1-{\rm e}^{-t/T}) ]$   ⇒   $\sigma_\nu= 10 \cdot (1-{\rm e}^{-\nu/10})$   ⇒   $\sigma_0=1$,  $\sigma_1=1.9$,  $\sigma_2=2.71$, ...
:*  For large  $\nu$  values, the (discrete-time) step response tends to the DC signal transmission factor  $H(f=0)= 10$:  $\sigma_{40}=9.867$,  $\sigma_{50}=9.954$,  $\sigma_\infty=10$.
:* The rectangular response  $〈\rho_ν^{(2, 8)}〉$  increases with a delay of  $2$  in the same way as  $〈\sigma_ν〉$.  In the area  $\nu \ge 8$  the  $\rho_ν$  values decrease exponentially.

{{BlaueBox|TEXT=
'''(5)'''  We continue to consider the filter with  $a_0=1$,  $b_1=0.9$,  $a_1=a_2=b_2=0$.  What is the output sequence  $〈y_ν〉$  for the input sequence  $〈x_ν〉= 〈1,\ 0,\ -0.5〉$?         ''Note'': The task can also be solved with this program, although the constellation considered here cannot be set directly.}}

:*  You can help yourself by setting the coefficient  $a_2=-0.5$  and reducing the input sequence  to $〈x_ν〉= 〈1,\ 0,\ 0,\ \text{ ...}〉$   ⇒   „Dirac delta function”.
:*  The actual impulse response of this filter  $($with  $a_2=0)$  was determined in task  '''(3)''':   $h_0= 1$,   $h_1= 0.9$,   $h_2= 0.81$,   $h_3= 0.729$,   $h_4= 0.646$.  
:*  The solution to this problem is:  $y_0 = h_0= 1$,   $y_1= h_1= 0.9$,   $y_2 =h_2-h_0/2= 0.31$,   $y_3 =h_3-h_1/2= 0.279$,   $y_4 =h_4-h_2/2= 0.251$.  
:*  Caution:  Step response and rectangular response now refer to the fictitious filter  $($with  $a_2=-0.5)$  and not to the actual filter  $($with  $a_2=0)$.

{{BlaueBox|TEXT=
'''(6)'''  Consider and interpret the impulse response and the step response for the filter coefficients  $a_0=1$,  $b_1=1$,  $a_1=a_2= b_2=0$.  }}

:*  '''The system is unstable''':   A discrete-time Dirac delta function at input $($at time  $t=0)$  causes an infinite number of Diracs of the same height in the output signal.
:*  A discrete-time step function at the input causes an infinite number of Diracs with monotonically increasing weights (to infinity) in the output signal.

{{BlaueBox|TEXT=
'''(7)'''  Consider and interpret the impulse response and step response for the filter coefficients  $a_0=1$,  $b_1=-1$,  $a_1=a_2= b_2=0$.  }}
:*  In contrast to exercise  '''(6)''', the weights of the impulse response  $〈h_ν〉$  are not constantly equal to  $1$, but alternating  $\pm 1$.  The system is unstable too.
:*  With the jump response  $〈\sigma_ν〉$, however, the weights alternate between  $0$  $($with even $\nu)$  and  $1$  $($with odd $\nu)$.

{{BlaueBox|TEXT=
'''(8)'''  We consider the "sine generator":  $a_1=0.5$,  $b_1=\sqrt{3}= 1.732$,  $b_2=-1.$  Compare the impulse response with the calculated values in  $\text{Example 4}$.         How do the parameters $a_1$ and $b_1$ influence the period duration  $T_0/T_{\rm A}$  and the amplitude  $A$  of the sine function?}}
:*  $〈x_ν〉=〈1, 0, 0, \text{...}〉$   ⇒   $〈y_ν〉=〈0, 0.5, 0.866, 1, 0.866, 0.5, 0, -0.5, -0.866, -1, -0.866, -0.5, 0, \text{...}〉$   ⇒   '''sine''',  period  $T_0/T_{\rm A}= 12$,  amplitude  $1$.
:*  The increase/decrease of $b_1$  leads to the larger/smaller period  $T_0/T_{\rm A}$  and the larger/smaller amplitude  $A$.  $b_1 < 2$ must apply.
:*  $a_1$  only affects the amplitude, not the period.  There is no value limit for  $a_1$. If  $a_1$  is negative, the minus sine function results.

{{BlaueBox|TEXT=
'''(9)'''  The basic setting is retained.  Which  $a_1$  and  $b_1$  result in a sine function with period  $T_0/T_{\rm A}=16$  and amplitude  $A=1$?}}
:*  Trying with  $b_1= 1.8478$  actually achieves the period duration  $T_0/T_{\rm A}=16$.  However, this increases the amplitude to  $A=1.307$.
:*  Adjusting the parameter  $a_1= 0.5/1.307=0.3826$  then leads to the desired amplitude  $A=1$.
:*  Or you can calculate this as in the example:  $b_1 = 2 \cdot \cos ( {2{\rm{\pi }}\cdot{T_{\rm A}}/{T_0 }})= 2 \cdot \cos (\pi/8)=1.8478$,     $a_1 = \sin (\pi/8)=0.3827$.

{{BlaueBox|TEXT=
'''(10)'''  We continue with the  "sine generator".  What modifications do you have to make to generate a  "cosine"?}}
:*  With  $a_1=0.3826$,  $b_1=1.8478$,  $b_2=-1$  and  $〈x_ν〉=〈1, 1, 1, \text{...}〉$  is the output sequence  $〈y_ν〉$  the discrete-time analogon of the step response  $\sigma(t)$.
:*  The step response is the integral over   $\sin(\pi\cdot\tau/8)$   within the limits of   $\tau=0$   to   $\tau=t$   ⇒   $\sigma(t)=-8/\pi\cdot\cos(\pi\cdot\tau/8)+1$.
:*  If you change   $a_1=0.3826$   on   $a_1=-0.3826\cdot\pi/8=-0.1502$, then   $\sigma(t)=\cos(\pi\cdot\tau/8)-1$   ⇒   Values between  $0$  and  $-2$.
:*  Would you still in the block diagram   $z_\nu=y_\nu+1$   add, then   $z_\nu$   a discrete-time cosine curve with   $T_0/T_{\rm A}=16$   and   $A=1$.

==Applet Manual==
 
'''Korrektur'''
==About the authors==
 
This interactive calculation tool was designed and implemented at the [http://www.lnt.ei.tum.de/startseite chair for communications engineering] at the [https://www.tum.de/ Technische Universität München].
*The first version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Andr.C3.A9_Schulz_.28Bachelorarbeit_LB_2020.29|André Schulz]] (Bachelor thesis LB, Supervisors: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_Übertragungstechnik#Benedikt_Leible.2C_M.Sc._.28bei_L.C3.9CT_seit_2017.29|Benedikt Leible]] and [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".

==Once again: Open Applet in new Tab==

{{LntAppletLink|digitalFilters_en}}

Applets:Digital Filters

2023-04-16T19:43:07Z

Noah:

{{LntAppletLinkEnDe|digitalFilters_en|digitalFilters}}

==Applet Description==
 
The applet should clarify the properties of digital filters, whereby we confine ourselves to filters of the order $M=2$. Both non-recursive filters $\rm (FIR$,  ''Finite Impulse Response''$)$  as well as recursive filters $\rm (IIR$,  ''Infinite Impulse Response''$)$.

The input signal $x(t)$ is represented by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$. The output sequence $〈y_ν〉$is calculated, i.e. the discrete-time representation of the output signal $y(t)$.

*$T_{\rm A}$ denotes the time interval between two samples.
*We also limit ourselves to causal signals and systems, which means that $x_ν \equiv 0$ and $y_ν \equiv 0$ for $ν \le 0$.

It should also be noted that we denote the initial sequence $〈y_ν〉$ as

'''(1)''' the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:         $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉,$

'''(2)''' the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:         $〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉,$

'''(3)''' the '''discrete-time rectangle response''' $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangle function” is present at the input:     $〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉;$         In quotation marks are the beginning of the ones $(2)$ and the position of the last ones $(4)$.

==Theoretical background==
 
===General block diagram===

Each signal $x(t)$ can only be represented on a computer by the sequence $〈x_ν〉$ of its samples, where $x_ν$ stands for $x(ν · T_{\rm A})$.
[[File:P_ID552__Sto_T_5_2_S1_neu.png|right |frame| Block diagram of a digital (IIR–) filter $M$–order]]
*The time interval $T_{\rm A}$ between two samples is limited by the [[https://en.lntwww.de/Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|"sampling theorem"]].
*We limit ourselves here to causal signals and systems, which means that $x_ν \equiv 0$ for $ν \le 0$.

*In order to determine the influence of a linear filter with frequency response $H(f)$ on the discrete-time input signal $〈x_ν〉$, it is advisable to describe the filter discrete-time. In the time domain, this happens with the discrete-time impulse response $〈h_ν〉$.
*On the right you can see the corresponding block diagram. The following therefore applies to the samples of the output signal $〈y_ν〉$ thus holds:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu } \cdot x_{\nu - \mu } + \sum\limits_{\mu = 1}^M {b_\mu } \cdot y_{\nu - \mu } .$$

The following should be noted here:
*The index $\nu$ refers to sequences, for example at the input $〈x_ν〉$ and output $〈y_ν〉$.
*On the other hand, we use the index $\mu$ to identify the $a$ and $b$ filter coefficients.
*The first sum describes the dependency of the current output $y_ν$ on the current input $x_ν$ and on the $M$ previous input values $x_{ν-1}$, ... , $x_{ν-M}$.
*The second sum indicates the influence of $y_ν$ by the previous values $y_{ν-1}$, ... , $y_{ν-M}$ at the filter output. It specifies the recursive part of the filter.
*The integer parameter $M$ is called the order of the digital filter. In the program, this value is limited to $M\le 2$.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

'''(1)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time impulse response''' $〈h_ν〉$ if the “discrete-time Dirac delta function” is present at the input:
:$$〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉 .$$
'''(2)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time step response''' $〈\sigma_ν〉$ if the “discrete-time step function” is present at the input:
:$$〈x_ν〉= 〈1,\ 1,\ 1,\ 1,\ 1,\ 1,\ 1, \text{...}〉 .$$
'''(3)'''  The output sequence $〈y_ν〉$ is called the '''discrete-time rectangle response'''  $〈\rho_ν^{(2, 4)}〉$ if the “discrete-time rectangular function” is present at the input:
:$$〈x_ν〉= 〈0,\ 0,\ 1,\ 1,\ 1,\ 0,\ 0, \text{...}〉 .$$
:The beginning of ones $(2)$ and the position of the last ones $(4)$ are given in single quotes.
}}

===Non-recursive filter   ⇒   FIR–filter ===

{{BlaueBox|TEXT=
[[File:P_ID553__Sto_T_5_2_S2_neu.png|right |frame| Non-recursive digital filter  $($FIR filter$)$  $M$ order]]
$\text{Definition:}$ If all feedback coefficients $b_{\mu} = 0$ , one speaks of one '''non-recursive filter'''. In the English language literature, the term '''FIR filter''' (''Finite Impulse Response'') is also used for this.

The following applies to the order $M$ applies:

*The output value $y_ν$ depends only on the current and the previous $M$ input values:
:$$y_\nu = \sum\limits_{\mu = 0}^M {a_\mu \cdot x_{\mu - \nu } } .$$
*Discrete-time impulse response with $〈x_ν〉= 〈1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 0, \text{...}〉$:
:$$〈h_\mu〉= 〈a_0,\ a_1,\ \text{...},\ a_M〉 .$$}}

{{GraueBox|TEXT=
$\text{Example 1:}$  A two-way channel where
*the signal on the main path arrives undamped compared to the input signal but is delayed by $2\ \rm µ s$ arrives with a delay, and
*at $4\ \rm µ s$ distance – so absolutely at time $t = 6\ \rm µ s$ – follows an echo with half the amplitude,

can be simulated by a non-recursive filter according to the sketch above, whereby the following parameter values must be set:
:$$M = 3,\quad T_{\rm A} = 2\;{\rm{µ s} },\quad a_{\rm 0} = 0,\quad a_{\rm 1} = 1, \quad a_{\rm 2} = 0, \quad a_{\rm 3} = 0.5.$$}}

{{GraueBox|TEXT=
$\text{Example 2:}$ Consider a non-recursive filter with the filter coefficients $a_0 = 1,\hspace{0.5cm} a_1 = 2,\hspace{0.5cm} a_2 = 1.$
[[File:P_ID608__Sto_Z_5_3.png|right|frame|Nichtrekursives Filter]]

'''(1)''' The conventional impulse response is: $h(t) = \delta (t) + 2 \cdot \delta ( {t - T_{\rm A} } ) + \delta ( {t - 2T_{\rm A} } ).$         ⇒   discrete-time impulse response: $〈h_\mu〉= 〈1,\ 2,\ 1〉 .$

'''(2)'''   The frequency response $H(f)$ is the Fourier transform of $h(t)$. By applying the displacement theorem:
:$$H(f) = 2\big [ {1 + \cos ( {2{\rm{\pi }\cdot }f \cdot T_{\rm A} } )} \big ] \cdot {\rm{e} }^{ - {\rm{j} }2{\rm{\pi } }fT_{\rm A} }\hspace{0.5cm}\Rightarrow \hspace{0.5cm}H(f = 0) = 4.$$

'''(3)'''   It follows that the '''discrete-time step response''' $〈\sigma_ν〉$ tends to become $4$ for large $\nu$.

'''(4)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;0,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;2,\;1,\;0,\;1,\;2,\;1,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$

'''(5)'''   The discrete-time convolution of the input sequence $\left\langle \hspace{0.05cm}{x_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;1,\;0,\;0,\;1,\;0,\;0,\;0,\;\text{...} } \hspace{0.05cm} \right\rangle$  with  $\left\langle \hspace{0.05cm}{h_\nu } \hspace{0.05cm}\right\rangle = \left\langle \hspace{0.05cm}{1, \ 2,\ 1 } \hspace{0.05cm}\right\rangle$  results
:$$\left\langle \hspace{0.05cm}{y_\nu } \hspace{0.05cm}\right\rangle = \left\langle {\;1,\;3,\;3,\;2,\;2,\;1,\;0,\;0,\;0,\;0,\;0,\; \text{...} \;} \right\rangle. $$}}

===Recursive filter   ⇒   IIR filter ===

{{BlaueBox|TEXT=
[[File:P_ID607__Sto_A_5_3.png|right|frame|First order recursive filter]]
$\text{Definition:}$ 
*If at least one of the feedback coefficients is $b_{\mu} \ne 0$, then this is referred to as a '''recursive filter''' (see graphic on the right). The term '''IIR filter'''  (''Infinite Impulse Response'') is also used for this, particularly in the English-language literature. This filter is dealt with in detail in the trial implementation.

*If all forward coefficients are also identical $a_\mu = 0$ with the exception of $a_0$, a '''purely recursive filter''' is available (see graphic on the left).

[[File:P_ID554__Sto_T_5_2_S3_neu.png|left|frame| Purely recursive first order filter]] }}

In the following we restrict ourselves to the special case “purely recursive filter of the first order”. This filter has the following properties:
*The output value $y_ν$ depends (indirectly) on an infinite number of input values:
:$$y_\nu = \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot x_{\nu - \mu } .}$$
*This shows the following calculation:
:$$y_\nu = a_0 \cdot x_\nu + b_1 \cdot y_{\nu - 1} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + {b_1} ^2 \cdot y_{\nu - 2} = a_0 \cdot x_\nu + a_0 \cdot b_1 \cdot x_{\nu - 1} + a_0 \cdot {b_1} ^2 \cdot x_{\nu - 2} + {b_1} ^3 \cdot y_{\nu - 3} = \text{...}. $$

*By definition, the discrete-time impulse response is the same as the output sequence if there is a single "one" at $t =0$ at the input.
:$$h(t)= \sum\limits_{\mu = 0}^\infty {a_0 \cdot {b_1} ^\mu \cdot \delta ( {t - \mu \cdot T_{\rm A} } )}\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉= 〈\hspace{0.05cm}a_0, \ a_0\cdot {b_1}, \ a_0\cdot {b_1}^2 \ \text{...} \hspace{0.05cm}〉.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  With a recursive filter, the (discrete-time) impulse response extends to infinity with $M = 1$:
*For reasons of stability, $b_1 < 1$ must apply.
*With $b_1 = 1$ the impulse response $h(t)$ would extend to infinity and with $b_1 > 1$ the variable $h(t)$ would even continue to infinity.
*With such a recursive filter of the first order, each individual Dirac delta line is exactly the factor $b_1$ smaller than the previous Dirac delta line:
:$$h_{\mu} = h(\mu \cdot T_{\rm A}) = {b_1} \cdot h_{\mu -1}.$$}}

{{GraueBox|TEXT=
[[File:Sto_T_5_2_S3_version2.png |frame| Discrete-time impulse response | rechts]]
$\text{Example 3:}$  The graphic opposite shows the discrete-time impulse response $〈\hspace{0.05cm}h_\mu\hspace{0.05cm}〉$ of a recursive filter of the first order with the parameters $a_0 = 1$ and $b_1 = 0.6$.
*The (discrete-time) course is exponentially falling and extends to infinity.
*The ratio of the weights of two successive Diracs is $b_1 = 0.6$.
}}

===Recursive filter as a sine generator===
[[File:EN_Sto_A_5_4_version2.png|right|frame|Proposed filter structure]]

The graphic shows a second-order digital filter that is suitable for generating a discrete-time sine function on a digital signal processor (DSP) if the input sequence $\left\langle \hspace{0.05cm} {x_\nu } \hspace{0.05cm}\right\rangle$  a (discrete-time) Dirac delta function is:
:$$\left\langle \hspace{0.05cm}{y_\nu }\hspace{0.05cm} \right\rangle = \left\langle {\, \sin ( {\nu \cdot T_{\rm A} \cdot \omega _0 } )\, }\right\rangle .$$

The five filter coefficients result from the:
[https://en.wikipedia.org/wiki/Z-transform "$Z$-transform"]:
:$$Z \big \{ {\sin ( {\nu T{\rm A}\cdot \omega _0 } )} \big \} = \frac{{z \cdot \sin \left( {\omega _0 \cdot T_{\rm A}} \right)}}{{z^2 - 2 \cdot z \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right) + 1}}.$$
After implementing this equation using a second-order recursive filter, the following filter coefficients are obtained:
:$$a_0 = 0,\quad a_1 = \sin \left( {\omega _0 \cdot T_{\rm A}} \right),\quad a_2 = 0, \quad b_1 = 2 \cdot \cos \left( {\omega _0 \cdot T_{\rm A}} \right),\quad b_2 = - 1.$$

*The filter coefficients $a_0$ and $a_2$ can be omitted and $b_2=-1$ has a fixed value.
*The angular frequency $\omega_0$ of the sine wave is therefore only determined by $a_0$ and $a_0$.

{{GraueBox|TEXT=
$\text{Example 3:}$  Let $a_1 = 0.5$, $b_1 = \sqrt 3$, $x_0 = 1$ and $x_{\nu \hspace{0.05cm}\ne\hspace{0.05cm} 0} = 0$.

'''(1)'''  Then the following applies to the initial values $y_\nu$ at times $\nu \ge 0$: 
:*  $y_0 = 0;$
:*  $y_1 = 0.5$                                                                                         ⇒  the "$1$" at the input only has an effect at time $\nu = 1$ because of $a_0= 0$ at the output;
:*  $y_2 = b_1 \cdot y_1 - y_0 = {\sqrt 3 }/{2} \approx 0.866$                             ⇒   with $\nu = 2$ the recursive part of the filter also takes effect;
:*  $y_3 = \sqrt 3 \cdot y_2 - y_1 = \sqrt 3 \cdot {\sqrt 3 }/{2} - {1}/{2} = 1$          ⇒  for  $\nu \ge 2$  the filter is purely recursive:     $y_\nu = b_1 \cdot y_{\nu - 1} - y_{\nu - 2}$;
:*  $y_4 = \sqrt 3 \cdot y_3 - y_2 = \sqrt 3 \cdot 1 - {\sqrt 3 }/{2} = {\sqrt 3 }/{2};$
:*  $y_5 = \sqrt 3 \cdot y_4 - y_3 = \sqrt 3 \cdot {\sqrt 3 }/{2} - 1 = 0.5;$
:*  $y_6 = \sqrt 3 \cdot y_5 - y_4 = \sqrt 3 \cdot {1}/{2} - {\sqrt 3 }/{2} = 0;$
:*  $y_7 = \sqrt 3 \cdot y_6 - y_5 = \sqrt 3 \cdot 0 - {1}/{2} = - 0.5.$

'''(2)'''  By continuing the recursive algorithm one gets for large $\nu$–values:     $y_\nu = y_{\nu - 12}$   ⇒   $T_0/T_{\rm A}= 12.$ }}

==Exercises==

[[File:Exercises_binomial_fertig.png|right]]
*First select the number '''1''' ... '''10''' of the task to be processed.
*A task description is displayed.  The parameter values are adjusted.
*Solution after pressing  "Sample Solution".
*The number '''0''' corresponds to a "reset":  Same setting as when the program was started.
 

{{BlaueBox|TEXT=
'''(1)'''  The filter coefficients are  $a_0=0.25$,  $a_1=0.5$,  $a_2=0.25$,  $b_1=b_2=0$.  Which filter is it?         Interpret the impulse response  $〈h_ν〉$,  the step response  $〈\sigma_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$, each in a discrete-time representation.}}

:*  Due to the missing  $b$ coefficients, it is a non-recursive digital filter ⇒   '''FIR filter''' (''Finite Impulse Response'').
:*  The impulse response consists of  $M+1=3$  Dirac delta lines according to the  $a$  coefficients:  $〈h_ν〉= 〈a_0, \ a_1,\ a_2〉= 〈0.25, \ 0.5,\ 0.25,\ 0, \ 0, \ 0,\text{...}〉 $.
:*  The step response is:  $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1,\text{...}〉 $.  The final value is equal to the DC signal transfer factor  $H(f=0)=a_0+a_1+a_2 = 1$.
:*  The distortions with rise and fall can also be seen from the rectangular response  $〈\rho_ν^{(2, 8)}〉= 〈0,\ 0, 0.25, \ 0.75,\ 1,\ 1, \ 1, \ 1, \ 1, \ 0.75, \ 0.25, \ \text{...}〉$.

{{BlaueBox|TEXT=
'''(2)'''  How do the results differ with  $a_2=-0.25$? }}

:*  Taking into account  $H(f=0)= 0.5$  there are comparable consequences   ⇒   Step response:    $〈\sigma_ν〉= 〈0.25, \ 0.75,\ 0.5,\ 0.5, \ 0.5, \ 0.5,\text{...}〉 $.

{{BlaueBox|TEXT=
'''(3)'''  Now let the filter coefficients  $a_0=1$,  $b_1=0.9$  and  $a_1=a_2= b_2=0$.  Which filter is it?  Interpret the impulse response  $〈h_ν〉$.}}

:*  It is a recursive digital filter   ⇒   '''IIR filter'''  (''Infinite Impulse Response'')  of the first order.  It is the discrete-time analogon of the RC low-pass.
:*  Starting from  $h_0= 1$ is $h_1= h_0 \cdot b_0= 0.9$,  $h_2= h_1 \cdot b_0= b_0^2=0.81$,  $h_3= h_2 \cdot b_0= b_0^3=0.729$,  and so on   ⇒   $〈h_ν〉$  extends to infinity.
:*  Impulse response  $h(t) = {\rm e}^{-t/T}$  with  $T$:   intersection $($Tangente bei  $t=0$, Abscissa$)$   ⇒   $h_\nu= h(\nu \cdot T_{\rm A}) = {\rm e}^{-\nu/(T/T_{\rm A})}$  with  $T/T_{\rm A} = 1/(h_0-h_1)= 10$.
:*  So:  The values of the continuous time differ from the discrete-time impulse response.  This results in the values  $1.0, \ 0.9048,\ 0.8187$ ...

{{BlaueBox|TEXT=
'''(4)'''  The filter setting is retained.  Interpret the step response  $〈h_ν〉$  and the rectangular response  $〈\rho_ν^{(2, 8)}〉$.  What is the value for  $H(f=0)$?}}

:*  The step response is the integral over the impulse response:   $\sigma(t) = T \cdot (1-{\rm e}^{-t/T}) ]$   ⇒   $\sigma_\nu= 10 \cdot (1-{\rm e}^{-\nu/10})$   ⇒   $\sigma_0=1$,  $\sigma_1=1.9$,  $\sigma_2=2.71$, ...
:*  For large  $\nu$  values, the (discrete-time) step response tends to the DC signal transmission factor  $H(f=0)= 10$:  $\sigma_{40}=9.867$,  $\sigma_{50}=9.954$,  $\sigma_\infty=10$.
:* The rectangular response  $〈\rho_ν^{(2, 8)}〉$  increases with a delay of  $2$  in the same way as  $〈\sigma_ν〉$.  In the area  $\nu \ge 8$  the  $\rho_ν$  values decrease exponentially.

{{BlaueBox|TEXT=
'''(5)'''  We continue to consider the filter with  $a_0=1$,  $b_1=0.9$,  $a_1=a_2=b_2=0$.  What is the output sequence  $〈y_ν〉$  for the input sequence  $〈x_ν〉= 〈1,\ 0,\ -0.5〉$?         ''Note'': The task can also be solved with this program, although the constellation considered here cannot be set directly.}}

:*  You can help yourself by setting the coefficient  $a_2=-0.5$  and reducing the input sequence  to $〈x_ν〉= 〈1,\ 0,\ 0,\ \text{ ...}〉$   ⇒   „Dirac delta function”.
:*  The actual impulse response of this filter  $($with  $a_2=0)$  was determined in task  '''(3)''':   $h_0= 1$,   $h_1= 0.9$,   $h_2= 0.81$,   $h_3= 0.729$,   $h_4= 0.646$.  
:*  The solution to this problem is:  $y_0 = h_0= 1$,   $y_1= h_1= 0.9$,   $y_2 =h_2-h_0/2= 0.31$,   $y_3 =h_3-h_1/2= 0.279$,   $y_4 =h_4-h_2/2= 0.251$.  
:*  Caution:  Step response and rectangular response now refer to the fictitious filter  $($with  $a_2=-0.5)$  and not to the actual filter  $($with  $a_2=0)$.

{{BlaueBox|TEXT=
'''(6)'''  Consider and interpret the impulse response and the step response for the filter coefficients  $a_0=1$,  $b_1=1$,  $a_1=a_2= b_2=0$.  }}

:*  '''The system is unstable''':   A discrete-time Dirac delta function at input $($at time  $t=0)$  causes an infinite number of Diracs of the same height in the output signal.
:*  A discrete-time step function at the input causes an infinite number of Diracs with monotonically increasing weights (to infinity) in the output signal.

{{BlaueBox|TEXT=
'''(7)'''  Consider and interpret the impulse response and step response for the filter coefficients  $a_0=1$,  $b_1=-1$,  $a_1=a_2= b_2=0$.  }}
:*  In contrast to exercise  '''(6)''', the weights of the impulse response  $〈h_ν〉$  are not constantly equal to  $1$, but alternating  $\pm 1$.  The system is unstable too.
:*  With the jump response  $〈\sigma_ν〉$, however, the weights alternate between  $0$  $($with even $\nu)$  and  $1$  $($with odd $\nu)$.

{{BlaueBox|TEXT=
'''(8)'''  We consider the "sine generator":  $a_1=0.5$,  $b_1=\sqrt{3}= 1.732$,  $b_2=-1.$  Compare the impulse response with the calculated values in  $\text{Example 4}$.         How do the parameters $a_1$ and $b_1$ influence the period duration  $T_0/T_{\rm A}$  and the amplitude  $A$  of the sine function?}}
:*  $〈x_ν〉=〈1, 0, 0, \text{...}〉$   ⇒   $〈y_ν〉=〈0, 0.5, 0.866, 1, 0.866, 0.5, 0, -0.5, -0.866, -1, -0.866, -0.5, 0, \text{...}〉$   ⇒   '''sine''',  period  $T_0/T_{\rm A}= 12$,  amplitude  $1$.
:*  The increase/decrease of $b_1$  leads to the larger/smaller period  $T_0/T_{\rm A}$  and the larger/smaller amplitude  $A$.  $b_1 < 2$ must apply.
:*  $a_1$  only affects the amplitude, not the period.  There is no value limit for  $a_1$. If  $a_1$  is negative, the minus sine function results.

{{BlaueBox|TEXT=
'''(9)'''  The basic setting is retained.  Which  $a_1$  and  $b_1$  result in a sine function with period  $T_0/T_{\rm A}=16$  and amplitude  $A=1$?}}
:*  Trying with  $b_1= 1.8478$  actually achieves the period duration  $T_0/T_{\rm A}=16$.  However, this increases the amplitude to  $A=1.307$.
:*  Adjusting the parameter  $a_1= 0.5/1.307=0.3826$  then leads to the desired amplitude  $A=1$.
:*  Or you can calculate this as in the example:  $b_1 = 2 \cdot \cos ( {2{\rm{\pi }}\cdot{T_{\rm A}}/{T_0 }})= 2 \cdot \cos (\pi/8)=1.8478$,     $a_1 = \sin (\pi/8)=0.3827$.

{{BlaueBox|TEXT=
'''(10)'''  We continue with the  "sine generator".  What modifications do you have to make to generate a  "cosine"?}}
:*  With  $a_1=0.3826$,  $b_1=1.8478$,  $b_2=-1$  and  $〈x_ν〉=〈1, 1, 1, \text{...}〉$  is the output sequence  $〈y_ν〉$  the discrete-time analogon of the step response  $\sigma(t)$.
:*  The step response is the integral over   $\sin(\pi\cdot\tau/8)$   within the limits of   $\tau=0$   to   $\tau=t$   ⇒   $\sigma(t)=-8/\pi\cdot\cos(\pi\cdot\tau/8)+1$.
:*  If you change   $a_1=0.3826$   on   $a_1=-0.3826\cdot\pi/8=-0.1502$, then   $\sigma(t)=\cos(\pi\cdot\tau/8)-1$   ⇒   Values between  $0$  and  $-2$.
:*  Would you still in the block diagram   $z_\nu=y_\nu+1$   add, then   $z_\nu$   a discrete-time cosine curve with   $T_0/T_{\rm A}=16$   and   $A=1$.

==Applet Manual==
 
'''Korrektur'''
==About the authors==
 
This interactive calculation tool was designed and implemented at the [http://www.lnt.ei.tum.de/startseite chair for communications engineering] at the [https://www.tum.de/ Technische Universität München].
*The first version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Andr.C3.A9_Schulz_.28Bachelorarbeit_LB_2020.29|André Schulz]] (Bachelor thesis LB, Supervisors: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_Übertragungstechnik#Benedikt_Leible.2C_M.Sc._.28bei_L.C3.9CT_seit_2017.29|Benedikt Leible]] and [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".

==Once again: Open Applet in new Tab==

{{LntAppletLink|digitalFilters_en}}

Applets:Complementary Gaussian Error Functions

2023-04-16T17:49:53Z

Noah:

{{LntAppletLinkEnDe|qfunction_en|qfunction}}

==Applet Description==
 
This applet allows the calculation and graphical representation of the (complementary) Gaussian error functions  ${\rm Q}(x)$  and  $1/2\cdot {\rm erfc}(x)$, which are of great importance for error probability calculation.
*Both the abscissa and the function value can be represented either linearly or logarithmically.
*For both functions an upper bound  $\rm (UB)$  and a lower bound  $\rm (LB)$  are given.

==Theoretical Background==
 
In the study of digital transmission systems, it is often necessary to determine the probability that a (zero mean) Gaussian distributed random variable  $x$  with variance  $σ^2$  exceeds a given value  $x_0$.  For this probability holds:
:$${\rm Pr}(x > x_0)={\rm Q}(\frac{x_0}{\sigma}) = 1/2 \cdot {\rm erfc}(\frac{x_0}{\sqrt{2} \cdot \sigma}).$$

===The function ${\rm Q}(x )$===

The function  ${\rm Q}(x)$  is called the  '''complementary Gaussian error integral'''.  The following calculation rule applies:
:$${\rm Q}(x ) = \frac{1}{\sqrt{2\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}/\hspace{0.05cm} 2}\,{\rm d} u .$$
*This integral cannot be solved analytically and must be taken from tables if one does not have this applet available.
*Specially for larger  $x$  values  (i.e., for small error probabilities), the bounds given below provide a useful estimate for  ${\rm Q}(x)$, which can also be calculated without tables.
*An upper bound  $\rm (UB)$  of this function is:
:$${\rm Q}_{\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ] = \frac{ 1}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{- x^{2}/\hspace{0.05cm}2} > {\rm Q}(x).$$
*Correspondingly, for the lower bound  $\rm (LB)$:
:$${\rm Q}_{\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ] =\frac{1-1/x^2}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{-x^ 2/\hspace{0.05cm}2} ={\rm Q}_{\rm UB}(x ) \cdot (1-1/x^2)< {\rm Q}(x).$$

However, in many program libraries, the function  ${\rm Q}(x )$  cannot be found.
 

===The function $1/2 \cdot {\rm erfc}(x )$===

On the other hand, in almost all program libraries, you can find the  '''complementary Gaussian error function''':
:$${\rm erfc}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u ,$$
which is related to  ${\rm Q}(x)$  as follows:   ${\rm Q}(x)=1/2\cdot {\rm erfc}(x/{\sqrt{2}}).$
*Since in almost all applications this function is used with the factor  $1/2$, in this applet exactly this function was realized:
:$$1/2 \cdot{\rm erfc}(x) = \frac{1}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u .$$

*Once again, an upper and lower bound can be specified for this function:
:$$\text{Upper Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ 1}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} ,$$
:$$\text{Lower Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ {1-1/(2x^2)}}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} .$$

===When which function offers advantages?===

{{GreyBox|TEXT=
$\text{Example 1:}$  We consider binary baseband transmission. Here, the bit error probability  $p_{\rm B} = {\rm Q}({s_0}/{\sigma_d})$, where the useful signal can take the values  $\pm s_0$  and the noise root-mean-square value  $\sigma_d$ .

It is assumed that tables are available listing the argument of the two Gaussian error functions at distance  $0.1$.  With  $s_0/\sigma_d = 4$  one obtains for the bit error probability according to the function  ${\rm Q}(x )$:
:$$p_{\rm B} = {\rm Q} (4) \approx 0.317 \cdot 10^{-4}\hspace{0.05cm}.$$
According to the second equation, we get:
:$$p_{\rm B} = {1}/{2} \cdot {\rm erfc} ( {4}/{\sqrt{2} })= {1}/{2} \cdot {\rm erfc} ( 2.828)\approx {1}/{2} \cdot {\rm erfc} ( 2.8)= 0.375 \cdot 10^{-4}\hspace{0.05cm}.$$
*The first value is more correct.  In the second method of calculation, one must round or – even better – interpolate, which is very difficult due to the strong nonlinearity of this function. 
*Accordingly, with the given numerical values, ${\rm Q}(x )$  is more suitable.  However, outside of exercise examples  $s_0/\sigma_d$  will usually have a "curvilinear" value.  In this case, of course,  ${\rm Q}(x)$  offers no advantage over  $1/2 \cdot{\rm erfc}(x)$. }}

{{GreyBox|TEXT=
$\text{Example 2:}$ 
With the energy per bit  $(E_{\rm B})$  and the noise power density  $(N_0)$  the bit error probability of ''Binary Phase Shift Keying''  (BPSK) is:
:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2 E_{\rm B} }/{N_0} }\right ) = {1}/{2} \cdot { \rm erfc} \left ( \sqrt{ {E_{\rm B} }/{N_0} }\right ) \hspace{0.05cm}.$$
For the numerical values  $E_{\rm B} = 16 \rm mWs$  and  $N_0 = 1 \rm mW/Hz$  we obtain:
:$$p_{\rm B} = {\rm Q} \left (4 \cdot \sqrt{ 2} \right ) = {1}/{2} \cdot {\rm erfc} \left ( 4\right ) \hspace{0.05cm}.$$
*The first way leads to the result  $p_{\rm B} = {\rm Q} (5.657) \approx {\rm Q} (5.7) = 0.6 \cdot 10^{-8}\hspace{0.01cm}$, while   $1/2 \cdot{\rm erfc}(x)$  yields the more correct value  $p_{\rm B} \approx 0.771 \cdot 10^{-8}$  here.
*As in the first example, however, you can see:   The functions  ${\rm Q}(x)$  and  $1/2 \cdot{\rm erfc}(x)$  are basically equally well suited.
*Advantages or disadvantages of one or the other function arise only for concrete numerical values.}}
 

==Exercises==

* First select the number  $(1, 2, \text{...})$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 

{{BlueBox|TEXT=
'''(1)'''   Find the values of the function  ${\rm Q}(x)$  for  $x=1$,  $x=2$,  $x=4$  and  $x=6$.  Interpret the graphs for linear and logarithmic ordinates.}}

*The applet returns the values  ${\rm Q}(1)=1.5866 \cdot 10^{-1}$,  ${\rm Q}(2)=2. 275 \cdot 10^{-2}$,  ${\rm Q}(4)=3.1671 \cdot 10^{-5}$  and  ${\rm Q}(6)=9.8659 \cdot 10^{-10}$.
*With linear ordinate, the values for  $x>3$  are indistinguishable from the zero line.  More interesting is the plot with logarithmic ordinate.

{{BlueBox|TEXT=
'''(2)'''   Evaluate the two bounds  ${\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ]$  and  ${\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ]$  for the  ${\rm Q}$  function. }}

*For  $x \ge 2$  the upper bound is only slightly above  ${\rm Q}(x)$  and the lower bound is only slightly below  ${\rm Q}(x)$. 
*For example:  ${\rm Q}(x=4)=3.1671 \cdot 10^{-5}$   ⇒   ${\rm LB}(x=4)=3.1366 \cdot 10^{-5}$,   ${\rm UB}(x=4)=3.3458 \cdot 10^{-5}$.
*The upper bound has greater significance for assessing a communications system than "LB",  since this corresponds to a "worst case" consideration.

{{BlueBox|TEXT=
'''(3)'''   Try to use the app to determine  ${\rm Q}(x=2 \cdot \sqrt{2} \approx 2.828)$  as accurately as possible despite the quantization of the input parameter. }}
*The program returns for  $x=2.8$  the too large result  $2.5551 \cdot 10^{-3}$  and for  $x=2.85$  the result  $2.186 \cdot 10^{-3}$.  The exact value lies in between.
*But it also holds:  ${\rm Q}(x=2 \cdot \sqrt{2})=0.5 \cdot {\rm erfc}(x=2)$.  This gives the exact value  ${\rm Q}(x=2 \cdot \sqrt{2})=2.3389 \cdot 10^{-3}$.

{{BlueBox|TEXT=
'''(4)'''   Find the values of the function  $0.5 \cdot {\rm erfc}(x)$  for  $x=1$,  $x=2$,  $x=3$  and  $x=4$.  Interpret the exact results and the bounds.}}

*The applet returns:  $0.5 \cdot {\rm erfc}(1)=7.865 \cdot 10^{-2}$,  $0.5 \cdot {\rm erfc}(2)=2. 3389 \cdot 10^{-3}$,  $0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$  and  $0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$.
*All the above statements about  ${\rm Q}(x)$  with respect to suitable representation type and upper and lower bounds also apply to the function  $0.5 \cdot {\rm erfc}(x)$.

{{BlueBox|TEXT=
'''(5)'''   The results of  '''(4)'''  are now to be converted for the case of a logarithmic abscissa.  The conversion is done according to  $\rho\big[{\rm dB}\big ] = 20 \cdot \lg(x)$. }}

* The linear abscissa value  $x=1$  leads to the logarithmic abscissa value  $\rho=0\ \rm dB$   ⇒   $0. 5 \cdot {\rm erfc}(\rho=0\ {\rm dB})={0.5 \cdot \rm erfc}(x=1)=7.865 \cdot 10^{-2}$.
*Similarly  $0.5 \cdot {\rm erfc}(\rho=6.021\ {\rm dB}) =0.5 \cdot {\rm erfc}(x=2)=2. 3389 \cdot 10^{-3}$,     $0.5 \cdot {\rm erfc}(\rho=9.542\ {\rm dB})=0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$,   
*$0.5 \cdot {\rm erfc}(\rho=12.041\ {\rm dB})= 0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$.
*As per right diagram:  $0.5 \cdot {\rm erfc}(\rho=6\ {\rm dB}) =2.3883 \cdot 10^{-3}$,     $0.5 \cdot {\rm erfc}(\rho=9. 5\ {\rm dB}) =1.2109 \cdot 10^{-5}$,     $0.5 \cdot {\rm erfc}(\rho=12\ {\rm dB}) =9.006 \cdot 10^{-9}$.

{{BlueBox|TEXT=
'''(6)'''   Find  ${\rm Q}(\rho=0\ {\rm dB})$,  ${\rm Q}(\rho=5\ {\rm dB})$  and  ${\rm Q}(\rho=10\ {\rm dB})$,  and establish the relationship between linear and logarithmic abscissa.}}
*The program returns for logarithmic abscissa  ${\rm Q}(\rho=0\ {\rm dB})=1. 5866 \cdot 10^{-1}$,  ${\rm Q}(\rho=5\ {\rm dB})=3.7679 \cdot 10^{-2}$,  ${\rm Q}(\rho=10\ {\rm dB})=7.827 \cdot 10^{-4}$.
*The conversion is done according to the equation  $x=10^{\hspace{0.05cm}0.05\hspace{0.05cm} \cdot\hspace{0.05cm} \rho[{\rm dB}]}$.  For  $\rho=0\ {\rm dB}$  we get  $x=1$   ⇒   ${\rm Q}(\rho=0\ {\rm dB})={\rm Q}(x=1) =1.5866 \cdot 10^{-1}$.
*For  $\rho=5\ {\rm dB}$  we get  $x=1.1778$   ⇒   ${\rm Q}(\rho=5\ {\rm dB})={\rm Q}(x=1. 778) =3.7679 \cdot 10^{-2}$.  From the left diagram:  ${\rm Q}(x=1.8) =3.593 \cdot 10^{-2}$.
*For  $\rho=10\ {\rm dB}$  we get  $x=3.162$   ⇒   ${\rm Q}(\rho=10\ {\rm dB})={\rm Q}(x=3. 162) =7.827 \cdot 10^{-4}$.  After "quantization":  ${\rm Q}(x=3.15) =8.1635 \cdot 10^{-4}$.

==Applet Manual==
 

[[File:Qfunction bedienung.png|right|550px]]
    '''(A)'''     Equations used in the example  ${\rm Q}(x)$

    '''(B)'''     Selection option for  ${\rm Q}(x)$  or  ${\rm 0.5 \cdot erfc}(x)$

    '''(C)'''     Bounds  ${\rm LB}$  and  ${\rm UB}$  are drawn

    '''(D)'''     Selection whether abscissa is linear  $\rm (lin)$  or logarithmic  $\rm (log)$ 

    '''(E)'''     Select whether ordinate is linear  $\rm (lin)$  or logarithmic  $\rm (log)$ 

    '''(F)'''     Numerical output using the example  ${\rm Q}(x)$  with linear abscissa

    '''(G)'''     Slider input of abscissa value  $x$  for linear abscissa

    '''(H)'''     Slider input of abscissa value  $\rho \ \rm [dB]$  for logarithmic abscissa

    '''(I)'''     Graphical output of function  ${\rm Q}(x)$  – here:  linear abscissa

    '''(J)'''     Graph output of function  ${\rm 0.5 \cdot erfc}(x)$  – here:  linear abscissa

    '''(K)'''     Variation possibility for the graphical representations

$\hspace{1.5cm}$"$+$" (zoom in),

$\hspace{1.5cm}$"$-$" (zoom out)

$\hspace{1.5cm}$ "$\rm o$" (Reset)

$\hspace{1.5cm}$ "$\leftarrow$" (Move left), etc.
 
==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2007 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Thomas_Gro.C3.9Fer_.28Diplomarbeit_LB_2006.2C_danach_freie_Mitarbeit_bis_2010.29|Thomas Großer]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2018 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.

==Once again: Open Applet in new Tab==
 
{{LntAppletLinkEnDe|qfunction_en|qfunction}}

Applets:Complementary Gaussian Error Functions

2023-04-16T17:49:23Z

Noah:

{{LntAppletLinkEnDe|qfunction_en|qfunction}}

==Applet Description==
 
This applet allows the calculation and graphical representation of the (complementary) Gaussian error functions  ${\rm Q}(x)$  and  $1/2\cdot {\rm erfc}(x)$, which are of great importance for error probability calculation.
*Both the abscissa and the function value can be represented either linearly or logarithmically.
*For both functions an upper bound  $\rm (UB)$  and a lower bound  $\rm (LB)$  are given.

==Theoretical Background==
 
In the study of digital transmission systems, it is often necessary to determine the probability that a (zero mean) Gaussian distributed random variable  $x$  with variance  $σ^2$  exceeds a given value  $x_0$.  For this probability holds:
:$${\rm Pr}(x > x_0)={\rm Q}(\frac{x_0}{\sigma}) = 1/2 \cdot {\rm erfc}(\frac{x_0}{\sqrt{2} \cdot \sigma}).$$

===The function ${\rm Q}(x )$===

The function  ${\rm Q}(x)$  is called the  '''complementary Gaussian error integral'''.  The following calculation rule applies:
:$${\rm Q}(x ) = \frac{1}{\sqrt{2\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}/\hspace{0.05cm} 2}\,{\rm d} u .$$
*This integral cannot be solved analytically and must be taken from tables if one does not have this applet available.
*Specially for larger  $x$  values  (i.e., for small error probabilities), the bounds given below provide a useful estimate for  ${\rm Q}(x)$, which can also be calculated without tables.
*An upper bound  $\rm (UB)$  of this function is:
:$${\rm Q}_{\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ] = \frac{ 1}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{- x^{2}/\hspace{0.05cm}2} > {\rm Q}(x).$$
*Correspondingly, for the lower bound  $\rm (LB)$:
:$${\rm Q}_{\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ] =\frac{1-1/x^2}{\sqrt{2\pi}\cdot x}\cdot {\rm e}^{-x^ 2/\hspace{0.05cm}2} ={\rm Q}_{\rm UB}(x ) \cdot (1-1/x^2)< {\rm Q}(x).$$

However, in many program libraries, the function  ${\rm Q}(x )$  cannot be found.
 

===The function $1/2 \cdot {\rm erfc}(x )$===

On the other hand, in almost all program libraries, you can find the  '''complementary Gaussian error function''':
:$${\rm erfc}(x) = \frac{2}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u ,$$
which is related to  ${\rm Q}(x)$  as follows:   ${\rm Q}(x)=1/2\cdot {\rm erfc}(x/{\sqrt{2}}).$
*Since in almost all applications this function is used with the factor  $1/2$, in this applet exactly this function was realized:
:$$1/2 \cdot{\rm erfc}(x) = \frac{1}{\sqrt{\pi}}\int_{x}^{ +\infty}\hspace{-0.2cm}{\rm e}^{-u^{2}}\,{\rm d} u .$$

*Once again, an upper and lower bound can be specified for this function:
:$$\text{Upper Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ 1}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} ,$$
:$$\text{Lower Bound }\big [1/2 \cdot{\rm erfc}(x) \big ] = \frac{ {1-1/(2x^2)}}{\sqrt{\pi}\cdot 2x}\cdot {\rm e}^{- x^{2}} .$$

===When which function offers advantages?===

{{GreyBox|TEXT=
$\text{Example 1:}$  We consider binary baseband transmission. Here, the bit error probability  $p_{\rm B} = {\rm Q}({s_0}/{\sigma_d})$, where the useful signal can take the values  $\pm s_0$  and the noise root-mean-square value  $\sigma_d$ .

It is assumed that tables are available listing the argument of the two Gaussian error functions at distance  $0.1$.  With  $s_0/\sigma_d = 4$  one obtains for the bit error probability according to the function  ${\rm Q}(x )$:
:$$p_{\rm B} = {\rm Q} (4) \approx 0.317 \cdot 10^{-4}\hspace{0.05cm}.$$
According to the second equation, we get:
:$$p_{\rm B} = {1}/{2} \cdot {\rm erfc} ( {4}/{\sqrt{2} })= {1}/{2} \cdot {\rm erfc} ( 2.828)\approx {1}/{2} \cdot {\rm erfc} ( 2.8)= 0.375 \cdot 10^{-4}\hspace{0.05cm}.$$
*The first value is more correct.  In the second method of calculation, one must round or – even better – interpolate, which is very difficult due to the strong nonlinearity of this function. 
*Accordingly, with the given numerical values, ${\rm Q}(x )$  is more suitable.  However, outside of exercise examples  $s_0/\sigma_d$  will usually have a "curvilinear" value.  In this case, of course,  ${\rm Q}(x)$  offers no advantage over  $1/2 \cdot{\rm erfc}(x)$. }}

{{GreyBox|TEXT=
$\text{Example 2:}$ 
With the energy per bit  $(E_{\rm B})$  and the noise power density  $(N_0)$  the bit error probability of ''Binary Phase Shift Keying''  (BPSK) is:
:$$p_{\rm B} = {\rm Q} \left ( \sqrt{ {2 E_{\rm B} }/{N_0} }\right ) = {1}/{2} \cdot { \rm erfc} \left ( \sqrt{ {E_{\rm B} }/{N_0} }\right ) \hspace{0.05cm}.$$
For the numerical values  $E_{\rm B} = 16 \rm mWs$  and  $N_0 = 1 \rm mW/Hz$  we obtain:
:$$p_{\rm B} = {\rm Q} \left (4 \cdot \sqrt{ 2} \right ) = {1}/{2} \cdot {\rm erfc} \left ( 4\right ) \hspace{0.05cm}.$$
*The first way leads to the result  $p_{\rm B} = {\rm Q} (5.657) \approx {\rm Q} (5.7) = 0.6 \cdot 10^{-8}\hspace{0.01cm}$, while   $1/2 \cdot{\rm erfc}(x)$  yields the more correct value  $p_{\rm B} \approx 0.771 \cdot 10^{-8}$  here.
*As in the first example, however, you can see:   The functions  ${\rm Q}(x)$  and  $1/2 \cdot{\rm erfc}(x)$  are basically equally well suited.
*Advantages or disadvantages of one or the other function arise only for concrete numerical values.}}
 

==Exercises==

* First select the number  $(1, 2, \text{...})$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 

{{BlueBox|TEXT=
'''(1)'''   Find the values of the function  ${\rm Q}(x)$  for  $x=1$,  $x=2$,  $x=4$  and  $x=6$.  Interpret the graphs for linear and logarithmic ordinates.}}

*The applet returns the values  ${\rm Q}(1)=1.5866 \cdot 10^{-1}$,  ${\rm Q}(2)=2. 275 \cdot 10^{-2}$,  ${\rm Q}(4)=3.1671 \cdot 10^{-5}$  and  ${\rm Q}(6)=9.8659 \cdot 10^{-10}$.
*With linear ordinate, the values for  $x>3$  are indistinguishable from the zero line.  More interesting is the plot with logarithmic ordinate.

{{BlueBox|TEXT=
'''(2)'''   Evaluate the two bounds  ${\rm UB}(x )=\text{Upper Bound }\big [{\rm Q}(x ) \big ]$  and  ${\rm LB}(x )=\text{Lower Bound }\big [{\rm Q}(x ) \big ]$  for the  ${\rm Q}$  function. }}

*For  $x \ge 2$  the upper bound is only slightly above  ${\rm Q}(x)$  and the lower bound is only slightly below  ${\rm Q}(x)$. 
*For example:  ${\rm Q}(x=4)=3.1671 \cdot 10^{-5}$   ⇒   ${\rm LB}(x=4)=3.1366 \cdot 10^{-5}$,   ${\rm UB}(x=4)=3.3458 \cdot 10^{-5}$.
*The upper bound has greater significance for assessing a communications system than "LB",  since this corresponds to a "worst case" consideration.

{{BlueBox|TEXT=
'''(3)'''   Try to use the app to determine  ${\rm Q}(x=2 \cdot \sqrt{2} \approx 2.828)$  as accurately as possible despite the quantization of the input parameter. }}
*The program returns for  $x=2.8$  the too large result  $2.5551 \cdot 10^{-3}$  and for  $x=2.85$  the result  $2.186 \cdot 10^{-3}$.  The exact value lies in between.
*But it also holds:  ${\rm Q}(x=2 \cdot \sqrt{2})=0.5 \cdot {\rm erfc}(x=2)$.  This gives the exact value  ${\rm Q}(x=2 \cdot \sqrt{2})=2.3389 \cdot 10^{-3}$.

{{BlueBox|TEXT=
'''(4)'''   Find the values of the function  $0.5 \cdot {\rm erfc}(x)$  for  $x=1$,  $x=2$,  $x=3$  and  $x=4$.  Interpret the exact results and the bounds.}}

*The applet returns:  $0.5 \cdot {\rm erfc}(1)=7.865 \cdot 10^{-2}$,  $0.5 \cdot {\rm erfc}(2)=2. 3389 \cdot 10^{-3}$,  $0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$  and  $0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$.
*All the above statements about  ${\rm Q}(x)$  with respect to suitable representation type and upper and lower bounds also apply to the function  $0.5 \cdot {\rm erfc}(x)$.

{{BlueBox|TEXT=
'''(5)'''   The results of  '''(4)'''  are now to be converted for the case of a logarithmic abscissa.  The conversion is done according to  $\rho\big[{\rm dB}\big ] = 20 \cdot \lg(x)$. }}

* The linear abscissa value  $x=1$  leads to the logarithmic abscissa value  $\rho=0\ \rm dB$   ⇒   $0. 5 \cdot {\rm erfc}(\rho=0\ {\rm dB})={0.5 \cdot \rm erfc}(x=1)=7.865 \cdot 10^{-2}$.
*Similarly  $0.5 \cdot {\rm erfc}(\rho=6.021\ {\rm dB}) =0.5 \cdot {\rm erfc}(x=2)=2. 3389 \cdot 10^{-3}$,     $0.5 \cdot {\rm erfc}(\rho=9.542\ {\rm dB})=0.5 \cdot {\rm erfc}(3)=1.1045 \cdot 10^{-5}$,   
*$0.5 \cdot {\rm erfc}(\rho=12.041\ {\rm dB})= 0.5 \cdot {\rm erfc}(4)=7.7086 \cdot 10^{-9}$.
*As per right diagram:  $0.5 \cdot {\rm erfc}(\rho=6\ {\rm dB}) =2.3883 \cdot 10^{-3}$,     $0.5 \cdot {\rm erfc}(\rho=9. 5\ {\rm dB}) =1.2109 \cdot 10^{-5}$,     $0.5 \cdot {\rm erfc}(\rho=12\ {\rm dB}) =9.006 \cdot 10^{-9}$.

{{BlueBox|TEXT=
'''(6)'''   Find  ${\rm Q}(\rho=0\ {\rm dB})$,  ${\rm Q}(\rho=5\ {\rm dB})$  and  ${\rm Q}(\rho=10\ {\rm dB})$,  and establish the relationship between linear and logarithmic abscissa.}}
*The program returns for logarithmic abscissa  ${\rm Q}(\rho=0\ {\rm dB})=1. 5866 \cdot 10^{-1}$,  ${\rm Q}(\rho=5\ {\rm dB})=3.7679 \cdot 10^{-2}$,  ${\rm Q}(\rho=10\ {\rm dB})=7.827 \cdot 10^{-4}$.
*The conversion is done according to the equation  $x=10^{\hspace{0.05cm}0.05\hspace{0.05cm} \cdot\hspace{0.05cm} \rho[{\rm dB}]}$.  For  $\rho=0\ {\rm dB}$  we get  $x=1$   ⇒   ${\rm Q}(\rho=0\ {\rm dB})={\rm Q}(x=1) =1.5866 \cdot 10^{-1}$.
*For  $\rho=5\ {\rm dB}$  we get  $x=1.1778$   ⇒   ${\rm Q}(\rho=5\ {\rm dB})={\rm Q}(x=1. 778) =3.7679 \cdot 10^{-2}$.  From the left diagram:  ${\rm Q}(x=1.8) =3.593 \cdot 10^{-2}$.
*For  $\rho=10\ {\rm dB}$  we get  $x=3.162$   ⇒   ${\rm Q}(\rho=10\ {\rm dB})={\rm Q}(x=3. 162) =7.827 \cdot 10^{-4}$.  After "quantization":  ${\rm Q}(x=3.15) =8.1635 \cdot 10^{-4}$.

==Applet Manual==
 

[[File:Qfunction bedienung.png|right|550px]]
    '''(A)'''     Equations used in the example  ${\rm Q}(x)$

    '''(B)'''     Selection option for  ${\rm Q}(x)$  or  ${\rm 0.5 \cdot erfc}(x)$

    '''(C)''''     Bounds  ${\rm LB}$  and  ${\rm UB}$  are drawn

    '''(D)''''     Selection whether abscissa is linear  $\rm (lin)$  or logarithmic  $\rm (log)$ 

    '''(E)''''     Select whether ordinate is linear  $\rm (lin)$  or logarithmic  $\rm (log)$ 

    '''(F)''''     Numerical output using the example  ${\rm Q}(x)$  with linear abscissa

    '''(G)'''     Slider input of abscissa value  $x$  for linear abscissa

    '''(H)'''     Slider input of abscissa value  $\rho \ \rm [dB]$  for logarithmic abscissa

    '''(I)''''     Graphical output of function  ${\rm Q}(x)$  – here:  linear abscissa

    '''(J)'''     Graph output of function  ${\rm 0.5 \cdot erfc}(x)$  – here:  linear abscissa

    '''(K)''''     Variation possibility for the graphical representations

$\hspace{1.5cm}$"$+$" (zoom in),

$\hspace{1.5cm}$"$-$" (zoom out)

$\hspace{1.5cm}$ "$\rm o$" (Reset)

$\hspace{1.5cm}$ "$\leftarrow$" (Move left), etc.
 
==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2007 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Thomas_Gro.C3.9Fer_.28Diplomarbeit_LB_2006.2C_danach_freie_Mitarbeit_bis_2010.29|Thomas Großer]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2018 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.

==Once again: Open Applet in new Tab==
 
{{LntAppletLinkEnDe|qfunction_en|qfunction}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-04-15T14:30:57Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}

===Signal reconstruction===

[[File:EN_Sig_T_5_1_S2_v2.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==
 
'''Korrektur'''
[[File:Anleitung_abtast.png|right|600px]]
    '''(A)'''     Selection:   Encoding                (binary,  quaternary,  AMI–code,  duobinary code).

    '''(B)'''     Selection:   Detection base pulse                 (by Gaussian–LP,  CRO–Nyquist,  by slit–LP}

    '''(C)'''     Parameter input to  '''(B)'''                  (cutoff frequency,  rolloff–factor,  rectangular wave duration)

    '''(D)'''     Eye diagram display control                  (start,  pause/continue,  single step,  total,  reset).

    '''(E)'''     Velocity of the eye diagram representation.

    '''(F)'''     Representation:  Detection ground momentum  $g_d(t)$

    '''(G)'''     Representation:  detection useful signal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Representation:  Eye diagram in the range  $\pm T$

    '''( I )''''     Numerical output:  $ö_{\rm norm}$  (normalized eye opening).

    '''(J)''''     Parameter input  $10 \cdot \lg \ E_{\rm B}/N_0$  for  '''(K)'''

    '''(K)''''     Numeric output:  $\sigma_{\rm norm}$  (normalized noise rms).

    '''(L)''''     Numeric output:  $p_{\rm U}$  (worst case error probability)

    '''(M)''''     Range for experimental performance:   Task selection.

    '''(N)''''     Range for the execution of experiments:   Task selection

    '''(O)''''     Range for carrying out the experiment:   Show sample solution
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Principle of 4B3T Coding

2023-04-15T13:34:34Z

Noah:

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

== Applet Description==
 
The applet illustrates the principle of  $\rm 4B3T$ coding.  Here, in each case a block of four binary symbols is replaced by a sequence of three ternary symbols.  This results in a relative code redundancy of just under  $16\%$,  which is used to achieve equal signal freedom.

The recoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be done according to a fixed code table. However, to further improve the spectral characteristics of these codes, the 4B3T codes always use multiple code tables, which are selected block by block according to the "running digital sum"  $(\rm RDS)$  .

In the applet, the corresponding code tables are given in the lower area, alternatively for
* the $\rm MS43$ code (from:   $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T–code), and
* the $\rm MMS43$ code (from:  $\rm M$odified $\rm MS43$).

Input parameters are, besides the desired code (MS43 or MMS43), the RDS start value  $\rm RDS_0$  and twelve binary source symbols  $q_\nu \in \{0,\ 1\}$,  either by hand, by default  $($source symbol sequence  $\rm A$,  $\rm B$,  $\rm C)$ or by random generator.  

Two different modes are offered by the program:

*In the "Step" mode, the three blocks are processed successively (in each case defining the three ternary symbols, updating the RDS value and thus defining the code table for the next block.

*In the "Total" mode, only the coding results are displayed, but simultaneously for the two possible codes and in each case for all four possible RDS ;start values.  The graphic and the RDS output block on the right refer to the settings made.

==Theoretical Background==

===Classification of various coding methods ===
 
We consider the digital transmission model shown.  As can be seen from this block diagram, depending on the target direction, a distinction is made between three different types of coding, each realized by the encoder at the transmitting end and the associated decoder at the receiving end:

[[File:EN_Inf_T_2_1_S1_v2.png|right|frame|Simplified model of a digital transmission system]]

*$\text{Source coding:}$  Removing (unnecessary) redundancy to store or transmit data as efficiently as possible   ⇒   Data compression.  Example:  Differential pulse code modulation  $\rm (DPCM)$  in image coding.

*$\text{Channel coding:}$  Targeted addition of (meaningful) redundancy, which can be used at the receiver for error detection or error detection.  Main representatives:  Block codes, convolutional codes, turbo codes.

*$\text{Line coding:}$  Recoding of source symbols to adapt the signal to the spectral characteristics of the channel and receiving equipment, for example to achieve a transmitted signal free of equal signals  $x(t)$  for a channel with  $H_{\rm K}(f = 0) = 0$ .  
 
In the case of line codes, a further distinction is made:
*$\text{Symbol-wise coding:}$  With each incoming binary symbol  $q_ν$  a multi-level (for example: ternary) code symbol  $c_ν$  is generated, which also depends on the previous binary symbols.   The symbol durations  $T_q$  and  $T_c$  are identical here.  Example:  Pseudo ternary codes (AMI code, duobinary code).

*$\text{Blockwise coding:}$  A block of  $m_q$  binary symbols  $(M_q = 2)$  is replaced by a sequence of  $m_c$  higher-level symbols  $(M_c > 2)$ .   A characteristic of this class of codes is  $T_c> T_q$.  Examples include redundancy-free multi-level codes  $(M_c$ is a power of two$)$  and the $\text{4B3T codes}$ considered here.

=== General description of 4B3T codes ===

The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =
3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks").

Such a 4B3T code has the following properties:
*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

*The 4B3T encoder signal can thus also be transmitted over a channel  (German:  "Kanal"   ⇒  subscript:  "K")  with the property  $H_{\rm K}(f= 0) = 0$  without noticeable degradation.

The encoding of the sixteen possible binary blocks into the corresponding ternary blocks could in principle be performed according to a fixed code table. To further improve the spectral properties of these codes, the common 4B3T codes, viz.

#the 4B3T code according to Jessop and Waters, 
#the MS43 code (from:  $\rm M$onitored $\rm S$um $\rm 4$B$\rm 3$T Code), 
#the FoMoT code (from:  $\rm Fo$ur $\rm Mo$de $\rm T$ernary), 

two or more code tables are used,  the selection of which is controlled by the  "running digital sum"  of the amplitude coefficients.  The principle is explained in the next section. 

== Running digital sum ==
 
After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
[[File:EN_Dig_T_2_3_S2.png|right|frame|Code tables for three 4B3T codes|class=fit]]
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot
\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.

The table shows the coding rules for the three 4B3T codes mentioned above.  To simplify the notation,
*  "+" stands for the amplitude coefficient "+1" and
*  "–" for the coefficient "–1". 

You can see from the graph:
#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$. 
#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
 
== ACF and PSD of the 4B3T codes==
 
The procedure for calculating the auto-correlation function  $\rm (ACF)$  and the power-spectral density  $\rm (PSD)$  is only outlined here in bullet points:
[[File:P_ID1335__Dig_T_2_3_S3_v1.png|right|frame|Markov diagram for the analysis of the 4B3T FoMoT code |class=fit]]

'''(1)'''   The transition of the running digital sum from  ${\it \Sigma}_l$  to  ${\it \Sigma}_{l+1}$  is described by a homogeneous stationary first-order Markov chain with six  $($Jessop–Waters$)$  or four states  $($MS43, FoMoT$)$.  For the FoMoT code, the Markov diagram sketched on the right applies. 

'''(2)'''   The values at the arrows denote the transition probabilities  ${\rm Pr}({\it \Sigma}_{l+1}|{\it \Sigma}_{l})$,  resulting from the respective code tables.  The colors correspond to the backgrounds of the table on the last section.  Due to the symmetry of the FoMoT Markov diagram,  the four probabilities are all the same:
:$${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}({\it \Sigma}_{l} = 3) = 1/4.$$

'''(3)'''   The auto-correlation function  $\varphi_a(\lambda) = {\rm E}\big [a_\nu \cdot a_{\nu+\lambda}\big ]$  of the amplitude coefficients can be determined from this diagram.  Simpler than the analytical calculation,  which requires a very large computational effort,  is the simulative determination of the ACF values by computer. 

Fourier transforming the ACF yields the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients corresponding to the following graph from  [TS87]<ref>Tröndle, K.; Söder, G.:  Optimization of Digital Transmission Systems.  Boston – London: Artech House, 1987,  ISBN:  0-89006-225-0.</ref>.  The outlined PSD was determined for the FoMoT code,  whose Markov diagram is shown above.  The differences between the individual 4B3T codes are not particularly pronounced.  Thus,  for the MS43 code  ${\rm E}\big [a_\nu^2 \big ] \approx 0.65$  and for the other two 4B3T codes  (Jessop/Waters, MS43)   ${\rm E}\big [a_\nu^2 \big ] \approx 0.69$. 
The statements of this graph can be summarized as follows:

[[File:EN_Dig_T_2_4_S3b_v23.png|right|frame|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding|class=fit]]

*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

*The PSD  ${\it \Phi}_s(f)$  including the transmission pulse is obtained by multiplying by  $1/T \cdot |G_s(f)|^2$   ⇒   ${\it \Phi}_a(f)$  must be multiplied by a  $\rm sinc^2$ function, if  $g_s(t)$  describes a rectangular pulse. 

*Redundancy-free binary or ternary coding results in a constant  ${\it \Phi}_a(f)$  in each case,  the magnitude of which depends on the number  $M$  of levels  (different signal power).

*In contrast,  the 4B3T power-spectral density has zeros at  $f = 0$  and multiples of  $f = 1/T$.  

*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers. 

*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous. 

*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation. 

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the filter impulse response  $h(t)$  are normalized, dimensionless and energy-limited ("time-limited pulses").
*All times, frequencies, and power values are to be understood normalized, too.

{{BlaueBox|TEXT=
'''(1)'''  Illustrate the 4B3T coding of the source symbol sequence  $\rm A$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 0, 1; \ 1, 0, 1, 1; \ 0, 1, 1, 0 \rangle $  according to the  $\rm MS43$  code ("Block–by–Block").         Let the RDS initial value be  ${\it \Sigma}_0= 0$.   ''Note'':  The source symbol sequence is already divided by semicolons into subsequences of four bits each. }}
* Starting from the RDS initial value  ${\it \Sigma}_0= 0$  you recognize the following coding of the first four bits (first block):  $(0, 1, 0, 1)\ \rightarrow\ (+,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_1= 2.$
*For the next four bits (second block), now assume  ${\it \Sigma}_1= 2$  $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3.$
*The encoding of bits 9 to 12 (third block) results:  ${\it \Sigma}_2= 3$  to  $(0, 1, 1, 0,)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(2)'''  Repeat this experiment with the other possible RDS initial values  ${\it \Sigma}_0= 1$,  ${\it \Sigma}_0= 2$  and  ${\it \Sigma}_0= 3.$  How do the coding results differ? }}

*${\it \Sigma}_0= 1$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 0$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 1$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 0.$
*${\it \Sigma}_0= 2$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 1$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 2$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 1.$
*${\it \Sigma}_0= 3$:     $(0, 1, 0, 1)\ \rightarrow\ (0,\ - ,\ 0) $   ⇒   ${\it \Sigma}_1= 2$:     $(1, 0, 1, 1)\ \rightarrow\ (+,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_2= 3$:     $(0, 1, 1, 0)\ \rightarrow\ (-,\ 0 ,\ 0) $   ⇒   ${\it \Sigma}_3= 2.$

{{BlaueBox|TEXT=
'''(3)'''  How many different code tables does the  $\rm MS43$ code use? }}

*From the previous experiments, we can see that the MS43 code uses at least two tables, switching between them according to the current RDS value.
*From the table given in the applet, it can be seen that three tables are actually used.  The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are in fact identical.

{{BlaueBox|TEXT=
'''(4)'''  Interpret the results of 4B3T coding for the source symbol sequence  $\rm B$   ⇒   $\langle q_\nu \rangle = \langle 1, 1, 1, 0; \ 0, 0, 1, 0; \ 1, 1, 1 \rangle $  and the MS43 code.}}

*For this source symbol sequence, the RDS value is not changed.   For each starting value  $(0$,  $1$,  $2$  and  $3)$  holds  ${\it \Sigma}_0 = {\it \Sigma}_1 ={\it \Sigma}_2 ={\it \Sigma}_3 $,  for example:
*${\it \Sigma}_0= 1$:     $(1, 1, 1, 0)\ \rightarrow\ (0,\ - ,\ +) $   ⇒   ${\it \Sigma}_1= 1$:     $(0, 0, 1, 0)\ \rightarrow\ (+,\ 0 ,\ -) $   ⇒   ${\it \Sigma}_2= 1$:     $(1, 1, 1, 1)\ \rightarrow\ (-,\ 0 ,\ +) $   ⇒   ${\it \Sigma}_3= 1.$
*The reason for this is that with this source symbol sequence, each ternary–triple contains exactly one "plus" and one "minus" after encoding.

{{BlaueBox|TEXT=
'''(5)'''  In contrast, how many different code tables does the modified MS43 code   ⇒   $\rm MMS43$ use? }}

*It can be seen from the table given in the applet that in the modified MS43 code all four tables are in fact different.  
*The entries for  ${\it \Sigma}_l= 1$  and  ${\it \Sigma}_l= 2$  are indeed largely the same.  They differ only for the binary sequences  $(0, 1, 1, 0)$  and  $(1, 0, 1, 0)$.
*The  $\rm MMS43$ code is used with  $\rm ISDN$  ("Integrated Services Digital Network")  on the local loop  $(U_{K0}$ interface$)$. 
*We do not know why the original MS43 code was modified during standardization.  We suspect a slightly more favorable power density spectrum.

{{BlaueBox|TEXT=
'''(6)'''  Compare the  $\rm MS43$  and  $\rm MMS43$  results for the source symbol sequences  $\rm A$  and  $\rm B$  and any RDS initial values.  Select "Overall View".}}
*For source symbol sequence  $\rm A$  there are two different  $\rm MS43$  code symbol sequences and three different   $\rm MMS43$  code symbol sequences.
*For the source symbol sequence  $\rm B$  the  $\rm MS43$  code symbol sequences are the same for all possible RDS initial values.  For  $\rm MMS43$:  two different coding results.

{{BlaueBox|TEXT=
'''(7)'''  Interpret the results for the sequence  $\rm C$   ⇒   $\langle q_\nu \rangle = \langle 0, 1, 1, 0; \ 0, 1, 1, 0; \ 0, 1, 1, 0 \rangle $  for both codes and all RDS initial values. Select "Overall View".}}
*The four input bits of each block are  $(0,\ 1,\ 1,\ 0)$.   With  $\rm MS43$  these are replaced by  $(0,\ +,\ +)$,  if  ${\it \Sigma}_l=0$;  resp.  $(-,\ 0,\ 0)$,  if  ${\it \Sigma}_l\ne0$.
*In the  $\rm MMS43$,  however, these are replaced by  $(-,\ +,\ +)$,  if  ${\it \Sigma}_l\le 1$;  resp.   $(-,\ -,\ +)$,  if  ${\it \Sigma}_l\ge 2$.  '''Only if you have enough time to spare:'''
*Try to make sense of this modification from  $\rm MS43$  to  $\rm MMS43$.  Our LNTww team did not succeed.

==Applet Manual==
 
[[Datei:Anleitung_4B3T.png|right|600px]] '''Korrektur'''
 
    '''(A)'''     Selection of one of four source signals

    '''(B)'''     Selection of parameters for source signal  $1$  (amplitude, frequency, phase)

    '''(C)'''     Output of the used program parameters.

    '''(D)'''     Parameter selection for sampling  $(f_{\rm G})$  and                 signal reconstruction  $(f_{\rm A},\ r)$

    '''(E)'''     Sketch of the receiver–frequency response  $H_{\rm E}(f)$

    '''(F)'''     Numerical output  $(P_x, \ P_{\rm \varepsilon}, \ 10 \cdot \lg(P_x/ P_{\rm \varepsilon})$

    '''(G)'''     Display selection for time domain

    '''(H)'''     Graphics area for time domain

    '''( I )'''     Display selection for frequency domain

    '''(J)'''     Graphics area for frequency domain

    '''(K)'''     Exercise area:  Exercise selection, questions, solution.
 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEnDe|4B3T_en|4B3T}}

Applets:Principle of Pseudo-Ternary Coding

2023-04-14T17:26:59Z

Noah:

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

==Applet Description==
 
The applet covers the properties of the best known pseudo-ternary codes, namely:
#  First-order bipolar code,  $\rm AMI$ code  (from: ''Alternate Mark Inversion''),  characterized by the parameters  $N_{\rm C} = 1, \ K_{\rm C} = +1$,
#  Duobinary code,  $(\rm DUOB)$,  code parameters:  $N_{\rm C} = 1, \ K_{\rm C} = -1$,
#  Second order bipolar code  $(\rm BIP2)$,  code parameters:  $N_{\rm C} = 2, \ K_{\rm C} = +1$.

At the input is the redundancy-free binary bipolar source symbol sequence  $\langle \hspace{0.05cm}q_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$   ⇒   rectangular signal  $q(t)$  an.  Illustrating the generation.
*of the binary–precoded sequence  $\langle \hspace{0.05cm}b_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$,  represented by the also redundancy-free binary bipolar rectangular signal  $b(t)$,
*the pseudo-ternary code sequence  $\langle \hspace{0.05cm}c_\nu \hspace{0.05cm}\rangle \ \in \{+1,\ 0, -1\}$,  represented by the redundant ternary bipolar rectangular signal  $c(t)$,
*the equally redundant ternary transmitted signal  $s(t)$, characterized by the amplitude coefficients  $a_\nu $,  and the (transmitted–) base impulse  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$

The base impulse  $g(t)$  –  in the applet  "Rectangle",  "Nyquist" and  "Root–Nyquist"  –  determines not only the shape of the transmitted signal, but also the course of
* of the auto-correlation function  $\rm (ACF)$  $\varphi_s (\tau)$  and
* of the associated power spectral density  $\rm (PSD)$  ${\it \Phi}_s (f)$.

The applet also shows that the total power spectral density  ${\it \Phi}_s (f)$ can be split into the part  ${\it \Phi}_a (f)$ that takes into account the statistical relations of the amplitude coefficients  $a_\nu$   and the energy spectral density $ {\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2 $, characterized by the shape  $g(t)$.

''Note''   In the applet, no distinction is made between the encoder symbols  $c_\nu \in \{+1,\ 0, -1\}$  and the amplitude coefficients  $a_\nu \in \{+1,\ 0, -1\}$  .   It should be remembered that the  $a_\nu$  are always numerical values, while for the encoder symbols also the notation  $c_\nu \in \{\text{plus},\ \text{zero},\ \text{minus}\}$  would be admissible.

==Theoretical Background==

== General description of the pseudo-multilevel codes ==
 
In symbolwise coding,  each incoming source symbol  $q_\nu$  generates an encoder symbol  $c_\nu$,  which depends not only on the current input symbol  $q_\nu$  but also on the  $N_{\rm C}$  preceding symbols  $q_{\nu-1}$, ... , $q_{\nu-N_{\rm C}} $.   $N_{\rm C}$  is referred to as the  "order"  of the code. 

Typical for symbolwise coding is that
*the symbol duration  $T$  of the encoded signal  (and of the transmitted signal)  matches the bit duration  $T_{\rm B}$  of the binary source signal,  and

*encoding and decoding do not lead to major time delays,  which are unavoidable when block codes are used. 

The  "pseudo-multilevel codes"  – better known as  "partial response codes"  –  are of special importance.  In the following,  only  "pseudo-ternary codes"   ⇒   level number  $M = 3$  are considered. 
*These can be described by the block diagram corresponding to the left graph. 
*In the right graph an equivalent circuit is given,  which is very suitable for an analysis of these codes.

[[File:EN_Dig_T_2_4_S1_v23.png|right|frame|Block diagram  (above)  and equivalent circuit  (below)  of a pseudo-ternary encoder|class=fit]]

One can see from the two representations:
*The pseudo-ternary encoder can be split into the  "non-linear pre-encoder"  and the  "linear coding network",  if the delay by  $N_{\rm C} \cdot T$  and the weighting by  $K_{\rm C}$  are drawn twice for clarity  – as shown in the right equivalent figure.

*The  "non-linear pre-encoder"  obtains the precoded symbols  $b_\nu$,  which are also binary,  by a modulo–2 addition  ("antivalence")  between the symbols  $q_\nu$  and  $K_{\rm C} \cdot b_{\nu-N_{\rm C}} $: 
:$$q_\nu \in \{-1, +1\},\hspace{0.1cm} K_{\rm C} \in \{-1,
+1\}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}b_\nu \in \{-1,
+1\}\hspace{0.05cm}.$$
*Like the source symbols  $q_\nu$,  the symbols  $b_\nu$  are statistically independent of each other.  Thus,  the pre-encoder does not add any redundancy.  However,  it allows a simpler realization of the decoder and prevents error propagation after a transmission error. 

*The actual encoding from binary  $(M_q = 2)$  to ternary  $(M = M_c = 3)$  is done by the  "linear coding network"  by the conventional subtraction
:$$c(t) ={1}/{2} \cdot \big [b(t) - K_{\rm C} \cdot b(t- N_{\rm
C}\cdot T)\big] \in \{-1, \ 0, +1\}\hspace{0.05cm},$$

:which can be described by the following  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|"impulse response"]]  resp.  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function|"transfer function"]]  with respect to the input signal  $b(t)$  and the output signal  $c(t)$: 
:$$h_{\rm C}(t) = {1}/{2} \cdot \big [\delta(t) - K_{\rm C} \cdot \delta(t- N_{\rm
C}\cdot T)\big] \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ H_{\rm C}(f) ={1}/{2} \cdot \left [1 - K_{\rm C} \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}N_{\rm C}\hspace{0.05cm}\cdot \hspace{0.05cm}T}\right]\hspace{0.05cm}. $$

*The relative redundancy is the same for all pseudo-ternary codes.  Substituting  $M_q=2$,  $M_c=3$  and  $T_c =T_q$  into the  [[Digital_Signal_Transmission/Redundancy-Free_Coding#Symbolwise_coding_vs._blockwise_coding|"general definition equation"]],  we obtain
:$$r_c = 1- \frac{R_q}{R_c} = 1- \frac{T_c}{T_q} \cdot \frac{{\rm log_2}\hspace{0.05cm} (M_q)}{{\rm log_2} \hspace{0.05cm}(M_c)} = 1- \frac{T_c}{T_q \cdot {\rm log_2} \hspace{0.05cm}(M_c)}\hspace{0.5cm}\Rightarrow \hspace{0.5cm} r_c = 1 -1/\log_2\hspace{0.05cm}(3) \approx 36.9 \%\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
The  $\text{transmitted signal of all pseudo-ternary codes}$  is always represented as follows:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$
*The property of the current pseudo-ternary code is reflected in the statistical bonds between the  $a_\nu$  In all cases  $a_\nu \in \{-1, \ 0, +1\}$.
*The basic transmitting pulse  $g(t)$  provides on the one hand the required energy, but has also influence on the statistical bonds within the signal.
*In addition to the NRZ rectangular pulse  $g_{\rm R}(t)$ can be selected in the program: 
:*the Nyquist impulse   ⇒   impulse response of the raised cosine low-pass with rolloff factor $r$:
:$$g_{\rm Nyq}(t)={\rm const.} \cdot \frac{\cos(\pi \cdot r\cdot t/T)}{1-(2\cdot r\cdot t/T)^2} \cdot {\rm si}(\pi \cdot t/T) \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\rm Nyq}(f),$$

:*the root Nyquist impulse   ⇒   impulse response of the root raised cosine low-pass with rolloff factor $r$:
:$$g_{\sqrt{\rm Nyq} }(t)\ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\sqrt{\rm Nyq} }(f)={\rm const.} \cdot \sqrt{G_{\rm Nyq}(f)} .$$ }}
 

== Properties of the AMI code==
 
The individual pseudo-ternary codes differ in the  $N_{\rm C}$  and  $K_{\rm C}$ parameters.  The best-known representative is the  '''first-order bipolar code'''  with the code parameters
:* $N_{\rm C} = 1$, 
:* $K_{\rm C} = 1$,

which is also known as  '''AMI code'''  (from: "Alternate Mark Inversion").  This is used e.g. with  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks")  on the so-called  $S_0$  interface.

[[File:EN_Dig_T_2_4_S2a.png|right|frame|Signals with AMI coding and HDB3 coding|class=fit]]
*The graph above shows the binary source signal  $q(t)$.

*The second and third diagrams show:
:* the likewise binary signal  $b(t)$  after the pre-encoder,  and

:* the encoded signal  $c(t) = s(t)$  of the AMI code.

One can see the simple AMI encoding principle:
#Each binary value  "–1"  of the source signal  $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary amplitude coefficient  $a_\nu = 0$.  
#The binary value  "+1"  of the source signal  $q(t)$   ⇒   symbol  $\rm H$  is alternately represented by  $a_\nu = +1$  and  $a_\nu = -1$.  

This ensures that the AMI encoded signal does not contain any   "long sequences" 
*   $ \langle c_\nu \rangle = \langle \text{...}, +1, +1, +1, +1, +1, \text{...}\rangle$   resp.
*   $ \langle c_\nu \rangle = \langle \text{...}, -1, -1, -1, -1, -1, \text{...}\rangle$,

which would lead to problems with a DC-free channel.

On the other hand,  the occurrence of long zero sequences is quite possible,  where no clock information is transmitted over a longer period of time.
 
To avoid this second problem,  some modified AMI codes have been developed, for example the  "B6ZS code"  and the  "HDB3 code":
*In the  '''HDB3 code'''  (green curve in the graphic),  four consecutive zeros in the AMI encoded signal are replaced by a subsequence that violates the AMI encoding rule. 

*In the gray shaded area,  this is the sequence  "$+\ 0\ 0\ +$",  since the last symbol before the replacement was a  "minus". 

*This limits the number of consecutive zeros to   $3$   for the HDB3 code and to   $5$   for the  [https://www.itwissen.info/en/bipolar-with-six-zero-substitution-B6ZS-121675.html#gsc.tab=0 "B6ZS code"]. 

*The decoder detects this code violation and replaces "$+\ 0\ 0\ +$" with "$0\ 0\ 0\ 0$" again. 
 
=== Calculating the ACF of a digital signal ===
In the execution of the experiment some quantities and correlations are used, which shall be briefly explained here:

*  The (time-unlimited) digital signal includes both the source statistics $($amplitude coefficients  $a_\nu$)  and the transmitted pulse shape  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g ( t - \nu \cdot T)\hspace{0.05cm}.$$

*  If  $s(t)$  is the pattern function of a stationary and ergodic random process, then for the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Random_processes|"Auto-Correlation Function"]]  $\rm (ACF)$:
:$$\varphi_s(\tau) = {\rm E}\big [s(t) \cdot s(t + \tau)\big ] = \sum_{\lambda = -\infty}^{+\infty}{1}/{T}
\cdot \varphi_a(\lambda)\cdot\varphi^{^{\bullet} }_{gs}(\tau -
\lambda \cdot T)\hspace{0.05cm}.$$

*  This equation describes the convolution of the discrete ACF  $\varphi_a(\lambda) = {\rm E}\big [ a_\nu \cdot a_{\nu + \lambda}\big]$  of the amplitude coefficients with the energy–ACF of the base impulse:

:$$\varphi^{^{\bullet} }_{g}(\tau) =
\int_{-\infty}^{+\infty} g ( t ) \cdot g ( t +
\tau)\,{\rm d} t \hspace{0.05cm}.$$

*The point is to indicate that  $\varphi^{^{\bullet} }_{g}(\tau)$  has the unit of an energy, while  $\varphi_s(\tau)$  indicates a power and  $\varphi_a(\lambda)$  is dimensionless.
 
=== Calculating the PSD of a digital signal ===
The corresponding quantity to the ACF in the frequency domain is the [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|"Power Spectral Density"]]  $\rm (PSD)$  ${\it \Phi}_s(f)$, which is fixedly related to  $\varphi_s(\tau)$  via the Fourier integral: 
:$$\varphi_s(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}_s(f) = \int_{-\infty}^{+\infty} \varphi_s(\tau) \cdot
{\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \tau}
\,{\rm d} \tau \hspace{0.05cm}.$$
*The power spectral density  ${\it \Phi}_s(f)$  can be represented as a product of two functions, taking into account the dimensional adjustment  $(1/T)$ :
:$${\it \Phi}_s(f) = {\it \Phi}_a(f) \cdot {1}/{T} \cdot
|G_s(f)|^2 \hspace{0.05cm}.$$
*The first term  ${\it \Phi}_a(f)$  is dimensionless and describes the spectral shaping of the transmitted signal by the statistical relations of the source: 
:$$\varphi_a(\lambda) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}{\it \Phi}_a(f) = \sum_{\lambda =
-\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm
j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda \hspace{0.02cm}T} =
\varphi_a(0) + 2 \cdot \sum_{\lambda =
1}^{\infty}\varphi_a(\lambda)\cdot\cos ( 2 \pi f
\lambda T) \hspace{0.05cm}.$$
*${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  takes into account the spectral shaping by  $g(t)$. The narrower this is, the wider  $\vert G(f) \vert^2$  and thus the larger the bandwidth requirement:
:$$\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2
\hspace{0.05cm}.$$
*The energy spectral density ${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  has the unit  $\rm Ws/Hz$  and the power spectral density  ${\it \Phi_{s}}(f)$  after division by the symbol spacing  $T$  has the unit  $\rm W/Hz$.

== Power-spectral density of the AMI code==
 
The frequency response of the linear code network of a pseudo-ternary code is generally:
:$$H_{\rm C}(f) = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
2\pi\hspace{0.03cm}\cdot \hspace{0.03cm}f \hspace{0.03cm}\cdot
\hspace{0.03cm} N_{\rm C}\hspace{0.03cm}\cdot \hspace{0.03cm}T}
\big] ={1}/{2} \cdot \big [1 - K \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
\alpha}
\big ]\hspace{0.05cm}.$$

This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients  $(K$  and  $\alpha$  are abbreviations according to the above equation$)$:
:$$ {\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = \frac{\big [1 - K \cos
(\alpha) + {\rm j}\cdot K \sin (\alpha) \big ] \big [1 - K \cos
(\alpha) - {\rm j}\cdot K \sin (\alpha) \big ] }{4} = \text{...} = {1}/{4} \cdot \big [2 - 2 \cdot K \cdot \cos
(\alpha) \big ] $$
[[File:P_ID1347__Dig_T_2_4_S2b_v2.png|right|frame|Power-spectral density of the AMI code|class=fit]]
:$$ \Rightarrow \hspace{0.3cm}{\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot \cos
(2\pi f N_{\rm C} T)\big ]
\hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}
\varphi_a(\lambda \cdot T)\hspace{0.05cm}.$$

In particular,  for the power-spectral density of the AMI code  $(N_{\rm C} = K_{\rm C} = 1)$,  we obtain:
:$${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos
(2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$$

The graph shows
*the PSD  ${\it \Phi}_a(f)$  of the amplitude coefficients  (red curve),  and 

*the PSD  ${\it \Phi}_s(f)$  of the total transmitted signal  (blue),  valid for NRZ rectangular pulses. 

One recognizes from this representation
*that the AMI code has no DC component,  since  ${\it \Phi}_a(f = 0) = {\it \Phi}_s(f = 0) = 0$,  

*the power  $P_{\rm S} = s_0^2/2$  of the AMI-coded transmitted signal  $($integral over  ${\it \Phi}_s(f)$  from  $- \infty$  to  $+\infty)$.

Notes:
*The PSD of the HDB3 and B6ZS codes differs only slightly from that of the AMI code. 

*You can use the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]]  to clarify the topic discussed here.

== Properties of the duobinary code ==
 
The  '''duobinary code'''  is defined by the code parameters  $N_{\rm C} = 1$  and  $K_{\rm C} = -1$.  This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients and the PSD of the transmitted signal:
[[File:P_ID1348__Dig_T_2_4_S3b_v1.png|right|frame|Power-spectral density of the duobinary code|right|class=fit]]

:$${\it \Phi}_a(f) ={1}/{2} \cdot \big [1 + \cos
(2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm},$$
:$$ {\it \Phi}_s(f) = s_0^2 \cdot T \cdot \cos^2
(\pi f T)\cdot {\rm si}^2
(\pi f T)= s_0^2 \cdot T \cdot {\rm si}^2
(2 \pi f T) \hspace{0.05cm}.$$

The graph shows the power-spectral density
*of the amplitude coefficients   ⇒   ${\it \Phi}_a(f)$  as a red curve, 

*of the total transmitted signal   ⇒   ${\it \Phi}_s(f)$  as a blue curve. 

In the second graph,  the signals  $q(t)$,  $b(t)$  and  $c(t) = s(t)$  are sketched. We refer here again to the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]],  which also clarifies the duobinary code.

[[File:P_ID1349__Dig_T_2_4_S3a_v2.png|left|frame|Signals in duobinary coding|class=fit]]
 From these illustrations it is clear:
*In the duobinary code,  any number of symbols with same polarity  ("+1"  or  "–1")  can directly succeed each other   ⇒   ${\it \Phi}_a(f = 0)=1$,  ${\it \Phi}_s(f = 0) = 1/2 \cdot s_0^2 \cdot T$. 

*In contrast,  for the duobinary code,  the alternating sequence  "... , +1, –1, +1, –1, +1, ..."  does not occur,  which is particularly disturbing with respect to intersymbol interference.  Therefore,  in the duobinary code:  ${\it \Phi}_s(f = 1/(2T) = 0$. 

*The power-spectral density  ${\it \Phi}_s(f)$  of the pseudo-ternary duobinary code is identical to the PSD with redundancy-free binary coding at half rate $($symbol duration  $2T)$. 
 

==Exercises==

 
*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  "$0$"  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".

{{BlueBox|TEXT=
'''(1)'''  Consider and interpret the binary pre–coding of the  $\text{AMI}$  code using the source symbol sequence  $\rm C$  assuming  $b_0 = +1$. }}
*The modulo–2 addition can also be taken as "antivalence".  It holds  $b_{\nu} = +1$,  if  $q_{\nu}$  and  $b_{\nu - 1}$  differ, otherwise set  $b_{\nu} = -1$ :
:  $b_1 = (q_1 = +1)\ {\rm XOR}\ (b_0= +1) = -1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (b_1= -1) = -1,\ \ b_3 = (q_3 = -1)\ {\rm XOR}\ (b_2= -1) = -1,$
:  $b_4 = (q_4 = +1)\ {\rm XOR}\ (b_3= -1) = +1,\ \ b_5 = (q_5 = +1)\ {\rm XOR}\ (b_4= +1) = -1,\ \ b_6 = (q_6 = +1)\ {\rm XOR}\ (b_5= -1) = +1,\ \ b_7 = b_8 = \text{...} = -1.$
*With the initial condition  $b_0 = -1$  we get the negated sequence:  $b_4 = b_6 =-1$.  All others  $b_\nu = +1$.

{{BlueBox|TEXT=
'''(2)'''  Let  $b_0 = +1$.  Consider the AMI encoded sequence  $\langle c_\nu \rangle$  of the source symbol sequence  $\rm C$  and give their amplitude coefficients  $a_\nu$.  }}

*It holds:  $a_1= 0.5 \cdot (b_1-b_0) = -1$,  $a_2= 0.5 \cdot (b_2-b_1) =0$,  $a_3= 0. 5 \cdot (b_3-b_2) =0$,  $a_4= +1$,  $a_5= -1$,  $a_6= +1$,  $a_7= -1$,  $a_8= a_9 = \text{...} = 0$.
*In contrast to the pre–coding, the conventional addition (subtraction) is to be applied here and not the modulo–2 addition.

{{BlueBox|TEXT=
'''(3)'''  Now consider the AMI coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$ ?}}

*Each binary value  "–1"  of   $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary coefficient  $a_\nu = 0$.  Any number of  $a_\nu = 0$  can be consecutive.
*The binary value  "+1"  of   $q(t)$   ⇒   symbol  $\rm H$  is represented alternatively with   $a_\nu = +1$  and  $a_\nu = -1$ , starting with  $a_\nu = -1$,  if  $b_0 = +1$.
*From the source symbol sequence  $\rm A$   ⇒   $\langle \hspace{0.05cm}q_\nu \equiv +1 \hspace{0.05cm}\rangle$  the code symbol sequence  $+1, -1, +1, -1, \text{...}$ . Long sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  or   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are shot out.

{{BlueBox|TEXT=
'''(4)'''  Continue with  AMI  coding.  Interpret the autocorrelation function  $\varphi_a(\lambda)$  of the amplitude coefficients and the power density spectrum  ${\it \Phi}_a(f)$. }}
*The discrete ACF  $\varphi_a(\lambda)$  of the amplitude coefficients is only defined for integer  $\lambda$  values.   With AMI coding  $(N_{\rm C}=1)$  holds:   For  $|\lambda| > 1$   ⇒   all  $\varphi_a(\lambda)= 0$.
*$\varphi_a(\lambda = 0)$  is equal to the root mean square of the amplitude coefficients   ⇒   $\varphi_a(\lambda = 0) = {\rm Pr}(a_\nu = +1) \cdot (+1)^2 + {\rm Pr}(a_\nu = -1) \cdot (-1)^2 = 0.5.$
*Only the combinations  $(+1, -1)$  and  $(-1, +1)$  contribute to the expected value  ${\rm E}\big [a_\nu \cdot a_{\nu+1}\big]$   Result:  $\varphi_a(\lambda = \pm 1)={\rm E}\big [a_\nu \cdot a_{\nu+1}\big]=-0.25.$
*The power density spectrum  ${\it \Phi}_a(f)$  is the Fourier transform of the discrete ACF  $\varphi_a(\lambda)$.  Result:  ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos (2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$
*From   ${\it \Phi}_a(f = 0) = 0$  follows:   The AMI code is especially interesting for channels over which no DC component can be transmitted.

{{BlueBox|TEXT=
'''(5)'''  We consider further AMI coding and rectangular pulses.  Interpret the ACF  $\varphi_s(\tau)$  of the transmission signal and the PDS  ${\it \Phi}_s(f)$. }}
*$\varphi_s(\tau)$  results from the convolution of the discrete AKF  $\varphi_a(\lambda)$  with  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$.   For rectangular pulses  $($duration $T)$:  The energy–AKF  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$  is a triangle of duration  $2T$.
*It holds  $\varphi_s(\tau = 0)= \varphi_a(\lambda = 0) =0.5, \ \varphi_s(\pm T)= \varphi_a( 1) =-0. 25,\ , \ \varphi_s( \pm 2T)= \varphi_a(2) =0.$  Between these discrete values, $\varphi_{s}(\tau)$  is always linear.
*The PDS  ${\it \Phi}_s(f)$  is obtained from  ${\it \Phi}_a(f) = \sin^2(\pi f T)$  by multiplying with  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = {\rm sinc}^2(f T).$  This does not change anything at the zeros of  ${\it \Phi}_a(f)$ .

{{BlueBox|TEXT=
'''(6)'''  What changes with respect to  $s(t)$,  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  with the  $\rm Nyquist$  pulse?  Vary the roll–off factor here in the range  $0 \le r \le 1$.}}

*A single Nyquist pulse can be represented with the source symbol sequence  $\rm B$  in the  $s(t)$  range.  You can see equidistant zero crossings in the distance  $T$.
*Also, for any AMI random sequence, the signal values  $s(t=\nu \cdot T)$  for each  $r$  correspond exactly to their nominal positions.  Outside these points, there are deviations.
*In the special case  $r=0$  the energy–LDS  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$.  Accordingly, the energy–ACF  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*On the other hand, for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are no longer equidistant, since although  $G(f)$  satisfies the first Nyquist criterion, it does not  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$.
*The main advantage of the Nyquist pulse is the much smaller bandwidth.  Here only the frequency range  $|f| < (1+r)/(2T)$  has to be provided.

{{BlueBox|TEXT=
'''(7)'''  Repeat the last experiment using the  $\text{Root raised cosine}$  pulse instead of the Nyquist pulse.  Interpret the results. }}

*In the special case  $r=0$  the results are as in  '''(6)'''. ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$  and outside zero;  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*Also for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are eqidistant  (but not  $\rm sinc$  shaped)   ⇒   ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$  satisfies the first Nyquist criterion.
*On the other hand, $G(f)$  does not satisfy the first Nyquist criterion  $($except for  $r=0)$.  Intersymbol interference occurs already at the transmitter   ⇒   signal  $s(t)$.
*But this is also not a fundamental problem.  By using an identically shaped reception filter like  $G(f)$  intersymbol interference at the decider is avoided.

{{BlueBox|TEXT=
'''(8)'''  Consider and check the pre–coding  $(b_\nu)$  and the amplitude coefficients  $(a_\nu)$  with the  $\rm duobinary$  coding   $($source symbol sequence  $\rm C$,  $b_0 = +1)$. }}
*$b_1 = (q_1 = +1)\ {\rm XOR}\ (\overline{b_0}= -1) = +1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (\overline{b_1}= -1) = -1,\ \ b_3 = \text{...} =b_7 = +1,$  $b_8 = b_{10} = \text{...} =-1$,  $b_9 =b_{11} = \text{...}= +1$.
*$a_1= 0.5 \cdot (b_1+b_0) = +1$,  $a_2= 0.5 \cdot (b_2+b_1) =0$,  $a_3= 0.5 \cdot (b_3+b_2) = 0$,  $a_4= \text{...}= a_7=+1$,  $a_8=a_9= \text{...}= 0$.
*With the starting condition  $b_0 = -1$  we get again the negated sequence:     $a_1= -1$,  $a_2= a_3= 0$,  $a_4= \text{...}= a_7=-1$,  $a_8=a_9= \text{...}= 0$.

{{BlueBox|TEXT=
'''(9)'''  Now consider the Duobinary coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$?}}

*The discrete ACF values are  $\varphi_a(\lambda = 0) = +0.5$,  $\varphi_a(\lambda = 1) = +0.25$,  $\varphi_a(\lambda = 2) = 0$   ⇒   ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 + \cos (2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm}.$
*Unlike the AMI coding, the encoded sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  and   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are possible here   ⇒   For the Duobinary code holds  ${\it \Phi}_a(f= 0) = 1 \ (\ne 0).$
*As with the AMI code, the "long zero sequence"   ⇒   $\langle \hspace{0.05cm}a_\nu \equiv 0 \hspace{0.05cm}\rangle$  is possible, which can again lead to synchronization problems.
*Excluded are the combinations  $a_\nu = +1, \ a_{\nu+1} = -1$  and   $a_\nu = -1, \ a_{\nu+1} = +1$,  recognizable by the PDS value  ${\it \Phi}_a(f= 1/(2T)) = 0.$
*Such direct transitions  $a_\nu = +1$   ⇒   $a_{\nu+1} = -1$  resp.   $a_\nu = -1$   ⇒   $a_{\nu+1} = +1$  lead to large intersymbol interference and thus to a higher error rate.

{{BlueBox|TEXT=
'''(10)'''  Compare the coding results of second order bipolar code  $\rm (BIP2)$  and first order bipolar code  $\rm (AMI)$  for different source symbol sequences. }}
*For a single rectangular pulse   ⇒   source symbol sequence  $\rm B$  both codes result in the same encoded sequence and the same encoded signal  $c(t)$   ⇒   also an isolated pulse.
*The "Permanent–One sequence"  $\rm A$  now results  $\langle c_\nu \rangle = \langle -1, -1, +1, +1, -1, -1, +1, +1, \text{. ..}\rangle $  instead of   $\langle c_\nu \rangle = \langle -1, +1, -1, +1, -1, +1, \text{...}\rangle $.
*The simple decoding algorithm of the AMI code  $($the ternary  $0$  becomes the binary  $-1$,  the ternary  $\pm 1$  becomes the binary  $+1)$  cannot be applied to  $\rm BIP2$ .

{{BlueBox|TEXT=
'''(11)'''  View and interpret the various ACF and LDS graphs of the  $\rm BIP2$  compared to the  $\rm AMI$  code.}}
*For  $\rm AMI$:   $\varphi_a(\lambda = \pm 1) = -0.25, \ \varphi_a(\lambda = \pm 2) = 0$.   For  $\rm BIP2$:   $\varphi_a(\lambda = \pm 1) = 0, \ \varphi_a(\lambda = \pm 2) = -0.25$.  In both cases:  $\varphi_a(\lambda = 0) = 0.5$.
*From the  $\rm AMI$ power density spectrum  ${\it \Phi}_a(f) = \sin^2 (\pi \cdot f T)$  follows for  $\rm BIP2$:  ${\it \Phi}_a(f) = \sin^2 (2\pi \cdot f T)$  by compression with respect to the  $f$–axis.
*Zero at  $f=0$:  At most two  $+1$  directly follow each other, and also at most only two  $-1$.  In the AMI code  $+1$  and  $-1$  occur only in isolation.
*Next zero at  $f=1/(2T)$:  The infinitely long  $(+1, -1)$  sequence is excluded in  $\rm BIP2$  as in the  $\rm Duobinary$  code.
*Consider and interpret also the functions  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  for the pulses "rectangle", "Nyquist" and "Root raised cosine".

==Applet Manual==
[[File:BS_Pseudoternär.png|right|600px|frame|Screenshot (German version, light background)]]
 
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Underlying block diagram

    '''(C)'''     Selection of the pseudoternary code:
                  AMI–code, duobinary code, 2nd order bipolar code.

    '''(D)'''     Selection of the base pulse  $g(t)$:
                   Rectangular pulse, Nyquist pulse, Root–Nyquist pulse.

    '''(E)'''     Rolloff–factor (frequency range) for "Nyquist" and "Root–Nyquist"

    '''(F)'''     Setting of  $3 \cdot 4 = 12$  Bit of source symbol sequence.

    '''(G)''''     Selection of one of three preset source symbol sequences.

    '''(H)'''     Random binary source symbol sequence.

    '''( I )'''     Stepwise elucidation of pseudoternary coding.

    '''(J)'''     Result of pseudoternary coding:  Signals  $q(t)$,  $b(t)$,  $c(t)$,  $s(t)$

    '''(K)'''     Deleting the signal waveforms in the graphics area  $\rm M$

    '''(L)'''     Sketches for autocorrelation function & power spectral density.

    '''(M)'''     Graphics area:   Source signal  $q(t)$, signal  $b(t)$  after precoding,                  encoder signal  $c(t)$  with rectangles, transmission signal  $s(t)$  according to  $g(t)$

    '''(N)'''     Area for exercises:  Exercise selection, questions, sample solution.
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]    as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.
 

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

Applets:Principle of Pseudo-Ternary Coding

2023-04-14T17:26:01Z

Noah:

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

==Applet Description==
 
The applet covers the properties of the best known pseudo-ternary codes, namely:
#  First-order bipolar code,  $\rm AMI$ code  (from: ''Alternate Mark Inversion''),  characterized by the parameters  $N_{\rm C} = 1, \ K_{\rm C} = +1$,
#  Duobinary code,  $(\rm DUOB)$,  code parameters:  $N_{\rm C} = 1, \ K_{\rm C} = -1$,
#  Second order bipolar code  $(\rm BIP2)$,  code parameters:  $N_{\rm C} = 2, \ K_{\rm C} = +1$.

At the input is the redundancy-free binary bipolar source symbol sequence  $\langle \hspace{0.05cm}q_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$   ⇒   rectangular signal  $q(t)$  an.  Illustrating the generation.
*of the binary–precoded sequence  $\langle \hspace{0.05cm}b_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$,  represented by the also redundancy-free binary bipolar rectangular signal  $b(t)$,
*the pseudo-ternary code sequence  $\langle \hspace{0.05cm}c_\nu \hspace{0.05cm}\rangle \ \in \{+1,\ 0, -1\}$,  represented by the redundant ternary bipolar rectangular signal  $c(t)$,
*the equally redundant ternary transmitted signal  $s(t)$, characterized by the amplitude coefficients  $a_\nu $,  and the (transmitted–) base impulse  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$

The base impulse  $g(t)$  –  in the applet  "Rectangle",  "Nyquist" and  "Root–Nyquist"  –  determines not only the shape of the transmitted signal, but also the course of
* of the auto-correlation function  $\rm (ACF)$  $\varphi_s (\tau)$  and
* of the associated power spectral density  $\rm (PSD)$  ${\it \Phi}_s (f)$.

The applet also shows that the total power spectral density  ${\it \Phi}_s (f)$ can be split into the part  ${\it \Phi}_a (f)$ that takes into account the statistical relations of the amplitude coefficients  $a_\nu$   and the energy spectral density $ {\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2 $, characterized by the shape  $g(t)$.

''Note''   In the applet, no distinction is made between the encoder symbols  $c_\nu \in \{+1,\ 0, -1\}$  and the amplitude coefficients  $a_\nu \in \{+1,\ 0, -1\}$  .   It should be remembered that the  $a_\nu$  are always numerical values, while for the encoder symbols also the notation  $c_\nu \in \{\text{plus},\ \text{zero},\ \text{minus}\}$  would be admissible.

==Theoretical Background==

== General description of the pseudo-multilevel codes ==
 
In symbolwise coding,  each incoming source symbol  $q_\nu$  generates an encoder symbol  $c_\nu$,  which depends not only on the current input symbol  $q_\nu$  but also on the  $N_{\rm C}$  preceding symbols  $q_{\nu-1}$, ... , $q_{\nu-N_{\rm C}} $.   $N_{\rm C}$  is referred to as the  "order"  of the code. 

Typical for symbolwise coding is that
*the symbol duration  $T$  of the encoded signal  (and of the transmitted signal)  matches the bit duration  $T_{\rm B}$  of the binary source signal,  and

*encoding and decoding do not lead to major time delays,  which are unavoidable when block codes are used. 

The  "pseudo-multilevel codes"  – better known as  "partial response codes"  –  are of special importance.  In the following,  only  "pseudo-ternary codes"   ⇒   level number  $M = 3$  are considered. 
*These can be described by the block diagram corresponding to the left graph. 
*In the right graph an equivalent circuit is given,  which is very suitable for an analysis of these codes.

[[File:EN_Dig_T_2_4_S1_v23.png|right|frame|Block diagram  (above)  and equivalent circuit  (below)  of a pseudo-ternary encoder|class=fit]]

One can see from the two representations:
*The pseudo-ternary encoder can be split into the  "non-linear pre-encoder"  and the  "linear coding network",  if the delay by  $N_{\rm C} \cdot T$  and the weighting by  $K_{\rm C}$  are drawn twice for clarity  – as shown in the right equivalent figure.

*The  "non-linear pre-encoder"  obtains the precoded symbols  $b_\nu$,  which are also binary,  by a modulo–2 addition  ("antivalence")  between the symbols  $q_\nu$  and  $K_{\rm C} \cdot b_{\nu-N_{\rm C}} $: 
:$$q_\nu \in \{-1, +1\},\hspace{0.1cm} K_{\rm C} \in \{-1,
+1\}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}b_\nu \in \{-1,
+1\}\hspace{0.05cm}.$$
*Like the source symbols  $q_\nu$,  the symbols  $b_\nu$  are statistically independent of each other.  Thus,  the pre-encoder does not add any redundancy.  However,  it allows a simpler realization of the decoder and prevents error propagation after a transmission error. 

*The actual encoding from binary  $(M_q = 2)$  to ternary  $(M = M_c = 3)$  is done by the  "linear coding network"  by the conventional subtraction
:$$c(t) ={1}/{2} \cdot \big [b(t) - K_{\rm C} \cdot b(t- N_{\rm
C}\cdot T)\big] \in \{-1, \ 0, +1\}\hspace{0.05cm},$$

:which can be described by the following  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|"impulse response"]]  resp.  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function|"transfer function"]]  with respect to the input signal  $b(t)$  and the output signal  $c(t)$: 
:$$h_{\rm C}(t) = {1}/{2} \cdot \big [\delta(t) - K_{\rm C} \cdot \delta(t- N_{\rm
C}\cdot T)\big] \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ H_{\rm C}(f) ={1}/{2} \cdot \left [1 - K_{\rm C} \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}N_{\rm C}\hspace{0.05cm}\cdot \hspace{0.05cm}T}\right]\hspace{0.05cm}. $$

*The relative redundancy is the same for all pseudo-ternary codes.  Substituting  $M_q=2$,  $M_c=3$  and  $T_c =T_q$  into the  [[Digital_Signal_Transmission/Redundancy-Free_Coding#Symbolwise_coding_vs._blockwise_coding|"general definition equation"]],  we obtain
:$$r_c = 1- \frac{R_q}{R_c} = 1- \frac{T_c}{T_q} \cdot \frac{{\rm log_2}\hspace{0.05cm} (M_q)}{{\rm log_2} \hspace{0.05cm}(M_c)} = 1- \frac{T_c}{T_q \cdot {\rm log_2} \hspace{0.05cm}(M_c)}\hspace{0.5cm}\Rightarrow \hspace{0.5cm} r_c = 1 -1/\log_2\hspace{0.05cm}(3) \approx 36.9 \%\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
The  $\text{transmitted signal of all pseudo-ternary codes}$  is always represented as follows:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$
*The property of the current pseudo-ternary code is reflected in the statistical bonds between the  $a_\nu$  In all cases  $a_\nu \in \{-1, \ 0, +1\}$.
*The basic transmitting pulse  $g(t)$  provides on the one hand the required energy, but has also influence on the statistical bonds within the signal.
*In addition to the NRZ rectangular pulse  $g_{\rm R}(t)$ can be selected in the program: 
:*the Nyquist impulse   ⇒   impulse response of the raised cosine low-pass with rolloff factor $r$:
:$$g_{\rm Nyq}(t)={\rm const.} \cdot \frac{\cos(\pi \cdot r\cdot t/T)}{1-(2\cdot r\cdot t/T)^2} \cdot {\rm si}(\pi \cdot t/T) \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\rm Nyq}(f),$$

:*the root Nyquist impulse   ⇒   impulse response of the root raised cosine low-pass with rolloff factor $r$:
:$$g_{\sqrt{\rm Nyq} }(t)\ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\sqrt{\rm Nyq} }(f)={\rm const.} \cdot \sqrt{G_{\rm Nyq}(f)} .$$ }}
 

== Properties of the AMI code==
 
The individual pseudo-ternary codes differ in the  $N_{\rm C}$  and  $K_{\rm C}$ parameters.  The best-known representative is the  '''first-order bipolar code'''  with the code parameters
:* $N_{\rm C} = 1$, 
:* $K_{\rm C} = 1$,

which is also known as  '''AMI code'''  (from: "Alternate Mark Inversion").  This is used e.g. with  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks")  on the so-called  $S_0$  interface.

[[File:EN_Dig_T_2_4_S2a.png|right|frame|Signals with AMI coding and HDB3 coding|class=fit]]
*The graph above shows the binary source signal  $q(t)$.

*The second and third diagrams show:
:* the likewise binary signal  $b(t)$  after the pre-encoder,  and

:* the encoded signal  $c(t) = s(t)$  of the AMI code.

One can see the simple AMI encoding principle:
#Each binary value  "–1"  of the source signal  $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary amplitude coefficient  $a_\nu = 0$.  
#The binary value  "+1"  of the source signal  $q(t)$   ⇒   symbol  $\rm H$  is alternately represented by  $a_\nu = +1$  and  $a_\nu = -1$.  

This ensures that the AMI encoded signal does not contain any   "long sequences" 
*   $ \langle c_\nu \rangle = \langle \text{...}, +1, +1, +1, +1, +1, \text{...}\rangle$   resp.
*   $ \langle c_\nu \rangle = \langle \text{...}, -1, -1, -1, -1, -1, \text{...}\rangle$,

which would lead to problems with a DC-free channel.

On the other hand,  the occurrence of long zero sequences is quite possible,  where no clock information is transmitted over a longer period of time.
 
To avoid this second problem,  some modified AMI codes have been developed, for example the  "B6ZS code"  and the  "HDB3 code":
*In the  '''HDB3 code'''  (green curve in the graphic),  four consecutive zeros in the AMI encoded signal are replaced by a subsequence that violates the AMI encoding rule. 

*In the gray shaded area,  this is the sequence  "$+\ 0\ 0\ +$",  since the last symbol before the replacement was a  "minus". 

*This limits the number of consecutive zeros to   $3$   for the HDB3 code and to   $5$   for the  [https://www.itwissen.info/en/bipolar-with-six-zero-substitution-B6ZS-121675.html#gsc.tab=0 "B6ZS code"]. 

*The decoder detects this code violation and replaces "$+\ 0\ 0\ +$" with "$0\ 0\ 0\ 0$" again. 
 
=== Calculating the ACF of a digital signal ===
In the execution of the experiment some quantities and correlations are used, which shall be briefly explained here:

*  The (time-unlimited) digital signal includes both the source statistics $($amplitude coefficients  $a_\nu$)  and the transmitted pulse shape  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g ( t - \nu \cdot T)\hspace{0.05cm}.$$

*  If  $s(t)$  is the pattern function of a stationary and ergodic random process, then for the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Random_processes|"Auto-Correlation Function"]]  $\rm (ACF)$:
:$$\varphi_s(\tau) = {\rm E}\big [s(t) \cdot s(t + \tau)\big ] = \sum_{\lambda = -\infty}^{+\infty}{1}/{T}
\cdot \varphi_a(\lambda)\cdot\varphi^{^{\bullet} }_{gs}(\tau -
\lambda \cdot T)\hspace{0.05cm}.$$

*  This equation describes the convolution of the discrete ACF  $\varphi_a(\lambda) = {\rm E}\big [ a_\nu \cdot a_{\nu + \lambda}\big]$  of the amplitude coefficients with the energy–ACF of the base impulse:

:$$\varphi^{^{\bullet} }_{g}(\tau) =
\int_{-\infty}^{+\infty} g ( t ) \cdot g ( t +
\tau)\,{\rm d} t \hspace{0.05cm}.$$

*The point is to indicate that  $\varphi^{^{\bullet} }_{g}(\tau)$  has the unit of an energy, while  $\varphi_s(\tau)$  indicates a power and  $\varphi_a(\lambda)$  is dimensionless.
 
=== Calculating the PSD of a digital signal ===
The corresponding quantity to the ACF in the frequency domain is the [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|"Power Spectral Density"]]  $\rm (PSD)$  ${\it \Phi}_s(f)$, which is fixedly related to  $\varphi_s(\tau)$  via the Fourier integral: 
:$$\varphi_s(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}_s(f) = \int_{-\infty}^{+\infty} \varphi_s(\tau) \cdot
{\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \tau}
\,{\rm d} \tau \hspace{0.05cm}.$$
*The power spectral density  ${\it \Phi}_s(f)$  can be represented as a product of two functions, taking into account the dimensional adjustment  $(1/T)$ :
:$${\it \Phi}_s(f) = {\it \Phi}_a(f) \cdot {1}/{T} \cdot
|G_s(f)|^2 \hspace{0.05cm}.$$
*The first term  ${\it \Phi}_a(f)$  is dimensionless and describes the spectral shaping of the transmitted signal by the statistical relations of the source: 
:$$\varphi_a(\lambda) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}{\it \Phi}_a(f) = \sum_{\lambda =
-\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm
j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda \hspace{0.02cm}T} =
\varphi_a(0) + 2 \cdot \sum_{\lambda =
1}^{\infty}\varphi_a(\lambda)\cdot\cos ( 2 \pi f
\lambda T) \hspace{0.05cm}.$$
*${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  takes into account the spectral shaping by  $g(t)$. The narrower this is, the wider  $\vert G(f) \vert^2$  and thus the larger the bandwidth requirement:
:$$\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2
\hspace{0.05cm}.$$
*The energy spectral density ${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  has the unit  $\rm Ws/Hz$  and the power spectral density  ${\it \Phi_{s}}(f)$  after division by the symbol spacing  $T$  has the unit  $\rm W/Hz$.

== Power-spectral density of the AMI code==
 
The frequency response of the linear code network of a pseudo-ternary code is generally:
:$$H_{\rm C}(f) = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
2\pi\hspace{0.03cm}\cdot \hspace{0.03cm}f \hspace{0.03cm}\cdot
\hspace{0.03cm} N_{\rm C}\hspace{0.03cm}\cdot \hspace{0.03cm}T}
\big] ={1}/{2} \cdot \big [1 - K \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
\alpha}
\big ]\hspace{0.05cm}.$$

This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients  $(K$  and  $\alpha$  are abbreviations according to the above equation$)$:
:$$ {\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = \frac{\big [1 - K \cos
(\alpha) + {\rm j}\cdot K \sin (\alpha) \big ] \big [1 - K \cos
(\alpha) - {\rm j}\cdot K \sin (\alpha) \big ] }{4} = \text{...} = {1}/{4} \cdot \big [2 - 2 \cdot K \cdot \cos
(\alpha) \big ] $$
[[File:P_ID1347__Dig_T_2_4_S2b_v2.png|right|frame|Power-spectral density of the AMI code|class=fit]]
:$$ \Rightarrow \hspace{0.3cm}{\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot \cos
(2\pi f N_{\rm C} T)\big ]
\hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}
\varphi_a(\lambda \cdot T)\hspace{0.05cm}.$$

In particular,  for the power-spectral density of the AMI code  $(N_{\rm C} = K_{\rm C} = 1)$,  we obtain:
:$${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos
(2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$$

The graph shows
*the PSD  ${\it \Phi}_a(f)$  of the amplitude coefficients  (red curve),  and 

*the PSD  ${\it \Phi}_s(f)$  of the total transmitted signal  (blue),  valid for NRZ rectangular pulses. 

One recognizes from this representation
*that the AMI code has no DC component,  since  ${\it \Phi}_a(f = 0) = {\it \Phi}_s(f = 0) = 0$,  

*the power  $P_{\rm S} = s_0^2/2$  of the AMI-coded transmitted signal  $($integral over  ${\it \Phi}_s(f)$  from  $- \infty$  to  $+\infty)$.

Notes:
*The PSD of the HDB3 and B6ZS codes differs only slightly from that of the AMI code. 

*You can use the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]]  to clarify the topic discussed here.

== Properties of the duobinary code ==
 
The  '''duobinary code'''  is defined by the code parameters  $N_{\rm C} = 1$  and  $K_{\rm C} = -1$.  This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients and the PSD of the transmitted signal:
[[File:P_ID1348__Dig_T_2_4_S3b_v1.png|right|frame|Power-spectral density of the duobinary code|right|class=fit]]

:$${\it \Phi}_a(f) ={1}/{2} \cdot \big [1 + \cos
(2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm},$$
:$$ {\it \Phi}_s(f) = s_0^2 \cdot T \cdot \cos^2
(\pi f T)\cdot {\rm si}^2
(\pi f T)= s_0^2 \cdot T \cdot {\rm si}^2
(2 \pi f T) \hspace{0.05cm}.$$

The graph shows the power-spectral density
*of the amplitude coefficients   ⇒   ${\it \Phi}_a(f)$  as a red curve, 

*of the total transmitted signal   ⇒   ${\it \Phi}_s(f)$  as a blue curve. 

In the second graph,  the signals  $q(t)$,  $b(t)$  and  $c(t) = s(t)$  are sketched. We refer here again to the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]],  which also clarifies the duobinary code.

[[File:P_ID1349__Dig_T_2_4_S3a_v2.png|left|frame|Signals in duobinary coding|class=fit]]
 From these illustrations it is clear:
*In the duobinary code,  any number of symbols with same polarity  ("+1"  or  "–1")  can directly succeed each other   ⇒   ${\it \Phi}_a(f = 0)=1$,  ${\it \Phi}_s(f = 0) = 1/2 \cdot s_0^2 \cdot T$. 

*In contrast,  for the duobinary code,  the alternating sequence  "... , +1, –1, +1, –1, +1, ..."  does not occur,  which is particularly disturbing with respect to intersymbol interference.  Therefore,  in the duobinary code:  ${\it \Phi}_s(f = 1/(2T) = 0$. 

*The power-spectral density  ${\it \Phi}_s(f)$  of the pseudo-ternary duobinary code is identical to the PSD with redundancy-free binary coding at half rate $($symbol duration  $2T)$. 
 

==Exercises==

 
*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  "$0$"  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".

{{BlueBox|TEXT=
'''(1)'''  Consider and interpret the binary pre–coding of the  $\text{AMI}$  code using the source symbol sequence  $\rm C$  assuming  $b_0 = +1$. }}
*The modulo–2 addition can also be taken as "antivalence".  It holds  $b_{\nu} = +1$,  if  $q_{\nu}$  and  $b_{\nu - 1}$  differ, otherwise set  $b_{\nu} = -1$ :
:  $b_1 = (q_1 = +1)\ {\rm XOR}\ (b_0= +1) = -1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (b_1= -1) = -1,\ \ b_3 = (q_3 = -1)\ {\rm XOR}\ (b_2= -1) = -1,$
:  $b_4 = (q_4 = +1)\ {\rm XOR}\ (b_3= -1) = +1,\ \ b_5 = (q_5 = +1)\ {\rm XOR}\ (b_4= +1) = -1,\ \ b_6 = (q_6 = +1)\ {\rm XOR}\ (b_5= -1) = +1,\ \ b_7 = b_8 = \text{...} = -1.$
*With the initial condition  $b_0 = -1$  we get the negated sequence:  $b_4 = b_6 =-1$.  All others  $b_\nu = +1$.

{{BlueBox|TEXT=
'''(2)'''  Let  $b_0 = +1$.  Consider the AMI encoded sequence  $\langle c_\nu \rangle$  of the source symbol sequence  $\rm C$  and give their amplitude coefficients  $a_\nu$.  }}

*It holds:  $a_1= 0.5 \cdot (b_1-b_0) = -1$,  $a_2= 0.5 \cdot (b_2-b_1) =0$,  $a_3= 0. 5 \cdot (b_3-b_2) =0$,  $a_4= +1$,  $a_5= -1$,  $a_6= +1$,  $a_7= -1$,  $a_8= a_9 = \text{...} = 0$.
*In contrast to the pre–coding, the conventional addition (subtraction) is to be applied here and not the modulo–2 addition.

{{BlueBox|TEXT=
'''(3)'''  Now consider the AMI coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$ ?}}

*Each binary value  "–1"  of   $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary coefficient  $a_\nu = 0$.  Any number of  $a_\nu = 0$  can be consecutive.
*The binary value  "+1"  of   $q(t)$   ⇒   symbol  $\rm H$  is represented alternatively with   $a_\nu = +1$  and  $a_\nu = -1$ , starting with  $a_\nu = -1$,  if  $b_0 = +1$.
*From the source symbol sequence  $\rm A$   ⇒   $\langle \hspace{0.05cm}q_\nu \equiv +1 \hspace{0.05cm}\rangle$  the code symbol sequence  $+1, -1, +1, -1, \text{...}$ . Long sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  or   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are shot out.

{{BlueBox|TEXT=
'''(4)'''  Continue with  AMI  coding.  Interpret the autocorrelation function  $\varphi_a(\lambda)$  of the amplitude coefficients and the power density spectrum  ${\it \Phi}_a(f)$. }}
*The discrete ACF  $\varphi_a(\lambda)$  of the amplitude coefficients is only defined for integer  $\lambda$  values.   With AMI coding  $(N_{\rm C}=1)$  holds:   For  $|\lambda| > 1$   ⇒   all  $\varphi_a(\lambda)= 0$.
*$\varphi_a(\lambda = 0)$  is equal to the root mean square of the amplitude coefficients   ⇒   $\varphi_a(\lambda = 0) = {\rm Pr}(a_\nu = +1) \cdot (+1)^2 + {\rm Pr}(a_\nu = -1) \cdot (-1)^2 = 0.5.$
*Only the combinations  $(+1, -1)$  and  $(-1, +1)$  contribute to the expected value  ${\rm E}\big [a_\nu \cdot a_{\nu+1}\big]$   Result:  $\varphi_a(\lambda = \pm 1)={\rm E}\big [a_\nu \cdot a_{\nu+1}\big]=-0.25.$
*The power density spectrum  ${\it \Phi}_a(f)$  is the Fourier transform of the discrete ACF  $\varphi_a(\lambda)$.  Result:  ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos (2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$
*From   ${\it \Phi}_a(f = 0) = 0$  follows:   The AMI code is especially interesting for channels over which no DC component can be transmitted.

{{BlueBox|TEXT=
'''(5)'''  We consider further AMI coding and rectangular pulses.  Interpret the ACF  $\varphi_s(\tau)$  of the transmission signal and the PDS  ${\it \Phi}_s(f)$. }}
*$\varphi_s(\tau)$  results from the convolution of the discrete AKF  $\varphi_a(\lambda)$  with  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$.   For rectangular pulses  $($duration $T)$:  The energy–AKF  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$  is a triangle of duration  $2T$.
*It holds  $\varphi_s(\tau = 0)= \varphi_a(\lambda = 0) =0.5, \ \varphi_s(\pm T)= \varphi_a( 1) =-0. 25,\ , \ \varphi_s( \pm 2T)= \varphi_a(2) =0.$  Between these discrete values, $\varphi_{s}(\tau)$  is always linear.
*The PDS  ${\it \Phi}_s(f)$  is obtained from  ${\it \Phi}_a(f) = \sin^2(\pi f T)$  by multiplying with  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = {\rm sinc}^2(f T).$  This does not change anything at the zeros of  ${\it \Phi}_a(f)$ .

{{BlueBox|TEXT=
'''(6)'''  What changes with respect to  $s(t)$,  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  with the  $\rm Nyquist$  pulse?  Vary the roll–off factor here in the range  $0 \le r \le 1$.}}

*A single Nyquist pulse can be represented with the source symbol sequence  $\rm B$  in the  $s(t)$  range.  You can see equidistant zero crossings in the distance  $T$.
*Also, for any AMI random sequence, the signal values  $s(t=\nu \cdot T)$  for each  $r$  correspond exactly to their nominal positions.  Outside these points, there are deviations.
*In the special case  $r=0$  the energy–LDS  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$.  Accordingly, the energy–ACF  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*On the other hand, for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are no longer equidistant, since although  $G(f)$  satisfies the first Nyquist criterion, it does not  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$.
*The main advantage of the Nyquist pulse is the much smaller bandwidth.  Here only the frequency range  $|f| < (1+r)/(2T)$  has to be provided.

{{BlueBox|TEXT=
'''(7)'''  Repeat the last experiment using the  $\text{Root raised cosine}$  pulse instead of the Nyquist pulse.  Interpret the results. }}

*In the special case  $r=0$  the results are as in  '''(6)'''. ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$  and outside zero;  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*Also for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are eqidistant  (but not  $\rm sinc$  shaped)   ⇒   ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$  satisfies the first Nyquist criterion.
*On the other hand, $G(f)$  does not satisfy the first Nyquist criterion  $($except for  $r=0)$.  Intersymbol interference occurs already at the transmitter   ⇒   signal  $s(t)$.
*But this is also not a fundamental problem.  By using an identically shaped reception filter like  $G(f)$  intersymbol interference at the decider is avoided.

{{BlueBox|TEXT=
'''(8)'''  Consider and check the pre–coding  $(b_\nu)$  and the amplitude coefficients  $(a_\nu)$  with the  $\rm duobinary$  coding   $($source symbol sequence  $\rm C$,  $b_0 = +1)$. }}
*$b_1 = (q_1 = +1)\ {\rm XOR}\ (\overline{b_0}= -1) = +1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (\overline{b_1}= -1) = -1,\ \ b_3 = \text{...} =b_7 = +1,$  $b_8 = b_{10} = \text{...} =-1$,  $b_9 =b_{11} = \text{...}= +1$.
*$a_1= 0.5 \cdot (b_1+b_0) = +1$,  $a_2= 0.5 \cdot (b_2+b_1) =0$,  $a_3= 0.5 \cdot (b_3+b_2) = 0$,  $a_4= \text{...}= a_7=+1$,  $a_8=a_9= \text{...}= 0$.
*With the starting condition  $b_0 = -1$  we get again the negated sequence:     $a_1= -1$,  $a_2= a_3= 0$,  $a_4= \text{...}= a_7=-1$,  $a_8=a_9= \text{...}= 0$.

{{BlueBox|TEXT=
'''(9)'''  Now consider the Duobinary coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$?}}

*The discrete ACF values are  $\varphi_a(\lambda = 0) = +0.5$,  $\varphi_a(\lambda = 1) = +0.25$,  $\varphi_a(\lambda = 2) = 0$   ⇒   ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 + \cos (2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm}.$
*Unlike the AMI coding, the encoded sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  and   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are possible here   ⇒   For the Duobinary code holds  ${\it \Phi}_a(f= 0) = 1 \ (\ne 0).$
*As with the AMI code, the "long zero sequence"   ⇒   $\langle \hspace{0.05cm}a_\nu \equiv 0 \hspace{0.05cm}\rangle$  is possible, which can again lead to synchronization problems.
*Excluded are the combinations  $a_\nu = +1, \ a_{\nu+1} = -1$  and   $a_\nu = -1, \ a_{\nu+1} = +1$,  recognizable by the PDS value  ${\it \Phi}_a(f= 1/(2T)) = 0.$
*Such direct transitions  $a_\nu = +1$   ⇒   $a_{\nu+1} = -1$  resp.   $a_\nu = -1$   ⇒   $a_{\nu+1} = +1$  lead to large intersymbol interference and thus to a higher error rate.

{{BlueBox|TEXT=
'''(10)'''  Compare the coding results of second order bipolar code  $\rm (BIP2)$  and first order bipolar code  $\rm (AMI)$  for different source symbol sequences. }}
*For a single rectangular pulse   ⇒   source symbol sequence  $\rm B$  both codes result in the same encoded sequence and the same encoded signal  $c(t)$   ⇒   also an isolated pulse.
*The "Permanent–One sequence"  $\rm A$  now results  $\langle c_\nu \rangle = \langle -1, -1, +1, +1, -1, -1, +1, +1, \text{. ..}\rangle $  instead of   $\langle c_\nu \rangle = \langle -1, +1, -1, +1, -1, +1, \text{...}\rangle $.
*The simple decoding algorithm of the AMI code  $($the ternary  $0$  becomes the binary  $-1$,  the ternary  $\pm 1$  becomes the binary  $+1)$  cannot be applied to  $\rm BIP2$ .

{{BlueBox|TEXT=
'''(11)'''  View and interpret the various ACF and LDS graphs of the  $\rm BIP2$  compared to the  $\rm AMI$  code.}}
*For  $\rm AMI$:   $\varphi_a(\lambda = \pm 1) = -0.25, \ \varphi_a(\lambda = \pm 2) = 0$.   For  $\rm BIP2$:   $\varphi_a(\lambda = \pm 1) = 0, \ \varphi_a(\lambda = \pm 2) = -0.25$.  In both cases:  $\varphi_a(\lambda = 0) = 0.5$.
*From the  $\rm AMI$ power density spectrum  ${\it \Phi}_a(f) = \sin^2 (\pi \cdot f T)$  follows for  $\rm BIP2$:  ${\it \Phi}_a(f) = \sin^2 (2\pi \cdot f T)$  by compression with respect to the  $f$–axis.
*Zero at  $f=0$:  At most two  $+1$  directly follow each other, and also at most only two  $-1$.  In the AMI code  $+1$  and  $-1$  occur only in isolation.
*Next zero at  $f=1/(2T)$:  The infinitely long  $(+1, -1)$  sequence is excluded in  $\rm BIP2$  as in the  $\rm Duobinary$  code.
*Consider and interpret also the functions  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  for the pulses "rectangle", "Nyquist" and "Root raised cosine".

==Applet Manual==
[[File:BS_Pseudoternär.png|right|600px|frame|Bildschirmabzug (deutsche Version, heller Hintergrund)]]
 
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Underlying block diagram

    '''(C)'''     Selection of the pseudoternary code:
                  AMI–code, duobinary code, 2nd order bipolar code.

    '''(D)'''     Selection of the base pulse  $g(t)$:
                   Rectangular pulse, Nyquist pulse, Root–Nyquist pulse.

    '''(E)'''     Rolloff–factor (frequency range) for "Nyquist" and "Root–Nyquist"

    '''(F)'''     Setting of  $3 \cdot 4 = 12$  Bit of source symbol sequence.

    '''(G)''''     Selection of one of three preset source symbol sequences.

    '''(H)'''     Random binary source symbol sequence.

    '''( I )'''     Stepwise elucidation of pseudoternary coding.

    '''(J)'''     Result of pseudoternary coding:  Signals  $q(t)$,  $b(t)$,  $c(t)$,  $s(t)$

    '''(K)'''     Deleting the signal waveforms in the graphics area  $\rm M$

    '''(L)'''     Sketches for autocorrelation function & power spectral density.

    '''(M)'''     Graphics area:   Source signal  $q(t)$, signal  $b(t)$  after precoding,                  encoder signal  $c(t)$  with rectangles, transmission signal  $s(t)$  according to  $g(t)$

    '''(N)'''     Area for exercises:  Exercise selection, questions, sample solution.
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]    as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.
 

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

Applets:Principle of Pseudo-Ternary Coding

2023-04-14T17:19:05Z

Noah:

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

==Applet Description==
 
The applet covers the properties of the best known pseudo-ternary codes, namely:
#  First-order bipolar code,  $\rm AMI$ code  (from: ''Alternate Mark Inversion''),  characterized by the parameters  $N_{\rm C} = 1, \ K_{\rm C} = +1$,
#  Duobinary code,  $(\rm DUOB)$,  code parameters:  $N_{\rm C} = 1, \ K_{\rm C} = -1$,
#  Second order bipolar code  $(\rm BIP2)$,  code parameters:  $N_{\rm C} = 2, \ K_{\rm C} = +1$.

At the input is the redundancy-free binary bipolar source symbol sequence  $\langle \hspace{0.05cm}q_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$   ⇒   rectangular signal  $q(t)$  an.  Illustrating the generation.
*of the binary–precoded sequence  $\langle \hspace{0.05cm}b_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$,  represented by the also redundancy-free binary bipolar rectangular signal  $b(t)$,
*the pseudo-ternary code sequence  $\langle \hspace{0.05cm}c_\nu \hspace{0.05cm}\rangle \ \in \{+1,\ 0, -1\}$,  represented by the redundant ternary bipolar rectangular signal  $c(t)$,
*the equally redundant ternary transmitted signal  $s(t)$, characterized by the amplitude coefficients  $a_\nu $,  and the (transmitted–) base impulse  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$

The base impulse  $g(t)$  –  in the applet  "Rectangle",  "Nyquist" and  "Root–Nyquist"  –  determines not only the shape of the transmitted signal, but also the course of
* of the auto-correlation function  $\rm (ACF)$  $\varphi_s (\tau)$  and
* of the associated power spectral density  $\rm (PSD)$  ${\it \Phi}_s (f)$.

The applet also shows that the total power spectral density  ${\it \Phi}_s (f)$ can be split into the part  ${\it \Phi}_a (f)$ that takes into account the statistical relations of the amplitude coefficients  $a_\nu$   and the energy spectral density $ {\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2 $, characterized by the shape  $g(t)$.

''Note''   In the applet, no distinction is made between the encoder symbols  $c_\nu \in \{+1,\ 0, -1\}$  and the amplitude coefficients  $a_\nu \in \{+1,\ 0, -1\}$  .   It should be remembered that the  $a_\nu$  are always numerical values, while for the encoder symbols also the notation  $c_\nu \in \{\text{plus},\ \text{zero},\ \text{minus}\}$  would be admissible.

==Theoretical Background==

== General description of the pseudo-multilevel codes ==
 
In symbolwise coding,  each incoming source symbol  $q_\nu$  generates an encoder symbol  $c_\nu$,  which depends not only on the current input symbol  $q_\nu$  but also on the  $N_{\rm C}$  preceding symbols  $q_{\nu-1}$, ... , $q_{\nu-N_{\rm C}} $.   $N_{\rm C}$  is referred to as the  "order"  of the code. 

Typical for symbolwise coding is that
*the symbol duration  $T$  of the encoded signal  (and of the transmitted signal)  matches the bit duration  $T_{\rm B}$  of the binary source signal,  and

*encoding and decoding do not lead to major time delays,  which are unavoidable when block codes are used. 

The  "pseudo-multilevel codes"  – better known as  "partial response codes"  –  are of special importance.  In the following,  only  "pseudo-ternary codes"   ⇒   level number  $M = 3$  are considered. 
*These can be described by the block diagram corresponding to the left graph. 
*In the right graph an equivalent circuit is given,  which is very suitable for an analysis of these codes.

[[File:EN_Dig_T_2_4_S1_v23.png|right|frame|Block diagram  (above)  and equivalent circuit  (below)  of a pseudo-ternary encoder|class=fit]]

One can see from the two representations:
*The pseudo-ternary encoder can be split into the  "non-linear pre-encoder"  and the  "linear coding network",  if the delay by  $N_{\rm C} \cdot T$  and the weighting by  $K_{\rm C}$  are drawn twice for clarity  – as shown in the right equivalent figure.

*The  "non-linear pre-encoder"  obtains the precoded symbols  $b_\nu$,  which are also binary,  by a modulo–2 addition  ("antivalence")  between the symbols  $q_\nu$  and  $K_{\rm C} \cdot b_{\nu-N_{\rm C}} $: 
:$$q_\nu \in \{-1, +1\},\hspace{0.1cm} K_{\rm C} \in \{-1,
+1\}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}b_\nu \in \{-1,
+1\}\hspace{0.05cm}.$$
*Like the source symbols  $q_\nu$,  the symbols  $b_\nu$  are statistically independent of each other.  Thus,  the pre-encoder does not add any redundancy.  However,  it allows a simpler realization of the decoder and prevents error propagation after a transmission error. 

*The actual encoding from binary  $(M_q = 2)$  to ternary  $(M = M_c = 3)$  is done by the  "linear coding network"  by the conventional subtraction
:$$c(t) ={1}/{2} \cdot \big [b(t) - K_{\rm C} \cdot b(t- N_{\rm
C}\cdot T)\big] \in \{-1, \ 0, +1\}\hspace{0.05cm},$$

:which can be described by the following  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|"impulse response"]]  resp.  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function|"transfer function"]]  with respect to the input signal  $b(t)$  and the output signal  $c(t)$: 
:$$h_{\rm C}(t) = {1}/{2} \cdot \big [\delta(t) - K_{\rm C} \cdot \delta(t- N_{\rm
C}\cdot T)\big] \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ H_{\rm C}(f) ={1}/{2} \cdot \left [1 - K_{\rm C} \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}N_{\rm C}\hspace{0.05cm}\cdot \hspace{0.05cm}T}\right]\hspace{0.05cm}. $$

*The relative redundancy is the same for all pseudo-ternary codes.  Substituting  $M_q=2$,  $M_c=3$  and  $T_c =T_q$  into the  [[Digital_Signal_Transmission/Redundancy-Free_Coding#Symbolwise_coding_vs._blockwise_coding|"general definition equation"]],  we obtain
:$$r_c = 1- \frac{R_q}{R_c} = 1- \frac{T_c}{T_q} \cdot \frac{{\rm log_2}\hspace{0.05cm} (M_q)}{{\rm log_2} \hspace{0.05cm}(M_c)} = 1- \frac{T_c}{T_q \cdot {\rm log_2} \hspace{0.05cm}(M_c)}\hspace{0.5cm}\Rightarrow \hspace{0.5cm} r_c = 1 -1/\log_2\hspace{0.05cm}(3) \approx 36.9 \%\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
The  $\text{transmitted signal of all pseudo-ternary codes}$  is always represented as follows:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$
*The property of the current pseudo-ternary code is reflected in the statistical bonds between the  $a_\nu$  In all cases  $a_\nu \in \{-1, \ 0, +1\}$.
*The basic transmitting pulse  $g(t)$  provides on the one hand the required energy, but has also influence on the statistical bonds within the signal.
*In addition to the NRZ rectangular pulse  $g_{\rm R}(t)$ can be selected in the program: 
:*the Nyquist impulse   ⇒   impulse response of the raised cosine low-pass with rolloff factor $r$:
:$$g_{\rm Nyq}(t)={\rm const.} \cdot \frac{\cos(\pi \cdot r\cdot t/T)}{1-(2\cdot r\cdot t/T)^2} \cdot {\rm si}(\pi \cdot t/T) \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\rm Nyq}(f),$$

:*the root Nyquist impulse   ⇒   impulse response of the root raised cosine low-pass with rolloff factor $r$:
:$$g_{\sqrt{\rm Nyq} }(t)\ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\sqrt{\rm Nyq} }(f)={\rm const.} \cdot \sqrt{G_{\rm Nyq}(f)} .$$ }}
 

== Properties of the AMI code==
 
The individual pseudo-ternary codes differ in the  $N_{\rm C}$  and  $K_{\rm C}$ parameters.  The best-known representative is the  '''first-order bipolar code'''  with the code parameters
:* $N_{\rm C} = 1$, 
:* $K_{\rm C} = 1$,

which is also known as  '''AMI code'''  (from: "Alternate Mark Inversion").  This is used e.g. with  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks")  on the so-called  $S_0$  interface.

[[File:EN_Dig_T_2_4_S2a.png|right|frame|Signals with AMI coding and HDB3 coding|class=fit]]
*The graph above shows the binary source signal  $q(t)$.

*The second and third diagrams show:
:* the likewise binary signal  $b(t)$  after the pre-encoder,  and

:* the encoded signal  $c(t) = s(t)$  of the AMI code.

One can see the simple AMI encoding principle:
#Each binary value  "–1"  of the source signal  $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary amplitude coefficient  $a_\nu = 0$.  
#The binary value  "+1"  of the source signal  $q(t)$   ⇒   symbol  $\rm H$  is alternately represented by  $a_\nu = +1$  and  $a_\nu = -1$.  

This ensures that the AMI encoded signal does not contain any   "long sequences" 
*   $ \langle c_\nu \rangle = \langle \text{...}, +1, +1, +1, +1, +1, \text{...}\rangle$   resp.
*   $ \langle c_\nu \rangle = \langle \text{...}, -1, -1, -1, -1, -1, \text{...}\rangle$,

which would lead to problems with a DC-free channel.

On the other hand,  the occurrence of long zero sequences is quite possible,  where no clock information is transmitted over a longer period of time.
 
To avoid this second problem,  some modified AMI codes have been developed, for example the  "B6ZS code"  and the  "HDB3 code":
*In the  '''HDB3 code'''  (green curve in the graphic),  four consecutive zeros in the AMI encoded signal are replaced by a subsequence that violates the AMI encoding rule. 

*In the gray shaded area,  this is the sequence  "$+\ 0\ 0\ +$",  since the last symbol before the replacement was a  "minus". 

*This limits the number of consecutive zeros to   $3$   for the HDB3 code and to   $5$   for the  [https://www.itwissen.info/en/bipolar-with-six-zero-substitution-B6ZS-121675.html#gsc.tab=0 "B6ZS code"]. 

*The decoder detects this code violation and replaces "$+\ 0\ 0\ +$" with "$0\ 0\ 0\ 0$" again. 
 
=== Calculating the ACF of a digital signal ===
In the execution of the experiment some quantities and correlations are used, which shall be briefly explained here:

*  The (time-unlimited) digital signal includes both the source statistics $($amplitude coefficients  $a_\nu$)  and the transmitted pulse shape  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g ( t - \nu \cdot T)\hspace{0.05cm}.$$

*  If  $s(t)$  is the pattern function of a stationary and ergodic random process, then for the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Random_processes|"Auto-Correlation Function"]]  $\rm (ACF)$:
:$$\varphi_s(\tau) = {\rm E}\big [s(t) \cdot s(t + \tau)\big ] = \sum_{\lambda = -\infty}^{+\infty}{1}/{T}
\cdot \varphi_a(\lambda)\cdot\varphi^{^{\bullet} }_{gs}(\tau -
\lambda \cdot T)\hspace{0.05cm}.$$

*  This equation describes the convolution of the discrete ACF  $\varphi_a(\lambda) = {\rm E}\big [ a_\nu \cdot a_{\nu + \lambda}\big]$  of the amplitude coefficients with the energy–ACF of the base impulse:

:$$\varphi^{^{\bullet} }_{g}(\tau) =
\int_{-\infty}^{+\infty} g ( t ) \cdot g ( t +
\tau)\,{\rm d} t \hspace{0.05cm}.$$

*The point is to indicate that  $\varphi^{^{\bullet} }_{g}(\tau)$  has the unit of an energy, while  $\varphi_s(\tau)$  indicates a power and  $\varphi_a(\lambda)$  is dimensionless.
 
=== Calculating the PSD of a digital signal ===
The corresponding quantity to the ACF in the frequency domain is the [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|"Power Spectral Density"]]  $\rm (PSD)$  ${\it \Phi}_s(f)$, which is fixedly related to  $\varphi_s(\tau)$  via the Fourier integral: 
:$$\varphi_s(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}_s(f) = \int_{-\infty}^{+\infty} \varphi_s(\tau) \cdot
{\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \tau}
\,{\rm d} \tau \hspace{0.05cm}.$$
*The power spectral density  ${\it \Phi}_s(f)$  can be represented as a product of two functions, taking into account the dimensional adjustment  $(1/T)$ :
:$${\it \Phi}_s(f) = {\it \Phi}_a(f) \cdot {1}/{T} \cdot
|G_s(f)|^2 \hspace{0.05cm}.$$
*The first term  ${\it \Phi}_a(f)$  is dimensionless and describes the spectral shaping of the transmitted signal by the statistical relations of the source: 
:$$\varphi_a(\lambda) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}{\it \Phi}_a(f) = \sum_{\lambda =
-\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm
j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda \hspace{0.02cm}T} =
\varphi_a(0) + 2 \cdot \sum_{\lambda =
1}^{\infty}\varphi_a(\lambda)\cdot\cos ( 2 \pi f
\lambda T) \hspace{0.05cm}.$$
*${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  takes into account the spectral shaping by  $g(t)$. The narrower this is, the wider  $\vert G(f) \vert^2$  and thus the larger the bandwidth requirement:
:$$\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2
\hspace{0.05cm}.$$
*The energy spectral density ${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  has the unit  $\rm Ws/Hz$  and the power spectral density  ${\it \Phi_{s}}(f)$  after division by the symbol spacing  $T$  has the unit  $\rm W/Hz$.

== Power-spectral density of the AMI code==
 
The frequency response of the linear code network of a pseudo-ternary code is generally:
:$$H_{\rm C}(f) = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
2\pi\hspace{0.03cm}\cdot \hspace{0.03cm}f \hspace{0.03cm}\cdot
\hspace{0.03cm} N_{\rm C}\hspace{0.03cm}\cdot \hspace{0.03cm}T}
\big] ={1}/{2} \cdot \big [1 - K \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
\alpha}
\big ]\hspace{0.05cm}.$$

This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients  $(K$  and  $\alpha$  are abbreviations according to the above equation$)$:
:$$ {\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = \frac{\big [1 - K \cos
(\alpha) + {\rm j}\cdot K \sin (\alpha) \big ] \big [1 - K \cos
(\alpha) - {\rm j}\cdot K \sin (\alpha) \big ] }{4} = \text{...} = {1}/{4} \cdot \big [2 - 2 \cdot K \cdot \cos
(\alpha) \big ] $$
[[File:P_ID1347__Dig_T_2_4_S2b_v2.png|right|frame|Power-spectral density of the AMI code|class=fit]]
:$$ \Rightarrow \hspace{0.3cm}{\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot \cos
(2\pi f N_{\rm C} T)\big ]
\hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}
\varphi_a(\lambda \cdot T)\hspace{0.05cm}.$$

In particular,  for the power-spectral density of the AMI code  $(N_{\rm C} = K_{\rm C} = 1)$,  we obtain:
:$${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos
(2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$$

The graph shows
*the PSD  ${\it \Phi}_a(f)$  of the amplitude coefficients  (red curve),  and 

*the PSD  ${\it \Phi}_s(f)$  of the total transmitted signal  (blue),  valid for NRZ rectangular pulses. 

One recognizes from this representation
*that the AMI code has no DC component,  since  ${\it \Phi}_a(f = 0) = {\it \Phi}_s(f = 0) = 0$,  

*the power  $P_{\rm S} = s_0^2/2$  of the AMI-coded transmitted signal  $($integral over  ${\it \Phi}_s(f)$  from  $- \infty$  to  $+\infty)$.

Notes:
*The PSD of the HDB3 and B6ZS codes differs only slightly from that of the AMI code. 

*You can use the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]]  to clarify the topic discussed here.

== Properties of the duobinary code ==
 
The  '''duobinary code'''  is defined by the code parameters  $N_{\rm C} = 1$  and  $K_{\rm C} = -1$.  This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients and the PSD of the transmitted signal:
[[File:P_ID1348__Dig_T_2_4_S3b_v1.png|right|frame|Power-spectral density of the duobinary code|right|class=fit]]

:$${\it \Phi}_a(f) ={1}/{2} \cdot \big [1 + \cos
(2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm},$$
:$$ {\it \Phi}_s(f) = s_0^2 \cdot T \cdot \cos^2
(\pi f T)\cdot {\rm si}^2
(\pi f T)= s_0^2 \cdot T \cdot {\rm si}^2
(2 \pi f T) \hspace{0.05cm}.$$

The graph shows the power-spectral density
*of the amplitude coefficients   ⇒   ${\it \Phi}_a(f)$  as a red curve, 

*of the total transmitted signal   ⇒   ${\it \Phi}_s(f)$  as a blue curve. 

In the second graph,  the signals  $q(t)$,  $b(t)$  and  $c(t) = s(t)$  are sketched. We refer here again to the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]],  which also clarifies the duobinary code.

[[File:P_ID1349__Dig_T_2_4_S3a_v2.png|left|frame|Signals in duobinary coding|class=fit]]
 From these illustrations it is clear:
*In the duobinary code,  any number of symbols with same polarity  ("+1"  or  "–1")  can directly succeed each other   ⇒   ${\it \Phi}_a(f = 0)=1$,  ${\it \Phi}_s(f = 0) = 1/2 \cdot s_0^2 \cdot T$. 

*In contrast,  for the duobinary code,  the alternating sequence  "... , +1, –1, +1, –1, +1, ..."  does not occur,  which is particularly disturbing with respect to intersymbol interference.  Therefore,  in the duobinary code:  ${\it \Phi}_s(f = 1/(2T) = 0$. 

*The power-spectral density  ${\it \Phi}_s(f)$  of the pseudo-ternary duobinary code is identical to the PSD with redundancy-free binary coding at half rate $($symbol duration  $2T)$. 
 

==Exercises==

 
*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  "$0$"  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".

{{BlueBox|TEXT=
'''(1)'''  Consider and interpret the binary pre–coding of the  $\text{AMI}$  code using the source symbol sequence  $\rm C$  assuming  $b_0 = +1$. }}
*The modulo–2 addition can also be taken as "antivalence".  It holds  $b_{\nu} = +1$,  if  $q_{\nu}$  and  $b_{\nu - 1}$  differ, otherwise set  $b_{\nu} = -1$ :
:  $b_1 = (q_1 = +1)\ {\rm XOR}\ (b_0= +1) = -1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (b_1= -1) = -1,\ \ b_3 = (q_3 = -1)\ {\rm XOR}\ (b_2= -1) = -1,$
:  $b_4 = (q_4 = +1)\ {\rm XOR}\ (b_3= -1) = +1,\ \ b_5 = (q_5 = +1)\ {\rm XOR}\ (b_4= +1) = -1,\ \ b_6 = (q_6 = +1)\ {\rm XOR}\ (b_5= -1) = +1,\ \ b_7 = b_8 = \text{...} = -1.$
*With the initial condition  $b_0 = -1$  we get the negated sequence:  $b_4 = b_6 =-1$.  All others  $b_\nu = +1$.

{{BlueBox|TEXT=
'''(2)'''  Let  $b_0 = +1$.  Consider the AMI encoded sequence  $\langle c_\nu \rangle$  of the source symbol sequence  $\rm C$  and give their amplitude coefficients  $a_\nu$.  }}

*It holds:  $a_1= 0.5 \cdot (b_1-b_0) = -1$,  $a_2= 0.5 \cdot (b_2-b_1) =0$,  $a_3= 0. 5 \cdot (b_3-b_2) =0$,  $a_4= +1$,  $a_5= -1$,  $a_6= +1$,  $a_7= -1$,  $a_8= a_9 = \text{...} = 0$.
*In contrast to the pre–coding, the conventional addition (subtraction) is to be applied here and not the modulo–2 addition.

{{BlueBox|TEXT=
'''(3)'''  Now consider the AMI coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$ ?}}

*Each binary value  "–1"  of   $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary coefficient  $a_\nu = 0$.  Any number of  $a_\nu = 0$  can be consecutive.
*The binary value  "+1"  of   $q(t)$   ⇒   symbol  $\rm H$  is represented alternatively with   $a_\nu = +1$  and  $a_\nu = -1$ , starting with  $a_\nu = -1$,  if  $b_0 = +1$.
*From the source symbol sequence  $\rm A$   ⇒   $\langle \hspace{0.05cm}q_\nu \equiv +1 \hspace{0.05cm}\rangle$  the code symbol sequence  $+1, -1, +1, -1, \text{...}$ . Long sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  or   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are shot out.

{{BlueBox|TEXT=
'''(4)'''  Continue with  AMI  coding.  Interpret the autocorrelation function  $\varphi_a(\lambda)$  of the amplitude coefficients and the power density spectrum  ${\it \Phi}_a(f)$. }}
*The discrete ACF  $\varphi_a(\lambda)$  of the amplitude coefficients is only defined for integer  $\lambda$  values.   With AMI coding  $(N_{\rm C}=1)$  holds:   For  $|\lambda| > 1$   ⇒   all  $\varphi_a(\lambda)= 0$.
*$\varphi_a(\lambda = 0)$  is equal to the root mean square of the amplitude coefficients   ⇒   $\varphi_a(\lambda = 0) = {\rm Pr}(a_\nu = +1) \cdot (+1)^2 + {\rm Pr}(a_\nu = -1) \cdot (-1)^2 = 0.5.$
*Only the combinations  $(+1, -1)$  and  $(-1, +1)$  contribute to the expected value  ${\rm E}\big [a_\nu \cdot a_{\nu+1}\big]$   Result:  $\varphi_a(\lambda = \pm 1)={\rm E}\big [a_\nu \cdot a_{\nu+1}\big]=-0.25.$
*The power density spectrum  ${\it \Phi}_a(f)$  is the Fourier transform of the discrete ACF  $\varphi_a(\lambda)$.  Result:  ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos (2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$
*From   ${\it \Phi}_a(f = 0) = 0$  follows:   The AMI code is especially interesting for channels over which no DC component can be transmitted.

{{BlueBox|TEXT=
'''(5)'''  We consider further AMI coding and rectangular pulses.  Interpret the ACF  $\varphi_s(\tau)$  of the transmission signal and the PDS  ${\it \Phi}_s(f)$. }}
*$\varphi_s(\tau)$  results from the convolution of the discrete AKF  $\varphi_a(\lambda)$  with  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$.   For rectangular pulses  $($duration $T)$:  The energy–AKF  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$  is a triangle of duration  $2T$.
*It holds  $\varphi_s(\tau = 0)= \varphi_a(\lambda = 0) =0.5, \ \varphi_s(\pm T)= \varphi_a( 1) =-0. 25,\ , \ \varphi_s( \pm 2T)= \varphi_a(2) =0.$  Between these discrete values, $\varphi_{s}(\tau)$  is always linear.
*The PDS  ${\it \Phi}_s(f)$  is obtained from  ${\it \Phi}_a(f) = \sin^2(\pi f T)$  by multiplying with  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = {\rm sinc}^2(f T).$  This does not change anything at the zeros of  ${\it \Phi}_a(f)$ .

{{BlueBox|TEXT=
'''(6)'''  What changes with respect to  $s(t)$,  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  with the  $\rm Nyquist$  pulse?  Vary the roll–off factor here in the range  $0 \le r \le 1$.}}

*A single Nyquist pulse can be represented with the source symbol sequence  $\rm B$  in the  $s(t)$  range.  You can see equidistant zero crossings in the distance  $T$.
*Also, for any AMI random sequence, the signal values  $s(t=\nu \cdot T)$  for each  $r$  correspond exactly to their nominal positions.  Outside these points, there are deviations.
*In the special case  $r=0$  the energy–LDS  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$.  Accordingly, the energy–ACF  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*On the other hand, for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are no longer equidistant, since although  $G(f)$  satisfies the first Nyquist criterion, it does not  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$.
*The main advantage of the Nyquist pulse is the much smaller bandwidth.  Here only the frequency range  $|f| < (1+r)/(2T)$  has to be provided.

{{BlueBox|TEXT=
'''(7)'''  Repeat the last experiment using the  $\text{Root raised cosine}$  pulse instead of the Nyquist pulse.  Interpret the results. }}

*In the special case  $r=0$  the results are as in  '''(6)'''. ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$  and outside zero;  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*Also for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are eqidistant  (but not  $\rm sinc$  shaped)   ⇒   ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$  satisfies the first Nyquist criterion.
*On the other hand, $G(f)$  does not satisfy the first Nyquist criterion  $($except for  $r=0)$.  Intersymbol interference occurs already at the transmitter   ⇒   signal  $s(t)$.
*But this is also not a fundamental problem.  By using an identically shaped reception filter like  $G(f)$  intersymbol interference at the decider is avoided.

{{BlueBox|TEXT=
'''(8)'''  Consider and check the pre–coding  $(b_\nu)$  and the amplitude coefficients  $(a_\nu)$  with the  $\rm duobinary$  coding   $($source symbol sequence  $\rm C$,  $b_0 = +1)$. }}
*$b_1 = (q_1 = +1)\ {\rm XOR}\ (\overline{b_0}= -1) = +1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (\overline{b_1}= -1) = -1,\ \ b_3 = \text{...} =b_7 = +1,$  $b_8 = b_{10} = \text{...} =-1$,  $b_9 =b_{11} = \text{...}= +1$.
*$a_1= 0.5 \cdot (b_1+b_0) = +1$,  $a_2= 0.5 \cdot (b_2+b_1) =0$,  $a_3= 0.5 \cdot (b_3+b_2) = 0$,  $a_4= \text{...}= a_7=+1$,  $a_8=a_9= \text{...}= 0$.
*With the starting condition  $b_0 = -1$  we get again the negated sequence:     $a_1= -1$,  $a_2= a_3= 0$,  $a_4= \text{...}= a_7=-1$,  $a_8=a_9= \text{...}= 0$.

{{BlueBox|TEXT=
'''(9)'''  Now consider the Duobinary coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$?}}

*The discrete ACF values are  $\varphi_a(\lambda = 0) = +0.5$,  $\varphi_a(\lambda = 1) = +0.25$,  $\varphi_a(\lambda = 2) = 0$   ⇒   ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 + \cos (2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm}.$
*Unlike the AMI coding, the encoded sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  and   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are possible here   ⇒   For the Duobinary code holds  ${\it \Phi}_a(f= 0) = 1 \ (\ne 0).$
*As with the AMI code, the "long zero sequence"   ⇒   $\langle \hspace{0.05cm}a_\nu \equiv 0 \hspace{0.05cm}\rangle$  is possible, which can again lead to synchronization problems.
*Excluded are the combinations  $a_\nu = +1, \ a_{\nu+1} = -1$  and   $a_\nu = -1, \ a_{\nu+1} = +1$,  recognizable by the PDS value  ${\it \Phi}_a(f= 1/(2T)) = 0.$
*Such direct transitions  $a_\nu = +1$   ⇒   $a_{\nu+1} = -1$  resp.   $a_\nu = -1$   ⇒   $a_{\nu+1} = +1$  lead to large intersymbol interference and thus to a higher error rate.

{{BlueBox|TEXT=
'''(10)'''  Compare the coding results of second order bipolar code  $\rm (BIP2)$  and first order bipolar code  $\rm (AMI)$  for different source symbol sequences. }}
*For a single rectangular pulse   ⇒   source symbol sequence  $\rm B$  both codes result in the same encoded sequence and the same encoded signal  $c(t)$   ⇒   also an isolated pulse.
*The "Permanent–One sequence"  $\rm A$  now results  $\langle c_\nu \rangle = \langle -1, -1, +1, +1, -1, -1, +1, +1, \text{. ..}\rangle $  instead of   $\langle c_\nu \rangle = \langle -1, +1, -1, +1, -1, +1, \text{...}\rangle $.
*The simple decoding algorithm of the AMI code  $($the ternary  $0$  becomes the binary  $-1$,  the ternary  $\pm 1$  becomes the binary  $+1)$  cannot be applied to  $\rm BIP2$ .

{{BlueBox|TEXT=
'''(11)'''  View and interpret the various ACF and LDS graphs of the  $\rm BIP2$  compared to the  $\rm AMI$  code.}}
*For  $\rm AMI$:   $\varphi_a(\lambda = \pm 1) = -0.25, \ \varphi_a(\lambda = \pm 2) = 0$.   For  $\rm BIP2$:   $\varphi_a(\lambda = \pm 1) = 0, \ \varphi_a(\lambda = \pm 2) = -0.25$.  In both cases:  $\varphi_a(\lambda = 0) = 0.5$.
*From the  $\rm AMI$ power density spectrum  ${\it \Phi}_a(f) = \sin^2 (\pi \cdot f T)$  follows for  $\rm BIP2$:  ${\it \Phi}_a(f) = \sin^2 (2\pi \cdot f T)$  by compression with respect to the  $f$–axis.
*Zero at  $f=0$:  At most two  $+1$  directly follow each other, and also at most only two  $-1$.  In the AMI code  $+1$  and  $-1$  occur only in isolation.
*Next zero at  $f=1/(2T)$:  The infinitely long  $(+1, -1)$  sequence is excluded in  $\rm BIP2$  as in the  $\rm Duobinary$  code.
*Consider and interpret also the functions  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  for the pulses "rectangle", "Nyquist" and "Root raised cosine".

==Applet Manual==
[[File:BS_Pseudoternär.png|right|600px|frame|Bildschirmabzug (deutsche Version, heller Hintergrund)]]
 
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Zugrundeliegendes Blockschaltbild

    '''(C)'''     Auswahl des Pseudoternörcodes:
                    AMI–Code, Duobinärcode, Bipolarcode 2. Ordnung

    '''(D)'''     Auswahl des Grundimpulses  $g(t)$:
                    Rechteckimpuls, Nyquistimpuls, Wurzel–Nyquistimpuls

    '''(E)'''     Rolloff–Faktor (Frequenzbereich) für "Nyquist" und "Wurzel–Nyquist"

    '''(F)'''     Einstellung von  $3 \cdot 4 = 12$  Bit der Quellensymbolfolge

    '''(G)'''     Auswahl einer drei voreingestellten Quellensymbolfolgen

    '''(H)'''     Zufällige binäre Quellensymbolfolge

    '''( I )'''     Schrittweise Verdeutlichung der Pseudoternärcodierung

    '''(J)'''     Ergebnis der Pseudoternärcodierung:  Signale  $q(t)$,  $b(t)$,  $c(t)$,  $s(t)$

    '''(K)'''     Löschen der Signalverläufe im Grafikbereich  $\rm M$

    '''(L)'''     Skizzen für Autokorrelationsfunktion & Leistungsdichtespektrum

    '''(M)'''     Grafikbereich:  Quellensignal  $q(t)$, Signal  $b(t)$  nach Vorcodierung,                    Codersignal  $c(t)$  mit Rechtecken, Sendesignal  $s(t)$  gemäß  $g(t)$

    '''(N)'''     Bereich für Übungen:  Aufgabenauswahl, Fragen, Musterlösung
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]    as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.
 

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-03-26T22:40:04Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity.}}

===Signal reconstruction===

[[File:P_ID1123__Sig_T_5_1_S5a_neu.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==
 
[[File:Anleitung_Auge.png|right|600px]]
    '''(A)'''     Auswahl:   Codierung                    (binär,  quaternär,  AMI–Code,  Duobinärcode)

    '''(B)'''     Auswahl:   Detektionsgrundimpuls                     (nach Gauß–TP,  CRO–Nyquist,  nach Spalt–TP}

    '''(C)'''     Prametereingabe zu  '''(B)'''                    (Grenzfrequenz,  Rolloff–Faktor,  Rechteckdauer)

    '''(D)'''     Steuerung der Augendiagrammdarstellung                    (Start,  Pause/Weiter,  Einzelschritt,  Gesamt,  Reset)

    '''(E)'''     Geschwindigkeit der Augendiagrammdarstellung

    '''(F)'''     Darstellung:  Detektionsgrundimpuls  $g_d(t)$

    '''(G)'''     Darstellung:  Detektionsnutzsignal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Darstellung:  Augendiagramm im Bereich  $\pm T$

    '''( I )'''     Numerikausgabe:  $ö_{\rm norm}$  (normierte Augenöffnung)

    '''(J)'''     Prametereingabe  $10 \cdot \lg \ E_{\rm B}/N_0$  für  '''(K)'''

    '''(K)'''     Numerikausgabe:  $\sigma_{\rm norm}$  (normierter Rauscheffektivwert)

    '''(L)'''     Numerikausgabe:  $p_{\rm U}$  (ungünstigste Fehlerwahrscheinlichkeit)

    '''(M)'''     Bereich für die Versuchsdurchführung:   Aufgabenauswahl

    '''(N)'''     Bereich für die Versuchsdurchführung:   Aufgabenstellung

    '''(O)'''     Bereich für die Versuchsdurchführung:   Musterlösung einblenden
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Physical Signal & Equivalent Lowpass Signal

2023-03-26T22:35:45Z

Noah:

{{LntAppletLink|physAnSignal_en}}         [https://www.lntwww.de/Applets:Physikalisches_Signal_%26_%C3%84quivalentes_TP-Signal '''English Applet with German WIKI description''']
==Applet Description==
 
This applet shows the relationship between the physical band-pass signal $x(t)$ and the associated equivalent low-pass signal $x_{\rm TP}(t)$. It is assumed that the band-pass signal $x(t)$ has a frequency-discrete spectrum $X(f)$:
:$$x(t) = x_{\rm T}(t) + x_{\rm O}(t)+ x_{\rm U}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right) + A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right). $$
The physical signal $x(t)$ is thus composed of three harmonic oscillations, a constellation that can be found, for example, in the ''Double-sideband Amplitude Modulation''
*of the message signal $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$   ⇒   in German:   '''N'''achrichtensignal
*with the carrier signal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$   ⇒   in German:   '''T'''rägersignal.

The nomenclature is also adapted to this case:
* $x_{\rm O}(t)$ denotes the "upper sideband"   (in German:   '''O'''beres Seitenband) with the amplitude $A_{\rm O}= A_{\rm N}/2$, the frequency $f_{\rm O} = f_{\rm T} + f_{\rm N}$ and the phase $\varphi_{\rm O} = \varphi_{\rm T} + \varphi_{\rm N}$.
*Similarly, for the "lower sideband"   (in German:   '''U'''nteres Seitenband) $x_{\rm U}(t)$ with $f_{\rm U} = f_{\rm T} - f_{\rm N}$, $A_{\rm U}= A_{\rm O}$ and $\varphi_{\rm U} = -\varphi_{\rm O}$.

The associated equivalent low-pass signal is $f_{\rm O}\hspace{0.01cm}' = f_{\rm O}- f_{\rm T} > 0$,   $f_{\rm U}\hspace{0.01cm}' = f_{\rm U}- f_{\rm T} < 0$  and  $f_{\rm T}\hspace{0.01cm}' = 0$:

:$$x_{\rm TP}(t) = x_\text{TP, T}(t) + x_\text{TP, O}(t) + x_\text{TP, U}(t) = A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} } \hspace{0.1cm}+ \hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{-{\rm j} \varphi_{\rm O} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}+ \hspace{0.1cm}
A_{\rm U}\cdot {\rm e}^{-{\rm j} \varphi_{\rm U} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t} . $$

[[File:Ortskurve_1.png|right|frame|Equivalent low-pass signal currently $t=0$ for cosinusoidal carrier   ⇒   $\varphi_{\rm T} = 0$]]
The program shows $x_{\rm TP}(t)$ as the vectorial sum of three rotation pointers as a violet dot (see figure for start time $t=0$ and cosinusoidal carrier):

*The (red) pointer of the carrier $x_\text{TP, T}(t)$ with the length $A_{\rm T}$ and the zero phase position $\varphi_{\rm T}=0$ is fixed in the complex plane. So it applies to all times $t$:   $x_{\rm TP}(t)= A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} }$.

*The (blue) pointer of the upper sideband $x_\text{TP, O}(t)$ with the length $A_{\rm O}$ and the zero phase position $\varphi_{\rm O}$ rotates at the angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'$ in mathematically positive direction (one revolution in time $1/f_{\rm O}\hspace{0.01cm}')$.

*The (green) pointer of the lower sideband $x_{\rm U+}(t)$ with the length $A_{\rm U}$ and the zero phase position $\varphi_{\rm U}$ rotates at the angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'$, because of $f_{\rm U}\hspace{0.01cm}'<0$ counterclockwise.

*With $f_{\rm U}\hspace{0.01cm}' = -f_{\rm O}\hspace{0.01cm}'$ the blue and green pointers will spin at the same speed but in different directions. Also, if $A_{\rm O} = A_{\rm U}$ and $\varphi_{\rm U} = -\varphi_{\rm O}$, then $x_{\rm TP}(t)$ moves on a straight line with a incline of $\varphi_{\rm T}$.

''Note:''   In the figure $\varphi_{\rm O} = +30^\circ$. From this follows for the start time $t=0$ the angle of the upper sideband (OSB, blue pointer) with respect to the coordinate system:   $\phi_{\rm O} = -\varphi_{\rm O} = -30^\circ$. Likewise, the zero phase position $\varphi_{\rm U} = -30^\circ$ of the lower sideband (USB, green pointer) follows for the phase angle to be considered in the complex plane:   $\phi_{\rm U} = +30^\circ$.

The temporal process of $x_{\rm TP}(t)$ is also referred to below as "locus". The relationship between $x_{\rm TP}(t)$ and the physical band-pass signal $x(t)$ is given in the section and the associated analytic signal is $x_+(t)$ :

:$$x_{\rm TP}(t) = x_{\rm +}(t)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t},$$
:$$x_{\rm +}(t) = x_{\rm TP}(t)\cdot {\rm e}^{+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t}.$$

==Theoretical Background==
 
===Description of Band-pass Signals===
[[File:Zeigerdiagramm_1a.png|right|frame|band-pass spectrum $X(f)$ |class=fit]]
We consider '''band-pass signals''' $x(t)$ with the property that their spectra $X(f)$ are not in the range around the frequency $f=0$, but around a carrier frequency $f_{\rm T}$. In most cases it can also be assumed that the bandwidth is $B \ll f_{\rm T}$.

The figure shows such a band-pass spectrum $X(f)$. Assuming that the associated $x(t)$ is a physical signal and thus real, the spectral function $X(f)$ has a symmetry with respect to the frequency $f = 0$, if $x(t)$ is an even function   ⇒   $x(-t)=x(t)$, $X(f)$ is real and even.

Beside the physical signal $x(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)$, one can also use the following descriptions of band-pass signals:
*the analytic signal $x_+(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_+(f)$, see applet [[Applets:Physical_Signal_%26_Analytic_Signal|"Physical Signal & Analytic Signal"]],
*the equivalent low-pass signal $x_{\rm TP}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_{\rm TP}(f)$, see next section
 

===Spectral Functions of the Analytic and the Equivalent Low-pass Signal===

The '''analytic signal''' $x_+(t)$ belonging to the physical signal $x(t)$ is the time function whose spectrum fulfills the following property:
[[File:Ortskurve_2.png|right|frame|spectral functions $X(f)$, $X_+(f)$ and $X_{\rm TP}(f)$ |class=fit]]
:$$X_+(f)=\big[1+{\rm sign}(f)\big] \cdot X(f) = \left\{ {2 \cdot
X(f) \; \hspace{0.2cm}\rm for\hspace{0.2cm} {\it f} > 0, \atop {\,\,\,\, \rm 0 \; \hspace{0.9cm}\rm for\hspace{0.2cm} {\it f} < 0.} }\right.$$

The ''Signum function'' is for positive values of $f$ equal to $+1$ and for negative $f$ values equal to $-1$.
*The (double-sided) limit returns $\sign(0) = 0$.
*The index „+” should make it clear that $X_+(f)$ only has parts at positive frequencies.

From the graph you can see the calculation rule for $X_+(f)$:   The actual band-pass spectrum $X(f)$ becomes
*doubled at the positive frequencies, and
*set to zero at the negative frequencies.

Due to the asymmetry of $X_+(f)$ with respect to the frequency $f = 0$, it can already be said that the time function $x_+(t)$ except for a trivial special case $x_+(t)= 0 \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \ X_+(f)= 0$ is always complex.

The spectrum $X_{\rm TP}(f)$ of the equivalent low-pass signal is obtained by shifting $X_+(f)$ to the left by the carrier frequency $f_{\rm T}$:
:$$X_{\rm TP}(f)= X_+(f+f_{\rm T}).$$

In the time domain this operation corresponds to the multiplication of $x_{\rm +}(t)$ with the complex exponential function with negative exponent:
:$$x_{\rm TP}(t) = x_{\rm +}(t)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \cdot f_{\rm T}\cdot \hspace{0.05cm}t}.$$

It can be seen that $x_{\rm TP}(t)$ is generally complex. But if $X_+(f)$ is symmetric about the carrier frequency $f_{\rm T}$, $X_{\rm TP}(f)$ is symmetric about the frequency $f=0$ and there is accordingly a real time function $x_{\rm TP}(t)$.
 

===$x_{\rm TP}(t)$ Representation of a Sum of Three Harmonic Oscillations===

In our applet, we always assume a set of three rotating pointers. The physical signal is:
:$$x(t) = x_{\rm T}(t) + x_{\rm O}(t) + x_{\rm U}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right) + A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right). $$
* Each of the three harmonic oscillations $x_{\rm T}(t)$, $x_{\rm U}(t)$ and $x_{\rm O}(t)$ is represented by an amplitude $(A)$, a frequency $(f)$ and a phase value $(\varphi)$.
*The indices are based on the modulation method [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation|"double-sideband amplitude modulation"]]. "T" stands for "carrier", "U" for "lower sideband" and "O" for "upper Sideband". Similarly, $f_{\rm U} < f_{\rm T}$ and $f_{\rm O} > f_{\rm T}$. There are no restrictions for the amplitudes and phases.

The associated equivalent low-pass signal is with $f_{\rm O}\hspace{0.01cm}' = f_{\rm O}- f_{\rm T} > 0$,   $f_{\rm U}\hspace{0.01cm}' = f_{\rm U}- f_{\rm T} < 0$  and  $f_{\rm T}\hspace{0.01cm}' = 0$:

:$$x_{\rm TP}(t) = x_\text{TP, T}(t) + x_\text{TP, O}(t) + x_\text{TP, U}(t) = A_{\rm T}\cdot {\rm e}^{-{\rm j} \varphi_{\rm T} } \hspace{0.1cm}+ \hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{-{\rm j} \varphi_{\rm O} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}+ \hspace{0.1cm}
A_{\rm U}\cdot {\rm e}^{-{\rm j} \varphi_{\rm U} } \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.01cm}'\hspace{0.05cm}\cdot \hspace{0.05cm}t} . $$

{{GraueBox|TEXT=
$\text{Example 1:}$ 
The constellation given here results, for example, in the [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation#AM-Signale_und_-Spektren_bei_harmonischem_Eingangssignal|"double-sideband amplitude modulation"]] of the message signal $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ with the carrier signal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$. This is discussed frequently in the exercises.

[[File:Ortskurve_5.png|center|frame|Spectrum $X_{\rm TP}(f)$ of the equivalent low–pass signal for different phase constellations |class=fit]]

There are some limitations to the program parameters in this approach:
* The frequencies are always $f\hspace{0.05cm}'_{\rm O} = f_{\rm N}$ and $f\hspace{0.05cm}'_{\rm U} = -f_{\rm N}$.
*Without distortion, the amplitude of the sidebands is $A_{\rm O}= A_{\rm U}= A_{\rm N}/2$.
*The respective phase relationships can be seen in the following graphic.

}}
 

===Representation of the Equivalent Low-pass Signal by Magnitude and Phase===

The generally complex equivalent low-pass signal
:$$x_{\rm TP}(t) = a(t) \cdot {\rm e}^{ {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \phi(t) }$$
can be split into a magnitude function $a(t)$ and a phase function $\phi(t)$ according to the equation given here, where:
:$$a(t) = \vert x_{\rm TP}(t)\vert = \sqrt{ {\rm Re}^2\big [x_{\rm TP}(t)\big ] + {\rm Im}^2\big [x_{\rm TP}(t)\big ] }\hspace{0.05cm},$$
:$$\phi(t) = \text{arc }x_{\rm TP}(t) = \arctan \frac{{\rm Im}\big [x_{\rm TP}(t)\big ]}{{\rm Re}\big [x_{\rm TP}(t)\big ]}.$$

The reason for this is that a band-pass signal $x(t)$ is usually described by the equivalent low-pass signal $x_{\rm TP}(t)$ that the functions $a(t)$ and $\phi(t)$ are interpretable in both representations:
*The magnitude $a(t)$ of the equivalent low-pass signal $x_{\rm TP}(t)$ indicates the (time-dependent) envelope of $x(t)$.
*The phase $\phi(t)$ of $x_{\rm TP}(t)$ denotes the location of the zero crossings of $x(t)$, where:
:–   For $\phi(t)>0$ the zero crossing is earlier than its nominal position   ⇒   the signal is leading here.
:–  When $\phi(t)<0$, the zero crossing is later than its target position   ⇒   the signal is trailing here.

{{GraueBox|TEXT=
$\text{Example 2:}$ 
The graph is intended to illustrate this relationship, assuming $A_{\rm U} > A_{\rm O}$   ⇒   the green pointer (for the lower sideband) is longer than the blue pointer (upper sideband). This is a snapshot at time $t_0$:

[[File:Ortskurve_3_neu.png|center|frame|band-pass spectrum $X(f)$ |class=fit]]

*For these system parameters, the top of the pointer cluster $x_{\rm TP}(t)$ – that is, the geometric sum of red, blue and green pointers – on an ellipse.
* The magnitude $a(t_0) = \vert x_{\rm TP}(t_0) \vert$ is drawn in black in the left-hand diagram and the phase value $\phi(t_0) = \text{arc }x_{\rm TP}(t_0) > 0$ is indicated in brown color.
*In the graph on the right, the magnitude $a(t_0) = \vert x_{\rm TP}(t_0) \vert$ of the equivalent low-pass signal indicates the envelope of the physical signal $x(t)$.
* At $\phi(t) \equiv 0$, all zero crossings of $x(t)$ would occur at equidistant intervals. Because of $\phi(t_0) > 0$, the signal is leading at the time $t_0$, that is: the zero crossings come earlier than the grid dictates. }}

==Exercises==
[[File:Exercises_verzerrungen.png|right]]
*First select the task number.
*A task description is displayed.
*Parameter values are adjusted.
*Solution after pressing "Hide solition".

The number "0" will reset the program and output a text with the further explanation of the applet.

In the following, $\rm Green$ denotes the lower sideband   ⇒   $\big (A_{\rm U}, f_{\rm U}, \varphi_{\rm U}\big )$,  
$\rm Red$ the carrier   ⇒   $\big (A_{\rm T}, f_{\rm T}, \varphi_{\rm T}\big )$ and
$\rm Blue$ the upper sideband   ⇒   $\big (A_{\rm O}, f_{\rm O}, \varphi_{\rm O}\big )$.

{{BlaueBox|TEXT=
'''(1)'''   Let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, f_{\rm T} = 100 \ \text{kHz}, \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0.4 \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \varphi_{\rm U} = -90^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, f_{\rm O} = 120 \ \text{kHz}, \varphi_{\rm O} = 90^\circ$.

:Consider and interpret the equivalent low-pass signal $x_{\rm TP}(t)$ and the physical signal $x(t)$. Which period $T_0$ recognizable?}}

:: The equivalent low-pass signal $x_{\rm TP}(t)$ takes from $x_{\rm TP}(t=0)=1\ \text{V}$ on the real axis values between $0.2\ \text{V}$ and $1.8\ \text{V}$   ⇒   phase $\phi(t) \equiv 0$.  The magnitude $|x_{\rm TP}(t)|$ indicates the envelope $a(t)$ of the physical signal $x(t)$. It holds $A_{\rm N} = 0.8\ \text{V}$ and $f_{\rm N} = 20\ \text{kHz}$:   $a(t) = A_{\rm T}+ A_{\rm N} \cdot \sin(2\pi\cdot f_{\rm N} \cdot t)$.  Both $x_{\rm TP}(t)$ and $x(t)$ are periodically with the period $T_0 = 1/f_{\rm N} = 50\ \rm µ s$.

{{BlaueBox|TEXT=
'''(2)'''   How do the ratios change to '''(1)''' with $f_{\rm U} = 99 \ \text{kHz}$ and $f_{\rm O} = 101 \ \text{kHz}$ ? How could $x(t)$ have arisen?}}

:: For the envelope $a(t)$ of the signal $x(t)$ we still have $a(t) = A_{\rm T}+ A_{\rm N} \cdot \sin(2\pi\cdot f_{\rm N} \cdot t)$, but now $f_{\rm N} = 1\ \text{kHz}$. Even though it can not be recognized:  $x_{\rm TP}(t)$ and $x(t)$ are still periodically:   $T_0 = 1\ \rm ms$. Example: Double-sideband Amplitude modulation '''(DSB–AM)''' of a sine signal with cosine carrier.

{{BlaueBox|TEXT=
'''(3)'''   Which settings have to be changed from '''(2)''' in order to arrive at the DSB–AM of a cosine signal with sine carrier. What changes over '''(2)'''?}}

::The carrier phase must be changed to $\varphi_{\rm T} = 90^\circ$   ⇒   sine carrier. Similarly, $\varphi_{\rm O} =\varphi_{\rm U} =\varphi_{\rm T} = 90^\circ$ must be set   ⇒   cosinusoidal message  The locus now lies on the imaginary axis  ⇒   $\phi(t) \equiv -90^\circ$. At the beginning $x_{\rm TP}(t=0)= - {\rm j} \cdot 1.8 \ \text{V}$.

{{BlaueBox|TEXT=
'''(4)'''   Now let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0.4 \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = 0^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 0^\circ$.

:What are the characteristics of this system "DSB–AM, where the message signal and carrier are respectively cosinusoidal"? What is the degree of modulation $m$?}}

:: The equivalent low-pass signal $x_{\rm TP}(t)$ takes from $x_{\rm TP}(t=0)=1.8\ \text{V}$ on the real axis values between $0.2\ \text{V}$ and $1.8\ \text{V}$   ⇒   phase $\phi(t) \equiv 0$.  Except for the start state $x_{\rm TP}(t=0)$ same behavior as at the setting '''(1)'''. The degree of modulation is $m = 0.8$.

{{BlaueBox|TEXT=
'''(5)'''   The parameters are still valid according to '''(4)''' with the exception of $A_{\rm T}= 0.6 \text{V}$. What is the degree of modulation $m$? What are the consequences?}}

:: There is now a DSB–AM with modulation degree $m = 1.333$. For $m > 1$, the simpler ''Envelope Demodulation'' is not applicable, since the phase function $\phi(t) \in \{ 0, \ \pm 180^\circ\}$ is no more constant and the envelope $a(t)$ no more matches the message signal. Rather, the complex ''Synchronous Demodulation'' must be used. Envelope detection would produce nonlinear distortions.

{{BlaueBox|TEXT=
'''(6)'''   The parameters are still valid according to '''(4)''' or '''(5)''' with the exception from $A_{\rm T}= 0$ on   ⇒   $m \to \infty$. Which modulation method is described in this way?}}

::It is a '''DSB–AM without carrier''' and a synchronous demodulation is required. The equivalent low-pass signal $x_{\rm TP}(t)$ is on the real axis, but not only in the right half-plane. Thus, the phase function $\phi(t) \in \{ 0, \ \pm 180^\circ\}$, also applies here, which means that ''Envelope Demodulation'' is not applicable.

{{BlaueBox|TEXT=
'''(7)'''   Now let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, f_{\rm T} = 100 \ \text{kHz}, \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0, \ f_{\rm U} = 80 \ \text{kHz}, \varphi_{\rm U} = -90^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.8\ \text{V}, f_{\rm O} = 120 \ \text{kHz}, \varphi_{\rm O} = 90^\circ$.

:Which constellation is described here? Which characteristics of this procedure can be recognized from the graphic?}}

::It is a [[Modulation_Methods/Einseitenbandmodulation|"single-sideband modulation"]] '''(SSB–AM)''', more specifically an '''OSB–AM''': the red carrier is fixed, the green pointer missing and the blue pointer (OSB) turns counterclockwise. The degree of modulation is $\mu = 0.8$ (in the case of SSB we denote the degree of modulation with $\mu$ instead of $m$). The carrier signal is cosinusoidal and the message signal sinusoidal. The locus is a circle. $x_{\rm TP}(t)$ moves in a mathematically positive direction. Because of $\phi(t) \ne \text{const.}$ the envelope demodulation is not applicable here:  This can be seen by the fact that the envelope $a(t)$ is not cosinusoidal. Rather, the lower half-wave is sharper than the upper one   ⇒   strong linear distortions.

{{BlaueBox|TEXT=
'''(8)'''   The parameters are still valid according to '''(7)''' with the exception of $A_{\rm O}= 0$ and $A_{\rm U}= 0.8 \text{ V}$. What differences arise opposite '''(7)'''?}}

::Now it is a '''LSB–AM''': The red carrier is fixed, the blue pointer is missing and the green pointer (LSB) rotates clockwise. All other statements of '''(7)''' apply here as well.

{{BlaueBox|TEXT=
'''(9)'''   The parameters according to '''(7)''' are still valid with the exception of $A_{\rm O} = 0.2 \text{ V} \ne A_{\rm U} = 0.4 \text{ V} $. What are the differences from '''(7)'''?}}

::The locus $x_{\rm TP}(t)$ is not a horizontal straight line, but an ellipse with the real part between $0.4 \text{ V}$ and $1.6 \text{ V}$ and the imaginary part in the range $\pm 0.2 \text{ V}$. Because of $\phi(t) \ne \text{const.}$ , Envelope demodulation would lead to non-linear distortions here too. The constellation simulated here describes the situation of '''(4)''', namely a DSB–AM with modulation degree $m = 0.8$, where the upper sideband is reduced to $50\%$ due to channel attenuation.

==Applet Manual==

[[File:Ortskurve_abzug3.png|right|frame|Screenshot]]

    '''(A)'''     Plot of the equivalent low-pass signal $x_{\rm TP}(t)$

    '''(B)'''     Plot of the physical signal $x(t)$

    '''(C)'''     Parameter input via slider:   amplitudes, frequencies, phase values

    '''(D)'''     Control elements:   Start – Step – Pause/Continue – Reset

    '''(E)'''     Speed of animation:   "Speed"   ⇒   Values: 1, 2 oder 3

    '''(F)'''     "Trace"   ⇒   On or Off, trace of equivalent low-pass signal   $x_{\rm TP}(t)$

    '''(G)'''     Numerical output:   time $t$, the signal values  ${\rm Re}[x_{\rm TP}(t)]$  and  ${\rm Im}[x_{\rm TP}(t)]$,

$\text{}\hspace{4.2cm}$   envelope $a(t) = |x_{\rm TP}(t)|$  and  phase $\phi(t) = {\rm arc} \ x_{\rm TP}(t)$

    '''(H)'''     Variations for the graphical representation

$\hspace{1.5cm}$Zoom–Functions "$+$" (Enlarge), "$-$" (Decrease) and $\rm o$ (Reset to default)

$\hspace{1.5cm}$Move with "$\leftarrow$" (Section to the left, ordinate to the right), "$\uparrow$" "$\downarrow$" "$\rightarrow$"

    '''(I)'''     Experiment section:   Task selection and task

    '''(J)'''     Experiment section:  solution
 

In all applets top right:    Changeable graphical interface design   ⇒   '''Theme''':
* Dark:   black background  (recommended by the authors).
* Bright:   white background  (recommended for beamers and printouts)
* Deuteranopia:   for users with pronounced green–visual impairment
* Protanopia:   for users with pronounced red–visual impairment
 
Note:
*Red parameters  $(A_{\rm T}, \ f_{\rm T}, \ \varphi_{\rm T})$  and the red pointer mark the "Carrier"  (German:  $\rm T$räger).  The red pointer does not turn.
* Green parameters  $(A_{\rm U}, \ f_{\rm U} < f_{\rm T}, \ \varphi_{\rm U})$  mark the "Lower sideband"  (German:  $\rm U$nteres Seitenband).  The green pointer rotates in a mathematically negative direction.
* Blue parameters  $(A_{\rm O}, \ f_{\rm O} > f_{\rm T}, \ \varphi_{\rm O})$  mark the "Upper sideband"  (German:  $\rm O$beres Seitenband).  The blue pointer turns counterclockwise.

==About the Authors==
This interactive calculation was designed and realized at the  [https://www.ei.tum.de/en/lnt/home//startseite Institute for Communications Engineering]  of the  [https://www.tum.de/ Technical University of Munich] .
*The original version was created in 2005 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her Diploma thesis using "FlashMX–Actionscript"  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]]  as part of her Bachelor's thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]).

==Once again: Open Applet in new Tab==

{{LntAppletLink|physAnSignal_en}}         [https://www.lntwww.de/Applets:Physikalisches_Signal_%26_%C3%84quivalentes_TP-Signal '''English Applet with German WIKI description''']

LNTwww:Applets

2023-03-26T22:32:23Z

Noah:

{{BlaueBox|TEXT=
*The selection list is organized by book.  In each book, the HTML5/JavaScript applets are listed first, followed by the SWF applets.
*At the end there are two more alphabetical lists of all HTML5/JS applets and all SWF applets.  Here you can also find information about their specifics.}}

{{Collapse|ID=signald|TITEL='''Applets about  "Signal Representation"'''|TEXT=
#  [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fouriertransform and Inverse]]
#  [[Applets:Frequency_&_Impulse_Responses|Frequency & Impulse Responses]]  
#  [[Applets:Graphical_Convolution|Graphical Convolution]]
#  [[Applets:Period_Duration_of_Periodic_Signals|Period Duration of Periodic_Signals]]
#  [[Applets:Physical_Signal_&_Analytic_Signal|Physical Signal & Analytic Signal]]
#  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal|Physical Signal & Equivalent Low-pass Signal]]
#  [[Applets:Pulses_and_Spectra|Pulses and Spectra]]
#  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]
}}
{{Collapse|ID=lzs|TITEL='''Applets about  "Linear Time–invariant Systems"'''|TEXT=
#  [[Applets:Attenuation_of_Copper_Cables|Attenuation of Copper Cables]]
#  [https://www.lntwww.de/Applets:Kausale_Systeme_und_Laplacetransformation Causal Systems and Laplace Transform]   ⇒   Note:  '''German language''' !
#  [[Applets:Frequency_&_Impulse_Responses|Frequency & Impulse Responses]]  
#  [[Applets:Graphical_Convolution|Graphical Convolution]]
#  [[Applets:Linear_Distortions_of_Periodic_Signals|Linear Distortions of Periodic Signals]]

==== SWF applets, based on Shockwave Flash ====
#  [[Applets:Bandbegrenzung|Einfluss einer Bandbegrenzung auf Sprache und Musik]]   ⇒   "Influence of a band limitation on speech and music"
#  [[Applets:Kausale_Systeme_-_Laplacetransformation|Kausale Systeme - Laplacetransformation]]   ⇒   "Causal systems - Laplace transform"
#  [[Applets:Laufzeit|Phasenlaufzeit & Gruppenlaufzeit]]   ⇒   "Phase delay & group delay"
#  [[Applets:Zeitverhalten_von_Kupferkabeln|Zeitverhalten von Kupferkabeln]]   ⇒   "Time behavior of copper cables"
}}
{{Collapse|ID=stosi|TITEL='''Applets about  "Theory of Stochastic Signals"|TEXT=
#  [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson Distribution]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [https://www.lntwww.de/Applets:Korrelation_und_Regressionsgerade Correlation and Regression Line]   ⇒   Note:  '''German language''' !
#  [[Applets:Digital_Filters|Digital Filters]]
#  [[Applets:Matched_Filter_Properties|Matched Filter Properties]]
#  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF, and Moments of Special Distributions]]
#  [[Applets:Two-dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]]
#  [https://www.lntwww.de/Applets:Zweidimensionale_Laplace-Zufallsgr%C3%B6%C3%9Fen_(Applet) Two-dimensional Laplace Random Variables]   ⇒   Note:  '''German language''' !

==== SWF applets, based on Shockwave Flash ====

#  [[Applets:Markovketten|Ereigniswahrscheinlichkeiten einer Markovkette]]   ⇒   "Event probabilities of a Markov chain"
#  [[Applets:Korrelationskoeffizient_%26_Regressionsgerade|Korrelationskoeffizient & Regressionsgerade]]   ⇒   "Correlation coefficient and regression line"
#  [[Applets:2D_Laplace_(SWF)|Zweidimensionale Laplaceverteilung]]   ⇒   "Two-dimensional Laplace distribution"
}}
{{Collapse|ID=infot|TITEL='''Applets about  "Information Theory"|TEXT=
#  [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson Distribution]]
#  [[Applets:Capacity_of_Memoryless_Digital_Channels|Capacity of Memoryless Digital Channels]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [https://www.lntwww.de/Applets:Entropie_und_N%C3%A4herungen_bin%C3%A4rer_Nachrichtenquellen Entropies and approximations of binary sources]   ⇒   Note:  '''German language''' !
#  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF, and Moments of Special Distributions]]
#  [[Applets:Principle_of_Pseudo-Ternary_Coding|Principle of Pseudo-Ternary Coding]]
#  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]
#  [[Applets:Two-dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]]

==== SWF applets, based on Shockwave Flash ====
#  [[Applets:Bandbegrenzung|Einfluss einer Bandbegrenzung auf Sprache und Musik]]   ⇒   "Influence of a band limitation on speech and music"
#  [[Applets:Entropien_von_Nachrichtenquellen|Entropien von Nachrichtenquellen]]   ⇒   "Entropy of binary and ternary sources"
#  [[Applets:Huffman_Shannon_Fano|Huffman-Shannon-Fano]]   ⇒   "Coding according to Huffman and Shannon/Fano"
#  [[Applets:Markovketten|Ereigniswahrscheinlichkeiten einer Markovkette]]   ⇒   "Event probabilities of a Markov chain"
#  [[Applets:Lempel-Ziv-Welch|Lempel-Ziv-Welch-Algorithmen]]   ⇒   "Lempel-Ziv-Welch algorithms"
#  [[Applets:QPSK_und_Offset-QPSK_(Applet)|QPSK und Offset-QPSK]]   ⇒   "QPSK and offset QPSK"
#  [[Applets:Qualität_verschiedener_Sprach–Codecs_(Applet)|Qualität verschiedener Sprach–Codecs]]   ⇒   "Quality of different speech codecs"
}}
{{Collapse|ID=modula|TITEL='''Applets about  "Modulation Methods"|TEXT=
#  [[Applets:Bessel_Functions_of_the_First_Kind|Bessel Functions of the First Kind]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fouriertransform and Inverse]]
#  [[Applets:Frequency_&_Impulse_Responses|Frequency & Impulse Responses]]
#  [[Applets:Generation_of_Walsh_functions|Generation of Walsh functions]]
#  [[Applets:Linear_Distortions_of_Periodic_Signals|Linear Distortions of Periodic Signals]]
#  [[Applets:Physical_Signal_&_Analytic_Signal|Physical Signal & Analytic Signal]]
#  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal|Physical Signal & Equivalent Low-pass Signal]]
#  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]

==== SWF applets, based on Shockwave Flash ====
#  [[Applets:DMT|DMT – Discrete Multiton Transmission]]
#  [[Applets:Synchrondemodulator|Eigenschaften des Synchrondemodulators]]   ⇒   "Features of the synchronous demodulator"
#  [[Applets:Bandbegrenzung|Einfluss einer Bandbegrenzung auf Sprache und Musik]]   ⇒   "Influence of a band limitation on speech and music"
#  [[Applets:Frequency_Shift_Keying_%26_Continuous_Phase_Modulation|Frequency Shift Keying & Continuous Phase Modulation]]
#  [[Applets:OFDM|OFDM - Spektrum & Signale]]   ⇒   "OFDM - Spectra and signals"
#  [[Applets:OVSF_codes_(Applet)|OVSF-Codes]]   ⇒   "OVSF codes"
#  [[Applets:DMT-Prinzip|Prinzip der Discrete Multitone Transmission]]   ⇒   "Principle of Discrete Multitone Transmission"
#  [[Applets:QPSK_und_Offset-QPSK_(Applet)|QPSK und Offset-QPSK]]   ⇒   "QPSK and offset QPSK"
}}
{{Collapse|ID=digsig|TITEL='''Applets about  "Digital Signal Transmission"|TEXT=
#  [[Applets:Attenuation_of_Copper_Cables|Attenuation of Copper Cables]]
#  [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson Distribution]]
#  [[Applets:Coherent_and_Non-Coherent_On-Off_Keying|Coherent and Non-Coherent On-Off Keying]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [[Applets:Eye_Pattern_and_Worst-Case_Error_Probability|Eye Pattern and Worst-Case Error Probability]]
#  [[Applets:Frequency_&_Impulse_Responses|Frequency & Impulse Responses]]
#  [https://www.lntwww.de/Applets:Das_Gram-Schmidt-Verfahren Gram–Schmidt method]   ⇒   Note:  '''German language''' !
#  [[Applets:Graphical_Convolution|Graphical Convolution]]
#  [[Applets:Matched_Filter_Properties|Matched Filter Properties]]
#  [[Applets:Principle_of_Pseudo-Ternary_Coding|Principle of Pseudo-Ternary Coding]]
#  [[Applets:Principle_of_4B3T_Coding|Principle of 4B3T Coding]]
#  [[Applets:Pulses_and_Spectra|Pulses and Spectra]]
#  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]
#  [[Applets:Two-dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]]
#  [https://www.lntwww.de/Applets:Zweidimensionale_Laplace-Zufallsgr%C3%B6%C3%9Fen_(Applet) Two-dimensional Laplace Random Variables]   ⇒   Note:  '''German language''' !

==== SWF applets, based on Shockwave Flash ====
#  [[Applets:Bandbegrenzung|Einfluss einer Bandbegrenzung auf Sprache und Musik]]   ⇒   "Influence of a band limitation on speech and music"
#  [[Applets:Entscheidungsrückkopplung|Applets:Entscheidungsrückkopplung]]   ⇒   "Decision feedback equalization"
#  [[Applets:Gram-Schmidt-Verfahren|Gram-Schmidt-Verfahren]]   ⇒   "Gram-Schmidt method"
#  [[Applets:Lineare_Nyquistentzerrung|Lineare Nyquistentzerrung]]   ⇒   "Linear Nyquist equalization"
#  [[Applets:MPSK_%26_Union-Bound(Applet)|Mehrstufige PSK & Union-Bound]]   ⇒   "Multi-stage PSK and Union Bound"
#  [[Applets:Regionen|Optimale Entscheidungsregionen]]   ⇒   "Optimal decision regions"
#  [[Applets:Prinzip_der_Quadratur-Amplitudenmodulation_(Applet)|Prinzip der Quadratur–Amplitudenmodulation]]   ⇒   "Principle of Quadrature Amplitude Modulation"
#  [[Applets:Fehlerwahrscheinlichkeit|Symbolfehlerwahrscheinlichkeit von Digitalsystemen]]   ⇒   "Symbol error probability of digital communications systems"
#  [[Applets:Viterbi|Viterbi–Entscheider für einen Vorläufer]]   ⇒   "Viterbi decider for one precursor"
#  [[Applets:Zeitverhalten_von_Kupferkabeln|Zeitverhalten von Kupferkabeln]]   ⇒   "Time behavior of copper cables"
#  [[Applets:2D_Laplace_(SWF)|Zweidimensionale Laplaceverteilung]]   ⇒   "Two-dimensional Laplace distribution"
}}
{{Collapse|ID=mobcomm|TITEL='''Applets about  "Mobile Communications"'''|TEXT=
#  [[Applets:Bessel_Functions_of_the_First_Kind|Bessel Functions of the First Kind]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [[Applets:Digital_Filters|Digital Filters]]
#  [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fouriertransform and Inverse]]
#  [[Applets:The_Doppler_Effect|Doppler Effect]]
#  [[Applets:Frequency_&_Impulse_Responses|Frequency & Impulse Responses]]  
#  [[Applets:Generation_of_Walsh_functions|Generation of Walsh functions]]  
#  [[Applets:Graphical_Convolution|Graphical Convolution]]
#  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF, and Moments of Special Distributions]]

==== SWF applets - based on Shockwave Flash ====
#  [[Applets:Mehrwegeausbreitung_und_Frequenzselektivit%C3%A4t_(Applet)| Mehrwegeausbreitung und Frequenzselektivität]]   ⇒   "Multipath propagation and frequency selectivity"
#  [[Applets:OFDM|OFDM - Spektrum & Signale]]   ⇒   "OFDM - Spectra and signals"
#  [[Applets:OVSF-Codes|OVSF-Codes]]   ⇒   "OVSF codes"
#  [[Applets:Prinzip_der_Quadratur-Amplitudenmodulation_(Applet)|Prinzip der Quadratur–Amplitudenmodulation]]   ⇒   "Principle of Quadrature Amplitude Modulation"
#  [[Applets:QPSK|QPSK und Offset–QPSK]]   ⇒   "QPSK and offset QPSK"
#  [[Applets:Sprachcodecs|Qualität verschiedener Sprachcodecs]]   ⇒   "Quality of different speech codecs"
}}
{{Collapse|ID=chancod|TITEL='''Applets about  "Channel Coding"|TEXT=
#  [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson Distribution]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [https://www.lntwww.de/Applets:Korrelation_und_Regressionsgerade Correlation and Regression Line]   ⇒   Note:  '''German language''' !
#  [[Applets:Digital_Filters|Digital Filters]]
#  [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fouriertransform and Inverse]]
#  [https://www.lntwww.de/Applets:Das_Gram-Schmidt-Verfahren Gram–Schmidt method]   ⇒   Note:  '''German language''' !
#  [[Applets:Graphical_Convolution|Graphical Convolution]]
#  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF, and Moments of Special Distributions]]

==== SWF applets - based on Shockwave Flash ====
#  [[Applets:Gram-Schmidt-Verfahren|Gram-Schmidt-Verfahren]]   ⇒   "Gram-Schmidt method"
#  [[Applets:Korrelationskoeffizient_%26_Regressionsgerade|Korrelationskoeffizient & Regressionsgerade]]   ⇒   "Correlation coefficient and regression line"
#  [[Applets:MPSK_%26_Union-Bound(Applet)|Mehrstufige PSK & Union-Bound]]   ⇒   "Multi-stage PSK and Union Bound"
#  [[Applets:Fehlerwahrscheinlichkeit|Symbolfehlerwahrscheinlichkeit von Digitalsystemen]]   ⇒   "Symbol error probability of digital communications systems"
#  [[Applets:Viterbi|Viterbi–Entscheider für einen Vorläufer]]   ⇒   "Viterbi decider for one precursor"
}}
{{Collapse|ID=nachrbeisp|TITEL='''Applets about  "Examples of Communication Systems"|TEXT=
#  [[Applets:Attenuation_of_Copper_Cables|Attenuation of Copper Cables]]
#  [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fouriertransform and Inverse]]
#  [[Applets:The_Doppler_Effect|Doppler Effect]]
#  [[Applets:Generation_of_Walsh_functions|Generation of Walsh functions]]  
#  [[Applets:Principle_of_Pseudo-Ternary_Coding|Principle of Pseudo-Ternary Coding]]  
#  [[Applets:Principle_of_4B3T_Coding|Principle of 4B3T Coding]]  
#  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]

==== SWF applets - based on Shockwave Flash ====
#  [[Applets:DMT|DMT – Discrete Multiton Transmission]]
#  [[Applets:Frequency_Shift_Keying_%26_Continuous_Phase_Modulation|Frequency Shift Keying & Continuous Phase Modulation]]
#  [[Applets:Mehrwegeausbreitung_und_Frequenzselektivit%C3%A4t_(Applet)| Mehrwegeausbreitung und Frequenzselektivität]]   ⇒   "Multipath propagation and frequency selectivity"
#  [[Applets:OFDM|OFDM - Spektrum & Signale]]   ⇒   "OFDM - Spectra and signals"
#  [[Applets:OVSF_codes_(Applet)|OVSF-Codes]]   ⇒   "OVSF codes"
#  [[Applets:DMT-Prinzip|Prinzip der Discrete Multitone Transmission]]   ⇒   "Principle of Discrete Multitone Transmission"
#  [[Applets:Prinzip_der_Quadratur-Amplitudenmodulation_(Applet)|Prinzip der Quadratur–Amplitudenmodulation]]   ⇒   "Principle of Quadrature Amplitude Modulation"
#  [[Applets:Qualität_verschiedener_Sprach–Codecs_(Applet)|Qualität verschiedener Sprach–Codecs]]   ⇒   "Quality of different speech codecs"
#  [[Applets:QPSK_und_Offset-QPSK_(Applet)|QPSK und Offset-QPSK]]   ⇒   "QPSK and offset QPSK"
#  [[Applets:Zeitverhalten_von_Kupferkabeln|Zeitverhalten von Kupferkabeln]]   ⇒   "Time behavior of copper cables"
}}
{{Collapse|ID=allhtml|TITEL=⇒   '''Alphabetical list of all HTML5/JS applets (English language)'''|TEXT=

{{BlaueBox|TEXT=
$\text{Notes and tips for the HTML5/JS applets:}$ 
*After selecting the desired applet, a wiki description page appears with a brief description of its contents and user interface. 

*At the top and bottom of this description page, there are links to the actual HTML5 applet.

*The HTML5/JS applets can be rendered by many browsers like Firefox, Chrome and Safari, as well as smartphones and tablets.

*The exercises and the corresponding sample solutions are also integrated into the applet. }}

#  [[Applets:Attenuation_of_Copper_Cables|Attenuation of Copper Cables]]
#  [[Applets:Bessel_Functions_of_the_First_Kind|Bessel Functions of the First Kind]]
#  [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson Distribution]]
#  [[Applets:Capacity_of_Memoryless_Digital_Channels|Capacity of Memoryless Digital Channels]]
#  [[Applets:Coherent_and_Non-Coherent_On-Off_Keying|Coherent and Non-Coherent On-Off Keying]]
#  [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]]
#  [[Applets:Digital_Filters|Digital Filters]]
#  [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fouriertransform and Inverse]]
#  [[Applets:The_Doppler_Effect|Doppler Effect]]
#  [[Applets:Eye_Pattern_and_Worst-Case_Error_Probability|Eye Pattern and Worst-Case Error Probability]]
#  [[Applets:Frequency_&_Impulse_Responses|Frequency & Impulse Responses]]  
#  [[Applets:Generation_of_Walsh_functions|Generation of Walsh functions]]  
#  [[Applets:Graphical_Convolution|Graphical Convolution]]
#  [[Applets:Linear_Distortions_of_Periodic_Signals|Linear Distortions of Periodic Signals]]
#  [[Applets:Matched_Filter_Properties|Matched Filter Properties]]  
#  [[Applets:Period_Duration_of_Periodic_Signals|Period Duration of Periodic Signals]]
#  [[Applets:Physical_Signal_&_Analytic_Signal|Physical Signal & Analytic Signal]]
#  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal|Physical Signal & Equivalent Low-pass Signal]]
#  [[Applets:PDF,_CDF_and_Moments_of_Special_Distributions|PDF, CDF, and Moments of Special Distributions]]
#  [[Applets:Principle_of_Pseudo-Ternary_Coding|Principle of Pseudo-Ternary Coding]]  
#  [[Applets:Principle_of_4B3T_Coding|Principle of 4B3T Coding]]  
#  [[Applets:Pulses_and_Spectra|Pulses and Spectra]]
#  [[Applets:Sampling_of_Analog_Signals_and_Signal_Reconstruction|Sampling of Analog Signals and Signal Reconstruction]]
#  [[Applets:Two-dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]]
}}
{{Collapse|ID=allhtmlGerman|TITEL=⇒   '''Alphabetical list of all HTML5/JS applets (German language)'''|TEXT=

{{BlaueBox|TEXT=
$\text{Notes and tips for the German HTML5/JS applets:}$ 
*After selecting the desired applet, a  '''German'''  wiki description page appears with a brief description of its contents and user interface. 

*At the top and bottom of this description page, there are links to the actual  '''German''' HTML5 applet.

*The HTML5/JS applets can be rendered by many browsers like Firefox, Chrome and Safari, as well as smartphones and tablets.

*The exercises and the corresponding sample solutions are also integrated into the applet. }}


#  [https://www.lntwww.de/Applets:Das_Gram-Schmidt-Verfahren Das Gram-Schmidt-Verfahren]   ⇒   "Gram–Schmidt method"
#  [https://www.lntwww.de/Applets:Entropie_und_N%C3%A4herungen_bin%C3%A4rer_Nachrichtenquellen Entropie und Näherungen binärer Nachrichtenquellen]   ⇒   "Entropies and approximations of binary sources"
#  [https://www.lntwww.de/Applets:Kausale_Systeme_und_Laplacetransformation Kausale Systeme und Laplacetransformation]   ⇒   "Causal Systems and Laplace Transform"
#  [https://www.lntwww.de/Applets:Korrelation_und_Regressionsgerade Korrelation und Regressionsgerade]   ⇒   "Correlation and Regression Line"
#  [https://www.lntwww.de/Applets:Zweidimensionale_Laplace-Zufallsgr%C3%B6%C3%9Fen_(Applet) Zweidimensionale Laplace-Zufallsgrößen]   ⇒   "Two-dimensional Laplace Random Variables"
}}
{{Collapse|ID=allSWL|TITEL=⇒   '''Alphabetical list of all SWF applets'''|TEXT=

{{BlaueBox|TEXT=
$\text{Notes and tips about SWF applets:}$ 
* Our previous SWF applications were programmed for Adobe Flash. 
*Since the Flashplayer browser plugin is no longer supported for security reasons, these applets must be opened with the "projector version". 
*You do not need to install this program and it is not integrated into your browser. There are no security gaps in this respect (as long as you trust the LNTwww).
* On the WIKI pages for the above SWF applets you will find the projector version of the flash player and of course the applet itself. 
* Unfortunately, these applets '''do not work on smartphones and tablets'''.
* All SWF applets '''use the German language'''.
}}

#  [[Applets:DMT|DMT – Discrete Multiton Transmission]]
#  [[Applets:Synchrondemodulator|Eigenschaften des Synchrondemodulators]]   ⇒   "Features of the synchronous demodulator"
#  [[Applets:Bandbegrenzung|Einfluss einer Bandbegrenzung auf Sprache und Musik]]   ⇒   "Influence of a band limitation on speech and music"
#  [[Applets:Entropien_von_Nachrichtenquellen|Entropien von Nachrichtenquellen]]   ⇒   "Entropy of binary and ternary sources"
#  [[Applets:Entscheidungsrückkopplung|Applets:Entscheidungsrückkopplung]]   ⇒   "Decision feedback equalization"
#  [[Applets:Markovketten|Ereigniswahrscheinlichkeiten einer Markovkette]]   ⇒   "Event probabilities of a Markov chain"
#  [[Applets:Frequency_Shift_Keying_%26_Continuous_Phase_Modulation|Frequency Shift Keying & Continuous Phase Modulation]]
#  [[Applets:Gram-Schmidt-Verfahren|Gram-Schmidt-Verfahren]]   ⇒   "Gram-Schmidt method"
#  [[Applets:Huffman_Shannon_Fano|Huffman-Shannon-Fano]]   ⇒   "Coding according to Huffman and Shannon/Fano"
#  [[Applets:Kausale_Systeme_-_Laplacetransformation|Kausale Systeme - Laplacetransformation]]   ⇒   "Causal systems - Laplace transform"
#  [[Applets:Korrelationskoeffizient_%26_Regressionsgerade|Korrelationskoeffizient & Regressionsgerade]]   ⇒   "Correlation coefficient and regression line"
#  [[Applets:Lempel-Ziv-Welch|Lempel-Ziv-Welch-Algorithmen]]   ⇒   "Lempel-Ziv-Welch algorithms"
#  [[Applets:Lineare_Nyquistentzerrung|Lineare Nyquistentzerrung]]   ⇒   "Linear Nyquist equalization"
#  [[Applets:MPSK_%26_Union-Bound(Applet)|Mehrstufige PSK & Union-Bound]]   ⇒   "Multi-stage PSK and Union Bound"
#  [[Applets:Mehrwegeausbreitung_und_Frequenzselektivit%C3%A4t_(Applet)| Mehrwegeausbreitung und Frequenzselektivität]]   ⇒   "Multipath propagation and frequency selectivity"
#  [[Applets:OFDM|OFDM - Spektrum & Signale]]   ⇒   "OFDM - Spectra and signals"
#  [[Applets:OVSF_codes_(Applet)|OVSF-Codes]]   ⇒   "OVSF codes"
#  [[Applets:Regionen|Optimale Entscheidungsregionen]]   ⇒   "Optimal decision regions"
#  [[Applets:Laufzeit|Phasenlaufzeit & Gruppenlaufzeit]]   ⇒   "Phase delay & group delay"
#  [[Applets:DMT-Prinzip|Prinzip der Discrete Multitone Transmission]]   ⇒   "Principle of Discrete Multitone Transmission"
#  [[Applets:Prinzip_der_Quadratur-Amplitudenmodulation_(Applet)|Prinzip der Quadratur–Amplitudenmodulation]]   ⇒   "Principle of Quadrature Amplitude Modulation"
#  [[Applets:Qualität_verschiedener_Sprach–Codecs_(Applet)|Qualität verschiedener Sprach–Codecs]]   ⇒   "Quality of different speech codecs"
#  [[Applets:QPSK_und_Offset-QPSK_(Applet)|QPSK und Offset-QPSK]]   ⇒   "QPSK and offset QPSK"
#  [[Applets:Fehlerwahrscheinlichkeit|Symbolfehlerwahrscheinlichkeit von Digitalsystemen]]   ⇒   "Symbol error probability of digital communications systems"
#  [[Applets:Viterbi|Viterbi–Entscheider für einen Vorläufer]]   ⇒   "Viterbi decider for one precursor"
#  [[Applets:Zeitverhalten_von_Kupferkabeln|Zeitverhalten von Kupferkabeln]]   ⇒   "Time behavior of copper cables"
#  [[Applets:2D_Laplace_(SWF)|Zweidimensionale Laplaceverteilung]]   ⇒   "Two-dimensional Laplace distribution"
}}

__NOTOC__

Applets:Physical Signal & Analytic Signal

2023-03-26T22:25:25Z

Noah:

{{LntAppletLink|physAnSignal_en}}         [https://www.lntwww.de/Applets:Physikalisches_Signal_%26_Analytisches_Signal '''English Applet with German WIKI description''']
==Applet Description==
 
This applet shows the relationship between the physical band-pass signal $x(t)$ and the associated analytic signal $x_+(t)$. It is assumed that the band-pass signal $x(t)$ has a frequency-discrete spectrum $X(f)$:
:$$x(t) = x_{\rm U}(t) + x_{\rm T}(t) + x_{\rm O}(t) = A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right)+A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right). $$
The physical signal $x(t)$ is thus composed of three harmonic oscillations, a constellation that can be found, for example, in the ''Double-sideband Amplitude Modulation''
*of the message signal $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$   ⇒   in German:   '''N'''achrichtensignal
*with the carrier signal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$   ⇒   in German:   '''T'''rägersignal.

The nomenclature is also adapted to this case:
* $x_{\rm O}(t)$ denotes the "upper sideband"   (in German:   '''O'''beres Seitenband) with the amplitude $A_{\rm O}= A_{\rm N}/2$, the frequency $f_{\rm O} = f_{\rm T} + f_{\rm N}$ and the phase $\varphi_{\rm O} = \varphi_{\rm T} + \varphi_{\rm N}$.
*Similarly, for the "lower sideband"   (in German:   '''U'''nteres Seitenband) $x_{\rm U}(t)$ with $f_{\rm U} = f_{\rm T} - f_{\rm N}$, $A_{\rm U}= A_{\rm O}$ and $\varphi_{\rm U} = -\varphi_{\rm O}$.

The associated analytic signal is:

:$$x_+(t) = x_{\rm U+}(t) + x_{\rm T+}(t) + x_{\rm O+}(t) = A_{\rm U}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm U})}
\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm T}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm T})}
\hspace{0.1cm}+\hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm O})}. $$

[[File:Zeigerdiagramm_2a_version2.png|right|frame|Analytic signal at the time $t=0$]]
The program displays $x_+(t)$ as the vectorial sum of three rotating pointers (all with counterclockwise) as a violet dot (see figure for start time $t=0$):

*The (red) pointer of the carrier $x_{\rm T+}(t)$ with length $A_{\rm T}$ and zero phase position $\varphi_{\rm T} = 0$ rotates at constant angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}$ (one revolution in time $1/f_{\rm T})$.

*The (blue) pointer of the upper sideband $x_{\rm O+}(t)$ with length $A_{\rm O}$ and zero phase position $\varphi_{\rm O}$ rotates at the angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}$, which is slightly faster than $x_{\rm T+}(t)$.

*The (green) pointer of the lower sideband $x_{\rm U+}(t)$ with length $A_{\rm U}$ and zero phase position $\varphi_{\rm U}$ rotates at the angular velocity $2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}$, which is slightly slower than $x_{\rm T+}(t)$.

The time trace of $x_+(t)$ is also referred to below as ''Pointer Diagram''. The relationship between the physical band-pass signal $x(t)$ and the associated analytic signal $x_+(t)$ is:

:$$x(t) = {\rm Re}\big [x_+(t)\big ].$$

''Note:''   In the figure $\varphi_{\rm O} = +30^\circ$. This leads to the angle with respect to the coordinate system at $t=0$:   $\phi_{\rm O}=-\varphi_{\rm O}=-30^\circ$. Similarly, the zero phase angle $\varphi_{\rm U}=-30^\circ$ of the lower sideband leads to the phase angle to be considered in the complex plane:   $\phi_{\rm U}=+30^\circ$.

==Theoretical Background==
 
===Description of Band-pass Signals===
[[File:Zeigerdiagramm_1a.png|right|frame|Band-pass spectrum $X(f)$ |class=fit]]
We consider '''band-pass signals''' $x(t)$ with the property that their spectra $X(f)$ are not in the range around the frequency $f=0$, but around a carrier frequency $f_{\rm T}$. In most cases it can also be assumed that the bandwidth is $B \ll f_{\rm T}$.

The figure shows such a band-pass spectrum $X(f)$. Assuming that the associated $x(t)$ is a physical signal and thus real, the spectral function $X(f)$ has a symmetry with respect to the frequency $f = 0$, if $x(t)$ is an even function   ⇒   $x(-t)=x(t)$, $X(f)$ is real and even.

Besides the physical signal $x(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X(f)$, one can also use the following descriptions of band-pass signals:
*the analytic signal $x_+(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_+(f)$, see next section,
*the equivalent low-pass signal   (in German:   äquivalentes '''T'''ief '''P'''ass–Signal) $x_{\rm TP}(t)\ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ X_{\rm TP}(f)$, see Applet [[Applets:Physical_Signal_%26_Equivalent_Low-pass_Signal|"Physical Signal & Equivalent Low-pass signal"]].
 

===Analytic Signal – Frequency Domain===

The '''analytic signal''' $x_+(t)$ belonging to the physical signal $x(t)$ is the time function whose spectrum fulfills the following property:
[[File:Zeigerdiagramm_3a.png|right|frame|Construction of the spectral function $X_+(f)$ |class=fit]]
:$$X_+(f)=\big[1+{\rm sign}(f)\big] \cdot X(f) = \left\{ {2 \cdot
X(f) \; \hspace{0.2cm}\rm for\hspace{0.2cm} {\it f} > 0, \atop {\,\,\,\, \rm 0 \; \hspace{0.9cm}\rm for\hspace{0.2cm} {\it f} < 0.} }\right.$$

The ''signum function'' is for positive values of $f$ equal to $+1$ and for negative $f$ values equal to $-1$.
* The (double-sided) limit returns $\sign(0)=0$.
* The index „+” should make it clear that $X_+(f)$ only has parts at positive frequencies.

From the graph you can see the calculation rule for $X_+(f)$:

The actual band-pass spectrum $X(f)$ becomes
* doubled at the positive frequencies, and
* set to zero at the negative frequencies.

Due to the asymmetry of $X_+(f)$ with respect to the frequency $f=0$, it can already be said that the time function $x_+(t)$ except for a trivial special case $x_+(t)=0 \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,X_+(f)=0$ is always complex.
 

===Analytic Signal – Time Domain===
At this point, it is necessary to briefly discuss another spectral transformation.

{{BlaueBox|TEXT=
$\text{Definition:}$ 
For the '''Hilbert transform''' $ {\rm H}\left\{x(t)\right\}$ of a time function $x(t)$ we have:

:$$y(t) = {\rm H}\left\{x(t)\right\} = \frac{1}{ {\rm \pi} } \cdot
\hspace{0.03cm}\int_{-\infty}^{+\infty}\frac{x(\tau)}{ {t -
\tau} }\hspace{0.15cm} {\rm d}\tau.$$

This particular integral is not solvable in a simple, conventional way, but must be evaluated using the [https://en.wikipedia.org/wiki/Cauchy_principal_value "Cauchy principal value theorem"].

Accordingly, in the frequency domain:
:$$Y(f) = {\rm -j \cdot sign}(f) \cdot X(f) \hspace{0.05cm} .$$}}

The above result can be summarized with this definition as follows:
* The analytic signal $x_+(t)$ is obtained from the physical band-pass signal $x(t)$ by adding an imaginary part to $x(t)$ according to the Hilbert transform:

:$$x_+(t) = x(t)+{\rm j} \cdot {\rm H}\left\{x(t)\right\} .$$

*$\text{H}\{x(t)\}$ disappears only for the case $x(t) = \rm const.$   ⇒   the same signal. For all other signal forms, the analytic signal $x_+(t)$ is complex.

* From the analytic signal $x_+(t)$, the physical band-pass signal can be easily determined by the following operation:
:$$x(t) = {\rm Re}\big[x_+(t)\big] .$$

{{GraueBox|TEXT=
$\text{Example 1:}$  The principle of the Hilbert transformation should be further clarified by the following graphic:
*After the left representation $\rm(A)$ one gets from the physical signal $x(t)$ to the analytic signal $x_+(t)$, by adding an imaginary part ${\rm j} \cdot y(t)$.
*Here $y(t) = {\rm H}\left\{x(t)\right\}$ is a real time function that can be indicated in the spectral domain by multiplying the spectrum $X(f)$ with ${\rm - j} \cdot \sign(f)$.

[[File:P_ID2729__Sig_T_4_2_S2b_neu.png|center|frame|To clarify the Hilbert transform]]

The right representation $\rm(B)$ is equivalent to $\rm(A)$. Now $x_+(t) = x(t) + z(t)$ stand with the purely imaginary function $z(t)$. A comparison of the two figures shows that in fact $z(t) = {\rm j} \cdot y(t)$.}}
 

===Representation of the Harmonic Oscillation as an Analytic Signal===

The spectral function $X(f)$ of a harmonic oscillation $x(t) = A\cdot\text{cos}(2\pi f_{\rm T}\cdot t - \varphi)$ is known to consist of two Dirac delta functions at the frequencies
* $+f_{\rm T}$ with the complex weight $A/2 \cdot \text{e}^{-\text{j}\hspace{0.05cm}\varphi}$,
* $-f_{\rm T}$ with the complex weight $A/2 \cdot \text{e}^{+\text{j}\hspace{0.05cm}\varphi}$.

Thus, the spectrum of the analytic signal (that is, without the Dirac delta function at the frequency $f =-f_{\rm T}$, but doubling at $f =+f_{\rm T}$):

:$$X_+(f) = A \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\varphi}\cdot\delta (f - f_{\rm
T}) .$$

The associated time function is obtained by applying the Displacement Law:

:$$x_+(t) = x_{\rm U+}(t) + x_{\rm T+}(t) + x_{\rm O+}(t) = A_{\rm U}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm U})}
\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm T}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm T})}
\hspace{0.1cm}+\hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm O})}. $$

This equation describes a pointer rotating at constant angular velocity $\omega_{\rm T} = 2\pi f_{\rm T}$.

{{GraueBox|TEXT=
$\text{Example 2:}$  Here the coordinate system is rotated by $90^\circ$ (real part up, imaginary part to the left) contrary to the usual representation.

[[File:P_ID712__Sig_T_4_2_S3.png|center|frame|Pointer diagram of a harmonic oscillation]]

Based on this graphic, the following statements are possible:
* At time $t = 0$, the pointer of length $A$ (signal amplitude) lies with the angle $-\varphi$ in the complex plane. In the example shown, $\varphi=45^\circ$.
* For times $t>0$, the constant angular velocity vector $\omega_{\rm T}$ rotates in a mathematically positive direction, that is, counterclockwise.
* The tip of the pointer is thus always on a circle with radius $A$ and needs exactly the time $T_0$, i.e. the period of the harmonic oscillation $x(t)$ for one revolution.
* The projection of the analytic signal $x_+(t)$ on the real axis, marked by red dots, gives the instantaneous values of $x(t)$.}}
 

===Analytic Signal Representation of a Sum of Three Harmonic Oscillations===

In our applet, we always assume a set of three rotating pointers. The physical signal is:
:$$x(t) = x_{\rm U}(t) + x_{\rm T}(t) + x_{\rm O}(t) = A_{\rm U}\cdot \cos\left(2\pi f_{\rm U}\cdot t- \varphi_{\rm U}\right)+A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t- \varphi_{\rm T}\right)+A_{\rm O}\cdot \cos\left(2\pi f_{\rm O}\cdot t- \varphi_{\rm O}\right). $$
* Each of the three harmonic oscillations $x_{\rm T}(t)$, $x_{\rm U}(t)$ and $x_{\rm O}(t)$ is represented by an amplitude $(A)$, a frequency $(f)$ and a phase value $(\varphi)$.
*The indices are based on the ''Double-sideband Amplitude Modulation'' method. "T" stands for "carrier", "U" for "lower sideband" and "O" for "upper Sideband".
*Accordingly, $f_{\rm U} < f_{\rm T}$ and $f_{\rm O} > f_{\rm T}$. There are no restrictions for the amplitudes and phases.

The associated analytic signal is:
:$$x_+(t) = x_{\rm U+}(t) + x_{\rm T+}(t) + x_{\rm O+}(t) = A_{\rm U}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm U}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm U})}
\hspace{0.1cm}+ \hspace{0.1cm}A_{\rm T}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm T}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm T})}
\hspace{0.1cm}+\hspace{0.1cm} A_{\rm O}\cdot {\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm O}\hspace{0.05cm}\cdot \hspace{0.05cm}t- \varphi_{\rm O})}. $$

{{GraueBox|TEXT=
$\text{Example 3:}$ 
Shown the constellation arises i.e. in the [https://en.wikipedia.org/wiki/Sideband "Double-sideband Amplitude Modulation"] (with carrier) of the message signal $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t- \varphi_{\rm N}\right)$ with the carrier signal $x_{\rm T}(t) = A_{\rm T}\cdot \cos\left(2\pi f_{\rm T}\cdot t - \varphi_{\rm T}\right)$. This is discussed frequently in the Exercises.

There are some limitations to the program parameters in this approach:
* For the frequencies, it always applies $f_{\rm O} = f_{\rm T} + f_{\rm N}$ and $f_{\rm U} = f_{\rm T} - f_{\rm N}$.

*Without distortions the amplitude of the sidebands are $A_{\rm O}= A_{\rm U}= A_{\rm N}/2$.
*The respective phase relationships can be seen in the following graphic.

[[File:Zeigerdiagramm_5.png|center|frame|Spectrum $X_+(f)$ of the analytic signal for different phase constellations |class=fit]]}}

==Exercises==
[[File:Zeigerdiagramm_aufgabe_2.png|right]]
*First select the task number.
*A task description is displayed.
*Parameter values are adjusted.
*Solution after pressing "Hide solition".

The number "0" will reset the program and output a text with further explanation of the applet.
 
In the following, $\rm Green$ denotes the lower sideband   ⇒   $\big (A_{\rm U}, f_{\rm U}, \varphi_{\rm U}\big )$,  
$\rm Red$ the carrier   ⇒   $\big (A_{\rm T}, f_{\rm T}, \varphi_{\rm T}\big )$ and
$\rm Blue$ the upper sideband   ⇒   $\big (A_{\rm O}, f_{\rm O}, \varphi_{\rm O}\big )$.
{{BlaueBox|TEXT=
'''(1)'''   Consider and interpret the analytic signal $x_+(t)$ for $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1.5\ \text{V}, \ f_{\rm T} = 50 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$, $A_{\rm U} = A_{\rm O} = 0$.

:Which signal values $x_+(t)$ result for $t = 0$, $t = 5 \ \rm µ s$ and $t = 20 \ \rm µ s$? What are the corresponding signal values for $x(t)$? }}

:: For a cosine signal, let $x_+(t= 0) = A_{\rm T} = 1.5\ \text{V}$. Then $x_+(t)$ rotates in a mathematically positive direction (one revolution per period $T_0 = 1/f_{\rm T}$):

:: $x_+(t= 20 \ {\rm µ s}) = x_+(t= 0) = 1.5\ \text{V}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}x(t= 20 \ {\rm µ s}) = 1.5\ \text{V,}$
:: $x_+(t= 5 \ {\rm µ s}) = {\rm j} \cdot 1.5\ \text{V}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}x(t= 5 \ {\rm µ s}) = {\rm Re}[x_+(t= 5 \ {\rm µ s})] = 0$.

{{BlaueBox|TEXT=
'''(2)'''   How do the ratios change for $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1.0\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 90^\circ$?}}

::The signal $x(t)$ is now a sine signal with a smaller amplitude. The analytic signal now starts because of $\varphi_{\rm T} = 90^\circ$   ⇒   $\phi_{\rm T} = -90^\circ$ at $x_+(t= 0) = -{\rm j} \cdot A_{\rm T}$. After that, $x_+(t)$ rotates again in a mathematically positive direction, but twice as fast because of $T_0 = 10 \ \rm µ s$ as in $\rm (1)$.

{{BlaueBox|TEXT=
'''(3)'''   Now   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0.4\ \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = 0^\circ$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.4\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 0^\circ$.

:Consider and interpret the physical signal $x(t)$ and the analytic signal $x_+(t)$.}}

::The Signal $x(t)$ results in the [https://en.wikipedia.org/wiki/Sideband Double-sideband Amplitude Modulation] (DSB–AM) of the message signal $A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t\right)$ with $A_{\rm N} = 0.8\ \text{V}$, $f_{\rm N} = 20\ \text{kHz}$. The carrier $x_{\rm T}(t)$ with $f_{\rm T} = 100\ \text{kHz}$ is also cosinusoidal. The degree of modulation is $m = A_{\rm N}/A_{\rm T} = 0.8$ and the period $T_{\rm 0} = 50\ \text{µs}$.

::In the phase diagram, the (red) carrier rotates faster than the (green) lower sideband and slower than the (blue) upper sideband. The analytic signal $x_+(t)$ results as the geometric sum of the three rotating pointers. It seems that the blue pointer is leading the carrier and the green pointer is following the carrier.

{{BlaueBox|TEXT=
'''(4)'''   The settings of task '''(3)''' still apply. Which signal values are obtained at $t=0$, $t=2.5 \ \rm µ s$, $t= 5 \ \rm µ s$ and $t=10 \ \rm µ s$? }}

::At time $t=0$, all pointers are in the direction of the real axis, so that $x(t=0) = {\rm Re}\big [x+(t= 0)\big] = A_{\rm U} + A_{\rm T} + A_{\rm O} = 1.8\ \text{V}$.

::Until the time $t=2.5 \ \rm µ s$, the red carrier has rotated by $90^\circ$, the blue one by $108^\circ$ and the green one by $72^\circ$. We have $x(t=2.5 \ \rm µ s) = {\rm Re}\big [x_+(t= 2.5 \ \rm µ s)\big] = 0$, because now the pointer group points in the direction of the imaginary axis. The other sought signal values are $x(t=5 \ \rm µ s) = {\rm Re}\big [x_+(t= 5 \ \rm µ s)\big] = -1.647\ \text{V}$ and $x(t=10 \ \rm µ s) = {\rm Re}\big [x_+(t= 10 \ \rm µ s)\big] = 1.247\ \text{V}$.
::For $x_+(t)$ a spiral shape results, alternating with a smaller radius and then with a larger radius.

{{BlaueBox|TEXT=
'''(5)'''   How should the phase parameters $\varphi_{\rm T}$, $\varphi_{\rm U}$ and $\varphi_{\rm O}$ be set if both the carrier $x_{\rm T}(t)$ and the message signal $x_{\rm N}(t)$ are sinusoidal?}}

::The parameter selection $\varphi_{\rm T} = \varphi_{\rm U} = \varphi_{\rm O}=90^\circ$ describes the signals $x_{\rm T}(t) = A_{\rm T}\cdot \sin\left(2\pi f_{\rm T}\cdot t\right)$ and $x_{\rm N}(t) = A_{\rm N}\cdot \cos\left(2\pi f_{\rm N}\cdot t\right)$. If, in addition, the message $x_{\rm N}(t)$ is sinusoidal, then $\varphi_{\rm O}=\varphi_{\rm T} - 90^\circ = 0$ and $\varphi_{\rm U}=\varphi_{\rm T} + 90^\circ = 180^\circ$ must be set.

{{BlaueBox|TEXT=
'''(6)'''   The settings of task '''(3)''' apply except $A_{\rm T} = 0.6\ \text{V}$. Which modulation method is described here?

: What are the consequences of this? What changes with $A_{\rm T} = 0$? }}

::It is a [https://en.wikipedia.org/wiki/Sideband Double-sideband Amplitude Modulation] (DSB–AM with carrier) with the modulation degree $m=0.8/0.6 = 1.333$. For $m > 1$, however, [https://www.radio-electronics.com/info/rf-technology-design/am-reception/synchronous-demodulator-demodulation-detector.php Synchronous Demodulation] is required. [https://en.wikipedia.org/wiki/Envelope_detector Envelope Detection] no longer works. One reason for this is that now the zero crossings of $x(t)$ are no longer equidistant from $5\ \rm µ s$   ⇒   additional phase modulation.

::With $A_{\rm T} = 0$   ⇒   $m \to \infty$ results in a [https://en.wikipedia.org/wiki/Double-sideband_suppressed-carrier_transmission ''DSB–AM without carrier'']. For this, one also needs coherent demodulation.

{{BlaueBox|TEXT=
'''(7)'''     Now let   $\text{Red:} \hspace{0.15cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.15cm} A_{\rm U} = 0$,   $\text{Blue:} \hspace{0.15cm} A_{\rm O} = 0.8\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = 90^\circ$.

:Which constellation is described here? Which figure is given for the equivalent low-pass signal $x_{\rm TP}(t)$?   ⇒   "locus"? What changes with $A_{\rm U} = 0.8\ \text{V}$ and $A_{\rm O} = 0$?}}

::In both cases, it is a [https://en.wikipedia.org/wiki/Single-sideband_modulation Single-sideband Amplitude Modulation] (SSB–AM) with the modulation degree $\mu = 0.8$ (in SSB we denote the degree of modulation with $\mu$ instead of $m$). The carrier signal is cosinusoidal and the message signal is sinusoidal. The equivalent low-pass signal $x_{\rm TP}(t)$ has a circular course in the complex plane.

:: $A_{\rm O} = 0.8\ \text{V}$, $A_{\rm U} = 0$ is an OSB modulation. The green pointer is missing and the blue pointer rotates faster compared to the red carrier.

:: $A_{\rm U} = 0.8\ \text{V}$, $A_{\rm O} = 0$ is a USB modulation. The blue pointer is missing and the green pointer rotates slower compared to the red carrier.

{{BlaueBox|TEXT=
'''(8)'''   Now  $\text{Red:} \hspace{0.05cm} A_{\rm T} = 1\ \text{V}, \ f_{\rm T} = 100 \ \text{kHz}, \ \varphi_{\rm T} = 0^\circ$,   $\text{Green:} \hspace{0.05cm} A_{\rm U} = 0.4\ \text{V}, \ f_{\rm U} = 80 \ \text{kHz}, \ \varphi_{\rm U} = -90^\circ$,   $\text{Blue:} \hspace{0.05cm} A_{\rm O} = 0.2\ \text{V}, \ f_{\rm O} = 120 \ \text{kHz}, \ \varphi_{\rm O} = +90^\circ$.

:Which constellation could be described here? Which shape results for the equivalent low-pass signal $x_{\rm TP}(t)$?}}

::It could be a DSB–AM of a sinusoidal signal with cosinusoidal carrier and modulation degree $m=0.8$, in which the upper sideband is attenuated by a factor of 2. The equivalent low-pass signal $x_{\rm TP}(t)$ has an elliptical trace in the complex plane.

==Applet Manual==
 
[[File:Zeigerdiagramm_abzug.png|right|Screenshot]]

* The red parameters $(A_{\rm T}, \ f_{\rm T}, \ \varphi_{\rm T})$ and the red pointer mark the ''Carrier'' (German:   '''T'''räger).
* The green parameters $(A_{\rm U}, \ f_{\rm U} < f_{\rm T}, \ \varphi_{\rm U})$ mark the ''Lower sideband'' (German:  '''U'''nteres Seitenband).
* The blue parameters $(A_{\rm O}, \ f_{\rm O} > f_{\rm T}, \ \varphi_{\rm O})$ mark the ''Upper sideband'' (German:  '''O'''beres Seitenband).
*All pointers rotate in a mathematically positive direction (counterclockwise).

Meaning of the letters in the adjacent graphic:

    '''(A)'''     Plot of the analytic signal $x_{\rm +}(t)$

    '''(B)'''     Plot of the physical signal $x(t)$

    '''(C)'''     Parameter input via slider: amplitudes, frequencies, phase values

    '''(D)'''     Control elements:   Start – Step – Pause/Continue – Reset

    '''(E)'''     Speed of animation:   "Speed"   ⇒   Values: 1, 2, 3

    '''(F)'''     "Trace"   ⇒   On or Off, trace of complex signal values $x_{\rm +}(t)$

    '''(G)'''     Numerical output of the time $t$ and the signal values  ${\rm Re}[x_{\rm +}(t)] = x(t)$  and  ${\rm Im}[x_{\rm +}(t)]$

    '''(H)'''     Variations for the graphical representation

$\hspace{1.5cm}$Zoom–Functions "$+$" (Enlarge), "$-$" (Decrease) and $\rm o$ (Reset to default)

$\hspace{1.5cm}$Move with "$\leftarrow$" (Section to the left, ordinate to the right), "$\uparrow$" "$\downarrow$" and "$\rightarrow$"

    '''(I)'''     Experiment section:  Task selection and task

    '''(J)'''     Experiment section:  solution

In all applets top right:    Changeable graphical interface design   ⇒   '''Theme''':
* Dark:   black background  (recommended by the authors).
* Bright:   white background  (recommended for beamers and printouts)
* Deuteranopia:   for users with pronounced green–visual impairment
* Protanopia:   for users with pronounced red–visual impairment
 

==About the Authors==
This interactive calculation was designed and realized at the  [https://www.ei.tum.de/en/lnt/home//startseite Institute for Communications Engineering]  of the  [https://www.tum.de/ Technical University of Munich] .
*The original version was created in 2005 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her Diploma thesis using "FlashMX–Actionscript"  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]]  as part of her Bachelor's thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]).

==Once again:  Open Applet in new Tab==

{{LntAppletLink|physAnSignal_en}}         [https://www.lntwww.de/Applets:Physikalisches_Signal_%26_Analytisches_Signal '''English Applet with German WIKI description''']

Applets:Linear Distortions of Periodic Signals

2023-03-26T22:12:21Z

Noah:

{{LntAppletLink|physAnSignal_en}}         [https://en.lntwww.de/Applets:Physical_Signal_%26_Analytic_Signal '''English Applet with English WIKI description''']

==Applet Description==
 
This applet illustrates the effects of linear distortions (attenuation distortions and phase distortions) with
[[File:Modell_version2.png|right|frame|Meanings of the used signals]]
*the input signal $x(t)$   ⇒   power $P_x$:
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
*the output signal $y(t)$   ⇒   power $P_y$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2),$$
*the matched output signal $z(t)$   ⇒   power $P_z$:
:$$z(t) = k_{\rm M} \cdot y(t-\tau_{\rm M}) + \alpha_2 \cdot x_2(t-\tau_2),$$
*the difference signal   $\varepsilon(t) = z(t) - x(t)$   ⇒   power $P_\varepsilon$.

The next block in the model above is ''Matching'': The output signal $y(t)$ is adjusted in amplitude and phase with equal variables $k_{\rm M}$ and $\tau_{\rm M}$ for all frequencies which means that this is not a frequency-dependent equalization. Using the signal $z(t)$, one can differentiate between:
*attenuation distortion and frequency–independent attenuation, as well as
*phase distortion and frequency–independent delay.

The ''Distortion Power'' $P_{\rm D}$ is used to measure the strength of the linear distortion and is defined as:
:$$P_{\rm D} = \min_{k_{\rm M}, \ \tau_{\rm M}} P_\varepsilon.$$

==Theoretical Background==
 
Distortions refer to generally unwanted alterations of a message signal through a transmission system. Together with the strong stochastic effects (noise, crosstalk, etc.), they are a crucial limitation for the quality and rate of transmission.

Just as the intensity of noise can be assessed through
*the ''Noise Power'' $P_{\rm N}$ and
*the ''Signal–to–Noise Ratio'' (SNR) $\rho_{\rm N}$,

distortions can be quantified through

*the ''Distortion Power'' $P_{\rm D}$ and
*the ''Signal–to–Distortion Ratio'' (SDR)
:$$\rho_{\rm D}=\frac{\rm Signal \ Power}{\rm Distortion \ Power} = \frac{P_x}{P_{\rm D} }.$$

=== Linear and Nonlinear Distortions ===
 
A distinction is made between linear and nonlinear distortions:
*'''Nonlinear distortions''' occur, if at all times $t$ the nonlinear correlation $y = g(x) \ne {\rm const.} \cdot x$ exists between the signal values $x = x(t)$ at the input and $y = y(t)$ at the output, whereby $y = g(x)$ is defined as the system's nonlinear characteristic. By creating a cosine signal at the input with frequency $f_0$ the output signal includes $f_0$, as well as multiple harmonic waves. We conclude that new frequencies arise through nonlinear distortion.

[[File:LZI_T_2_2_S3_vers2.png|center|frame|For clarification of nonlinear distortions |class=fit]]

[[File:P_ID899__LZI_T_2_3_S1_neu.png|right |frame| Description of a linear system|class=fit]]
*'''Linear distortions''' occur, if the transmission channel is characterized by a frequency response $H(f) \ne \rm const.$ Various frequencies are attenuated and delayed differently. Characteristic of this is that although frequencies can disappear (for example, through a low–pass, a high–pass, or a band–pass), no new frequencies can arise.

In this applet only linear distortions are considered.

=== Description Forms for the Frequency Response ===
 
The generally complex valued frequency response can be represented as follows:
:$$H(f) = |H(f)| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot
\hspace{0.05cm} b(f)} = {\rm e}^{-a(f)}\cdot {\rm e}^{-{\rm j}
\hspace{0.05cm} \cdot \hspace{0.05cm} b(f)}.$$

This results in the following description variables:
*The absolute value $|H(f)|$ is called '''amplitude response''' and in logarithmic form '''attenuation function''':
:$$a(f) = - \ln |H(f)|\hspace{0.2cm}{\rm in \hspace{0.1cm}Neper
\hspace{0.1cm}(Np) } = - 20 \cdot \lg |H(f)|\hspace{0.2cm}{\rm in
\hspace{0.1cm}Decibel \hspace{0.1cm}(dB) }.$$
*The '''phase function''' $b(f)$ indicates the negative frequency–dependent angle of $H(f)$ in the complex plane based on the real axis:
:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) \hspace{0.2cm}{\rm in
\hspace{0.1cm}Radians \hspace{0.1cm}(rad)}.$$

=== Low–pass of Order N ===
 
[[File:Tiefpass_version2.png|right|frame|Attenuation function $a(f)$ and phase function $b(f)$ of a low–Pass of order $N$]]
The frequency response of a realizable low–pass (LP) of order $N$ is:
:$$H(f) = \left [\frac{1}{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
For example the RC low–pass is a first order low–pass. Consequently we can obtain
*the attenuation function:
:$$a(f) =N/2 \cdot \ln [1+( f/f_0)^2] \hspace{0.05cm},$$
*the phase function:
:$$b(f) =N \cdot \arctan( f/f_0) \hspace{0.05cm},$$
*the attenuation factor for the frequency $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_i/f_0)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*the phase delay for the frequency $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2 \pi f_i} = \frac{N \cdot \arctan( f_i/f_0)}{2 \pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

=== High–pass of Order N ===
 
[[File:Hochpass_version2.png|right|frame|Attenuation function $a(f)$ and phase function $b(f)$ of a high–pass of order $N$]]
The frequency response of a realizable high–pass (HP) of order $N$ is:
:$$H(f) = \left [\frac{ {\rm j}\cdot f/f_0 }{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
For example the LC high-pass is a first order high-pass. Consequently we can obtain
*the attenuation function:
:$$a(f) =N/2 \cdot \ln [1+( f_0/f)^2] \hspace{0.05cm},$$
*the phase function:
:$$b(f) =-N \cdot \arctan( f_0/f) \hspace{0.05cm},$$
*the attenuation factor for the frequency $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_0/f_i)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*the phase delay for the frequency $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2\pi f_i} = \frac{-N \cdot \arctan( f_0/f_i)}{2\pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

[[File:Verzerrungen_HP_TP_1_englisch.png|right|frame|Phase function $b(f)$ of high–pass and low–pass]]
{{GraueBox|TEXT=
$\text{Example:}$ 
This graphic shows the phase function $b(f)$ with the cutoff frequency $f_0 = 1\ \rm kHz$ and order $N=1$
* of a low–pass (green curve),
* of a high–pass (violet curve).

The input signal is sinusoidal with frequency $f_{\rm S} = 1.25\ {\rm kHz}$ whereby this signal is only turned on at $t=0$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.75cm}0 \\ \sin(2\pi \cdot f_{\rm S} \cdot t ) \\ \end{array} \right.\quad\begin{array}{l} (t < 0), \\ (t>0). \\ \end{array}$$

The left graphic shows the signal $x(t)$. The dashed line marks the first zero at $t = T_0 = 0.8\ {\rm ms}$. The other two graphics show the output signals $y_{\rm LP}(t)$ und $y_{\rm HP}(t)$ of low–pass and high–pass, whereby the change in amplitude was balanced in both cases.

[[File:Verzerrungen_HP_TP_2_version2.png|center|frame|Input signal $x(t)$ (enframed in blue) as well as output signals $y_{\rm LP}(t)$ ⇒   green and $y_{\rm HP}(t)$ ⇒   magenta]]

*The first zero of the signal $y_{\rm LP}(t)$ after the low–pass is delayed by $\tau_{\rm LP} = 0.9/(2\pi) \cdot T_0 \approx 0.115 \ {\rm ms}$ compared to the first zero of $x(t)$   ⇒   marked with green arrow, whereby $b_{\rm LP}(f/f_{\rm S} = 0.9 \ {\rm rad})$ was considered.
* In contrast, the phase delay of the high–pass is negative: $\tau_{\rm HP} = -0.67/(2\pi) \cdot T_0 \approx -0.085 \ {\rm ms}$ and therefore the first zero of $y_{\rm HP}(t)$ occurs before the dashed line.
*Following this transient response, in both cases the zero crossings again come in the raster of the period duration $T_0 = 0.8 \ {\rm ms}.$

''Remark:'' The shown signals were created using the interactive applet [[Applets:Kausale_Systeme_-_Laplacetransformation|"Causal systems – Laplace transform"]]. }}

=== Attenuation and Phase Distortions ===
 
[[File:P_ID900__LZI_T_2_3_S2_neu.png|frame| Requirements for a non–distorting channel|right|class=fit]]
The adjacent figure shows
*the even attenuation function $a(f)$   ⇒   $a(-f) = a(f)$, and
*the uneven function curve $b(f)$   ⇒   $b(-f) = -b(- f)$

of a non–distorting channel. One can see:
*In a distortion–free system the attenuation function $a(f)$ must be constant between$f_{\rm U}$ and $f_{\rm O}$ around the carrier frequency $f_{\rm T}$, where the input signal exists   ⇒   $X(f) \ne 0$.
*From the specified constant attenuation value $6 \ \rm dB$ follows for the amplitude response $|H(f)| = 0.5$   ⇒   the signal values of all frequencies are thus halved by the system   ⇒   '''no attenuation distortions'''.
*In addition, in such a system, the phase function $b(f)$ between $f_{\rm U}$ and $f_{\rm O}$ must increase linearly with the frequency. As a result, all frequency components are delayed by the same phase delay $τ$   ⇒   '''no phase distortion'''.
*The delay $τ$ is fixed by the slope of $b(f)$. The phase function $b(f) \equiv 0$ would result in a delay–less system   ⇒   $τ = 0$.

The following summary considers that – in this applet – the input signal is always the sum of two harmonic oscillations,
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
and therefore the channel influence is fully described by the attenuation factors $\alpha_1$ and $\alpha_2$ as well as the phase delays $\tau_1$ and $\tau_2$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2).$$

{{BlaueBox|TEXT=
$\text{Summary:}$ 
*A signal $y(t)$ is only '''distortion–free''' compared to $x(t)$ if $\alpha_1 = \alpha_2= \alpha$   and   $\tau_1 = \tau_2= \tau$   ⇒   $y(t) = \alpha \cdot x(t-\tau)$.
* '''Attenuation distortions''' occur when $\alpha_1 \ne \alpha_2$. If $\alpha_1 \ne \alpha_2$ and $\tau_1 = \tau_2$, then there are exclusively attenuation distortions.
* '''Phase distortions''' occur when $\tau_1 \ne \tau_2$. If $\tau_1 \ne \tau_2$ and $\alpha_1 = \alpha_2$, then there are exclusively phase distortions. }}

==Exercises==
[[File:Exercises_verzerrungen.png|right]]
*First choose an exercise number.
*An exercise description is displayed.
*Parameter values are adjusted to the respective exercises.
*Click "Hide solution" to display the solution.

Number "0" is a "Reset" button:
*Sets parameters to initial values (when loading the page).
*Displays a "Reset text" to describe the applet further.

{{BlaueBox|TEXT=
'''(1)'''   We consider the parameters $A_1 = 0.8\ {\rm V}, \ A_2 = 0.6\ {\rm V}, \ f_1 = 0.5\ {\rm kHz}, \ f_2 = 1.5\ {\rm kHz}, \ \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ$ for the input signal $x(t)$.
:Calculate the signal's period duration $T_0$ and power $P_x$. Can you read the value for $P_x$ off the applet? }}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}T_0 = \big [\hspace{-0.1cm}\text{ greatest common divisor }(0.5 \ {\rm kHz}, \ 1.5 \ {\rm kHz})\big ]^{-1}\hspace{0.15cm}\underline{ = 2.0 \ {\rm ms}};$

$\hspace{1.85cm} P_x = A_1^2/2 + A_2^2/2 \hspace{0.15cm}\underline{= 0.5 \ {\rm V^2}} = P_\varepsilon\text{, if }\hspace{0.15cm}\underline{k_{\rm M} = 0} \ \Rightarrow \ z(t) \equiv 0$.

{{BlaueBox|TEXT=
'''(2)'''  Vary $\varphi_2$ between $\pm 180^\circ$ while keeping all other parameters from Exercise (1). How does the value of $T_0$ and $P_x$ change?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes:}\hspace{0.2cm}\hspace{0.15cm}\underline{ T_0 = 2.0 \ {\rm ms}; \hspace{0.2cm} P_x = 0.5 \ {\rm V^2}}$.

{{BlaueBox|TEXT=
'''(3)'''   Vary $f_2$ between $0 \le f_2 \le 10\ {\rm kHz}$ while keeping all other parameters from Exercise (1). How does the value of $P_x$ change?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes if }f_2 \ne 0\text{ and } f_2 \ne f_1\text{:}\hspace{0.3cm} \hspace{0.15cm}\underline{P_x = 0.5 \ {\rm V^2}}\text{.} \hspace{0.2cm} T_0 \text{ changes if }f_2\text{is not a multiple of }f_1$.

$\hspace{1.85cm}\text{If }f_2 = 0\text{:}\hspace{0.2cm} P_x = A_1^2/2 + A_2^2\hspace{0.15cm}\underline{ = 0.68 \ {\rm V^2}}$. $\hspace{3cm}$

$\hspace{1.85cm}\text{If }f_2 = f_1\text{:}\hspace{0.2cm} P_x = [A_1\cos(\varphi_1) + A_2\cos(\varphi_2)]^2/2 + [A_1\sin(\varphi_1) + A_2\sin(\varphi_2)]^2/2 \text{.} $

$\hspace{1.85cm}\text{With } \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ\text{:}\hspace{0.3cm}\hspace{0.15cm}\underline{ P_x = 0.74 \ {\rm V^2}}\text{.} $

{{BlaueBox|TEXT=
'''(4)'''   Keeping the previous input signal $x(t)$, set following parameters
: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$, $k_{\rm M} = 1 \text{ and } \tau_{\rm M} = 0$ .
:Are there linear distortions? Calculate the received power $P_y$ and the power $P_\varepsilon$ of the differential signal $\varepsilon(t) = z(t) - x(t)$. }}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{ y(t) = 0.5 \cdot x(t- 1\ {\rm ms})}\text{ is only attenuated and delayed, but not distorted.}$

$\hspace{1.85cm}\text{Received power:}\hspace{0.2cm} P_y = (A_1/2)^2/2 + (A_2/2)^2/2\hspace{0.15cm}\underline{ = 0.125 \ {\rm V^2}}\text{. } P_\varepsilon \text{ is significantly larger:} \hspace{0.1cm} \hspace{0.15cm}\underline{P_\varepsilon = 0.625 \ {\rm V^2}}.$

{{BlaueBox|TEXT=
'''(5)'''   With the same settings as in Exercise (4), vary the matching parameters $k_{\rm M} \text{ and } \tau_{\rm M}$. How big is the distortion power $P_{\rm D}$?}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D}\text{ is equal to }P_\varepsilon \text{ when using the ideal matching parameters:} \hspace{0.2cm}k_{\rm M} = 2 \text{ and } \tau_{\rm M}=T_0 - 0.5\ {\rm ms} = 1.5\ {\rm ms}$

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}z(t) = x(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon(t) = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm D}\hspace{0.15cm}\underline{ = P_\varepsilon = 0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\text{Neither attenuation nor phase distortion.}$

{{BlaueBox|TEXT=
'''(6)'''   The channel parameters are now set to: $\alpha_1 = 0.5, \hspace{0.15cm}\underline{\alpha_2 = 0.2}, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the Signal-to-Distortion ratio $(\rm SDR) \ \rho_{\rm D}$.}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{ when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 2.24} \text{ and } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.5\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.059 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Attenuation distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 8.5}$.

{{BlaueBox|TEXT=
'''(7)'''   The channel parameters are now set to: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 \hspace{0.15cm}\underline{= 2\ {\rm ms} }, \ \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the the Signal-to-Distortion ratio $\rho_{\rm D}$.}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.84} \text{ and } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.071 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Phase distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 7}$.

{{BlaueBox|TEXT=
'''(8)'''   The channel parameters are now set to: $\hspace{0.15cm}\underline{\alpha_1 = 0.5} , \hspace{0.15cm}\underline{\alpha_2 = 0.2} , \ \hspace{0.15cm}\underline{\tau_1= 0.5\ {\rm ms} }, \ \hspace{0.15cm}\underline{\tau_2 = 0.3\ {\rm ms} }$. Are there attenuation distortions? Are there phase distortions? How can $y(t)$ be approximated? ''Hint:'' $\cos(3x) = 4 \cdot \cos^3(x) - 3\cdot \cos(x).$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Both attenuation and phase distortions, because }\alpha_1 \ne \alpha_2\text{ and }\tau_1 \ne \tau_2$.

$\hspace{1.85cm}y(t) = y_1(t) + y_2(t)\ \Rightarrow \ y_1(t) = A_1 \cdot \alpha_1 \cdot \sin[2\pi f_1\ (t- 0.5\ \rm ms)] = -0.4 \ {\rm V} \cdot \cos(2\pi f_1 t)$

$\hspace{1.85cm} y_2(t) = \alpha_2 \cdot x_2(t- \tau_2) \text{ mit }x_2(t) = A_2 \cdot \cos[2\pi f_2\ (t- 30^\circ)] \approx A_2 \cdot \cos[2\pi f_2\ (t- 1/36 \ \rm ms)]$

$\hspace{1.85cm} \Rightarrow \ y_2(t) = 0.12 \ {\rm V} \cdot \cos[2\pi f_2\ (t- 0.328 \ {\rm ms})] \approx -0.12 \ { \rm V} \cdot \cos[2\pi f_2t] $.

$\hspace{1.85cm} \Rightarrow \ y(t) = y_1(t) + y_2(t) \approx -0.4 \ {\rm V} \cdot [\cos(2\pi \cdot f_1\cdot t) + 1/3 \cdot \cos(2\pi \cdot 3 f_1 \cdot t) = -0.533 \ {\rm V} \cdot \cos^3(2\pi f_1 t)$.

{{BlaueBox|TEXT=
'''(9)'''   Using the parameters from Exercise (8), calculate the distortion power $P_{\rm D}$ and the the Signal-to-Distortion ratio $\rho_{\rm D}$.}}

$\hspace{1.0cm}\text{Best possible adaptation:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.96} \text{, } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.65\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.15 \ {\rm V^2} },\hspace{0.1cm}\hspace{0.15cm}\underline{\rho_{\rm D} = 0.500/0.15 \approx 3.3}$.

{{BlaueBox|TEXT=
'''(10)'''  Now we set $A_2 = 0$ and $A_1 = 1\ {\rm V}, \ f_1 = 1\ {\rm kHz}, \ \varphi_1 = 0^\circ$. The channel is a Low-pass of order 1 $\underline{(f_0 = 1\ {\rm kHz})}$. Are there any attenuation and/or phase distortions? Calculate the channel coefficients $\alpha_1$ and $\tau_1$.}}

$\hspace{1.0cm}\text{At only one frequency there are neither attenuation nor phase distortions.}$
$\hspace{1.0cm}\text{Attenuation factor for }f_1=f_0\text{ and }N=1\text{: }\alpha_1 =|H(f = f_1)| = [1+( f_1/f_0)^2]^{-N/2} = 2^{-1/2}= 1/\sqrt{2}\hspace{0.15cm}\underline{=0.707},$
$\hspace{1.0cm}\text{Phase factor for }f_1=f_0\text{ and }N=1\text{: }\tau_1 = N \cdot \arctan( f_1/f_0)/(2 \pi f_1)=\arctan( 1)/(2 \pi f_1) =1/(8f_1) \hspace{0.15cm}\underline{=0.125 \ \rm ms}.$

{{BlaueBox|TEXT=
'''(11)'''   How do the channel parameters change when using a Low-pass of order 2 compared to a Low-pass of order 1 $(f_0 = 1\ {\rm kHz})$?}}

$\hspace{1.0cm}\alpha_1 = 0.707^2 = 0.5$ and $\tau_1 = 2 \cdot 0.125 = 0.25 \ {\rm ms}$.

$\hspace{1.0cm}\text{The signal }y(t)\text{ is only half as big as }x(t)\text{ and is retarded: The cosine turns into a sine function}$.

{{BlaueBox|TEXT=
'''(12)'''   What differences arise when using a High-pass of order 2 compared to a Low-pass of order 2 $(f_0 = 1\ {\rm kHz})$? }}

$\hspace{1.0cm}\text{Since }f_1 = f_0\text{ the attenuation factor }\alpha_1 = 0.5\text{ stays the same and }\tau_1 = -0.25 \ {\rm ms}\text{ which means:}$

$\hspace{1.0cm}\text{The signal }y(t)\text{ is also only half as big as }x(t)\text{ and precedes it: The cosine turns into the Minus–sine function}$.

{{BlaueBox|TEXT=
'''(13)'''   What differences at the signal $y(t)$ can be observed between the Low-pass and the High-pass of order 2 $(f_0 = 1\ {\rm kHz})$ when you start with the initial input signal according to Exercise (1) and continuously raise $f_2$ up to $10 \ \rm kHz$ ? }}

$\hspace{1.0cm}\text{With the Low-pass the second term is increasingly suppressed. For }f_2 = 10 \ {\rm kHz}\text{: }y_{\rm LP}(t) \approx 0.8 \cdot x_1(t-0.3 \ \rm ms).$

$\hspace{1.0cm}\text{With the High-pass however the second term dominates. For }f_2 = 10 \ {\rm kHz}\text{: }y_{\rm HP}(t) \approx 0.2 \cdot x_1(t+0.7 \ {\rm ms}) + x_2(t).$

==Applet Manual==
[[File:Handhabung_verzerrungen.png|center]]
 
    '''(A)'''     Parameter selection for input signal $x(t)$ per slider: Amplitude, frequency, phase values

    '''(B)'''     Preselection for channel parameters per slider: Low-pass or High-pass

    '''(C)'''     Selction of channel parameters per slider: Dämpfungsfaktoren und Phasenlaufzeiten

    '''(D)'''     Selection of channel parameters for High and Low pass: Order$n$, cutoff frequency $f_0$

    '''(E)'''     Selection of matching parameters $k_{\rm M}$ and $\varphi_{\rm M}$

    '''(F)'''     Selection of the signals to be displayed: $x(t)$, $y(t)$, $z(t)$, $\varepsilon(t)$, $\varepsilon^2(t)$

    '''(G)'''     Graphic display of the signals

    '''(H)'''     Enter the time $t_*$ for the numeric output

    '''( I )'''     numeric output of the signal values $x(t_*)$, $y(t_*)$, $z(t_*)$ and $\varepsilon(t_*)$

    '''(J)'''     Numeric output of the main result $P_\varepsilon$

    '''(K)'''     Save and reall parameters

    '''(L)'''     Exercises: Exercise selection, description and solution

    '''(M)'''     Variation possibilities for the graphic display

$\hspace{1.5cm}$Zoom–functions "$+$" (scale up), "$-$" (scale down) und $\rm o$ (reset)

$\hspace{1.5cm}$Move with "$\leftarrow$" (section to the left, ordinate to the right), "$\uparrow$" "$\downarrow$" und "$\rightarrow$"

$\hspace{1.5cm}$'''Other options'':

$\hspace{1.5cm}$Hold shift and scroll: Zoom in on/out of coordinate system,

$\hspace{1.5cm}$Hold shift and left click: Move the coordinate system.

==About the Authors==
This interactive calculation was designed and realized at the [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] of the [https://www.tum.de/ Technische Universität München].
*The original version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her Diploma thesis using "FlashMX–Actionscript" (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]] ).
*In 2018 this Applet was redesigned and updated to "HTML5" by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]] as part of his Bachelor's thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) .

==Once again: Open Applet in new Tab==

{{LntAppletLinkEn|linDistortions_en}}         [https://www.lntwww.de/Applets:Lineare_Verzerrungen_periodischer_Signale '''English Applet with German WIKI description''']

Applets:Linear Distortions of Periodic Signals

2023-03-26T22:09:15Z

Noah:

{{LntAppletLink|physAnSignal_en}}         [https://en.lntwww.de/Applets:Physical_Signal_%26_Analytic_Signal '''English Applet with English WIKI description''']

==Applet Description==
 
This applet illustrates the effects of linear distortions (attenuation distortions and phase distortions) with
[[File:Modell_version2.png|right|frame|Meanings of the used signals]]
*the input signal $x(t)$   ⇒   power $P_x$:
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
*the output signal $y(t)$   ⇒   power $P_y$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2),$$
*the matched output signal $z(t)$   ⇒   power $P_z$:
:$$z(t) = k_{\rm M} \cdot y(t-\tau_{\rm M}) + \alpha_2 \cdot x_2(t-\tau_2),$$
*the difference signal   $\varepsilon(t) = z(t) - x(t)$   ⇒   power $P_\varepsilon$.

The next block in the model above is ''Matching'': The output signal $y(t)$ is adjusted in amplitude and phase with equal variables $k_{\rm M}$ and $\tau_{\rm M}$ for all frequencies which means that this is not a frequency-dependent equalization. Using the signal $z(t)$, one can differentiate between:
*attenuation distortion and frequency–independent attenuation, as well as
*phase distortion and frequency–independent delay.

The ''Distortion Power'' $P_{\rm D}$ is used to measure the strength of the linear distortion and is defined as:
:$$P_{\rm D} = \min_{k_{\rm M}, \ \tau_{\rm M}} P_\varepsilon.$$

==Theoretical Background==
 
Distortions refer to generally unwanted alterations of a message signal through a transmission system. Together with the strong stochastic effects (noise, crosstalk, etc.), they are a crucial limitation for the quality and rate of transmission.

Just as the intensity of noise can be assessed through
*the ''Noise Power'' $P_{\rm N}$ and
*the ''Signal–to–Noise Ratio'' (SNR) $\rho_{\rm N}$,

distortions can be quantified through

*the ''Distortion Power'' $P_{\rm D}$ and
*the ''Signal–to–Distortion Ratio'' (SDR)
:$$\rho_{\rm D}=\frac{\rm Signal \ Power}{\rm Distortion \ Power} = \frac{P_x}{P_{\rm D} }.$$

=== Linear and Nonlinear Distortions ===
 
A distinction is made between linear and nonlinear distortions:
*'''Nonlinear distortions''' occur, if at all times $t$ the nonlinear correlation $y = g(x) \ne {\rm const.} \cdot x$ exists between the signal values $x = x(t)$ at the input and $y = y(t)$ at the output, whereby $y = g(x)$ is defined as the system's nonlinear characteristic. By creating a cosine signal at the input with frequency $f_0$ the output signal includes $f_0$, as well as multiple harmonic waves. We conclude that new frequencies arise through nonlinear distortion.

[[File:LZI_T_2_2_S3_vers2.png|center|frame|For clarification of nonlinear distortions |class=fit]]

[[File:P_ID899__LZI_T_2_3_S1_neu.png|right |frame| Description of a linear system|class=fit]]
*'''Linear distortions''' occur, if the transmission channel is characterized by a frequency response $H(f) \ne \rm const.$ Various frequencies are attenuated and delayed differently. Characteristic of this is that although frequencies can disappear (for example, through a low–pass, a high–pass, or a band–pass), no new frequencies can arise.

In this applet only linear distortions are considered.

=== Description Forms for the Frequency Response ===
 
The generally complex valued frequency response can be represented as follows:
:$$H(f) = |H(f)| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot
\hspace{0.05cm} b(f)} = {\rm e}^{-a(f)}\cdot {\rm e}^{-{\rm j}
\hspace{0.05cm} \cdot \hspace{0.05cm} b(f)}.$$

This results in the following description variables:
*The absolute value $|H(f)|$ is called '''amplitude response''' and in logarithmic form '''attenuation function''':
:$$a(f) = - \ln |H(f)|\hspace{0.2cm}{\rm in \hspace{0.1cm}Neper
\hspace{0.1cm}(Np) } = - 20 \cdot \lg |H(f)|\hspace{0.2cm}{\rm in
\hspace{0.1cm}Decibel \hspace{0.1cm}(dB) }.$$
*The '''phase function''' $b(f)$ indicates the negative frequency–dependent angle of $H(f)$ in the complex plane based on the real axis:
:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) \hspace{0.2cm}{\rm in
\hspace{0.1cm}Radian \hspace{0.1cm}(rad)}.$$

=== Low–pass of Order N ===
 
[[File:Tiefpass_version2.png|right|frame|Attenuation function $a(f)$ and phase function $b(f)$ of a low–Pass of order $N$]]
The frequency response of a realizable low–pass (LP) of order $N$ is:
:$$H(f) = \left [\frac{1}{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
For example the RC low–pass is a first order low–pass. Consequently we can obtain
*the attenuation function:
:$$a(f) =N/2 \cdot \ln [1+( f/f_0)^2] \hspace{0.05cm},$$
*the phase function:
:$$b(f) =N \cdot \arctan( f/f_0) \hspace{0.05cm},$$
*the attenuation factor for the frequency $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_i/f_0)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*the phase delay for the frequency $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2 \pi f_i} = \frac{N \cdot \arctan( f_i/f_0)}{2 \pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

=== High–pass of Order N ===
 
[[File:Hochpass_version2.png|right|frame|Attenuation function $a(f)$ and phase function $b(f)$ of a high–pass of order $N$]]
The frequency response of a realizable high–pass (HP) of order $N$ is:
:$$H(f) = \left [\frac{ {\rm j}\cdot f/f_0 }{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
For example the LC high-pass is a first order high-pass. Consequently we can obtain
*the attenuation function:
:$$a(f) =N/2 \cdot \ln [1+( f_0/f)^2] \hspace{0.05cm},$$
*the phase function:
:$$b(f) =-N \cdot \arctan( f_0/f) \hspace{0.05cm},$$
*the attenuation factor for the frequency $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_0/f_i)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*the phase delay for the frequency $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2\pi f_i} = \frac{-N \cdot \arctan( f_0/f_i)}{2\pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

[[File:Verzerrungen_HP_TP_1_englisch.png|right|frame|Phase function $b(f)$ of high–pass and low–pass]]
{{GraueBox|TEXT=
$\text{Example:}$ 
This graphic shows the phase function $b(f)$ with the cutoff frequency $f_0 = 1\ \rm kHz$ and order $N=1$
* of a low–pass (green curve),
* of a high–pass (violet curve).

The input signal is sinusoidal with frequency $f_{\rm S} = 1.25\ {\rm kHz}$ whereby this signal is only turned on at $t=0$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.75cm}0 \\ \sin(2\pi \cdot f_{\rm S} \cdot t ) \\ \end{array} \right.\quad\begin{array}{l} (t < 0), \\ (t>0). \\ \end{array}$$

The left graphic shows the signal $x(t)$. The dashed line marks the first zero at $t = T_0 = 0.8\ {\rm ms}$. The other two graphics show the output signals $y_{\rm LP}(t)$ und $y_{\rm HP}(t)$ of low–pass and high–pass, whereby the change in amplitude was balanced in both cases.

[[File:Verzerrungen_HP_TP_2_version2.png|center|frame|Input signal $x(t)$ (enframed in blue) as well as output signals $y_{\rm LP}(t)$ ⇒   green and $y_{\rm HP}(t)$ ⇒   magenta]]

*The first zero of the signal $y_{\rm LP}(t)$ after the low–pass is delayed by $\tau_{\rm LP} = 0.9/(2\pi) \cdot T_0 \approx 0.115 \ {\rm ms}$ compared to the first zero of $x(t)$   ⇒   marked with green arrow, whereby $b_{\rm LP}(f/f_{\rm S} = 0.9 \ {\rm rad})$ was considered.
* In contrast, the phase delay of the high–pass is negative: $\tau_{\rm HP} = -0.67/(2\pi) \cdot T_0 \approx -0.085 \ {\rm ms}$ and therefore the first zero of $y_{\rm HP}(t)$ occurs before the dashed line.
*Following this transient response, in both cases the zero crossings again come in the raster of the period duration $T_0 = 0.8 \ {\rm ms}.$

''Remark:'' The shown signals were created using the interactive applet [[Applets:Kausale_Systeme_-_Laplacetransformation|"Causal systems – Laplace transform"]]. }}

=== Attenuation and Phase Distortions ===
 
[[File:P_ID900__LZI_T_2_3_S2_neu.png|frame| Requirements for a non–distorting channel|right|class=fit]]
The adjacent figure shows
*the even attenuation function $a(f)$   ⇒   $a(-f) = a(f)$, and
*the uneven function curve $b(f)$   ⇒   $b(-f) = -b(- f)$

of a non–distorting channel. One can see:
*In a distortion–free system the attenuation function $a(f)$ must be constant between$f_{\rm U}$ and $f_{\rm O}$ around the carrier frequency $f_{\rm T}$, where the input signal exists   ⇒   $X(f) \ne 0$.
*From the specified constant attenuation value $6 \ \rm dB$ follows for the amplitude response $|H(f)| = 0.5$   ⇒   the signal values of all frequencies are thus halved by the system   ⇒   '''no attenuation distortions'''.
*In addition, in such a system, the phase function $b(f)$ between $f_{\rm U}$ and $f_{\rm O}$ must increase linearly with the frequency. As a result, all frequency components are delayed by the same phase delay $τ$   ⇒   '''no phase distortion'''.
*The delay $τ$ is fixed by the slope of $b(f)$. The phase function $b(f) \equiv 0$ would result in a delay–less system   ⇒   $τ = 0$.

The following summary considers that – in this applet – the input signal is always the sum of two harmonic oscillations,
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
and therefore the channel influence is fully described by the attenuation factors $\alpha_1$ and $\alpha_2$ as well as the phase delays $\tau_1$ and $\tau_2$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2).$$

{{BlaueBox|TEXT=
$\text{Summary:}$ 
*A signal $y(t)$ is only '''distortion–free''' compared to $x(t)$ if $\alpha_1 = \alpha_2= \alpha$   and   $\tau_1 = \tau_2= \tau$   ⇒   $y(t) = \alpha \cdot x(t-\tau)$.
* '''Attenuation distortions''' occur when $\alpha_1 \ne \alpha_2$. If $\alpha_1 \ne \alpha_2$ and $\tau_1 = \tau_2$, then there are exclusively attenuation distortions.
* '''Phase distortions''' occur when $\tau_1 \ne \tau_2$. If $\tau_1 \ne \tau_2$ and $\alpha_1 = \alpha_2$, then there are exclusively phase distortions. }}

==Exercises==
[[File:Exercises_verzerrungen.png|right]]
*First choose an exercise number.
*An exercise description is displayed.
*Parameter values are adjusted to the respective exercises.
*Click "Hide solution" to display the solution.

Number "0" is a "Reset" button:
*Sets parameters to initial values (when loading the page).
*Displays a "Reset text" to describe the applet further.

{{BlaueBox|TEXT=
'''(1)'''   We consider the parameters $A_1 = 0.8\ {\rm V}, \ A_2 = 0.6\ {\rm V}, \ f_1 = 0.5\ {\rm kHz}, \ f_2 = 1.5\ {\rm kHz}, \ \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ$ for the input signal $x(t)$.
:Calculate the signal's period duration $T_0$ and power $P_x$. Can you read the value for $P_x$ off the applet? }}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}T_0 = \big [\hspace{-0.1cm}\text{ greatest common divisor }(0.5 \ {\rm kHz}, \ 1.5 \ {\rm kHz})\big ]^{-1}\hspace{0.15cm}\underline{ = 2.0 \ {\rm ms}};$

$\hspace{1.85cm} P_x = A_1^2/2 + A_2^2/2 \hspace{0.15cm}\underline{= 0.5 \ {\rm V^2}} = P_\varepsilon\text{, if }\hspace{0.15cm}\underline{k_{\rm M} = 0} \ \Rightarrow \ z(t) \equiv 0$.

{{BlaueBox|TEXT=
'''(2)'''  Vary $\varphi_2$ between $\pm 180^\circ$ while keeping all other parameters from Exercise (1). How does the value of $T_0$ and $P_x$ change?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes:}\hspace{0.2cm}\hspace{0.15cm}\underline{ T_0 = 2.0 \ {\rm ms}; \hspace{0.2cm} P_x = 0.5 \ {\rm V^2}}$.

{{BlaueBox|TEXT=
'''(3)'''   Vary $f_2$ between $0 \le f_2 \le 10\ {\rm kHz}$ while keeping all other parameters from Exercise (1). How does the value of $P_x$ change?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes if }f_2 \ne 0\text{ and } f_2 \ne f_1\text{:}\hspace{0.3cm} \hspace{0.15cm}\underline{P_x = 0.5 \ {\rm V^2}}\text{.} \hspace{0.2cm} T_0 \text{ changes if }f_2\text{is not a multiple of }f_1$.

$\hspace{1.85cm}\text{If }f_2 = 0\text{:}\hspace{0.2cm} P_x = A_1^2/2 + A_2^2\hspace{0.15cm}\underline{ = 0.68 \ {\rm V^2}}$. $\hspace{3cm}$

$\hspace{1.85cm}\text{If }f_2 = f_1\text{:}\hspace{0.2cm} P_x = [A_1\cos(\varphi_1) + A_2\cos(\varphi_2)]^2/2 + [A_1\sin(\varphi_1) + A_2\sin(\varphi_2)]^2/2 \text{.} $

$\hspace{1.85cm}\text{With } \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ\text{:}\hspace{0.3cm}\hspace{0.15cm}\underline{ P_x = 0.74 \ {\rm V^2}}\text{.} $

{{BlaueBox|TEXT=
'''(4)'''   Keeping the previous input signal $x(t)$, set following parameters
: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$, $k_{\rm M} = 1 \text{ and } \tau_{\rm M} = 0$ .
:Are there linear distortions? Calculate the received power $P_y$ and the power $P_\varepsilon$ of the differential signal $\varepsilon(t) = z(t) - x(t)$. }}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{ y(t) = 0.5 \cdot x(t- 1\ {\rm ms})}\text{ is only attenuated and delayed, but not distorted.}$

$\hspace{1.85cm}\text{Received power:}\hspace{0.2cm} P_y = (A_1/2)^2/2 + (A_2/2)^2/2\hspace{0.15cm}\underline{ = 0.125 \ {\rm V^2}}\text{. } P_\varepsilon \text{ is significantly larger:} \hspace{0.1cm} \hspace{0.15cm}\underline{P_\varepsilon = 0.625 \ {\rm V^2}}.$

{{BlaueBox|TEXT=
'''(5)'''   With the same settings as in Exercise (4), vary the matching parameters $k_{\rm M} \text{ and } \tau_{\rm M}$. How big is the distortion power $P_{\rm D}$?}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D}\text{ is equal to }P_\varepsilon \text{ when using the ideal matching parameters:} \hspace{0.2cm}k_{\rm M} = 2 \text{ and } \tau_{\rm M}=T_0 - 0.5\ {\rm ms} = 1.5\ {\rm ms}$

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}z(t) = x(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon(t) = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm D}\hspace{0.15cm}\underline{ = P_\varepsilon = 0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\text{Neither attenuation nor phase distortion.}$

{{BlaueBox|TEXT=
'''(6)'''   The channel parameters are now set to: $\alpha_1 = 0.5, \hspace{0.15cm}\underline{\alpha_2 = 0.2}, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the Signal-to-Distortion ratio $(\rm SDR) \ \rho_{\rm D}$.}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{ when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 2.24} \text{ and } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.5\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.059 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Attenuation distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 8.5}$.

{{BlaueBox|TEXT=
'''(7)'''   The channel parameters are now set to: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 \hspace{0.15cm}\underline{= 2\ {\rm ms} }, \ \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the the Signal-to-Distortion ratio $\rho_{\rm D}$.}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.84} \text{ and } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.071 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Phase distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 7}$.

{{BlaueBox|TEXT=
'''(8)'''   The channel parameters are now set to: $\hspace{0.15cm}\underline{\alpha_1 = 0.5} , \hspace{0.15cm}\underline{\alpha_2 = 0.2} , \ \hspace{0.15cm}\underline{\tau_1= 0.5\ {\rm ms} }, \ \hspace{0.15cm}\underline{\tau_2 = 0.3\ {\rm ms} }$. Are there attenuation distortions? Are there phase distortions? How can $y(t)$ be approximated? ''Hint:'' $\cos(3x) = 4 \cdot \cos^3(x) - 3\cdot \cos(x).$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Both attenuation and phase distortions, because }\alpha_1 \ne \alpha_2\text{ and }\tau_1 \ne \tau_2$.

$\hspace{1.85cm}y(t) = y_1(t) + y_2(t)\ \Rightarrow \ y_1(t) = A_1 \cdot \alpha_1 \cdot \sin[2\pi f_1\ (t- 0.5\ \rm ms)] = -0.4 \ {\rm V} \cdot \cos(2\pi f_1 t)$

$\hspace{1.85cm} y_2(t) = \alpha_2 \cdot x_2(t- \tau_2) \text{ mit }x_2(t) = A_2 \cdot \cos[2\pi f_2\ (t- 30^\circ)] \approx A_2 \cdot \cos[2\pi f_2\ (t- 1/36 \ \rm ms)]$

$\hspace{1.85cm} \Rightarrow \ y_2(t) = 0.12 \ {\rm V} \cdot \cos[2\pi f_2\ (t- 0.328 \ {\rm ms})] \approx -0.12 \ { \rm V} \cdot \cos[2\pi f_2t] $.

$\hspace{1.85cm} \Rightarrow \ y(t) = y_1(t) + y_2(t) \approx -0.4 \ {\rm V} \cdot [\cos(2\pi \cdot f_1\cdot t) + 1/3 \cdot \cos(2\pi \cdot 3 f_1 \cdot t) = -0.533 \ {\rm V} \cdot \cos^3(2\pi f_1 t)$.

{{BlaueBox|TEXT=
'''(9)'''   Using the parameters from Exercise (8), calculate the distortion power $P_{\rm D}$ and the the Signal-to-Distortion ratio $\rho_{\rm D}$.}}

$\hspace{1.0cm}\text{Best possible adaptation:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.96} \text{, } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.65\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.15 \ {\rm V^2} },\hspace{0.1cm}\hspace{0.15cm}\underline{\rho_{\rm D} = 0.500/0.15 \approx 3.3}$.

{{BlaueBox|TEXT=
'''(10)'''  Now we set $A_2 = 0$ and $A_1 = 1\ {\rm V}, \ f_1 = 1\ {\rm kHz}, \ \varphi_1 = 0^\circ$. The channel is a Low-pass of order 1 $\underline{(f_0 = 1\ {\rm kHz})}$. Are there any attenuation and/or phase distortions? Calculate the channel coefficients $\alpha_1$ and $\tau_1$.}}

$\hspace{1.0cm}\text{At only one frequency there are neither attenuation nor phase distortions.}$
$\hspace{1.0cm}\text{Attenuation factor for }f_1=f_0\text{ and }N=1\text{: }\alpha_1 =|H(f = f_1)| = [1+( f_1/f_0)^2]^{-N/2} = 2^{-1/2}= 1/\sqrt{2}\hspace{0.15cm}\underline{=0.707},$
$\hspace{1.0cm}\text{Phase factor for }f_1=f_0\text{ and }N=1\text{: }\tau_1 = N \cdot \arctan( f_1/f_0)/(2 \pi f_1)=\arctan( 1)/(2 \pi f_1) =1/(8f_1) \hspace{0.15cm}\underline{=0.125 \ \rm ms}.$

{{BlaueBox|TEXT=
'''(11)'''   How do the channel parameters change when using a Low-pass of order 2 compared to a Low-pass of order 1 $(f_0 = 1\ {\rm kHz})$?}}

$\hspace{1.0cm}\alpha_1 = 0.707^2 = 0.5$ and $\tau_1 = 2 \cdot 0.125 = 0.25 \ {\rm ms}$.

$\hspace{1.0cm}\text{The signal }y(t)\text{ is only half as big as }x(t)\text{ and is retarded: The cosine turns into a sine function}$.

{{BlaueBox|TEXT=
'''(12)'''   What differences arise when using a High-pass of order 2 compared to a Low-pass of order 2 $(f_0 = 1\ {\rm kHz})$? }}

$\hspace{1.0cm}\text{Since }f_1 = f_0\text{ the attenuation factor }\alpha_1 = 0.5\text{ stays the same and }\tau_1 = -0.25 \ {\rm ms}\text{ which means:}$

$\hspace{1.0cm}\text{The signal }y(t)\text{ is also only half as big as }x(t)\text{ and precedes it: The cosine turns into the Minus–sine function}$.

{{BlaueBox|TEXT=
'''(13)'''   What differences at the signal $y(t)$ can be observed between the Low-pass and the High-pass of order 2 $(f_0 = 1\ {\rm kHz})$ when you start with the initial input signal according to Exercise (1) and continuously raise $f_2$ up to $10 \ \rm kHz$ ? }}

$\hspace{1.0cm}\text{With the Low-pass the second term is increasingly suppressed. For }f_2 = 10 \ {\rm kHz}\text{: }y_{\rm LP}(t) \approx 0.8 \cdot x_1(t-0.3 \ \rm ms).$

$\hspace{1.0cm}\text{With the High-pass however the second term dominates. For }f_2 = 10 \ {\rm kHz}\text{: }y_{\rm HP}(t) \approx 0.2 \cdot x_1(t+0.7 \ {\rm ms}) + x_2(t).$

==Applet Manual==
[[File:Handhabung_verzerrungen.png|center]]
 
    '''(A)'''     Parameter selection for input signal $x(t)$ per slider: Amplitude, frequency, phase values

    '''(B)'''     Preselection for channel parameters per slider: Low-pass or High-pass

    '''(C)'''     Selction of channel parameters per slider: Dämpfungsfaktoren und Phasenlaufzeiten

    '''(D)'''     Selection of channel parameters for High and Low pass: Order$n$, cutoff frequency $f_0$

    '''(E)'''     Selection of matching parameters $k_{\rm M}$ and $\varphi_{\rm M}$

    '''(F)'''     Selection of the signals to be displayed: $x(t)$, $y(t)$, $z(t)$, $\varepsilon(t)$, $\varepsilon^2(t)$

    '''(G)'''     Graphic display of the signals

    '''(H)'''     Enter the time $t_*$ for the numeric output

    '''( I )'''     numeric output of the signal values $x(t_*)$, $y(t_*)$, $z(t_*)$ and $\varepsilon(t_*)$

    '''(J)'''     Numeric output of the main result $P_\varepsilon$

    '''(K)'''     Save and reall parameters

    '''(L)'''     Exercises: Exercise selection, description and solution

    '''(M)'''     Variation possibilities for the graphic display

$\hspace{1.5cm}$Zoom–functions "$+$" (scale up), "$-$" (scale down) und $\rm o$ (reset)

$\hspace{1.5cm}$Move with "$\leftarrow$" (section to the left, ordinate to the right), "$\uparrow$" "$\downarrow$" und "$\rightarrow$"

$\hspace{1.5cm}$'''Other options'':

$\hspace{1.5cm}$Hold shift and scroll: Zoom in on/out of coordinate system,

$\hspace{1.5cm}$Hold shift and left click: Move the coordinate system.

==About the Authors==
This interactive calculation was designed and realized at the [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] of the [https://www.tum.de/ Technische Universität München].
*The original version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her Diploma thesis using "FlashMX–Actionscript" (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]] ).
*In 2018 this Applet was redesigned and updated to "HTML5" by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]] as part of his Bachelor's thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) .

==Once again: Open Applet in new Tab==

{{LntAppletLinkEn|linDistortions_en}}         [https://www.lntwww.de/Applets:Lineare_Verzerrungen_periodischer_Signale '''English Applet with German WIKI description''']

Applets:Linear Distortions of Periodic Signals

2023-03-26T22:08:45Z

Noah:

{{LntAppletLink|physAnSignal_en}}         [https://en.lntwww.de/Applets:Physical_Signal_%26_Analytic_Signal '''English Applet with English WIKI description''']

==Applet Description==
 
This applet illustrates the effects of linear distortions (attenuation distortions and phase distortions) with
[[File:Modell_version2.png|right|frame|Meanings of the used signals]]
*the input signal $x(t)$   ⇒   power $P_x$:
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
*the output signal $y(t)$   ⇒   power $P_y$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2),$$
*the matched output signal $z(t)$   ⇒   power $P_z$:
:$$z(t) = k_{\rm M} \cdot y(t-\tau_{\rm M}) + \alpha_2 \cdot x_2(t-\tau_2),$$
*the difference signal   $\varepsilon(t) = z(t) - x(t)$   ⇒   power $P_\varepsilon$.

The next block in the model above is ''Matching'': The output signal $y(t)$ is adjusted in amplitude and phase with equal variables $k_{\rm M}$ and $\tau_{\rm M}$ for all frequencies which means that this is not a frequency-dependent equalization. Using the signal $z(t)$, one can differentiate between:
*attenuation distortion and frequency–independent attenuation, as well as
*phase distortion and frequency–independent delay.

The ''Distortion Power'' $P_{\rm D}$ is used to measure the strength of the linear distortion and is defined as:
:$$P_{\rm D} = \min_{k_{\rm M}, \ \tau_{\rm M}} P_\varepsilon.$$

==Theoretical Background==
 
Distortions refer to generally unwanted alterations of a message signal through a transmission system. Together with the strong stochastic effects (noise, crosstalk, etc.), they are a crucial limitation for the quality and rate of transmission.

Just as the intensity of noise can be assessed through
*the ''Noise Power'' $P_{\rm N}$ and
*the ''Signal–to–Noise Ratio'' (SNR) $\rho_{\rm N}$,

distortions can be quantified through

*the ''Distortion Power'' $P_{\rm D}$ and
*the ''Signal–to–Distortion Ratio'' (SDR)
:$$\rho_{\rm D}=\frac{\rm Signal \ Power}{\rm Distortion \ Power} = \frac{P_x}{P_{\rm D} }.$$

=== Linear and Nonlinear Distortions ===
 
A distinction is made between linear and nonlinear distortions:
*'''Nonlinear distortions''' occur, if at all times $t$ the nonlinear correlation $y = g(x) \ne {\rm const.} \cdot x$ exists between the signal values $x = x(t)$ at the input and $y = y(t)$ at the output, whereby $y = g(x)$ is defined as the system's nonlinear characteristic. By creating a cosine signal at the input with frequency $f_0$ the output signal includes $f_0$, as well as multiple harmonic waves. We conclude that new frequencies arise through nonlinear distortion.

[[File:LZI_T_2_2_S3_vers2.png|center|frame|For clarification of nonlinear distortions |class=fit]]

[[File:P_ID899__LZI_T_2_3_S1_neu.png|right |frame| Description of a linear system|class=fit]]
*'''Linear distortions''' occur, if the transmission channel is characterized by a frequency response $H(f) \ne \rm const.$ Various frequencies are attenuated and delayed differently. Characteristic of this is that although frequencies can disappear (for example, through a low–pass, a high–pass, or a band–pass), no new frequencies can arise.

In this applet only linear distortions are considered.

=== Description Forms for the Frequency Response ===
 
The generally complex valued frequency response can be represented as follows:
:$$H(f) = |H(f)| \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot
\hspace{0.05cm} b(f)} = {\rm e}^{-a(f)}\cdot {\rm e}^{-{\rm j}
\hspace{0.05cm} \cdot \hspace{0.05cm} b(f)}.$$

This results in the following description variables:
*The absolute value $|H(f)|$ is called '''amplitude response''' and in logarithmic form '''attenuation function''':
:$$a(f) = - \ln |H(f)|\hspace{0.2cm}{\rm in \hspace{0.1cm}Neper
\hspace{0.1cm}(Np) } = - 20 \cdot \lg |H(f)|\hspace{0.2cm}{\rm in
\hspace{0.1cm}Decibel \hspace{0.1cm}(dB) }.$$
*The '''phase function''' $b(f)$ indicates the negative frequency–dependent angle of $H(f)$ in the complex plane based on the real axis:
:$$b(f) = - {\rm arc} \hspace{0.1cm}H(f) \hspace{0.2cm}{\rm in
\hspace{0.1cm}Radian \hspace{0.1cm}(rad)}.$$

=== Low–pass of Order N ===
 
[[File:Tiefpass_version2.png|right|frame|Attenuation function $a(f)$ and phase function $b(f)$ of a low–Pass of order $N$]]
The frequency response of a realizable low–pass (LP) of order $N$ is:
:$$H(f) = \left [\frac{1}{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
For example the RC low–pass is a first order low–pass. Consequently we can obtain
*the attenuation function:
:$$a(f) =N/2 \cdot \ln [1+( f/f_0)^2] \hspace{0.05cm},$$
*the phase function:
:$$b(f) =N \cdot \arctan( f/f_0) \hspace{0.05cm},$$
*the attenuation factor for the frequency $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_i/f_0)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*the phase delay for the frequency $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2 \pi f_i} = \frac{N \cdot \arctan( f_i/f_0)}{2 \pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

=== High–pass of Order N ===
 
[[File:Hochpass_version2.png|right|frame|Attenuation function $a(f)$ and phase function $b(f)$ of a high–pass of order $N$]]
The frequency response of a realizable high–pass (HP) of order $N$ is:
:$$H(f) = \left [\frac{ {\rm j}\cdot f/f_0 }{1 + {\rm j}\cdot f/f_0 }\right ]^N\hspace{0.05cm}.$$
For example the LC high pass is a first order high pass. Consequently we can obtain
*the attenuation function:
:$$a(f) =N/2 \cdot \ln [1+( f_0/f)^2] \hspace{0.05cm},$$
*the phase function:
:$$b(f) =-N \cdot \arctan( f_0/f) \hspace{0.05cm},$$
*the attenuation factor for the frequency $f=f_i$:
:$$\alpha_i =|H(f = f_i)| = [1+( f_0/f_i)^2]^{-N/2}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)= \alpha_i \cdot A_i\cdot \cos(2\pi f_i t)\hspace{0.05cm},$$
*the phase delay for the frequency $f=f_i$:
:$$\tau_i =\frac{b(f_i)}{2\pi f_i} = \frac{-N \cdot \arctan( f_0/f_i)}{2\pi f_i}$$
:$$\Rightarrow \hspace{0.3cm} x(t)= A_i\cdot \cos(2\pi f_i t) \hspace{0.1cm}\rightarrow \hspace{0.1cm} y(t)=A_i\cdot \cos(2\pi f_i (t- \tau_i))\hspace{0.05cm}.$$

[[File:Verzerrungen_HP_TP_1_englisch.png|right|frame|Phase function $b(f)$ of high–pass and low–pass]]
{{GraueBox|TEXT=
$\text{Example:}$ 
This graphic shows the phase function $b(f)$ with the cutoff frequency $f_0 = 1\ \rm kHz$ and order $N=1$
* of a low–pass (green curve),
* of a high–pass (violet curve).

The input signal is sinusoidal with frequency $f_{\rm S} = 1.25\ {\rm kHz}$ whereby this signal is only turned on at $t=0$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.75cm}0 \\ \sin(2\pi \cdot f_{\rm S} \cdot t ) \\ \end{array} \right.\quad\begin{array}{l} (t < 0), \\ (t>0). \\ \end{array}$$

The left graphic shows the signal $x(t)$. The dashed line marks the first zero at $t = T_0 = 0.8\ {\rm ms}$. The other two graphics show the output signals $y_{\rm LP}(t)$ und $y_{\rm HP}(t)$ of low–pass and high–pass, whereby the change in amplitude was balanced in both cases.

[[File:Verzerrungen_HP_TP_2_version2.png|center|frame|Input signal $x(t)$ (enframed in blue) as well as output signals $y_{\rm LP}(t)$ ⇒   green and $y_{\rm HP}(t)$ ⇒   magenta]]

*The first zero of the signal $y_{\rm LP}(t)$ after the low–pass is delayed by $\tau_{\rm LP} = 0.9/(2\pi) \cdot T_0 \approx 0.115 \ {\rm ms}$ compared to the first zero of $x(t)$   ⇒   marked with green arrow, whereby $b_{\rm LP}(f/f_{\rm S} = 0.9 \ {\rm rad})$ was considered.
* In contrast, the phase delay of the high–pass is negative: $\tau_{\rm HP} = -0.67/(2\pi) \cdot T_0 \approx -0.085 \ {\rm ms}$ and therefore the first zero of $y_{\rm HP}(t)$ occurs before the dashed line.
*Following this transient response, in both cases the zero crossings again come in the raster of the period duration $T_0 = 0.8 \ {\rm ms}.$

''Remark:'' The shown signals were created using the interactive applet [[Applets:Kausale_Systeme_-_Laplacetransformation|"Causal systems – Laplace transform"]]. }}

=== Attenuation and Phase Distortions ===
 
[[File:P_ID900__LZI_T_2_3_S2_neu.png|frame| Requirements for a non–distorting channel|right|class=fit]]
The adjacent figure shows
*the even attenuation function $a(f)$   ⇒   $a(-f) = a(f)$, and
*the uneven function curve $b(f)$   ⇒   $b(-f) = -b(- f)$

of a non–distorting channel. One can see:
*In a distortion–free system the attenuation function $a(f)$ must be constant between$f_{\rm U}$ and $f_{\rm O}$ around the carrier frequency $f_{\rm T}$, where the input signal exists   ⇒   $X(f) \ne 0$.
*From the specified constant attenuation value $6 \ \rm dB$ follows for the amplitude response $|H(f)| = 0.5$   ⇒   the signal values of all frequencies are thus halved by the system   ⇒   '''no attenuation distortions'''.
*In addition, in such a system, the phase function $b(f)$ between $f_{\rm U}$ and $f_{\rm O}$ must increase linearly with the frequency. As a result, all frequency components are delayed by the same phase delay $τ$   ⇒   '''no phase distortion'''.
*The delay $τ$ is fixed by the slope of $b(f)$. The phase function $b(f) \equiv 0$ would result in a delay–less system   ⇒   $τ = 0$.

The following summary considers that – in this applet – the input signal is always the sum of two harmonic oscillations,
:$$x(t) = x_1(t) + x_2(t) = A_1\cdot \cos\left(2\pi f_1\cdot t- \varphi_1\right)+A_2\cdot \cos\left(2\pi f_2\cdot t- \varphi_2\right), $$
and therefore the channel influence is fully described by the attenuation factors $\alpha_1$ and $\alpha_2$ as well as the phase delays $\tau_1$ and $\tau_2$:
:$$y(t) = \alpha_1 \cdot x_1(t-\tau_1) + \alpha_2 \cdot x_2(t-\tau_2).$$

{{BlaueBox|TEXT=
$\text{Summary:}$ 
*A signal $y(t)$ is only '''distortion–free''' compared to $x(t)$ if $\alpha_1 = \alpha_2= \alpha$   and   $\tau_1 = \tau_2= \tau$   ⇒   $y(t) = \alpha \cdot x(t-\tau)$.
* '''Attenuation distortions''' occur when $\alpha_1 \ne \alpha_2$. If $\alpha_1 \ne \alpha_2$ and $\tau_1 = \tau_2$, then there are exclusively attenuation distortions.
* '''Phase distortions''' occur when $\tau_1 \ne \tau_2$. If $\tau_1 \ne \tau_2$ and $\alpha_1 = \alpha_2$, then there are exclusively phase distortions. }}

==Exercises==
[[File:Exercises_verzerrungen.png|right]]
*First choose an exercise number.
*An exercise description is displayed.
*Parameter values are adjusted to the respective exercises.
*Click "Hide solution" to display the solution.

Number "0" is a "Reset" button:
*Sets parameters to initial values (when loading the page).
*Displays a "Reset text" to describe the applet further.

{{BlaueBox|TEXT=
'''(1)'''   We consider the parameters $A_1 = 0.8\ {\rm V}, \ A_2 = 0.6\ {\rm V}, \ f_1 = 0.5\ {\rm kHz}, \ f_2 = 1.5\ {\rm kHz}, \ \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ$ for the input signal $x(t)$.
:Calculate the signal's period duration $T_0$ and power $P_x$. Can you read the value for $P_x$ off the applet? }}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}T_0 = \big [\hspace{-0.1cm}\text{ greatest common divisor }(0.5 \ {\rm kHz}, \ 1.5 \ {\rm kHz})\big ]^{-1}\hspace{0.15cm}\underline{ = 2.0 \ {\rm ms}};$

$\hspace{1.85cm} P_x = A_1^2/2 + A_2^2/2 \hspace{0.15cm}\underline{= 0.5 \ {\rm V^2}} = P_\varepsilon\text{, if }\hspace{0.15cm}\underline{k_{\rm M} = 0} \ \Rightarrow \ z(t) \equiv 0$.

{{BlaueBox|TEXT=
'''(2)'''  Vary $\varphi_2$ between $\pm 180^\circ$ while keeping all other parameters from Exercise (1). How does the value of $T_0$ and $P_x$ change?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes:}\hspace{0.2cm}\hspace{0.15cm}\underline{ T_0 = 2.0 \ {\rm ms}; \hspace{0.2cm} P_x = 0.5 \ {\rm V^2}}$.

{{BlaueBox|TEXT=
'''(3)'''   Vary $f_2$ between $0 \le f_2 \le 10\ {\rm kHz}$ while keeping all other parameters from Exercise (1). How does the value of $P_x$ change?}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{No changes if }f_2 \ne 0\text{ and } f_2 \ne f_1\text{:}\hspace{0.3cm} \hspace{0.15cm}\underline{P_x = 0.5 \ {\rm V^2}}\text{.} \hspace{0.2cm} T_0 \text{ changes if }f_2\text{is not a multiple of }f_1$.

$\hspace{1.85cm}\text{If }f_2 = 0\text{:}\hspace{0.2cm} P_x = A_1^2/2 + A_2^2\hspace{0.15cm}\underline{ = 0.68 \ {\rm V^2}}$. $\hspace{3cm}$

$\hspace{1.85cm}\text{If }f_2 = f_1\text{:}\hspace{0.2cm} P_x = [A_1\cos(\varphi_1) + A_2\cos(\varphi_2)]^2/2 + [A_1\sin(\varphi_1) + A_2\sin(\varphi_2)]^2/2 \text{.} $

$\hspace{1.85cm}\text{With } \varphi_1 = 90^\circ, \ \varphi_2 = 30^\circ\text{:}\hspace{0.3cm}\hspace{0.15cm}\underline{ P_x = 0.74 \ {\rm V^2}}\text{.} $

{{BlaueBox|TEXT=
'''(4)'''   Keeping the previous input signal $x(t)$, set following parameters
: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$, $k_{\rm M} = 1 \text{ and } \tau_{\rm M} = 0$ .
:Are there linear distortions? Calculate the received power $P_y$ and the power $P_\varepsilon$ of the differential signal $\varepsilon(t) = z(t) - x(t)$. }}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{ y(t) = 0.5 \cdot x(t- 1\ {\rm ms})}\text{ is only attenuated and delayed, but not distorted.}$

$\hspace{1.85cm}\text{Received power:}\hspace{0.2cm} P_y = (A_1/2)^2/2 + (A_2/2)^2/2\hspace{0.15cm}\underline{ = 0.125 \ {\rm V^2}}\text{. } P_\varepsilon \text{ is significantly larger:} \hspace{0.1cm} \hspace{0.15cm}\underline{P_\varepsilon = 0.625 \ {\rm V^2}}.$

{{BlaueBox|TEXT=
'''(5)'''   With the same settings as in Exercise (4), vary the matching parameters $k_{\rm M} \text{ and } \tau_{\rm M}$. How big is the distortion power $P_{\rm D}$?}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D}\text{ is equal to }P_\varepsilon \text{ when using the ideal matching parameters:} \hspace{0.2cm}k_{\rm M} = 2 \text{ and } \tau_{\rm M}=T_0 - 0.5\ {\rm ms} = 1.5\ {\rm ms}$

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm}z(t) = x(t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\varepsilon(t) = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}P_{\rm D}\hspace{0.15cm}\underline{ = P_\varepsilon = 0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\text{Neither attenuation nor phase distortion.}$

{{BlaueBox|TEXT=
'''(6)'''   The channel parameters are now set to: $\alpha_1 = 0.5, \hspace{0.15cm}\underline{\alpha_2 = 0.2}, \ \tau_1 = \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the Signal-to-Distortion ratio $(\rm SDR) \ \rho_{\rm D}$.}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{ when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 2.24} \text{ and } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.5\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.059 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Attenuation distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 8.5}$.

{{BlaueBox|TEXT=
'''(7)'''   The channel parameters are now set to: $\alpha_1 = \alpha_2 = 0.5, \ \tau_1 \hspace{0.15cm}\underline{= 2\ {\rm ms} }, \ \tau_2 = 0.5\ {\rm ms}$. Calculate the distortion power $P_{\rm D}$ and the the Signal-to-Distortion ratio $\rho_{\rm D}$.}}

$\hspace{1.0cm}\Rightarrow \hspace{0.3cm} P_{\rm D} = P_\varepsilon \text{when using the best matching parameters:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.84} \text{ and } \tau_{\rm M}\hspace{0.15cm}\underline{ = 0.15\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.071 \ {\rm V^2}}$.

$\hspace{1.85cm}\text{Phase distortions only.} \hspace{0.3cm}\text{Signal-to-Distortion-Ratio}\ \hspace{0.15cm}\underline{\rho_{\rm D} = P_x/P_\varepsilon \approx 7}$.

{{BlaueBox|TEXT=
'''(8)'''   The channel parameters are now set to: $\hspace{0.15cm}\underline{\alpha_1 = 0.5} , \hspace{0.15cm}\underline{\alpha_2 = 0.2} , \ \hspace{0.15cm}\underline{\tau_1= 0.5\ {\rm ms} }, \ \hspace{0.15cm}\underline{\tau_2 = 0.3\ {\rm ms} }$. Are there attenuation distortions? Are there phase distortions? How can $y(t)$ be approximated? ''Hint:'' $\cos(3x) = 4 \cdot \cos^3(x) - 3\cdot \cos(x).$}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Both attenuation and phase distortions, because }\alpha_1 \ne \alpha_2\text{ and }\tau_1 \ne \tau_2$.

$\hspace{1.85cm}y(t) = y_1(t) + y_2(t)\ \Rightarrow \ y_1(t) = A_1 \cdot \alpha_1 \cdot \sin[2\pi f_1\ (t- 0.5\ \rm ms)] = -0.4 \ {\rm V} \cdot \cos(2\pi f_1 t)$

$\hspace{1.85cm} y_2(t) = \alpha_2 \cdot x_2(t- \tau_2) \text{ mit }x_2(t) = A_2 \cdot \cos[2\pi f_2\ (t- 30^\circ)] \approx A_2 \cdot \cos[2\pi f_2\ (t- 1/36 \ \rm ms)]$

$\hspace{1.85cm} \Rightarrow \ y_2(t) = 0.12 \ {\rm V} \cdot \cos[2\pi f_2\ (t- 0.328 \ {\rm ms})] \approx -0.12 \ { \rm V} \cdot \cos[2\pi f_2t] $.

$\hspace{1.85cm} \Rightarrow \ y(t) = y_1(t) + y_2(t) \approx -0.4 \ {\rm V} \cdot [\cos(2\pi \cdot f_1\cdot t) + 1/3 \cdot \cos(2\pi \cdot 3 f_1 \cdot t) = -0.533 \ {\rm V} \cdot \cos^3(2\pi f_1 t)$.

{{BlaueBox|TEXT=
'''(9)'''   Using the parameters from Exercise (8), calculate the distortion power $P_{\rm D}$ and the the Signal-to-Distortion ratio $\rho_{\rm D}$.}}

$\hspace{1.0cm}\text{Best possible adaptation:} \hspace{0.2cm}\hspace{0.15cm}\underline{k_{\rm M} = 1.96} \text{, } \hspace{0.15cm}\underline{\tau_{\rm M} = 1.65\ {\rm ms} }\text{:} \hspace{0.2cm}\hspace{0.15cm}\underline{P_{\rm D} = 0.15 \ {\rm V^2} },\hspace{0.1cm}\hspace{0.15cm}\underline{\rho_{\rm D} = 0.500/0.15 \approx 3.3}$.

{{BlaueBox|TEXT=
'''(10)'''  Now we set $A_2 = 0$ and $A_1 = 1\ {\rm V}, \ f_1 = 1\ {\rm kHz}, \ \varphi_1 = 0^\circ$. The channel is a Low-pass of order 1 $\underline{(f_0 = 1\ {\rm kHz})}$. Are there any attenuation and/or phase distortions? Calculate the channel coefficients $\alpha_1$ and $\tau_1$.}}

$\hspace{1.0cm}\text{At only one frequency there are neither attenuation nor phase distortions.}$
$\hspace{1.0cm}\text{Attenuation factor for }f_1=f_0\text{ and }N=1\text{: }\alpha_1 =|H(f = f_1)| = [1+( f_1/f_0)^2]^{-N/2} = 2^{-1/2}= 1/\sqrt{2}\hspace{0.15cm}\underline{=0.707},$
$\hspace{1.0cm}\text{Phase factor for }f_1=f_0\text{ and }N=1\text{: }\tau_1 = N \cdot \arctan( f_1/f_0)/(2 \pi f_1)=\arctan( 1)/(2 \pi f_1) =1/(8f_1) \hspace{0.15cm}\underline{=0.125 \ \rm ms}.$

{{BlaueBox|TEXT=
'''(11)'''   How do the channel parameters change when using a Low-pass of order 2 compared to a Low-pass of order 1 $(f_0 = 1\ {\rm kHz})$?}}

$\hspace{1.0cm}\alpha_1 = 0.707^2 = 0.5$ and $\tau_1 = 2 \cdot 0.125 = 0.25 \ {\rm ms}$.

$\hspace{1.0cm}\text{The signal }y(t)\text{ is only half as big as }x(t)\text{ and is retarded: The cosine turns into a sine function}$.

{{BlaueBox|TEXT=
'''(12)'''   What differences arise when using a High-pass of order 2 compared to a Low-pass of order 2 $(f_0 = 1\ {\rm kHz})$? }}

$\hspace{1.0cm}\text{Since }f_1 = f_0\text{ the attenuation factor }\alpha_1 = 0.5\text{ stays the same and }\tau_1 = -0.25 \ {\rm ms}\text{ which means:}$

$\hspace{1.0cm}\text{The signal }y(t)\text{ is also only half as big as }x(t)\text{ and precedes it: The cosine turns into the Minus–sine function}$.

{{BlaueBox|TEXT=
'''(13)'''   What differences at the signal $y(t)$ can be observed between the Low-pass and the High-pass of order 2 $(f_0 = 1\ {\rm kHz})$ when you start with the initial input signal according to Exercise (1) and continuously raise $f_2$ up to $10 \ \rm kHz$ ? }}

$\hspace{1.0cm}\text{With the Low-pass the second term is increasingly suppressed. For }f_2 = 10 \ {\rm kHz}\text{: }y_{\rm LP}(t) \approx 0.8 \cdot x_1(t-0.3 \ \rm ms).$

$\hspace{1.0cm}\text{With the High-pass however the second term dominates. For }f_2 = 10 \ {\rm kHz}\text{: }y_{\rm HP}(t) \approx 0.2 \cdot x_1(t+0.7 \ {\rm ms}) + x_2(t).$

==Applet Manual==
[[File:Handhabung_verzerrungen.png|center]]
 
    '''(A)'''     Parameter selection for input signal $x(t)$ per slider: Amplitude, frequency, phase values

    '''(B)'''     Preselection for channel parameters per slider: Low-pass or High-pass

    '''(C)'''     Selction of channel parameters per slider: Dämpfungsfaktoren und Phasenlaufzeiten

    '''(D)'''     Selection of channel parameters for High and Low pass: Order$n$, cutoff frequency $f_0$

    '''(E)'''     Selection of matching parameters $k_{\rm M}$ and $\varphi_{\rm M}$

    '''(F)'''     Selection of the signals to be displayed: $x(t)$, $y(t)$, $z(t)$, $\varepsilon(t)$, $\varepsilon^2(t)$

    '''(G)'''     Graphic display of the signals

    '''(H)'''     Enter the time $t_*$ for the numeric output

    '''( I )'''     numeric output of the signal values $x(t_*)$, $y(t_*)$, $z(t_*)$ and $\varepsilon(t_*)$

    '''(J)'''     Numeric output of the main result $P_\varepsilon$

    '''(K)'''     Save and reall parameters

    '''(L)'''     Exercises: Exercise selection, description and solution

    '''(M)'''     Variation possibilities for the graphic display

$\hspace{1.5cm}$Zoom–functions "$+$" (scale up), "$-$" (scale down) und $\rm o$ (reset)

$\hspace{1.5cm}$Move with "$\leftarrow$" (section to the left, ordinate to the right), "$\uparrow$" "$\downarrow$" und "$\rightarrow$"

$\hspace{1.5cm}$'''Other options'':

$\hspace{1.5cm}$Hold shift and scroll: Zoom in on/out of coordinate system,

$\hspace{1.5cm}$Hold shift and left click: Move the coordinate system.

==About the Authors==
This interactive calculation was designed and realized at the [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik] of the [https://www.tum.de/ Technische Universität München].
*The original version was created in 2005 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Bettina_Hirner_.28Diplomarbeit_LB_2005.29|Bettina Hirner]] as part of her Diploma thesis using "FlashMX–Actionscript" (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]] ).
*In 2018 this Applet was redesigned and updated to "HTML5" by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]] as part of his Bachelor's thesis (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) .

==Once again: Open Applet in new Tab==

{{LntAppletLinkEn|linDistortions_en}}         [https://www.lntwww.de/Applets:Lineare_Verzerrungen_periodischer_Signale '''English Applet with German WIKI description''']

Applets:Graphical Convolution

2023-03-26T21:41:04Z

Noah:

{{LntAppletLinkEnDe|convolution_en|convolution}}

== Applet Description==
 
This applet illustrates the convolution operation in the time domain
*between an input pulse  $x(t)$   ⇒   rectangle, triangle, Gaussian, exponential function
*and the impulse response  $h(t)$  of an LTI system with low–pass character  ⇒   slit low–pass, first or second order low–pass, Gaussian low–pass.

For the output signal  $y(t)$  corresponding to the block diagram in  $\text{Example 1}$,  then, as stated in the chapter  [[Applets:Graphical_Convolution#Graphical_Convolution|"Graphical Convolution"]]: 
:$$y( t ) = x(t) * h( t ) = \int_{ - \infty }^{ + \infty } \hspace{-0.15cm}{x( \tau )} \cdot h( {t - \tau } )\hspace{0.1cm}{\rm d}\tau .$$

For causal systems   ⇒    $h(t) \equiv 0$  for  $t < 0$  (examples: slit low–pass as well as first and second order low–pass)   can be written for this also:

:$$y( t ) = \int_{ - \infty }^{ t } \hspace{-0.15cm}{x( \tau )} \cdot h( {t - \tau } )\hspace{0.1cm}{\rm d}\tau .$$

Please note:
*All quantities – also the time $t$ – are to be understood normalized (dimensionless).
*The time functions  $x(t)$,  $h(t)$  and  $y(t)$  cannot assume negative signal values in the program.
*The ''absolute duration''  of a pulse  $y(t)$  is the (continuous) time range for which  $y(t) > 0$. 
*The ''equivalent duration''  of a pulse can be calculated by the rectangle of equal area.

==Theoretical Background==

===Convolution in the time domain===

The  [[Signal_Representation/The_Convolution_Theorem_and_Operation|"convolution theorem"]]  is one of the most important laws of the Fourier transform. We first consider the convolution theorem in the time domain and assume that the spectra of two time functions  $x_1(t)$  and  $x_2(t)$  are known:

:$$X_1 ( f )\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.15cm}x_1( t ),\quad X_2 ( f )\hspace{0.1cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.1cm}x_2 ( t ).$$

Then for the time function of the product  $X_1(f) \cdot X_2(f)$ holds:

:$$X_1 ( f ) \cdot X_2 ( f )\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.15cm}\int_{ - \infty }^{ + \infty } {x_1 ( \tau )} \cdot x_2 ( {t - \tau } )\hspace{0.1cm}{\rm d}\tau.$$

Here  $\tau$  is a formal integration variable with the dimension of a time.

{{BlaueBox|TEXT=
$\text{Definition:}$  The above connection of the time function  $x_1(t)$  and  $x_2(t)$  is called  '''convolution'''  and represents this functional connection with a star:

:$$x_{\rm{1} } (t) * x_{\rm{2} } (t) = \int_{ - \infty }^{ + \infty } {x_1 ( \tau ) } \cdot x_2 ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$

Thus, above Fourier correspondence can also be written as follows:

:$$X_1 ( f ) \cdot X_2 ( f )\hspace{0.15cm}\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\hspace{0.15cm}{ {x} }_{\rm{1} } ( t ) * { {x} }_{\rm{2} } (t ).$$

[[Signal_Representation/The_Convolution_Theorem_and_Operation#Proof_of_the_Convolution_Theorem|$\text{"Proof"}$]]}}

''Note'':   The convolution is  '''commutative'''   ⇒   The order of the operands is interchangeable:   ${ {x}}_{\rm{1}} ( t ) * { {x}}_{\rm{2}} (t ) ={ {x}}_{\rm{2}} ( t ) * { {x}}_{\rm{1}} (t ) $.

[[File:P_ID579__Sig_T_3_4_S1_neu.png|right|frame|For the calculation of signal and spectrum at the LTI output]]
{{GraueBox|TEXT=
$\text{Example 1:}$  Any linear time-invariant (LTI) system can be described both by the frequency response  $H(f)$  and by the impulse response  $h(t)$,  the relationship between these two system quantities also being given by the Fourier transform.

If a signal  $x(t)$  with the spectrum  $X(f)$  is applied to the input, the following applies to the spectrum of the output signal:

:$$Y(f) = X(f) \cdot H(f)\hspace{0.05cm}.$$

With the convolution theorem it is now possible to calculate the output signal also directly in the time domain:

:$$y( t ) = x(t) * h( t ) = \int_{ - \infty }^{ + \infty } \hspace{-0.15cm}{x( \tau )} \cdot h( {t - \tau } )\hspace{0.1cm}{\rm d}\tau = \int_{ - \infty }^{ + \infty } \hspace{-0.15cm} {h( \tau )} \cdot x( {t - \tau } )\hspace{0.1cm}{\rm d}\tau = h(t) * x( t ).$$

From this equation, it is again clear that the convolution operation is  ''commutative''.  }}

===Convolution in the frequency domain===

The duality between the time and frequency domains also allows us to make statements regarding the spectrum of a product signal:

:$$x_1 ( t ) \cdot x_2 ( t )\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,X_1 (f) * X_2 (f) = \int_{ - \infty }^{ + \infty } {X_1 ( \nu )} \cdot X_2 ( {f - \nu })\hspace{0.1cm}{\rm d}\nu.$$

This result can be proved similarly to the  [[Applets:Graphical_Convolution#Convolution_in_the_time_domain|"convolution theorem in the time domain"]].  However, the integration variable  $\nu$  now has the dimension of a frequency.

[[File:P_ID580__Sig_T_3_4_S2_neu.png|right|frame|Convolution in the frequency domain]]
{{GraueBox|TEXT=
$\text{Example 2:}$  The  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_time_domain|"double-sideband amplitude modulation"]]  (DSB-AM) without carrier is described by the sketched model.
*In the time domain representation (blue), the modulated signal  $s(t)$  is the product of the message signal  $q(t)$  and the (normalized) carrier signal  $z(t)$.
*According to the convolution theorem, it follows for the frequency domain (red) that the output spectrum  $S(f)$  is equal to the convolution product of  $Q(f)$  and  $Z(f)$.  }}

===Convolution of a function with a Dirac delta function===

The convolution operation becomes very simple if one of the two operands is a  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal#Dirac_.28delta.29_function_in_frequency_domain|"Dirac delta function"]].  This is equally true for convolution in the time and frequency domain.

As an example, we consider the convolution of a function  $x_1(t)$  with the function

:$$x_2 ( t ) = \alpha \cdot \delta ( {t - T} ) \quad \circ\,\!\!\!-\!\!\!-\!\!\!-\!\!\bullet \quad X_2 ( f )= \alpha \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.03cm}2\hspace{0.03cm}{\rm{\pi }}\hspace{0.01cm}f\hspace{0.01cm}T}.$$

Then for the spectral function of the signal  $y(t) = x_1(t) \ast x_2(t)$  holds:

:$$Y( f ) = X_1 ( f ) \cdot X_2 ( f ) = X_1 ( f ) \cdot \alpha \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.03cm}2\hspace{0.03cm}{\rm{\pi }}\hspace{0.01cm}f\hspace{0.01cm}T} .$$

The complex exponential function leads to a shift by  $T$   ⇒   [[Signal_Representation/Fourier_Transform_Theorems#Shifting_Theorem|"shifting theorem"]], the factor  $\alpha$  to an attenuation  $(\alpha < 1)$  or an amplification  $(\alpha > 1)$. It follows that:

:$$x_1 (t) * x_2 (t) = \alpha \cdot x_1 ( {t - T} ).$$

{{BlaueBox|TEXT=
$\text{In words: }$  The convolution of an arbitrary function with a Dirac delta function at  $t = T$  results in the function shifted to the right by  $T$,  whereby the weighting of the Dirac delta function by the factor  $\alpha$  must still be taken into account.}}

{{GraueBox|TEXT=
$\text{Example 3:}$  A rectangular signal  $x(t)$  is delayed by a running time  $\tau = 3\,\text{ ms}$  and attenuated by a factor  $\alpha = 0.5$  by an LTI system.

[[File:P_ID522__Sig_T_3_4_S3_neu.png|center|frame|Convolution of a rectangle with a Dirac delta function]]

Shift and attenuation can be seen in the output signal  $y(t)$  as well as in the impulse response  $h(t)$.}}

===Graphical Convolution===

This applet assumes the following convolution operation:
[[File:P_ID2723__Sig_T_3_4_programm.png|right|frame|Screen capture of the program "Graphical Convolution" (former version)]]
:$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {x ( \tau )} \cdot h ( {t - \tau } )\hspace{0.1cm}{\rm d}\tau.$$

The solution of the convolution integral is to be done graphically. It is assumed that  $x(t)$  and  $h(t)$  are continuous-time signals.

Then the following steps are required:
#    '''Change''' the  '''time variables'''  of the two functions:       $x(t) \to x(\tau)$,   $h(t) \to h(\tau)$.
#  '''Mirror''' second '''function''':   $h(\tau) \to h(-\tau)$.
#  '''Shift''' mirrored '''function''' by  $t$:   $h(-\tau) \to h(t-\tau)$.
#  '''Multiplication''' of the two functions  $x(\tau)$  and  $h(t-\tau)$.
#  '''Integration'''  over the product with respect to  $\tau$  in the limits from  $-\infty$  to  $+\infty$.

Since the convolution is commutative,  $h(\tau)$  can be mirrored instead of  $x(\tau)$. 

 
The accompanying graphic shows a screen shot of an older version of this applet.
 

[[File:P_ID582__Sig_T_3_4_S4_neu.png|right|frame|Example of a convolution operation: Jump function convolved with exponential function]]
{{GraueBox|TEXT=
$\text{Example 4:}$ 
The procedure of graphical convolution is now explained with a detailed example:
*Let a jump function  $x(t) = \gamma(t)$  be applied to the input of a filter.
*Let the impulse response of the RC low-pass filter be  $h( t ) = {1}/{T} \cdot {\rm{e} }^{ - t/T}.$

The graph shows the input signal  $x(\tau)$ in red, the impulse response  $h(\tau)$ in blue, and the output signal  $y(\tau)$ in gray.
The time axis has already been renamed  $\tau$. 

The output signal can be calculated, for example, according to the following equation:

:$$y(t) = h(t) * x(t) = \int_{ - \infty }^{ + \infty } {h( \tau )} \cdot x( {t - \tau } )\hspace{0.1cm}{\rm d}\tau.$$

A few more remarks on the graphical convolution:
*The output value at  $t = 0$  is obtained by mirroring the input signal  $x(\tau)$,  multiplying this mirrored signal  $x(-\tau)$  by the impulse response  $h(\tau)$,  and integrating over it.
*Since there is no time interval here where both the blue curve  $h(\tau)$  and at the same time the red dashed mirroring  $x(-\tau)$  is not equal to zero, it follows that  $y(t=0)=0$.
*For any other time  $t$,  the input signal must be shifted   ⇒   $x(t-\tau)$, for example corresponding to the green dashed curve for  $t=T$.
*Since in this example also  $x(t-\tau)$  can only take the values  $0$  or  $1$,  the integration  $($generally from  $\tau_1$  to  $\tau_2)$  becomes simple and we obtain with  $\tau_1 = 0$  and  $\tau_2 = t$ :
:$$y( t) = \int_0^{\hspace{0.05cm} t} {h( \tau)}\hspace{0.1cm} {\rm d}\tau = \frac{1}{T}\cdot\int_0^{\hspace{0.05cm} t} {{\rm{e}}^{ - \tau /T } }\hspace{0.1cm} {\rm d}\tau = 1 - {{\rm{e}}^{ - t /T } }.$$

The sketch is valid for  $t=T$  and leads to the initial value  $y(t=T) = 1 – 1/\text{e} \approx 0.632$.}}

==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the exercise to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
* An exercise description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*Both the input signal  $x(t)$  and the impulse response  $h(t)$  of the filter are are normalized, dimensionless and energy-limited ("time-limited pulses").

{{BlueBox|TEXT=
'''(1)'''   Select the following parameters:  $\text{Gaussian pulse: }A_x = 1, \ \Delta t_x= 1, \ \tau_x = 1;     \text{ Impulse response according to 2nd order low-pass: } \Delta t_h= 1$.
          Interpret the displayed graphs.  What is the maximum output value  $y_{\rm max}$?  At what time  $t_{\rm max}$  does  $y_{\rm max}$  occur? }}

* After renaming:  Input signal  $x(\tau)$   ⇒   red curve,   impulse response  $h(\tau)$   ⇒   blue curve,  after mirroring  $h(-\tau)$   ⇒   green curve.
* Shifting the green curve by  $t$  to the right, we get  $h(t-\tau)$.  The output signal  $y(t)$  is obtained by multiplication and integration with respect to  $\tau$:

:$$y (t) = \int_{ - \infty }^{ +\infty } {x ( \tau ) } \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau = \int_{ - \infty }^{ t } {x ( \tau ) } \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau .$$
* The output pulse  $y(t)$  is asymmetric in the present case;  the maximum output value  $y_{\rm max}\approx 0.67$  occurs at  $t_{\rm max}\approx 1.5$.

{{BlueBox|TEXT=
'''(2)'''   What changes if we increase the equivalent pulse duration of  $h(t)$  to  $\Delta t_h= 1.5$? }}

* $y_{\rm max}\approx 0.53$  now occurs at  $t_{\rm max}\approx 1.75$.  Due to the less favorable (wider) impulse response, the input pulse is more deformed.
* In a digital communication system, this would result in stronger  "intersymbol interferences".

{{BlueBox|TEXT=
'''(3)'''   Now select the symmetric  $\text{rectangular pulse: }A_x = 1, \ \Delta t_x= 1, \ \tau_x = 0$  and the  $\text{rectangular impulse response}$  of the low-pass filter:  $\Delta t_h= 1$.
          Interpret the convolution result.  What is the maximum output value  $y_{\rm max}$?  At what times is  $y(t)>0$?  Does  $h(t)$  describe a causal system? }}

* The convolution of two rectangles with respective durations  $1$  yields a triangle with absolute duration  $2$  ⇒   equivalent pulse duration  $\Delta t_y= 1$.
* $y(t)$  is different from zero in the range from  $-0.5$  to  $+1.5$ .  The pulse maximum  $y_{\rm max} = 1$  is at  $t_{\rm max} = +0.5$.
* $h(t)$  describes a causal system, since  $h(t) \equiv 0$  for  $t < 0$.  That means:  The "effect"  $y(t)$  does not come before the "cause"  $x(t)$.

{{BlueBox|TEXT=
'''(4)'''   What changes if we increase the equivalent pulse duration of  $h(t)$  to  $\Delta t_h= 2$ ? }}

* The convolution of two rectangles of different widths results in a trapezoid, here between  $-0.5$  and  $+2.5$ ⇒   equivalent pulse duration  $\Delta t_y= 2$.
* The maximum  $y_{\rm max} = 0.5$  occurs in the range  $0.5 \le t \le 1.5$.  Nothing changes with respect to causality.

{{BlueBox|TEXT=
'''(5)'''   Now select the (unsymmetrical)  $\text{rectangular pulse: }A_x = 1, \ \Delta t_x= 1, \ \tau_x = 0.5$  and the  $\text{impulse response of a 1st order low-pass: }\Delta t_h= 1$.
          Interpret the results.  What is the value of  $y_{\rm max}$?  At what times is  $y(t)>0$ ?  Does  $h(t)$  describe a causal system? }}

* $h(t)$  has an exponentially decreasing curve for  $t > 0$.  It always applies:   $y(t>0) > 0$,  but the signal values can become very small.
* $y_{\rm max} = 0.63$  occurs for  $t_{\rm max} = +1$.  For  $ t < t_{\rm max}$ the progression is exponentially increasing, for  $ t > t_{\rm max}$  exponentially decreasing.
* The 1st order low-pass can be realized with a resistor and a capacitor.  Any realizable system is causal per se.

{{BlueBox|TEXT=
'''(6)'''   Select as in  $(3)$  the  $\text{rectangular impulse response}$  of the low-pass filter:  $\Delta t_h= 1$.  With which  $x(t)$  results the same  $y(t)$  as for  $(5)$?}}

* The signal  $y(t)$  in  $(5)$  resulted as a convolution between the rectangular input  $x(t)$  and the exponential function  $h(t)$.
* Since the convolution operation is commutative, the same result is obtained with the exponential function  $x(t)$  and the rectangular function  $h(t)$.
* Thus, the correct setting for the input signal  $x(t)$  is the  $\text{exponential pulse }$with  $ A_x = 1, \ \Delta t_x= 1, \ \tau_x = 0$ .

{{BlueBox|TEXT=
'''(7)'''   For the remainder of the exercises, we consider the Gaussian low-pass.  The equivalent duration of the impulse response  $h(t)$  should first be  $\Delta t_h= 0.8$.
        Analyze and interpret this  "system"  in terms of causality and the resulting distortions for the rectangular pulse. }}

* The low-pass is not causal (and therefore non–realizable):  For  $t < 0$:  $h(t) \equiv 0$  does not hold. Suitable model if infinite delay is ignored.
* The larger  $\Delta t_h$  is, the wider the output pulse and the stronger the degradation of a digital system due to intersymbol interference.
* The frequency response  $H(f)$  is the Fourier transform of  $h(t)$. The larger  $\Delta t_h$  is, the smaller  $\Delta f_h = 1/\Delta t_h$   ⇒   The system is more narrowband.

{{BlueBox|TEXT=
'''(8)'''   Now select  $\text{Gaussian pulse: }A_x = 1, \ \Delta t_x= 1.5, \ \tau_x = 0$  and  $\text{Gaussian low-pass: }\Delta t_h= 2$.  What is the course of the output pulse  $y(t)$?
          What are the equivalent duration  $\Delta t_y$  of the output pulse and the maximum output value  $y_{\rm max}$?  At what time  $t_{\rm max}$  does  $y_{\rm max}$  occur? }}

* $y(t)$  is (exactly) Gaussian, too.   Mnemonic:  $\text{"Gaussian convolved with Gaussian gives always Gaussian"}$.
* Equivalent duration:  $\Delta t_y =\sqrt{\Delta t_x^2+ \Delta t_h^2} = 2.5$.  Pulse maximum  $($at  $t=0)$:  $y_{\rm max} = A_x \cdot \Delta t_x/\Delta t_y = 1 \cdot 1.5/2.5 = 0.6$.

{{BlueBox|TEXT=
'''(9)'''   Now select  $\text{Triangular pulse: }A_x = 1, \ \Delta t_x= 1.5, \ \tau_x = 0$  and  $\text{Gaussian low–pass: }\Delta t_h= 2$.  What is the course of the output pulse  $y(t)$?
        What is the equivalent duration  $\Delta t_y$  of the output pulse and the maximum output value  $y_{\rm max}$?  At what time  $t_{\rm max}$  does  $y_{\rm max}$  occur?}}

* $y(t)$  is nearly Gaussian, but not exactly.  Mnemonic:  $\text{"Gaussian convolved with non–Gaussian never gives exactly Gaussian"}$.
* The characteristics of the output pulse  $y(t)$  differ only slightly from  $(8)$:  $\Delta t_y \approx 2.551$,  $y_{\rm max} \approx 0.588$.

==Applet Manual==
 
[[File:Anleitung_Faltung_2.png|right|frame|Screen shot  (German version)]]

    '''(A)'''     Selection:   Shape of the input pulse  $x(t)$

    '''(B)'''     Parameter input for the input pulse  $x(t)$

    '''(C)'''     Selection:   Shape of the impulse response  $h(t)$  of the low-pass system

    '''(D)'''     Parameter input for the impulse response  $h(t)$

    '''(E)'''     Control panel  (Start;   Pause/Continue   ;  Step >   ;  Step <  ;  Reset)

    '''(F)'''     Output the initial value  $y(t)$  at the continuous time  $t$

    '''(G)'''     Maximum value  $y_{\rm max} = y(t_{\rm max})$  and equivalent width $\Delta\hspace{0.03cm} t_y$

                 After renaming the abscissa:   $t \ \to \ \tau$:

    '''(H)'''     Representation of  $x(\tau)$  ⇒   red static curve.

    '''(I)'''       Representation of  $h(\tau)$  ⇒  blue curve  and   $h(t-\tau)$  ⇒   green curve                 $($this is shifted to the right with the motion parameter   $t$ $)$

    '''(J)'''       Plot of  $x(\tau) \cdot h(t - \tau)$  ⇒   purple curve, dynamic with  $t$

    '''(K)'''      Successive representation of the output signal  $y(t)$  ⇒   brown curve

    '''(L)'''      Area for exercise execution:   Exercise selection.

    '''(M)'''      Exercise execution:   Area for exercise description

    '''(N)'''      Exercise execution:   Area for the sample solution
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2006 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Markus_Elsberger_.28Diplomarbeit_LB_2006.29|Markus Elsberger]]  as part of his bachelor thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  Many thanks.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|convolution_en|convolution}}

Applets:Eye Pattern and Worst-Case Error Probability

2023-03-26T21:34:31Z

Noah:

{{LntAppletLinkEnDe|eyeDiagram_en|eyeDiagram}}

==Applet Description==
 
The applet illustrates the eye pattern for different encodings 
* binary  (redundancy-free), 
*quaternary  (redundancy-free),
*pseudo–ternary:  (AMI and duobinary) 

and for various reception concepts 
*Matched Filter receiver, 
*CRO Nyquist system, 
*Gaussian low-pass filter.

The last reception concept leads to intersymbol interference, that is:  Neighboring symbols interfere with each other in symbol decision.

Such intersymbol interferences and their influence on the error probability can be captured and quantified very easily by the "eye pattern".  But also for the other two (without intersymbol interference) systems important insights can be gained from the graphs.

Furthermore, the most unfavorable ("worst case") error probability  
:$$p_{\rm U} = {\rm Q}\left[ö_{\rm norm}/\sigma_{\rm norm} \right ]$$

is output, which for binary Nyquist systems is identical to the mean error probability  $p_{\rm M}$  and represents a suitable upper bound for the other system variants:  $p_{\rm U} \ge p_{\rm M}$.

In the  $p_{\rm U}$–equation mean:
*${\rm Q}(x)$  is the  [[Applets:Complementary_Gaussian_Error_Functions|"Complementary Gaussian Error Function"]].  The normalized eye opening can have values between  $0 \le ö_{\rm norm} \le 1$  .
*The maximum value  $(ö_{\rm norm} = 1)$  applies to the binary Nyquist system and  $ö_{\rm norm}=0$  represents a "closed eye".
*The normalized detection noise rms value  $\sigma_{\rm norm}$  depends on the adjustable parameter  $10 \cdot \lg \ E_{\rm B}/N_0$  but also on the coding and the receiver concept.

==Theoretical Background==
 
 
=== System description and prerequisites===

The binary baseband transmission model outlined below applies to this applet. First, the following prerequisites apply:
*The transmission is binary, bipolar, and redundancy-free with bit rate  $R_{\rm B} = 1/T$, where  $T$  is the symbol duration.
*The transmitted signal  $s(t)$  is equal to  $ \pm s_0$   at all times  $t$  ⇒   The basic transmission pulse  $g_s(t)$  is NRZ–rectangular with amplitude  $s_0$  and pulse duration  $T$.

*Let the received signal be  $r(t) = s(t) + n(t)$, where the AWGN term  $n(t)$  is characterized by the (one-sided) noise power density  $N_0$. 
*Let the channel frequency response be best possible (ideal) and need not be considered further:  $H_{\rm K}(f) =1$.
*The receiver filter with the impulse response  $h_{\rm E}(t)$  forms the detection signal  $d(t) = d_{\rm S}(t)+ d_{\rm N}(t)$ from  $r(t)$. 
* This is evaluated by the decision with the decision threshold  $E = 0$  at the equidistant times  $\nu \cdot T$. 
*A distinction is made between the signal component  $d_{\rm S}(t)$  – originating from  $s(t)$  – and the noise component  $d_{\rm N}(t)$,  whose cause is the AWGN noise  $n(t)$. 
*$d_{\rm S}(t)$  can be represented as a weighted sum of weighted basic detection pulses  $T$,  each shifted by  $g_d(t) = g_s(t) \star h_{\rm E}(t)$. 

*To calculate the (average) error probability, one further needs the variance  $\sigma_d^2 = {\rm E}\big[d_{\rm N}(t)^2\big]$  of the detection noise component (for AWGN noise).

===Optimal intersymbol interference-free system – matched filter receiver===

The minimum error probability results for the case considered here  $H_{\rm K}(f) =1$  with the matched filter receiver, i.e. when  $h_{\rm E}(t)$  is equal in shape to the NRZ basic transmission pulse  $g_s(t)$.  The rectangular impulse response  $h_{\rm E}(t)$  then has duration  $T_{\rm E} = T$  and height  $1/T$.

[[File:Auge_1neu.png|center|frame|Binary baseband transmission system;  the sketch for  $h_{\rm E}(t)$  applies only to the matched filter receiver ]]
[[File:EN_Dig_T_1_4_S1_v2.png|center|frame|Binary baseband transmission system;  the sketch for  $h_{\rm E}(t)$  applies only to the matched filter receiver ]]

*The basic detection pulse  $g_d(t)$  is triangular with maximum  $s_0$  at  $t=0$ ;  $g_d(t)=0$  for  $|t| \ge T$. Due to this tight temporal constraint, there is no intersymbol interference   ⇒   $d_{\rm S}(t = \nu \cdot T) = \pm s_0$   ⇒   the distance of all useful samples from the threshold  $E = 0$  is always  $|d_{\rm S}(t = \nu \cdot T)| = s_0$.
*The detection noise power for this constellation is:
:$$\sigma_d^2 = N_0/2 \cdot \int_{-\infty}^{+\infty} |h_{\rm E}(t)|^2 {\rm d}t = N_0/(2T)=\sigma_{\rm MF}^2.$$
*For the (average) error probability, using the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|"Complementary Gaussian Error Function"]]  ${\rm Q}(x)$ :
:$$p_{\rm M} = {\rm Q}\left[\sqrt{{s_0^2}/{\sigma_d^2}}\right ] = {\rm Q}\left[\sqrt{{2 \cdot s_0^2 \cdot T}/{N_0}}\right ] = {\rm Q}\left[\sqrt{2 \cdot E_{\rm B}/ N_0}\right ].$$

The applet considers this case with the settings  "after gap–low-pass"  as well as  $T_{\rm E}/T = 1$. The output values are with regard to later constellations
*the normalized eye opening  $ö_{\rm norm} =1$   ⇒   this is the maximum possible value,
*the normalized detection noise rms value (equal to the square root of the detection noise power)  $\sigma_{\rm norm} =\sqrt{1/(2 \cdot E_{\rm B}/ N_0)}$  as well as
*the worst-case error probability  $p_{\rm U} = {\rm Q}\left[ö_{\rm norm}/\sigma_{\rm norm} \right ]$   ⇒   for intersymbol interference-free systems,   $p_{\rm M}$  and   $p_{\rm U}$  agree.

$\text{Differences in the multi-level systems}$
*There are  $M\hspace{-0.1cm}-\hspace{-0.1cm}1$ eyes and just as many thresholds   ⇒   $ö_{\rm norm} =1/(M\hspace{-0.1cm}-\hspace{-0.1cm}1)$  ⇒   $M=4$:  quaternary system,  $M=3$:  AMI code, duobinary code.
*The normalized detection noise rms value  $\sigma_{\rm norm}$  is smaller by a factor of  $\sqrt{5/9} \approx 0.745$  for the quaternary system than for the binary system.
*For the AMI code and the duobinary code, this improvement factor, which goes back to the smaller  $E_{\rm B}/ N_0$,  has the value  $\sqrt{1/2} \approx 0.707$.

 
===Nyquist system with raised cosine overall frequency response===

[[File:Auge_2_neu.png|right|frame|Raised cosine overall frequency response ]]

We assume that the overall frequency response between the Dirac-shaped source to the decision has the shape of a  [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Raised-cosine_low-pass_filter|"raised cosine low-pass"]]    ⇒   $H_{\rm S}(f)\cdot H_{\rm E}(f) = H_{\rm CRO}(f)$ .
*The rolloff of  $H_{\rm CRO}(f)$  is point symmetric about the Nyquist frequency  $1/(2T)$. The larger the rolloff factor  $r_{ \hspace {-0.05cm}f}$,  ithe flatter the Nyquist slope.
*The basic detection pulse  $g_d(t) = s_0 \cdot T \cdot {\mathcal F}^{-1}\big[H_{\rm CRO}(f)\big]$  has zeros at times  $\nu \cdot T$  independent of  $r_{ \hspace {-0.05cm}f}$.  There are further zero crossings depending on  $r_{ \hspace {-0.05cm}f}$.  For the pulse holds:
:$$g_d(t) = s_0 \hspace{-0.05cm}\cdot\hspace{-0.05cm} {\rm sinc}( t/T )\hspace{-0.05cm}\cdot\hspace{-0.05cm}\frac {\cos(\pi \cdot r_{\hspace{-0.05cm}f} \cdot t/T )}{1 - (2 \cdot
r_{\hspace{-0.05cm}f} \cdot t/T)^2}.$$
*It follows:  As with the matched filter receiver, the eye is maximally open   ⇒   $ö_{\rm norm} =1$.

[[File:EN_Dig_T_1_4_S6.png|right|frame|Optimizing the rolloff factor]]
Let us now consider the noise power before the decision. For this holds:

:$$\sigma_d^2 = N_0/2 \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 {\rm d}f = N_0/2 \cdot \int_{-\infty}^{+\infty} \frac{|H_{\rm CRO}(f)|^2}{|H_{\rm S}(f)|^2} {\rm d}f.$$

The graph shows the power transfer function  $|H_{\rm E}(f)|^2$  for three different rolloff factors

* $r_{ \hspace {-0.05cm}f}=0$   ⇒   green curve,
* $r_{ \hspace {-0.05cm}f}=1$   ⇒   red curve,
* $r_{ \hspace {-0.05cm}f}=0.8$   ⇒   blue curve.

The areas under these curves are each a measure of the noise power  $\sigma_d^2$.  The rectangle with a gray background marks the smallest value  $\sigma_d^2 =\sigma_{\rm MF}^2$, which also resulted with the matched filter receiver.
 
One can see from this plot:
*The rolloff factor  $r_{\hspace{-0.05cm}f} = 0$  (rectangular frequency response) leads to  $\sigma_d^2 =K \cdot \sigma_{\rm MF}^2$  with  $K \approx 1.5$ despite the very narrow receiver filter, since  $|H_{\rm E}(f)|^2$  increases steeply as  $f$  increases. The reason for this noise power increase is the  $\rm sinc^2(f T)$  function in the denominator, which is required to compensate for the  $|H_{\rm S}(f)|^2$–decay. 
* Since the area under the red curve is smaller than that under the green curve,  $r_{\hspace{-0.05cm}f} = 1$  leads to a smaller noise power despite a spectrum twice as wide:  $K \approx 1.23$.  For  $r_{\hspace{-0.05cm}f} \approx 0.8$, a slightly better value results. For this, the best possible compromise between bandwidth and excess noise is achieved.
*The normalized detection noise rms value is thus for the rolloff factor  $r_{ \hspace {-0.05cm}f}$:   $\sigma_{\rm norm} =\sqrt{K(r_f)/(2 \cdot E_{\rm B}/ N_0)}$. 
*Again, the worst-case error probability  $p_{\rm U} = {\rm Q}\left[ö_{\rm norm}/\sigma_{\rm norm} \right ]$   coincides exactly with the mean error probability  $p_{\rm M}$. 

$\text{Differences in the multi-level systems}$

All remarks in section $2.2$ apply in the same way to the "Nyquist system with raised cosine total frequency response".

===Intersymbol interference system with Gaussian receiver filter===

[[File:Auge_4.png|right|frame|System with Gaussian receiver filter]]

We start from the block diagram sketched on the right. Further it shall be valid:
*Rectangular NRZ basic transmission pulse  $g_s(t)$  with height  $s_0$  and duration  $T$:
:$$H_{\rm S}(f) = {\rm sinc}(f T).$$
*Gaussian receiver filter with cutoff frequency  $f_{\rm G}$:
:$$H_{\rm E}(f) = H_{\rm G}(f) = {\rm e}^{- \pi \hspace{0.05cm}\cdot \hspace{0.03cm} f^2/(2\hspace{0.05cm}\cdot \hspace{0.03cm}f_{\rm G})^2 } \hspace{0.2cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ
\hspace{0.2cm}h_{\rm E}(t) = h_{\rm G}(t) = {\rm e}^{- \pi \cdot (2\hspace{0.05cm}\cdot \hspace{0.03cm}
f_{\rm G}\hspace{0.05cm}\cdot \hspace{0.02cm} t)^2}
\hspace{0.05cm}.$$

Based on the assumptions made here, the following applies to the basic detection pulse:

[[File:Auge_5_neu.png|right|frame|Frequency response and impulse response of the receiver filter]]
:$$g_d(t) = s_0 \cdot T \cdot \big [h_{\rm S}(t) \star h_{\rm G}(t)\big ] = 2 f_{\rm G} \cdot s_0 \cdot \int_{t-T/2}^{t+T/2}
{\rm e}^{- \pi \hspace{0.05cm}\cdot\hspace{0.05cm} (2 \hspace{0.05cm}\cdot\hspace{0.02cm}
f_{\rm G}\hspace{0.05cm}\cdot\hspace{0.02cm} \tau )^2} \,{\rm d} \tau \hspace{0.05cm}.$$

The integration leads to the result:

:$$g_d(t) = s_0 \cdot \big [ {\rm Q} \left ( 2 \cdot \sqrt {2 \pi}
\cdot f_{\rm G}\cdot ( t - {T}/{2})\right )- {\rm Q} \left (
2 \cdot \sqrt {2 \pi} \cdot f_{\rm G}\cdot ( t + {T}/{2}
)\right ) \big ],$$

using the complementary Gaussian error function

:$${\rm Q} (x) = \frac{\rm 1}{\sqrt{\rm 2\pi}}\int_{\it
x}^{+\infty}\rm e^{\it -u^{\rm 2}/\rm 2}\,d {\it u}
\hspace{0.05cm}.$$

The module  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]]  provides the numerical values of  ${\rm Q} (x)$. 
*This basic detection pulse causes  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Definition_of_the_term_.22Intersymbol_Interference.22|"intersymbol interference"]].
*This is understood to mean that the symbol decision is influenced by the spurs of neighboring pulses. While in intersymbol interference free transmission systems each symbol is falsified with the same probability – namely the mean error probability  $p_{\rm M}$  – there are favorable symbol combinations with the falsification probability  ${\rm Pr}(v_{\nu} \ne q_{\nu}) < p_{\rm M}$.
*In contrast, other symbol combinations increase the falsification probability significantly.

[[File:Auge_6.png|right|frame|Binary eye $($Gaussian low-pass,  $f_{\rm G}/R_{\rm B} = 0.35)$.]]
The intersymbol interferences can be captured and analyzed very easily by the so-called  '''eye diagram'''.  These are the focus of this applet. All important information can be found  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram|"here"]].
*The eye diagram is created by drawing all sections of the detection useful signal  $d_{\rm S}(t)$  of length  $2T$  on top of each other. You can visualize the formation in the program with "single step".

* A measure for the strength of the intersymbol interference is the ''vertical eye opening''. For the symmetric binary case, with  $g_\nu = g_d(\pm \nu \cdot T)$  and appropriate normalization:
:$$ ö_{\rm norm} = g_0 -2 \cdot (|g_1| + |g_2| + \text{...}).$$
* With larger cutoff frequency, the pulses interfere less and  $ ö_{\rm norm}$  increases continuously. At the same time, with larger  $f_{\rm G}/R_{\rm B}$,  the (normalized) detection noise rms value also becomes larger:
:$$ \sigma_{\rm norm} = \sqrt{\frac{f_{\rm G}/R_{\rm B}}{\sqrt{2} \cdot E_{\rm B}/N_{\rm 0}}}.$$
*The worst-case error probability  $p_{\rm U} = {\rm Q}\left[ö_{\rm norm}/\sigma_{\rm norm} \right ]$   ⇒   "Worst Case" is usually significantly higher than the mean error probability  $p_{\rm M}$.

$\text{Differences in the redundancy-free quaternary system}$
*For  $M=4$,  other basic pulse values result. ''Example'':     With  $M=4, \ f_{\rm G}/R_{\rm B}=0.4$  basic pulse values  $g_0 = 0.955, \ g_1 = 0.022$  are identical with  $M=2, \ f_{\rm G}/R_{\rm B}=0.8$.
* There are now three eye openings and just as many thresholds.  The equation for the normalized eye opening is now:   $ ö_{\rm norm} = g_0/3 -2 \cdot (|g_1| + |g_2| + \text{...}).$
*The normalized detection noise rms  $\sigma_{\rm norm}$  is again a factor of  $\sqrt{5/9} \approx 0.745$  smaller for the quaternary system than for the binary system.

===Pseudo-ternary codes===

In symbolwise coding, each incoming source symbol  $q_\nu$  generates an encoder symbol  $c_\nu$  that depends not only on the current input symbol  $q_\nu$  but also on the  $N_{\rm C}$  preceding symbols  $q_{\nu-1}$, ... , $q_{\nu-N_{\rm C}} $.   $N_{\rm C}$  is referred to as the ''order''  of the code.  It is typical for a symbolwise coding that
[[File:Dig_T_2_4_S1_v1_neu.png|right|frame|Block diagram and equivalent circuit of a pseudo-ternary encoder|class=fit]]

*the symbol duration  $T$  of the encoded signal (and of the transmitted signal) coincides with the bit duration  $T_{\rm B}$  of the binary source signal, and
*coding and decoding do not lead to major time delays, which are unavoidable when block codes are used. 

Special importance has ''pseudo-ternary codes''   ⇒   level number  $M = 3$, which can be described by the block diagram according to the left graphic. In the right graphic an equivalent circuit is given, which is very suitable for an analysis of these codes. More details can be found in the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes|"$\rm LNTwww$ theory section"]].  Conclusion:

*Recoding from binary  $(M_q = 2)$  to ternary  $(M = M_c = 3)$:
:$$q_\nu \in \{-1, +1\},\hspace{0.5cm} c_\nu \in \{-1, \ 0, +1\}\hspace{0.05cm}.$$

*The relative code redundancy is the same for all pseudo-ternary codes:
:$$ r_c = 1 -1/\log_2\hspace{0.05cm}(3) \approx 36.9 \%\hspace{0.05cm}.$$

Based on the code parameter  $K_{\rm C}$,  different first-order pseudo-ternary codes  $(N_{\rm C} = 1)$  are characterized.

[[File:Auge_16.png|right|frame|Signals in AMI coding|class=fit]]
$\Rightarrow \ \ K_{\rm C} = 1\text{: AMI code}$  (from:   ''Alternate Mark Inversion'')

The graph shows the binary source signal  $q(t)$ at the top. Below are shown:
* the likewise binary signal  $b(t)$  after the pre-encoder, and
* the encoded signal  $c(t) = s(t)$  of the AMI code.

One can see the simple AMI coding principle:
*Each binary value  "–1"  of $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary amplitude coefficient  $a_\nu = 0$.  
*The binary value  "+1"  of  $q(t)$   ⇒   symbol  $\rm H$  is alternately represented by  $a_\nu = +1$  and  $a_\nu = -1$.  

This ensures that there are no long  "+1"–  or  "–1" sequences in the AMI-encoded signal, which would be problematic for an DC signal-free channel. 
 
[[File:Auge_16a.png|left|frame|class=fit]]

The eye diagram is shown on the left.
::* There are two eye openings and two thresholds.
::* The normalized eye opening is  $ö_{\rm norm}= 1/2 \cdot (g_0 -3 \cdot g_1)$, where  $g_0 = g_d(t=0)$  denotes the main value of the basic detection pulse and  $g_1 = g_d(t=\pm T)$  denotes the relevant precursors and postcursors that vertically limit the eye.

::* The normalized eye opening is thus significantly smaller than for the comparable binary system   ⇒   $ö_{\rm norm}= g_0 -2 \cdot g_1$.
::* The normalized noise rms  $\sigma_{\rm norm}$  is smaller than for the comparable binary system by a factor of  $\sqrt{1/2} \approx 0.707$. 
 
[[File:Auge_17.png|right|frame|Signals in duobinary coding|class=fit]]

$\Rightarrow \ \ K_{\rm C} = -1\text{: duobinary code}$ 

From the right graph with the signal curves one recognizes:
*Here, any number of symbols of the same polarity  ("+1" or "–1")  can directly follow each other   ⇒   the duobinary code is not free of DC signals. 
*In contrast, the alternating sequence  " ... , +1, –1, +1, –1, +1, ... "  does not occur, which is particularly disturbing with regard to intersymbol interference.
* Also the duobinary encoded sequence consists to 50% of zeros. The enhancement factor due to the smaller  $E_{\rm B}/ N_0$  is equal to  $\sqrt{1/2} \approx 0.707$, as in the AMI code.

[[File:Auge_17a.png|left|frame|class=fit]]
 
The eye diagram is shown on the left.
::* There are again two "eyes" and two thresholds.
::* The eye opening is   $ö_{\rm norm}= 1/2 \cdot (g_0 - g_1)$.
*$ö_{\rm norm}$  is thus larger than in the AMI code and also as in the comparable binary system.
*A disadvantage compared to the AMI code, however, is that it is not DC signal-free.

==Exercises==

* First select the number  $(1,\ 2, \text{...})$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 

{{BlueBox|TEXT=
'''(1)'''  Explain the occurrence of the eye pattern for  $M=2 \text{, Gaussian low-pass, }f_{\rm G}/R_{\rm B} = 0.48$. For this, select "step–by–step". }}

::* The eye pattern is obtained by dividing the "useful" signal  $d_{\rm S}(t)$  (without noise) into pieces of duration  $2T$  and drawing these pieces on top of each other.
::* In  $d_{\rm S}(t)$  all  "five bit combinations"  must be contained   ⇒   at least  $2^5 = 32$  pieces   ⇒   at most  $32$  distinguishable lines.
::* The eye pattern evaluates the transient response of the signal.  The larger the (normalized) eye opening, the smaller are the intersymbol interferences.

{{BlueBox|TEXT=
'''(2)'''  Same setting as in  $(1)$. In addition,  $10 \cdot \lg \ E_{\rm B}/N_0 = 10 \ {\rm dB}$.  Evaluate the output characteristics  $ö_{\rm norm}$,  $\sigma_{\rm norm}$,  and  $p_{\rm U}$.}}

::* $ö_{\rm norm}= 0.542$  indicates that symbol detection is affected by adjacent pulses. For binary systems without intersymbol interference:  $ö_{\rm norm}= 1$.
::* The eye opening indicates only the signal  $d_{\rm S}(t)$  without noise.  The noise influence is captured by  $\sigma_{\rm norm}= 0.184$ . This value should be as small as possible.
::* The error probability  $p_{\rm U} = {\rm Q}(ö_{\rm norm}/\sigma_{\rm norm}\approx 0.16\%)$  refers solely to the "worst-case sequences", for Gaussian low–pass e.g.  $\text{...}\ , -1, -1, +1, -1, -1, \text{...}$.
::* Other sequences are less distorted   ⇒   the mean error probability  $p_{\rm M}$  is (usually) significantly smaller than $p_{\rm U}$  (describing the worst case).

{{BlueBox|TEXT=
'''(3)'''  The last settings remain.  With which  $f_{\rm G}/R_{\rm B}$  value does the worst case error probability  $p_{\rm U}$  become minimal?  Consider also the eye pattern.}}

::* The minimum value  $p_{\rm U, \ min} \approx 0.65 \cdot 10^{-4}$  is obtained for  $f_{\rm G}/R_{\rm B} \approx 0.8$, and this is almost independent of the setting of  $10 \cdot \lg \ E_{\rm B}/N_0$.
::* The normalized noise rms value does increase compared to the experiment  $(2)$  from  $\sigma_{\rm norm}= 0.168$  to  $\sigma_{\rm norm}= 0.238$.
::* However, this is more than compensated by the larger eye opening  $ö_{\rm norm}= 0.91$  compared to  $ö_{\rm norm}= 0.542$  $($magnification factor $\approx 1.68)$.

{{BlueBox|TEXT=
'''(4)'''  Which cutoff frequencies  $(f_{\rm G}/R_{\rm B})$  result in a completely inadequate error probability  $p_{\rm U} \approx 50\%$ ? Look at the eye pattern again  ("Overall view").}}

::* For  $f_{\rm G}/R_{\rm B}<0.28$  we get a "closed eye"  $(ö_{\rm norm}= 0)$  and thus a worst case error probability on the order of  $50\%$.
::* The decision on unfavorably framed bits must then be random, even with low noise  $(10 \cdot \lg \ E_{\rm B}/N_0 = 16 \ {\rm dB})$.

{{BlueBox|TEXT=
'''(5)'''  Now select the settings  $M=2 \text{, Matched Filter receiver, }T_{\rm E}/T = 1$,  $10 \cdot \lg \ E_{\rm B}/N_0 = 10 \ {\rm dB}$  and  "Overall view". Interpret the results. }}

::* The basic detection impulse  $g_d(t)$  is triangular and the eye is "fully open".  Consequently, the normalized eye opening is  $ö_{\rm norm}= 1.$
::* From  $10 \cdot \lg \ E_{\rm B}/N_0 = 10 \ {\rm dB}$  it follows $E_{\rm B}/N_0 = 10$   ⇒   $\sigma_{\rm norm} =\sqrt{1/(2\cdot E_{\rm B}/ N_0)} = \sqrt{0.05} \approx 0.224 $  ⇒   $p_{\rm U} = {\rm Q}(4.47) \approx 3.9 \cdot 10^{-6}.$
::* This  $p_{\rm U}$ value is by a factor  $15$  better than in  $(3)$.   But:  For  $H_{\rm K}(f) \ne 1$  this so–called "Matched Filter Receiver" is not applicable.

{{BlueBox|TEXT=
'''(6)'''  Same settings as in  $(5)$.  Now vary  $T_{\rm E}/T$  in the range between  $0.5$  and  $1.5$.  Interpret the results.}}

::* For  $T_{\rm E}/T < 1$ ,  $ö_{\rm norm}= 1$  still holds.  But  $\sigma_{\rm norm}$  becomes larger, for example  $\sigma_{\rm norm} = 0.316$  for  $T_{\rm E}/T =0.5$   ⇒   the filter is too broadband!
::* $T_{\rm E}/T > 1$  results in a smaller  $\sigma_{\rm norm}$  compared to  $(5)$.  But the "eye" is no longer open, e.g.  $T_{\rm E}/T =1.25$:   $ö_{\rm norm}= g_0 - 2 \cdot g_1 = 0.6$.

{{BlueBox|TEXT=
'''(7)'''  Now select the settings  $M=2 \text{, CRO Nyquist system, }r_f = 0.2$  and  "Overall view". Interpret the eye pattern, also for other  $r_f$ values. }}

::* Unlike  $(6)$  here the basic detection impulse is not zero for  $|t|>T$,  but  $g_d(t)$  has equidistant zero crossings:  $g_0 = 1, \ g_1 = g_2 = 0$   ⇒   '''Nyquist system'''.
::* All  $32$  eye lines pass through only two points at  $t=0$.  The vertical eye opening is maximum for all  $r_f$    ⇒    $ö_{\rm norm}= 1$.
::* In contrast, the horizontal eye opening increases with  $r_f$  and is for  $r_f = 1$  maximum equal to  $T$   ⇒   the phase jitter has no influence in this case.

{{BlueBox|TEXT=
'''(8)'''  Same setting as in  $(7)$.  Now vary  $r_f$  with respect to minimum error probability.  Interpret the results.}}
::*$ö_{\rm norm}= 1$  always holds.  In contrast,  $\sigma_{\rm norm}$  shows a slight dependence on  $r_f$.  The minimum  $\sigma_{\rm norm}=0.236$  results for  $r_f = 0.9$   ⇒   $p_{\rm U} \approx 1.1 \cdot 10^{-5}.$
::* Compared to the best possible case according to  $(5)$   ⇒   "Matched Filter Receiver"  $p_{\rm U}$  is three times larger, although  $\sigma_{\rm norm}$  is only larger by about  $5\%$.
::* The larger  $\sigma_{\rm norm}$ value is due to the exaggeration of the noise PDS to compensate for the drop through the transmitter frequency response  $H_{\rm S}(f)$.

{{BlueBox|TEXT=
'''(9)'''  Select the settings  $M=4 \text{, Matched Filter receiver, }T_{\rm E}/T = 1$,  $10 \cdot \lg \ E_{\rm B}/N_0 = 10 \ {\rm dB}$  and  $12 \ {\rm dB}$.  Interpret the results. }}

::* Now there are three eye openings.  Compared to  $(5)$   $ö_{\rm norm}$  is thus smaller by a factor of  $3$.  $\sigma_{\rm norm}$  on the other hand, only by a factor of  $\sqrt{5/9)} \approx 0.75$.
::* For  $10 \cdot \lg \ E_{\rm B}/N_0 = 10 \ {\rm dB}$  the  (worst–case)  error probability is  $p_{\rm U} \approx 2.27\%$  and for  $10 \cdot \lg \ E_{\rm B}/N_0 = 12 \ {\rm dB}$  approx.  $0.59\%$.

{{BlueBox|TEXT=
'''(10)'''  For the remaining tasks, always  $10 \cdot \lg \ E_{\rm B}/N_0 = 12 \ {\rm dB}$. Consider the eye pattern  ("overall view")  for  $M=4 \text{, CRO Nyquist system, }r_f = 0.5$. }}

::* In the analyzed  $d_{\rm S}(t)$  region all  "five symbol combinations"  must be contained   ⇒   minimum  $4^5 = 1024$  parts   ⇒   maximum  $1024$  distinguishable lines.
::* All  $1024$  eye lines pass through only four points at  $t=0$ :   $ö_{\rm norm}= 0.333$.  $\sigma_{\rm norm} = 0.143$  is slightly larger than in  $(9)$  ⇒   $p_{\rm U} \approx 1\%$.

{{BlueBox|TEXT=
'''(11)'''  Select the settings  $M=4 \text{, Gaussian low-pass, }f_{\rm G}/R_{\rm B} = 0.48$  and vary  $f_{\rm G}/R_{\rm B}$.  Interpret the results. }}

::* $f_{\rm G}/R_{\rm B}=0.48$  leads to the minimum error probability  $p_{\rm U} \approx 0.21\%$.  $\text{Compromise between}$  $ö_{\rm norm}= 0.312$  and  $\sigma_{\rm norm}= 0.109$.
::* If the cutoff frequency is too small, intersymbol interference dominates.  Example:  $f_{\rm G}/R_{\rm B}= 0.3$:  $ö_{\rm norm}= 0.157; $ $\sigma_{\rm norm}= 0.086$  ⇒    $p_{\rm U} \approx 3.5\%$.
::* If the cutoff frequency is too high, noise dominates.  Example:  $f_{\rm G}/R_{\rm B}= 1.0$:  $ö_{\rm norm}= 0.333; $ $\sigma_{\rm norm}= 0.157$  ⇒    $p_{\rm U} \approx 1.7\%$.
::* From the comparison with  $(9)$  one can see:  $\text{With quaternary coding it is more convenient to allow intersymbol interference}$.

{{BlueBox|TEXT=
'''(12)'''  What differences does the eye pattern show for  $M=3 \text{ (AMI code), Gaussian low-pass, }f_{\rm G}/R_{\rm B} = 0.48$  compared to the binary system  $(1)$? Interpretation. }}
::* The basic detection impulse  $g_d(t)$  is the same in both cases.  The sample values are respectively  $g_0 = 0.771, \ g_1 = 0.114$.
::* With the AMI code, there are two eye openings with each  $ö_{\rm norm}= 1/2 \cdot (g_0 -3 \cdot g_1) = 0.214$.  With the binary code:  $ö_{\rm norm}= g_0 -2 \cdot g_1 = 0.543$.
::* The AMI sequence consists of  $50\%$  zeros.  The symbols  $+1$  and  $-1$  alternate   ⇒   there is no long  $+1$  sequence and no long  $-1$  sequence.
::* Therein lies the only advantage of the AMI code:  This can also be applied to a channel with  $H_{\rm K}(f= 0)=0$  ⇒   a DC signal is suppressed.

{{BlueBox|TEXT=
'''(13)'''  Same setting as in  $(12)$.  Select additionally  $10 \cdot \lg \ E_{\rm B}/N_0 = 12 \ {\rm dB}$.  Analyze the worst-case error probability of the AMI code. }}
::* Despite smaller  $\sigma_{\rm norm} = 0.103$  the AMI code has higher error probability  $p_{\rm U} \approx 2\%$  than the binary code:  $\sigma_{\rm norm} = 0.146, \ p_{\rm U} \approx \cdot 10^{-4}.$
::* $f_{\rm G}/R_{\rm B}<0.34$  results in a closed eye  $(ö_{\rm norm}= 0)$  ⇒    $p_{\rm U} =50\%$.  With binary coding:  For  $f_{\rm G}/R_{\rm B}>0.34$  the eye is open.

{{BlueBox|TEXT=
'''(14)'''  What differences does the eye pattern show for  $M=3 \text{ (Duobinary code), Gaussian low-pass, }f_{\rm G}/R_{\rm B} = 0.30$  compared to the binary system  '''(1)'''? }}
::* With redundancy-free binary code:  $ö_{\rm norm}= 0.096, \ \sigma_{\rm norm} = 0.116 \ p_{\rm U} \approx 20\% $.   With Duobinary code:  $ö_{\rm norm}= 0.167, \ \sigma_{\rm norm} = 0.082 \ p_{\rm U} \approx 2\% $.
::*In particular, with small  $f_{\rm G}/R_{\rm B}$  the Duobinary code gives good results, since the transitions from  $+1$  to  $-1$  (and vice versa) are absent in the eye pattern.
::*Even with  $f_{\rm G}/R_{\rm B}=0.2$  the eye is open.  But in contrast to AMI  the Duobinary code is not applicable with a DC-free channel   ⇒   $H_{\rm K}(f= 0)=0$.

==Applet Manual==
 
[[File:Anleitung_Auge.png|right|600px|frame|Screenshot of the German version]]
    '''(A)'''     Selection:   Coding                    (binary,  quaternary,  AMI code,  duobinary code)

    '''(B)'''     Selection:   Basic detection pulse                     (according to Gauss–TP,  CRO–Nyquist,  according to gap–TP}

    '''(C)'''     Parameter input for  '''(B)'''                    (cutoff frequency,  rolloff factor,  rectangular duration)

    '''(D)'''     Control of the eye diagram display                    (start,  pause/continue,  single step,  total,  reset)

    '''(E)'''     Speed of the eye diagram display

    '''(F)'''     Display:  basic detection pulse  $g_d(t)$

    '''(G)'''     Display:  detection useful signal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Display:  eye diagram in the range  $\pm T$

    '''( I )'''     Numerical output:  $ö_{\rm norm}$  (normalized eye opening)

    '''(J)'''     Parameter input  $10 \cdot \lg \ E_{\rm B}/N_0$  for  '''(K)'''

    '''(K)'''     Numerical output:  $\sigma_{\rm norm}$  (normalized noise rms)

    '''(L)'''     Numerical output:  $p_{\rm U}$  (worst-case error probability)

    '''(M)'''     Range for experimental performance:   task selection

    '''(N)'''     Range for experimental performance:   task description

    '''(O)'''     Range for experimental performance:   Show sample solution
 
==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Thomas_Gro.C3.9Fer_.28Diplomarbeit_LB_2006.2C_danach_freie_Mitarbeit_bis_2010.29|Thomas Großer]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|eyeDiagram_en|eyeDiagram}}

Applets:The Doppler Effect

2023-03-26T21:05:52Z

Noah:

{{LntAppletLinkEnDe|dopplereffect_en|dopplereffect}}

==Applet Description==
 
The applet is intended to illustrate the "Doppler effect", named after the Austrian mathematician, physicist and astronomer Christian Andreas Doppler.  This predicts the change in the perceived frequency of waves of any kind, which occurs when the source (transmitter) and observer (receiver) move relative to each other.  Because of this, the reception frequency $f_{\rm E}$  differs from the transmission frequency $f_{\rm S}$.  The Doppler frequency $f_{\rm D}=f_{\rm E}-f_{\rm S}$  is positive if the observer and the source approach each other, otherwise the observer perceives a lower frequency than which was actually transmitted.

The exact equation for the reception frequency $f_{\rm E}$  considering the theory of relativity is:
:$$f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2} }{1 - v/c \cdot \cos(\alpha)}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\text{ Exact equation}}.$$
*Here is  $v$  the relative speed between transmitter and receiver, while  $c = 3 \cdot 10^8 \, {\rm m/s}$  indicates the speed of light.
*$\alpha$  is the angle between the direction of movement and the connecting line between transmitter and receiver.
*$\varphi$  denotes the angle between the direction of movement and the horizontal in the applet. In general,  $\alpha \ne \varphi$.

At realistic speeds  $(v/c \ll 1)$  the following approximation is sufficient, ignoring the effects of relativity:
:$$f_{\rm E} \approx f_{\rm S} \cdot \big [1 +{v}/{c} \cdot \cos(\alpha) \big ] \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\text{ Approximation}}\hspace{0.05cm}.$$
For example, in the case of mobile communications, the deviations between  $f_{\rm E}$  and  $f_{\rm S}$  – the Doppler frequency $f_{\rm D}$  – is only a fraction of the transmission frequency. 

==Theoretical Background==
 
=== Phenomenological description of the Doppler effect===

{{BlaueBox|TEXT=
$\text{Definition:}$  The  $\rm Doppler\:effect$  is the change in the perceived frequency of waves of any kind that occurs when the source (transmitter) and observer (receiver) move relative to each other. This was theoretically predicted by the Austrian mathematician, physicist and astronomer  [https://en.wikipedia.org/wiki/Christian_Doppler "Christian Andreas Doppler"]  in the middle of the 19th century and named after him.}}

Qualitatively, the Doppler effect can be described as follows:
*If the observer and the source approach each other, the frequency increases from the observer's point of view, regardless of whether the observer is moving or the source or both. 

*If the source moves away from the observer or the observer moves away from the source, the observer perceives a lower frequency than was actually transmitted. 

{{GraueBox|TEXT=
$\text{Example 1:}$  We look at the change in pitch of the "Martinhorn" of an ambulance. As long as the vehicle is approaching, the observer hears a higher tone than when the vehicle is stationary.  If the ambulance moves away, a lower tone is perceived. 

The same effect can be seen in a car race.  The frequency changes and the "sound" are all the clearer the faster the cars go. }} 

[[File:P ID2113 Mob T 1 3 S2a v1.png|right|frame|Starting position: $\rm (S)$ and $\rm (E)$ do not move|class=fit]]
{{GraueBox|TEXT=
$\text{Example 2:}$ 
Some properties of this effect, which may be still known from physics lessons, are now to be shown on the basis of screen shots from an earlier version of the present applet, with the dynamic program properties of course being lost. 

The first graphic shows the initial situation:
*The stationary transmitter  $\rm (S)$  emits the constant frequency $f_{\rm S}$.
*The wave propagation is illustrated in the graphic by concentric circles around  $\rm (S)$.
*The receiver   $\rm (E)$ , which is also at rest, receives naturally the frequency $f_{\rm E} = f_{\rm S}$.}}
 

{{GraueBox|TEXT=
$\text{Example 3:}$  In this snapshot, the transmitter  $\rm (S)$  has moved from its starting point  $\rm (S_0)$  to the receiver  $\rm (E)$  at a constant speed.

[[File:P ID2114 Mob T 1 3 S2b v2.png|right|frame|Doppler effect: $\rm (S)$ moves towards the resting $\rm (E)$]]

*The diagram on the right shows that the frequency $f_{\rm E}$ perceived by the receiver (blue oscillation) is about $20\%$ greater than the frequency $f_{\rm S}$ at the transmitter (red oscillation).
*Due to the movement of the transmitter, the circles are no longer concentric.

[[File:P ID2115 Mob T 1 3 S2c v2.png|left|frame|Doppler effect: $\rm (S)$ moves away from resting $\rm (E)$ ]]
 
* The left scenario is the result when the transmitter moves away from the receiver:
* Then the reception frequency $f_{\rm E}$  (blue oscillation)  is about  $20\%$  lower than the transmission frequency $f_{\rm S}$. }}
 

===Doppler frequency as a function of speed and angle of the connecting line===

We agree:  The frequency $f_{\rm S}$  is sent and the frequency $f_{\rm E}$  is received.  The Doppler frequency is the difference $f_{\rm D} = f_{\rm E} - f_{\rm S}$  due to the relative movement between the transmitter (source) and receiver (observer).

*A positive Doppler frequency  $(f_{\rm E} > f_{\rm S})$  arises when transmitter and receiver move (relatively) towards each other.
*A negative Doppler frequency  $(f_{\rm E} < f_{\rm S})$  means that transmitter and receiver are moving apart  (directly or at an angle). 

The exact equation for the reception frequency $f_{\rm E}$  including an angle  $\alpha$  between the direction of movement and the connecting line between transmitter and receiver is:
::<math>f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2} }{1 - v/c \cdot \cos(\alpha)}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\text{ Exact equation}}.</math>
Here  $v$  denotes the relative speed between transmitter and receiver, while  $c = 3 \cdot 10^8 \, {\rm m/s}$  indicates the velocity of light.

*The graphics in  $\text{Example 3}$  apply to the unrealistically high speed  $v = c/5 = 60000\, {\rm km/s}$, which lead to the Doppler frequencies $f_{\rm D} = \pm 0.2\cdot f_{\rm S}$.

*In the case of mobile communications, the deviations between $f_{\rm S}$  and $f_{\rm E}$  are usually only a fraction of the transmission frequency.  At such realistic velocities  $(v \ll c)$  one can start from the following approximation, which does not take into account the effects described by the [https://en.wikipedia.org/wiki/Theory_of_relativity "theory of Relativity"]:
::<math>f_{\rm E} \approx f_{\rm S} \cdot \big [1 +{v}/{c} \cdot \cos(\alpha) \big ] \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\text{ Approach}}\hspace{0.05cm}.</math>

{{GraueBox|TEXT=
$\text{Example 4:}$  We are assuming a fixed station here.  The receiver approaches the transmitter at an angle $\alpha = 0$. 

Different speeds are to be examined:
* an unrealistically high speed  $v_1 = 0.6 \cdot c = 1.8 \cdot 10^8 \ {\rm m/s}$ $\hspace{0.3cm}\Rightarrow\hspace{0.3cm}v_1/c = 0.6$,
* the maximum speed  $v_2 = 3 \ {\rm km/s} \ \ (10800 \ {\rm km/h})$  for an unmanned space flight  $\hspace{0.3cm}\Rightarrow\hspace{0.3cm}v_2/c = 10^{-5}$,
* approximately the top speed  $v_3 = 30 \ {\rm m/s} = 108 \ \rm km/h$  on federal roads  $\hspace{0.3cm}\Rightarrow\hspace{0.3cm}v_3/c = 10^{-7}$.

'''(1)'''  According to the exact, relativistic first equation:
:$$f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2} }{1 - v/c }
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm D} = f_{\rm E} - f_{\rm S} = f_{\rm S} \cdot \left [ \frac{\sqrt{1 - (v/c)^2} }{1 - v/c } - 1 \right ]\hspace{0.3cm}
\Rightarrow \hspace{0.3cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - (v/c)^2} }{1 - v/c } - 1 \hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_1/c = 0.6\text{:}\hspace{0.5cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - 0.6^2} }{1 - 0.6 } - 1 = \frac{0.8}{0.4 } - 1 = 1
\hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E} }/{f_{\rm S} } = 2
\hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_2/c = 10^{\rm -5}\text{:}\hspace{0.5cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - (10^{-5})^2} }{1 - (10^{-5}) } - 1 \approx 1 + 10^{-5} - 1 = 10^{-5} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E} }/{f_{\rm S} } = 1.00001
\hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_3/c = 10^{\rm -7}\text{:}\hspace{0.5cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - (10^{-7})^2} }{1 - (10^{-7}) } - 1 \approx 1 + 10^{-7} - 1 = 10^{-7} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E} }/{f_{\rm S} } = 1.0000001
\hspace{0.05cm}.$$

'''(2)'''  On the other hand, according to the approximation, i.e. without taking into account the theory of relativity:
:$$f_{\rm E} = f_{\rm S} \cdot \big [ 1 + {v}/{c} \big ]
\Rightarrow \hspace{0.3cm}{f_{\rm D} }/{f_{\rm S} } = {v}/{c} \hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_1/c = 0.6\text{:}\hspace{0.7cm} f_{\rm D}/f_{\rm S} \ \ = \ 0.6 \hspace{0.5cm} ⇒ \ \ \ f_{\rm E}/f_{\rm S} = 1.6,$$
:$$\Rightarrow\hspace{0.3cm}v_2/c = 10^{-5}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ = \ 10^{-5} \ \ \ ⇒ \ \ \ f_{\rm E}/f_{\rm S} = 1.00001,$$
:$$\Rightarrow\hspace{0.3cm}v_3/c = 10^{-7}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ = \ 10^{-5} \ \ \ ⇒ \ \ \ f_{\rm E}/f_{\rm S} = 1.0000001.$$}}

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
#  For "low speeds", the approximation to the accuracy of a calculator gives the same result as the relativistic equation.
#  The numerical values show that we can also rate the speed  $v_2 = \ 10800 \ {\rm km/h}$  as "low" in this respect.}}

{{GraueBox|TEXT=
$\text{Example 5:}$  The same requirements apply as in the last example with the difference: Now the receiver moves away from the transmitter $(\alpha = 180^\circ)$.

'''(1)'''  According to the exact, relativistic first equation with  ${\rm cos}(\alpha) = -1$:

:$$f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2} }{1 + v/c }
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm D} = f_{\rm E} - f_{\rm S} = f_{\rm S} \cdot \left [ \frac{\sqrt{1 - (v/c)^2} }{1 + v/c } - 1 \right ] \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - (v/c)^2} }{1 + v/c } - 1 \hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_1/c = 0.6\text{:}\hspace{0.5cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - 0.6^2} }{1 + 0.6 } - 1 = \frac{0.8}{1.6 } - 1 =-0.5
\hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E} }/{f_{\rm S} } = 0.5
\hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_2/c = 10^{\rm -5}\text{:}\hspace{0.5cm}{f_{\rm D} }/{f_{\rm S} } = \frac{\sqrt{1 - (10^{-5})^2} }{1 + (10^{-5}) } - 1 \approx - 10^{-5}
\hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E} }/{f_{\rm S} } = 0.99999
\hspace{0.05cm}.$$
'''(2)'''  On the other hand, according to the approximation, i.e. without taking into account the theory of relativity:
:$$f_{\rm E} = f_{\rm S} \cdot \big [ 1 - {v}/{c} \big ] \hspace{0.3cm}
\Rightarrow \hspace{0.3cm}{f_{\rm D} }/{f_{\rm S} } = - {v}/{c} \hspace{0.05cm}.$$

:$$\Rightarrow\hspace{0.3cm}v_1/c = 0.6\text{:}\hspace{0.7cm} f_{\rm D}/f_{\rm S} \ \underline {= \ 0.6} \ \ \ ⇒ \ \ \ f_{\rm E}/f_{\rm S} = 0.4,$$
:$$\Rightarrow\hspace{0.3cm}v_2/c = 10^{-5}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ = \ - 10^{-5} \ \ \ ⇒ \ \ \ f_{\rm E}/f_{\rm S} = 0.99999.$$}}

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
#  The reception frequency  $f_{\rm E}$  is now lower than the transmission frequency  $f_{\rm S}$  and the Doppler frequency   $f_{\rm D}$  is negative.
#  Using the approximation, the Doppler frequencies for the two directions of movement differ only in the sign   ⇒   $f_{\rm E} = f_{\rm S} \pm f_{\rm D}$.
#  This symmetry does not exist with the exact, relativistic equation. }}

{{GraueBox|TEXT=
$\text{Example 6:}$  Now let's look at the speed that is also realistic for mobile communications  $v = 30 \ {\rm m/s} = 108 \ \rm km/h$   ⇒   $v/c=10^{-7}$. 

[[File:P_ID2118__Mob_Z_1_4.png|right|frame|Directions  $\rm (A)$,  $\rm (B)$, $\rm (C)$, $\rm (D)$]]

*This allows us to limit ourselves to the non-relativistic approximation:   $f_{\rm D} = f_{\rm E} - f_{\rm S} = f_{\rm S} \cdot {v}/{c} \cdot \cos(\alpha) \hspace{0.05cm}.$
*As in the previous examples, the transmitter is fixed. The transmission frequency is  $f_{\rm S} = 2 \ {\rm GHz}$.

The graphic shows possible directions of movement of the receiver. 
* The direction  $\rm (A)$  was used in $\text{Example 4}$ .  With the current parameter values

:$$f_{\rm D} = 2 \cdot 10^{9}\,\,{\rm Hz} \cdot \frac{30\,\,{\rm m/s} }{3 \cdot 10^{8}\,\,{\rm m/s} } = 200\,{\rm Hz}.$$

* For the direction  $\rm (B)$  you get the same numerical value with negative sign according to  $\text{Example 5}$:  
:$$f_{\rm D} = -200\,{\rm Hz}.$$

* The direction of travel  $\rm (C)$  is perpendicular  $(\alpha = 90^\circ)$  to the connecting line between transmitter and receiver.  In this case there is no Doppler shift:
:$$f_{\rm D} = 0.$$
* The direction of movement  $\rm (D)$  is characterized by  $\alpha = \ -135^\circ$.  This results:
:$$f_{\rm D} = 200 \,{\rm Hz} \cdot \cos(-135^{\circ}) \approx -141\,\,{\rm Hz} \hspace{0.05cm}.$$
}}
 

=== Doppler frequency and its distribution===

We briefly summarize the statements in the last section, while we proceed with the second, the non – relativistic equation:
*A relative movement between transmitter (source) and receiver (observer) results in a shift by the Doppler frequency  $f_{\rm D} = f_{\rm E} - f_{\rm S}$.

*A positive Doppler frequency  $(f_{\rm E} > f_{\rm S})$  results when the transmitter and receiver move (relatively) towards each other.
*A negative Doppler frequency  $(f_{\rm E} < f_{\rm S})$  means, that the sender and receiver are moving apart (directly or at an angle). 

*The maximum frequency shift occurs when the transmitter and receiver move directly towards each other   ⇒   angle  $\alpha = 0^\circ$.  This maximum value depends in the first approximation on the transmission frequency  $ f_{\rm S}$  and the speed  $v$  $(c = 3 \cdot 10^8 \, {\rm m/s}$  indicates the velocity of light$)$:  $f_{\rm D, \hspace{0.05cm} max} = f_{\rm S} \cdot {v}/{c} \hspace{0.05cm}.$

*If the relative movement occurs at any angle  $\alpha$  to the transmitter-receiver connection line, the Doppler shift is

::<math>f_{\rm D} = f_{\rm E} - f_{\rm S} = f_{\rm D, \hspace{0.05cm} max} \cdot \cos(\alpha)
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} - \hspace{-0.05cm}f_{\rm D, \hspace{0.05cm} max} \le f_{\rm D} \le + \hspace{-0.05cm}f_{\rm D, \hspace{0.05cm} max}
\hspace{0.05cm}.</math>

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Assuming equally probable directions of movement   $($uniform distribution for the angle  $\alpha$  in the area  $- \pi \le \alpha \le +\pi)$  results for the probability density function  $($referred to here as "pdf"$)$  the Doppler frequency in the range  $- f_\text{D, max} \le f_{\rm D} \le + f_\text{D, max}$:

::<math>{\rm pdf}(f_{\rm D}) = \frac{1}{2\pi \cdot f_{\rm D, \hspace{0.05cm} max} \cdot \sqrt {1 - (f_{\rm D}/f_{\rm D, \hspace{0.05cm} max})^2 } }
\hspace{0.05cm}.</math>

Outside the range between  $-f_{\rm D}$  and  $+f_{\rm D}$ , the probability density function is always zero.

[[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution|$\text{"Derivation"}$]]  about the “nonlinear transformation of random quantities”.}} 

=== Power density spectrum in Rayleigh fading ===

We now presuppose an antenna radiating equally in all directions.  Then the Doppler– $ \rm PDS $  $($Power Density Spectrum$)$  has the same shape as the  $ \rm PDF $  $($Probability Density Function$)$  of the Doppler frequencies.

*For the in-phase component  ${\it \Phi}_x(f_{\rm D})$  of the PDS, the PDF must still be multiplied by the variance  $\sigma^2$  of the Gaussian process.
*For the resulting PDS  ${\it \Phi}_z(f_{\rm D})$  of the complex factor  $z(t) = x(t) + {\rm j} \cdot y(t) $  applies after doubling:

::<math>{\it \Phi}_z(f_{\rm D}) =
\left\{ \begin{array}{c} (2\sigma^2)/( \pi \cdot f_{\rm D, \hspace{0.05cm} max}) \cdot \left [ 1 - (f_{\rm D}/f_{\rm D, \hspace{0.05cm} max})^2 \right ]^{-0.5} \\
0 \end{array} \right.\quad
\begin{array}{*{1}c} {\rm f\ddot{u}r}\hspace{0.15cm} |f_{\rm D}| \le f_{\rm D, \hspace{0.05cm} max}
\\ {\rm other} \\ \end{array}
\hspace{0.05cm}.</math>

This course is called  '''Jakes spectrum'''  named after  [http://ethw.org/William_C._Jakes,_Jr. "William C. Jakes Jr."].  The doubling is necessary, because so far only the contribution of the real part  $x(t)$  has been considered. 

[[File:P_ID2117__Mob_T_1_3_S4_v2.png|right|frame|Doppler PDS and time function (magnitude in dB) for Rayleigh fading with Doppler effect]]
{{GraueBox|TEXT=
$\text{Example 7:}$  The Jakes spectrum is shown on the left
*for $f_{\rm D, \hspace{0.05cm} max} = 50 \ \rm Hz$  (blue curve) bzw.
*for $f_{\rm D, \hspace{0.05cm} max} = 100 \ \rm Hz$  (red curve).

In the  [[Examples_of_Communication_Systems/General_Description_of_GSM#Cellular_structure_of_GSM|"GSM–D network"]]  $(f_{\rm S} = 900 \ \rm MHz)$  these values correspond to the speeds  $v = 60 \ \rm km/h$  and  $v = 120 \ \rm km/h$  respectively.

For the GSM–E network $(f_{\rm S} = 1.8 \ \rm GHz)$  these values apply to speeds that are half as high:   $v = 30 \ \rm km/h$  and  $v = 60 \ \rm km/h$  respectively.

The right picture shows the logarithmic magnitude of  $z(t)$:
*You can see that the red curve is fading twice as fast as the blue one.
*The Rayleigh – PDF (amplitude distribution) is independent of  $f_{\rm D, \hspace{0.05cm} max}$  and is therefore the same for both cases.}} 

==Exercises==

* First select the number  $(1, 2, \text{...})$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
*In the following descriptions, $f_{\rm S}$, $f_{\rm E}$  and$f_{\rm D}$  are each normalized to the reference frequency $f_{\rm 0}$.
 
{{BlaueBox|TEXT=
'''(1)'''  First we consider the relativistic setting "Exact".  The transmitter moves with  $v/c = 0.8$,  the transmission frequency is $f_{\rm S}= 1$.          Which reception frequencies  $f_{\rm E}$  result in both directions of movement?  What is the Doppler frequency $f_{\rm D}$?}}

:*  If the transmitter approaches the receiver under the angle  $\varphi=0^\circ$ , the reception frequency is $f_{\rm E}= 3$   ⇒   $f_{\rm D}= f_{\rm E} - f_{\rm S}= 2$.
:*  If the transmitter moves away from the receiver  $($for  $\varphi=0^\circ$, if it overtakes it, or  $\varphi=180^\circ)$, then:  $f_{\rm E}= 0.333$   ⇒   $f_{\rm D}= -0.667$.
:* The same result with the transmitter at rest and the receiver moving:  If both come closer, then  $f_{\rm D}= 2$  applies, otherwise  $f_{\rm D}= -0.667$.

{{BlaueBox|TEXT=
'''(2)'''  The settings are largely retained.  How do the results change compared to  $(1)$  with the transmission frequency $f_{\rm S}= 1.5$?         Tip for a time-saving experiment:  Switch alternately between "right" and "left".}}

:* Direction of movement $\varphi=0^\circ$: $f_{\rm E}= 4.5$ ⇒   $f_{\rm D}= f_{\rm E} - f_{\rm S}= 3$.  Thus: $f_{\rm E}/f_{\rm S}= 3$, $f_{\rm D}/f_{\rm S}= 2$   ⇒   Both as in  $(1)$.
:* Direction of movement $\varphi=180^\circ$:  $f_{\rm E}= 0.5$   ⇒   $f_{\rm D}= -1$.  Thus: $f_{\rm E}/f_{\rm S}= 0.333$,  $f_{\rm D}/f_{\rm S}= -0.667$  ⇒   Both as in  $(1)$.

{{BlaueBox|TEXT=
'''(3)'''  Still relativistic setting "Exact".  The transmitter is now moving at a speed of  $v/c = 0.4$  and the transmission frequency is $f_{\rm S}= 2$.         Which frequencies $f_{\rm D}$  and  $f_{\rm E}$  result in both directions of movement?  Alternately select "Right" or "Left" again.}}

:* Direction of movement $\varphi=0^\circ$:  Reception frequency  $f_{\rm E}= 3.055$   ⇒   Doppler frequency  $f_{\rm D}= 1.055$.   ⇒   $f_{\rm E}/f_{\rm S}= 1.528$,  $f_{\rm D}/f_{\rm S}= 0.528$.
:* Direction of movement $\varphi=180^\circ$:  Reception frequency $f_{\rm E}= 1.309$   ⇒   Doppler frequency  $f_{\rm D}= -0.691$.   ⇒   $f_{\rm E}/f_{\rm S}= 0.655$,  $f_{\rm D}/f_{\rm S}= -0.346$.

{{BlaueBox|TEXT=
'''(4)'''  The previous conditions still apply, but now the "Approximation" setting.  What are the differences compared to $(3)$?}}

:* Direction of movement $\varphi=0^\circ$:  Reception frequency  $f_{\rm E}= 2.8$   ⇒   Doppler frequency  $f_{\rm D}= f_{\rm E} - f_{\rm S}= 0.8$   ⇒   $f_{\rm E}/f_{\rm S}= 1.4$,  $f_{\rm D}/f_{\rm S}= 0.4$.
:* Direction of movement $\varphi=180^\circ$:  Reception frequency  $f_{\rm E}= 1.2$   ⇒   Doppler frequency  $f_{\rm D}= -0.8$.   ⇒   $f_{\rm E}/f_{\rm S}= 0.6$,  $f_{\rm D}/f_{\rm S}= -0.4$.
:* With “Approximation”:  For both,  $f_{\rm D}$  has the same numerical values with different signs.  This symmetry does not exist with "Exact".

{{BlaueBox|TEXT=
'''(5)'''  $f_{\rm S}= 2$ still apply.  Up to what speed  $(v/c)$  is the relative error between "Approximation" and "Exact" less than $\pm5\%$?}}

:* With  $v/c =0.08$  and "Exact" one obtains for the Doppler frequencies  $f_{\rm D}= 0.167$  respectively  $f_{\rm D}= -0.154$  and with "Approximation" $f_{\rm D}= \pm0.16$.
:* Thus the relative deviation “(Approximation - Exact)/Exact” is equal to  $0.16/0.167-1=-4.2\%$  and   $(-0.16)/(-0.154)-1=+3.9\%$ respectively.
:* With  $v/c =0.1$ , the deviations are greater than  $\pm 5\%$.  For  $v < 0.08 \cdot c = 24\hspace{0.05cm}000$  km/s  the Doppler approximation is sufficient.

{{BlaueBox|TEXT=
'''(6)'''  The following should apply here and in the following tasks: $f_{\rm S}= 1$,  $v/c= 0.4$   ⇒   $f_{\rm D} = f_{\rm S} \cdot v/c \cdot \cos(\alpha)$.  With  $\cos(\alpha) = \pm 1$:     $f_{\rm D}/f_{\rm S} =\pm 0.4$.         Which normalized Doppler frequencies result from the set start coordinates $(0,\ 150)$  and the direction of movement $\varphi=-45^\circ$?}}

:* Here the transmitter moves directly to the receiver to $(\alpha=0^\circ)$  or moves away from it $(\alpha=180^\circ)$.
:* Same constellation as with the starting point $(0,\ 0)$ and  $\varphi=0^\circ$.  Therefore, the following also applies to the Doppler frequency:  $f_{\rm D}/f_{\rm S} =\pm 0.4$.
:* After the transmitter has been “reflected” on a boundary, any angles  $\alpha$  and correspondingly more Doppler frequencies are possible.

{{BlaueBox|TEXT=
'''(7)'''  The transmitter is fixed at $(S_x = 0,\ S_y =10),$ the receiver moves horizontally left and right $(v/c = 0.4, \hspace{0.3cm}\varphi=0^\circ)$.         Observe and interpret the temporal change in the Doppler frequency  $f_{\rm D}$. }}

:* As in  $(6)$, only values between $f_{\rm D}=0.4$  and  $f_{\rm D}=-0.4$  are possible, but now all intermediate values $(-0.4 \le f_{\rm D} \le +0.4)$.
:* With "Step" you can see:  $f_{\rm D}\equiv0$  only occurs if the receiver is exactly below the transmitter $(\alpha=\pm 90^\circ$, depending on the direction of travel$)$.
:* Doppler frequencies at the edges are much more common:  $|f_{\rm D}| = 0.4 -\varepsilon$, where  $\varepsilon$  indicates a small positive size.
:* The basic course of Doppler – PDF and Doppler – PDS   ⇒   "Jakes spectrum" can be explained from this experiment alone.

{{BlaueBox|TEXT=
'''(8)'''  What changes if the transmitter is fixed at the top of the graphic area in the middle with the same settings $(0,\ 200) $? }}

:* The Doppler values  $f_{\rm D} \approx0$  become more frequent, those at the edges less frequent.  No values  $|f_{\rm D}| > 0.325$  due to limited drawing space.

{{BlaueBox|TEXT=
'''(9)'''  The transmitter is $S_x = 300,\ S_y =200)$, the receiver moves with $v/c = 0.4$  under the angle $\varphi=60^\circ$.         Think about the relationship between $\varphi$ and $\alpha$.}}

:* Model solutions are still missing

==Applet Manual==
[[File:Anleitung_Doppler.png|right|600px|frame|Screenshot]]
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Start position of the transmitter   ⇒   $(S_x,\ S_Y)$

    '''(C)'''     Input parameters
:* Direction  $\varphi$  of movement of transmitter/receiver
:* (Normalized) velocity  $(v/c)$  of transmitter/receiver
:* (Normalized) transmission frequency  $(f_{\rm S}/f_0)$ 

    '''(D)'''     Equation used for the reception frequency
:* Exact  (considering the Relativity Theory)
:* Approximation  (sufficient for mobile radio)

    '''(E)'''     Graphic field: Motion and wave propagation

    '''(F)'''     Graphic field:  Transmission & reception frequency (time domain)

    '''(G)'''     Graphic field:  Transmission & reception frequency (frequency domain)

    '''(H)'''     Control panel 1
:* Transmitter (or receiver) is moving
:* Movement to the right or left  (movement up or down)

    '''(I)'''     Control panel 2  (Start, Stop, Step, Continue, Reset)

    '''(J)'''     Output parameters
:* Angle  $\alpha$  between movement and transmitter/receiver connecting line
:* (Normalized) Doppler frequency  $(f_{\rm D}/f_0)$ 
:* (Normalized) reception frequency  $(f_{\rm E}/f_0)$ 

    '''(K)'''     Selection of the exercise according to the numbers

    '''(L)'''     Task description and questions

    '''(M)'''     Show and hide sample solution

==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2009 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Alexander_Happach_.28Diplomarbeit_EI_2009.29|Alexander Happach]] as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Andr.C3.A9_Schulz_.28Bachelorarbeit_LB_2020.29|André Schulz]] (Bachelor thesis LB, Supervisors: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_Übertragungstechnik#Benedikt_Leible.2C_M.Sc._.28bei_L.C3.9CT_seit_2017.29|Benedikt Leible]] and [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|dopplereffect_en|dopplereffect}}

Applets:Discrete Fouriertransform and Inverse

2023-03-26T20:50:39Z

Noah:

{{LntAppletLinkEnDe|dft_en|dft}}

==Applet Description==
 
The conventional Fourier Transform  $\rm (FT)$  allows the calculation of the spectral function  $X(f)$  of a time-continuous signal  $x(t)$. 

In contrast, the Discrete Fourier Transform  $\rm (DFT)$  is limited to a time discrete signal, represented by  $N$  time domain coefficients   $d(\nu)$  with indices  $\nu = 0, \text{...} , N\hspace{-0.1cm}-\hspace{-0.1cm}1$, which can be interpreted as equidistant samples of the time continuous signal  $x(t)$.

If the  [[Signal_Representation/Discrete-Time_Signal_Representation#Sampling_theorem|"sampling theorem"]]  is fulfilled, the DFT algorithm likewise allows the calculation of  $N$  frequency domain coefficients  $D(\mu)$  with indices  $\mu = 0, \text{...} , N\hspace{-0.1cm}-\hspace{-0.1cm}1$.  These are equidistant samples of the frequency continuous spectrum  $X(f)$.

*The applet illustrates the properties of the  $\text{DFT:}\hspace{0.3cm}d(\nu)\hspace{0.1cm} \Rightarrow \hspace{0.1cm} D(\mu)$  by using the example  $N=16$.  The default   $d(\nu)$ assignments for the DFT are:

:(a)  According to the input field,  (b)  Constant signal,  (c)  Complex exponential function (of time),  (d)  Harmonic oscillation (with  $($Phase  $\varphi = 45^\circ)$,
:(e)  Cosine signal (one period),  (f)  Sinusoidal signal (one period),  (g)  Cosine signal (two periods), (h)  Alternating time coefficients, (i)  Dirac delta impulse, 
: (j)  Rectangular pulse ,  (k)  Triangular pulse,  (l)  Gaussian pulse.

*Possible  $D(\mu)$ assignments for the Inverse Discrete Fourier Transform   ⇒   $\text{IDFT:}\hspace{0.3cm}D(\mu)\hspace{0.1cm} \Rightarrow \hspace{0.1cm} d(\nu)$  are:

:(A)  According to the input field,  (B)  Constant spectrum,  (C)  Complex exponential function (of frequency),  (D)  Equivalent to setting (d) in the time domain,
:(E)  Cosine spectrum (one frequency period),  (F)  Sinusoidal spectrum (one frequency period),  (G)  Cosine spectrum (two frequency periods), 
:(H)  Alternating spectral coefficients, (I)  Dirac delta spectrum,  (J)  Rectangular spectrum,  (K)  Triangular spectrum,  (L)  Gaussian spectrum.

The applet uses the framework  [https://en.wikipedia.org/wiki/Plotly "Plot.ly"].

==Theoretical Background==
 
===Arguments for the discrete realization of the Fourier transform===

The  '''Fourier transform'''  according to the conventional description for continuous-time signals has an infinitely high selectivity due to the unlimited extension of the integration interval and is therefore an ideal theoretical tool for spectral analysis.

If the spectral components  $X(f)$  of a time function  $x(t)$  are to be determined numerically, the general transformation equations

:$$\begin{align*}X(f) & = \int_{-\infty
}^{+\infty}x(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi f t}\hspace{0.1cm} {\rm d}t\hspace{0.5cm} \Rightarrow\hspace{0.5cm} \text{Forward transformation}\hspace{0.7cm} \Rightarrow\hspace{0.5cm} \text{First Fourier integral}
\hspace{0.05cm},\\
x(t) & = \int_{-\infty
}^{+\infty}\hspace{-0.15cm}X(f) \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi f t}\hspace{0.1cm} {\rm d}f\hspace{0.35cm} \Rightarrow\hspace{0.5cm}
\text{Backward transformation}\hspace{0.4cm} \Rightarrow\hspace{0.5cm} \text{Second Fourier integral}
\hspace{0.05cm}\end{align*}$$

are unsuitable for two reasons:
*The equations apply exclusively to continuous-time signals. With digital computers or signal processors, however, only discrete-time signals can be processed.
*For a numerical evaluation of the two Fourier integrals it is necessary to limit the respective integration interval to a finite value.

{{BlaueBox|TEXT=
$\text{This results in the following consequence:}$ 

A  '''continuous signal'''  must undergo two processes before the numerical determination of its spectral properties, viz.
*  '''sampling'''  for discretization, and
*  '''windowing'''  to limit the integration interval.}}

In the following, starting from an aperiodic time function  $x(t)$  and the corresponding Fourier spectrum  $X(f)$,  a discrete-time and discrete-frequency description suitable for computer processing is presented.

===Time discretization – periodization in the frequency domain===

The following graphs uniformly show the time domain on the left and the frequency domain on the right. Without restriction of generality,  $x(t)$  and  $X(f)$  are real and Gaussian, respectively.

[[File:P_ID1132__Sig_T_5_1_S2_neu.png|center|frame|Discretization in the time domain – periodization in the frequency domain]]

One can describe the sampling of the time signal  $x(t)$  by multiplication with a Dirac delta pulse  $p_{\delta}(t)$.  This results in the time signal sampled at distance  $T_{\rm A}$ 

:$${\rm A}\{x(t)\} = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

We now transform this sampled signal  $\text{A}\{ x(t)\}$  into the frequency domain. The multiplication of the Dirac delta pulse  $p_{\delta}(t)$  with  $x(t)$  corresponds in the frequency domain to the convolution of  $P_{\delta}(f)$  with  $X(f)$. The periodized spectrum  $\text{P}\{ X(f)\}$ is obtained, where  $f_{\rm P}$  is the frequency period of the function  $\text{P}\{ X(f)\}$: 

:$${\rm A}\{x(t)\} \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} {\rm P}\{X(f)\} = \sum_{\mu = - \infty }^{+\infty}
X (f- \mu \cdot f_{\rm P} )\hspace{0.5cm} {\rm with }\hspace{0.5cm}f_{\rm
P}= {1}/{T_{\rm A}}\hspace{0.05cm}.$$

*We call the sampled signal  $\text{A}\{ x(t)\}$.
*The  '''frequency period'''  is denoted by  $f_{\rm P}$ = $1/T_{\rm A}$. 

The graph above shows the functional relationship described here. It should be noted:
*The frequency period  $f_{\rm P}$  was deliberately chosen small here so that the overlap of the spectra to be summed can be clearly seen.
*In practice, due to the sampling theorem,  $f_{\rm P}$  should be at least twice as large as the largest frequency contained in the signal  $x(t)$. 
*If this is not fulfilled,  '''aliasing'''  must be expected.

===Frequency discretization – periodization in the time domain===

The discretization of  $X(f)$  can also be described by a multiplication with a Dirac delta pulse. The result is the spectrum sampled at distance  $f_{\rm A}$: 

:$${\rm A}\{X(f)\} = X(f) \cdot \sum_{\mu = - \infty }^{+\infty}
f_{\rm A} \cdot \delta (f- \mu \cdot f_{\rm A } ) = \sum_{\mu = - \infty }^{+\infty}
f_{\rm A} \cdot X(\mu \cdot f_{\rm A } ) \cdot\delta (f- \mu \cdot f_{\rm A } )\hspace{0.05cm}.$$

Transforming the frequency Dirac delta pulse used here $($with pulse weights  $f_{\rm A})$  into the time domain, we obtain with  $T_{\rm P} = 1/f_{\rm A}$:

:$$\sum_{\mu = - \infty }^{+\infty}
f_{\rm A} \cdot \delta (f- \mu \cdot f_{\rm A } ) \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm}
\sum_{\nu = - \infty }^{+\infty}
\delta (t- \nu \cdot T_{\rm P } ) \hspace{0.05cm}.$$

The multiplication with  $X(f)$  corresponds in the time domain to the convolution with  $x(t)$. The signal  $\text{P}\{ x(t)\}$ periodized at distance  $T_{\rm P}$  is obtained:

:$${\rm A}\{X(f)\} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm}
{\rm P}\{x(t)\} = x(t) \star \sum_{\nu = - \infty }^{+\infty}
\delta (t- \nu \cdot T_{\rm P } )= \sum_{\nu = - \infty }^{+\infty}
x (t- \nu \cdot T_{\rm P } ) \hspace{0.05cm}.$$

[[File:P_ID1134__Sig_T_5_1_S3_neu.png|right|frame|Discretization in the frequency domain – periodization in the time domain]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
This relationship is illustrated in the graphic:
*Due to the coarse frequency rastering, this example results in a relatively small value for the time period  $T_{\rm P}$. 

*Therefore, the (blue) periodized time signal  $\text{P}\{ x(t)\}$  differs significantly from  $x(t)$ due to overlaps.}}

===Finite signal representation===

[[File:P_ID1135__Sig_T_5_1_S4_neu.png|right|frame|Finite signals of the Discrete Fourier Transform]]
One arrives at the so-called  ''finite signal representation'' 
*when both the time function  $x(t)$ and
*the spectral function  $X(f)$

are specified exclusively by their sample values.
 
The graph is to be interpreted as follows:
*In the left picture the function  $\text{A}\{ \text{P}\{ x(t)\}\}$ is drawn in blue. It is obtained by sampling the periodized time function  $\text{P}\{ x(t)\}$  with equidistant Dirac delta pulses in the distance  $T_{\rm A} = 1/f_{\rm P}$.
*In the right picture the function  $\text{P}\{ \text{A}\{ X(f)\}\}$ is drawn in green. This results from periodization $($with  $f_{\rm P})$  of the sampled spectral function  $\{ \text{A}\{ X(f)\}\}$.
*There is also a Fourier correspondence between the blue finite signal and the green finite signal, as follows:

:$${\rm A}\{{\rm P}\{x(t)\}\} \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} {\rm P}\{{\rm A}\{X(f)\}\} \hspace{0.05cm}.$$

{{BlaueBox|TEXT=
However, the Dirac delta lines of the periodic continuation  $\text{P}\{ \text{A}\{ X(f)\}\}$  of the sampled spectral function fall into the same frequency grid as those of  $\text{A}\{ X(f)\}$ only if the frequency period  $f_{\rm P}$  is an integer multiple  $(N)$  of the frequency sampling interval  $f_{\rm A}$. 

*When using the finite signal representation, the following condition must always be fulfilled, where in practice a power of two is usually used for the natural number  $N$  (the graph above is based on the value  $N = 8$ ):

:$$f_{\rm P} = N \cdot f_{\rm A} \hspace{0.5cm} \Rightarrow\hspace{0.5cm} {1}/{T_{\rm A} }= N \cdot f_{\rm A} \hspace{0.5cm} \Rightarrow\hspace{0.5cm}
N \cdot f_{\rm A}\cdot T_{\rm A} = 1\hspace{0.05cm}.$$}}

If the condition  $N \cdot f_{\rm A} \cdot T_{\rm A} = 1$  is satisfied, the order of periodization and sampling can be interchanged. Thus:

:$${\rm A}\{{\rm P}\{x(t)\}\} = {\rm P}\{{\rm A}\{x(t)\}\}\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
{\rm P}\{{\rm A}\{X(f)\}\} = {\rm A}\{{\rm P}\{X(f)\}\}\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$ 
*The time function  $\text{P}\{ \text{A}\{ x(t)\}\}$  has the period  $T_{\rm P} = N \cdot T_{\rm A}$.
*The period in the frequency domain is  $f_{\rm P} = N \cdot f_{\rm A}$.
*For the description of the discretized time and frequency course  $N$  '''complex numerical values''' in the form of pulse weights are sufficient in each case.}}

{{GraueBox|TEXT=
$\text{Example 2:}$ 
A time-limited (pulse-like) signal  $x(t)$  is present in sampled form, where the distance between two samples is  $T_{\rm A} = 1\, {\rm µ s}$: 
*After a discrete Fourier transformation with  $N = 512$  the spectrum  $X(f)$  is available as samples with the distance  $f_{\rm A} = (N \cdot T_{\rm A})^{–1} \approx 1.953\,\text{kHz} $. 
*Increasing the DFT parameter to  $N= 2048$ results in a finer frequency grid with  $f_{\rm A} \approx 488\,\text{Hz}$.}}

===Discrete Fourier Transform===

From the conventional  "first Fourier integral"

:$$X(f) =\int_{-\infty
}^{+\infty}x(t) \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi \hspace{0.05cm}\cdot \hspace{0.05cm} f \hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm} {\rm d}t$$

discretization  $(\text{d}t \to T_{\rm A}$,  $t \to \nu \cdot T_{\rm A}$,  $f \to \mu \cdot f_{\rm A}$,  $T_{\rm A} \cdot f_{\rm A} = 1/N)$  yields the sampled and periodized spectral function

:$${\rm P}\{X(\mu \cdot f_{\rm A})\} = T_{\rm A} \cdot \sum_{\nu = 0 }^{N-1}
{\rm P}\{x(\nu \cdot T_{\rm A})\}\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi \hspace{0.05cm} \cdot \hspace{0.05cm}\nu \hspace{0.05cm}
\cdot \hspace{0.05cm}\mu /N} \hspace{0.05cm}.$$

It is taken into account that due to the discretization the periodized functions have to be used in each case.

For reasons of a simplified notation we now make the following substitutions:
*Let the  $N$  '''time-domain coefficients'''  be associated with the indexing variable  $\nu = 0$, ... , $N - 1$:
:$$d(\nu) =
{\rm P}\left\{x(t)\right\}{\big|}_{t \hspace{0.05cm}= \hspace{0.05cm}\nu \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm A}}\hspace{0.05cm}.$$
*Let the  $N$  '''frequency-domain coefficients'''  be associated with the indexing variable  $\mu = 0,$ ... , $N$ – 1:
:$$D(\mu) = f_{\rm A} \cdot
{\rm P}\left\{X(f)\right\}{\big|}_{f \hspace{0.05cm}= \hspace{0.05cm}\mu \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm A}}\hspace{0.05cm}.$$
*Abbreviation for the  '''complex rotation factor'''  depending on  $N$  is written:
:$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N}
= \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right)
\hspace{0.05cm}.$$

[[File:P_ID2730__Sig_T_5_1_S5_neu.png|right|frame|On the definition of the Discrete Fourier Transform (DFT) with  $N=8$]]
{{BlaueBox|TEXT=
$\text{Definition:}$ 

The term  '''Discrete Fourier Transform'''  ('''DFT''' for short)  means the calculation of the  $N$  spectral coefficients  $D(\mu)$  from the  $N$  signal coefficients  $d(\nu)$:

:$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}
d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.03cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}. $$

In the graph you can see in an example
*the  $N = 8$  signal coefficients  $d(\nu)$  at the blue filling,
*the  $N = 8$  spectral coefficients  $D(\mu)$  at the green filling.}}

===Inverse Discrete Fourier Transform===

The Inverse Discrete Fourier Transform (IDFT) describes the  "second Fourier integral"

:$$\begin{align*}x(t) & = \int_{-\infty
}^{+\infty}X(f) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi \hspace{0.05cm}\cdot \hspace{0.05cm} f \hspace{0.05cm}\cdot \hspace{0.05cm}
t}\hspace{0.1cm} {\rm d}f\end{align*}$$

in discretized form:   $d(\nu) =
{\rm P}\left\{x(t)\right\}{\big|}_{t \hspace{0.05cm}= \hspace{0.05cm}\nu \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm
A}}\hspace{0.01cm}.$

[[File:P_ID2731__Sig_T_5_1_S6_neu.png|right|frame|For the definition of the IDFT with  $N=8$]]
{{BlaueBox|TEXT=
$\text{Definition:}$ 

The term  '''Inverse Discrete Fourier Transform'''  ('''IDFT''' for short)  refers to the calculation of the signal coefficients  $d(\nu)$  from the spectral coefficients  $D(\mu)$:

:$$d(\nu) = \sum_{\mu = 0 }^{N-1}
D(\mu) \cdot {w}^{-\nu \hspace{0.03cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

With the indexing variables  $\nu = 0, \hspace{0.05cm}\text{...} \hspace{0.05cm}, N-1$  und  $\mu = 0, \hspace{0.05cm}\text{...} \hspace{0.05cm}, N-1$  also holds here:
:$$d(\nu) =
{\rm P}\left\{x(t)\right\}{\big \vert}_{t \hspace{0.05cm}= \hspace{0.05cm}\nu \hspace{0.05cm}\cdot \hspace{0.05cm}T_{\rm
A} }\hspace{0.01cm},$$

:$$D(\mu) = f_{\rm A} \cdot
{\rm P}\left\{X(f)\right\}{\big \vert}_{f \hspace{0.05cm}= \hspace{0.05cm}\mu \hspace{0.05cm}\cdot \hspace{0.05cm}f_{\rm A} }
\hspace{0.01cm},$$

:$$w = {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N}
\hspace{0.01cm}.$$}}

A comparison between DFT and IDFT shows that exactly the same algorithm can be used. The only differences of the IDFT compared to the DFT are:
*The exponent of the rotation factor must be applied with a different sign.
*With the IDFT the division by  $N$ is omitted.

===Interpretation of DFT and IDFT===
 
The graph shows the discrete coefficients in the time and frequency domain together with the periodized continuous-time functions.

[[File:P_ID1136__Sig_T_5_1_S7_neu.png|center|frame|Time and frequency domain coefficients of the DFT]]

When using DFT or IDFT, it should be noted:
*According to the above definitions, the DFT coefficients  $d(ν)$  and  $D(\mu)$  always have the unit of the time function.
*Dividing  $D(\mu)$  by  $f_{\rm A}$ gives the spectral value  $X(\mu \cdot f_{\rm A})$.
*The spectral coefficients  $D(\mu)$  must always be set complex to be able to consider also odd time functions.
*In order to be able to transform band–pass signals in the equivalent lown&dash;pass range, complex time coefficients  $d(\nu)$ are usually used.
*The basic interval for  $\nu$  and  $\mu$  is usually defined as the range from  $0$  to  $N - 1$, as in the above diagram.
*With the complex-valued number sequences  $\langle \hspace{0.1cm}d(\nu)\hspace{0.1cm}\rangle = \langle \hspace{0.1cm}d(0), \hspace{0.05cm}\text{...} \hspace{0.05cm} , d(N-1) \hspace{0.1cm}\rangle$   as well as   $\langle \hspace{0.1cm}D(\mu)\hspace{0.1cm}\rangle = \langle \hspace{0.1cm}D(0), \hspace{0.05cm}\text{...} \hspace{0.05cm} , D(N-1) \hspace{0.1cm}\rangle$  DFT and IDFT are symbolized similar to the conventional Fourier transform:
:$$\langle \hspace{0.1cm} D(\mu)\hspace{0.1cm}\rangle \hspace{0.2cm}\bullet\!\!-\!\!\!-(N)\!-\!\!\!-\!\!\hspace{0.05cm}\circ\, \hspace{0.2cm} \langle \hspace{0.1cm} d(\nu) \hspace{0.1cm}\rangle \hspace{0.05cm}.$$
*If the time function  $x(t)$  is already limited to the range  $0 \le t \lt N \cdot T_{\rm A}$,  then the time coefficients output by the IDFT directly indicate the samples of the time function:   $d(\nu) = x(\nu \cdot T_{\rm A}).$
*If  $x(t)$  is shifted with respect to the basic interval, one has to choose the assignment between  $x(t)$  and the coefficients  $d(\nu)$  shown in  $\text{Example 3}$. 

{{GraueBox|TEXT=
$\text{Example 3:}$ 
The upper graph shows the asymmetric triangular pulse  $x(t)$ whose absolute width is smaller than  $T_{\rm P} = N \cdot T_{\rm A}$.

[[File:P_ID1139__Sig_T_5_1_S7b_neu.png|right|frame|For the assignment of the DFT coefficients with  $N=8$]]

The sketch below shows the assigned DFT coefficients valid for  $N = 8$

*For  $\nu = 0,\hspace{0.05cm}\text{...} \hspace{0.05cm} , N/2 = 4$,    $d(\nu) = x(\nu \cdot T_{\rm A})$ is valid:

:$$d(0) = x (0)\hspace{0.05cm}, \hspace{0.15cm}
d(1) = x (T_{\rm A})\hspace{0.05cm}, \hspace{0.15cm}
d(2) = x (2T_{\rm A})\hspace{0.05cm}, $$
:$$d(3) = x (3T_{\rm A})\hspace{0.05cm}, \hspace{0.15cm}
d(4) = x (4T_{\rm A})\hspace{0.05cm}.$$
*On the other hand, the coefficients  $d(5)$,  $d(6)$  and  d$(7)$  are to be set as follows:

:$$d(\nu) = x \big ((\nu\hspace{-0.05cm} - \hspace{-0.05cm} N ) \cdot T_{\rm A}\big ) $$

:$$ \Rightarrow \hspace{0.2cm}d(5) = x (-3T_{\rm A})\hspace{0.05cm}, \hspace{0.35cm}
d(6) = x (-2T_{\rm A})\hspace{0.05cm}, \hspace{0.35cm}
d(7) = x (-T_{\rm A})\hspace{0.05cm}.$$ }}

 

==Exercises==
 
[[File:Aufgaben_2D-Gauss.png|right]]
* First select the number (1,...) of the exercise. 
* A description of the exercise will be displayed.
*The parameter values are adjusted. 
*Solution after pressing "Show solution". 
*The number 0 corresponds to a "Reset":
:Same setting as at program start.
:Output of a "reset text" with further explanations about the applet.

{{BlaueBox|TEXT=
'''(1)'''  New setting:  DFT of signal  $\rm (b)$:  Constant signal.  Interpret the result in the frequency domain.  What is the analogon of the conventional Fourier transform?}}

:* All coefficients in the time domain are  $d(\nu)=1$.  Thus all  $D(\mu)=0$  with the exception of  $\textrm{Re}[D(0)]=1$. 
:* This corresponds to the conventional (time-continuous) Fourier Transform:  $x(t)=A\hspace{0.15cm} \circ\!\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} X(f)=A\cdot \delta (f=0)$  with  $A=1$.

{{BlaueBox|TEXT=
'''(2)'''  Assume the obtained  $D(\mu)$  field and shift all coefficients one entry down.  Which time function does the IDFT provide?
}}

:* Now all  $D(\mu)=0$,  except for  $\textrm{Re}[D(1)]=1$.  The result in the time domain is a complex exponential function.
:* The real part of the  $d(\nu)$  field shows a cosine and the imaginary part a sine function.  For each function one can see one period respectively.

{{BlaueBox|TEXT=
'''(3)'''  Add the following coefficient to the current  $D(\mu)$ field:  $\textrm{Im}[D(1)]=1$.  What are the differences compared to  '''(2)'''  in the time domain?
 }}
:* On the one hand, a phase shift of two support values can now be detected for the real and the imaginary parts.  This corresponds to the phase  $\varphi = 45^\circ$.
:* On the other hand, the amplitudes of the real and the imaginary part were each increased by the factor  $\sqrt{2}$.

{{BlaueBox|TEXT=
'''(4)'''  Set the  $D(\mu)$ field  to zero except for  $\textrm{Re}[D(1)]=1$.  Which additional  $D(\mu)$  coefficient yields a real  $d(\nu)$  field?
}}

:* By trial and error, one can see that  $\textrm{Re}[D(15)]=1$  must apply additionally.  Then the  $d(\nu)$  field describes a cosine.
:* The following applies to the conventional (time continuous) Fourier Transform:  $x(t)=2\cdot \cos(2\pi \cdot f_0 \cdot t)\hspace{0.15cm}\circ\!\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}
X(f)=\delta (f-f_0)+\delta (f+f_0)$.
:* The entry  $D(1)$  is representative of the frequency  $f_0$  and due to the periodicity with  $N=16$  the frequency  $-f_0$  is expressed by  $D(15)=D(-1)$.

{{BlaueBox|TEXT=
'''(5)'''  According to the IDFT in the  $d(\nu)$  field, by which  $D(\mu)$  field does one obtain a real cosine function with the amplitude  $A=1$?}}

:* Like the conventional Fourier transform the discrete Fourier Transform is linear  ⇒   $D(1)=D(15)=0.5$.

{{BlaueBox|TEXT=
'''(6)'''  New setting:  DFT of signal  $\rm (e)$:  Cosine signal and subsequent signal shifts.  What are the effects of these shifts in the frequency domain?
}}

:* A shift in the time domain changes the cosine signal to a  "harmonic oscillation"  with arbitrary phase.
:* The  $D(\mu)$  field is still zero except for  $D(1)$  and  $D(15)$.  The absolute values  $|D(1)|$  and  $|D(15)|$  also remain the same.
:* The only change concerns the phase,  i.e. the different distribution of the absolute values between the real and imaginary part.

{{BlaueBox|TEXT=
'''(7)'''  New setting:  DFT of signal  $\rm (f)$:  Sinusoidal signal.  Interpret the result in the frequency domain.  What is the analogon of the conventional Fourier Transform?
}}

:* The sine signal results from the cosine signal by applying four time shifts.  Therefore all statements of  '''(6)'''  are still valid.
:* For the conventional (time continuous) Fourier transform it holds that  $x(t)= \sin(2\pi \cdot f_0 \cdot t)\hspace{0.15cm}\circ\!\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}
X(f)=j/2 \cdot [\delta (f+f_0)-\delta (f-f_0)]$.
:* The coefficient  $D(1)$   $\Rightarrow$  $($frequency:  $+f_0)$  is imaginary and has the imaginary part  $-0.5$.  Accordingly,  $\textrm{Im}[D(15)]=+0.5$   ⇒   $($frequency:  $-f_0)$  applies.

{{BlaueBox|TEXT=
'''(8)'''  New setting:  DFT of signal  $\rm (g)$:  Cosine signal (two periods).  Interpret the result in comparison to exercise  '''(5)'''.
}}

:* Here the time continuous Fourier transform reads  $x(t)=\cos(2\pi \cdot (2 f_0) \cdot t)\hspace{0.15cm}\circ\!\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}X(f)=0.5 \cdot \delta (f- 2 f_0)+0.5 \cdot \delta (f+ 2 f_0)$.
:* $D(2)$  is representative of the frequency  $2 f_0$.  Due to the periodicity,  $D(14)=D(-2)$:   $D(2)=D(14)=0.5$ is representative of the frequency  $-2 f_0$.

{{BlaueBox|TEXT=
'''(9)'''  Now examine the case DFT of a sinodial signal (two periods).  Which modifications do you need to make in the time domain?  Interpret the result.
}}

:* The desired signal can be obtained from the DFT of signal  $\rm (g)$:  Cosine signal (two periods) with two shifts.  With the result of  '''(7)''':  Four shifts.
:* The DFT result is accordingly  $\textrm{Im}[D(2)]=-0.5$  and  $\textrm{Im}[D(14)]=+0.5$.

{{BlaueBox|TEXT=
'''(10)'''  New setting: DFT of signal  $\rm (h)$:  Alternating time coefficients.  Interpret the DFT result.}}

:* Here, the time continuous Fourier transform is given by:  $x(t)=\cos(2\pi \cdot (8 f_0) \cdot t)\hspace{0.15cm}\circ\!\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}
X(f)=0.5 \cdot \delta (f- 8 f_0)+0.5 \cdot \delta (f+ 8 f_0)$.
:* $8 f_0$ is the highest frequency that can be displayed with  $N=16$  in the DFT.  There are only two sampled values per period, namely $+1$ and $-1$.
:* Difference to exercise  '''(5)''':  $D(1)=0.5$  now becomes  $D(8)=0.5$.  Likewise,  $D(15)=0.5$  is shifted to  $D(8)=0.5$.  Final result:  $D(8)=1$.

{{BlaueBox|TEXT=
'''(11)'''  What are the differences between the two settings DFT from signal  $\rm (i)$:  Dirac delta impulse and   IDFT from spectrum  $\rm (I)$:  Dirac delta spectrum?
}}

:* None! In the first case, all coefficients are  $D(\mu)=1$ (real);  in the second case, however, equivalently  $d(\nu)=1$ (real).

{{BlaueBox|TEXT=
'''(12)'''  Are there differences in shifting the real  "$1$"  in the according input fields by one place at a time, that is for  $d(\nu = 1)=1$  and  $D(\mu = 1)=1$?
}}

:* The first case  $\Rightarrow$  $\textrm{Re}[d(\nu = 1)]=1$  results in the complex exponential function in the frequency domain given by  $X(f)= \textrm{e}^{-{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi\hspace{0.05cm}\cdot\hspace{0.05cm} f/f_0}$  with negative sign.
:* The second case  $\Rightarrow$  $\textrm{Re}[D(\mu = 1)]=1$ results in the complex exponential function in the time domain given by  $x(t)= \textrm{e}^{+{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi\hspace{0.05cm}\cdot\hspace{0.05cm} f_0\cdot t}$  with positive sign.
:* Note:  With  $\textrm{Re}[D(\mu=15)]=1$  the result in the time domain would also be a complex exponential function  $x(t)= \textrm{e}^{-{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi\hspace{0.05cm}\cdot\hspace{0.05cm} f_0\hspace{0.05cm}\cdot\hspace{0.05cm} t}$ with negative sign.

{{BlaueBox|TEXT=
'''(13)'''  New setting: DFT of signal  $\rm (k)$:  Triangle pulse.  Interpret the  $d(\nu)$  assignment under the assumption  $T_\textrm{A} = 1$ ms.}}

:* Change the display to "absolute value".  $x(t)$ is symmetrical around  $t=0$  and extends from  $-8 \cdot T_\textrm{A} = -8$  ms to  $+8 \cdot T_\textrm{A}= +8$  ms.
:* $d(\nu)$  assignment:  $d(0)=x(0)=1$,  $d(1)=x(T_\textrm{A})=0.875$, ... ,  $d(8)=x(8 T_\textrm{A})=0$,  $d(9)=x(-7 T_\textrm{A})=0.125$, ... ,  $d(15)=x(-T_\textrm{A})=0.875$.

{{BlaueBox|TEXT=
'''(14)'''  Same setting as  '''(13)'''.  Interpret the DFT result, especially the coefficients  $D(0)$,  $D(1)$,  $D(2)$  and  $D(15)$.}}

:* In the frequency range  $D(0)$  stands for the frequency  $f=0$  and  $D(1)$  and  $D(15)$  for the frequencies  $\pm f_\textrm{A}$.  It holds that  $f_\textrm{A}= 1/ (N\cdot T_\textrm{A})=62.5$  Hz.
:* For the value of the continuous spectrum at $f=0$ the following applies:   $X(f=0)=D(0)/f_\textrm{A} = 0.5/ (0.0625$ kHz$)=8\cdot \textrm{kHz}^{-1}$.
:* The first zero of the  $\textrm{si}^2$–shaped spectrum  $X(f)$  occurs at  $2\cdot f_\textrm{A} = 125$ Hz.  The other zeros are equidistant.

{{BlaueBox|TEXT=
'''(15)'''  New setting: DFT of signal  $\rm (i)$:  Rectangular pulse.  Interpret the displayed results.}}
:* The set (symmetrical) rectangle extends over  $\pm 4 \cdot T_\textrm{A}$.  At the edges, the time coefficients are only half as large:  $d(4)=d(12)=0.5$.
:* The further statements of  '''(14)'''  also apply to this  $\textrm{si}$–shaped spectrum  $X(f)$.

{{BlaueBox|TEXT=
'''(16)'''  Same setting as for  '''(15)'''.  Which modifications need to be made in the  $d(\nu)$  field,
to have the duration of the rectangle   $\Rightarrow$   $\pm 2 \cdot T_\textrm{A}$.
}}
:* $d(0) = d(1) = d(15) =1, \ d(2) = d(14) = 0.5$. All other time coefficients zero  ⇒   first zero of the  ${\rm si}$ spectrum at  $4 \cdot f_{\rm A}= 250\text{ Hz}$.

{{BlaueBox|TEXT=
'''(17)'''  New setting:  IDFT of spectrum  $\rm (L)$:  Gaussian spectrum.  Interpret the result in the time domain.}}
:* Here, the time function  $x(t)$  is Gaussian with the maximum  $x(t=0)=4$.  For the spectrum the following applies:  $X(f=0)=D(0)/f_\textrm{A} = 16 \cdot \textrm{kHz}^{-1}$.
:* The equivalent duration of the pulse is  $\Delta t = X(f=0)/x(t=0)=4\text{ ms}$.  The inverse value gives the equivalent bandwidth  $\Delta f = 1/\Delta t = 250\text{ Hz}$.

==Applet Manual==
 
[[File:Anleitung_DFT_endgültig.png|left|600px|frame|Screenshot of the German version]]
    '''(A)'''     Time domain (input and result field)

    '''(B)'''     '''(A)''' representation numerical, graphical, magnitude

    '''(C)'''     frequency domain (input and result field)

    '''(D)'''     '''(C)''' representation numerical, graphical, magnitude

    '''(E)'''     Selection: DFT  $(t \to f)$  or IDFT  $(f \to t)$

    '''(F)'''     Given  $d(\nu)$ assignments (if DFT), or

                    Given  $D(\mu)$ assignments (if IDFT)

    '''(G)'''     Set input field to zero

    '''(H)'''     Move input field cyclically down (or up)

    '''( I )'''     Range for experiment execution:   exercise selection

    '''(J)'''     Range for experiment execution:   exercise definition

    '''(K)'''     Range for experiment execution:   show sample solution
 
*Given  $d(\nu)$ assignments (for DFT):

:(a)  corresponding number field,  (b)  DC signal,  (c)  Complex exponential function of time,  (d)  Harmonic oscillation  $($phase  $\varphi = 45^\circ)$,
:(e)  Cosine signal (one period),  (f)  Sine signal (one period),  (g)  Cosine signal (two periods), (h)  Alternating time coefficients,
:  (i)  Dirac delta pulse,  (j)  Rectangular pulse,  (k)  Triangular pulse,  (l)  Gaussian pulse.

*Given  $D(\mu)$ assignments (for IDFT):

:(A)  corresponding number field,  (B)  Constant spectrum,  (C)  Complex exponential function of frequency,  (D)  equivalent to setting (d) in time domain ,
:(E)  Cosine signal (one frequency period),  (F)  Sine signal (one frequency period),  (G)  Cosine signal (two frequency periods),  (H)  Alternating spectral coefficients,
:(I)  Dirac delta spectrum,  (J)  Rectangular spectrum,  (K)  Triangular spectrum,  (L)  Gaussian spectrum.

==About the Authors==

This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Thomas_Gro.C3.9Fer_.28Diplomarbeit_LB_2006.2C_danach_freie_Mitarbeit_bis_2010.29|»Thomas Großer«]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|»Carolin Mirschina«]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants") of the TUM Faculty EI. We thank them.

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|dft_en|dft}}

Applets:Binomial and Poisson Distribution (Applet)

2023-03-26T20:40:48Z

Noah:

{{LntAppletLinkEn|binomPoissonDistributions_en}}         [https://www.lntwww.de/Applets:Binomial-_und_Poissonverteilung_(Applet) '''English Applet with German WIKI description''']

==Applet Description==

This applet allows the calculation and graphical display of
*the probabilities ${\rm Pr}(z=\mu)$ of a discrete random variable $z \in \{\mu \} = \{0, 1, 2, 3, \text{...} \}$, that determine its ''Probability Density Function'' (PDF) – here representation with Dirac delta functions ${\rm \delta}( z-\mu)$:
:$$f_{z}(z)=\sum_{\mu=1}^{M}{\rm Pr}(z=\mu)\cdot {\rm \delta}( z-\mu),$$
*the probabilities ${\rm Pr}(z \le \mu)$ of the ''Cumulative Distribution Function'' (CDF):
:$$F_{z}(\mu)={\rm Pr}(z\le\mu).$$

Discrete distributions are available in two sets of parameters:
* the Binomial distribution with the parameters $I$ and $p$   ⇒   $z \in \{0, 1, \text{...} \ , I \}$   ⇒   $M = I+1$ possible values,
*the Poisson distribution with the parameter $\lambda$   ⇒   $z \in \{0, 1, 2, 3, \text{...}\}$   ⇒   $M \to \infty$.

In the exercises below you will be able to compare:
* two Binomial distributions with different sets of parameters $I$ and $p$,
* two Poisson distributions with different rates $\lambda$,
*a Binomial distribution with a Poisson distribution.

==Theoretical Background==

===Properties of the Binomial Distribution===
The ''Binomial distribution'' represents an important special case for the likelihood of occurence of a discrete random variable. For the derivation we assume, that $I$ binary and statistically independent random variables $b_i \in \{0, 1 \}$ can take
*the value $1$ with the probability ${\rm Pr}(b_i = 1) = p$, and
*the value $0$ with the probability ${\rm Pr}(b_i = 0) = 1-p$.

The sum
:$$z=\sum_{i=1}^{I}b_i$$
is also a discrete random variable with symbols from the set $\{0, 1, 2, \cdots\ , I\}$ with size $M = I + 1$ and is called "binomially distributed".

'''Probabilities of the Binomial Distribution'''

The probabilities to find $z = \mu$ for $μ = 0, \text{...}\ , I$ are given as
:$$p_\mu = {\rm Pr}(z=\mu)={I \choose \mu}\cdot p^\mu\cdot ({\rm 1}-p)^{I-\mu},$$
with the number of combinations $(I \text{ over }\mu)$:
:$${I \choose \mu}=\frac{I !}{\mu !\cdot (I-\mu) !}=\frac{ {I\cdot (I- 1) \cdot \ \cdots \ \cdot (I-\mu+ 1)} }{ 1\cdot 2\cdot \ \cdots \ \cdot \mu}.$$

'''Moments of the Binomial Distribution'''

Consider a binomially distributed random variable $z$ and its expected value of order $k$:
:$$m_k={\rm E}[z^k]=\sum_{\mu={\rm 0}}^{I}\mu^k\cdot{I \choose \mu}\cdot p^\mu\cdot ({\rm 1}-p)^{I-\mu}.$$

We can derive the formulas for
*the linear average:   $m_1 = I\cdot p,$
*the second moment:   $m_2 = (I^2-I)\cdot p^2+I\cdot p,$
*the variance and standard deviation:   $\sigma^2 = {m_2 - m_1^2} = {I \cdot p\cdot (1-p)} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
\sigma = \sqrt{I \cdot p\cdot (1-p)}.$

'''Applications of the Binomial Distribution'''

The Binomial distribution has a variety of uses in telecommunications as well as in other disciplines:
*It characterizes the distribution of rejected parts (Ausschussstücken) in statistical quality control.
*The simulated bit error rate of a digital transmission system is technically a binomially distributed random variable.
*The binomial distribution can be used to calculate the residual error probability with blockwise coding, as the following example shows.

{{GraueBox|TEXT=
$\text{Example 1:}$ 
When transfering blocks of $I =5$ binary symbols through a channel, that
*distorts a symbol with probability $p = 0.1$   ⇒   random variable $e_i = 1$, and
*transfers the symbol undistorted with probability $1 - p = 0.9$   ⇒   random variable $e_i = 0$,

the new random variable $f$ ("error per block") calculates to:
:$$f=\sum_{i=1}^{I}e_i.$$

$f$ can now take integer values between $\mu = 0$ (all symbols are correct) and $\mu = I = 5$ (all five symbols are erroneous). We describe the probability of $\mu$ errors as $p_μ = {\rm Pr}(f = \mu)$.
*The case that all five symbols are transmitted correctly occurs with the probability of $p_0 = 0.9^{5} ≈ 0.5905$. This can also be seen from the binomial formula for $μ = 0$ , considering the definition $5\text{ over } 0 = 1$.
*A single error $(f = 1)$ occurs with the probability $p_1 = 5\cdot 0.1\cdot 0.9^4\approx 0.3281$. The first factor indicates, that there are $5\text{ over } 1 = 5$ possibe error positions. The other two factors take into account, that one symbol was erroneous and the other four are correct when $f =1$.
*For $f =2$ there are $5\text{ over } 2 = (5 \cdot 4)/(1 \cdot 2) = 10$ combinations and you get a probability of $p_2 = 10\cdot 0.1^2\cdot 0.9^3\approx 0.0729$.

If a block code can correct up to two errors, the residual error probability is $p_{\rm R} = 1-p_{\rm 0}-p_{\rm 1}-p_{\rm 2}\approx 0.85\%$.
A second calculation option would be $p_{\rm R} = p_{3} + p_{4} + p_{5}$ with the approximation $p_{\rm R} \approx p_{3} = 0.81\%.$

The average number of errors in a block is $m_f = 5 \cdot 0.1 = 0.5$ and the variance of the random variable $f$ is $\sigma_f^2 = 5 \cdot 0.1 \cdot 0.9= 0.45$   ⇒   standard deviation $\sigma_f \approx 0.671.$}}

===Properties of the Poisson Distribution===

The ''Poisson distribution'' is a special case of the Binomial distribution, where
* $I → \infty$ and $p →0$.
*Additionally, the parameter $λ = I · p$ must be finite.

The parameter $λ$ indicates the average number of "ones" in a specified time unit and is called ''rate''.

Unlike the Binomial distribution where $0 ≤ μ ≤ I$, here, the random variable can assume arbitrarily large non-negative integers, which means that the number of possible values is not countable. However, since no intermediate values can occur, the Poisson distribution is still a "discrete distribution".

'''Probabilities of the Poisson Distribution'''

With the limits $I → \infty$ and $p →0$, the likelihood of occurence of the Poisson distributed random variable $z$ can be derived from the probabilities of the Binomial distribution:
:$$p_\mu = {\rm Pr} ( z=\mu ) = \lim_{I\to\infty} \cdot \frac{I !}{\mu ! \cdot (I-\mu )!} \cdot (\frac{\lambda}{I} )^\mu \cdot ( 1-\frac{\lambda}{I})^{I-\mu}.$$
After some algebraic transformations we finally obtain
:$$p_\mu = \frac{ \lambda^\mu}{\mu!}\cdot {\rm e}^{-\lambda}.$$

'''Moments of the Poisson Distribution'''

The moments of the Poisson distribution can be derived directly from the corresponding equations of the Binomial distribution by taking the limits again:
:$$m_1 =\lim_{\left.{I\hspace{0.05cm}\to\hspace{0.05cm}\infty, \hspace{0.2cm} {p\hspace{0.05cm}\to\hspace{0.05cm} 0}}\right.} \hspace{0.2cm} I \cdot p= \lambda,\hspace{0.8cm}
\sigma =\lim_{\left.{I\hspace{0.05cm}\to\hspace{0.05cm}\infty, \hspace{0.2cm} {p\hspace{0.05cm}\to\hspace{0.05cm} 0}}\right.} \hspace{0.2cm} \sqrt{I \cdot p \cdot (1-p)} = \sqrt {\lambda}.$$

We can see that for the Poisson distribution $\sigma^2 = m_1 = \lambda$ always holds. In contrast, the moments of the Binomial distribution always fulfill $\sigma^2 < m_1$.

[[File: P_ID616__Sto_T_2_4_S2neu.png |frame| Moments of Poisson Distribution]]
{{GraueBox|TEXT=
$\text{Example 2:}$ 
We now compare the Binomial distribution with parameters $I =6$ und $p = 0.4$ with the Poisson distribution with $λ = 2.4$:
*Both distributions have the same linear average $m_1 = 2.4$.
*The standard deviation of the Poisson distribution (marked red in the figure) is $σ ≈ 1.55$.
*The standard deviation of the Binomial distribution (marked blue) is $σ = 1.2$.
}}

'''Applications of the Poisson Distribution'''

The Poisson distribution is the result of a so-called ''Poisson point process'' which is often used as a model for a series of events that may occur at random times. Examples of such events are
* failure of devices - an important task in reliability theory,
* shot noise in the optical transmission simulations, and
* the start of conversations in a telephone relay center („Teletraffic engineering”).

{{GraueBox|TEXT=
$\text{Example 3:}$ 
A telephone relay receives ninety requests per minute on average $(λ = 1.5 \text{ per second})$. The probabilities $p_µ$, that in an arbitrarily large time frame exactly $\mu$ requests are received, is:
:$$p_\mu = \frac{1.5^\mu}{\mu!}\cdot {\rm e}^{-1.5}.$$

The resulting numerical values are $p_0 = 0.223$, $p_1 = 0.335$, $p_2 = 0.251$, etc.

From this, additional parameters can be derived:
* The distance $τ$ between two requests satisfies the "exponential distribution",
* The mean time span between two requests is ${\rm E}[τ] = 1/λ ≈ 0.667 \ \rm s$.
}}

=== Comparison of Binomial and Poisson Distribution ===
This section deals with the similarities and differences between Binomial and Poisson distributions.

[[File: EN_Sto_T_2_4_S3.png |frame| Binomial vs. Poisson distribution]]

The '''Binomial distribution''' is used to describe stochastic events, that have a fixed period $T$. For example the period of an ISDN (''Integrated Services Digital Network'') network with $64 \ \rm kbit/s$ is $T \approx 15.6 \ \rm \mu s$.
* Binary events such as the error-free $(e_i = 0)$/ faulty $(e_i = 1)$ transmission of individual symbols only occur in this time frame.
* With the Binomial distribution, it is possible to make statistical statements about the number of expected erros in a period $T_{\rm I} = I · T$, as is shown in the time figure above (marked blue).
* For very large values of $I$ and very small values of $p$, the Binomial distribution can be approximated by the ''Poisson distribution'' with rate $\lambda = I \cdot p$.
* If at the same time $I · p \gg 1$, the Poisson distribution as well as the Binomial distribution turn into a discrete Gaussian distribution according to the ''de Moivre-Laplace Theorem''.

The '''Poisson distribution''' can also be used to make statements about the number of occuring binary events in a finite time interval.

By assuming the same observation period $T_{\rm I}$ and increasing the number of partial periods $I$, the period $T$, in which a new event ($0$ or $1$) can occur, gets smaller and smaller. In the limit where $T$ goes to zero, this means:
* With the Poisson distribution binary events can not only occur at certain given times, but at any time, which is illustrated in the second time chart.
* In order to get the same number of "ones" in the period $T_{\rm I}$ - in average - as in the Binomial distribution (six pulses in the example), the characteristic probability $p = {\rm Pr}( e_i = 1)$ for an infinitesimal small time interval $T$ must go to zero.

==Exercises==

* First select the number  $(1,\text{...}, 7)$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
*In these exercises, the term  '''Blue'''  refers to distribution function 1 (marked blue in the applet) and the term  '''Red'''  refers to distribution function 2 (marked red in applet).

{{BlaueBox|TEXT=
'''(1)'''  Set '''Blue''' to Binomial distribution $(I=5, \ p=0.4)$ and '''Red''' to Binomial distribution $(I=10, \ p=0.2)$.
:What are the probabilities ${\rm Pr}(z=0)$ and ${\rm Pr}(z=1)$?
}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{Blue: }{\rm Pr}(z=0)=0.6^5=7.78\%, \hspace{0.3cm}{\rm Pr}(z=1)=0.4 \cdot 0.6^4=25.92\%;$

$\hspace{1.85cm}\text{Red: }{\rm Pr}(z=0)=0.8^{10}=10.74\%, \hspace{0.3cm}{\rm Pr}(z=1)=0.2 \cdot 0.8^9=26.84\%.$

{{BlaueBox|TEXT=
'''(2)'''  Using the same settings as in '''(1)''', what are the probabilities ${\rm Pr}(3 \le z \le 5)$?
}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{Note that }{\rm Pr}(3 \le z \le 5) = {\rm Pr}(z=3) + {\rm Pr}(z=4) + {\rm Pr}(z=5)\text{, or }
{\rm Pr}(3 \le z \le 5) = {\rm Pr}(z \le 5) - {\rm Pr}(z \le 2)$

$\hspace{1.85cm}\text{Blue: }{\rm Pr}(3 \le z \le 5) = 0.2304+ 0.0768 + 0.0102 =1 - 0.6826 = 0.3174;$

$\hspace{1.85cm}\text{Red: }{\rm Pr}(3 \le z \le 5) = 0.2013 + 0.0881 + 0.0264 = 0.9936 - 0.6778 = 0.3158.$

{{BlaueBox|TEXT=
'''(3)'''  Using the same settings as in '''(1)''', what are the differences in the linear average $m_1$ and the standard deviation $\sigma$ between the two Binomial distributions?
}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{Average:}\hspace{0.2cm}m_\text{1} = I \cdot p\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m_\text{1, Blue} = 5 \cdot 0.4\underline{ = 2 =} \ m_\text{1, Red} = 10 \cdot 0.2; $

$\hspace{1.85cm}\text{Standard deviation:}\hspace{0.4cm}\sigma = \sqrt{I \cdot p \cdot (1-p)} = \sqrt{m_1 \cdot (1-p)}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} \sigma_{\rm Blue} = \sqrt{2 \cdot 0.6} =1.095 < \sigma_{\rm Red} = \sqrt{2 \cdot 0.8} = 1.265.$

{{BlaueBox|TEXT=
'''(4)'''  Set '''Blue''' to Binomial distribution $(I=15, p=0.3)$ and '''Red''' to Poisson distribution $(\lambda=4.5)$.
:What differences arise between both distributions regarding the average $m_1$ and variance $\sigma^2$?
}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm}\text{Both distributions have the same average:}\hspace{0.2cm}m_\text{1, Blue} = I \cdot p\ = 15 \cdot 0.3\hspace{0.15cm}\underline{ = 4.5 =} \ m_\text{1, Red} = \lambda$;

$\hspace{1.85cm} \text{Binomial distribution: }\hspace{0.2cm} \sigma_\text{Blue}^2 = m_\text{1, Blue} \cdot (1-p)\hspace{0.15cm}\underline { = 3.15} < \text{Poisson distribution: }\hspace{0.2cm} \sigma_\text{Red}^2 = \lambda\hspace{0.15cm}\underline { = 4.5}$;

{{BlaueBox|TEXT=
'''(5)'''  Using the same settings as in '''(4)''', what are the probabilities ${\rm Pr}(z \gt 10)$ and ${\rm Pr}(z \gt 15)$?
}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Binomial: }\hspace{0.2cm} {\rm Pr}(z \gt 10) = 1 - {\rm Pr}(z \le 10) = 1 - 0.9993 = 0.0007;\hspace{0.3cm} {\rm Pr}(z \gt 15) = 0 \ {\rm (exactly)}$.

$\hspace{1.85cm}\text{Poisson: }\hspace{0.2cm} {\rm Pr}(z \gt 10) = 1 - 0.9933 = 0.0067;\hspace{0.3cm}{\rm Pr}(z \gt 15) \gt 0\hspace{0.2cm}( \approx 0)$;

$\hspace{1.85cm}\text{Approximation: }\hspace{0.2cm}{\rm Pr}(z \gt 15) \ge {\rm Pr}(z = 16) = \lambda^{16} /{16!}\approx 2 \cdot 10^{-22}$

{{BlaueBox|TEXT=
'''(6)'''  Using the same settings as in '''(4)''', which parameters lead to a symmetric distribution around $m_1$?
}}

$\hspace{1.0cm}\Rightarrow\hspace{0.3cm} \text{Binomial distribution with }p = 0.5\text{: }p_\mu = {\rm Pr}(z = \mu)\text{ symmetric around } m_1 = I/2 = 7.5 \ ⇒ \ p_μ = p_{I–μ}\ ⇒ \ p_8 = p_7, \ p_9 = p_6, \text{etc.}$

$\hspace{1.85cm}\text{In contrast, the Poisson distribution is never symmetric, since it extends to infinity!}$

==Applet Manual==
[[File:Handhabung_binomial.png|left|600px]]
    '''(A)'''     Preselection for blue parameter set

    '''(B)'''     Parameter input: Sliders $I$ and $p$

    '''(C)'''     Preselection for Red parameter set

    '''(D)'''     Parameter input: Slider $\lambda$

    '''(E)'''     Graphic display of the Distribution

    '''(F)'''     Output of moments for blue parameter set

    '''(G)'''     Output of moments for redparameter set

    '''(H)'''     Variation possibilities for the graphic display

$\hspace{1.5cm}$"$+$" (Zoom in),

$\hspace{1.5cm}$ "$-$" (Zoom out)

$\hspace{1.5cm}$ "$\rm o$" (Reset)

$\hspace{1.5cm}$ "$\leftarrow$" (Move left), etc.

    '''( I )'''     Output of ${\rm Pr} (z = \mu)$ and ${\rm Pr} (z \le \mu)$

    '''(J)'''     Exercises: Exercise selection, description and solution
 
 '''Other options for graphic display''':
*Hold shift and scroll: Zoom in on/out of coordinate system,
*Hold shift and left click: Move the coordinate system.

==About the Authors==
This interactive calculation was designed and realized at the  [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik]  of the  [https://www.tum.de/ Technische Universität München].
*The original version was created in 2003 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]] as part of her Diploma thesis using "FlashMX–Actionscript"  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]]  as part of his Bachelor's thesis (Supervisor:  [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) .

==Once again: Open Applet in new Tab==

{{LntAppletLinkEn|binomPoissonDistributions_en}}         [https://www.lntwww.de/Applets:Binomial-_und_Poissonverteilung_(Applet) '''English Applet with German WIKI description''']

Applets:Bessel Functions of the First Kind

2023-03-26T20:20:48Z

Noah:

{{LntAppletLinkEnDe|besselFuns_en|besselFuns}}
==Applet Description==
 
This applet allows the calculation and graphical representation of the Bessel functions of the first kind and  $n$–th order according to the series representation:

:$${\rm J}_n (x) = \sum\limits_{k=0}^{\infty}\frac{(-1)^k \cdot (x/2)^{n \hspace{0.05cm} + \hspace{0.05cm} 2
\hspace{0.05cm}\cdot \hspace{0.05cm}k}}{k! \cdot (n+k)!} \hspace{0.05cm}.$$

*The functions  ${\rm J}_n (x)$  can be represented graphically for the order  $n=0$  to  $n=9$  in different colors.
*The left output provides the function values  ${\rm J}_0 (x = x_1)$, ... , ${\rm J}_9 (x = x_1)$  for a slider-settable value  $x_1$  in the range  $0 \le x_1 \le 15$  with increment  $0.5$.
*The right output provides the function values  ${\rm J}_0 (x = x_2)$, ... , ${\rm J}_9 (x = x_2)$  for a slider-settable value  $x_2$  (same range and value increment like on the left).

==Theoretical Background==
 
===General Information about the Bessel Functions===
Bessel functions (or cylinder functions) are solutions of the Bessel differential equation of the form

:$$x^2 \cdot \frac{ {\rm d}^2}{{\rm d}x^2}\ {\rm J}_n (x) \ + \ x \cdot \frac{ {\rm d}}{{\rm d}x}\ {\rm J}_n (x) \ + \ (x^2 - n^2)
\cdot {\rm J}_n (x)= 0. $$

This is an ordinary second-order linear differential equation.  The parameter  $ n $  is usually integer, also in this program.  These mathematical functions, introduced by  [https://en.wikipedia.org/wiki/Friedrich_Bessel "Friedrich Wilhelm Bessel"]  in 1844, can also be represented in closed form as integrals:

:$${\rm J}_n (x) = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}[\hspace{0.05cm}x \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha \hspace{0.05cm}]}}\hspace{0.1cm}{\rm d}\alpha
\hspace{0.05cm}.$$

The functions  ${\rm J}_n (x)$  belong to the class of Bessel functions of the first kind.  The parameter  $n$  is called the "order".

''Note:''   There are a number of modifications of the Bessel functions, including the Bessel functions of the second kind named  ${\rm Y}_n (x)$.  For integer  $n$,  ${\rm Y}_n (x)$  can be replaced by  ${\rm J}_n (x)$  functions.  However, in this applet, only the first kind Bessel functions  ${\rm J}_n (x)$  are considered.
 

===Properties of the Bessel Functions===

{{BlaueBox|TEXT=
$\text{Property (A):}$   If the function values for  $n = 0$  and  $n = 1$  are known, then the Bessel function for  $n ≥ 2$  can be determined iteratively:
:$${\rm J}_n (x) ={2 \cdot (n-1)}/{x} \cdot {\rm J}_{n-1} (x) - {\rm J}_{n-2} (x) \hspace{0.05cm}.$$
}}

{{GraueBox|TEXT=
$\text{Example (A):}$   Let  ${\rm J}_0 (x = 2) = 0.22389$  and  ${\rm J}_1 (x= 2) = 0.57672$.  From this it can be calculated iteratively:
:$${\rm J}_2 (x= 2) ={2 \cdot 1}/{2} \cdot {\rm J}_{1} (x= 2) - {\rm J}_{0} (x= 2) = 0.57672 - 0.22389 = \hspace{0.15cm}\underline{0.35283}\hspace{0.05cm},$$
:$${\rm J}_3 (x= 2) ={2 \cdot 2}/{2} \cdot {\rm J}_{2} (x= 2) - {\rm J}_{1} (x= 2) = 2 \cdot 0.35283 - 0.57672 = \hspace{0.15cm}\underline{0.12894}\hspace{0.05cm},$$
:$${\rm J}_4 (x= 2) ={2 \cdot 3}/{2} \cdot {\rm J}_{3} (x= 2) - {\rm J}_{2} (x= 2) = 3 \cdot 0.12894 - 0.35283 = \hspace{0.15cm}\underline{0.03400}\hspace{0.05cm}.$$
}}

{{BlaueBox|TEXT=
$\text{Property (B):}$   The symmetry relationship applies  ${\rm J}_{–n}(x) = (–1)^n · {\rm J}_n(x)$:
:$${\rm J}_{-1}(x) = - {\rm J}_{1}(x), \hspace{0.3cm}{\rm J}_{-2}(x) = {\rm J}_{2}(x), \hspace{0.3cm}{\rm J}_{-3}(x) = - {\rm J}_{3}(x).$$
}}

{{GraueBox|TEXT=
$\text{Example (B):}$   For the spectrum of the analytic signal, the following applies to the case  "phase modulation of a sinusoidal signal":
[[File:Mod_T_3_1_S4_version2.png|right|frame|Phase Modulation:   Spectrum of the Analytic Signal]]
:$$S_{\rm +}(f) = A_{\rm T} \cdot \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot \delta \big[f - (f_{\rm T}+ n \cdot f_{\rm N})\big]\hspace{0.05cm}.$$
Denote this
*$f_{\rm T}$  the carrier frequency   (German:   $\rm T$rägerfrequenz),
*$f_{\rm N}$  the frequency of the source signal   (German:   $\rm N$achrichtenfrequenz),
* $A_{\rm T}$  the carrier amplitude   (German:   $\rm T$rägeramplitude).

Parameter of the Bessel functions in this application is the modulation index  $\eta$.

Based on the graphic, the following statements are possible:
*$S_+(f)$  consists here of an infinite number of discrete lines at a distance of $f_{\rm N}$.
*Thus the sectrum is in principle infinitely extended.
*The weights of the spectral lines at $f_{\rm T} + n · f_{\rm N}$  $(n$ integer$)$  are determined by the modulation index  $η$  over the Bessel functions  ${\rm J}_n(η)$.
*The spectral lines are real in the case of a sinusoidal source signal and a cosinusoidal carrier and symmetrical about $f_{\rm T}$  for even  $n$.
*In the case of odd  $n$, a change of sign corresponding to  $\text{Property (B)}$  must be taken into account.
*The phase modulation of another phase oscillation of source and / or carrier signal provides the same magnitude spectrum.}}
 

===Applications of the Bessel Functions===

The applications of the Bessel functions in nature and engineering are diverse and play an important role in physics, for example:
* Electromagnetic waves in a cylindrical waveguide,
* Solutions to the radial Schroedinger equation,
* Pressure amplitudes of inviscid rotational flows,
* Heat conduction in a cylindrical object,
* Diffusion problems on a lattice,
* Dynamics of floating bodies,
*Frequency-dependent friction in circular pipelines,
* Angular resolution.

The Bessel functions belong to the "Special Functions" because of their many applications in mathematical physics.

In the following, we restrict ourselves to a few areas that are addressed in our tutorial  $\rm LNTwww$.

{{GraueBox|TEXT=
$\text{Example (C):} \hspace{0.5cm} \text{Use in Spectral Analysis} \ \Rightarrow \ \text{Kaiser-Bessel filter}$

The  '''Spectral Leakage Effect'''  is the falsification of the spectrum of a periodic and thereby time unlimited signal due to the implicit time limitation of the discrete Fourier transform (DFT).  This will be done, for example, by a spectrum analyzer
* in the time signal not existing frequency components simulated, and/or
* actually existing spectral components are obscured by sidelobes.

The task of the  [[Signal_Representation/Spectrum_Analysis|"Spectrum Analysis"]]  is to limit the influence of the spectral leakage effect by providing suitable window functions.

Such a window function provides, for example, the Kaiser–Bessel window   ⇒   see section  [[Signal_Representation/Spectrum_Analysis#Special_window_functions|"Special Window Functions"]].  Its time-discrete window function with the Bessel function zero order   ⇒   ${\rm J}_0(x)$, the parameter  $\alpha=3.5$  and the window length  $N$:
:$$w_\nu = \frac{ {\rm J}_0\big(\pi \cdot \alpha \cdot \sqrt{1 - (2\nu/N)^2}\big)}{ {\rm J}_0\big(\pi \cdot \alpha \big)}.$$
In the section  [[Signal_Representation/Spectrum_Analysis#Quality_criteria_of_window_functions|"Quality Criteria of Window Functions"]]  the parameters of the Kaiser-Bessel window are given.
*The large "Minimum Distance between the main lobe and side lobes" and the desired small "maximum Scaling Error" are favourable.
*Due to the very large "Equivalent Noise Width" the Kaiser-Bessel window cuts in the most important comparison criterion "Maximum Process Loss" but worse than the established Hamming and Hanning windows.}}

{{GraueBox|TEXT=
$\text{Example (D):} \hspace{0.5cm} \text{Rice Fading Channel Model}$

The  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading#A_very_general_description_of_the_mobile_communication_channel| "Rayleigh distribution"]]  describes the mobile channel on the assumption that there is no direct path and thus the multiplicative factor  $z(t) = x(t) + {\rm j} \cdot y(t)$  is composed solely of diffusely scattered components.

In the case of a direct component  (Line of Sight, LoS) one has to add in the model DC components  $x_0$  and/or  $y_0$  for the zero mean Gaussian processes  $x(t)$  and  $y(t)$ :

[[File:EN_Mob_T_1_4_S1.png|right|frame|Rice fading channel model|class=fit]]
:$$x(t) \hspace{0.1cm} \Rightarrow \hspace{0.1cm} x(t) +x_0 \hspace{0.05cm}, $$
:$$y(t) \hspace{0.1cm} \Rightarrow \hspace{0.1cm} y(t) +y_0\hspace{0.05cm},$$

:$$z(t) = x(t) + {\rm j} \cdot y(t)\hspace{0.1cm}\Rightarrow \hspace{0.1cm} z(t) +z_0 \hspace{0.05cm},$$
:$$ z_0 = x_0 + {\rm j} \cdot y_0\hspace{0.05cm}.$$

The graph shows the Rice fading channel model.  It can be summarized as follows:
*The real part  $x(t)$  is Gaussian with mean  $x_0$  and variance  $\sigma ^2$.
*The imaginary part  $y(t)$  is also Gaussian  $($mean  $y_0$,  equal variance  $\sigma ^2)$  and independent of  $x(t)$. 

*For  $z_0 \ne 0$, the magnitude  $\vert z(t)\vert$  is Rice distributed, from which the term "Rice Fading" arises.

*To simplify the notation, we set  $\vert z(t)\vert = a(t)$.  For $a < 0$ , the pdf is  $f_a(a) \equiv 0$, for  $a \ge 0$  the following equation holds, where  ${\rm I_0}(x)$  is the modified Bessel function of zero order:
:$$f_a(a) = \frac{a}{\sigma^2} \cdot {\rm e}^{ - (a^2 + \vert z_0 \vert ^2)/(2\sigma^2)} \cdot {\rm I}_0 \left [ \frac{a \cdot \vert z_0 \vert}{\sigma^2} \right ] \hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm I }_0 (u) = {\rm J }_0 ({\rm j} \cdot u) =
\sum_{k = 0}^{\infty} \frac{ (u/2)^{2k} }{k! \cdot \Gamma (k+1)}
\hspace{0.05cm}.$$
*Between the modified Bessel function  ${\rm J_0}(x)$  and the traditional Bessel function  ${\rm I_0}(x)$  of first kind the following relation exists:  ${\rm I }_0 (u) = {\rm J }_0 ({\rm j} \cdot u)$.}}

{{GraueBox|TEXT=
$\text{Example (E):} \hspace{0.5cm} \text{Analysis of the Frequency Spectrum of Frequency Modulated Signals}$

$\text{Example (B)}$  has already been shown that the angle modulation of a harmonic oscillation of the frequency $f_{\rm N}$  leads to a line spectrum.  The spectral lines are around the carrier frequency $f_{\rm T}$  at  $f_{\rm T} + n \cdot f_{\rm N}$  with  $n \in \{ \ \text{...}, -2, -1, \ 0, +1, +2, \text{...} \ \}$.  The weights of the Dirac delta lines are  ${\rm J }_n(\eta)$, depending on the modulation index  $\eta$.

The graph shows the magnitude spectrum  $\vert S_{\rm +}(f) \vert$  of the analytic signal in phase modulation (PM) and frequency modulation (FM), two different forms of angle modulation.  Bessel lines with values less than  $0.03$  are neglected in both cases.

[[File:P_ID1095__Mod_T_3_2_S4_neu.png|right|frame|Discrete spectra with phase modulation (left) and frequency modulation (right)]]

For the upper half of the graphic, the modulator parameters are chosen so that for each $f_{\rm N} = 5 \ \rm kHz$  here is a Bessel spectrum with the modulation index  $η = 1.5$.  Disregarding the phase relationships, the same spectra and the same signals result for both systems.

The lower graphics apply with otherwise identical settings for the same message frequency $f_{\rm N} = 3 \ \rm kHz$.  One notices:
*In phase modulation, the spectral function is narrower than  $f_{\rm N} = 5 \ \rm kHz$, because the distance of the Bessel lines is now only  $3 \ \rm kHz$.  Since the PN modulation index is independent of $f_{\rm N}$, the same bessel weights result as in $f_{\rm N} = 5 \ \rm kHz$.
*Also in the frequency modulation, the Bessel lines now occur at a distance of  $3 \ \rm kHz$.  However, since FM has a modulation index inversely proportional to $f_{\rm N}$, there are now significantly more Bessel lines at the bottom due to the larger modulation index  $η = 2.5$ than in the upper right chart  $($for  $η = 1.5$  valid$)$.}}

==Applet Manual==
[[File:Bessel_abzug3.png|left|600px]]
    '''(A)'''     Sum formula of the Bessel functions  ${\rm J}_n(x)$

    '''(B)'''     Selection of the order  $n$  for the graphical representation

    '''(C)'''     Plot area of the Bessel functions

    '''(D)'''     Variation of the graphic representation

$\hspace{1.5cm}$"$+$" (Enlarge),

$\hspace{1.5cm}$ "$-$" (Decrease)

$\hspace{1.5cm}$ "$\rm o$" (Reset to default)

$\hspace{1.5cm}$ "$\leftarrow$" (pushed to the left), usw.

    '''(E)'''     Selection of the abscissa value  $x_1$  for the left numeric output

    '''(F)'''     Numerical output of the Bessel function values  ${\rm J}_n(x_1)$

    '''(G)'''     Selection of the abscissa value  $x_2$ for  the right numeric output

    '''(F)'''     Numerical output of the Bessel function values  ${\rm J}_n(x_2)$
 

==About the Authors==
This interactive applet was designed and realized at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].

*The original version was created in 2005 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Markus_Elsberger_.28Diplomarbeit_LB_2006.29|Markus Elsberger]]  and  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]] as part of thesis using "FlashMX–Actionscript" (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Xiaohan_Liu_.28Bachelorarbeit_2018.29|Xiaohan Liu]]  as part of her Bachelor's thesis (Supervisor:  [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]).

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|besselFuns_en|besselFuns}}

Applets:Attenuation of Copper Cables

2023-03-25T21:04:29Z

Noah:

{{LntAppletLinkEn|attenuationCopperCables_en}}         [https://www.lntwww.de/Applets:Dämpfung_von_Kupferkabeln '''English Applet with German WIKI description''']

==Applet Description==
 
This applet calculates the attenuation function $a_{\rm K}(f)$ of conducted transmission media (with cable length $l$):
*For coaxial cables one usually uses the equation $a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l$.
*In contrast, two-wire lines are often displayed in the form $a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l$.
*The conversion of the $(k_1, \ k_2, \ k_3)$ parameters to the $(\alpha_0, \ \alpha_1, \ \alpha_2)$ parameters for $B = 30 \ \rm MHz$ is realized as well as the other way around.

Aside from the attenuation function $a_{\rm K}(f)$ the applet can display:
*the associated magnitude frequency response $\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20},$
*the equalizer frequency response $\left | H_{\rm E}(f)\right | = \left | H_{\rm CRO}(f) / H_{\rm K}(f)\right | $, that leads to a Nyquist total frequency response $ H_{\rm CRO}(f) $,
*the corresponding squared magnitude frequency response $\left | H_{\rm E}(f)\right |^2 $.

The integral over $\left | H_{\rm E}(f)\right |^2 $ is a measure of the noise exaggeration of the selected Nyquist total frequency response and thus also for the expected error probability. From this, the ''total efficiency''  $\eta_\text{K+E}$ for channel (ger.:'''K'''anal) and equalizer (ger.:'''E'''ntzerrer) is calculated, which is output in the applet in $\rm dB$.

Through optimization of the roll-off factor $r$ of the raised cosine frequency response $ H_{\rm CRO}(f) $ one gets the ''channel efficiency''  $ \eta_\text{K}$. This therefore indicates the deterioration of the overall system due to the attenuation function $ a _ {\ rm K} (f) $ of the transmission medium.

==Theoretical Background==
 
===Magnitude Frequency Response and Attenuation Function===
Following relationship exists between the magnitude frequency response and the attenuation function:
:$$\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20} = {\rm e}^{-a_\text{K, Np}(f)}.$$
*The index "K" makes it clear, that the considered LTI system is a cable (German : '''K'''abel).
*For the first calculation rule, the attenuation function $a_\text{K}(f)$ must be used in $\rm dB$ (decibel).
*For the second calculation rule, the attenuation function $a_\text{K, Np}(f)$ must be used in $\rm Np$ (Neper).
* The following conversions apply: $\rm 1 \ dB = 0.05 \cdot \ln (10) \ Np= 0.1151 \ Np$ or $\rm 1 \ Np = 20 \cdot \lg (e) \ dB= 8.6859 \ dB$.
* This applet exclusively uses dB values.

===Attenuation Function of a Coaxial Cable===
According to [Wel77]<ref name ='Wel77'>Wellhausen, H. W.: Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren. Frequenz 31, S. 23-28, 1977.</ref> the Attenuation Function of a Coaxial Cable of length $l$ is given as follows:
:$$a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l.$$
*It is important to note the difference between $a_{\rm K}(f)$ in $\rm dB$ and the "alpha" coefficient with other pseudo–units.
*The attenuation function $a_{\rm K}(f)$ is directly proportional to the cable length $l$; $\alpha_{\rm K}(f)= a_{\rm K}(f)/l$ is referred to as the "attenuation factor" or "kilometric attenuation".
*The frequency-independent component $α_0$ of the attenuation factor takes into account the Ohmic losses.
*The frequency-proportional portion $α_1 · f$ of the attenuation factor is due to the derivation losses ("crosswise loss") .
*The dominant portion $α_2$ goes back to [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable|"Skin effect"]], which causes a lower current density inside the conductor compared to its surface. As a result, the resistance of an electric line increases with the square root of the frequency.

The constants for the ''standard coaxial cable'' with a 2.6 mm inner diameter and a 9.5 mm outer diameter   ⇒  short '''Coax (2.6/9.5 mm)''' are:
:$$\alpha_0 = 0.014\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.0038\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 2.36\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

The same applies to the ''small coaxial cable''   ⇒  short '''Coax (1.2/4.4 mm)''':
:$$\alpha_0 = 0.068\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}
\alpha_1 = 0.0039\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 =5.2\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

These values can be calculated from the cables' geometric dimensions and have been confirmed by measurements at the Fernmeldetechnisches Zentralamt in Darmstadt – see [Wel77]<ref name ='Wel77'>Wellhausen, H. W.: Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren. Frequenz 31, S. 23-28, 1977.</ref> . They are valid for a temperature of 20° C (293 K) and frequencies greater than 200 kHz.

===Attenuation Function of a Two–wired Line===
According to [PW95]<ref name ='PW95'>Pollakowski, M.; Wellhausen, H.W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Mitteilung aus dem Forschungs- und Technologiezentrum der Deutschen Telekom AG, Darmstadt, Verlag für Wissenschaft und Leben Georg Heidecker, 1995.</ref> the attenuation function of a Two–wired Line of length $l$ is given as follows:
:$$a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l.$$
This function is not directly interpretable, but is a phenomenological description.

[PW95]<ref name ='PW95'>Pollakowski, M.; Wellhausen, H.W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Mitteilung aus dem Forschungs- und Technologiezentrum der Deutschen Telekom AG, Darmstadt, Verlag für Wissenschaft und Leben Georg Heidecker, 1995.</ref>also provides the constants determined by measurement results:
* $d = 0.35 \ {\rm mm}$:   $k_1 = 7.9 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 15.1 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.62$,
* $d = 0.40 \ {\rm mm}$:   $k_1 = 5.1 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 14.3 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.59$,
* $d = 0.50 \ {\rm mm}$:   $k_1 = 4.4 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 10.8 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.60$,
* $d = 0.60 \ {\rm mm}$:   $k_1 = 3.8 \ {\rm dB/km}, \hspace{0.2cm}k_2 = \hspace{0.25cm}9.2 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.61$.

From these numerical values one recognizes:
*The attenuation factor $α(f)$ and the attenuation function $a_{\rm K}(f) = α(f) · l$ depend significantly on the pipe diameter. The cables laid since 1994 with $d = 0.35 \ \rm mm$ and $d = 0.5\ \rm mm$ have a 10% greater attenuation factor than the older lines with $d = 0.4\ \rm mm$ or $d= 0.6\ \rm mm$.
*However, this smaller diameter, which is based on the manufacturing and installation costs, significantly reduces the range $l_{\rm max}$ of the transmission systems used on these lines, so that in the worst case scenario expensive intermediate regenerators have to be used.
*The current transmission methods for copper lines prove only a relatively narrow frequency band, for example $120\ \rm kHz$ with [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|ISDN]] and $\approx 1100 \ \rm kHz$ with [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_DSL|DSL]]. For $f = 1 \ \rm MHz$ the attenuation factor of a 0.4 mm cable is around $20 \ \rm dB/km$, so that even with a cable length of $l = 4 \ \rm km$ the attenuation does not exceed $80 \ \rm dB$.

===Conversion between $k$ and $\alpha$ parameters===
The $k$–parameters of the attenuation factor   ⇒   $\alpha_{\rm I} (f)$ can be converted into corresponding $\alpha$–parameters   ⇒   $\alpha_{\rm II} (f)$:
:$$\alpha_{\rm I} (f) = k_1 + k_2 \cdot (f/f_0)^{k_3}\hspace{0.05cm}, \hspace{0.2cm}{\rm with} \hspace{0.15cm} f_0 = 1\,{\rm MHz},$$
:$$\alpha_{\rm II} (f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}.$$

As a criterion of this conversion, we assume that the quadratic deviation of these two functions is minimal within a bandwidth $B$:
:$$\int_{0}^{B} \left [ \alpha_{\rm I} (f) - \alpha_{\rm II} (f)\right ]^2 \hspace{0.1cm}{\rm d}f \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Minimum} \hspace{0.05cm} .$$
It is obvious that $α_0 = k_1$. The parameters $α_1$ and $α_2$ are dependent on the underlying bandwidth $B$ and are:
:$$\begin{align*}\alpha_1 & = 15 \cdot (B/f_0)^{k_3 -1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{ {f_0} }\hspace{0.05cm} ,\\ \alpha_2 & = 10 \cdot (B/f_0)^{k_3 -0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{\sqrt{f_0} }\hspace{0.05cm} .\end{align*}$$

In the opposite direction the conversion rule for the exponent is:

:$$k_3 = \frac{A + 0.5} {A +1}, \hspace{0.2cm}\text{Auxiliary variable: }A = \frac{2} {3} \cdot \frac{\alpha_1 \cdot \sqrt{f_0}}{\alpha_2} \cdot \sqrt{B/f_0}.$$

With this result you can specify $ k_2 $ with each of the above equations.

{{GraueBox|TEXT=
$\text{Example 1:}$ 
*For $k_3 = 1$ (frequency-proportional attenuation factor) we get   $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_1 = {k_2}/{ {f_0} }\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = 0\hspace{0.05cm} .$
*For $k_3 = 0.5$ (Skin effect) we get the coefficients:   $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm}\alpha_1 = 0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = {k_2}/{\sqrt{f_0} }\hspace{0.05cm}.$
*For $k_3 < 0.5$ we get a negative $\alpha_1$. Conversion is only possible for $0.5 \le k_3 \le 1$.
*For $0.5 \le k_3 \le$ we get the coefficients $\alpha_1 > 0$ and $\alpha_2 > 0$, which are also dependent on $B/f_0$.
*From $\alpha_1 = 0.3\, {\rm dB}/ ({\rm km \cdot MHz}) \hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 3\, {\rm dB}/ ({\rm km \cdot \sqrt{MHz} })\hspace{0.05cm},\hspace{0.2cm}B = 30 \ \rm MHz$ folgt $k_3 = 0.63$ und $k_2 = 2.9 \ \rm dB/km$.}}

===Channel Influence on the Binary Nyquistent Equalization===
[[File:Applet_Kabeldaempfung_1_version_englisch.png|right|frame|Simplified block diagram of the optimal Nyquistent equalizer|class=fit]]
Going by the block diagram: Between the Dirac source and the (threshold) decider are the frequency responses for the transmitter (German: $\rm S$ender)  ⇒  $H_{\rm S}(f)$, channel (German: $\rm K$anal)  ⇒  $H_{\rm K}(f)$ and receiver (German: $\rm E$mpfänger)   ⇒  $H_{\rm E}(f)$.

In this applet
*we neglect the influence of the transmitted pulse form   ⇒   $H_{\rm S}(f) \equiv 1$   ⇒   dirac shaped transmission signal $s(t)$, and
*presuppose a binary Nyquist system with raised cosine around the Nyquist frequency $f_{\rm Nyq} = [f_1 + f_2]/2 =1(2T)$ :
:$$H_{\rm K}(f) · H_{\rm E}(f) = H_{\rm CRO}(f).$$

[[File:Applet_Kabeldaempfung_2_version2.png|right|frame|Frequency Response with raised cosine|class=fit]]

This means: The [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_frequency_domain|"first Nyquist criterion"]] is met ⇒   Timely successive impulses do not disturb each other ⇒   there are no [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|"Intersymbol Interferences"]].

In the case of white Gaussian noise, the transmission quality is thus determined solely by the noise power in front of the receiver:

:$$P_{\rm N} =\frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \ {\rm d}f\hspace{1cm}\text{with}\hspace{1cm}|H_{\rm E}(f)|^2 = \frac{|H_{\rm CRO}(f)|^2}{|H_{\rm K}(f)|^2}.$$

The lowest possible noise performance results with an ideal channel   ⇒   $H_{\rm K}(f) \equiv 1$ and a rectangular $H_{\rm CRO}(f) \equiv 1$ in $|f| \le f_{\rm Nyq}$:

:$$P_\text{N, min} = P_{\rm N} \ \big [\text{optimal system: }H_{\rm K}(f) \equiv 1, \ r=r_{\rm opt} =1 \big ] = N_0 \cdot f_{\rm Nyq} .$$

{{BlaueBox|TEXT=
$\text{Definitions:}$ 
*As a quality criterion for a given system we use the '''total efficiency''' with respect to the channel $\rm (K)$ and the receiver $\rm (E)$:

:$$\eta_\text{K+E} = \frac{P_{\rm N} \ \big [\text{Optimal system: Channel }H_{\rm K}(f) \equiv 1,\ \text{Roll-off factor } r=r_{\rm opt} =1 \big ]}{P_{\rm N} \ \big [\text{Given system: Channel }H_{\rm K}(f), \ \text{Roll-off factor }r \big ]} =\left [ \frac{1}{3/4 \cdot f_{\rm Nyq} } \cdot \int_{0}^{+\infty} \vert H_{\rm E}(f) \vert^2 \ {\rm d}f \right ]^{-1}\le 1.$$

This quality criterion is specified in the applet for both parameter sets in logarithm form:   $10 \cdot \lg \ \eta_\text{K+E} \le 0 \ \rm dB$.

*Through variation and optimization of the receiver   ⇒   roll-off factor $r$ we get the '''channel efficiency''':

:$$\eta_\text{K} = \min_{0 \le r \le 1} \ \eta_\text{K+E} .$$}}

[[File:Applet_Kabeldaempfung_3_version2.png|right|frame|Square value frequency response $\left \vert H_{\rm E}(f)\right \vert ^2 $|class=fit]]
{{GraueBox|TEXT=
$\text{Example 2:}$ 
The graph shows the square value frequency response $\left \vert H_{\rm E}(f)\right \vert ^2 $ with $\left \vert H_{\rm E}(f)\right \vert = H_{\rm CRO}(f) / \left \vert H_{\rm K}(f)\right \vert$ for the following boundary conditions:
*Attenuation function of the channel:   $a_{\rm K}(f) = 1 \ {\rm dB} \cdot \sqrt{f/\ {\rm MHz} }$,
*Nyquist frequency:   $f_{\rm Nyq} = 20 \ {\rm MHz}$, Roll-off factor $r = 0.5$

This results in the following consequences:
*In the area up to $f_{1} = 10 \ \text{MHz: }$ $H_{\rm CRO}(f) = 1$   ⇒   $\left \vert H_{\rm E}(f)\right \vert ^2 = \left \vert H_{\rm K}(f)\right \vert ^{-2}$ (see yellow area).
* The flank of $H_{\rm CRO}(f)$ is only effective from $f_{1}$ to $f_{2} = 30 \ {\rm MHz}$ and $\left \vert H_{\rm E}(f)\right \vert ^2$ decreases more and more.
*The maximum of $\left \vert H_{\rm E}(f_{\rm max})\right \vert ^2$ at $f_{\rm max} \approx 11.5 \ {\rm MHz}$ is twice the value of $\left \vert H_{\rm E}(f = 0)\right \vert ^2 = 1$.
*The integral over $\left \vert H_{\rm E}(f)\right \vert ^2$ is a measure of the effective noise power. In the current example this is $4.6$ times bigger than the minimal noise power (for $a_{\rm K}(f) = 0 \ {\rm dB}$ and $r=1$)   ⇒   $10 \cdot \lg \ \eta_\text{K+E} \approx - 6.6 \ {\rm dB}.$}}

==Exercises==

[[File:Applet_Kabeldaempfung_6_version1.png|right]]
*First choose an exercise number $1$ ... $11$.
*An exercise description is displayed.
*Parameter values are adjusted to the respective exercises.
*Click "Show solution" to display the solution.
*Exercise description and solution are in English.

Number "0" is a "Reset" button:
*Sets parameters to initial values (like after loading the section).
*Displays a "Reset text" to further describe the applet.

In the following desctiption '''Blue''' means the left parameter set (blue in the applet), and '''Red''' means the right parameter set (red in the applet). For parameters that are marked with an apostrophe the unit is not displayed. For example we write ${\alpha_2}' =2$   for   $\alpha_2 =2\, {\rm dB} / ({\rm km \cdot \sqrt{MHz} })$.

{{BlaueBox|TEXT=
'''(1)'''  First set '''Blue''' to $\text{Coax (1.2/4.4 mm)}$ and then to $\text{Coax (2.6/9.5 mm)}$. The cable length is $l_{\rm Blue}= 5\ \rm km$.
:Interpret $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$, in particular the functional values $a_{\rm K}(f = f_\star = 30 \ \rm MHz)$ and $\vert H_{\rm K}(f = 0) \vert$.}}

$\Rightarrow\hspace{0.3cm}\text{The attenuation function increases approximately with }\sqrt{f}\text{ and the magnitude frequency response decreases similarly to an exponential function};$
$\hspace{1.15cm}\text{Coax (1.2/4.4 mm): }a_{\rm K}(f = f_\star) = 143.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.96.$

$\hspace{1.15cm}\text{Coax (2.6/9.5 mm): }a_{\rm K}(f = f_\star) = 65.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.99;$

{{BlaueBox|TEXT=
'''(2)'''  Set '''Blue''' to $\text{Coax (2.6/9.5 mm)}$ and $l_{\rm Blue} = 5\ \rm km$. How is $a_{\rm K}(f =f_\star = 30 \ \rm MHz)$ affected by $\alpha_0$, $\alpha_1$ und $\alpha_2$?}}

$\Rightarrow\hspace{0.3cm}\alpha_2\text{ is dominant due to the skin effect. The contributions of } \alpha_0\text{ (ca. 0.1 dB) and }\alpha_1 \text{ (ca. 0.6 dB) are comparatively small.}$

{{BlaueBox|TEXT=
'''(3)'''  Additionally, set '''Red''' to $\text{Two–wired Line (0.5 mm)}$ and $l_{\rm Red} = 1\ \rm km$. What is the resulting value for $a_{\rm K}(f =f_\star= 30 \ \rm MHz)$?
:Up to what length $l_{\rm Red}$ does the red attenuation function stay under the blue one?}}

$\Rightarrow\hspace{0.3cm}\text{Red curve: }a_{\rm K}(f = f_\star) = 87.5 {\ \rm dB} \text{. The condition above is fulfilled for }l_{\rm Red} = 0.7\ {\rm km} \ \Rightarrow \ a_{\rm K}(f = f_\star) = 61.3 {\ \rm dB}.$

{{BlaueBox|TEXT=
'''(4)'''  Set '''Red''' to ${k_1}' = 0, {k_2}' = 10, {k_3}' = 0.75, {l_{\rm red} } = 1 \ \rm km$ and vary the Parameter $0.5 \le k_3 \le 1$.
:How do the parameters affect $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$? }}

$\Rightarrow\hspace{0.3cm}\text{With }k_2\text {being constant, }a_{\rm K}(f)\text{ increases with bigger values of }k_3\text{ and }\vert H_{\rm K}(f) \vert \text{ decreases faster and faster. With }k_3 =1: a_{\rm K}(f)\text{ rises linearly.}$

$\hspace{1.15cm}\text{With }k_3 \to 0.5, \text{ the attenuation function is more and more determined by the skin effect, same as in the coaxial cable.}$

{{BlaueBox|TEXT=
'''(5)'''  Set '''Red''' to $\text{Two–wired Line (0.5 mm)}$ and '''Blue''' to $\text{Conversion of Red}$. For the length use $l_{\rm Red} = l_{\rm Blue} = 1\ \rm km$.
:Analyse and interpret the displayed functions $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$.}}

$\Rightarrow\hspace{0.3cm}\text{Very good approximation of the two-wire line through the blue parameter set, both with regard to }a_{\rm K}(f) \text{, as well as }\vert H_{\rm K}(f) \vert.$

$\hspace{1.15cm}\text{The resulting parameters from the conversion are }{\alpha_0}' = {k_1}' = 4.4, \ {\alpha_1}' = 0.76, \ {\alpha_2}' = 11.12.$

{{BlaueBox|TEXT=
'''(6)'''  We assume the settings of '''(5)'''. Which parts of the attenuation function are due to ohmic loss, lateral losses and skin effect? }}

$\Rightarrow\hspace{0.3cm}\text{Solution based on '''Blue''': }a_{\rm K}(f = f_\star= 30 \ {\rm MHz}) = 88.1\ {\rm dB}, \hspace{0.2cm}\text{without }\alpha_0\text{: }83.7\ {\rm dB}, \hspace{0.2cm}\text{without }\alpha_0 \text{ and } \alpha_1\text{: }60.9\ {\rm dB}.$

$\hspace{1.15cm}\text{For a two-wire cable, the influence of the longitudinal and transverse losses is significantly greater than for a coaxial cable.}$

{{BlaueBox|TEXT=
'''(7)'''  Set '''Blue''' to ${\alpha_0}' = {\alpha_1}' ={\alpha_2}' = 0$ and '''Red''' to ${k_1}' = 2, {k_2}' = 0, {l_{\rm red} } = 1 \ \rm km$. Additionally, set ${f_{\rm Nyq} }' =15$ and $r= 0.5$.
:How big are the total efficiency $\eta_\text{K+E}$ and the channel efficiency $\eta_\text{K}$?}}

$\Rightarrow\hspace{0.3cm}10 \cdot \lg \ \eta_\text{K+E} = -0.7\ \ {\rm dB}\text{ (Blue: ideal system) and }10 \cdot \lg \ \eta_\text{K+E} = -2.7\ \ {\rm dB}\text{ (Red: DC signal attenuation only)}$.

$\hspace{0.95cm}\text{The best possible rolloff factor is }r = 1.\text{ Therefore }10 \cdot \lg \ \eta_\text{K} = 0 \ {\rm dB}\text{ (Blue) or }10 \cdot \lg \ \eta_\text{K} = -2\ {\rm dB}\text{ (Red)}.$

{{BlaueBox|TEXT=
'''(8)'''  The same settings apply as in '''(7)'''. Under what transmission power $P_{\rm red}$ with respect to $P_{\rm blue}$ do both systems achieve the same error probability? }}

$\Rightarrow\hspace{0.3cm}\text{We need to achieve }10 \cdot \lg {P_{\rm Red}}/{P_{\rm Blue}} = 2 \ {\rm dB} \ \ \Rightarrow \ \ {P_{\rm Red}}/{P_{\rm Blue}} = 10^{0.2} = 1.585.$

{{BlaueBox|TEXT=
'''(9)'''  Set '''Blue''' to ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 2$ and '''Red''' to "Inactive". Additionally set ${f_{\rm Nyq} }' =15$ and $r= 0.7$.
:How does $\vert H_{\rm E}(f) \vert$ look like? Calculate the total efficiency $\eta_\text{K+E}$ and the channel efficiency$\eta_\text{K}$.}}

$\Rightarrow\hspace{0.3cm}\text{For} f < 7.5 {\ \rm MHz: } \vert H_{\rm E}(f) \vert = \vert H_{\rm K}(f) \vert ^{-1}.\text{ For } f > 25 {\ \rm MHz: }\vert H_{\rm E}(f) \vert = 0.\text{ In between, the effect of the CRO edge can be observed.}$

$\hspace{0.95cm}\text{The best possible rolloff factor }r = 0.7 \text{ is already set }\Rightarrow \ 10 \cdot \lg \ \eta_\text{K+E} = 10 \cdot \lg \ \eta_\text{K} \approx - 18.1 \ {\rm dB}.$

{{BlaueBox|TEXT=
'''(10)'''  Set '''Blue''' to ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 8$ and '''Red''' to "Inactive". Additionally, set ${f_{\rm Nyq} }' =15$ and $r= 0.7$.
:How big is $\vert H_{\rm E}(f = 0) \vert$? What is the maximum value of $\vert H_{\rm E}(f) \vert$? Calculate the channel efficiency $\eta_\text{K}$.}}

$\Rightarrow\hspace{0.3cm}\vert H_{\rm E}(f = 0) \vert = \vert H_{\rm E}(f = 0) \vert ^{-1}= 1 \text{ and the maximum value } \vert H_{\rm E}(f) \vert \text{ is approximately }37500\text{ for }r=0.7 \Rightarrow 10 \cdot \lg \ \eta_\text{K+E} \approx -89.2 \ {\rm dB},$

$\hspace{0.95cm}\text{because the integral over }\vert H_{\rm E}(f) \vert^2\text{is huge. After the optimization }r=0.17 \text{ we get }10 \cdot \lg \ \eta_\text{K} \approx -82.6 \ {\rm dB}.$

{{BlaueBox|TEXT=
'''(11)''' The same settings apply as in '''(10)''' and $r= 0.17$. Vary the cable length up to $l_{\rm blue} = 10 \ \rm km$.
:How much do the maximum value of $\vert H_{\rm E}(f) \vert$, the channel efficiency $\eta_\text{K}$ and the optimal rolloff factor $r_{\rm opt}$ change?}}

$\Rightarrow\hspace{0.3cm}\text{The maximum value of } \vert H_{\rm E}(f) \vert \text{ increases and }10 \cdot \lg \ \eta_\text{K} \text{ decreases more and more.}$

$\hspace{0.95cm}\text{At 10 km length } 10 \cdot \lg \ \eta_\text{K} \approx -104.9 \ {\rm dB} \text{ and } r_{\rm opt}=0.14\text{. For }f_\star \approx 14.5\ {\rm MHz} \Rightarrow \vert H_{\rm E}(f = f_\star) \vert = 352000 \approx \vert H_{\rm E}(f =0)\vert$.

==Applet Manual==
[[File:Applet_Kabeldaempfung_5_version2.png|left|600px]]
    '''(A)'''     Preselection for blue parameter set

    '''(B)'''     Input of the $\alpha$ parameters via sliders

    '''(C)'''     Preselection for red parameter set

    '''(D)'''     Input of the $k$ parameters via sliders

    '''(E)'''     Input of the parameters $f_{\rm Nyq}$ and $r$

    '''(F)'''     Selection for the graphic display

    '''(G)'''     Display $a_\text{K}(f)$, $|H_\text{K}(f)|$, $|H_\text{E}(f)|$, ...

    '''(H)'''     Scaling factor $H_0$ for $|H_\text{E}(f)|$, $|H_\text{E}(f)|^2$

    '''(I)'''     Selection of the frequency $f_\star$ for numeric values

    '''(J)'''     Numeric values for blue parameter set

    '''(K)'''     Numeric values for red parameter set

    '''(L)'''     Output system efficiency $\eta_\text{K+E}$ in dB

    '''(M)'''     Store & Recall of settings

    '''(N)'''     Exercise section

    '''(O)'''     Variation of the graphic display:$\hspace{0.5cm}$"$+$" (Zoom in),

$\hspace{0.5cm}$ "$-$" (Zoom out)

$\hspace{0.5cm}$ "$\rm o$" (Reset)

$\hspace{0.5cm}$ "$\leftarrow$" (Move left), etc.

'''Other options for graphic display''':
*Hold shift and scroll: Zoom in on/out of coordinate system,
*Hold shift and left click: Move the coordinate system.

==About the Authors==
This interactive calculation was designed and realized at the  [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik]  of the  [https://www.tum.de/ Technische Universität München].
*The original version was created in 2009 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Sebastian_Seitz_.28Diplomarbeit_LB_2009.29|Sebastian Seitz]]  as part of his Diploma thesis using "FlashMX–Actionscript" (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]] ).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]]  as part of his Bachelor's thesis (Supervisor:  [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) .

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEn|attenuationCopperCables_en}}

Applets:Attenuation of Copper Cables

2023-03-25T21:00:20Z

Noah:

{{LntAppletLinkEn|attenuationCopperCables_en}}         [https://www.lntwww.de/Applets:Dämpfung_von_Kupferkabeln '''English Applet with German WIKI description''']

==Applet Description==
 
This applet calculates the attenuation function $a_{\rm K}(f)$ of conducted transmission media (with cable length $l$):
*For coaxial cables one usually uses the equation $a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l$.
*In contrast, two-wire lines are often displayed in the form $a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l$.
*The conversion of the $(k_1, \ k_2, \ k_3)$ parameters to the $(\alpha_0, \ \alpha_1, \ \alpha_2)$ parameters for $B = 30 \ \rm MHz$ is realized as well as the other way around.

Aside from the attenuation function $a_{\rm K}(f)$ the applet can display:
*the associated magnitude frequency response $\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20},$
*the equalizer frequency response $\left | H_{\rm E}(f)\right | = \left | H_{\rm CRO}(f) / H_{\rm K}(f)\right | $, that leads to a Nyquist total frequency response $ H_{\rm CRO}(f) $,
*the corresponding squared magnitude frequency response $\left | H_{\rm E}(f)\right |^2 $.

The integral over $\left | H_{\rm E}(f)\right |^2 $ is a measure of the noise exaggeration of the selected Nyquist total frequency response and thus also for the expected error probability. From this, the ''total efficiency''  $\eta_\text{K+E}$ for channel (ger.:'''K'''anal) and equalizer (ger.:'''E'''ntzerrer) is calculated, which is output in the applet in $\rm dB$.

Through optimization of the roll-off factor $r$ of the raised cosine frequency response $ H_{\rm CRO}(f) $ one gets the ''channel efficiency''  $ \eta_\text{K}$. This therefore indicates the deterioration of the overall system due to the attenuation function $ a _ {\ rm K} (f) $ of the transmission medium.

==Theoretical Background==
 
===Magnitude Frequency Response and Attenuation Function===
Following relationship exists between the magnitude frequency response and the attenuation function:
:$$\left | H_{\rm K}(f)\right |=10^{-a_\text{K}(f)/20} = {\rm e}^{-a_\text{K, Np}(f)}.$$
*The index "K" makes it clear, that the considered LTI system is a cable (German : '''K'''abel).
*For the first calculation rule, the attenuation function $a_\text{K}(f)$ must be used in $\rm dB$ (decibel).
*For the second calculation rule, the attenuation function $a_\text{K, Np}(f)$ must be used in $\rm Np$ (Neper).
* The following conversions apply: $\rm 1 \ dB = 0.05 \cdot \ln (10) \ Np= 0.1151 \ Np$ or $\rm 1 \ Np = 20 \cdot \lg (e) \ dB= 8.6859 \ dB$.
* This applet exclusively uses dB values.

===Attenuation Function of a Coaxial Cable===
According to [Wel77]<ref name ='Wel77'>Wellhausen, H. W.: Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren. Frequenz 31, S. 23-28, 1977.</ref> the Attenuation Function of a Coaxial Cable of length $l$ is given as follows:
:$$a_{\rm K}(f)=(\alpha_0+\alpha_1\cdot f+\alpha_2\cdot \sqrt{f}) \cdot l.$$
*It is important to note the difference between $a_{\rm K}(f)$ in $\rm dB$ and the "alpha" coefficient with other pseudo–units.
*The attenuation function $a_{\rm K}(f)$ is directly proportional to the cable length $l$; $\alpha_{\rm K}(f)= a_{\rm K}(f)/l$ is referred to as the "attenuation factor" or "kilometric attenuation".
*The frequency-independent component $α_0$ of the attenuation factor takes into account the Ohmic losses.
*The frequency-proportional portion $α_1 · f$ of the attenuation factor is due to the derivation losses ("crosswise loss") .
*The dominant portion $α_2$ goes back to [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable|"Skin effect"]], which causes a lower current density inside the conductor compared to its surface. As a result, the resistance of an electric line increases with the square root of the frequency.

The constants for the ''standard coaxial cable'' with a 2.6 mm inner diameter and a 9.5 mm outer diameter   ⇒  short '''Coax (2.6/9.5 mm)''' are:
:$$\alpha_0 = 0.014\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.0038\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 2.36\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

The same applies to the ''small coaxial cable''   ⇒  short '''Coax (1.2/4.4 mm)''':
:$$\alpha_0 = 0.068\, \frac{ {\rm dB} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}
\alpha_1 = 0.0039\, \frac{ {\rm dB} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 =5.2\, \frac{ {\rm dB} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}.$$

These values can be calculated from the cables' geometric dimensions and have been confirmed by measurements at the Fernmeldetechnisches Zentralamt in Darmstadt – see [Wel77]<ref name ='Wel77'>Wellhausen, H. W.: Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren. Frequenz 31, S. 23-28, 1977.</ref> . They are valid for a temperature of 20° C (293 K) and frequencies greater than 200 kHz.

===Attenuation Function of a Two–wired Line===
According to [PW95]<ref name ='PW95'>Pollakowski, M.; Wellhausen, H.W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Mitteilung aus dem Forschungs- und Technologiezentrum der Deutschen Telekom AG, Darmstadt, Verlag für Wissenschaft und Leben Georg Heidecker, 1995.</ref> the attenuation function of a Two–wired Line of length $l$ is given as follows:
:$$a_{\rm K}(f)=(k_1+k_2\cdot (f/{\rm MHz})^{k_3}) \cdot l.$$
This function is not directly interpretable, but is a phenomenological description.

[PW95]<ref name ='PW95'>Pollakowski, M.; Wellhausen, H.W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Mitteilung aus dem Forschungs- und Technologiezentrum der Deutschen Telekom AG, Darmstadt, Verlag für Wissenschaft und Leben Georg Heidecker, 1995.</ref>also provides the constants determined by measurement results:
* $d = 0.35 \ {\rm mm}$:   $k_1 = 7.9 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 15.1 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.62$,
* $d = 0.40 \ {\rm mm}$:   $k_1 = 5.1 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 14.3 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.59$,
* $d = 0.50 \ {\rm mm}$:   $k_1 = 4.4 \ {\rm dB/km}, \hspace{0.2cm}k_2 = 10.8 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.60$,
* $d = 0.60 \ {\rm mm}$:   $k_1 = 3.8 \ {\rm dB/km}, \hspace{0.2cm}k_2 = \hspace{0.25cm}9.2 \ {\rm dB/km}, \hspace{0.2cm}k_3 = 0.61$.

From these numerical values one recognizes:
*The attenuation factor $α(f)$ and the attenuation function $a_{\rm K}(f) = α(f) · l$ depend significantly on the pipe diameter. The cables laid since 1994 with $d = 0.35 \ \rm mm$ and $d = 0.5\ \rm mm$ have a 10% greater attenuation factor than the older lines with $d = 0.4\ \rm mm$ or $d= 0.6\ \rm mm$.
*However, this smaller diameter, which is based on the manufacturing and installation costs, significantly reduces the range $l_{\rm max}$ of the transmission systems used on these lines, so that in the worst case scenario expensive intermediate regenerators have to be used.
*The current transmission methods for copper lines prove only a relatively narrow frequency band, for example $120\ \rm kHz$ with [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|ISDN]] and $\approx 1100 \ \rm kHz$ with [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_DSL|DSL]]. For $f = 1 \ \rm MHz$ the attenuation factor of a 0.4 mm cable is around $20 \ \rm dB/km$, so that even with a cable length of $l = 4 \ \rm km$ the attenuation does not exceed $80 \ \rm dB$.

===Conversion between $k$ and $\alpha$ parameters===
The $k$–parameters of the attenuation factor   ⇒   $\alpha_{\rm I} (f)$ can be converted into corresponding $\alpha$–parameters   ⇒   $\alpha_{\rm II} (f)$:
:$$\alpha_{\rm I} (f) = k_1 + k_2 \cdot (f/f_0)^{k_3}\hspace{0.05cm}, \hspace{0.2cm}{\rm with} \hspace{0.15cm} f_0 = 1\,{\rm MHz},$$
:$$\alpha_{\rm II} (f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}.$$

As a criterion of this conversion, we assume that the quadratic deviation of these two functions is minimal within a bandwidth $B$:
:$$\int_{0}^{B} \left [ \alpha_{\rm I} (f) - \alpha_{\rm II} (f)\right ]^2 \hspace{0.1cm}{\rm d}f \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Minimum} \hspace{0.05cm} .$$
It is obvious that $α_0 = k_1$. The parameters $α_1$ and $α_2$ are dependent on the underlying bandwidth $B$ and are:
:$$\begin{align*}\alpha_1 & = 15 \cdot (B/f_0)^{k_3 -1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{ {f_0} }\hspace{0.05cm} ,\\ \alpha_2 & = 10 \cdot (B/f_0)^{k_3 -0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 + 2)}\cdot {k_2}/{\sqrt{f_0} }\hspace{0.05cm} .\end{align*}$$

In the opposite direction the conversion rule for the exponent is:

:$$k_3 = \frac{A + 0.5} {A +1}, \hspace{0.2cm}\text{Auxiliary variable: }A = \frac{2} {3} \cdot \frac{\alpha_1 \cdot \sqrt{f_0}}{\alpha_2} \cdot \sqrt{B/f_0}.$$

With this result you can specify $ k_2 $ with each of the above equations.

{{GraueBox|TEXT=
$\text{Example 1:}$ 
*For $k_3 = 1$ (frequency-proportional attenuation factor) we get   $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_1 = {k_2}/{ {f_0} }\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = 0\hspace{0.05cm} .$
*For $k_3 = 0.5$ (Skin effect) we get the coefficients:   $\alpha_0 = k_0\hspace{0.05cm} ,\hspace{0.2cm}\alpha_1 = 0\hspace{0.05cm} ,\hspace{0.2cm} \alpha_2 = {k_2}/{\sqrt{f_0} }\hspace{0.05cm}.$
*For $k_3 < 0.5$ we get a negative $\alpha_1$. Conversion is only possible for $0.5 \le k_3 \le 1$.
*For $0.5 \le k_3 \le$ we get the coefficients $\alpha_1 > 0$ and $\alpha_2 > 0$, which are also dependent on $B/f_0$.
*From $\alpha_1 = 0.3\, {\rm dB}/ ({\rm km \cdot MHz}) \hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 3\, {\rm dB}/ ({\rm km \cdot \sqrt{MHz} })\hspace{0.05cm},\hspace{0.2cm}B = 30 \ \rm MHz$ folgt $k_3 = 0.63$ und $k_2 = 2.9 \ \rm dB/km$.}}

===Channel Influence on the Binary Nyquistent Equalization===
[[File:Applet_Kabeldaempfung_1_version_englisch.png|right|frame|Simplified block diagram of the optimal Nyquistent equalizer|class=fit]]
Going by the block diagram: Between the Dirac source and the (threshold) decider are the frequency responses for the transmitter (German: $\rm S$ender)  ⇒  $H_{\rm S}(f)$, channel (German: $\rm K$anal)  ⇒  $H_{\rm K}(f)$ and receiver (German: $\rm E$mpfänger)   ⇒  $H_{\rm E}(f)$.

In this applet
*we neglect the influence of the transmitted pulse form   ⇒   $H_{\rm S}(f) \equiv 1$   ⇒   dirac shaped transmission signal $s(t)$, and
*presuppose a binary Nyquist system with cosine–roll–off around the Nyquist frequency $f_{\rm Nyq} = [f_1 + f_2]/2 =1(2T)$ :
:$$H_{\rm K}(f) · H_{\rm E}(f) = H_{\rm CRO}(f).$$

[[File:Applet_Kabeldaempfung_2_version2.png|right|frame|Frequency Response with cosine–roll–off|class=fit]]

This means: The [[Digital_Signal_Transmission/Properties_of_Nyquist_Systems#First_Nyquist_criterion_in_the_frequency_domain|"first Nyquist criterion"]] is met ⇒   Timely successive impulses do not disturb each other ⇒   there are no [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|"Intersymbol Interferences"]].

In the case of white Gaussian noise, the transmission quality is thus determined solely by the noise power in front of the receiver:

:$$P_{\rm N} =\frac{N_0}{2} \cdot \int_{-\infty}^{+\infty} |H_{\rm E}(f)|^2 \ {\rm d}f\hspace{1cm}\text{with}\hspace{1cm}|H_{\rm E}(f)|^2 = \frac{|H_{\rm CRO}(f)|^2}{|H_{\rm K}(f)|^2}.$$

The lowest possible noise performance results with an ideal channel   ⇒   $H_{\rm K}(f) \equiv 1$ and a rectangular $H_{\rm CRO}(f) \equiv 1$ in $|f| \le f_{\rm Nyq}$:

:$$P_\text{N, min} = P_{\rm N} \ \big [\text{optimal system: }H_{\rm K}(f) \equiv 1, \ r=r_{\rm opt} =1 \big ] = N_0 \cdot f_{\rm Nyq} .$$

{{BlaueBox|TEXT=
$\text{Definitions:}$ 
*As a quality criterion for a given system we use the '''total efficiency''' with respect to the channel $\rm (K)$ and the receiver $\rm (E)$:

:$$\eta_\text{K+E} = \frac{P_{\rm N} \ \big [\text{Optimal system: Channel }H_{\rm K}(f) \equiv 1,\ \text{Roll-off factor } r=r_{\rm opt} =1 \big ]}{P_{\rm N} \ \big [\text{Given system: Channel }H_{\rm K}(f), \ \text{Roll-off factor }r \big ]} =\left [ \frac{1}{3/4 \cdot f_{\rm Nyq} } \cdot \int_{0}^{+\infty} \vert H_{\rm E}(f) \vert^2 \ {\rm d}f \right ]^{-1}\le 1.$$

This quality criterion is specified in the applet for both parameter sets in logarithm form:   $10 \cdot \lg \ \eta_\text{K+E} \le 0 \ \rm dB$.

*Through variation and optimization of the receiver   ⇒   roll-off factor $r$ we get the '''channel efficiency''':

:$$\eta_\text{K} = \min_{0 \le r \le 1} \ \eta_\text{K+E} .$$}}

[[File:Applet_Kabeldaempfung_3_version2.png|right|frame|Square value frequency response $\left \vert H_{\rm E}(f)\right \vert ^2 $|class=fit]]
{{GraueBox|TEXT=
$\text{Example 2:}$ 
The graph shows the square value frequency response $\left \vert H_{\rm E}(f)\right \vert ^2 $ with $\left \vert H_{\rm E}(f)\right \vert = H_{\rm CRO}(f) / \left \vert H_{\rm K}(f)\right \vert$ for the following boundary conditions:
*Attenuation function of the channel:   $a_{\rm K}(f) = 1 \ {\rm dB} \cdot \sqrt{f/\ {\rm MHz} }$,
*Nyquist frequency:   $f_{\rm Nyq} = 20 \ {\rm MHz}$, Roll-off factor $r = 0.5$

This results in the following consequences:
*In the area up to $f_{1} = 10 \ \text{MHz: }$ $H_{\rm CRO}(f) = 1$   ⇒   $\left \vert H_{\rm E}(f)\right \vert ^2 = \left \vert H_{\rm K}(f)\right \vert ^{-2}$ (see yellow area).
* The flank of $H_{\rm CRO}(f)$ is only effective from $f_{1}$ to $f_{2} = 30 \ {\rm MHz}$ and $\left \vert H_{\rm E}(f)\right \vert ^2$ decreases more and more.
*The maximum of $\left \vert H_{\rm E}(f_{\rm max})\right \vert ^2$ at $f_{\rm max} \approx 11.5 \ {\rm MHz}$ is twice the value of $\left \vert H_{\rm E}(f = 0)\right \vert ^2 = 1$.
*The integral over $\left \vert H_{\rm E}(f)\right \vert ^2$ is a measure of the effective noise power. In the current example this is $4.6$ times bigger than the minimal noise power (for $a_{\rm K}(f) = 0 \ {\rm dB}$ and $r=1$)   ⇒   $10 \cdot \lg \ \eta_\text{K+E} \approx - 6.6 \ {\rm dB}.$}}

==Exercises==

[[File:Applet_Kabeldaempfung_6_version1.png|right]]
*First choose an exercise number $1$ ... $11$.
*An exercise description is displayed.
*Parameter values are adjusted to the respective exercises.
*Click "Show solution" to display the solution.
*Exercise description and solution are in English.

Number "0" is a "Reset" button:
*Sets parameters to initial values (like after loading the section).
*Displays a "Reset text" to further describe the applet.

In the following desctiption '''Blue''' means the left parameter set (blue in the applet), and '''Red''' means the right parameter set (red in the applet). For parameters that are marked with an apostrophe the unit is not displayed. For example we write ${\alpha_2}' =2$   for   $\alpha_2 =2\, {\rm dB} / ({\rm km \cdot \sqrt{MHz} })$.

{{BlaueBox|TEXT=
'''(1)'''  First set '''Blue''' to $\text{Coax (1.2/4.4 mm)}$ and then to $\text{Coax (2.6/9.5 mm)}$. The cable length is $l_{\rm Blue}= 5\ \rm km$.
:Interpret $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$, in particular the functional values $a_{\rm K}(f = f_\star = 30 \ \rm MHz)$ and $\vert H_{\rm K}(f = 0) \vert$.}}

$\Rightarrow\hspace{0.3cm}\text{The attenuation function increases approximately with }\sqrt{f}\text{ and the magnitude frequency response decreases similarly to an exponential function};$
$\hspace{1.15cm}\text{Coax (1.2/4.4 mm): }a_{\rm K}(f = f_\star) = 143.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.96.$

$\hspace{1.15cm}\text{Coax (2.6/9.5 mm): }a_{\rm K}(f = f_\star) = 65.3\text{ dB;}\hspace{0.5cm}\vert H_{\rm K}(f = 0) \vert = 0.99;$

{{BlaueBox|TEXT=
'''(2)'''  Set '''Blue''' to $\text{Coax (2.6/9.5 mm)}$ and $l_{\rm Blue} = 5\ \rm km$. How is $a_{\rm K}(f =f_\star = 30 \ \rm MHz)$ affected by $\alpha_0$, $\alpha_1$ und $\alpha_2$?}}

$\Rightarrow\hspace{0.3cm}\alpha_2\text{ is dominant due to the skin effect. The contributions of } \alpha_0\text{ (ca. 0.1 dB) and }\alpha_1 \text{ (ca. 0.6 dB) are comparatively small.}$

{{BlaueBox|TEXT=
'''(3)'''  Additionally, set '''Red''' to $\text{Two–wired Line (0.5 mm)}$ and $l_{\rm Red} = 1\ \rm km$. What is the resulting value for $a_{\rm K}(f =f_\star= 30 \ \rm MHz)$?
:Up to what length $l_{\rm Red}$ does the red attenuation function stay under the blue one?}}

$\Rightarrow\hspace{0.3cm}\text{Red curve: }a_{\rm K}(f = f_\star) = 87.5 {\ \rm dB} \text{. The condition above is fulfilled for }l_{\rm Red} = 0.7\ {\rm km} \ \Rightarrow \ a_{\rm K}(f = f_\star) = 61.3 {\ \rm dB}.$

{{BlaueBox|TEXT=
'''(4)'''  Set '''Red''' to ${k_1}' = 0, {k_2}' = 10, {k_3}' = 0.75, {l_{\rm red} } = 1 \ \rm km$ and vary the Parameter $0.5 \le k_3 \le 1$.
:How do the parameters affect $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$? }}

$\Rightarrow\hspace{0.3cm}\text{With }k_2\text {being constant, }a_{\rm K}(f)\text{ increases with bigger values of }k_3\text{ and }\vert H_{\rm K}(f) \vert \text{ decreases faster and faster. With }k_3 =1: a_{\rm K}(f)\text{ rises linearly.}$

$\hspace{1.15cm}\text{With }k_3 \to 0.5, \text{ the attenuation function is more and more determined by the skin effect, same as in the coaxial cable.}$

{{BlaueBox|TEXT=
'''(5)'''  Set '''Red''' to $\text{Two–wired Line (0.5 mm)}$ and '''Blue''' to $\text{Conversion of Red}$. For the length use $l_{\rm Red} = l_{\rm Blue} = 1\ \rm km$.
:Analyse and interpret the displayed functions $a_{\rm K}(f)$ and $\vert H_{\rm K}(f) \vert$.}}

$\Rightarrow\hspace{0.3cm}\text{Very good approximation of the two-wire line through the blue parameter set, both with regard to }a_{\rm K}(f) \text{, as well as }\vert H_{\rm K}(f) \vert.$

$\hspace{1.15cm}\text{The resulting parameters from the conversion are }{\alpha_0}' = {k_1}' = 4.4, \ {\alpha_1}' = 0.76, \ {\alpha_2}' = 11.12.$

{{BlaueBox|TEXT=
'''(6)'''  We assume the settings of '''(5)'''. Which parts of the attenuation function are due to ohmic loss, lateral losses and skin effect? }}

$\Rightarrow\hspace{0.3cm}\text{Solution based on '''Blue''': }a_{\rm K}(f = f_\star= 30 \ {\rm MHz}) = 88.1\ {\rm dB}, \hspace{0.2cm}\text{without }\alpha_0\text{: }83.7\ {\rm dB}, \hspace{0.2cm}\text{without }\alpha_0 \text{ and } \alpha_1\text{: }60.9\ {\rm dB}.$

$\hspace{1.15cm}\text{For a two-wire cable, the influence of the longitudinal and transverse losses is significantly greater than for a coaxial cable.}$

{{BlaueBox|TEXT=
'''(7)'''  Set '''Blue''' to ${\alpha_0}' = {\alpha_1}' ={\alpha_2}' = 0$ and '''Red''' to ${k_1}' = 2, {k_2}' = 0, {l_{\rm red} } = 1 \ \rm km$. Additionally, set ${f_{\rm Nyq} }' =15$ and $r= 0.5$.
:How big are the total efficiency $\eta_\text{K+E}$ and the channel efficiency $\eta_\text{K}$?}}

$\Rightarrow\hspace{0.3cm}10 \cdot \lg \ \eta_\text{K+E} = -0.7\ \ {\rm dB}\text{ (Blue: ideal system) and }10 \cdot \lg \ \eta_\text{K+E} = -2.7\ \ {\rm dB}\text{ (Red: DC signal attenuation only)}$.

$\hspace{0.95cm}\text{The best possible rolloff factor is }r = 1.\text{ Therefore }10 \cdot \lg \ \eta_\text{K} = 0 \ {\rm dB}\text{ (Blue) or }10 \cdot \lg \ \eta_\text{K} = -2\ {\rm dB}\text{ (Red)}.$

{{BlaueBox|TEXT=
'''(8)'''  The same settings apply as in '''(7)'''. Under what transmission power $P_{\rm red}$ with respect to $P_{\rm blue}$ do both systems achieve the same error probability? }}

$\Rightarrow\hspace{0.3cm}\text{We need to achieve }10 \cdot \lg {P_{\rm Red}}/{P_{\rm Blue}} = 2 \ {\rm dB} \ \ \Rightarrow \ \ {P_{\rm Red}}/{P_{\rm Blue}} = 10^{0.2} = 1.585.$

{{BlaueBox|TEXT=
'''(9)'''  Set '''Blue''' to ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 2$ and '''Red''' to "Inactive". Additionally set ${f_{\rm Nyq} }' =15$ and $r= 0.7$.
:How does $\vert H_{\rm E}(f) \vert$ look like? Calculate the total efficiency $\eta_\text{K+E}$ and the channel efficiency$\eta_\text{K}$.}}

$\Rightarrow\hspace{0.3cm}\text{For} f < 7.5 {\ \rm MHz: } \vert H_{\rm E}(f) \vert = \vert H_{\rm K}(f) \vert ^{-1}.\text{ For } f > 25 {\ \rm MHz: }\vert H_{\rm E}(f) \vert = 0.\text{ In between, the effect of the CRO edge can be observed.}$

$\hspace{0.95cm}\text{The best possible rolloff factor }r = 0.7 \text{ is already set }\Rightarrow \ 10 \cdot \lg \ \eta_\text{K+E} = 10 \cdot \lg \ \eta_\text{K} \approx - 18.1 \ {\rm dB}.$

{{BlaueBox|TEXT=
'''(10)'''  Set '''Blue''' to ${\alpha_0}' = {\alpha_1}' = 0, \ {\alpha_2}' = 3, \ {l_{\rm blue} }' = 8$ and '''Red''' to "Inactive". Additionally, set ${f_{\rm Nyq} }' =15$ and $r= 0.7$.
:How big is $\vert H_{\rm E}(f = 0) \vert$? What is the maximum value of $\vert H_{\rm E}(f) \vert$? Calculate the channel efficiency $\eta_\text{K}$.}}

$\Rightarrow\hspace{0.3cm}\vert H_{\rm E}(f = 0) \vert = \vert H_{\rm E}(f = 0) \vert ^{-1}= 1 \text{ and the maximum value } \vert H_{\rm E}(f) \vert \text{ is approximately }37500\text{ for }r=0.7 \Rightarrow 10 \cdot \lg \ \eta_\text{K+E} \approx -89.2 \ {\rm dB},$

$\hspace{0.95cm}\text{because the integral over }\vert H_{\rm E}(f) \vert^2\text{is huge. After the optimization }r=0.17 \text{ we get }10 \cdot \lg \ \eta_\text{K} \approx -82.6 \ {\rm dB}.$

{{BlaueBox|TEXT=
'''(11)''' The same settings apply as in '''(10)''' and $r= 0.17$. Vary the cable length up to $l_{\rm blue} = 10 \ \rm km$.
:How much do the maximum value of $\vert H_{\rm E}(f) \vert$, the channel efficiency $\eta_\text{K}$ and the optimal rolloff factor $r_{\rm opt}$ change?}}

$\Rightarrow\hspace{0.3cm}\text{The maximum value of } \vert H_{\rm E}(f) \vert \text{ increases and }10 \cdot \lg \ \eta_\text{K} \text{ decreases more and more.}$

$\hspace{0.95cm}\text{At 10 km length } 10 \cdot \lg \ \eta_\text{K} \approx -104.9 \ {\rm dB} \text{ and } r_{\rm opt}=0.14\text{. For }f_\star \approx 14.5\ {\rm MHz} \Rightarrow \vert H_{\rm E}(f = f_\star) \vert = 352000 \approx \vert H_{\rm E}(f =0)\vert$.

==Applet Manual==
[[File:Applet_Kabeldaempfung_5_version2.png|left|600px]]
    '''(A)'''     Preselection for blue parameter set

    '''(B)'''     Input of the $\alpha$ parameters via sliders

    '''(C)'''     Preselection for red parameter set

    '''(D)'''     Input of the $k$ parameters via sliders

    '''(E)'''     Input of the parameters $f_{\rm Nyq}$ and $r$

    '''(F)'''     Selection for the graphic display

    '''(G)'''     Display $a_\text{K}(f)$, $|H_\text{K}(f)|$, $|H_\text{E}(f)|$, ...

    '''(H)'''     Scaling factor $H_0$ for $|H_\text{E}(f)|$, $|H_\text{E}(f)|^2$

    '''(I)'''     Selection of the frequency $f_\star$ for numeric values

    '''(J)'''     Numeric values for blue parameter set

    '''(K)'''     Numeric values for red parameter set

    '''(L)'''     Output system efficiency $\eta_\text{K+E}$ in dB

    '''(M)'''     Store & Recall of settings

    '''(N)'''     Exercise section

    '''(O)'''     Variation of the graphic display:$\hspace{0.5cm}$"$+$" (Zoom in),

$\hspace{0.5cm}$ "$-$" (Zoom out)

$\hspace{0.5cm}$ "$\rm o$" (Reset)

$\hspace{0.5cm}$ "$\leftarrow$" (Move left), etc.

'''Other options for graphic display''':
*Hold shift and scroll: Zoom in on/out of coordinate system,
*Hold shift and left click: Move the coordinate system.

==About the Authors==
This interactive calculation was designed and realized at the  [http://www.lnt.ei.tum.de/startseite Lehrstuhl für Nachrichtentechnik]  of the  [https://www.tum.de/ Technische Universität München].
*The original version was created in 2009 by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Sebastian_Seitz_.28Diplomarbeit_LB_2009.29|Sebastian Seitz]]  as part of his Diploma thesis using "FlashMX–Actionscript" (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]] ).
*In 2018 this Applet was redesigned and updated to "HTML5" by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Jimmy_He_.28Bachelorarbeit_2018.29|Jimmy He]]  as part of his Bachelor's thesis (Supervisor:  [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]]) .

==Once again:  Open Applet in new Tab==

{{LntAppletLinkEn|attenuationCopperCables_en}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-03-25T19:04:02Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulation"]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity}}.

===Signal reconstruction===

[[File:P_ID1123__Sig_T_5_1_S5a_neu.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==
 
[[File:Anleitung_Auge.png|right|600px]]
    '''(A)'''     Auswahl:   Codierung                    (binär,  quaternär,  AMI–Code,  Duobinärcode)

    '''(B)'''     Auswahl:   Detektionsgrundimpuls                     (nach Gauß–TP,  CRO–Nyquist,  nach Spalt–TP}

    '''(C)'''     Prametereingabe zu  '''(B)'''                    (Grenzfrequenz,  Rolloff–Faktor,  Rechteckdauer)

    '''(D)'''     Steuerung der Augendiagrammdarstellung                    (Start,  Pause/Weiter,  Einzelschritt,  Gesamt,  Reset)

    '''(E)'''     Geschwindigkeit der Augendiagrammdarstellung

    '''(F)'''     Darstellung:  Detektionsgrundimpuls  $g_d(t)$

    '''(G)'''     Darstellung:  Detektionsnutzsignal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Darstellung:  Augendiagramm im Bereich  $\pm T$

    '''( I )'''     Numerikausgabe:  $ö_{\rm norm}$  (normierte Augenöffnung)

    '''(J)'''     Prametereingabe  $10 \cdot \lg \ E_{\rm B}/N_0$  für  '''(K)'''

    '''(K)'''     Numerikausgabe:  $\sigma_{\rm norm}$  (normierter Rauscheffektivwert)

    '''(L)'''     Numerikausgabe:  $p_{\rm U}$  (ungünstigste Fehlerwahrscheinlichkeit)

    '''(M)'''     Bereich für die Versuchsdurchführung:   Aufgabenauswahl

    '''(N)'''     Bereich für die Versuchsdurchführung:   Aufgabenstellung

    '''(O)'''     Bereich für die Versuchsdurchführung:   Musterlösung einblenden
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-03-25T18:51:04Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulatio"n]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem".]] This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity}}.

===Signal reconstruction===

[[File:P_ID1123__Sig_T_5_1_S5a_neu.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==
 
[[File:Anleitung_Auge.png|right|600px]]
    '''(A)'''     Auswahl:   Codierung                    (binär,  quaternär,  AMI–Code,  Duobinärcode)

    '''(B)'''     Auswahl:   Detektionsgrundimpuls                     (nach Gauß–TP,  CRO–Nyquist,  nach Spalt–TP}

    '''(C)'''     Prametereingabe zu  '''(B)'''                    (Grenzfrequenz,  Rolloff–Faktor,  Rechteckdauer)

    '''(D)'''     Steuerung der Augendiagrammdarstellung                    (Start,  Pause/Weiter,  Einzelschritt,  Gesamt,  Reset)

    '''(E)'''     Geschwindigkeit der Augendiagrammdarstellung

    '''(F)'''     Darstellung:  Detektionsgrundimpuls  $g_d(t)$

    '''(G)'''     Darstellung:  Detektionsnutzsignal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Darstellung:  Augendiagramm im Bereich  $\pm T$

    '''( I )'''     Numerikausgabe:  $ö_{\rm norm}$  (normierte Augenöffnung)

    '''(J)'''     Prametereingabe  $10 \cdot \lg \ E_{\rm B}/N_0$  für  '''(K)'''

    '''(K)'''     Numerikausgabe:  $\sigma_{\rm norm}$  (normierter Rauscheffektivwert)

    '''(L)'''     Numerikausgabe:  $p_{\rm U}$  (ungünstigste Fehlerwahrscheinlichkeit)

    '''(M)'''     Bereich für die Versuchsdurchführung:   Aufgabenauswahl

    '''(N)'''     Bereich für die Versuchsdurchführung:   Aufgabenstellung

    '''(O)'''     Bereich für die Versuchsdurchführung:   Musterlösung einblenden
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-03-25T18:50:27Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|"Puls code modulatio"n]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Dirac delta lines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Dirac delta lines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal pulse weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the pulse weights of the Dirac delta functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

The spectrum of the sampled signal  $x_{\rm A}(t)$  is obtained by applying the  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Convolution Theorem"]]. This states that multiplication in the time domain corresponds to convolution in the spectral domain:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

If one develops the  Dirac delta pulse  $p_{\delta}(t)$   (in the time domain)   into a  [[Signal_Representation/Fourier_Series|"Fourier Series"]]  and transforms it using the  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Shifting Theorem"]]  into the frequency domain, the following correspondence   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"proof"]] results with the distance  $f_{\rm A} = 1/T_{\rm A}$  of two adjacent dirac delta lines in the frequency domain:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Dirac delta pulse in time and frequency domain with  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
The result states:
*The Dirac delta pulse  $p_{\delta}(t)$  in the time domain consists of infinitely many Dirac delta pulses, each at the same distance  $T_{\rm A}$  and all with the same pulse weight  $T_{\rm A}$.
*The Fourier transform of  $p_{\delta}(t)$  again gives a Dirac delta pulse, but now in the frequency domain   ⇒   $P_{\delta}(f)$.
*Also  $P_{\delta}(f)$  consists of infinitely many Dirac delta pulses, now in the respective spacing  $f_{\rm A} = 1/T_{\rm A}$  and all with pulse weight  $1$.
*The distances of the Dirac delta lines in time and frequency domain thus follow the  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity Theorem"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

From this follows:   From the spectrum  $X(f)$  is obtained by convolution with the Dirac delta line shifted by  $\mu \cdot f_{\rm A}$ :

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Applying this result to all Dirac delta lines of the Dirac delta pulse, we finally obtain:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Conclusion:}$  Sampling the analog time signal  $x(t)$  at equidistant intervals  $T_{\rm A}$  results in the spectral domain in a  '''periodic continuation'''  of  $X(f)$  with frequency spacing  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spectrum of the sampled signal]]
{{GraueBox|TEXT=
$\text{Example 1:}$ 
The upper graph shows  '''(schematic!)'''  the spectrum  $X(f)$  of an analog signal  $x(t)$, which contains frequencies up to  $5 \text{ kHz}$ .

Sampling the signal at the sampling rate  $f_{\rm A}\,\text{ = 20 kHz}$, i.e., at the respective spacing  $T_{\rm A}\, = {\rm 50 \, µs}$  yields the periodic spectrum  $X_{\rm A}(f)$ sketched below.
*Since the Dirac delta functions are infinitely narrow, the sampled signal  $x_{\rm A}(t)$  also contains arbitrary high frequency components.
*Correspondingly, the spectral function  $X_{\rm A}(f)$  of the sampled signal is extended to infinity}}.

===Signal reconstruction===

[[File:P_ID1123__Sig_T_5_1_S5a_neu.png|right|frame|Joint model of "signal sampling" and "signal reconstruction"]]
Signal sampling is not an end in itself in a digital transmission system, but it must be reversed at some point  For example, consider the following system:
*The analog signal  $x(t)$  with bandwidth  $B_{\rm NF}$  is sampled as described above.
*At the output of an ideal transmission system, the also discrete-time signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is present.
*The question now is how the block   '''signal reconstruction'''   has to be designed so that also  $y(t) = x(t)$  holds.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequency domain representation of the "signal reconstruction"]]
 The solution is simple if you look at the spectral functions:  

One obtains from  $Y_{\rm A}(f)$  the spectrum  $Y(f) = X(f)$  by a low-pass filter with the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_. E2.80.93_Transfer_function|"Frequency response"]]  $H_{\rm E}(f)$, which 

*passes the low frequencies unaltered:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*completely suppresses the high frequencies:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{for}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Further, it can be seen from the accompanying graph:   As long as the above two conditions are satisfied,  $H_{\rm E}(f)$  can be arbitrarily shaped in the range from  $B_{\rm NF}$  to  $f_{\rm A}-B_{\rm NF}$  ,
*for example linearly descending (dashed line)
*or also rectangular.

===The Sampling Theorem===

The complete reconstruction of the analog signal  $y(t)$  from the sampled signal  $y_{\rm A}(t) = x_{\rm A}(t)$  is only possible if the sampling rate  $f_{\rm A}$  corresponding to the bandwidth  $B_{\rm NF}$  of the message signal has been chosen correctly.

From the above graph, it can be seen that the following condition must be satisfied:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Sampling theorem:}$  If an analog signal  $x(t)$  has only spectral components in the range  $\vert f \vert < B_{\rm NF}$, it can be completely reconstructed from its sampled signal  $x_{\rm A}(t)$  only if the sampling rate is sufficiently large:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Accordingly, the following must apply to the distance between two samples:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

If the largest possible value   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  is used for sampling,
*so, for signal reconstruction of the analog signal from its samples.
*an ideal, rectangular low-pass filter with cut off frequency  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  must be used.

{{GraueBox|TEXT=
$\text{Example 2:}$  The graph shows above the spectrum  $\pm\text{ 5 kHz}$  of an analog signal limited to  $X(f)$  below the spectrum  $X_{\rm A}(f)$  of the signal sampled at distance  $T_{\rm A} =\,\text{ 100 µs}$  ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Sampling theorem in the frequency domain]]
Additionally drawn is the frequency response  $H_{\rm E}(f)$  of the low-pass receiving filter for signal reconstruction, whose cutoff frequency must be exactly  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$ .

*With any other  $f_{\rm G}$ value, there would be  $Y(f) \neq X(f)$.
*For  $f_{\rm G} < 5\,\text{ kHz}$  the upper  $X(f)$ portions are missing.
* At  $f_{\rm G} > 5\,\text{ kHz}$  there are unwanted spectral components in  $Y(f)$ due to convolution products.

If at the transmitter the sampling had been done with a sampling rate  $f_{\rm A} < 10\,\text{ kHz}$    ⇒   $T_{\rm A} >100 \ {\rm µ s}$, the analog signal  $y(t) = x(t)$  would not be reconstructible from the samples  $y_{\rm A}(t)$  in any case. }}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==
 
[[File:Anleitung_Auge.png|right|600px]]
    '''(A)'''     Auswahl:   Codierung                    (binär,  quaternär,  AMI–Code,  Duobinärcode)

    '''(B)'''     Auswahl:   Detektionsgrundimpuls                     (nach Gauß–TP,  CRO–Nyquist,  nach Spalt–TP}

    '''(C)'''     Prametereingabe zu  '''(B)'''                    (Grenzfrequenz,  Rolloff–Faktor,  Rechteckdauer)

    '''(D)'''     Steuerung der Augendiagrammdarstellung                    (Start,  Pause/Weiter,  Einzelschritt,  Gesamt,  Reset)

    '''(E)'''     Geschwindigkeit der Augendiagrammdarstellung

    '''(F)'''     Darstellung:  Detektionsgrundimpuls  $g_d(t)$

    '''(G)'''     Darstellung:  Detektionsnutzsignal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Darstellung:  Augendiagramm im Bereich  $\pm T$

    '''( I )'''     Numerikausgabe:  $ö_{\rm norm}$  (normierte Augenöffnung)

    '''(J)'''     Prametereingabe  $10 \cdot \lg \ E_{\rm B}/N_0$  für  '''(K)'''

    '''(K)'''     Numerikausgabe:  $\sigma_{\rm norm}$  (normierter Rauscheffektivwert)

    '''(L)'''     Numerikausgabe:  $p_{\rm U}$  (ungünstigste Fehlerwahrscheinlichkeit)

    '''(M)'''     Bereich für die Versuchsdurchführung:   Aufgabenauswahl

    '''(N)'''     Bereich für die Versuchsdurchführung:   Aufgabenstellung

    '''(O)'''     Bereich für die Versuchsdurchführung:   Musterlösung einblenden
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Pulses and Spectra

2023-03-25T18:20:32Z

Noah:

{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

==Applet Description==
 
Time-limited symmetric signals   ⇒   "pulses"  $x(t)$  and the corresponding spectral functions  $X(f)$  are considered, namely

*Gaussian pulse, 
*rectangular pulse,  
*triangular pulse, 
*trapezoidal pulse, 
*raised cosine pulse,
*cosine square pulse.

Further it is to be noted:
* The functions  $x(t)$  resp.  $X(f)$  are shown for up to two parameter sets in one diagram each.
* The red curves and numbers apply to the left parameter set, the blue ones to the right parameter set.
* The abscissas  $t$  (time) and  $f$  (frequency) as well as the ordinates  $x(t)$  (signal values) and  $X(f)$  (spectral values) are normalized.

==Theoretical Background==
 
===Relationship $x(t)\Leftrightarrow X(f)$===
*The relationship between the time function  $x(t)$  and the spectrum  $X(f)$  is given by the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|"first Fourier integral"]] :
:$$X(f)={\rm FT} [x(t)] = \int_{-\infty}^{+\infty}x(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm}
\rm FT\hspace{-0.1cm}: \ Fourier \ transform.$$

*In order to calculate the time function  $x(f)$  from the spectral function  $x(t)$  one needs the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|"second Fourier integral"]]:
:$$x(t)={\rm IFT} [X(f)] = \int_{-\infty}^{+\infty}X(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm}
{\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fourier \ transform.$$

*In all examples we use real and even functions.  Thus:
:$$x(t)=\int_{-\infty}^{+\infty}X(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ X(f)=\int_{-\infty}^{+\infty}x(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$
*$x(t)$  and  $X(f)$  have different units, for example  $x(t)$  in  $\rm V$,  $X(f)$  in  $\rm V/Hz$.
*The relationship between this module and the similarly constructed applet  [[Applets:Frequenzgang_und_Impulsantwort|"Frequency response & Impulse response"]]  is based on the  [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem|"Duality Theorem"]].
*All times are normalized to a time  $T$  and all frequencies are normalized to  $1/T$   ⇒   the spectral values  $X(f)$  still have to be multiplied by the normalization time  $T$ .

{{GraueBox|TEXT=
$\text{Example:}$   If one sets a rectangular pulse with amplitude  $A_1 = 1$  and equivalent pulse duration  $\Delta t_1 = 1$  then  $x_1(t)$  in the range  $-0.5 < t < +0. 5$  equal to one and outside this range equal to zero.  The spectral function  $X_1(f)$  proceeds  $\rm si$–shaped with  $X_1(f= 0) = 1$  and the first zero at  $f=1$.

*If a rectangular pulse with  $A = K = 3 \ \rm V$  and  $\delta t = T = 2 \ \rm ms$  is to be simulated with this setting, then all signal values with  $K = 3 \ \rm V$  and all spectral values with  $K \cdot T = 0. 006 \ \rm V/Hz$  to be multiplied by.
*The maximum spectral value is then  $X(f= 0) = 0.006 \ \rm V/Hz$  and the first zero is at  $f=1/T = 0.5 \ \rm kHz$.}}

===Gaussian Pulse ===

*The time function of the Gaussian pulse with height  $K$  and (equivalent) duration  $\Delta t$  is:
:$$x(t)=K\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/\Delta t)^2}.$$
*The equivalent time duration  $\Delta t$  is obtained from the rectangle of equal area.
*The value at  $t = \Delta t/2$  is smaller than the value at  $t=0$ by the factor  $0.456$ .
*For the spectral function we get according to the Fourier transform:
:$$X(f)=K\cdot \Delta t \cdot {\rm e}^{-\pi(f\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t)^2} .$$
*The smaller the equivalent time duration  $\Delta t$  is, the wider and lower the spectrum   ⇒   [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity law of bandwidth and pulse duration"]].
*Both  $x(t)$  and  $X(f)$  are not exactly zero at any  $f$–  or  $t$–value, respectively.
*For practical applications, however, the Gaussian pulse can be assumed to be limited in time and frequency.  For example,  $x(t)$  has already dropped to less than  $0.1\% $  of the maximum at  $t=1.5 \delta t$  .

===Rectangular Pulse ===
*The time function of the rectangular pulse with height  $K$  and (equivalent) duration  $\Delta t$  is:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < T/2,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| = T/2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| > T/2.} \\ \end{array}$$

*The $\pm \Delta t/2$ value lies midway between the left- and right-hand limits.
*For the spectral function one obtains according to the laws of the Fourier transform (1st Fourier integral):
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f) \quad \text{with} \ {\rm si}(x)=\frac{\sin(x)}{x}.$$
*The spectral value at  $f=0$  is equal to the rectangular area of the time function.
*The spectral function has zeros at equidistant distances  $1/\delta t$.
*The integral over the spectral function  $X(f)$  is equal to the signal value at time  $t=0$, i.e. the pulse height  $K$.

===Triangular Pulse===
*The time function of the triangular pulse with height  $K$  and (equivalent) duration  $\Delta t$  is:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot (1-|t|/{\Delta t}) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*The absolute time duration is  $2 \cdot \Delta t$;  this is twice as large as that of the rectangle.
*For the spectral function, we obtain according to the Fourier transform:
:$$X(f)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta t \cdot f) \quad \text{with} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*The above time function is equal to the convolution of two rectangular pulses, each with width  $\delta t$.
*From this follows:  $X(f)$  contains instead of the  ${\rm si}$-function the  ${\rm si}^2$-function.
*$X(f)$  thus also has zeros at equidistant intervals  $1/\rm f$ .
*The asymptotic decay of  $X(f)$  occurs here with  $1/f^2$, while for comparison the rectangular pulse decays with  $1/f$ .

===Trapezoidal Pulse ===
The time function of the trapezoidal pulse with height  $K$  and time parameters  $t_1$  and  $t_2$  is:
:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{t_2-|t|}{t_2-t_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}}\quad \\ {\rm{for}}\quad \\ {\rm{for}} \quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*For the equivalent pulse duration (rectangle of equal area) holds:   $\Delta t = t_1+t_2$.
*The rolloff factor (in the time domain) characterizes the slope:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*The special case  $r=0$  corresponds to the rectangular pulse and the special case  $r=1$  to the triangular pulse.
*For the spectral function one obtains according to the Fourier transform:
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f)\cdot {\rm si}(\pi \cdot r \cdot \Delta t \cdot f) \quad \text{with} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*The asymptotic decay of  $X(f)$  lies between  $1/f$  $($for rectangle,  $r=0)$  and  $1/f^2$  $($for triangle,  $r=1)$.

===Raised cosine Pulse ===
The time function of the raised cosine pulse with height  $K$  and time parameters  $t_1$  and  $t_2$  is:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|t|-t_1}{t_2-t_1}\cdot {\pi}/{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}}\quad \\ {\rm{for}}\quad \\ {\rm{for}}\quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*For the equivalent pulse duration (rectangle of equal area) holds:   $\Delta t = t_1+t_2$.
*The rolloff factor (in the time domain) characterizes the slope:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*The special case  $r=0$  corresponds to the square pulse and the special case  $r=1$  to the cosine square pulse.
*For the spectral function one obtains according to the Fourier transform:
:$$X(f)=K\cdot \Delta t \cdot \frac{\cos(\pi \cdot r\cdot \Delta t \cdot f)}{1-(2\cdot r\cdot \Delta t \cdot f)^2} \cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*The larger the rolloff factor  $r$  is, the faster  $X(f)$  decreases asymptotically with  $f$ .

===Cosine square Pulse ===
*This is a special case of the raised cosine pulse and results for  $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}t_1=0, \ t_2= \Delta t$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|t|\hspace{0.05cm}\cdot \hspace{0.05cm} \pi}{2\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*For the spectral function, we obtain according to the Fourier transform:
:$$X(f)=K\cdot \Delta f \cdot \frac{\pi}{4}\cdot \big [{\rm si}(\pi(\Delta t\cdot f +0.5))+{\rm si}(\pi(\Delta t\cdot f -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Because of the last  ${\rm si}$-function is  $X(f)=0$  for all multiples of  $F=1/\delta t$.  The equidistant zero crossings of the raised cosine pulse are preserved.
*Because of the bracket expression,  $X(f)$  now exhibits further zero crossings at  $f=\pm1.5 F$,  $\pm2.5 F$,  $\pm3.5 F$, ... .
*For frequency  $f=\pm F/2$  the spectral values  $K\cdot \Delta t/2$ are obtained.
*The asymptotic decay of  $X(f)$  runs in this special case with  $1/f^3$.

==Exercises==
 

* First select the number  $(1,\text{...}, 7)$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
* "Red" refers to the first parameter set ⇒ $x_1(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_1(f)$,  "Blue" refers to the second parameter set ⇒ $x_2(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_2(f)$.
*Values with magnitude less than  $0.0005$  are output in the program as "zero".

{{BlaueBox|TEXT=
'''(1)'''   Compare the  red Gaussian pulse  $(A_1 = 1, \Delta t_1 = 1)$  with the  blue rectangular pulse  $(A_2 = 1, \Delta t_2 = 1)$ ⇒ default setting.
           What are the differences in the time and frequency domain?}}

* The Gaussian pulse theoretically reaches infinity in the time as well as in the frequency domain. 
* Practically  $x_1(t)$  for  $|t| > 1.5$  and  $X_1(f)$  for  $|f| > 1.5$  are almost zero. 
* The rectangle is strictly limited in time:  $x_2(|t| > 0.5) \equiv 0$.  $X_2(f)$  has shares in a much larger range than  $X_1(f)$. 
* It holds  $X_1(f = 0) = X_2(f = 0)$  since the integral over the Gaussian pulse  $x_1(t)$  is equal to the integral over the rectangular pulse  $x_2(t)$.

{{BlaueBox|TEXT=
'''(2)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2)$.           Vary the equivalent pulse duration  $\Delta t_2$  between  $0.5$  and  $2$.  Interpret the displayed graphs.}}

* One can recognize the reciprocity law of bandwidth and pulse duration.  The greater  $\Delta t_2$, the higher and narrower the spectral function  $X_2(f)$. 
* For each setting of  $\Delta t_2$,  $x_1(t=0)$  and  $x_2(t=0)$  are equal   ⇒   Also, the integrals over  $X_1(f)$  and  $X_2(f)$  are identical.

{{BlaueBox|TEXT=
'''(3)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2 = 0.5)$.           Vary  $\Delta t_2$  between  $0.05$  and  $2$.  Interpret the displayed graphs and extrapolate the result.}}

* The blue spectrum is now twice as wide as the red one, but only half as high.  First zero of  $X_1(f)$  at  $f = 1$, of  $X_2(f)$  at  $f = 2$. 
* Reduction of  $\Delta t_2$:  $X_2(f)$  lower and wider.  Very flat course at  $\Delta t_2 = 0.05$:  $X_2(f = 0)= 0.05$,  $X_2(f = \pm 3)= 0.048$. 
* If one choose  $\Delta t_2 = \varepsilon \to 0$  (not possible in the program),  the result would be the almost constant, very small spectrum  $X_2(f)=A \cdot \varepsilon \to 0$. 
* Increasing the amplitude to  $A=1/\varepsilon$  results in the constant spectral function  $X_2(f) = 1$  of the Dirac function  $\delta(t)$.  That means: 
* $\delta(t)$  is approximated by a rectangle  $($width  $\Delta t = \varepsilon \to 0$,  height  $A = 1/\varepsilon \to \infty)$.  The weight of the Dirac function is one:  $x(t) = 1 \cdot \delta (t)$.

{{BlaueBox|TEXT=
'''(4)'''   Compare the  rectangular pulse  $(A_1 = 1,   \Delta t_1 = 1)$  with the  triangular pulse  $(A_2 = 1,   \Delta t_2 = 1)$.  Interpret the spectral functions.}}

* The (normalized) spectrum of the rectangle  $x_1(t)$  with the (normalized) parameters  $A_1 = 1, \ \ \Delta t_1 = 1$  is:  $X_1(f)= {\rm si}(\pi\cdot f)= {\rm sinc}(f)$. 
* The convolution of the rectangle  $x_1(t)$  with itself gives the triangle  $x_2(t) = x_1(t) \star x_1(t)$.  By the convolution theorem:   $X_2(f) = X_1(f)^2 $. 
* By squaring the  ${\rm sinc}(f)$–shaped spectral function  $X_1(f)$  the zeros of  $X_2(f)$  remain unchanged.  But now it holds that: $X_2(f) \ge 0$.

{{BlaueBox|TEXT=
'''(5)'''   Compare the  trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, r_1 = 0.5)$  with the 
triangular pulse $(A_2 = 1,   \Delta t_2 = 1)$.          Vary  $r_1$  between  $0$  and  $1$.  Interpret the spectral function  $X_1(f)$.}}

* The trapezoidal pulse with roll–off factor  $r_1= 0$  is identical to the rectangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}(f)$. 
* The trapezoidal pulse with roll–off factor  $r_1= 1$  is identical to the triangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}^2(f)$. 
* In both cases  $X_1(f)$  has equidistant zeros at  $\pm 1$,  $\pm 2$, ...  (none else);   $0 < r_1 < 1$:  depending on  $r_1$  further zeros.

{{BlaueBox|TEXT=
'''(6)'''   Compare this  trapezoidal pulse  with the cosine roll-off pulse 
$(A_2 = 1,\ \Delta t_2 = 1.0,\ r_2 = 0.5)$.          Vary  $r_2$  between  $0$  and  $1$.  Interpret the spectral function  $X_2(f)$  for  $r_2 = 0.7$.}}

* With the same  $r= 0.5$  the cosine roll-off pulse  $X_2(f)$ is for  $f > 1$  greater in magnitude than the trapezoidal pulse. 
* With the same roll-off factor  $(r_1 = r_2= 0.5)$  the drop of  $X_2(f)$  around the frequency  $f = 0.5$  is steeper than the drop of  $X_1(f)$. 
* With  $r_1 = 0.5$  and  $r_2 = 0.7$  $x_1(t) \approx x_2(t)$  is valid and therefore also  $X_1(f) \approx X_2(f)$.  Comparable edge steepness.

{{BlaueBox|TEXT=
'''(7)'''   Compare the  red trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, \ r_1 = 1)$  with the  blue cosine roll-off pulse  $(A_2 = 1,\ \Delta t_2 = 1.0, \ r_2 = 1)$.           Interpret the time function  $x_2(t)$  and the spectral function  $X_2(f)$  system theoretically.}}

* $x_2(t) = \cos^2(|t|\cdot \pi/2) \ \ \ \text{for} \ |t| \le 1$  is the  cosine square pulse.  Zeros at  $f = \pm 1$,  $\pm 2$, ... 
* For the frequency  $f=\pm 0.5$  one obtains the spectral values  $X_2(f)=0.5$.  The asymptotic decline is shown here with  $1/f^3$.

==Applet Manual==
 
[[File:Exercise_impuls.png |right|frame|Screenshot]]
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Preselection for pulse shape  $x_1(t)$  (red curve)

    '''(C)'''     Parameter definition for  $x_1(t)$ 

    '''(D)'''     Numeric output for  $x_1(t_*)$  and  $X_1(f_*)$

    '''(E)'''     Preselection for pulse shape  $x_2(t)$  (blue curve)

    '''(F)'''     Parameter definition for  $x_2(t)$ 

    '''(G)'''     Numeric output for  $x_2(t_*)$  and  $X_2(f_*)$

    '''(H)'''     Setting the time  $t_*$  for the numeric output

    '''(I)'''     Setting the frequency  $f_*$  for the numeric output

    '''(J)'''     Graphic field for the time domain

    '''(K)'''     Graphic field for the frequency domain

    '''(L)'''     Selection of the exercise according to the numbers

    '''(M)'''     Task description and questions

    '''(N)'''     Show and hide sample solution

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2005 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2017 the program was redesigned by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]]  (Ingenieurspraxis_Math, Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2020 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

Applets:Principle of Pseudo-Ternary Coding

2023-03-25T18:19:59Z

Noah:

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

==Applet Description==
 
The applet covers the properties of the best known pseudo-ternary codes, namely:
#  First-order bipolar code,  $\rm AMI$ code  (from: ''Alternate Mark Inversion''),  characterized by the parameters  $N_{\rm C} = 1, \ K_{\rm C} = +1$,
#  Duobinary code,  $(\rm DUOB)$,  code parameters:  $N_{\rm C} = 1, \ K_{\rm C} = -1$,
#  Second order bipolar code  $(\rm BIP2)$,  code parameters:  $N_{\rm C} = 2, \ K_{\rm C} = +1$.

At the input is the redundancy-free binary bipolar source symbol sequence  $\langle \hspace{0.05cm}q_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$   ⇒   rectangular signal  $q(t)$  an.  Illustrating the generation.
*of the binary–precoded sequence  $\langle \hspace{0.05cm}b_\nu \hspace{0.05cm}\rangle \ \in \{+1, -1\}$,  represented by the also redundancy-free binary bipolar rectangular signal  $b(t)$,
*the pseudo-ternary code sequence  $\langle \hspace{0.05cm}c_\nu \hspace{0.05cm}\rangle \ \in \{+1,\ 0, -1\}$,  represented by the redundant ternary bipolar rectangular signal  $c(t)$,
*the equally redundant ternary transmitted signal  $s(t)$, characterized by the amplitude coefficients  $a_\nu $,  and the (transmitted–) base impulse  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$

The base impulse  $g(t)$  –  in the applet  "Rectangle",  "Nyquist" and  "Root–Nyquist"  –  determines not only the shape of the transmitted signal, but also the course of
* of the auto-correlation function  $\rm (ACF)$  $\varphi_s (\tau)$  and
* of the associated power spectral density  $\rm (PSD)$  ${\it \Phi}_s (f)$.

The applet also shows that the total power spectral density  ${\it \Phi}_s (f)$ can be split into the part  ${\it \Phi}_a (f)$ that takes into account the statistical relations of the amplitude coefficients  $a_\nu$   and the energy spectral density $ {\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2 $, characterized by the shape  $g(t)$.

''Note''   In the applet, no distinction is made between the encoder symbols  $c_\nu \in \{+1,\ 0, -1\}$  and the amplitude coefficients  $a_\nu \in \{+1,\ 0, -1\}$  .   It should be remembered that the  $a_\nu$  are always numerical values, while for the encoder symbols also the notation  $c_\nu \in \{\text{plus},\ \text{zero},\ \text{minus}\}$  would be admissible.

==Theoretical Background==

== General description of the pseudo-multilevel codes ==
 
In symbolwise coding,  each incoming source symbol  $q_\nu$  generates an encoder symbol  $c_\nu$,  which depends not only on the current input symbol  $q_\nu$  but also on the  $N_{\rm C}$  preceding symbols  $q_{\nu-1}$, ... , $q_{\nu-N_{\rm C}} $.   $N_{\rm C}$  is referred to as the  "order"  of the code. 

Typical for symbolwise coding is that
*the symbol duration  $T$  of the encoded signal  (and of the transmitted signal)  matches the bit duration  $T_{\rm B}$  of the binary source signal,  and

*encoding and decoding do not lead to major time delays,  which are unavoidable when block codes are used. 

The  "pseudo-multilevel codes"  – better known as  "partial response codes"  –  are of special importance.  In the following,  only  "pseudo-ternary codes"   ⇒   level number  $M = 3$  are considered. 
*These can be described by the block diagram corresponding to the left graph. 
*In the right graph an equivalent circuit is given,  which is very suitable for an analysis of these codes.

[[File:EN_Dig_T_2_4_S1_v23.png|right|frame|Block diagram  (above)  and equivalent circuit  (below)  of a pseudo-ternary encoder|class=fit]]

One can see from the two representations:
*The pseudo-ternary encoder can be split into the  "non-linear pre-encoder"  and the  "linear coding network",  if the delay by  $N_{\rm C} \cdot T$  and the weighting by  $K_{\rm C}$  are drawn twice for clarity  – as shown in the right equivalent figure.

*The  "non-linear pre-encoder"  obtains the precoded symbols  $b_\nu$,  which are also binary,  by a modulo–2 addition  ("antivalence")  between the symbols  $q_\nu$  and  $K_{\rm C} \cdot b_{\nu-N_{\rm C}} $: 
:$$q_\nu \in \{-1, +1\},\hspace{0.1cm} K_{\rm C} \in \{-1,
+1\}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}b_\nu \in \{-1,
+1\}\hspace{0.05cm}.$$
*Like the source symbols  $q_\nu$,  the symbols  $b_\nu$  are statistically independent of each other.  Thus,  the pre-encoder does not add any redundancy.  However,  it allows a simpler realization of the decoder and prevents error propagation after a transmission error. 

*The actual encoding from binary  $(M_q = 2)$  to ternary  $(M = M_c = 3)$  is done by the  "linear coding network"  by the conventional subtraction
:$$c(t) ={1}/{2} \cdot \big [b(t) - K_{\rm C} \cdot b(t- N_{\rm
C}\cdot T)\big] \in \{-1, \ 0, +1\}\hspace{0.05cm},$$

:which can be described by the following  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|"impulse response"]]  resp.  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function|"transfer function"]]  with respect to the input signal  $b(t)$  and the output signal  $c(t)$: 
:$$h_{\rm C}(t) = {1}/{2} \cdot \big [\delta(t) - K_{\rm C} \cdot \delta(t- N_{\rm
C}\cdot T)\big] \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ H_{\rm C}(f) ={1}/{2} \cdot \left [1 - K_{\rm C} \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}N_{\rm C}\hspace{0.05cm}\cdot \hspace{0.05cm}T}\right]\hspace{0.05cm}. $$

*The relative redundancy is the same for all pseudo-ternary codes.  Substituting  $M_q=2$,  $M_c=3$  and  $T_c =T_q$  into the  [[Digital_Signal_Transmission/Redundancy-Free_Coding#Symbolwise_coding_vs._blockwise_coding|"general definition equation"]],  we obtain
:$$r_c = 1- \frac{R_q}{R_c} = 1- \frac{T_c}{T_q} \cdot \frac{{\rm log_2}\hspace{0.05cm} (M_q)}{{\rm log_2} \hspace{0.05cm}(M_c)} = 1- \frac{T_c}{T_q \cdot {\rm log_2} \hspace{0.05cm}(M_c)}\hspace{0.5cm}\Rightarrow \hspace{0.5cm} r_c = 1 -1/\log_2\hspace{0.05cm}(3) \approx 36.9 \%\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
The  $\text{transmitted signal of all pseudo-ternary codes}$  is always represented as follows:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g( t - \nu \cdot T)\hspace{0.05cm}.$$
*The property of the current pseudo-ternary code is reflected in the statistical bonds between the  $a_\nu$  In all cases  $a_\nu \in \{-1, \ 0, +1\}$.
*The basic transmitting pulse  $g(t)$  provides on the one hand the required energy, but has also influence on the statistical bonds within the signal.
*In addition to the NRZ rectangular pulse  $g_{\rm R}(t)$ can be selected in the program: 
:*the Nyquist impulse   ⇒   impulse response of the raised cosine low-pass with rolloff factor $r$:
:$$g_{\rm Nyq}(t)={\rm const.} \cdot \frac{\cos(\pi \cdot r\cdot t/T)}{1-(2\cdot r\cdot t/T)^2} \cdot {\rm si}(\pi \cdot t/T) \ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\rm Nyq}(f),$$

:*the root Nyquist impulse   ⇒   impulse response of the root raised cosine low-pass with rolloff factor $r$:
:$$g_{\sqrt{\rm Nyq} }(t)\ \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ G_{\sqrt{\rm Nyq} }(f)={\rm const.} \cdot \sqrt{G_{\rm Nyq}(f)} .$$ }}
 

== Properties of the AMI code==
 
The individual pseudo-ternary codes differ in the  $N_{\rm C}$  and  $K_{\rm C}$ parameters.  The best-known representative is the  '''first-order bipolar code'''  with the code parameters
:* $N_{\rm C} = 1$, 
:* $K_{\rm C} = 1$,

which is also known as  '''AMI code'''  (from: "Alternate Mark Inversion").  This is used e.g. with  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN|"ISDN"]]  ("Integrated Services Digital Networks")  on the so-called  $S_0$  interface.

[[File:EN_Dig_T_2_4_S2a.png|right|frame|Signals with AMI coding and HDB3 coding|class=fit]]
*The graph above shows the binary source signal  $q(t)$.

*The second and third diagrams show:
:* the likewise binary signal  $b(t)$  after the pre-encoder,  and

:* the encoded signal  $c(t) = s(t)$  of the AMI code.

One can see the simple AMI encoding principle:
#Each binary value  "–1"  of the source signal  $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary amplitude coefficient  $a_\nu = 0$.  
#The binary value  "+1"  of the source signal  $q(t)$   ⇒   symbol  $\rm H$  is alternately represented by  $a_\nu = +1$  and  $a_\nu = -1$.  

This ensures that the AMI encoded signal does not contain any   "long sequences" 
*   $ \langle c_\nu \rangle = \langle \text{...}, +1, +1, +1, +1, +1, \text{...}\rangle$   resp.
*   $ \langle c_\nu \rangle = \langle \text{...}, -1, -1, -1, -1, -1, \text{...}\rangle$,

which would lead to problems with a DC-free channel.

On the other hand,  the occurrence of long zero sequences is quite possible,  where no clock information is transmitted over a longer period of time.
 
To avoid this second problem,  some modified AMI codes have been developed, for example the  "B6ZS code"  and the  "HDB3 code":
*In the  '''HDB3 code'''  (green curve in the graphic),  four consecutive zeros in the AMI encoded signal are replaced by a subsequence that violates the AMI encoding rule. 

*In the gray shaded area,  this is the sequence  "$+\ 0\ 0\ +$",  since the last symbol before the replacement was a  "minus". 

*This limits the number of consecutive zeros to   $3$   for the HDB3 code and to   $5$   for the  [https://www.itwissen.info/en/bipolar-with-six-zero-substitution-B6ZS-121675.html#gsc.tab=0 "B6ZS code"]. 

*The decoder detects this code violation and replaces "$+\ 0\ 0\ +$" with "$0\ 0\ 0\ 0$" again. 
 
=== Calculating the ACF of a digital signal ===
In the execution of the experiment some quantities and correlations are used, which shall be briefly explained here:

*  The (time-unlimited) digital signal includes both the source statistics $($amplitude coefficients  $a_\nu$)  and the transmitted pulse shape  $g(t)$:
:$$s(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g ( t - \nu \cdot T)\hspace{0.05cm}.$$

*  If  $s(t)$  is the pattern function of a stationary and ergodic random process, then for the  [[Theory_of_Stochastic_Signals/Auto-Correlation_Function#Random_processes|"Auto-Correlation Function"]]  $\rm (ACF)$:
:$$\varphi_s(\tau) = {\rm E}\big [s(t) \cdot s(t + \tau)\big ] = \sum_{\lambda = -\infty}^{+\infty}{1}/{T}
\cdot \varphi_a(\lambda)\cdot\varphi^{^{\bullet} }_{gs}(\tau -
\lambda \cdot T)\hspace{0.05cm}.$$

*  This equation describes the convolution of the discrete ACF  $\varphi_a(\lambda) = {\rm E}\big [ a_\nu \cdot a_{\nu + \lambda}\big]$  of the amplitude coefficients with the energy–ACF of the base impulse:

:$$\varphi^{^{\bullet} }_{g}(\tau) =
\int_{-\infty}^{+\infty} g ( t ) \cdot g ( t +
\tau)\,{\rm d} t \hspace{0.05cm}.$$

*The point is to indicate that  $\varphi^{^{\bullet} }_{g}(\tau)$  has the unit of an energy, while  $\varphi_s(\tau)$  indicates a power and  $\varphi_a(\lambda)$  is dimensionless.
 
=== Calculating the PSD of a digital signal ===
The corresponding quantity to the ACF in the frequency domain is the [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|"Power Spectral Density"]]  $\rm (PSD)$  ${\it \Phi}_s(f)$, which is fixedly related to  $\varphi_s(\tau)$  via the Fourier integral: 
:$$\varphi_s(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}_s(f) = \int_{-\infty}^{+\infty} \varphi_s(\tau) \cdot
{\rm e}^{- {\rm j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \tau}
\,{\rm d} \tau \hspace{0.05cm}.$$
*The power spectral density  ${\it \Phi}_s(f)$  can be represented as a product of two functions, taking into account the dimensional adjustment  $(1/T)$ :
:$${\it \Phi}_s(f) = {\it \Phi}_a(f) \cdot {1}/{T} \cdot
|G_s(f)|^2 \hspace{0.05cm}.$$
*The first term  ${\it \Phi}_a(f)$  is dimensionless and describes the spectral shaping of the transmitted signal by the statistical relations of the source: 
:$$\varphi_a(\lambda) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}{\it \Phi}_a(f) = \sum_{\lambda =
-\infty}^{+\infty}\varphi_a(\lambda)\cdot {\rm e}^{- {\rm
j}\hspace{0.05cm} 2 \pi f \hspace{0.02cm} \lambda \hspace{0.02cm}T} =
\varphi_a(0) + 2 \cdot \sum_{\lambda =
1}^{\infty}\varphi_a(\lambda)\cdot\cos ( 2 \pi f
\lambda T) \hspace{0.05cm}.$$
*${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  takes into account the spectral shaping by  $g(t)$. The narrower this is, the wider  $\vert G(f) \vert^2$  and thus the larger the bandwidth requirement:
:$$\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau) \hspace{0.4cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet \hspace{0.4cm}
{\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = |G(f)|^2
\hspace{0.05cm}.$$
*The energy spectral density ${\it \Phi^{^{\hspace{0.08cm}\bullet}}_{g}}(f)$  has the unit  $\rm Ws/Hz$  and the power spectral density  ${\it \Phi_{s}}(f)$  after division by the symbol spacing  $T$  has the unit  $\rm W/Hz$.

== Power-spectral density of the AMI code==
 
The frequency response of the linear code network of a pseudo-ternary code is generally:
:$$H_{\rm C}(f) = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
2\pi\hspace{0.03cm}\cdot \hspace{0.03cm}f \hspace{0.03cm}\cdot
\hspace{0.03cm} N_{\rm C}\hspace{0.03cm}\cdot \hspace{0.03cm}T}
\big] ={1}/{2} \cdot \big [1 - K \cdot {\rm
e}^{-{\rm j}\hspace{0.03cm}\cdot \hspace{0.03cm}
\alpha}
\big ]\hspace{0.05cm}.$$

This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients  $(K$  and  $\alpha$  are abbreviations according to the above equation$)$:
:$$ {\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = \frac{\big [1 - K \cos
(\alpha) + {\rm j}\cdot K \sin (\alpha) \big ] \big [1 - K \cos
(\alpha) - {\rm j}\cdot K \sin (\alpha) \big ] }{4} = \text{...} = {1}/{4} \cdot \big [2 - 2 \cdot K \cdot \cos
(\alpha) \big ] $$
[[File:P_ID1347__Dig_T_2_4_S2b_v2.png|right|frame|Power-spectral density of the AMI code|class=fit]]
:$$ \Rightarrow \hspace{0.3cm}{\it \Phi}_a(f) = | H_{\rm C}(f)|^2 = {1}/{2} \cdot \big [1 - K_{\rm C} \cdot \cos
(2\pi f N_{\rm C} T)\big ]
\hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm}
\varphi_a(\lambda \cdot T)\hspace{0.05cm}.$$

In particular,  for the power-spectral density of the AMI code  $(N_{\rm C} = K_{\rm C} = 1)$,  we obtain:
:$${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos
(2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$$

The graph shows
*the PSD  ${\it \Phi}_a(f)$  of the amplitude coefficients  (red curve),  and 

*the PSD  ${\it \Phi}_s(f)$  of the total transmitted signal  (blue),  valid for NRZ rectangular pulses. 

One recognizes from this representation
*that the AMI code has no DC component,  since  ${\it \Phi}_a(f = 0) = {\it \Phi}_s(f = 0) = 0$,  

*the power  $P_{\rm S} = s_0^2/2$  of the AMI-coded transmitted signal  $($integral over  ${\it \Phi}_s(f)$  from  $- \infty$  to  $+\infty)$.

Notes:
*The PSD of the HDB3 and B6ZS codes differs only slightly from that of the AMI code. 

*You can use the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]]  to clarify the topic discussed here.

== Properties of the duobinary code ==
 
The  '''duobinary code'''  is defined by the code parameters  $N_{\rm C} = 1$  and  $K_{\rm C} = -1$.  This gives the power-spectral density  $\rm (PSD)$  of the amplitude coefficients and the PSD of the transmitted signal:
[[File:P_ID1348__Dig_T_2_4_S3b_v1.png|right|frame|Power-spectral density of the duobinary code|right|class=fit]]

:$${\it \Phi}_a(f) ={1}/{2} \cdot \big [1 + \cos
(2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm},$$
:$$ {\it \Phi}_s(f) = s_0^2 \cdot T \cdot \cos^2
(\pi f T)\cdot {\rm si}^2
(\pi f T)= s_0^2 \cdot T \cdot {\rm si}^2
(2 \pi f T) \hspace{0.05cm}.$$

The graph shows the power-spectral density
*of the amplitude coefficients   ⇒   ${\it \Phi}_a(f)$  as a red curve, 

*of the total transmitted signal   ⇒   ${\it \Phi}_s(f)$  as a blue curve. 

In the second graph,  the signals  $q(t)$,  $b(t)$  and  $c(t) = s(t)$  are sketched. We refer here again to the  (German language)  SWF applet  [[Applets:Pseudoternaercodierung|"Signals, ACF, and PSD of pseudo-ternary codes"]],  which also clarifies the duobinary code.

[[File:P_ID1349__Dig_T_2_4_S3a_v2.png|left|frame|Signals in duobinary coding|class=fit]]
 From these illustrations it is clear:
*In the duobinary code,  any number of symbols with same polarity  ("+1"  or  "–1")  can directly succeed each other   ⇒   ${\it \Phi}_a(f = 0)=1$,  ${\it \Phi}_s(f = 0) = 1/2 \cdot s_0^2 \cdot T$. 

*In contrast,  for the duobinary code,  the alternating sequence  "... , +1, –1, +1, –1, +1, ..."  does not occur,  which is particularly disturbing with respect to intersymbol interference.  Therefore,  in the duobinary code:  ${\it \Phi}_s(f = 1/(2T) = 0$. 

*The power-spectral density  ${\it \Phi}_s(f)$  of the pseudo-ternary duobinary code is identical to the PSD with redundancy-free binary coding at half rate $($symbol duration  $2T)$. 
 

==Exercises==

 
*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  "$0$"  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".

{{BlueBox|TEXT=
'''(1)'''  Consider and interpret the binary pre–coding of the  $\text{AMI}$  code using the source symbol sequence  $\rm C$  assuming  $b_0 = +1$. }}
*The modulo–2 addition can also be taken as "antivalence".  It holds  $b_{\nu} = +1$,  if  $q_{\nu}$  and  $b_{\nu - 1}$  differ, otherwise set  $b_{\nu} = -1$ :
:  $b_1 = (q_1 = +1)\ {\rm XOR}\ (b_0= +1) = -1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (b_1= -1) = -1,\ \ b_3 = (q_3 = -1)\ {\rm XOR}\ (b_2= -1) = -1,$
:  $b_4 = (q_4 = +1)\ {\rm XOR}\ (b_3= -1) = +1,\ \ b_5 = (q_5 = +1)\ {\rm XOR}\ (b_4= +1) = -1,\ \ b_6 = (q_6 = +1)\ {\rm XOR}\ (b_5= -1) = +1,\ \ b_7 = b_8 = \text{...} = -1.$
*With the initial condition  $b_0 = -1$  we get the negated sequence:  $b_4 = b_6 =-1$.  All others  $b_\nu = +1$.

{{BlueBox|TEXT=
'''(2)'''  Let  $b_0 = +1$.  Consider the AMI encoded sequence  $\langle c_\nu \rangle$  of the source symbol sequence  $\rm C$  and give their amplitude coefficients  $a_\nu$.  }}

*It holds:  $a_1= 0.5 \cdot (b_1-b_0) = -1$,  $a_2= 0.5 \cdot (b_2-b_1) =0$,  $a_3= 0. 5 \cdot (b_3-b_2) =0$,  $a_4= +1$,  $a_5= -1$,  $a_6= +1$,  $a_7= -1$,  $a_8= a_9 = \text{...} = 0$.
*In contrast to the pre–coding, the conventional addition (subtraction) is to be applied here and not the modulo–2 addition.

{{BlueBox|TEXT=
'''(3)'''  Now consider the AMI coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$ ?}}

*Each binary value  "–1"  of   $q(t)$   ⇒   symbol  $\rm L$  is encoded by the ternary coefficient  $a_\nu = 0$.  Any number of  $a_\nu = 0$  can be consecutive.
*The binary value  "+1"  of   $q(t)$   ⇒   symbol  $\rm H$  is represented alternatively with   $a_\nu = +1$  and  $a_\nu = -1$ , starting with  $a_\nu = -1$,  if  $b_0 = +1$.
*From the source symbol sequence  $\rm A$   ⇒   $\langle \hspace{0.05cm}q_\nu \equiv +1 \hspace{0.05cm}\rangle$  the code symbol sequence  $+1, -1, +1, -1, \text{...}$ . Long sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  or   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are shot out.

{{BlueBox|TEXT=
'''(4)'''  Continue with  AMI  coding.  Interpret the autocorrelation function  $\varphi_a(\lambda)$  of the amplitude coefficients and the power density spectrum  ${\it \Phi}_a(f)$. }}
*The discrete ACF  $\varphi_a(\lambda)$  of the amplitude coefficients is only defined for integer  $\lambda$  values.   With AMI coding  $(N_{\rm C}=1)$  holds:   For  $|\lambda| > 1$   ⇒   all  $\varphi_a(\lambda)= 0$.
*$\varphi_a(\lambda = 0)$  is equal to the root mean square of the amplitude coefficients   ⇒   $\varphi_a(\lambda = 0) = {\rm Pr}(a_\nu = +1) \cdot (+1)^2 + {\rm Pr}(a_\nu = -1) \cdot (-1)^2 = 0.5.$
*Only the combinations  $(+1, -1)$  and  $(-1, +1)$  contribute to the expected value  ${\rm E}\big [a_\nu \cdot a_{\nu+1}\big]$   Result:  $\varphi_a(\lambda = \pm 1)={\rm E}\big [a_\nu \cdot a_{\nu+1}\big]=-0.25.$
*The power density spectrum  ${\it \Phi}_a(f)$  is the Fourier transform of the discrete ACF  $\varphi_a(\lambda)$.  Result:  ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 - \cos (2\pi f T)\big ] = \sin^2
(\pi f T)\hspace{0.05cm}.$
*From   ${\it \Phi}_a(f = 0) = 0$  follows:   The AMI code is especially interesting for channels over which no DC component can be transmitted.

{{BlueBox|TEXT=
'''(5)'''  We consider further AMI coding and rectangular pulses.  Interpret the ACF  $\varphi_s(\tau)$  of the transmission signal and the PDS  ${\it \Phi}_s(f)$. }}
*$\varphi_s(\tau)$  results from the convolution of the discrete AKF  $\varphi_a(\lambda)$  with  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$.   For rectangular pulses  $($duration $T)$:  The energy–AKF  $\varphi^{^{\hspace{0.05cm}\bullet}}_{g}(\tau)$  is a triangle of duration  $2T$.
*It holds  $\varphi_s(\tau = 0)= \varphi_a(\lambda = 0) =0.5, \ \varphi_s(\pm T)= \varphi_a( 1) =-0. 25,\ , \ \varphi_s( \pm 2T)= \varphi_a(2) =0.$  Between these discrete values, $\varphi_{s}(\tau)$  is always linear.
*The PDS  ${\it \Phi}_s(f)$  is obtained from  ${\it \Phi}_a(f) = \sin^2(\pi f T)$  by multiplying with  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f) = {\rm sinc}^2(f T).$  This does not change anything at the zeros of  ${\it \Phi}_a(f)$ .

{{BlueBox|TEXT=
'''(6)'''  What changes with respect to  $s(t)$,  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  with the  $\rm Nyquist$  pulse?  Vary the roll–off factor here in the range  $0 \le r \le 1$.}}

*A single Nyquist pulse can be represented with the source symbol sequence  $\rm B$  in the  $s(t)$  range.  You can see equidistant zero crossings in the distance  $T$.
*Also, for any AMI random sequence, the signal values  $s(t=\nu \cdot T)$  for each  $r$  correspond exactly to their nominal positions.  Outside these points, there are deviations.
*In the special case  $r=0$  the energy–LDS  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$.  Accordingly, the energy–ACF  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*On the other hand, for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are no longer equidistant, since although  $G(f)$  satisfies the first Nyquist criterion, it does not  ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$.
*The main advantage of the Nyquist pulse is the much smaller bandwidth.  Here only the frequency range  $|f| < (1+r)/(2T)$  has to be provided.

{{BlueBox|TEXT=
'''(7)'''  Repeat the last experiment using the  $\text{Root raised cosine}$  pulse instead of the Nyquist pulse.  Interpret the results. }}

*In the special case  $r=0$  the results are as in  '''(6)'''. ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)$  is constant in the range  $|f|<1/2T$  and outside zero;  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  has a  $\rm sinc$  shape.
*Also for larger  $r$  the zeros of  ${\it \varphi}^{^{\hspace{0.08cm}\bullet}}_{g}(\tau)$  are eqidistant  (but not  $\rm sinc$  shaped)   ⇒   ${\it \Phi}^{^{\hspace{0.08cm}\bullet}}_{g}(f)= [G(f)]^2$  satisfies the first Nyquist criterion.
*On the other hand, $G(f)$  does not satisfy the first Nyquist criterion  $($except for  $r=0)$.  Intersymbol interference occurs already at the transmitter   ⇒   signal  $s(t)$.
*But this is also not a fundamental problem.  By using an identically shaped reception filter like  $G(f)$  intersymbol interference at the decider is avoided.

{{BlueBox|TEXT=
'''(8)'''  Consider and check the pre–coding  $(b_\nu)$  and the amplitude coefficients  $(a_\nu)$  with the  $\rm duobinary$  coding   $($source symbol sequence  $\rm C$,  $b_0 = +1)$. }}
*$b_1 = (q_1 = +1)\ {\rm XOR}\ (\overline{b_0}= -1) = +1,\ \ b_2 = (q_2 = -1)\ {\rm XOR}\ (\overline{b_1}= -1) = -1,\ \ b_3 = \text{...} =b_7 = +1,$  $b_8 = b_{10} = \text{...} =-1$,  $b_9 =b_{11} = \text{...}= +1$.
*$a_1= 0.5 \cdot (b_1+b_0) = +1$,  $a_2= 0.5 \cdot (b_2+b_1) =0$,  $a_3= 0.5 \cdot (b_3+b_2) = 0$,  $a_4= \text{...}= a_7=+1$,  $a_8=a_9= \text{...}= 0$.
*With the starting condition  $b_0 = -1$  we get again the negated sequence:     $a_1= -1$,  $a_2= a_3= 0$,  $a_4= \text{...}= a_7=-1$,  $a_8=a_9= \text{...}= 0$.

{{BlueBox|TEXT=
'''(9)'''  Now consider the Duobinary coding for several random sequences.  What rules can be derived from these experiments for the amplitude coefficients  $a_\nu$?}}

*The discrete ACF values are  $\varphi_a(\lambda = 0) = +0.5$,  $\varphi_a(\lambda = 1) = +0.25$,  $\varphi_a(\lambda = 2) = 0$   ⇒   ${\it \Phi}_a(f) = {1}/{2} \cdot \big [1 + \cos (2\pi f T)\big ] = \cos^2
(\pi f T)\hspace{0.05cm}.$
*Unlike the AMI coding, the encoded sequences  $\langle \hspace{0.05cm}a_\nu \equiv +1 \hspace{0.05cm}\rangle$  and   $\langle \hspace{0.05cm}a_\nu \equiv -1 \hspace{0.05cm}\rangle$  are possible here   ⇒   For the Duobinary code holds  ${\it \Phi}_a(f= 0) = 1 \ (\ne 0).$
*As with the AMI code, the "long zero sequence"   ⇒   $\langle \hspace{0.05cm}a_\nu \equiv 0 \hspace{0.05cm}\rangle$  is possible, which can again lead to synchronization problems.
*Excluded are the combinations  $a_\nu = +1, \ a_{\nu+1} = -1$  and   $a_\nu = -1, \ a_{\nu+1} = +1$,  recognizable by the PDS value  ${\it \Phi}_a(f= 1/(2T)) = 0.$
*Such direct transitions  $a_\nu = +1$   ⇒   $a_{\nu+1} = -1$  resp.   $a_\nu = -1$   ⇒   $a_{\nu+1} = +1$  lead to large intersymbol interference and thus to a higher error rate.

{{BlueBox|TEXT=
'''(10)'''  Compare the coding results of second order bipolar code  $\rm (BIP2)$  and first order bipolar code  $\rm (AMI)$  for different source symbol sequences. }}
*For a single rectangular pulse   ⇒   source symbol sequence  $\rm B$  both codes result in the same encoded sequence and the same encoded signal  $c(t)$   ⇒   also an isolated pulse.
*The "Permanent–One sequence"  $\rm A$  now results  $\langle c_\nu \rangle = \langle -1, -1, +1, +1, -1, -1, +1, +1, \text{. ..}\rangle $  instead of   $\langle c_\nu \rangle = \langle -1, +1, -1, +1, -1, +1, \text{...}\rangle $.
*The simple decoding algorithm of the AMI code  $($the ternary  $0$  becomes the binary  $-1$,  the ternary  $\pm 1$  becomes the binary  $+1)$  cannot be applied to  $\rm BIP2$ .

{{BlueBox|TEXT=
'''(11)'''  View and interpret the various ACF and LDS graphs of the  $\rm BIP2$  compared to the  $\rm AMI$  code.}}
*For  $\rm AMI$:   $\varphi_a(\lambda = \pm 1) = -0.25, \ \varphi_a(\lambda = \pm 2) = 0$.   For  $\rm BIP2$:   $\varphi_a(\lambda = \pm 1) = 0, \ \varphi_a(\lambda = \pm 2) = -0.25$.  In both cases:  $\varphi_a(\lambda = 0) = 0.5$.
*From the  $\rm AMI$ power density spectrum  ${\it \Phi}_a(f) = \sin^2 (\pi \cdot f T)$  follows for  $\rm BIP2$:  ${\it \Phi}_a(f) = \sin^2 (2\pi \cdot f T)$  by compression with respect to the  $f$–axis.
*Zero at  $f=0$:  At most two  $+1$  directly follow each other, and also at most only two  $-1$.  In the AMI code  $+1$  and  $-1$  occur only in isolation.
*Next zero at  $f=1/(2T)$:  The infinitely long  $(+1, -1)$  sequence is excluded in  $\rm BIP2$  as in the  $\rm Duobinary$  code.
*Consider and interpret also the functions  $\varphi_s(\tau)$  and  ${\it \Phi}_s(f)$  for the pulses "rectangle", "Nyquist" and "Root raised cosine".

==Applet Manual==
[[File:BS_Pseudoternär.png|right|600px|frame|Bildschirmabzug (deutsche Version, heller Hintergrund)]]
 
    '''(A)'''     Theme (veränderbare grafische Oberflächengestaltung)
:* Dark:   schwarzer Hintergrund  (wird von den Autoren empfohlen)
:* Bright:   weißer Hintergrund  (empfohlen für Beamer und Ausdrucke)
:* Deuteranopia:   für Nutzer mit ausgeprägter Grün–Sehschwäche
:* Protanopia:   für Nutzer mit ausgeprägter Rot–Sehschwäche

    '''(B)'''     Zugrundeliegendes Blockschaltbild

    '''(C)'''     Auswahl des Pseudoternörcodes:
                    AMI–Code, Duobinärcode, Bipolarcode 2. Ordnung

    '''(D)'''     Auswahl des Grundimpulses  $g(t)$:
                    Rechteckimpuls, Nyquistimpuls, Wurzel–Nyquistimpuls

    '''(E)'''     Rolloff–Faktor (Frequenzbereich) für "Nyquist" und "Wurzel–Nyquist"

    '''(F)'''     Einstellung von  $3 \cdot 4 = 12$  Bit der Quellensymbolfolge

    '''(G)'''     Auswahl einer drei voreingestellten Quellensymbolfolgen

    '''(H)'''     Zufällige binäre Quellensymbolfolge

    '''( I )'''     Schrittweise Verdeutlichung der Pseudoternärcodierung

    '''(J)'''     Ergebnis der Pseudoternärcodierung:  Signale  $q(t)$,  $b(t)$,  $c(t)$,  $s(t)$

    '''(K)'''     Löschen der Signalverläufe im Grafikbereich  $\rm M$

    '''(L)'''     Skizzen für Autokorrelationsfunktion & Leistungsdichtespektrum

    '''(M)'''     Grafikbereich:  Quellensignal  $q(t)$, Signal  $b(t)$  nach Vorcodierung,                    Codersignal  $c(t)$  mit Rechtecken, Sendesignal  $s(t)$  gemäß  $g(t)$

    '''(N)'''     Bereich für Übungen:  Aufgabenauswahl, Fragen, Musterlösung
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2010 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Stefan_M.C3.BCller_.28Diplomarbeit_LB_2010.29|Stefan Müller]]    as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2020 the program was redesigned by  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  as part of her bachelor thesis  (Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ "Studienzuschüsse"]  (Faculty EI of the TU Munich).  We thank.
 

==Once again: Open Applet in new Tab==

{{LntAppletLinkEnDe|pseudoternarycodes_en|pseudoternarycodes}}

Applets:Sampling of Analog Signals and Signal Reconstruction

2023-03-25T17:58:00Z

Noah:

{{LntAppletLinkEnDe|sampling_en|sampling}}

== Applet Description==
 
The applet deals with the system components  "sampling"  and  "signal reconstruction", two components that are of great importance for understanding the  [[Modulation_Methods/Pulscodemodulation|Pulscodemodulation]]  $({\rm PCM})$  for example.   The upper graphic shows the model on which this applet is based.  Below it are the samples  $x(\nu \cdot T_{\rm A})$  of the time continuous signal  $x(t)$. The (infinite) sum over all these samples is called the sampled signal  $x_{\rm A}(t)$.

[[File:EN_Abtastung_1.png|center|frame|Top:    Underlying model for sampling and signal reconstruction Bottom:   Example for time discretization of the continuous–time signal  $x(t)$]]

*At the transmitter, the time discrete (sampled) signal  $x_{\rm A}(t)$  is obtained from the continuous–time signal  $x(t)$.  This process is called  '''sampling'''   or  '''A/D conversion'''.
*The corresponding program parameter for the transmitter is the sampling rate  $f_{\rm A}= 1/T_{\rm A}$.  The lower graphic shows the sampling distance  $T_{\rm A}$ .
*In the receiver, the discrete-time received signal  $y_{\rm A}(t)$  is used to generate the continuous-time sink signal  $y(t)$    ⇒   '''signal reconstruction'''  or  '''D/A conversion'''  corresponding to the receiver frequency response  $H_{\rm E}(f)$.

The applet does not consider the PCM blocks  "Quantization"and  "encoding/decoding".   The digital transmission channel is assumed to be ideal. 

[[File:Abtastung_2_neu.png|right|frame|Receiver frequency response  $H_{\rm E}(f)$]]

The following consequences result from this:
*In the program simplifying  $y_{\rm A}(t) = x_{\rm A}(t)$  is set.
* With suitable system parameters, the error signal   $\varepsilon(t) = y(t)-x(t)\equiv 0$  is therefore also possible.

The sampling theorem and the signal reconstruction can be better explained in the frequency domain.  Therefore all spectral functions are displayed in the program;

             $X(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x(t)$,  $X_{\rm A}(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ x_{\rm A}(t)$,  $Y(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ y(t)$,  $E(f)\ \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,\ \varepsilon(t).$ 

Parameters for the receiver frequency response  $H_{\rm E}(f)$  are the cut–off frequency and the rolloff factor  (see lower graph):
:$$f_{\rm G} = \frac{f_2 +f_1}{2},\hspace{1cm}r = \frac{f_2 -f_1}{f_2 +f_1}.$$

''Notes:''

'''(1)'''   All signal values are normalized to  $\pm 1$.

'''(2)'''   The power calculation is done by integration over the respective period duration  $T_0$:
:$$P_x = \frac{1}{T_0} \cdot \int_0^{T_0} x^2(t)\ {\rm d}t,\hspace{0.8cm}P_\varepsilon = \frac{1}{T_0} \cdot \int_0^{T_0} \varepsilon^2(t).$$

'''(3)'''   The signal power  $P_x$  and the distortion power  $P_\varepsilon$  are also output in normalized form, which implicitly assumes the reference resistance  $R = 1\, \rm \Omega$ ;

'''(4)'''   From these the signal–distortion–distance  $10 \cdot \lg \ (P_x/P_\varepsilon)$  can be calculated.

'''(5)'''   Does the spectral function  $X(f)$  for positive frequencies consists of  $I$  Diraclines with the (possibly complex) weights  $X_1$, ... , $X_I$,           so applies to the transmission power taking into account the mirror-image lines at the negative frequencies:

:$$P_x = 2 \cdot \sum_{i=1}^I |X_k|^2.$$

'''(6)'''   Correspondingly, the following applies to the distortion power if the spectral function  $E(f)$  in the range  $f>0$  has  $J$  Diraclines with weights  $E_1$, ... , $E_J$:

:$$P_\varepsilon = 2 \cdot \sum_{j=1}^J |E_j|^2.$$

==Theoretical Background==

===Description of sampling in the time domain===

[[File:P_ID1120__Sig_T_5_1_S1_neu.png|center|frame|For the time discretization of the continuous-time signal  $x(t)$]]

In the following, we use the following nomenclature to describe the sampling:
*let the continuous-time signal be  $x(t)$.
*Let the time-discretized signal sampled at equidistant intervals  $T_{\rm A}$  be  $x_{\rm A}(t)$.
*Out of the sampling time points  $\nu \cdot T_{\rm A}$  always holds  $x_{\rm A}(t) \equiv 0$.
*The run variable  $\nu$  be an  [[Signal_Representation/Calculating_with_Complex_Numbers#The_set_of_real_numbers|"integer"]]:     $\nu \in \mathbb{Z} = \{\hspace{0.05cm} \text{...}\hspace{0.05cm} , –3, –2, –1, \hspace{0.2cm}0, +1, +2, +3, \text{...} \hspace{0.05cm}\} $.
*In contrast, at the equidistant sampling times with the constant  $K$, the result is:

:$$x_{\rm A}(\nu \cdot T_{\rm A}) = K \cdot x(\nu \cdot T_{\rm A})\hspace{0.05cm}.$$

The constant depends on the type of time discretization. For the above sketch $K = 1$ is valid.
 
===Description of sampling with the Dirac delta pulse===

In the following, we assume a slightly different form of description.  The following pages will show that these equations, which take some getting used to, do lead to useful results if they are applied consistently.

{{BlaueBox|TEXT=
$\text{Definitions:}$ 

* By  '''sampling'''  we mean here the multiplication of the time-continuous signal  $x(t)$  by a  '''Dirac delta pulse''':

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.05cm}.$$

*The  '''Dirac delta pulse (in the time domain)'''  consists of infinitely many Dirac delta pulses, each equally spaced  $T_{\rm A}$  and all with equal momentum weight  $T_{\rm A}$:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$}}

Based on this definition, the following properties result for the sampled signal:
:$$x_{\rm A}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot x(\nu \cdot T_{\rm A})\cdot
\delta (t- \nu \cdot T_{\rm A}
)\hspace{0.05cm}.$$

*The sampled signal at the considered time  $(\nu \cdot T_{\rm A})$  ist gleich  $T_{\rm A} \cdot x(\nu \cdot T_{\rm A}) · \delta (0)$.
*Since  $\delta (t)$  at time  $t = 0$  is infinite, actually all signal values  $x_{\rm A}(\nu \cdot T_{\rm A})$  are also infinite and also the factor  $K$ introduced above.
*Two samples  $x_{\rm A}(\nu_1 \cdot T_{\rm A})$  and  $x_{\rm A}(\nu_2 \cdot T_{\rm A})$  however, differ in the same proportion as the signal values  $x(\nu_1 \cdot T_{\rm A})$  and  $x(\nu_2 \cdot T_{\rm A})$.
*The samples of  $x(t)$  appear in the momentum weights of the Dirac functions:
*The additional multiplication by  $T_{\rm A}$  is necessary so that  $x(t)$  and  $x_{\rm A}(t)$  have the same unit.  Note here that  $\delta (t)$  itself has the unit "1/s".

===Description of sampling in the frequency domain===

Zum Spektrum des abgetasteten Signals  $x_{\rm A}(t)$  kommt man durch Anwendung des  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_frequency_domain|"Faltungssatzes"]]. Dieser besagt, dass der Multiplikation im Zeitbereich die Faltung im Spektralbereich entspricht:

:$$x_{\rm A}(t) = x(t) \cdot p_{\delta}(t)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
X_{\rm A}(f) = X(f) \star P_{\delta}(f)\hspace{0.05cm}.$$

Entwickelt man den  Diracpuls  $p_{\delta}(t)$   (im Zeitbereich)   in eine  [[Signal_Representation/Fourier_Series|"Fourierreihe"]]  und transformiert diese unter Anwendung des  [[Signal_Representation/The_Fourier_Transform_Theorems#Shifting_Theorem|"Verschiebungssatzes"]]  in den Frequenzbereich, so ergibt sich mit dem Abstand  $f_{\rm A} = 1/T_{\rm A}$  zweier benachbarter Diraclinien im Frequenzbereich folgende Korrespondenz   ⇒   [[Signal_Representation/Discrete-Time_Signal_Representation#Dirac_comb_in_time_and_frequency_domain|"Beweis"]]:

:$$p_{\delta}(t) = \sum_{\nu = - \infty }^{+\infty} T_{\rm A} \cdot
\delta(t- \nu \cdot T_{\rm A}
)\hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm} P_{\delta}(f) = \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A} ).$$

[[File:P_ID1121__Sig_T_5_1_S3_NEU.png|right|frame|Diracpuls im Zeit- und Frequenzbereich mit  $T_{\rm A} = 50\ {\rm µs}$  und  $f_{\rm A} = 1/T_{\rm A} = 20\ \text{kHz}$]]
Das Ergebnis besagt:
*Der Diracpuls  $p_{\delta}(t)$  im Zeitbereich besteht aus unendlich vielen Diracimpulsen, jeweils im gleichen Abstand  $T_{\rm A}$  und alle mit gleichem Impulsgewicht  $T_{\rm A}$.
*Die Fouriertransformierte von  $p_{\delta}(t)$  ergibt wiederum einen Diracpuls, aber nun im Frequenzbereich   ⇒   $P_{\delta}(f)$.
*Auch  $P_{\delta}(f)$  besteht aus unendlich vielen Diracimpulsen, nun im jeweiligen Abstand  $f_{\rm A} = 1/T_{\rm A}$  und alle mit dem Impulsgewicht  $1$.
*Die Abstände der Diraclinien in Zeit– und Frequenzbereich folgen demnach dem  [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reziprozitätsgesetz"]]:   $T_{\rm A} \cdot f_{\rm A} = 1 \hspace{0.05cm}.$

Daraus folgt:   Aus dem Spektrum  $X(f)$  wird durch Faltung mit der um  $\mu \cdot f_{\rm A}$  verschobenen Diraclinie:

:$$X(f) \star \delta
(f- \mu \cdot f_{\rm A}
)= X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

Wendet man dieses Ergebnis auf alle Diraclinien des Diracpulses an, so erhält man schließlich:

:$$X_{\rm A}(f) = X(f) \star \sum_{\mu = - \infty }^{+\infty} \delta
(f- \mu \cdot f_{\rm A}
) = \sum_{\mu = - \infty }^{+\infty} X (f- \mu \cdot f_{\rm A}
)\hspace{0.05cm}.$$

{{BlaueBox|TEXT=
$\text{Fazit:}$  Die Abtastung des analogen Zeitsignals  $x(t)$  in äquidistanten Abständen  $T_{\rm A}$  führt im Spektralbereich zu einer  '''periodischen Fortsetzung'''  von  $X(f)$  mit dem Frequenzabstand  $f_{\rm A} = 1/T_{\rm A}$.}}

[[File:P_ID1122__Sig_T_5_1_S4_neu.png|right|frame|Spektrum des abgetasteten Signals]]
{{GraueBox|TEXT=
$\text{Beispiel 1:}$ 
Die obere Grafik zeigt  '''(schematisch!)'''  das Spektrum  $X(f)$  eines Analogsignals  $x(t)$, das Frequenzen bis  $5 \text{ kHz}$  beinhaltet.

Tastet man das Signal mit der Abtastrate  $f_{\rm A}\,\text{ = 20 kHz}$, also im jeweiligen Abstand  $T_{\rm A}\, = {\rm 50 \, µs}$  ab, so erhält man das unten skizzierte periodische Spektrum  $X_{\rm A}(f)$.
*Da die Diracfunktionen unendlich schmal sind, beinhaltet das abgetastete Signal  $x_{\rm A}(t)$  auch beliebig hochfrequente Anteile.
*Dementsprechend ist die Spektralfunktion  $X_{\rm A}(f)$  des abgetasteten Signals bis ins Unendliche ausgedehnt.}}

===Signal reconstruction===

[[File:P_ID1123__Sig_T_5_1_S5a_neu.png|right|frame|Gemeinsames Modell von "Signalabtastung" und "Signalrekonstruktion"]]
Die Signalabtastung ist bei einem digitalen Übertragungssystem kein Selbstzweck, sondern sie muss irgendwann wieder rückgängig gemacht werden.  Betrachten wir zum Beispiel das folgende System:
*Das Analogsignal  $x(t)$  mit der Bandbreite  $B_{\rm NF}$  wird wie oben beschrieben abgetastet.
*Am Ausgang eines idealen Übertragungssystems liegt das ebenfalls zeitdiskrete Signal  $y_{\rm A}(t) = x_{\rm A}(t)$  vor.
*Die Frage ist nun, wie der Block   '''Signalrekonstruktion'''   zu gestalten ist, damit auch  $y(t) = x(t)$  gilt.

[[File:P_ID1124__Sig_T_5_1_S5b_neu.png|right|frame|Frequenzbereichsdarstellung der "Signalrekonstruktion"]]
 Die Lösung ist einfach, wenn man die Spektralfunktionen betrachtet:  

Man erhält aus  $Y_{\rm A}(f)$  das Spektrum  $Y(f) = X(f)$  durch ein Tiefpass Filter mit dem  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function|"Frequenzgang"]]  $H_{\rm E}(f)$, der 

*die tiefen Frequenzen unverfälscht durchlässt:
:$$H_{\rm E}(f) = 1 \hspace{0.3cm}{\rm{f\ddot{u}r}} \hspace{0.3cm} |f| \le B_{\rm
NF}\hspace{0.05cm},$$
*die hohen Frequenzen vollständig unterdrückt:
:$$H_{\rm E}(f) = 0 \hspace{0.3cm}{\rm{f\ddot{u}r}} \hspace{0.3cm} |f| \ge f_{\rm A} - B_{\rm
NF}\hspace{0.05cm}.$$

Weiter ist aus der nebenstehenden Grafik zu erkennen:   Solange die beiden oben genannten Bedingungen erfüllt sind, kann  $H_{\rm E}(f)$  im Bereich von  $B_{\rm NF}$  bis  $f_{\rm A}–B_{\rm NF}$  beliebig geformt sein kann,
*beispielsweise linear abfallend (gestrichelter Verlauf)
*oder auch rechteckförmig,

===The Sampling Theorem===

Die vollständige Rekonstruktion des Analogsignals  $y(t)$  aus dem abgetasteten Signal  $y_{\rm A}(t) = x_{\rm A}(t)$  ist nur möglich, wenn die Abtastrate  $f_{\rm A}$  entsprechend der Bandbreite  $B_{\rm NF}$  des Nachrichtensignals richtig gewählt wurde.

Aus der obigen Grafik erkennt man, dass folgende Bedingung erfüllt sein muss:   $f_{\rm A} - B_{\rm NF} > B_{\rm NF} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A} > 2 \cdot B_{\rm NF}\hspace{0.05cm}.$

{{BlaueBox|TEXT=
$\text{Abtasttheorem:}$  Besitzt ein Analogsignal  $x(t)$  nur Spektralanteile im Bereich  $\vert f \vert < B_{\rm NF}$, so kann dieses aus seinem abgetasteten Signal  $x_{\rm A}(t)$  nur dann vollständig rekonstruiert werden, wenn die Abtastrate hinreichend groß ist:
:$$f_{\rm A} ≥ 2 \cdot B_{\rm NF}.$$

Für den Abstand zweier Abtastwerte muss demnach gelten:

:$$T_{\rm A} \le \frac{1}{ 2 \cdot B_{\rm NF} }\hspace{0.05cm}.$$}}

Wird bei der Abtastung der größtmögliche Wert   ⇒   $T_{\rm A} = 1/(2B_{\rm NF})$  herangezogen,
*so muss zur Signalrekonstruktion des Analogsignals aus seinen Abtastwerten
*ein idealer, rechteckförmiger Tiefpass mit der Grenzfrequenz  $f_{\rm G} = f_{\rm A}/2 = 1/(2T_{\rm A})$  verwendet werden.

{{GraueBox|TEXT=
$\text{Beispiel 2:}$  Die Grafik zeigt oben das auf  $\pm\text{ 5 kHz}$  begrenzte Spektrum  $X(f)$  eines Analogsignals, unten das Spektrum  $X_{\rm A}(f)$  des im Abstand  $T_{\rm A} =\,\text{ 100 µs}$  abgetasteten Signals   ⇒   $f_{\rm A}=\,\text{ 10 kHz}$.
[[File:P_ID1125__Sig_T_5_1_S6_neu.png|right|frame|Abtasttheorem im Frequenzbereich]]
Zusätzlich eingezeichnet ist der Frequenzgang  $H_{\rm E}(f)$  des tiefpassartigen Empfangsfilters zur Signalrekonstruktion, dessen Grenzfrequenz exakt  $f_{\rm G} = f_{\rm A}/2 = 5\,\text{ kHz}$  betragen muss.

*Mit jedem anderen  $f_{\rm G}$–Wert ergäbe sich  $Y(f) \neq X(f)$.
*Bei  $f_{\rm G} < 5\,\text{ kHz}$  fehlen die oberen  $X(f)$–Anteile.
* Bei  $f_{\rm G} > 5\,\text{ kHz}$  kommt es aufgrund von Faltungsprodukten zu unerwünschten Spektralanteilen in  $Y(f)$.

Wäre am Sender die Abtastung mit einer Abtastrate  $f_{\rm A} < 10\ \text{ kHz}$  erfolgt   ⇒   $T_{\rm A} >100 \ {\rm µ s}$, so wäre das Analogsignal  $y(t) = x(t)$  aus den Abtastwerten  $y_{\rm A}(t)$  auf keinen Fall rekonstruierbar.}}
 
==Exercises==

*First, select the number  $(1,\ 2, \text{...} \ )$  of the task to be processed.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show Solution".
*All signal values are to be understood as normalized to  $\pm 1$.  Powers are normalized values, too.

{{BlaueBox|TEXT=
'''(1)'''  Source signal:  $x(t) = A \cdot \cos (2\pi \cdot f_0 \cdot t -\varphi)$  with  $f_0 = \text{4 kHz}$.   Sampling with  $f_{\rm A} = \text{10 kHz}$.  Rectanglular low-pass;  cut-off frequency:  $f_{\rm G} = \text{5 kHz}$.             Interpret the shown graphics and evaluate the present signal reconstruction for all permitted parameter values of $A$  and $\varphi$. }}

* The spectrum  $X(f)$  consists of two Dirac functions at  $\pm \text{4 kHz}$, each with impulse weight  $0.5$.
* By the periodic continuation  $X_{\rm A}(f)$  has lines of equal height at  $\pm \text{4 kHz}$,  $\pm \text{6 kHz}$,  $\pm \text{14 kHz}$,  $\pm \text{16 kHz}$,  $\pm \text{24 kHz}$,  $\pm \text{26 kHz}$,  etc.
* The rectangular low-pass with the cut-off frequency  $f_{\rm G} = \text{5 kHz}$  removes all lines except the two at  $\pm \text{4 kHz}$  ⇒  $Y(f) =X(f)$  ⇒  $y(t) =x(t)$  ⇒   $P_\varepsilon = 0$.
* The signal reconstruction works here perfectly  $(P_\varepsilon = 0)$  for all amplitudes $A$  and any phases $\varphi$.

{{BlaueBox|TEXT=
'''(2)'''  Continue with  $A=1$,  $f_0 = \text{4 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$.   What is the influence of the rolloff–factors  $r=0.2$,  $r=0.5$  and   $r=1$?           Specify the power values  $P_x$  and  $P_\varepsilon$ .   For which  $r$–values is  $P_\varepsilon= 0$?  Do these results also apply to other  $A$  and  $\varphi$? }}

:* With  $|X(f = \pm \text{4 kHz})|=0.5$  the signal power is  $P_x = 2\cdot 0.5^2 = 0.5$.  The distortion power  $P_\varepsilon$  depends significantly on the rolloff–factor  $r$ .
:* $P_\varepsilon$  is zero for  $r \le 0.2$.  Then the  $X_{\rm A}(f)$ line at  $f_0 = \text{4 kHz}$  is not changed by the low-pass and the unwanted  line at  $\text{6 kHz}$  is fully suppressed.
:* $r = 0.5$ :  $Y(f = \text{4 kHz}) = 0.35$,  $Y(f = \text{6 kHz}) = 0.15$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 15$  ⇒  $P_\varepsilon = 0.09$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=7.45\ \rm dB$.
:*$r = 1.0$ :  $Y(f = \text{4 kHz}) = 0.3$,  $Y(f = \text{6 kHz}) = 0.2$  ⇒   $|E(f = \text{4 kHz})| = |E(f = \text{6 kHz})|= 0. 2$  ⇒  $P_\varepsilon = 0.16$  ⇒  $10 \cdot \lg \ (P_x/P_\varepsilon)=4.95\ \rm dB$.
:* For all  $r$  the distortion power $P_\varepsilon$  is independent of  $\varphi$.   The amplitude  $A$  affects  $P_x$  and  $P_\varepsilon$  in the same way   ⇒   the quotient is independent of  $A$.

{{BlaueBox|TEXT=
'''(3)'''  Now apply  $A=1$,  $f_0 = \text{5 kHz}$,  $\varphi=0$,  $f_{\rm A} = \text{10 kHz}$,  $f_{\rm G} = \text{5 kHz}$,  $r=0$  $($rectangular low–pass$)$.   Interpret the result of the signal reconstruction.}}

:*  $X(f)$  consists of two Dirac delta lines at  $\pm \text{5 kHz}$  $($weight  $0.5)$.  By periodic continuation  $X_{\rm A}(f)$  has lines at  $\pm \text{5 kHz}$,  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$,  etc.
:*  The  rectanglular low-pass  removes the lines at  $\pm \text{15 kHz}$,  $\pm \text{25 kHz}$.  The lines at  $\pm \text{5 kHz}$  are halved because of  $H_{\rm E}(\pm f_{\rm G}) = H_{\rm E}(\pm \text{5 kHz}) = 0.5$.
:*   ⇒   $\text{Weights of }X(f = \pm \text{5 kHz})$:  $0.5$   #   $\text{Weights of }X(f_{\rm A} = \pm \text{5 kHz})$:  $1. 0$;     #   $\text{Weights of }Y(f = \pm \text{5 kHz})$:  $0.5$   ⇒   $Y(f)=X(f)$.
:* So the signal reconstruction works perfectly here too  $(P_\varepsilon = 0)$.  The same is true for the phase  $\varphi=180^\circ$   ⇒   $x(t) = -A \cdot \cos (2\pi \cdot f_0 \cdot t)$.

{{BlaueBox|TEXT=
'''(4)'''  The settings of  $(3)$  continue to apply except for  $\varphi=30^\circ$.  Interpret the differences from the setting  $(3)$   ⇒   $\varphi=0^\circ$.}}

:* Phase relations are lost.  The sink signal  $y(t)$  is cosine-shaped  $(\varphi_y=0^\circ)$  with by the factor  $\cos(\varphi_x)$  smaller amplitude than the source signal  $x(t)$.
:* Justification in the frequency domain:  In the periodic continuation of  $X(f)$  ⇒  $X_{\rm A}(f)$  only the real parts are to be added.  The imaginary parts cancel out.
:* The Dirac delta line of  $X(f)$  at frequency  $f_0$   ⇒   $X(f_0)$  is complex,   $Y(f_0)$  is real, and  $E(f_0)$  is imaginary   ⇒   $\varepsilon(t)$  is minus–sinusoidal   ⇒   $P_\varepsilon = 0. 125$.

{{BlueBox|TEXT=
'''(5)'''  Illustrate again the result of  $(4)$  compared to the settings  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* With this setting, the spectrum  $X_{\rm A}(f)$  also has a positive imaginary part at  $\text{5 kHz}$  and a negative imaginary part of the same magnitude at  $\text{6 kHz}$.
:* The rectangular low-pass with cutoff frequency  $\text{5.5 kHz}$  removes this second component.  Thus, with the new setting  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Any $f_0$ oscillation of arbitrary phase is error-free reconstructible from its samples if  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($any small $\mu>0)$.
:* For value–continuous spectrum with   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$no diraclines at $\pm f_0 \big ]$  the sampling rate  $f_{\rm A} = 2 \cdot f_{\rm 0}$  is sufficient in principle.

{{BlaueBox|TEXT=
'''(5)'''  Verdeutlichen Sie sich nochmals das Ergebnis von  '''(4)'''  im Vergleich zu den Einstellungen  $f_0 = \text{5 kHz}$,  $\varphi=30^\circ$,  $f_{\rm A} = \text{11 kHz}$,  $f_{\rm G} = \text{5.5 kHz}$.}}
:* Bei dieser Einstellung hat das  $X_{\rm A}(f)$–Spektrum auch einen positiven Imaginärteil bei  $\text{5 kHz}$  und einen negativen Imaginärteil gleicher Höhe bei  $\text{6 kHz}$.
:* Der Rechteck–Tiefpass mit der Grenzfrequenz  $\text{5.5 kHz}$  entfernt diesen zweiten Anteil.  Somit ist bei dieser Einstellung  $Y(f) =X(f)$   ⇒   $P_\varepsilon = 0$.
:* Jede  $f_0$–Schwingung beliebiger Phase ist fehlerfrei aus seinen Abtastwerten rekonstruierbar, falls  $f_{\rm A} = 2 \cdot f_{\rm 0} + \mu, \ f_{\rm G}= f_{\rm A}/2$  $($beliebig kleines $\mu>0)$.
:* Bei wertkontinuierlichem Spektrum mit   $X(|f|> f_0) \equiv 0$  ⇒   $\big[$keine Diraclinien bei $\pm f_0 \big ]$ genügt grundsätzlich die Abtastrate  $f_{\rm A} = 2 \cdot f_{\rm 0}$.

{{BlueBox|TEXT=
'''(6)'''  The settings of  $(3)$  and  $(4)$  continue to apply except for  $\varphi=90^\circ$.  Interpret the plots in the time and frequency domain.}}
:* The source signal is sampled exactly at its zero crossings   ⇒   $x_{\rm A}(t) \equiv 0$   ⇒     $y(t) \equiv 0$   ⇒   $\varepsilon(t)=-x(t)$   ⇒   $P_\varepsilon = P_x$   ⇒   $10 \cdot \lg \ (P_x/P_\varepsilon)=0\ \rm dB$.
:* Description in the frequency domain:   As in  $(4)$  the imaginary parts of  $X_{\rm A}(f)$  cancel out.  Also the real parts of  $X_{\rm A}(f)$  are zero because of the sinusoid.

{{BlueBox|TEXT=
'''(7)'''  Now consider the  $\text {Source Signal 2}$.  Let the other parameters be  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2.5 kHz}$,  $r=0$.  Interpret the results.}}
:* The source signal has spectral components up to  $\pm \text{2 kHz}$.  The signal power is $P_x = 2 \cdot \big[0.1^2 + 0.25^2+0.15^2\big]= 0.19 $. 
:* With the sampling rate  $f_{\rm A} = \text{5 kHz}$  and the receiver parameters  $f_{\rm G} = \text{2.5 kHz}$  and  $r=0$, the signal reconstruction works perfectly:  $P_\varepsilon = 0$.
:* Likewise with the trapezoidal low–pass with  $f_{\rm G} = \text{2.5 kHz}$, if for the rolloff factor holds:  $r \le 0.2$.

{{BlueBox|TEXT=
'''(8)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{1.5 kHz}$  of the rectangular low–pass filter is too small?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=-0.3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t -60^\circ)=0. 3 \cdot \cos(2\pi \cdot \text{2 kHz} \cdot t +120^\circ)$  is equal to the (negated) signal component at  $\text{2 kHz}$. 
:* The distortion power is  $P_\varepsilon=2 \cdot 0.15^2= 0.045$  and the signal–to–distortion ratio  $10 \cdot \lg \ (P_x/P_\varepsilon)=10 \cdot \lg \ (0.19/0.045)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(9)'''  What happens if the cutoff frequency  $f_{\rm G} = \text{3.5 kHz}$  of the rectangular low–pass filter is too large?  In particular, interpret the error signal  $\varepsilon(t)=y(t)-x(t)$.}}
:* The error signal  $\varepsilon(t)=0.3 \cdot \cos(2\pi \cdot \text{3 kHz} \cdot t +60^\circ)$  is now equal to the  $\text{3 kHz}$  portion of the sink signal  $y(t)$  not removed by the low-pass filter.
:* Compared to the subtask  $(8)$  the frequency changes from  $\text{2 kHz}$  to  $\text{3 kHz}$  and also the phase relationship.
:* The amplitude of this  $\text{3 kHz}$ error signal is equal to the amplitude of the  $\text{2 kHz}$ portion of  $x(t)$.  Again  $P_\varepsilon= 0.045$,  $10 \cdot \lg \ (P_x/P_\varepsilon)= 6.26\ \rm dB$.

{{BlueBox|TEXT=
'''(10)'''  Finally, we consider the  $\text {source signal 4}$  $($portions until  $\pm \text{4 kHz})$, as well as  $f_{\rm A} = \text{5 kHz}$,  $f_{\rm G} = \text{2. 5 kHz}$,  $0 \le r\le 1$.  Interpretation of results.}}
:* Up to  $r=0.2$  the signal reconstruction works perfectly  $(P_\varepsilon = 0)$.  If one increases  $r$, then  $P_\varepsilon$  increases continuously and  $10 \cdot \lg \ (P_x/P_\varepsilon)$  decreases.
:* With  $r=1$  the signal frequencies  $\text{0.5 kHz}$,  ...,  $\text{4 kHz}$  are attenuated, the more the higher the frequency is,  for example  $H_{\rm E}(f=\text{4 kHz}) = 0.6$.
:* Similarly,  $Y(f)$  also includes components at frequencies  $\text{6 kHz}$,  $\text{7 kHz}$,  $\text{8 kHz}$,  $\text{9 kHz}$  and  $\text{9.5 kHz}$ due to periodic continuation.
:* At the sampling times  $t\hspace{0.05cm}' = n \cdot T_{\rm A}$, the signals  $x(t\hspace{0.05cm}')$  and  $y(t\hspace{0.05cm}')$  agree exactly  ⇒   $\varepsilon(t\hspace{0.05cm}') = 0$.  In between, not  ⇒   small distortion power   $P_\varepsilon = 0.008$.

==Applet Manual==
 
[[File:Anleitung_Auge.png|right|600px]]
    '''(A)'''     Auswahl:   Codierung                    (binär,  quaternär,  AMI–Code,  Duobinärcode)

    '''(B)'''     Auswahl:   Detektionsgrundimpuls                     (nach Gauß–TP,  CRO–Nyquist,  nach Spalt–TP}

    '''(C)'''     Prametereingabe zu  '''(B)'''                    (Grenzfrequenz,  Rolloff–Faktor,  Rechteckdauer)

    '''(D)'''     Steuerung der Augendiagrammdarstellung                    (Start,  Pause/Weiter,  Einzelschritt,  Gesamt,  Reset)

    '''(E)'''     Geschwindigkeit der Augendiagrammdarstellung

    '''(F)'''     Darstellung:  Detektionsgrundimpuls  $g_d(t)$

    '''(G)'''     Darstellung:  Detektionsnutzsignal  $d_{\rm S}(t - \nu \cdot T)$

    '''(H)'''     Darstellung:  Augendiagramm im Bereich  $\pm T$

    '''( I )'''     Numerikausgabe:  $ö_{\rm norm}$  (normierte Augenöffnung)

    '''(J)'''     Prametereingabe  $10 \cdot \lg \ E_{\rm B}/N_0$  für  '''(K)'''

    '''(K)'''     Numerikausgabe:  $\sigma_{\rm norm}$  (normierter Rauscheffektivwert)

    '''(L)'''     Numerikausgabe:  $p_{\rm U}$  (ungünstigste Fehlerwahrscheinlichkeit)

    '''(M)'''     Bereich für die Versuchsdurchführung:   Aufgabenauswahl

    '''(N)'''     Bereich für die Versuchsdurchführung:   Aufgabenstellung

    '''(O)'''     Bereich für die Versuchsdurchführung:   Musterlösung einblenden
 
==About the Authors==
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2008 by [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Studierende#Slim_Lamine_.28Studienarbeit_EI_2006.29|Slim Lamine]]  as part of his diploma thesis with “FlashMX – Actionscript” (Supervisor: [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).

*Last revision and English version 2020/2021 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

The conversion of this applet to HTML 5 was financially supported by  [https://www.ei.tum.de/studium/studienzuschuesse/ Studienzuschüsse]  ("study grants")  of the TUM Faculty EI.  We thank.

==Once again:  Open Applet in new Tab==
{{LntAppletLinkEnDe|sampling_en|sampling}}

Applets:Pulses and Spectra

2023-03-25T17:44:44Z

Noah:

{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

==Applet Description==
 
Time-limited symmetric signals   ⇒   "pulses"  $x(t)$  and the corresponding spectral functions  $X(f)$  are considered, namely

*Gaussian pulse, 
*rectangular pulse,  
*triangular pulse, 
*trapezoidal pulse, 
*raised cosine pulse,
*cosine square pulse.

Further it is to be noted:
* The functions  $x(t)$  resp.  $X(f)$  are shown for up to two parameter sets in one diagram each.
* The red curves and numbers apply to the left parameter set, the blue ones to the right parameter set.
* The abscissas  $t$  (time) and  $f$  (frequency) as well as the ordinates  $x(t)$  (signal values) and  $X(f)$  (spectral values) are normalized.

==Theoretical Background==
 
===Relationship $x(t)\Leftrightarrow X(f)$===
*The relationship between the time function  $x(t)$  and the spectrum  $X(f)$  is given by the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|"first Fourier integral"]] :
:$$X(f)={\rm FT} [x(t)] = \int_{-\infty}^{+\infty}x(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm}
\rm FT\hspace{-0.1cm}: \ Fourier \ transform.$$

*In order to calculate the time function  $x(f)$  from the spectral function  $x(t)$  one needs the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|"second Fourier integral"]]:
:$$x(t)={\rm IFT} [X(f)] = \int_{-\infty}^{+\infty}X(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm}
{\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fourier \ transform.$$

*In all examples we use real and even functions.  Thus:
:$$x(t)=\int_{-\infty}^{+\infty}X(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ X(f)=\int_{-\infty}^{+\infty}x(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$
*$x(t)$  and  $X(f)$  have different units, for example  $x(t)$  in  $\rm V$,  $X(f)$  in  $\rm V/Hz$.
*The relationship between this module and the similarly constructed applet  [[Applets:Frequenzgang_und_Impulsantwort|"Frequency response & Impulse response"]]  is based on the  [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem|"Duality Theorem"]].
*All times are normalized to a time  $T$  and all frequencies are normalized to  $1/T$   ⇒   the spectral values  $X(f)$  still have to be multiplied by the normalization time  $T$ .

{{GraueBox|TEXT=
$\text{Example:}$   If one sets a rectangular pulse with amplitude  $A_1 = 1$  and equivalent pulse duration  $\Delta t_1 = 1$  then  $x_1(t)$  in the range  $-0.5 < t < +0. 5$  equal to one and outside this range equal to zero.  The spectral function  $X_1(f)$  proceeds  $\rm si$–shaped with  $X_1(f= 0) = 1$  and the first zero at  $f=1$.

*If a rectangular pulse with  $A = K = 3 \ \rm V$  and  $\delta t = T = 2 \ \rm ms$  is to be simulated with this setting, then all signal values with  $K = 3 \ \rm V$  and all spectral values with  $K \cdot T = 0. 006 \ \rm V/Hz$  to be multiplied by.
*The maximum spectral value is then  $X(f= 0) = 0.006 \ \rm V/Hz$  and the first zero is at  $f=1/T = 0.5 \ \rm kHz$.}}

===Gaussian Pulse ===

*The time function of the Gaussian pulse with height  $K$  and (equivalent) duration  $\Delta t$  is:
:$$x(t)=K\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/\Delta t)^2}.$$
*The equivalent time duration  $\Delta t$  is obtained from the rectangle of equal area.
*The value at  $t = \Delta t/2$  is smaller than the value at  $t=0$ by the factor  $0.456$ .
*For the spectral function we get according to the Fourier transform:
:$$X(f)=K\cdot \Delta t \cdot {\rm e}^{-\pi(f\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t)^2} .$$
*The smaller the equivalent time duration  $\Delta t$  is, the wider and lower the spectrum   ⇒   [[Signal_Representation/The_Fourier_Transform_Theorems#Reciprocity_Theorem_of_time_duration_and_bandwidth|"Reciprocity law of bandwidth and pulse duration"]].
*Both  $x(t)$  and  $X(f)$  are not exactly zero at any  $f$–  or  $t$–value, respectively.
*For practical applications, however, the Gaussian pulse can be assumed to be limited in time and frequency.  For example,  $x(t)$  has already dropped to less than  $0.1\% $  of the maximum at  $t=1.5 \delta t$  .

===Rectangular Pulse ===
*The time function of the rectangular pulse with height  $K$  and (equivalent) duration  $\Delta t$  is:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < T/2,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| = T/2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| > T/2.} \\ \end{array}$$

*The $\pm \Delta t/2$ value lies midway between the left- and right-hand limits.
*For the spectral function one obtains according to the laws of the Fourier transform (1st Fourier integral):
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f) \quad \text{with} \ {\rm si}(x)=\frac{\sin(x)}{x}.$$
*The spectral value at  $f=0$  is equal to the rectangular area of the time function.
*The spectral function has zeros at equidistant distances  $1/\delta t$.
*The integral over the spectral function  $X(f)$  is equal to the signal value at time  $t=0$, i.e. the pulse height  $K$.

===Triangular Pulse===
*The time function of the triangular pulse with height  $K$  and (equivalent) duration  $\Delta t$  is:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot (1-|t|/{\Delta t}) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*The absolute time duration is  $2 \cdot \Delta t$;  this is twice as large as that of the rectangle.
*For the spectral function, we obtain according to the Fourier transform:
:$$X(f)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta t \cdot f) \quad \text{with} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*The above time function is equal to the convolution of two rectangular pulses, each with width  $\delta t$.
*From this follows:  $X(f)$  contains instead of the  ${\rm si}$-function the  ${\rm si}^2$-function.
*$X(f)$  thus also has zeros at equidistant intervals  $1/\rm f$ .
*The asymptotic decay of  $X(f)$  occurs here with  $1/f^2$, while for comparison the rectangular momentum decays with  $1/f$ .

===Trapezoidal Pulse ===
The time function of the trapezoidal pulse with height  $K$  and time parameters  $t_1$  and  $t_2$  is:
:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{t_2-|t|}{t_2-t_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}}\quad \\ {\rm{for}}\quad \\ {\rm{for}} \quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*For the equivalent pulse duration (rectangle of equal area) holds:   $\Delta t = t_1+t_2$.
*The rolloff factor (in the time domain) characterizes the slope:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*The special case  $r=0$  corresponds to the rectangular pulse and the special case  $r=1$  to the triangular pulse.
*For the spectral function one obtains according to the Fourier transform:
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f)\cdot {\rm si}(\pi \cdot r \cdot \Delta t \cdot f) \quad \text{with} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*The asymptotic decay of  $X(f)$  lies between  $1/f$  $($for rectangle,  $r=0)$  and  $1/f^2$  $($for triangle,  $r=1)$.

===Raised cosine Pulse ===
The time function of the raised cosine pulse with height  $K$  and time parameters  $t_1$  and  $t_2$  is:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|t|-t_1}{t_2-t_1}\cdot {\pi}/{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}}\quad \\ {\rm{for}}\quad \\ {\rm{for}}\quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*For the equivalent pulse duration (rectangle of equal area) holds:   $\Delta t = t_1+t_2$.
*The rolloff factor (in the time domain) characterizes the slope:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*The special case  $r=0$  corresponds to the square pulse and the special case  $r=1$  to the cosine square pulse.
*For the spectral function one obtains according to the Fourier transform:
:$$X(f)=K\cdot \Delta t \cdot \frac{\cos(\pi \cdot r\cdot \Delta t \cdot f)}{1-(2\cdot r\cdot \Delta t \cdot f)^2} \cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*The larger the rolloff factor  $r$  is, the faster  $X(f)$  decreases asymptotically with  $f$ .

===Cosine square Pulse ===
*This is a special case of the raised cosine pulse and results for  $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}t_1=0, \ t_2= \Delta t$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|t|\hspace{0.05cm}\cdot \hspace{0.05cm} \pi}{2\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*For the spectral function, we obtain according to the Fourier transform:
:$$X(f)=K\cdot \Delta f \cdot \frac{\pi}{4}\cdot \big [{\rm si}(\pi(\Delta t\cdot f +0.5))+{\rm si}(\pi(\Delta t\cdot f -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Because of the last  ${\rm si}$-function is  $X(f)=0$  for all multiples of  $F=1/\delta t$.  The equidistant zero crossings of the raised cosine pulse are preserved.
*Because of the bracket expression,  $X(f)$  now exhibits further zero crossings at  $f=\pm1.5 F$,  $\pm2.5 F$,  $\pm3.5 F$, ... .
*For frequency  $f=\pm F/2$  the spectral values  $K\cdot \Delta t/2$ are obtained.
*The asymptotic decay of  $X(f)$  runs in this special case with  $1/f^3$.

==Exercises==
 

* First select the number  $(1,\text{...}, 7)$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
* "Red" refers to the first parameter set ⇒ $x_1(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_1(f)$,  "Blue" refers to the second parameter set ⇒ $x_2(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_2(f)$.
*Values with magnitude less than  $0.0005$  are output in the program as "zero".

{{BlaueBox|TEXT=
'''(1)'''   Compare the  red Gaussian pulse  $(A_1 = 1, \Delta t_1 = 1)$  with the  blue rectangular pulse  $(A_2 = 1, \Delta t_2 = 1)$ ⇒ default setting.
           What are the differences in the time and frequency domain?}}

* The Gaussian pulse theoretically reaches infinity in the time as well as in the frequency domain. 
* Practically  $x_1(t)$  for  $|t| > 1.5$  and  $X_1(f)$  for  $|f| > 1.5$  are almost zero. 
* The rectangle is strictly limited in time:  $x_2(|t| > 0.5) \equiv 0$.  $X_2(f)$  has shares in a much larger range than  $X_1(f)$. 
* It holds  $X_1(f = 0) = X_2(f = 0)$  since the integral over the Gaussian pulse  $x_1(t)$  is equal to the integral over the rectangular pulse  $x_2(t)$.

{{BlaueBox|TEXT=
'''(2)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2)$.           Vary the equivalent pulse duration  $\Delta t_2$  between  $0.5$  and  $2$.  Interpret the displayed graphs.}}

* One can recognize the reciprocity law of bandwidth and pulse duration.  The greater  $\Delta t_2$, the higher and narrower the spectral function  $X_2(f)$. 
* For each setting of  $\Delta t_2$,  $x_1(t=0)$  and  $x_2(t=0)$  are equal   ⇒   Also, the integrals over  $X_1(f)$  and  $X_2(f)$  are identical.

{{BlaueBox|TEXT=
'''(3)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2 = 0.5)$.           Vary  $\Delta t_2$  between  $0.05$  and  $2$.  Interpret the displayed graphs and extrapolate the result.}}

* The blue spectrum is now twice as wide as the red one, but only half as high.  First zero of  $X_1(f)$  at  $f = 1$, of  $X_2(f)$  at  $f = 2$. 
* Reduction of  $\Delta t_2$:  $X_2(f)$  lower and wider.  Very flat course at  $\Delta t_2 = 0.05$:  $X_2(f = 0)= 0.05$,  $X_2(f = \pm 3)= 0.048$. 
* If one choose  $\Delta t_2 = \varepsilon \to 0$  (not possible in the program),  the result would be the almost constant, very small spectrum  $X_2(f)=A \cdot \varepsilon \to 0$. 
* Increasing the amplitude to  $A=1/\varepsilon$  results in the constant spectral function  $X_2(f) = 1$  of the Dirac function  $\delta(t)$.  That means: 
* $\delta(t)$  is approximated by a rectangle  $($width  $\Delta t = \varepsilon \to 0$,  height  $A = 1/\varepsilon \to \infty)$.  The weight of the Dirac function is one:  $x(t) = 1 \cdot \delta (t)$.

{{BlaueBox|TEXT=
'''(4)'''   Compare the  rectangular pulse  $(A_1 = 1,   \Delta t_1 = 1)$  with the  triangular pulse  $(A_2 = 1,   \Delta t_2 = 1)$.  Interpret the spectral functions.}}

* The (normalized) spectrum of the rectangle  $x_1(t)$  with the (normalized) parameters  $A_1 = 1, \ \ \Delta t_1 = 1$  is:  $X_1(f)= {\rm si}(\pi\cdot f)= {\rm sinc}(f)$. 
* The convolution of the rectangle  $x_1(t)$  with itself gives the triangle  $x_2(t) = x_1(t) \star x_1(t)$.  By the convolution theorem:   $X_2(f) = X_1(f)^2 $. 
* By squaring the  ${\rm sinc}(f)$–shaped spectral function  $X_1(f)$  the zeros of  $X_2(f)$  remain unchanged.  But now it holds that: $X_2(f) \ge 0$.

{{BlaueBox|TEXT=
'''(5)'''   Compare the  trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, r_1 = 0.5)$  with the 
triangular pulse $(A_2 = 1,   \Delta t_2 = 1)$.          Vary  $r_1$  between  $0$  and  $1$.  Interpret the spectral function  $X_1(f)$.}}

* The trapezoidal pulse with roll–off factor  $r_1= 0$  is identical to the rectangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}(f)$. 
* The trapezoidal pulse with roll–off factor  $r_1= 1$  is identical to the triangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}^2(f)$. 
* In both cases  $X_1(f)$  has equidistant zeros at  $\pm 1$,  $\pm 2$, ...  (none else);   $0 < r_1 < 1$:  depending on  $r_1$  further zeros.

{{BlaueBox|TEXT=
'''(6)'''   Compare this  trapezoidal pulse  with the cosine roll-off pulse 
$(A_2 = 1,\ \Delta t_2 = 1.0,\ r_2 = 0.5)$.          Vary  $r_2$  between  $0$  and  $1$.  Interpret the spectral function  $X_2(f)$  for  $r_2 = 0.7$.}}

* With the same  $r= 0.5$  the cosine roll-off pulse  $X_2(f)$ is for  $f > 1$  greater in magnitude than the trapezoidal pulse. 
* With the same roll-off factor  $(r_1 = r_2= 0.5)$  the drop of  $X_2(f)$  around the frequency  $f = 0.5$  is steeper than the drop of  $X_1(f)$. 
* With  $r_1 = 0.5$  and  $r_2 = 0.7$  $x_1(t) \approx x_2(t)$  is valid and therefore also  $X_1(f) \approx X_2(f)$.  Comparable edge steepness.

{{BlaueBox|TEXT=
'''(7)'''   Compare the  red trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, \ r_1 = 1)$  with the  blue cosine roll-off pulse  $(A_2 = 1,\ \Delta t_2 = 1.0, \ r_2 = 1)$.           Interpret the time function  $x_2(t)$  and the spectral function  $X_2(f)$  system theoretically.}}

* $x_2(t) = \cos^2(|t|\cdot \pi/2) \ \ \ \text{for} \ |t| \le 1$  is the  cosine square pulse.  Zeros at  $f = \pm 1$,  $\pm 2$, ... 
* For the frequency  $f=\pm 0.5$  one obtains the spectral values  $X_2(f)=0.5$.  The asymptotic decline is shown here with  $1/f^3$.

==Applet Manual==
 
[[File:Exercise_impuls.png |right|frame|Screenshot]]
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Preselection for pulse shape  $x_1(t)$  (red curve)

    '''(C)'''     Parameter definition for  $x_1(t)$ 

    '''(D)'''     Numeric output for  $x_1(t_*)$  and  $X_1(f_*)$

    '''(E)'''     Preselection for pulse shape  $x_2(t)$  (blue curve)

    '''(F)'''     Parameter definition for  $x_2(t)$ 

    '''(G)'''     Numeric output for  $x_2(t_*)$  and  $X_2(f_*)$

    '''(H)'''     Setting the time  $t_*$  for the numeric output

    '''(I)'''     Setting the frequency  $f_*$  for the numeric output

    '''(J)'''     Graphic field for the time domain

    '''(K)'''     Graphic field for the frequency domain

    '''(L)'''     Selection of the exercise according to the numbers

    '''(M)'''     Task description and questions

    '''(N)'''     Show and hide sample solution

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2005 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2017 the program was redesigned by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]]  (Ingenieurspraxis_Math, Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2020 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

Applets:Pulses and Spectra

2023-03-25T16:48:54Z

Noah:

{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

==Applet Description==
 
Time-limited symmetric signals   ⇒   "pulses"  $x(t)$  and the corresponding spectral functions  $X(f)$  are considered, namely

*Gaussian pulse, 
*rectangular pulse,  
*triangular pulse, 
*trapezoidal pulse, 
*raised cosine pulse,
*cosine square pulse.

Further it is to be noted:
* The functions  $x(t)$  resp.  $X(f)$  are shown for up to two parameter sets in one diagram each.
* The red curves and numbers apply to the left parameter set, the blue ones to the right parameter set.
* The abscissas  $t$  (time) and  $f$  (frequency) as well as the ordinates  $x(t)$  (signal values) and  $X(f)$  (spectral values) are normalized.

==Theoretical Background==
 
===Relationship $x(t)\Leftrightarrow X(f)$===
*The relationship between the time function  $x(t)$  and the spectrum  $X(f)$  is given by the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|"first Fourier integral"]] :
:$$X(f)={\rm FT} [x(t)] = \int_{-\infty}^{+\infty}x(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm}
\rm FT\hspace{-0.1cm}: \ Fourier \ transform.$$

*In order to calculate the time function  $x(f)$  from the spectral function  $x(t)$  one needs the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|"second Fourier integral"]]:
:$$x(t)={\rm IFT} [X(f)] = \int_{-\infty}^{+\infty}X(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm}
{\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fourier \ transform.$$

*In all examples we use real and even functions.  Thus:
:$$x(t)=\int_{-\infty}^{+\infty}X(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ X(f)=\int_{-\infty}^{+\infty}x(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$
*$x(t)$  and  $X(f)$  have different units, for example  $x(t)$  in  $\rm V$,  $X(f)$  in  $\rm V/Hz$.
*The relationship between this module and the similarly constructed applet  [[Applets:Frequenzgang_und_Impulsantwort|"Frequency response & Impulse response"]]  is based on the  [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem|"Duality Theorem"]].
*All times are normalized to a time  $T$  and all frequencies are normalized to  $1/T$   ⇒   the spectral values  $X(f)$  still have to be multiplied by the normalization time  $T$ .

{{GraueBox|TEXT=
$\text{Example:}$   If one sets a rectangular pulse with amplitude  $A_1 = 1$  and equivalent pulse duration  $\Delta t_1 = 1$  then  $x_1(t)$  in the range  $-0.5 < t < +0. 5$  equal to one and outside this range equal to zero.  The spectral function  $X_1(f)$  proceeds  $\rm si$–shaped with  $X_1(f= 0) = 1$  and the first zero at  $f=1$.

*If a rectangular pulse with  $A = K = 3 \ \rm V$  and  $\delta t = T = 2 \ \rm ms$  is to be simulated with this setting, then all signal values with  $K = 3 \ \rm V$  and all spectral values with  $K \cdot T = 0. 006 \ \rm V/Hz$  to be multiplied by.
*The maximum spectral value is then  $X(f= 0) = 0.006 \ \rm V/Hz$  and the first zero is at  $f=1/T = 0.5 \ \rm kHz$.}}

===Gaussian Pulse ===

*Die Zeitfunktion des Gaußimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:
:$$x(t)=K\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/\Delta t)^2}.$$
*Die äquivalente Zeitdauer  $\Delta t$  ergibt sich aus dem flächengleichen Rechteck.
*Der Wert bei  $t = \Delta t/2$  ist um den Faktor  $0.456$  kleiner als der Wert bei  $t=0$.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot {\rm e}^{-\pi(f\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t)^2} .$$
*Je kleiner die äquivalente Zeitdauer  $\Delta t$  ist, um so breiter und niedriger ist das Spektrum   ⇒   [[Signal_Representation/Fourier_Transform_Laws#Reziprozit.C3.A4tsgesetz_von_Zeitdauer_und_Bandbreite|Reziprozitätsgesetz von Bandbreite und Impulsdauer]].
*Sowohl  $x(t)$  als auch  $X(f)$  sind zu keinem  $f$–  bzw.  $t$–Wert exakt gleich Null.
*Für praktische Anwendungen kann der Gaußimpuls jedoch in Zeit und Frequenz als begrenzt angenommen werden.  Zum Beispiel ist  $x(t)$  bereits bei  $t=1.5 \Delta t$  auf weniger als  $0.1\% $  des Maximums abgefallen.

===Rectangular Pulse ===
*Die Zeitfunktion des Rechteckimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < T/2,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| = T/2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| > T/2.} \\ \end{array}$$

*Der  $\pm \Delta t/2$–Wert liegt mittig zwischen links- und rechtsseitigem Grenzwert.
*Für die Spektralfunktion erhält man entsprechend den Gesetzmäßigkeiten der Fouriertransformation (1. Fourierintegral):
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f) \quad \text{mit} \ {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Der Spektralwert bei  $f=0$  ist gleich der Rechteckfläche der Zeitfunktion.
*Die Spektralfunktion besitzt Nullstellen in äquidistanten Abständen  $1/\Delta t$.
*Das Integral über der Spektralfunktion  $X(f)$  ist gleich dem Signalwert zum Zeitpunkt  $t=0$, also der Impulshöhe  $K$.

===Triangular Pulse===
*Die Zeitfunktion des Dreieckimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot (1-|t|/{\Delta t}) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*Die absolute Zeitdauer ist  $2 \cdot \Delta t$;  diese ist doppelt so groß als die des Rechtecks.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta t \cdot f) \quad \text{mit} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Obige Zeitfunktion ist gleich der Faltung zweier Rechteckimpulse, jeweils mit Breite  $\Delta t$.
*Daraus folgt:  $X(f)$  beinhaltet anstelle der  ${\rm si}$-Funktion die  ${\rm si}^2$-Funktion.
*$X(f)$  weist somit ebenfalls Nullstellen im äquidistanten Abständen  $1/\Delta f$  auf.
*Der asymptotische Abfall von  $X(f)$  erfolgt hier mit  $1/f^2$, während zum Vergleich der Rechteckimpuls mit  $1/f$  abfällt.

===Trapezoidal Pulse ===
Die Zeitfunktion des Trapezimpulses mit der Höhe  $K$  und den Zeitparametern  $t_1$  und  $t_2$  lautet:
:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{t_2-|t|}{t_2-t_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}} \quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*Für die äquivalente Impulsdauer (flächengleiches Rechteck) gilt:   $\Delta t = t_1+t_2$.
*Der Rolloff-Faktor (im Zeitbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*Der Sonderfall  $r=0$  entspricht dem Rechteckimpuls und der Sonderfall  $r=1$  dem Dreieckimpuls.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f)\cdot {\rm si}(\pi \cdot r \cdot \Delta t \cdot f) \quad \text{mit} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Der asymptotische Abfall von  $X(f)$  liegt zwischen  $1/f$  $($für Rechteck,  $r=0)$  und  $1/f^2$  $($für Dreieck,  $r=1)$.

===Cosine roll-off Pulse ===
Die Zeitfunktion des Cosinus-Rolloff-Impulses mit der Höhe  $K$  und den Zeitparametern  $t_1$  und  $t_2$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|t|-t_1}{t_2-t_1}\cdot {\pi}/{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*Für die äquivalente Impulsdauer (flächengleiches Rechteck) gilt:   $\Delta t = t_1+t_2$.
*Der Rolloff-Faktor (im Zeitbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*Der Sonderfall  $r=0$  entspricht dem Rechteckimpuls und der Sonderfall  $r=1$  dem Cosinus-Quadrat-Impuls.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot \frac{\cos(\pi \cdot r\cdot \Delta t \cdot f)}{1-(2\cdot r\cdot \Delta t \cdot f)^2} \cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Je größer der Rolloff-Faktor  $r$  ist, desto schneller nimmt  $X(f)$  asymptotisch mit  $f$  ab.

===Cosine square Pulse ===
*Dies ist ein Sonderfall des Cosinus-Rolloff-Impulses und ergibt sich für  $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}t_1=0, \ t_2= \Delta t$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|t|\hspace{0.05cm}\cdot \hspace{0.05cm} \pi}{2\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta f \cdot \frac{\pi}{4}\cdot \big [{\rm si}(\pi(\Delta t\cdot f +0.5))+{\rm si}(\pi(\Delta t\cdot f -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Wegen der letzten  ${\rm si}$-Funktion ist  $X(f)=0$  für alle Vielfachen von  $F=1/\Delta t$.  Die äquidistanten Nulldurchgänge des Cos-Rolloff-Impulses bleiben erhalten.
*Aufgrund des Klammerausdrucks weist  $X(f)$  nun weitere Nulldurchgänge bei  $f=\pm1.5 F$,  $\pm2.5 F$,  $\pm3.5 F$, ... auf.
*Für die Frequenz  $f=\pm F/2$  erhält man die Spektralwerte  $K\cdot \Delta t/2$.
*Der asymptotische Abfall von  $X(f)$  verläuft in diesem Sonderfall mit  $1/f^3$.

==Exercises==
 

* First select the number  $(1,\text{...}, 7)$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
* "Red" refers to the first parameter set ⇒ $x_1(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_1(f)$,  "Blue" refers to the second parameter set ⇒ $x_2(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_2(f)$.
*Values with magnitude less than  $0.0005$  are output in the program as "zero".

{{BlaueBox|TEXT=
'''(1)'''   Compare the  red Gaussian pulse  $(A_1 = 1, \Delta t_1 = 1)$  with the  blue rectangular pulse  $(A_2 = 1, \Delta t_2 = 1)$ ⇒ default setting.
           What are the differences in the time and frequency domain?}}

* The Gaussian pulse theoretically reaches infinity in the time as well as in the frequency domain. 
* Practically  $x_1(t)$  for  $|t| > 1.5$  and  $X_1(f)$  for  $|f| > 1.5$  are almost zero. 
* The rectangle is strictly limited in time:  $x_2(|t| > 0.5) \equiv 0$.  $X_2(f)$  has shares in a much larger range than  $X_1(f)$. 
* It holds  $X_1(f = 0) = X_2(f = 0)$  since the integral over the Gaussian pulse  $x_1(t)$  is equal to the integral over the rectangular pulse  $x_2(t)$.

{{BlaueBox|TEXT=
'''(2)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2)$.           Vary the equivalent pulse duration  $\Delta t_2$  between  $0.5$  and  $2$.  Interpret the displayed graphs.}}

* One can recognize the reciprocity law of bandwidth and pulse duration.  The greater  $\Delta t_2$, the higher and narrower the spectral function  $X_2(f)$. 
* For each setting of  $\Delta t_2$,  $x_1(t=0)$  and  $x_2(t=0)$  are equal   ⇒   Also, the integrals over  $X_1(f)$  and  $X_2(f)$  are identical.

{{BlaueBox|TEXT=
'''(3)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2 = 0.5)$.           Vary  $\Delta t_2$  between  $0.05$  and  $2$.  Interpret the displayed graphs and extrapolate the result.}}

* The blue spectrum is now twice as wide as the red one, but only half as high.  First zero of  $X_1(f)$  at  $f = 1$, of  $X_2(f)$  at  $f = 2$. 
* Reduction of  $\Delta t_2$:  $X_2(f)$  lower and wider.  Very flat course at  $\Delta t_2 = 0.05$:  $X_2(f = 0)= 0.05$,  $X_2(f = \pm 3)= 0.048$. 
* If one choose  $\Delta t_2 = \varepsilon \to 0$  (not possible in the program),  the result would be the almost constant, very small spectrum  $X_2(f)=A \cdot \varepsilon \to 0$. 
* Increasing the amplitude to  $A=1/\varepsilon$  results in the constant spectral function  $X_2(f) = 1$  of the Dirac function  $\delta(t)$.  That means: 
* $\delta(t)$  is approximated by a rectangle  $($width  $\Delta t = \varepsilon \to 0$,  height  $A = 1/\varepsilon \to \infty)$.  The weight of the Dirac function is one:  $x(t) = 1 \cdot \delta (t)$.

{{BlaueBox|TEXT=
'''(4)'''   Compare the  rectangular pulse  $(A_1 = 1,   \Delta t_1 = 1)$  with the  triangular pulse  $(A_2 = 1,   \Delta t_2 = 1)$.  Interpret the spectral functions.}}

* The (normalized) spectrum of the rectangle  $x_1(t)$  with the (normalized) parameters  $A_1 = 1, \ \ \Delta t_1 = 1$  is:  $X_1(f)= {\rm si}(\pi\cdot f)= {\rm sinc}(f)$. 
* The convolution of the rectangle  $x_1(t)$  with itself gives the triangle  $x_2(t) = x_1(t) \star x_1(t)$.  By the convolution theorem:   $X_2(f) = X_1(f)^2 $. 
* By squaring the  ${\rm sinc}(f)$–shaped spectral function  $X_1(f)$  the zeros of  $X_2(f)$  remain unchanged.  But now it holds that: $X_2(f) \ge 0$.

{{BlaueBox|TEXT=
'''(5)'''   Compare the  trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, r_1 = 0.5)$  with the 
triangular pulse $(A_2 = 1,   \Delta t_2 = 1)$.          Vary  $r_1$  between  $0$  and  $1$.  Interpret the spectral function  $X_1(f)$.}}

* The trapezoidal pulse with roll–off factor  $r_1= 0$  is identical to the rectangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}(f)$. 
* The trapezoidal pulse with roll–off factor  $r_1= 1$  is identical to the triangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}^2(f)$. 
* In both cases  $X_1(f)$  has equidistant zeros at  $\pm 1$,  $\pm 2$, ...  (none else);   $0 < r_1 < 1$:  depending on  $r_1$  further zeros.

{{BlaueBox|TEXT=
'''(6)'''   Compare this  trapezoidal pulse  with the cosine roll-off pulse 
$(A_2 = 1,\ \Delta t_2 = 1.0,\ r_2 = 0.5)$.          Vary  $r_2$  between  $0$  and  $1$.  Interpret the spectral function  $X_2(f)$  for  $r_2 = 0.7$.}}

* With the same  $r= 0.5$  the cosine roll-off pulse  $X_2(f)$ is for  $f > 1$  greater in magnitude than the trapezoidal pulse. 
* With the same roll-off factor  $(r_1 = r_2= 0.5)$  the drop of  $X_2(f)$  around the frequency  $f = 0.5$  is steeper than the drop of  $X_1(f)$. 
* With  $r_1 = 0.5$  and  $r_2 = 0.7$  $x_1(t) \approx x_2(t)$  is valid and therefore also  $X_1(f) \approx X_2(f)$.  Comparable edge steepness.

{{BlaueBox|TEXT=
'''(7)'''   Compare the  red trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, \ r_1 = 1)$  with the  blue cosine roll-off pulse  $(A_2 = 1,\ \Delta t_2 = 1.0, \ r_2 = 1)$.           Interpret the time function  $x_2(t)$  and the spectral function  $X_2(f)$  system theoretically.}}

* $x_2(t) = \cos^2(|t|\cdot \pi/2) \ \ \ \text{for} \ |t| \le 1$  is the  cosine square pulse.  Zeros at  $f = \pm 1$,  $\pm 2$, ... 
* For the frequency  $f=\pm 0.5$  one obtains the spectral values  $X_2(f)=0.5$.  The asymptotic decline is shown here with  $1/f^3$.

==Applet Manual==
 
[[File:Exercise_impuls.png |right|frame|Screenshot]]
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Preselection for pulse shape  $x_1(t)$  (red curve)

    '''(C)'''     Parameter definition for  $x_1(t)$ 

    '''(D)'''     Numeric output for  $x_1(t_*)$  and  $X_1(f_*)$

    '''(E)'''     Preselection for pulse shape  $x_2(t)$  (blue curve)

    '''(F)'''     Parameter definition for  $x_2(t)$ 

    '''(G)'''     Numeric output for  $x_2(t_*)$  and  $X_2(f_*)$

    '''(H)'''     Setting the time  $t_*$  for the numeric output

    '''(I)'''     Setting the frequency  $f_*$  for the numeric output

    '''(J)'''     Graphic field for the time domain

    '''(K)'''     Graphic field for the frequency domain

    '''(L)'''     Selection of the exercise according to the numbers

    '''(M)'''     Task description and questions

    '''(N)'''     Show and hide sample solution

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2005 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2017 the program was redesigned by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]]  (Ingenieurspraxis_Math, Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2020 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

Applets:Pulses and Spectra

2023-03-24T19:31:16Z

Noah:

{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

==Applet Description==
 
Time-limited symmetric signals   ⇒   "pulses"  $x(t)$  and the corresponding spectral functions  $X(f)$  are considered, namely

*Gaussian pulse, 
*rectangular pulse,  
*triangular pulse, 
*trapezoidal pulse, 
*raised cosine pulse,
*cosine square pulse.

Further it is to be noted:
* The functions  $x(t)$  resp.  $X(f)$  are shown for up to two parameter sets in one diagram each.
* The red curves and numbers apply to the left parameter set, the blue ones to the right parameter set.
* The abscissas  $t$  (time) and  $f$  (frequency) as well as the ordinates  $x(t)$  (signal values) and  $X(f)$  (spectral values) are normalized.

==Theoretical Background==
 
===Relationship $x(t)\Leftrightarrow X(f)$===
*The relationship between the time function  $x(t)$  and the spectrum  $X(f)$  is given by the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|"first Fourier integral"]] :
:$$X(f)={\rm FT} [x(t)] = \int_{-\infty}^{+\infty}x(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm}
\rm FT\hspace{-0.1cm}: \ Fouriertransformation.$$

*Um aus der Spektralfunktion  $X(f)$  die Zeitfunktion  $x(t)$  berechnen zu können, benötigt man das  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|"zweite Fourierintegral"]]:
:$$x(t)={\rm IFT} [X(f)] = \int_{-\infty}^{+\infty}X(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm}
{\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fouriertransformation.$$

*In allen Beispielen verwenden wir reelle und gerade Funktionen.  Somit gilt:
:$$x(t)=\int_{-\infty}^{+\infty}X(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ X(f)=\int_{-\infty}^{+\infty}x(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$
*$x(t)$  und  $X(f)$  haben unterschiedliche Einheiten, beispielsweise  $x(t)$  in  $\rm V$,  $X(f)$  in  $\rm V/Hz$.
*Der Zusammenhang zwischen diesem Modul und dem ähnlich aufgebauten Applet  [[Applets:Frequenzgang_und_Impulsantwort|"Frequenzgang & Impulsantwort"]]  basiert auf dem  [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem|"Duality Theorem"]].
*Alle Zeiten sind auf eine Zeit  $T$  normiert und alle Frequenzen auf  $1/T$   ⇒   die Spektralwerte  $X(f)$  müssen noch mit der Normierungszeit  $T$  multipliziert werden.

{{GraueBox|TEXT=
$\text{Beispiel:}$   Stellt man einen Rechteckimpuls mit Amplitude  $A_1 = 1$  und äquivalenter Impulsdauer  $\Delta t_1 = 1$  ein, so ist  $x_1(t)$  im Bereich  $-0.5 < t < +0.5$  gleich Eins und außerhalb dieses Bereichs gleich Null.  Die Spektralfunktion  $X_1(f)$  verläuft  $\rm si$–förmig mit  $X_1(f= 0) = 1$  und der ersten Nullstelle bei  $f=1$.

*Soll mit dieser Einstellung ein Rechteckimpuls mit  $A = K = 3 \ \rm V$  und  $\Delta t = T = 2 \ \rm ms$  nachgebildet werden, dann sind alle Signalwerte mit  $K = 3 \ \rm V$  und alle Spektralwerte mit  $K \cdot T = 0.006 \ \rm V/Hz$  zu multiplizieren.
*Der maximale Spektralwert ist dann  $X(f= 0) = 0.006 \ \rm V/Hz$  und die erste Nullstelle liegt bei  $f=1/T = 0.5 \ \rm kHz$.}}

===Gaussian Pulse ===

*Die Zeitfunktion des Gaußimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:
:$$x(t)=K\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/\Delta t)^2}.$$
*Die äquivalente Zeitdauer  $\Delta t$  ergibt sich aus dem flächengleichen Rechteck.
*Der Wert bei  $t = \Delta t/2$  ist um den Faktor  $0.456$  kleiner als der Wert bei  $t=0$.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot {\rm e}^{-\pi(f\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t)^2} .$$
*Je kleiner die äquivalente Zeitdauer  $\Delta t$  ist, um so breiter und niedriger ist das Spektrum   ⇒   [[Signal_Representation/Fourier_Transform_Laws#Reziprozit.C3.A4tsgesetz_von_Zeitdauer_und_Bandbreite|Reziprozitätsgesetz von Bandbreite und Impulsdauer]].
*Sowohl  $x(t)$  als auch  $X(f)$  sind zu keinem  $f$–  bzw.  $t$–Wert exakt gleich Null.
*Für praktische Anwendungen kann der Gaußimpuls jedoch in Zeit und Frequenz als begrenzt angenommen werden.  Zum Beispiel ist  $x(t)$  bereits bei  $t=1.5 \Delta t$  auf weniger als  $0.1\% $  des Maximums abgefallen.

===Rectangular Pulse ===
*Die Zeitfunktion des Rechteckimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < T/2,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| = T/2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| > T/2.} \\ \end{array}$$

*Der  $\pm \Delta t/2$–Wert liegt mittig zwischen links- und rechtsseitigem Grenzwert.
*Für die Spektralfunktion erhält man entsprechend den Gesetzmäßigkeiten der Fouriertransformation (1. Fourierintegral):
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f) \quad \text{mit} \ {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Der Spektralwert bei  $f=0$  ist gleich der Rechteckfläche der Zeitfunktion.
*Die Spektralfunktion besitzt Nullstellen in äquidistanten Abständen  $1/\Delta t$.
*Das Integral über der Spektralfunktion  $X(f)$  ist gleich dem Signalwert zum Zeitpunkt  $t=0$, also der Impulshöhe  $K$.

===Triangular Pulse===
*Die Zeitfunktion des Dreieckimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot (1-|t|/{\Delta t}) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*Die absolute Zeitdauer ist  $2 \cdot \Delta t$;  diese ist doppelt so groß als die des Rechtecks.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta t \cdot f) \quad \text{mit} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Obige Zeitfunktion ist gleich der Faltung zweier Rechteckimpulse, jeweils mit Breite  $\Delta t$.
*Daraus folgt:  $X(f)$  beinhaltet anstelle der  ${\rm si}$-Funktion die  ${\rm si}^2$-Funktion.
*$X(f)$  weist somit ebenfalls Nullstellen im äquidistanten Abständen  $1/\Delta f$  auf.
*Der asymptotische Abfall von  $X(f)$  erfolgt hier mit  $1/f^2$, während zum Vergleich der Rechteckimpuls mit  $1/f$  abfällt.

===Trapezoidal Pulse ===
Die Zeitfunktion des Trapezimpulses mit der Höhe  $K$  und den Zeitparametern  $t_1$  und  $t_2$  lautet:
:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{t_2-|t|}{t_2-t_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}} \quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*Für die äquivalente Impulsdauer (flächengleiches Rechteck) gilt:   $\Delta t = t_1+t_2$.
*Der Rolloff-Faktor (im Zeitbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*Der Sonderfall  $r=0$  entspricht dem Rechteckimpuls und der Sonderfall  $r=1$  dem Dreieckimpuls.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f)\cdot {\rm si}(\pi \cdot r \cdot \Delta t \cdot f) \quad \text{mit} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Der asymptotische Abfall von  $X(f)$  liegt zwischen  $1/f$  $($für Rechteck,  $r=0)$  und  $1/f^2$  $($für Dreieck,  $r=1)$.

===Cosine roll-off Pulse ===
Die Zeitfunktion des Cosinus-Rolloff-Impulses mit der Höhe  $K$  und den Zeitparametern  $t_1$  und  $t_2$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|t|-t_1}{t_2-t_1}\cdot {\pi}/{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*Für die äquivalente Impulsdauer (flächengleiches Rechteck) gilt:   $\Delta t = t_1+t_2$.
*Der Rolloff-Faktor (im Zeitbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*Der Sonderfall  $r=0$  entspricht dem Rechteckimpuls und der Sonderfall  $r=1$  dem Cosinus-Quadrat-Impuls.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot \frac{\cos(\pi \cdot r\cdot \Delta t \cdot f)}{1-(2\cdot r\cdot \Delta t \cdot f)^2} \cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Je größer der Rolloff-Faktor  $r$  ist, desto schneller nimmt  $X(f)$  asymptotisch mit  $f$  ab.

===Cosine square Pulse ===
*Dies ist ein Sonderfall des Cosinus-Rolloff-Impulses und ergibt sich für  $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}t_1=0, \ t_2= \Delta t$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|t|\hspace{0.05cm}\cdot \hspace{0.05cm} \pi}{2\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta f \cdot \frac{\pi}{4}\cdot \big [{\rm si}(\pi(\Delta t\cdot f +0.5))+{\rm si}(\pi(\Delta t\cdot f -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Wegen der letzten  ${\rm si}$-Funktion ist  $X(f)=0$  für alle Vielfachen von  $F=1/\Delta t$.  Die äquidistanten Nulldurchgänge des Cos-Rolloff-Impulses bleiben erhalten.
*Aufgrund des Klammerausdrucks weist  $X(f)$  nun weitere Nulldurchgänge bei  $f=\pm1.5 F$,  $\pm2.5 F$,  $\pm3.5 F$, ... auf.
*Für die Frequenz  $f=\pm F/2$  erhält man die Spektralwerte  $K\cdot \Delta t/2$.
*Der asymptotische Abfall von  $X(f)$  verläuft in diesem Sonderfall mit  $1/f^3$.

==Exercises==
 

* First select the number  $(1,\text{...}, 7)$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
* "Red" refers to the first parameter set ⇒ $x_1(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_1(f)$,  "Blue" refers to the second parameter set ⇒ $x_2(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_2(f)$.
*Values with magnitude less than  $0.0005$  are output in the program as "zero".

{{BlaueBox|TEXT=
'''(1)'''   Compare the  red Gaussian pulse  $(A_1 = 1, \Delta t_1 = 1)$  with the  blue rectangular pulse  $(A_2 = 1, \Delta t_2 = 1)$ ⇒ default setting.
           What are the differences in the time and frequency domain?}}

* The Gaussian pulse theoretically reaches infinity in the time as well as in the frequency domain. 
* Practically  $x_1(t)$  for  $|t| > 1.5$  and  $X_1(f)$  for  $|f| > 1.5$  are almost zero. 
* The rectangle is strictly limited in time:  $x_2(|t| > 0.5) \equiv 0$.  $X_2(f)$  has shares in a much larger range than  $X_1(f)$. 
* It holds  $X_1(f = 0) = X_2(f = 0)$  since the integral over the Gaussian pulse  $x_1(t)$  is equal to the integral over the rectangular pulse  $x_2(t)$.

{{BlaueBox|TEXT=
'''(2)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2)$.           Vary the equivalent pulse duration  $\Delta t_2$  between  $0.5$  and  $2$.  Interpret the displayed graphs.}}

* One can recognize the reciprocity law of bandwidth and pulse duration.  The greater  $\Delta t_2$, the higher and narrower the spectral function  $X_2(f)$. 
* For each setting of  $\Delta t_2$,  $x_1(t=0)$  and  $x_2(t=0)$  are equal   ⇒   Also, the integrals over  $X_1(f)$  and  $X_2(f)$  are identical.

{{BlaueBox|TEXT=
'''(3)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2 = 0.5)$.           Vary  $\Delta t_2$  between  $0.05$  and  $2$.  Interpret the displayed graphs and extrapolate the result.}}

* The blue spectrum is now twice as wide as the red one, but only half as high.  First zero of  $X_1(f)$  at  $f = 1$, of  $X_2(f)$  at  $f = 2$. 
* Reduction of  $\Delta t_2$:  $X_2(f)$  lower and wider.  Very flat course at  $\Delta t_2 = 0.05$:  $X_2(f = 0)= 0.05$,  $X_2(f = \pm 3)= 0.048$. 
* If one choose  $\Delta t_2 = \varepsilon \to 0$  (not possible in the program),  the result would be the almost constant, very small spectrum  $X_2(f)=A \cdot \varepsilon \to 0$. 
* Increasing the amplitude to  $A=1/\varepsilon$  results in the constant spectral function  $X_2(f) = 1$  of the Dirac function  $\delta(t)$.  That means: 
* $\delta(t)$  is approximated by a rectangle  $($width  $\Delta t = \varepsilon \to 0$,  height  $A = 1/\varepsilon \to \infty)$.  The weight of the Dirac function is one:  $x(t) = 1 \cdot \delta (t)$.

{{BlaueBox|TEXT=
'''(4)'''   Compare the  rectangular pulse  $(A_1 = 1,   \Delta t_1 = 1)$  with the  triangular pulse  $(A_2 = 1,   \Delta t_2 = 1)$.  Interpret the spectral functions.}}

* The (normalized) spectrum of the rectangle  $x_1(t)$  with the (normalized) parameters  $A_1 = 1, \ \ \Delta t_1 = 1$  is:  $X_1(f)= {\rm si}(\pi\cdot f)= {\rm sinc}(f)$. 
* The convolution of the rectangle  $x_1(t)$  with itself gives the triangle  $x_2(t) = x_1(t) \star x_1(t)$.  By the convolution theorem:   $X_2(f) = X_1(f)^2 $. 
* By squaring the  ${\rm sinc}(f)$–shaped spectral function  $X_1(f)$  the zeros of  $X_2(f)$  remain unchanged.  But now it holds that: $X_2(f) \ge 0$.

{{BlaueBox|TEXT=
'''(5)'''   Compare the  trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, r_1 = 0.5)$  with the 
triangular pulse $(A_2 = 1,   \Delta t_2 = 1)$.          Vary  $r_1$  between  $0$  and  $1$.  Interpret the spectral function  $X_1(f)$.}}

* The trapezoidal pulse with roll–off factor  $r_1= 0$  is identical to the rectangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}(f)$. 
* The trapezoidal pulse with roll–off factor  $r_1= 1$  is identical to the triangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}^2(f)$. 
* In both cases  $X_1(f)$  has equidistant zeros at  $\pm 1$,  $\pm 2$, ...  (none else);   $0 < r_1 < 1$:  depending on  $r_1$  further zeros.

{{BlaueBox|TEXT=
'''(6)'''   Compare this  trapezoidal pulse  with the cosine roll-off pulse 
$(A_2 = 1,\ \Delta t_2 = 1.0,\ r_2 = 0.5)$.          Vary  $r_2$  between  $0$  and  $1$.  Interpret the spectral function  $X_2(f)$  for  $r_2 = 0.7$.}}

* With the same  $r= 0.5$  the cosine roll-off pulse  $X_2(f)$ is for  $f > 1$  greater in magnitude than the trapezoidal pulse. 
* With the same roll-off factor  $(r_1 = r_2= 0.5)$  the drop of  $X_2(f)$  around the frequency  $f = 0.5$  is steeper than the drop of  $X_1(f)$. 
* With  $r_1 = 0.5$  and  $r_2 = 0.7$  $x_1(t) \approx x_2(t)$  is valid and therefore also  $X_1(f) \approx X_2(f)$.  Comparable edge steepness.

{{BlaueBox|TEXT=
'''(7)'''   Compare the  red trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, \ r_1 = 1)$  with the  blue cosine roll-off pulse  $(A_2 = 1,\ \Delta t_2 = 1.0, \ r_2 = 1)$.           Interpret the time function  $x_2(t)$  and the spectral function  $X_2(f)$  system theoretically.}}

* $x_2(t) = \cos^2(|t|\cdot \pi/2) \ \ \ \text{for} \ |t| \le 1$  is the  cosine square pulse.  Zeros at  $f = \pm 1$,  $\pm 2$, ... 
* For the frequency  $f=\pm 0.5$  one obtains the spectral values  $X_2(f)=0.5$.  The asymptotic decline is shown here with  $1/f^3$.

==Applet Manual==
 
[[File:Exercise_impuls.png |right|frame|Screenshot]]
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Preselection for pulse shape  $x_1(t)$  (red curve)

    '''(C)'''     Parameter definition for  $x_1(t)$ 

    '''(D)'''     Numeric output for  $x_1(t_*)$  and  $X_1(f_*)$

    '''(E)'''     Preselection for pulse shape  $x_2(t)$  (blue curve)

    '''(F)'''     Parameter definition for  $x_2(t)$ 

    '''(G)'''     Numeric output for  $x_2(t_*)$  and  $X_2(f_*)$

    '''(H)'''     Setting the time  $t_*$  for the numeric output

    '''(I)'''     Setting the frequency  $f_*$  for the numeric output

    '''(J)'''     Graphic field for the time domain

    '''(K)'''     Graphic field for the frequency domain

    '''(L)'''     Selection of the exercise according to the numbers

    '''(M)'''     Task description and questions

    '''(N)'''     Show and hide sample solution

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2005 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2017 the program was redesigned by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]]  (Ingenieurspraxis_Math, Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2020 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

Applets:Pulses and Spectra

2023-03-24T18:48:40Z

Noah:

{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}

==Applet Description==
 
Time-limited symmetric signals   ⇒   "pulses"  $x(t)$  and the corresponding spectral functions  $X(f)$  are considered, namely

*Gaussian pulse, 
*rectangular pulse,  
*triangular pulse, 
*trapezoidal pulse, 
*raised cosine pulse,
*cosine square pulse.

Further it is to be noted:
* The functions  $x(t)$  resp.  $X(f)$  are shown for up to two parameter sets in one diagram each.
* The red curves and numbers apply to the left parameter set, the blue ones to the right parameter set.
* The abscissas  $t$  (time) and  $f$  (frequency) as well as the ordinates  $x(t)$  (signal values) and  $X(f)$  (spectral values) are normalized.

==Theoretical Background==
 
===Relationship $x(t)\Leftrightarrow X(f)$===
*Der Zusammenhang zwischen der Zeitfunktion  $x(t)$  und dem Spektrum  $X(f)$  ist durch das  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|"erste Fourierintegral"]]  gegeben:
:$$X(f)={\rm FT} [x(t)] = \int_{-\infty}^{+\infty}x(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm}
\rm FT\hspace{-0.1cm}: \ Fouriertransformation.$$

*Um aus der Spektralfunktion  $X(f)$  die Zeitfunktion  $x(t)$  berechnen zu können, benötigt man das  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|"zweite Fourierintegral"]]:
:$$x(t)={\rm IFT} [X(f)] = \int_{-\infty}^{+\infty}X(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm}
{\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fouriertransformation.$$

*In allen Beispielen verwenden wir reelle und gerade Funktionen.  Somit gilt:
:$$x(t)=\int_{-\infty}^{+\infty}X(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ \ \ X(f)=\int_{-\infty}^{+\infty}x(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$
*$x(t)$  und  $X(f)$  haben unterschiedliche Einheiten, beispielsweise  $x(t)$  in  $\rm V$,  $X(f)$  in  $\rm V/Hz$.
*Der Zusammenhang zwischen diesem Modul und dem ähnlich aufgebauten Applet  [[Applets:Frequenzgang_und_Impulsantwort|"Frequenzgang & Impulsantwort"]]  basiert auf dem  [[Signal_Representation/The_Fourier_Transform_Theorems#Duality_Theorem|"Duality Theorem"]].
*Alle Zeiten sind auf eine Zeit  $T$  normiert und alle Frequenzen auf  $1/T$   ⇒   die Spektralwerte  $X(f)$  müssen noch mit der Normierungszeit  $T$  multipliziert werden.

{{GraueBox|TEXT=
$\text{Beispiel:}$   Stellt man einen Rechteckimpuls mit Amplitude  $A_1 = 1$  und äquivalenter Impulsdauer  $\Delta t_1 = 1$  ein, so ist  $x_1(t)$  im Bereich  $-0.5 < t < +0.5$  gleich Eins und außerhalb dieses Bereichs gleich Null.  Die Spektralfunktion  $X_1(f)$  verläuft  $\rm si$–förmig mit  $X_1(f= 0) = 1$  und der ersten Nullstelle bei  $f=1$.

*Soll mit dieser Einstellung ein Rechteckimpuls mit  $A = K = 3 \ \rm V$  und  $\Delta t = T = 2 \ \rm ms$  nachgebildet werden, dann sind alle Signalwerte mit  $K = 3 \ \rm V$  und alle Spektralwerte mit  $K \cdot T = 0.006 \ \rm V/Hz$  zu multiplizieren.
*Der maximale Spektralwert ist dann  $X(f= 0) = 0.006 \ \rm V/Hz$  und die erste Nullstelle liegt bei  $f=1/T = 0.5 \ \rm kHz$.}}

===Gaussian Pulse ===

*Die Zeitfunktion des Gaußimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:
:$$x(t)=K\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(t/\Delta t)^2}.$$
*Die äquivalente Zeitdauer  $\Delta t$  ergibt sich aus dem flächengleichen Rechteck.
*Der Wert bei  $t = \Delta t/2$  ist um den Faktor  $0.456$  kleiner als der Wert bei  $t=0$.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot {\rm e}^{-\pi(f\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t)^2} .$$
*Je kleiner die äquivalente Zeitdauer  $\Delta t$  ist, um so breiter und niedriger ist das Spektrum   ⇒   [[Signal_Representation/Fourier_Transform_Laws#Reziprozit.C3.A4tsgesetz_von_Zeitdauer_und_Bandbreite|Reziprozitätsgesetz von Bandbreite und Impulsdauer]].
*Sowohl  $x(t)$  als auch  $X(f)$  sind zu keinem  $f$–  bzw.  $t$–Wert exakt gleich Null.
*Für praktische Anwendungen kann der Gaußimpuls jedoch in Zeit und Frequenz als begrenzt angenommen werden.  Zum Beispiel ist  $x(t)$  bereits bei  $t=1.5 \Delta t$  auf weniger als  $0.1\% $  des Maximums abgefallen.

===Rectangular Pulse ===
*Die Zeitfunktion des Rechteckimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < T/2,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| = T/2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| > T/2.} \\ \end{array}$$

*Der  $\pm \Delta t/2$–Wert liegt mittig zwischen links- und rechtsseitigem Grenzwert.
*Für die Spektralfunktion erhält man entsprechend den Gesetzmäßigkeiten der Fouriertransformation (1. Fourierintegral):
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f) \quad \text{mit} \ {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Der Spektralwert bei  $f=0$  ist gleich der Rechteckfläche der Zeitfunktion.
*Die Spektralfunktion besitzt Nullstellen in äquidistanten Abständen  $1/\Delta t$.
*Das Integral über der Spektralfunktion  $X(f)$  ist gleich dem Signalwert zum Zeitpunkt  $t=0$, also der Impulshöhe  $K$.

===Triangular Pulse===
*Die Zeitfunktion des Dreieckimpulses mit der Höhe  $K$  und der (äquivalenten) Dauer  $\Delta t$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot (1-|t|/{\Delta t}) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*Die absolute Zeitdauer ist  $2 \cdot \Delta t$;  diese ist doppelt so groß als die des Rechtecks.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta t \cdot f) \quad \text{mit} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Obige Zeitfunktion ist gleich der Faltung zweier Rechteckimpulse, jeweils mit Breite  $\Delta t$.
*Daraus folgt:  $X(f)$  beinhaltet anstelle der  ${\rm si}$-Funktion die  ${\rm si}^2$-Funktion.
*$X(f)$  weist somit ebenfalls Nullstellen im äquidistanten Abständen  $1/\Delta f$  auf.
*Der asymptotische Abfall von  $X(f)$  erfolgt hier mit  $1/f^2$, während zum Vergleich der Rechteckimpuls mit  $1/f$  abfällt.

===Trapezoidal Pulse ===
Die Zeitfunktion des Trapezimpulses mit der Höhe  $K$  und den Zeitparametern  $t_1$  und  $t_2$  lautet:
:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{t_2-|t|}{t_2-t_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}} \quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*Für die äquivalente Impulsdauer (flächengleiches Rechteck) gilt:   $\Delta t = t_1+t_2$.
*Der Rolloff-Faktor (im Zeitbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*Der Sonderfall  $r=0$  entspricht dem Rechteckimpuls und der Sonderfall  $r=1$  dem Dreieckimpuls.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot {\rm si}(\pi\cdot \Delta t \cdot f)\cdot {\rm si}(\pi \cdot r \cdot \Delta t \cdot f) \quad \text{mit} \quad {\rm si}(x)=\frac{\sin(x)}{x}.$$
*Der asymptotische Abfall von  $X(f)$  liegt zwischen  $1/f$  $($für Rechteck,  $r=0)$  und  $1/f^2$  $($für Dreieck,  $r=1)$.

===Cosine roll-off Pulse ===
Die Zeitfunktion des Cosinus-Rolloff-Impulses mit der Höhe  $K$  und den Zeitparametern  $t_1$  und  $t_2$  lautet:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|t|-t_1}{t_2-t_1}\cdot {\pi}/{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ {\rm{f\ddot{u}r}}\quad \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| \le t_1,} \\ {t_1\le \left| \hspace{0.05cm}t\hspace{0.05cm} \right| \le t_2,} \\ {\left|\hspace{0.05cm} t \hspace{0.05cm} \right| \ge t_2.} \\ \end{array}$$

*Für die äquivalente Impulsdauer (flächengleiches Rechteck) gilt:   $\Delta t = t_1+t_2$.
*Der Rolloff-Faktor (im Zeitbereich) kennzeichnet die Flankensteilheit:
:$$r=\frac{t_2-t_1}{t_2+t_1}.$$
*Der Sonderfall  $r=0$  entspricht dem Rechteckimpuls und der Sonderfall  $r=1$  dem Cosinus-Quadrat-Impuls.
*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta t \cdot \frac{\cos(\pi \cdot r\cdot \Delta t \cdot f)}{1-(2\cdot r\cdot \Delta t \cdot f)^2} \cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Je größer der Rolloff-Faktor  $r$  ist, desto schneller nimmt  $X(f)$  asymptotisch mit  $f$  ab.

===Cosine square Pulse ===
*Dies ist ein Sonderfall des Cosinus-Rolloff-Impulses und ergibt sich für  $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}t_1=0, \ t_2= \Delta t$:

:$$x(t) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|t|\hspace{0.05cm}\cdot \hspace{0.05cm} \pi}{2\hspace{0.05cm}\cdot \hspace{0.05cm} \Delta t}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} t\hspace{0.05cm} \right| < \Delta t,} \\ {\left| \hspace{0.05cm}t\hspace{0.05cm} \right| \ge \Delta t.} \\ \end{array}$$

*Für die Spektralfunktion erhält man gemäß der Fouriertransformation:
:$$X(f)=K\cdot \Delta f \cdot \frac{\pi}{4}\cdot \big [{\rm si}(\pi(\Delta t\cdot f +0.5))+{\rm si}(\pi(\Delta t\cdot f -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta t \cdot f).$$
*Wegen der letzten  ${\rm si}$-Funktion ist  $X(f)=0$  für alle Vielfachen von  $F=1/\Delta t$.  Die äquidistanten Nulldurchgänge des Cos-Rolloff-Impulses bleiben erhalten.
*Aufgrund des Klammerausdrucks weist  $X(f)$  nun weitere Nulldurchgänge bei  $f=\pm1.5 F$,  $\pm2.5 F$,  $\pm3.5 F$, ... auf.
*Für die Frequenz  $f=\pm F/2$  erhält man die Spektralwerte  $K\cdot \Delta t/2$.
*Der asymptotische Abfall von  $X(f)$  verläuft in diesem Sonderfall mit  $1/f^3$.

==Exercises==
 

* First select the number  $(1,\text{...}, 7)$  of the exercise.  The number  $0$  corresponds to a "Reset":  Same setting as at program start.
*A task description is displayed.  The parameter values are adjusted.  Solution after pressing "Show solution". 
* "Red" refers to the first parameter set ⇒ $x_1(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_1(f)$,  "Blue" refers to the second parameter set ⇒ $x_2(t) \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\ X_2(f)$.
*Values with magnitude less than  $0.0005$  are output in the program as "zero".

{{BlaueBox|TEXT=
'''(1)'''   Compare the  red Gaussian pulse  $(A_1 = 1, \Delta t_1 = 1)$  with the  blue rectangular pulse  $(A_2 = 1, \Delta t_2 = 1)$ ⇒ default setting.
           What are the differences in the time and frequency domain?}}

* The Gaussian pulse theoretically reaches infinity in the time as well as in the frequency domain. 
* Practically  $x_1(t)$  for  $|t| > 1.5$  and  $X_1(f)$  for  $|f| > 1.5$  are almost zero. 
* The rectangle is strictly limited in time:  $x_2(|t| > 0.5) \equiv 0$.  $X_2(f)$  has shares in a much larger range than  $X_1(f)$. 
* It holds  $X_1(f = 0) = X_2(f = 0)$  since the integral over the Gaussian pulse  $x_1(t)$  is equal to the integral over the rectangular pulse  $x_2(t)$.

{{BlaueBox|TEXT=
'''(2)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2)$.           Vary the equivalent pulse duration  $\Delta t_2$  between  $0.5$  and  $2$.  Interpret the displayed graphs.}}

* One can recognize the reciprocity law of bandwidth and pulse duration.  The greater  $\Delta t_2$, the higher and narrower the spectral function  $X_2(f)$. 
* For each setting of  $\Delta t_2$,  $x_1(t=0)$  and  $x_2(t=0)$  are equal   ⇒   Also, the integrals over  $X_1(f)$  and  $X_2(f)$  are identical.

{{BlaueBox|TEXT=
'''(3)'''   Compare the  red Gaussian pulse  $(A_1 = 1,  \Delta t_1 = 1)$ with the  blue rectangular pulse  $(A_2 = 1,  \Delta t_2 = 0.5)$.           Vary  $\Delta t_2$  between  $0.05$  and  $2$.  Interpret the displayed graphs and extrapolate the result.}}

* The blue spectrum is now twice as wide as the red one, but only half as high.  First zero of  $X_1(f)$  at  $f = 1$, of  $X_2(f)$  at  $f = 2$. 
* Reduction of  $\Delta t_2$:  $X_2(f)$  lower and wider.  Very flat course at  $\Delta t_2 = 0.05$:  $X_2(f = 0)= 0.05$,  $X_2(f = \pm 3)= 0.048$. 
* If one choose  $\Delta t_2 = \varepsilon \to 0$  (not possible in the program),  the result would be the almost constant, very small spectrum  $X_2(f)=A \cdot \varepsilon \to 0$. 
* Increasing the amplitude to  $A=1/\varepsilon$  results in the constant spectral function  $X_2(f) = 1$  of the Dirac function  $\delta(t)$.  That means: 
* $\delta(t)$  is approximated by a rectangle  $($width  $\Delta t = \varepsilon \to 0$,  height  $A = 1/\varepsilon \to \infty)$.  The weight of the Dirac function is one:  $x(t) = 1 \cdot \delta (t)$.

{{BlaueBox|TEXT=
'''(4)'''   Compare the  rectangular pulse  $(A_1 = 1,   \Delta t_1 = 1)$  with the  triangular pulse  $(A_2 = 1,   \Delta t_2 = 1)$.  Interpret the spectral functions.}}

* The (normalized) spectrum of the rectangle  $x_1(t)$  with the (normalized) parameters  $A_1 = 1, \ \ \Delta t_1 = 1$  is:  $X_1(f)= {\rm si}(\pi\cdot f)= {\rm sinc}(f)$. 
* The convolution of the rectangle  $x_1(t)$  with itself gives the triangle  $x_2(t) = x_1(t) \star x_1(t)$.  By the convolution theorem:   $X_2(f) = X_1(f)^2 $. 
* By squaring the  ${\rm sinc}(f)$–shaped spectral function  $X_1(f)$  the zeros of  $X_2(f)$  remain unchanged.  But now it holds that: $X_2(f) \ge 0$.

{{BlaueBox|TEXT=
'''(5)'''   Compare the  trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, r_1 = 0.5)$  with the 
triangular pulse $(A_2 = 1,   \Delta t_2 = 1)$.          Vary  $r_1$  between  $0$  and  $1$.  Interpret the spectral function  $X_1(f)$.}}

* The trapezoidal pulse with roll–off factor  $r_1= 0$  is identical to the rectangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}(f)$. 
* The trapezoidal pulse with roll–off factor  $r_1= 1$  is identical to the triangular pulse.  The "normalized spectrum" is  $X_1(f)= {\rm sinc}^2(f)$. 
* In both cases  $X_1(f)$  has equidistant zeros at  $\pm 1$,  $\pm 2$, ...  (none else);   $0 < r_1 < 1$:  depending on  $r_1$  further zeros.

{{BlaueBox|TEXT=
'''(6)'''   Compare this  trapezoidal pulse  with the cosine roll-off pulse 
$(A_2 = 1,\ \Delta t_2 = 1.0,\ r_2 = 0.5)$.          Vary  $r_2$  between  $0$  and  $1$.  Interpret the spectral function  $X_2(f)$  for  $r_2 = 0.7$.}}

* With the same  $r= 0.5$  the cosine roll-off pulse  $X_2(f)$ is for  $f > 1$  greater in magnitude than the trapezoidal pulse. 
* With the same roll-off factor  $(r_1 = r_2= 0.5)$  the drop of  $X_2(f)$  around the frequency  $f = 0.5$  is steeper than the drop of  $X_1(f)$. 
* With  $r_1 = 0.5$  and  $r_2 = 0.7$  $x_1(t) \approx x_2(t)$  is valid and therefore also  $X_1(f) \approx X_2(f)$.  Comparable edge steepness.

{{BlaueBox|TEXT=
'''(7)'''   Compare the  red trapezoidal pulse  $(A_1 = 1, \Delta t_1 = 1, \ r_1 = 1)$  with the  blue cosine roll-off pulse  $(A_2 = 1,\ \Delta t_2 = 1.0, \ r_2 = 1)$.           Interpret the time function  $x_2(t)$  and the spectral function  $X_2(f)$  system theoretically.}}

* $x_2(t) = \cos^2(|t|\cdot \pi/2) \ \ \ \text{for} \ |t| \le 1$  is the  cosine square pulse.  Zeros at  $f = \pm 1$,  $\pm 2$, ... 
* For the frequency  $f=\pm 0.5$  one obtains the spectral values  $X_2(f)=0.5$.  The asymptotic decline is shown here with  $1/f^3$.

==Applet Manual==
 
[[File:Exercise_impuls.png |right|frame|Screenshot]]
    '''(A)'''     Theme (changeable graphical user interface design)
:* Dark:   dark background  (recommended by the authors)
:* Bright:   white background  (recommended for beamers and printouts)
:* Deuteranopia:   for users with pronounced green visual impairment
:* Protanopia:   for users with pronounced red visual impairment

    '''(B)'''     Preselection for pulse shape  $x_1(t)$  (red curve)

    '''(C)'''     Parameter definition for  $x_1(t)$ 

    '''(D)'''     Numeric output for  $x_1(t_*)$  and  $X_1(f_*)$

    '''(E)'''     Preselection for pulse shape  $x_2(t)$  (blue curve)

    '''(F)'''     Parameter definition for  $x_2(t)$ 

    '''(G)'''     Numeric output for  $x_2(t_*)$  and  $X_2(f_*)$

    '''(H)'''     Setting the time  $t_*$  for the numeric output

    '''(I)'''     Setting the frequency  $f_*$  for the numeric output

    '''(J)'''     Graphic field for the time domain

    '''(K)'''     Graphic field for the frequency domain

    '''(L)'''     Selection of the exercise according to the numbers

    '''(M)'''     Task description and questions

    '''(N)'''     Show and hide sample solution

==About the Authors==
 
This interactive calculation tool was designed and implemented at the  [https://www.ei.tum.de/en/lnt/home/ Institute for Communications Engineering]  at the  [https://www.tum.de/en Technical University of Munich].
*The first version was created in 2005 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Ji_Li_.28Bachelorarbeit_EI_2003.2C_Diplomarbeit_EI_2005.29|Ji Li]]  as part of her diploma thesis with “FlashMX – Actionscript”  (Supervisor:  [[Biographies_and_Bibliographies/An_LNTwww_beteiligte_Mitarbeiter_und_Dozenten#Prof._Dr.-Ing._habil._G.C3.BCnter_S.C3.B6der_.28am_LNT_seit_1974.29|Günter Söder]]).
*In 2017 the program was redesigned by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#David_Jobst_.28Ingenieurspraxis_Math_2017.29|David Jobst]]  (Ingenieurspraxis_Math, Supervisor: [[Biographies_and_Bibliographies/Beteiligte_der_Professur_Leitungsgebundene_%C3%9Cbertragungstechnik#Tasn.C3.A1d_Kernetzky.2C_M.Sc._.28bei_L.C3.9CT_seit_2014.29|Tasnád Kernetzky]] ) via "HTML5".
*Last revision and English version 2020 by  [[Biografien_und_Bibliografien/An_LNTwww_beteiligte_Studierende#Carolin_Mirschina_.28Ingenieurspraxis_Math_2019.2C_danach_Werkstudentin.29|Carolin Mirschina]]  in the context of a working student activity. 

==Once again: Open Applet in new Tab==
{{LntAppletLinkEnDe|pulsesAndSpectra_en|pulsesAndSpectra}}