Difference between revisions of "Aufgaben:Exercise 1.3: Rectangular Functions for Transmitter and Receiver"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Fehlerwahrscheinlichkeit bei Basisbandübertragung
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission
 
}}
 
}}
  
[[File: P_ID1267__Dig_A_1_3.png|right|frame|Drei verschiedene Systemkonzepte]]
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[[File: P_ID1267__Dig_A_1_3.png|right|frame|Three different system concepts]]
Wir betrachten hier drei Varianten eines binären bipolaren AWGN–Übertragungssystems, die sich hinsichtlich des Sendegrundimpulses  $g_{s}(t)$  sowie der Impulsantwort  $h_{\rm E}(t)$  des Empfangsfilters unterscheiden:
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We consider here three variants of a binary bipolar AWGN transmission system which differ with respect to the basic transmission pulse  $g_{s}(t)$  as well as the impulse response  $h_{\rm E}(t)$  of the receiver filter:
*Beim  $\text{System A}$  sind sowohl  $g_{s}(t)$  als auch  $h_{\rm E}(t)$  rechteckförmig, lediglich die Impulshöhen  $(s_{\rm 0}$  bzw.  $1/T)$  sind unterschiedlich.
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*For  $\text{System A}$,  both  $g_{s}(t)$  and  $h_{\rm E}(t)$  are rectangular, only the pulse heights  $(s_{\rm 0}$  and  $1/T)$  are different.
*Das  $\text{System B}$  unterscheidet sich vom  $\text{System A}$  durch einen dreieckförmigen Sendegrundimpuls mit  $g_{s}(t=0) = s_{\rm 0}$.
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* $\text{System B}$  differs from  $\text{System A}$  by having a triangular-shaped basic transmission pulse with  $g_{s}(t=0) = s_{\rm 0}$.
*Das  $\text{System C}$  hat den gleichen Sendegrundimpuls wie  $\text{System A}$, während die Impulsantwort mit  $h_{\rm E}(t=0) = 1/T$  dreieckförmig verläuft.
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* $\text{System C}$  has the same basic transmission pulse as  $\text{System A}$, while the impulse response is triangular with  $h_{\rm E}(t=0) = 1/T$.   
  
  
Die absolute Breite der hier betrachteten Rechteck– und Dreieckfunktionen beträgt jeweils  $T = 10 \ \rm µ s$. Die Bitrate ist  $R = 100 \ \rm kbit/s$. Die weiteren Systemparameter sind wie folgt gegeben:
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The absolute width of the rectangular and triangular functions considered here is  $T = 10 \ \rm µ s$ each. The bit rate is  $R = 100 \ \rm kbit/s$. The other system parameters are given as follows:
 
:$$s_0 = 6 \,\,\sqrt{W}\hspace{0.05cm},\hspace{0.3cm}  N_{\rm 0} = 2 \cdot 10^{-5} \,\,{\rm W/Hz}\hspace{0.05cm}.$$
 
:$$s_0 = 6 \,\,\sqrt{W}\hspace{0.05cm},\hspace{0.3cm}  N_{\rm 0} = 2 \cdot 10^{-5} \,\,{\rm W/Hz}\hspace{0.05cm}.$$
  
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''Hinweise:''  
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''Notes:''  
*Die Aufgabe gehört zum  Kapitel   [[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_bei_Basisbandübertragung| Fehlerwahrscheinlichkeit bei Basisbandübertragung]].
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*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Error_Probability_for_Baseband_Transmission|Error Probability for Baseband Transmission]].
 
   
 
   
*Zur Bestimmung von Fehlerwahrscheinlichkeiten können Sie das interaktive Applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Komplementäre Gaußsche Fehlerfunktionen]]  verwenden.
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*You can use the interactive applet  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Komplementäre Gaußsche Fehlerfunktionen]]  to determine error probabilities.
*Berücksichtigen Sie bei der Berechnung der Detektionsstörleistung das  [[Theory_of_Stochastic_Signals/Leistungsdichtespektrum_(LDS)#Theorem_von_Wiener-Chintchine|Theorem von Wiener–Chintchine]]:
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*Consider  [[Theory_of_Stochastic_Signals/Power-Spectral_Density#Wiener-Khintchine_Theorem|Wiener-Chintchine's theorem]] when calculating the detection noise power:
 
:$$ \sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
:$$ \sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
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===Fragebogen===
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===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
  
{Berechnen Sie für &nbsp;$\text{System A}$&nbsp; den Detektionsgrundimpuls &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$. <br>Welcher Wert &nbsp;$g_0 = g_{d}(t=0)$&nbsp; ergibt sich zum  Zeitpunkt &nbsp;$t = 0$?
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{Calculate for &nbsp;$\text{System A}$&nbsp; the basic detection pulse &nbsp;$g_{d}(t) =  g_{ s}(t) \star h_{\rm E}(t)$. <br>What value &nbsp;$g_0 = g_{d}(t=0)$&nbsp; results at time &nbsp;$t = 0$?
 
|type="{}"}
 
|type="{}"}
 
$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$
 
$g_0 \hspace{0.28cm} = \ $ { 6 3% } $\ \rm W^{1/2}$
  
{Berechnen Sie daraus die Detektionsstörleistung &nbsp;$σ_{d}^2$.
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{From this, calculate the detection noise power &nbsp;$σ_{d}^2$.
 
|type="{}"}
 
|type="{}"}
 
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $ { 1 3% } $\ \rm W$
 
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $ { 1 3% } $\ \rm W$
  
{Welche Bitfehlerwahrscheinlichkeit &nbsp;$p_{\rm B}$&nbsp; ergibt sich somit für das &nbsp;$\text{System A}$?
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{Thus, what is the bit error probability &nbsp;$p_{\rm B}$&nbsp; for &nbsp;$\text{System A}$?
 
|type="{}"}
 
|type="{}"}
 
$p_{\rm B} \hspace{0.2cm} = \ $ { 0.987 10% } $\ \cdot 10^{-9}$
 
$p_{\rm B} \hspace{0.2cm} = \ $ { 0.987 10% } $\ \cdot 10^{-9}$
  
{Ermitteln Sie die entsprechenden Größen für das &nbsp;$\text{System B}$&nbsp;.
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{Determine the corresponding quantities for &nbsp;$\text{System B}$&nbsp;.
 
|type="{}"}
 
|type="{}"}
 
$g_0 \hspace{0.28cm} = \ $ { 3 3% } $\ \rm W^{1/2}$
 
$g_0 \hspace{0.28cm} = \ $ { 3 3% } $\ \rm W^{1/2}$
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$p_{\rm B} \hspace{0.2cm} = \ $ { 0.135 10% } $\ \cdot 10^{-2}$
 
$p_{\rm B} \hspace{0.2cm} = \ $ { 0.135 10% } $\ \cdot 10^{-2}$
  
{Wie lauten die Kenngrößen für das &nbsp;$\text{System C}$&nbsp;?
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{What are the characteristics for &nbsp;$\text{System C}$&nbsp;?
 
|type="{}"}
 
|type="{}"}
 
$g_0 \hspace{0.28cm} = \ $ { 3 3% } $\ \rm W^{1/2}$
 
$g_0 \hspace{0.28cm} = \ $ { 3 3% } $\ \rm W^{1/2}$
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''&nbsp; Beim '''System A''' führt die Faltung der beiden gleich breiten Rechteckfunktionen $g_{s}(t)$ und $h_{\rm E}(t)$ zu einem dreieckförmigen Detektionsgrundimpuls mit dem Maximum bei $t = 0$:
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'''1.'''&nbsp; For '''System A''', the convolution of the two equal-width rectangular functions $g_{s}(t)$ and $h_{\rm E}(t)$ leads to a triangular basic detection pulse with the maximum at $t = 0$:
 
:$$g_d (t = 0)  =  \int_{ - T/2}^{
 
:$$g_d (t = 0)  =  \int_{ - T/2}^{
 
+ T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0
 
+ T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0
 
\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
 
\cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm
 
W}}}\hspace{0.05cm}.$$
 
W}}}\hspace{0.05cm}.$$
Es gibt keine Impulsinterferenzen, da für $| t |\ge T$ der Detektionsimpuls $g_{d}(t) = 0$ ist.
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There is no intersymbol interfering because for $| t |\ge T$ the detection pulse is $g_{d}(t) = 0$.
  
  
'''2.'''&nbsp; Die Varianz des Detektionsstörsignals &ndash; hier als Detektionsstörleistung bezeichnet &ndash; kann sowohl im Zeit&ndash; als auch im Frequenzbereich berechnet werden.  
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'''2.'''&nbsp; The variance of the noise component of the detection signal &ndash; referred to here as the detection noise power &ndash; can be calculated in both the time and frequency domains.
*Bei der vorliegenden Rechteckform führt die Berechnung im Zeitbereich schneller zum Ergebnis:
+
*For the present rectangular waveform, calculation in the time domain yields faster results:
 
:$$\sigma _d ^2  \ = \ \frac{N_0 }{2} \cdot \int_{ -
 
:$$\sigma _d ^2  \ = \ \frac{N_0 }{2} \cdot \int_{ -
 
\infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2
 
\infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2
Line 77: Line 77:
 
\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm
 
\,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm
 
W}}\hspace{0.05cm}.$$
 
W}}\hspace{0.05cm}.$$
*Die Frequenzbereichsberechnung würde mit $H_{\rm E}(f) = {\rm si}(&pi;fT)$ wie folgt aussehen:
+
*The frequency domain calculation would be as follows with $H_{\rm E}(f) = {\rm si}(&pi;fT)$:
 
:$$\sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
:$$\sigma _d ^2  = \frac{N_0 }{2} \cdot \int_{ - \infty }^{
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
 
+ \infty } {\left| {H_{\rm E}( f )} \right|^2
Line 85: Line 85:
  
  
'''3.'''&nbsp; Aufgrund der zeitlich begrenzten Impulsform (das bedeutet: keine Impulsinterferenzen!) ergibt sich bei der hier vorausgesetzten bipolaren Betrachtungsweise:
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'''3.'''&nbsp; Due to the time-limited pulse shape (this means: no intersymbol interfering!), the bipolar approach assumed here yields:
 
:$$p_{\rm B} =  {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)
 
:$$p_{\rm B} =  {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)
 
  = {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$
 
  = {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$
'''System A''' stellt die Matched&ndash;Filter&ndash;Realisierung des optimalen Binärempfängers dar, so dass auch folgende Gleichungen anwendbar wären:
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'''System A''' represents the matched filter realization of the optimal binary receiver, so the following equations would also be applicable:
 
:$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}
 
:$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm}
 
\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)
 
\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right)
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'''4.'''&nbsp; Da bei '''System B''' das gleiche Empfangsfilter wie bei '''System A''' verwendet wird, erhält man auch die gleiche Detektionsstörleistung $&sigma;_{d}^2 = 1 \ \rm W$.  
+
'''4.'''&nbsp; Since '''System B''' uses the same receiver filter as '''System A''', the same detection noise power $&sigma;_{d}^2 = 1 \ \rm W$ is also obtained.
*Der Detektionsgrundimpuls ist nun aber nicht mehr dreieckförmig, sondern weist eine spitzere Form auf. Zum Zeitpunkt $t = 0$ gilt:
+
*However, the basic detection pulse is now no longer triangular, but has a more pointed shape. At time $t = 0$ applies:
 
:$$g_d (t = 0)  = \frac{1}{T} \cdot  \int_{ - T/2}^{
 
:$$g_d (t = 0)  = \frac{1}{T} \cdot  \int_{ - T/2}^{
 
+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
 
+ T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot
 
\frac{s_0 }{2}  \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 
\frac{s_0 }{2}  \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 
W}}\hspace{0.05cm}.$$
 
W}}\hspace{0.05cm}.$$
*Auch das '''System B''' ist impulsinterferenzfrei. Man erhält deshalb für die Bitfehlerwahrscheinlichkeit:
+
*'''System B''' is also free of intersymbol interfering. Therefore, one obtains for the bit error probability:
 
:$$p_{\rm B} =  {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)
 
:$$p_{\rm B} =  {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right)
 
  = {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
 
  = {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
*Nicht anwendbar ist dagegen hier der folgende Rechengang:
+
*On the other hand, the following calculation is not applicable here:
 
:$$E_{\rm B} =    \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm
 
:$$E_{\rm B} =    \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm
 
  d}t =  2\cdot s_0^2 \cdot  \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm
 
  d}t =  2\cdot s_0^2 \cdot  \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm
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  ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}
 
  ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4}
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
*Man würde so eine zu niedrige Bitfehlerwahrscheinlichkeit berechnen, da die implizit getroffene Annahme eines angepassten Filters nicht zutrifft.
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*One would thus compute a bit error probability that is too low, since the implicit assumption of a matched filter does not hold.
  
  
'''5.'''&nbsp; Bei rechteckförmigem Sendegrundimpuls und dreieckförmiger Impulsantwort &nbsp; &rArr; &nbsp; '''System C''' erhält man den gleichen Detektionsgrundimpuls wie bei dreieckförmigem $g_{\rm s}(t)$ und rechteckförmigem $h_{\rm E}(t)$.  
+
'''5.'''&nbsp; For rectangular basic transmission pulse and triangular impulse response &nbsp; &rArr; &nbsp; '''System C''', the same basic detection pulse is obtained as for triangular $g_{\rm s}(t)$ and rectangular $h_{\rm E}(t)$.  
*Wie beim '''System B''' gilt deshalb:
+
*Therefore, as in '''System B''':
 
:$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 
:$$g_d (t = 0)  =  \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm
 
W}}\hspace{0.05cm}.$$
 
W}}\hspace{0.05cm}.$$
*Dagegen ist nun die Detektionsstörleistung kleiner als bei den Systemen A und B:
+
*In contrast, the detection noise power is now smaller than in systems A and B:
 
:$$\sigma _d ^2 =  \frac{N_0}{2}  \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm
 
:$$\sigma _d ^2 =  \frac{N_0}{2}  \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm
 
  d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
 
  d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
*Damit erhält man nun für die Bitfehlerwahrscheinlichkeit:
+
*This now gives us for the bit error probability:
 
:$$p_{\rm B} =    {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)
 
:$$p_{\rm B} =    {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right)
 
  \approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx  10^{-7} } \hspace{0.05cm}.$$
 
  \approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx  10^{-7} } \hspace{0.05cm}.$$
*Der gegenüber Teilfrage '''(3)''' erkennbare Anstieg der Fehlerwahrscheinlichkeit um etwa den Faktor $100$ ist auf die gravierende Fehlanpassung gegenüber dem Matched&ndash;Filter zurückzuführen.  
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*The apparent increase in error probability by a factor of about $100$ compared to subtask '''(3)''' is due to the severe mismatch compared to the matched filter.
*Die Verbesserung gegenüber Teilaufgabe '''(4)''' geht auf die höhere Signalenergie zurück.
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*The improvement over subtask '''(4)''' is due to the higher signal energy.
  
 
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{{ML-Fuß}}

Revision as of 17:46, 2 March 2022

Three different system concepts

We consider here three variants of a binary bipolar AWGN transmission system which differ with respect to the basic transmission pulse  $g_{s}(t)$  as well as the impulse response  $h_{\rm E}(t)$  of the receiver filter:

  • For  $\text{System A}$,  both  $g_{s}(t)$  and  $h_{\rm E}(t)$  are rectangular, only the pulse heights  $(s_{\rm 0}$  and  $1/T)$  are different.
  •  $\text{System B}$  differs from  $\text{System A}$  by having a triangular-shaped basic transmission pulse with  $g_{s}(t=0) = s_{\rm 0}$.
  •  $\text{System C}$  has the same basic transmission pulse as  $\text{System A}$, while the impulse response is triangular with  $h_{\rm E}(t=0) = 1/T$. 


The absolute width of the rectangular and triangular functions considered here is  $T = 10 \ \rm µ s$ each. The bit rate is  $R = 100 \ \rm kbit/s$. The other system parameters are given as follows:

$$s_0 = 6 \,\,\sqrt{W}\hspace{0.05cm},\hspace{0.3cm} N_{\rm 0} = 2 \cdot 10^{-5} \,\,{\rm W/Hz}\hspace{0.05cm}.$$



Notes:

$$ \sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {H_{\rm E}( f )} \right|^2 \hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2 \hspace{0.1cm}{\rm{d}}t}\hspace{0.05cm}.$$


Questions

1

Calculate for  $\text{System A}$  the basic detection pulse  $g_{d}(t) = g_{ s}(t) \star h_{\rm E}(t)$.
What value  $g_0 = g_{d}(t=0)$  results at time  $t = 0$?

$g_0 \hspace{0.28cm} = \ $

$\ \rm W^{1/2}$

2

From this, calculate the detection noise power  $σ_{d}^2$.

$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $

$\ \rm W$

3

Thus, what is the bit error probability  $p_{\rm B}$  for  $\text{System A}$?

$p_{\rm B} \hspace{0.2cm} = \ $

$\ \cdot 10^{-9}$

4

Determine the corresponding quantities for  $\text{System B}$ .

$g_0 \hspace{0.28cm} = \ $

$\ \rm W^{1/2}$
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $

$\ \rm W$
$p_{\rm B} \hspace{0.2cm} = \ $

$\ \cdot 10^{-2}$

5

What are the characteristics for  $\text{System C}$ ?

$g_0 \hspace{0.28cm} = \ $

$\ \rm W^{1/2}$
$σ_{d}^{\hspace{0.02cm}2} \hspace{0.2cm} = \ $

$\ \rm W$
$p_{\rm B} \hspace{0.2cm} = \ $

$\ \cdot 10^{-7}$


Solution

1.  For System A, the convolution of the two equal-width rectangular functions $g_{s}(t)$ and $h_{\rm E}(t)$ leads to a triangular basic detection pulse with the maximum at $t = 0$:

$$g_d (t = 0) = \int_{ - T/2}^{ + T/2} { g_s(t) \cdot h_{\rm E}( t )} \hspace{0.1cm}{\rm{d}}t =s_0 \cdot \frac{1 }{T} \cdot T = s_0 \hspace{0.1cm}\underline { = 6 \,\,\sqrt{{\rm W}}}\hspace{0.05cm}.$$

There is no intersymbol interfering because for $| t |\ge T$ the detection pulse is $g_{d}(t) = 0$.


2.  The variance of the noise component of the detection signal – referred to here as the detection noise power – can be calculated in both the time and frequency domains.

  • For the present rectangular waveform, calculation in the time domain yields faster results:
$$\sigma _d ^2 \ = \ \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {h_{\rm E}( t )} \right|^2 \hspace{0.1cm}{\rm{d}}t} =\frac{N_0 }{2} \cdot \int_{ - T/2 }^{ + T/2 } {\left| {h_{\rm E}( t )} \right|^2 \hspace{0.1cm}{\rm{d}}t} = \ \frac{N_0 }{2} \cdot\frac{1 }{T^2} \cdot T = \frac{N_0 }{2T} = \frac{2 \cdot 10^{-5} \,\,{\rm W/Hz}}{2 \cdot 10^{-5} \,\,{\rm s}} \hspace{0.1cm}\underline {= 1\,{\rm W}}\hspace{0.05cm}.$$
  • The frequency domain calculation would be as follows with $H_{\rm E}(f) = {\rm si}(πfT)$:
$$\sigma _d ^2 = \frac{N_0 }{2} \cdot \int_{ - \infty }^{ + \infty } {\left| {H_{\rm E}( f )} \right|^2 \hspace{0.1cm}{\rm{d}}f} = \frac{N_0 }{2} \cdot \int_{- \infty }^{ \infty } {\rm si}^2(\pi f T)\hspace{0.1cm}{\rm{d}}f = \frac{N_0 }{2T} \hspace{0.05cm}.$$


3.  Due to the time-limited pulse shape (this means: no intersymbol interfering!), the bipolar approach assumed here yields:

$$p_{\rm B} = {\rm Q} \left( \frac{s_0}{\sigma_d}\right)= {\rm Q} \left( \frac{ 6 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right) = {\rm Q}(6) \hspace{0.1cm}\underline {= 0.987 \cdot 10^{-9}} \hspace{0.05cm}.$$

System A represents the matched filter realization of the optimal binary receiver, so the following equations would also be applicable:

$$E_{\rm B} = s_0^2 \cdot T = 36\, {\rm W} \cdot 10^{-5} {\rm s}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right) ={\rm Q} \left( \sqrt{\frac{2 \cdot 36 \cdot 10^{-5}\,\, {\rm Ws}}{2 \cdot 10^{-5} \,\, {\rm Ws}}}\right)={\rm Q}(6) \hspace{0.05cm}.$$


4.  Since System B uses the same receiver filter as System A, the same detection noise power $σ_{d}^2 = 1 \ \rm W$ is also obtained.

  • However, the basic detection pulse is now no longer triangular, but has a more pointed shape. At time $t = 0$ applies:
$$g_d (t = 0) = \frac{1}{T} \cdot \int_{ - T/2}^{ + T/2} { g_s(t) } \hspace{0.1cm}{\rm{d}}t = \frac{1}{T} \cdot \frac{s_0 }{2} \cdot T = \frac{s_0 }{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm W}}\hspace{0.05cm}.$$
  • System B is also free of intersymbol interfering. Therefore, one obtains for the bit error probability:
$$p_{\rm B} = {\rm Q} \left( \frac{g_d (t = 0)}{\sigma_d}\right)= {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{1 \,\sqrt{\rm W}}\right) = {\rm Q}(3) \hspace{0.1cm}\underline {= 0.135 \cdot 10^{-2}} \hspace{0.05cm}.$$
  • On the other hand, the following calculation is not applicable here:
$$E_{\rm B} = \int^{+\infty} _{-\infty} g_s^2(t)\,{\rm d}t = 2\cdot s_0^2 \cdot \int ^{+T/2} _{0} \left( 1- \frac{2t}{T}\right)^2\,{\rm d}t = \frac{s_0^2 \cdot T }{3} = 12 \cdot 10^{-5} \,{\rm Ws}$$
$$\Rightarrow \hspace{0.3cm} p_{\rm B} = {\rm Q} \left( \sqrt{\frac{2 \cdot E_{\rm B}}{N_0}}\right) ={\rm Q} \left( \sqrt{12}\right)={\rm Q}(3.464) \approx 3 \cdot 10^{-4} \hspace{0.05cm}.$$
  • One would thus compute a bit error probability that is too low, since the implicit assumption of a matched filter does not hold.


5.  For rectangular basic transmission pulse and triangular impulse response   ⇒   System C, the same basic detection pulse is obtained as for triangular $g_{\rm s}(t)$ and rectangular $h_{\rm E}(t)$.

  • Therefore, as in System B:
$$g_d (t = 0) = \frac{s_0}{2}\hspace{0.1cm}\underline {= 3 \,\,\sqrt{\rm W}}\hspace{0.05cm}.$$
  • In contrast, the detection noise power is now smaller than in systems A and B:
$$\sigma _d ^2 = \frac{N_0}{2} \cdot \frac{1}{T^2} \cdot \int^{+T/2} _{-T/2} \left( 1- \frac{2t}{T}\right)^2\,{\rm d}t = \frac{N_0}{6T}\hspace{0.1cm}\underline { = 0.333 \,{\rm W}}.$$
  • This now gives us for the bit error probability:
$$p_{\rm B} = {\rm Q} \left( \frac{ 3 \,\sqrt{\rm W}}{0.577 \,\sqrt{\rm W}}\right) \approx {\rm Q}(5.2)\hspace{0.1cm}\underline { \approx 10^{-7} } \hspace{0.05cm}.$$
  • The apparent increase in error probability by a factor of about $100$ compared to subtask (3) is due to the severe mismatch compared to the matched filter.
  • The improvement over subtask (4) is due to the higher signal energy.