Difference between revisions of "Aufgaben:Exercise 4.7: Spectra of ASK and BPSK"

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{{quiz-Header|Buchseite=Modulationsverfahren/Lineare digitale Modulationsverfahren
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{{quiz-Header|Buchseite=Modulation_Methods/Linear_Digital_Modulation
 
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[[File:P_ID1701__Mod_A_4_6.png|right|]]
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[[File:P_ID1701__Mod_A_4_6.png|right|frame|Power-spectral densities of  $q(t)$  and  $s(t)$   – valid for ASK and BPSK]]
Die Sendesignale von ASK (Amplitude Shift Keying) und BPSK (Binary Phase Shift Keying) können beide in der Form $s(t) = q(t) · z(t)$ dargestellt werden, wobei z(t) eine harmonische Schwingung mit der Frequenz $f_T$ und der Amplitude 1 darstellt. Die Trägerphase $ϕ_T$ ist für die hier betrachteten Leistungsdichtespektren nicht von Bedeutung.
+
The transmitted signals of  $\rm ASK$  ("Amplitude Shift Keying")  and  $\rm BPSK$  ("Binary Phase Shift Keying")  can both be expressed in the form
 +
:$$s(t) = q(t) · z(t),$$
 +
where the carrier  $z(t)$  represents a harmonic oscillation with frequency  $f_{\rm T}$  and amplitude  $1$.  The carrier phase  $ϕ_{\rm T}$  is not important for the power-spectral densities considered here.
  
Bei ASK sind unipolare Amplitudenkoeffizienten – das heißt: $a_ν ∈ {0, 1}$ – des Quellensignals
+
*In each case,  the source is redundancy-free,  which means that the two possible symbols are equally probable and the symbols are statistically independent of each other.
$$ q(t) = \sum_{\nu = - \infty}^{+\infty}a_\nu \cdot g_q (t - \nu \cdot T)$$
+
*For ASK,  the unipolar amplitude coefficients – that is:  $a_ν ∈ \{0, 1\}$  – of the source signal are
anzusetzen, während im Fall der BPSK $a_ν$ ∈ {–1, +1} zu berücksichtigen ist. Die Quelle ist jeweils redundanzfrei, was bedeutet, dass die beiden möglichen Symbole ±1 gleichwahrscheinlich sind und die Symbole statistisch voneinander unabhängig.
+
:$$ q(t) = \sum_{\nu = - \infty}^{+\infty}a_\nu \cdot g_q (t - \nu \cdot T)$$
 +
:while in the case of BPSK  $a_ν ∈ \{-1, +1\}$  has to be considered.
  
In der Grafik sind die Leistungsdichtespektren $Φ_q(f)$ und $Φ_s(f)$ von Quellensignal und Sendesignal angegeben, die sich bei einem NRZ–Rechteckimpuls $g_q(t)$ mit der Amplitude $s_0 = 2 V$ und der Dauer $T = 1 μs$ ergeben. Damit lautet die Spektralfunktion:
 
$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$
 
Zu bestimmen sind in dieser Aufgabe die Konstanten A, B, C und D für ASK und BPSK.
 
  
'''Hinweis:''' Die Aufgabe bezieht sich auf das [http://en.lntwww.de/Modulationsverfahren/Lineare_digitale_Modulationsverfahren Kapitel 4.2] dieses Buches sowie auf das Kapitel 2.1 im Buch „Digitalsignalübertragung”.
+
In the diagram,  the power-spectral densities  ${\it Φ}_q(f)$  and  ${\it Φ}_s(f)$  of the source signal and the transmitted signal are given,  respectively,  for an NRZ rectangular pulse  $g_q(t)$  with amplitude  $s_0 = 2 \ \rm V$  and duration  $T = 1 \ \rm µ s$.  Thus the spectral function is:
===Fragebogen===
+
:$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$
 +
The constants  $A$,  $B$,  $C$  and  $D$  for the  $\rm ASK$  and  $\rm BPSK$ modulation methods are to be determined.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Notes:
 +
*The exercise belongs to the chapter  [[Modulation_Methods/Linear_Digital_Modulation|Linear Digital Modulation]].
 +
*Reference is also made to the chapter   [[Digital_Signal_Transmission/Grundlagen_der_codierten_Übertragung|Basics of Coded Transmission]]   in the book  "Digital Signal Transmission".
 +
*The powers are to be specified in   $\rm V^2$;  they thus refer to the reference resistance  $R = 1 \ \rm \Omega$.
 +
 +
 
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß sind der Parameter $A = Φ_q(f = 0)$ und das Diracgewicht B bei ASK?
+
{Which values result with ASK for the parameters &nbsp;$A = {\it Φ}_q(f = 0)$&nbsp; and &nbsp;$B$&nbsp; $($Dirac weight at &nbsp;$f = 0)$?
 
|type="{}"}
 
|type="{}"}
$ASK: A$ = { 1 3% } $10^{-6}$ $V^2$
+
$A \ = \ $ { 1 3% } $\ \cdot 10^{-6} \  \rm V^2/Hz$
$B$ = { 1 35 } $V^2$  
+
$B \ = \ $ { 1 3% } $\ \rm V^2$  
  
  
{Bestimmen Sie die Parameter $C = Φ_s(f = f_T)$ und D des ASK–Sendesignals.
+
{Determine the parameters &nbsp;$C = {\it Φ}_s(f = f_{\rm T})$&nbsp; and &nbsp;$D$&nbsp;  $($Dirac weight at $f = f_{\rm T})$ for the ASK transmitted signal.
 
|type="{}"}
 
|type="{}"}
$ASK:  C$ = { 0.25 3% } $10^{-6}$ $V^2/Hz$  
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$C \ = \ $ { 0.25 3% } $\ \cdot 10^{-6} \  \rm V^2/Hz$
$D$ = { 0.25 3% } $V^2$  
+
$D \ = \ $ { 0.25 3% } $\ \rm V^2$  
  
{Wie groß sind die Parameter $A = Φ_q(f = 0)$ und B bei BPSK?
+
{What are the values for parameters &nbsp;$A$&nbsp; and &nbsp;$B$&nbsp; for BPSK?
 
|type="{}"}
 
|type="{}"}
$BPSK:  A$ = { 4 3% }  $10^{-6}$ $V^2/Hz$  
+
$A \ = \ $ { 4 3% }  $\ \cdot 10^{-6} \  \rm V^2/Hz$
$B$ = { 0 3% } $V^2$  
+
$B \ = \ $ { 0. } $\ \rm V^2$  
  
{Bestimmen Sie die Parameter $C = Φ_s(f = f_T)$ und D des BPSK–Sendedsignals.
+
{What are the values of parameters &nbsp;$C$&nbsp; and &nbsp;$D$&nbsp; for BPSK?
 
|type="{}"}
 
|type="{}"}
$BPSK:  C$ = { 1 3% } $10^{-6}$ $V^2/Hz$  
+
$C \ = \ $ { 1 3% } $\ \cdot 10^{-6} \  \rm V^2/Hz$  
$D$ = { 0 3% }  $V^2$  
+
$D \ = \ $ { 0. }  $\ \rm V^2$
  
{Welche Aussagen treffen zu, auch wenn $g_q(t)$ kein NRZ–Rechteckimpuls ist?
+
{Which statements are always true  for BPSK even if &nbsp;$g_q(t)$&nbsp; is not an NRZ rectangular pulse?
|type="{}"}
+
|type="[]"}
+ Der kontinuierliche Anteil von $Φ_q(f)$ ist formgleich mit $|Gq(f)|^2$.
+
+ The continuous part of &nbsp;$ {\it Φ}_q(f)$&nbsp; is equal in form to &nbsp;$|G_q(f)|^2$.
- $Φ_q(f)$ beinhaltet bei ASK genau eine Diraclinie bei $f = 0$.
+
- ${\it Φ}_q(f)$&nbsp; contains a single Dirac line at ASK&nbsp; for BPSK $($at $f = 0)$.
- $Φ_q(f)$ beinhaltet bei BPSK genau eine Diraclinie bei $f = 0$.
+
- ${\it Φ}_q(f)$ contains a single Dirac line at BPSK&nbsp; for BPSK $($at $f = 0)$.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.''' Der Gleichanteil des unipolaren redundanzfreien Quellensignals beträgt $m_q = s_0/2$. Das Diracgewicht ist somit $B = m_q^2 = s_0^2/4 = 1 V^2$. Ohne diesen Gleichanteil ergibt sich das stochastische Rechtecksignal $q(t) m_q$ ∈ {$+s_0/2$, $–s_0/2$}. Dieses gleichsignalfreie Signal besitzt den kontinuierlichen LDS–Anteil $(s_0/2)^2 · T · si^2(πfT)$, woraus der gesuchte Wert bei der Frequenz f = 0 ermittelt werden kann:
+
'''(1)'''&nbsp; The DC component of the unipolar redundancy-free source signal is&nbsp; $m_q = s_0/2$.&nbsp; Thus,&nbsp; the Dirac weight is&nbsp; $B = m_q^2 = s_0^2/4\hspace{0.15cm}\underline{ = 1 \ \rm V^2}$.  
 +
 
 +
*Without this DC component,&nbsp; the stochastic  signal&nbsp; $q(t) - m_q ∈ \{+s_0/2, -s_0/2\}$ would be obtained.
 +
*This DC-free signal has the continuous PSD component&nbsp; $(s_0/2)^2 · T · {\rm sinc}^2(fT)$.
 +
*From this,&nbsp; the value we are looking for at frequency&nbsp; $f = 0$&nbsp; can be determined:
 +
:$$A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot
 +
10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.$$
 +
 
 +
 
  
 +
'''(2)'''&nbsp; The spectrum&nbsp; $Z(f)$&nbsp; of a cosine signal&nbsp; $z(t)$&nbsp; consists of two Dirac delta functions at&nbsp; $\pm f_{\rm T}$,&nbsp; each with weight&nbsp; $1/2$.
 +
*The power-spectral density&nbsp; ${\it Φ}_z(f)$&nbsp; also consists of the two Dirac delta  functions,&nbsp; but now with respective weights&nbsp; $1/4$.
 +
*The convolution&nbsp; ${\it Φ}_q(f) ∗ {\it Φ}_z(f)$&nbsp; gives the power-spectral density&nbsp; ${\it Φ}_s(f)$&nbsp; of the transmitted signal.&nbsp; It follows that:
 +
:$$C =  {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm
 +
V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.$$
  
$ A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.$
+
Note: &nbsp; The power per bit is obtained as the integral over&nbsp; ${\it Φ}_s(f)$:
 +
:$$P_{\rm S}  = \int_{ - \infty }^\infty \hspace{-0.3cm}  {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f
 +
= 2 \cdot \int_{ 0 }^\infty \hspace{-0.3cm}  {\left [ C \cdot {\rm sinc}^2(f T) + D \cdot \delta (f - f_{\rm T}]\right ]}\hspace{0.1cm} {\rm
 +
d}f= 2 \cdot \left [ \frac{C}{T}  + D \right ] =
 +
2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm
 +
V^{2}/Hz}}{10^{-6} \,{\rm s}} + 0.25 \,{\rm V^{2}} \right ] \hspace{0.15cm}\underline {= 1
 +
\,{\rm V^{2}}}.$$
  
  
'''2.''' Das Spektrum $Z(f)$ eines Cosinussignals $z(t)$ besteht aus zwei Diracfunktionen bei $± f_T$, jeweils mit dem Gewicht 1/2. Das Leistungsdichtespektrum $Φ_z(f)$ besteht ebenfalls aus den beiden Diracfunktionen, nun aber mit jeweiligem Gewicht 1/4. Die Faltung $Φ_q(f) ∗ Φ_z(f)$ ergibt das Leistungsdichtespektrum $Φ_s(f)$ des Sendesignals. Daraus folgt:
 
  
'''Anmerkung:''' Die Leistung pro Bit ergibt sich als das Integral über $Φ_s(f)$:
+
'''(3)'''&nbsp; For BPSK, the source signal&nbsp; $q(t)$&nbsp; is to be bipolar.
 +
*Therefore,&nbsp; the Dirac line is missing in the power-spectral density&nbsp;  ⇒  &nbsp; $\underline{B = 0}$.
 +
*  The continuous PSD component is four times larger than with ASK:
 +
:$$A =  {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.$$
  
  
  
'''3.''' Bei BPSK ist das Quellensignal $q(t)$ bipolar anzusetzen. Im Leistungsdichtespektrum fehlt deshalb die Diraclinie  ⇒  B = 0   und der kontinuierliche LDS–Anteil ist viermal so groß als bei der ASK:
+
'''(4)'''&nbsp; For the PSD parameters of the BPSK transmitted signal,&nbsp; the following applies analogously to the ASK:
 +
:$$C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D =
 +
\frac {B}{4} \hspace{0.15cm}\underline {= 0}.$$
  
  
'''4.''' Für die LDS–Parameter des BPSK–Sendesignals gilt analog zur ASK:
 
  
'''5.'''  Richtig ist nur die erste Aussage. Bei BPSK (bipolares Quellensignal) beinhaltet $Φ_q(f)$ auch dann keine einzige Diraclinie, wenn $g_q(t)$ von der Rechteckform abweicht. Dagegen beinhaltet das unipolare ASK–Quellensignal unendlich viele Diraclinien bei allen Vielfachen von 1/T. Weitere Informationen hierzu finden Sie bei AKF und LDS bei unipolaren Binärsignalen im Buch „Digitalsignalübertragung”.
+
'''(5)'''&nbsp; Only the&nbsp; <u>first statement</u>&nbsp; is correct:
 +
* For BPSK,&nbsp; ${\it Φ}_q(f)$&nbsp; does not contain a single Dirac line even if&nbsp; $g_q(t)$&nbsp; deviates from the rectangular form&nbsp; (assuming equally probable symbols).  
 +
*In contrast, the unipolar ASK source signal contains infinitely many Dirac lines at all multiples of&nbsp; $1/T$.  
 +
*For more information on this topic, see the page&nbsp;  "ACF and PSD for unipolar binary signals" in the book "Digital Signal Transmission".
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Modulationsverfahren|^4.2 Lineare digitale Modulationsverfahren^]]
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[[Category:Modulation Methods: Exercises|^4.2 Linear Digital Modulation^]]

Latest revision as of 14:31, 12 April 2022

Power-spectral densities of  $q(t)$  and  $s(t)$  – valid for ASK and BPSK

The transmitted signals of  $\rm ASK$  ("Amplitude Shift Keying")  and  $\rm BPSK$  ("Binary Phase Shift Keying")  can both be expressed in the form

$$s(t) = q(t) · z(t),$$

where the carrier  $z(t)$  represents a harmonic oscillation with frequency  $f_{\rm T}$  and amplitude  $1$.  The carrier phase  $ϕ_{\rm T}$  is not important for the power-spectral densities considered here.

  • In each case,  the source is redundancy-free,  which means that the two possible symbols are equally probable and the symbols are statistically independent of each other.
  • For ASK,  the unipolar amplitude coefficients – that is:  $a_ν ∈ \{0, 1\}$  – of the source signal are
$$ q(t) = \sum_{\nu = - \infty}^{+\infty}a_\nu \cdot g_q (t - \nu \cdot T)$$
while in the case of BPSK  $a_ν ∈ \{-1, +1\}$  has to be considered.


In the diagram,  the power-spectral densities  ${\it Φ}_q(f)$  and  ${\it Φ}_s(f)$  of the source signal and the transmitted signal are given,  respectively,  for an NRZ rectangular pulse  $g_q(t)$  with amplitude  $s_0 = 2 \ \rm V$  and duration  $T = 1 \ \rm µ s$.  Thus the spectral function is:

$$G_q(f) = s_0 \cdot T \cdot {\rm si}(\pi f T)\hspace{0.05cm}.$$

The constants  $A$,  $B$,  $C$  and  $D$  for the  $\rm ASK$  and  $\rm BPSK$ modulation methods are to be determined.




Notes:

  • The exercise belongs to the chapter  Linear Digital Modulation.
  • Reference is also made to the chapter  Basics of Coded Transmission  in the book  "Digital Signal Transmission".
  • The powers are to be specified in  $\rm V^2$;  they thus refer to the reference resistance  $R = 1 \ \rm \Omega$.



Questions

1

Which values result with ASK for the parameters  $A = {\it Φ}_q(f = 0)$  and  $B$  $($Dirac weight at  $f = 0)$?

$A \ = \ $

$\ \cdot 10^{-6} \ \rm V^2/Hz$
$B \ = \ $

$\ \rm V^2$

2

Determine the parameters  $C = {\it Φ}_s(f = f_{\rm T})$  and  $D$  $($Dirac weight at $f = f_{\rm T})$ for the ASK transmitted signal.

$C \ = \ $

$\ \cdot 10^{-6} \ \rm V^2/Hz$
$D \ = \ $

$\ \rm V^2$

3

What are the values for parameters  $A$  and  $B$  for BPSK?

$A \ = \ $

$\ \cdot 10^{-6} \ \rm V^2/Hz$
$B \ = \ $

$\ \rm V^2$

4

What are the values of parameters  $C$  and  $D$  for BPSK?

$C \ = \ $

$\ \cdot 10^{-6} \ \rm V^2/Hz$
$D \ = \ $

$\ \rm V^2$

5

Which statements are always true for BPSK even if  $g_q(t)$  is not an NRZ rectangular pulse?

The continuous part of  $ {\it Φ}_q(f)$  is equal in form to  $|G_q(f)|^2$.
${\it Φ}_q(f)$  contains a single Dirac line at ASK  for BPSK $($at $f = 0)$.
${\it Φ}_q(f)$ contains a single Dirac line at BPSK  for BPSK $($at $f = 0)$.


Solution

(1)  The DC component of the unipolar redundancy-free source signal is  $m_q = s_0/2$.  Thus,  the Dirac weight is  $B = m_q^2 = s_0^2/4\hspace{0.15cm}\underline{ = 1 \ \rm V^2}$.

  • Without this DC component,  the stochastic signal  $q(t) - m_q ∈ \{+s_0/2, -s_0/2\}$ would be obtained.
  • This DC-free signal has the continuous PSD component  $(s_0/2)^2 · T · {\rm sinc}^2(fT)$.
  • From this,  the value we are looking for at frequency  $f = 0$  can be determined:
$$A = \frac {s_0^2 \cdot T }{4} = \frac {(2\,{\rm V})^2 \cdot 10^{-6} \,{\rm s}}{4}\hspace{0.15cm}\underline {= 10^{-6} \,{\rm V^{2}/Hz}}.$$


(2)  The spectrum  $Z(f)$  of a cosine signal  $z(t)$  consists of two Dirac delta functions at  $\pm f_{\rm T}$,  each with weight  $1/2$.

  • The power-spectral density  ${\it Φ}_z(f)$  also consists of the two Dirac delta functions,  but now with respective weights  $1/4$.
  • The convolution  ${\it Φ}_q(f) ∗ {\it Φ}_z(f)$  gives the power-spectral density  ${\it Φ}_s(f)$  of the transmitted signal.  It follows that:
$$C = {A}/{4} \hspace{0.15cm}\underline { = 0.25 \cdot 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = {B}/{4}\hspace{0.15cm}\underline { = 0.25 \,{\rm V^{2}}}.$$

Note:   The power per bit is obtained as the integral over  ${\it Φ}_s(f)$:

$$P_{\rm S} = \int_{ - \infty }^\infty \hspace{-0.3cm} {{\it \Phi}_s(f)}\hspace{0.1cm} {\rm d}f = 2 \cdot \int_{ 0 }^\infty \hspace{-0.3cm} {\left [ C \cdot {\rm sinc}^2(f T) + D \cdot \delta (f - f_{\rm T}]\right ]}\hspace{0.1cm} {\rm d}f= 2 \cdot \left [ \frac{C}{T} + D \right ] = 2 \cdot \left [ \frac{0.25 \cdot 10^{-6} \,{\rm V^{2}/Hz}}{10^{-6} \,{\rm s}} + 0.25 \,{\rm V^{2}} \right ] \hspace{0.15cm}\underline {= 1 \,{\rm V^{2}}}.$$


(3)  For BPSK, the source signal  $q(t)$  is to be bipolar.

  • Therefore,  the Dirac line is missing in the power-spectral density  ⇒   $\underline{B = 0}$.
  • The continuous PSD component is four times larger than with ASK:
$$A = {s_0^2 \cdot T }\hspace{0.15cm}\underline { = 4 \cdot 10^{-6} \,{\rm V^{2}/Hz}}.$$


(4)  For the PSD parameters of the BPSK transmitted signal,  the following applies analogously to the ASK:

$$C = \frac {A}{4}\hspace{0.15cm}\underline { = 10^{-6} \,{\rm V^{2}/Hz}},\hspace{0.2cm}D = \frac {B}{4} \hspace{0.15cm}\underline {= 0}.$$


(5)  Only the  first statement  is correct:

  • For BPSK,  ${\it Φ}_q(f)$  does not contain a single Dirac line even if  $g_q(t)$  deviates from the rectangular form  (assuming equally probable symbols).
  • In contrast, the unipolar ASK source signal contains infinitely many Dirac lines at all multiples of  $1/T$.
  • For more information on this topic, see the page  "ACF and PSD for unipolar binary signals" in the book "Digital Signal Transmission".