Difference between revisions of "Aufgaben:Exercise 2.6Z: Synchronous Demodulator"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Lineare Verzerrungen
+
{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Linear_Distortions
 
}}
 
}}
  
[[File:P_ID913__LZI_Z_2_6_neu.png|right|frame|AM–Modulator (oben) & Synchrondemodulator (unten)]]
+
[[File:P_ID913__LZI_Z_2_6_neu.png|right|frame|Amplitude modulator (top), <br>synchronous demodulator]]
Das dargestellte Blockschaltbild zeigt ein Übertragungssystem mit [[Modulationsverfahren/Zweiseitenband-Amplitudenmodulation|Zweiseitenband-Amplitudenmodulation]] (ZSB-AM) und [[Modulationsverfahren/Synchrondemodulation|Synchrondemodulator]] (SD). Das Quellensignal bestehe aus zwei harmonischen Schwingungen mit den Frequenzen $f_2 = 2 \ \rm kHz$ und $f_5 = 5 \ \rm kHz$:
+
The depicted block diagram shows a transmission system
 +
*with&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]]&nbsp; $\rm(DSB\hspace{0.03cm}&ndash;\hspace{-0.1cm}AM)$
 +
*and&nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]]&nbsp; $\rm (SD)$.  
 +
 
 +
 
 +
Let the source signal consist of two harmonic oscillations with frequencies&nbsp; $f_2 = 2 \ \rm kHz$&nbsp; and&nbsp; $f_5 = 5 \ \rm kHz$:
 
:$$q(t) = {2 \, \rm V} \cdot {\rm cos}(\omega_2  t )+ {1 \, \rm V}
 
:$$q(t) = {2 \, \rm V} \cdot {\rm cos}(\omega_2  t )+ {1 \, \rm V}
 
\cdot {\rm sin}(\omega_5  t ) .$$
 
\cdot {\rm sin}(\omega_5  t ) .$$
  
*Dieses Signal wird mit dem dimensionslosen Trägersignal $z(t) = \cos(\omega_{\rm T} \cdot T)$ der Trägerfrequenz $f_{\rm T} = 50 \ \rm kHz$ multipliziert. Bei ZSB&ndash;AM ist der gestrichelt eingezeichnete Block unerheblich, so dass für das Sendesignal gilt:
+
*This signal is multiplied by the dimensionless carrier signal&nbsp; $z(t) = \cos(\omega_{\rm T} \cdot T)$&nbsp; of carrier frequency&nbsp; $f_{\rm T} = 50 \ \rm kHz$. <br>For DSB&ndash;AM, the dashed block is irrelevant so that the following holds for the transmission signal:
 
:$$s(t) = q(t) \cdot  {\rm cos}(\omega_{\rm T}  t ) .$$
 
:$$s(t) = q(t) \cdot  {\rm cos}(\omega_{\rm T}  t ) .$$
  
*Im Synchrondemodulator wird das Empfängersignal $r(t)$, das bei idealem Kanal identisch mit $s(t)$ ist, mit dem empfangsseitigem Trägersignal $z_{\rm E}(t)$ multipliziert, wobei gilt:
+
*In the synchronous demodulator, the received signal&nbsp; $r(t)$&nbsp; &ndash; in an ideal channel identical to the signal&nbsp; $s(t)$&nbsp; &ndash; is multiplied by the receive-site carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; where the following applies:
 
:$$z_{\rm E}(t) = K \cdot  {\rm cos}(\omega_{\rm T}  t - \Delta \varphi ) .$$
 
:$$z_{\rm E}(t) = K \cdot  {\rm cos}(\omega_{\rm T}  t - \Delta \varphi ) .$$
  
*Dieses Signal sollte nicht nur frequenzsynchron mit $z(t)$ sein, sondern auch phasensynchron  &ndash; daher der Name &bdquo;Synchrondemodulator&rdquo;. Der obige Ansatz berücksichtigt einen Phasenversatz zwischen  $z(t)$ und $z_{\rm E}(t)$, der idealerweise $\Delta \varphi = 0$ sein sollte, sich bei realen Systemen aber oft nicht vermeiden lässt.
+
*This signal should not only be frequency-synchronous with&nbsp; $z(t)$&nbsp; but also phase-synchronous &nbsp; &rArr; &nbsp; hence the name "synchronous demodulator".  
 +
*The above approach takes into account a phase shift between&nbsp; $z(t)$&nbsp; and&nbsp; $z_{\rm E}(t)$, which should ideally be&nbsp; $\Delta \varphi = 0$&nbsp; but often cannot be avoided in real systems.
  
*Das Ausgangssignal $b(t)$ des zweiten Multiplizierers beinhaltet neben dem gewünschten NF-Anteil auch Anteile um die doppelte Trägerfrequenz. Durch einen idealen Tiefpass &ndash; z.B. mit der Grenzfrequenz $f_{\rm T}$ &ndash; lässt sich das Sinkensignal $v(t)$ gewinnen, das im Idealfall gleich dem Quellensignal $q(t)$ sein sollte.
+
*The output signal&nbsp; $b(t)$&nbsp;  of the second multiplier includes components around twice the carrier frequency in addition to the desired low-frequency component.  
 +
*Using an ideal low-pass filter&nbsp; $\rm (LP)$&nbsp; (e.g. with cut-off frequency&nbsp; $f_{\rm T}$)&nbsp; the sink signal&nbsp; $v(t)$,&nbsp; which ideally should be equal to the source signal&nbsp; $q(t)$,&nbsp; can be obtained.
  
  
Die Multiplikation beim Sender mit dem Trägersignal $z(t)$ führt im Allgemeinen zu zwei Seitenbändern. Bei der [[Modulationsverfahren/Einseitenbandmodulation|Einseitenbandmodulation]] (ESB&ndash;AM) wird nur eines der beiden Bänder übertragen, zum Beispiel das untere Seitenband (USB). Damit erhält man bei idealem Kanal:
+
For the transmitter, multiplication by the carrier signal&nbsp; $z(t)$&nbsp; generally results in two sidebands. In&nbsp; [[Modulation_Methods/Single-Sideband_Modulation|Single-Sideband Modulation]]&nbsp; (ESB&ndash;AM), only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:
:$$r(t) = s(t)=  {1 \, \rm V} \cdot {\rm cos}((\omega_{\rm T} -
+
:$$r(t) = s(t)=  {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} -
\omega_2  )t ) - {0.5 \, \rm V} \cdot {\rm sin}((\omega_{\rm T} -
+
\omega_2  )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} -
\omega_5  )t ) .$$
+
\omega_5  )\cdot t \big ] .$$
  
*Hier führt die Synchrondemodulation unter Berücksichtigung eines Phasenversatzes $\Delta \varphi$, der Konstante $K = 4$ sowie des nachgeschalteten Tiefpasses zu folgendem verfälschten Sinkensignal:
+
*Here, synchronous demodulation results in the following distorted sink signal considering a phase shift&nbsp; $\Delta \varphi$, the constant&nbsp; $K = 4$&nbsp; and the downstream low-pass filter:
 
:$$v(t)=  {1 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm cos}(
 
:$$v(t)=  {1 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm cos}(
 
\omega_2 t - \Delta \varphi)+ {0.5 \, \rm V} \cdot
 
\omega_2 t - \Delta \varphi)+ {0.5 \, \rm V} \cdot
Line 32: Line 39:
 
\omega_5 t - \Delta \varphi)$$
 
\omega_5 t - \Delta \varphi)$$
  
:Im Idealfall phasensynchroner Demodulation $(\Delta \varphi = 0)$ gilt wieder $v(t) = q(t).$
+
*In the ideal case of phase-synchronous demodulation&nbsp; $(\Delta \varphi = 0)$,&nbsp; &nbsp; $v(t) = q(t)$ holds again.
 
 
  
  
  
  
 +
Please note:
 +
*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
 +
*The topic&nbsp; "amplitude modulation/synchronous demodulator"&nbsp; is discussed in detail in the book&nbsp; [[Modulation_Methods]].
 +
*The following trigonometric relationships are given:
 +
:$$\cos^2(\alpha) =  {1}/{2} \cdot \big [ 1 +
 +
\cos(2\alpha) \big ] \hspace{0.05cm}, $$
 +
:$$\cos(\alpha) \cdot \cos(\beta)  =  {1}/{2} \cdot \big[ \cos(\alpha -
 +
\beta)+ \cos(\alpha + \beta) \big],$$
 +
:$$ \sin(\alpha) \cdot \cos(\beta)  =  {1}/{2} \cdot \big[ \sin(\alpha -
 +
\beta)+ \sin(\alpha + \beta)
 +
\big] \hspace{0.05cm}.$$
 +
*The signal designations result from the German original of this exercise.&nbsp; '''Here again as a listing''' &nbsp;  $q(t)$ &nbsp; &rArr; &nbsp; source signal, &nbsp; $v(t)$ &nbsp; &rArr; &nbsp; sink signal, &nbsp; $z(t)$ &nbsp; &rArr; &nbsp; transmit-site carrier signal, &nbsp; $s(t)$ &nbsp; &rArr; &nbsp; transmission signal (BP),&nbsp; $r(t)$ &nbsp; &rArr; &nbsp; received signal  (BP), &nbsp; $z_{\rm E}(t)$ &nbsp; &rArr; &nbsp; receive-site carrier signal, &nbsp; $b(t)$ &nbsp; &rArr; &nbsp; BP signal before low-pass.
  
''Hinweise:''
 
*Die Aufgabe gehört zum Kapitel  [[Lineare_zeitinvariante_Systeme/Lineare_Verzerrungen|Lineare Verzerrungen]].
 
*Die Thematik &bdquo;Amplitudenmodulation/Synchrondemodulator&rdquo; wird im Buch [[Modulationsverfahren]] noch ausführlich diskutiert.
 
*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
 
*Gegeben sind die folgenden trigonometrischen Zusammenhänge:
 
:$$\cos^2(\alpha) =  {1}/{2} \cdot \left [ 1 +
 
\cos(2\alpha) \right ] \hspace{0.05cm}, $$
 
:$$\cos(\alpha) \cdot \cos(\beta)  =  {1}/{2} \cdot \left[ \cos(\alpha -
 
\beta)+ \cos(\alpha + \beta) \right],$$
 
:$$ \sin(\alpha) \cdot \cos(\beta)  =  {1}/{2} \cdot \left[ \sin(\alpha -
 
\beta)+ \sin(\alpha + \beta)
 
\right] \hspace{0.05cm}.$$
 
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie lautet das Sinkensignal $v(t)$ bei ZSB-AM und phasensynchroner Synchrondemodulation  &nbsp; &rArr; &nbsp; $\Delta \varphi = 0$? <br>Wie ist $K$ zu wählen, damit  $v(t) = q(t)$ gilt?
+
{What is the sink signal&nbsp; $v(t)$&nbsp; for DSB-AM and phase-synchronous modulation&nbsp; &rArr; &nbsp; $\Delta \varphi = 0$? <br>How should&nbsp; $K$&nbsp; be chosen such that &nbsp;$v(t) = q(t)$&nbsp; holds?
 
|type="{}"}
 
|type="{}"}
 
$K \ = \ $ { 2 3% }
 
$K \ = \ $ { 2 3% }
  
  
{Es gelte $K = 2$. Geben Sie das Sinkensignal  $v(t)$ unter Berücksichtigung eines Phasenversatzes $\Delta \varphi$ an. <br>Welche der folgenden Aussagen treffen zu?
+
{The following holds: &nbsp;$K = 2$.&nbsp; Specify the sink signal&nbsp; $v(t)$&nbsp; considering a phase shift&nbsp;$\Delta \varphi$.&nbsp; Which of the following statements are true?
 
|type="[]"}
 
|type="[]"}
- Unabhängig von $\Delta \varphi$ gilt $v(t) = q(t)$.
+
- $v(t) = q(t)$&nbsp; holds independently of&nbsp; $\Delta \varphi$&nbsp;.
+ $\Delta \varphi \ne 0$ führt zu einer frequenzunabhängigen Dämpfung.
+
+ $\Delta \varphi \ne 0$&nbsp; results in frequency-independent attenuation.
- Ein Phasenversatz $\Delta \varphi \ne 0$ führt zu Dämpfungsverzerrungen.
+
- A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in attenuation distortions.
- Ein Phasenversatz $\Delta \varphi \ne 0$ führt zu Phasenverzerrungen.
+
- A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in phase distortions.
+ Mit $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$ gilt $v(t) = q(t)/2$.
+
+ $v(t) = q(t)/2$&nbsp; holds with&nbsp; $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$.
  
  
{Welche Aussagen gelten bei Synchrondemodulation des ESB&ndash;Signals, wenn ein Phasenversatz um $\Delta \varphi$ berücksichtigt wird?
+
{Which statements hold for synchronous demodulation of the SSB signal if a phase shift of&nbsp;$\Delta \varphi$&nbsp; is considered?
 
|type="[]"}
 
|type="[]"}
- Unabhängig von $\Delta \varphi$ gilt $v(t) = q(t)$.
+
- Regardless of&nbsp; $\Delta \varphi$,&nbsp; $v(t) = q(t)$&nbsp; holds.
- $\Delta \varphi \ne 0$ führt zu einer frequenzunabhängigen Dämpfung.
+
- $\Delta \varphi \ne 0$&nbsp; results in frequency-independent attenuation.
- Ein Phasenversatz $\Delta \varphi \ne 0$ führt zu Dämpfungsverzerrungen.
+
- A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in attenuation distortions.
+ Ein Phasenversatz $\Delta \varphi \ne 0$ führt zu Phasenverzerrungen.
+
+ A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in phase distortions.
- Mit $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$ gilt $v(t) = q(t)/2$.
+
- $v(t) = q(t)/2$&nbsp; holds with&nbsp; $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$&nbsp;.
 +
 
  
  
Line 82: Line 89:
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Für das Bandpass&ndash;Signal nach dem zweiten Multiplizierer gilt:
+
'''(1)'''&nbsp; The following holds for the band-pass signal after the second multiplier:
$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rm
+
:$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rm
 
E}(t)= K \cdot q(t)\cdot
 
E}(t)= K \cdot q(t)\cdot
 
  \cos^2(\omega_{\rm T} t).$$
 
  \cos^2(\omega_{\rm T} t).$$
  
Mit der trigonometrischen Beziehung $\cos^2(\omega_{\rm T} t)  =  {1}/{2} \cdot\left[ 1  +
+
*Using the trigonometric relation&nbsp; $\cos^2(\omega_{\rm T} t)  =  {1}/{2} \cdot\big[ 1  +
  \cos(2\omega_{\rm T} t)\right]$ erhält man
+
  \cos(2\omega_{\rm T} t)\big]$,&nbsp; one obtains:
$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot
+
:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot
  \cos(2\omega_{\rm T} t).$$
+
  \cos(2\omega_{\rm T} t).$$  
 +
 
 +
*The second component is located at around twice the carrier frequency&nbsp; &rArr; &nbsp; $2 f_{\rm T}$.&nbsp;
 +
*This is removed by the low-pass filter&nbsp; $($with the cut-off frequency&nbsp; $  f_{\rm G} = f_{\rm T})$&nbsp;.
 +
*Hence, the following is obtained: &nbsp; $v(t) = {K}/{2} \cdot q(t) .$
 +
*An ideal demodulation&nbsp; &rArr; &nbsp; $v(t) =  q(t)$&nbsp; is obtained with&nbsp; $\underline {K = 2}$&nbsp;.
  
Der zweite Anteil liegt um die doppelte Trägerfrequenz &nbsp;&rArr;&nbsp; $2 f_{\rm T}$ und wird durch den Tiefpass (zum Beispiel mit der Grenzfrequenz $  f_{\rm G} = f_{\rm T}$) entfernt. Damit erhält man: $v(t) = {K}/{2} \cdot q(t) .$ Mit $\underline {K = 2}$ ergibt sich eine ideale Demodulation &nbsp;&rArr;&nbsp; $v(t) =  q(t) .$
 
  
  
'''(2)'''&nbsp; Unter Berücksichtigung der Beziehung
+
'''(2)'''&nbsp; Considering the relation
 
:$$\cos(\omega_{\rm T} t) \cdot \cos(\omega_{\rm T} t - \Delta \varphi)  =  {1}/{2} \cdot
 
:$$\cos(\omega_{\rm T} t) \cdot \cos(\omega_{\rm T} t - \Delta \varphi)  =  {1}/{2} \cdot
   \left[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \right]$$
+
   \big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$
 +
 
 +
and the downstream low-pass filter,&nbsp; which removes the component at around twice the carrier frequency, the following is obtained here with&nbsp; $ {K = 2}$:
 +
:$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$
  
sowie des nachgeschalteten Tiefpasses, der wieder den Anteil um die doppelte Trägerfrequenz entfernt, erhält man hier mit $ {K = 2}$:
+
<u>Proposed solutions 2 and 5</u>&nbsp; are correct:
$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$
+
*A phase shift&nbsp;$\Delta \varphi$&nbsp; only results in frequency-independent attenuation and not in attenuation distortions or phase distortions.  
 +
*A phase shift by&nbsp;$\varphi =\pm 60^\circ$&nbsp; results in halving of the signal amplitude.
  
Richtig sind die <u>Lösungsvorschläge 2 und 5</u>:
 
*Ein Phasenversatz $\Delta \varphi$ führt hier nur zu einer frequenzunabhängigen Dämpfung und nicht zu Dämpfungs&ndash; oder Phasenverzerrungen.
 
*Ein Phasenversatz um $\varphi =\pm 60^\circ$ hat jeweils eine Halbierung des Signals zur Folge.
 
  
  
'''(3)'''&nbsp; Richtig ist hier <u>nur der Lösungsvorschlag 4</u>.  
+
'''(3)'''&nbsp; Here, <u>proposed solution 4</u>&nbsp; is correct.  
*Bei beiden Summanden tritt genau der gleiche Phasenversatz $\Delta \varphi$ auf, und es kommt hier zu Phasenverzerrungen:
+
*Exactly the same phase shift&nbsp; $\Delta \varphi$&nbsp; occurs for both summands,&nbsp; and phase distortions occur here:
:$$v(t)=  {2 \, \rm V}  \cdot{\rm cos}( \omega_2 \cdot (t - \tau_2))+
+
:$$v(t)=  {2 \, \rm V}  \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+
{1 \, \rm V}  \cdot{\rm sin}( \omega_5 t \cdot (t - \tau_5)),$$
+
{1 \, \rm V}  \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$
:$${\rm wobei}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2}
+
:$${\rm where}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2}
 
\hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta
 
\hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta
 
\varphi}{\omega_5}.$$
 
\varphi}{\omega_5}.$$
  
*Ein Phasenversatz von $\varphi =60^\circ$ entsprechend $\pi/3$ führt hier zu den Verzögerungszeiten:
+
*A phase shift of&nbsp; $\varphi =60^\circ$ corresponding to&nbsp; $\pi/3$&nbsp; leads to the following delay times here:
 
:$$\tau_2  =  \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm  kHz }} \approx
 
:$$\tau_2  =  \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm  kHz }} \approx
83.3\,{\rm \mu s }, \hspace{0.5cm}
+
83.3\,{\rm &micro; s }, \hspace{0.5cm}
 
\tau_5  =  \frac{\pi/3}{2 \pi \cdot 5\,\,{\rm  kHz }} \approx
 
\tau_5  =  \frac{\pi/3}{2 \pi \cdot 5\,\,{\rm  kHz }} \approx
33.3\,{\rm \mu s }.$$
+
33.3\,{\rm &micro; s }.$$
  
*Das niederfrequentere Signal wird also stärker verzögert.
+
*The lower-frequency signal is thus delayed more.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^2.3 Lineare Verzerrungen^]]
+
[[Category:Linear and Time-Invariant Systems: Exercises|^2.3 Linear Distortions^]]

Latest revision as of 14:29, 6 October 2021

Amplitude modulator (top),
synchronous demodulator

The depicted block diagram shows a transmission system


Let the source signal consist of two harmonic oscillations with frequencies  $f_2 = 2 \ \rm kHz$  and  $f_5 = 5 \ \rm kHz$:

$$q(t) = {2 \, \rm V} \cdot {\rm cos}(\omega_2 t )+ {1 \, \rm V} \cdot {\rm sin}(\omega_5 t ) .$$
  • This signal is multiplied by the dimensionless carrier signal  $z(t) = \cos(\omega_{\rm T} \cdot T)$  of carrier frequency  $f_{\rm T} = 50 \ \rm kHz$.
    For DSB–AM, the dashed block is irrelevant so that the following holds for the transmission signal:
$$s(t) = q(t) \cdot {\rm cos}(\omega_{\rm T} t ) .$$
  • In the synchronous demodulator, the received signal  $r(t)$  – in an ideal channel identical to the signal  $s(t)$  – is multiplied by the receive-site carrier signal  $z_{\rm E}(t)$  where the following applies:
$$z_{\rm E}(t) = K \cdot {\rm cos}(\omega_{\rm T} t - \Delta \varphi ) .$$
  • This signal should not only be frequency-synchronous with  $z(t)$  but also phase-synchronous   ⇒   hence the name "synchronous demodulator".
  • The above approach takes into account a phase shift between  $z(t)$  and  $z_{\rm E}(t)$, which should ideally be  $\Delta \varphi = 0$  but often cannot be avoided in real systems.
  • The output signal  $b(t)$  of the second multiplier includes components around twice the carrier frequency in addition to the desired low-frequency component.
  • Using an ideal low-pass filter  $\rm (LP)$  (e.g. with cut-off frequency  $f_{\rm T}$)  the sink signal  $v(t)$,  which ideally should be equal to the source signal  $q(t)$,  can be obtained.


For the transmitter, multiplication by the carrier signal  $z(t)$  generally results in two sidebands. In  Single-Sideband Modulation  (ESB–AM), only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:

$$r(t) = s(t)= {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} - \omega_2 )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} - \omega_5 )\cdot t \big ] .$$
  • Here, synchronous demodulation results in the following distorted sink signal considering a phase shift  $\Delta \varphi$, the constant  $K = 4$  and the downstream low-pass filter:
$$v(t)= {1 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm cos}( \omega_2 t - \Delta \varphi)+ {0.5 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm sin}( \omega_5 t - \Delta \varphi)$$
$$\Rightarrow \hspace{0.5cm}v(t)= {2 \, \rm V} \cdot{\rm cos}( \omega_2 t - \Delta \varphi)+ {1 \, \rm V} \cdot{\rm sin}( \omega_5 t - \Delta \varphi)$$
  • In the ideal case of phase-synchronous demodulation  $(\Delta \varphi = 0)$,    $v(t) = q(t)$ holds again.



Please note:

  • The exercise belongs to the chapter  Linear Distortions.
  • The topic  "amplitude modulation/synchronous demodulator"  is discussed in detail in the book  Modulation Methods.
  • The following trigonometric relationships are given:
$$\cos^2(\alpha) = {1}/{2} \cdot \big [ 1 + \cos(2\alpha) \big ] \hspace{0.05cm}, $$
$$\cos(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$
$$ \sin(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \sin(\alpha - \beta)+ \sin(\alpha + \beta) \big] \hspace{0.05cm}.$$
  • The signal designations result from the German original of this exercise.  Here again as a listing   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (BP),  $r(t)$   ⇒   received signal (BP),   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   BP signal before low-pass.


Questions

1

What is the sink signal  $v(t)$  for DSB-AM and phase-synchronous modulation  ⇒   $\Delta \varphi = 0$?
How should  $K$  be chosen such that  $v(t) = q(t)$  holds?

$K \ = \ $

2

The following holds:  $K = 2$.  Specify the sink signal  $v(t)$  considering a phase shift $\Delta \varphi$.  Which of the following statements are true?

$v(t) = q(t)$  holds independently of  $\Delta \varphi$ .
$\Delta \varphi \ne 0$  results in frequency-independent attenuation.
A phase shift  $\Delta \varphi \ne 0$  results in attenuation distortions.
A phase shift  $\Delta \varphi \ne 0$  results in phase distortions.
$v(t) = q(t)/2$  holds with  $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$.

3

Which statements hold for synchronous demodulation of the SSB signal if a phase shift of $\Delta \varphi$  is considered?

Regardless of  $\Delta \varphi$,  $v(t) = q(t)$  holds.
$\Delta \varphi \ne 0$  results in frequency-independent attenuation.
A phase shift  $\Delta \varphi \ne 0$  results in attenuation distortions.
A phase shift  $\Delta \varphi \ne 0$  results in phase distortions.
$v(t) = q(t)/2$  holds with  $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$ .


Solution

(1)  The following holds for the band-pass signal after the second multiplier:

$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rm E}(t)= K \cdot q(t)\cdot \cos^2(\omega_{\rm T} t).$$
  • Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 + \cos(2\omega_{\rm T} t)\big]$,  one obtains:
$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot \cos(2\omega_{\rm T} t).$$
  • The second component is located at around twice the carrier frequency  ⇒   $2 f_{\rm T}$. 
  • This is removed by the low-pass filter  $($with the cut-off frequency  $ f_{\rm G} = f_{\rm T})$ .
  • Hence, the following is obtained:   $v(t) = {K}/{2} \cdot q(t) .$
  • An ideal demodulation  ⇒   $v(t) = q(t)$  is obtained with  $\underline {K = 2}$ .


(2)  Considering the relation

$$\cos(\omega_{\rm T} t) \cdot \cos(\omega_{\rm T} t - \Delta \varphi) = {1}/{2} \cdot \big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$

and the downstream low-pass filter,  which removes the component at around twice the carrier frequency, the following is obtained here with  $ {K = 2}$:

$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$

Proposed solutions 2 and 5  are correct:

  • A phase shift $\Delta \varphi$  only results in frequency-independent attenuation and not in attenuation distortions or phase distortions.
  • A phase shift by $\varphi =\pm 60^\circ$  results in halving of the signal amplitude.


(3)  Here, proposed solution 4  is correct.

  • Exactly the same phase shift  $\Delta \varphi$  occurs for both summands,  and phase distortions occur here:
$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+ {1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$
$${\rm where}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2} \hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta \varphi}{\omega_5}.$$
  • A phase shift of  $\varphi =60^\circ$ corresponding to  $\pi/3$  leads to the following delay times here:
$$\tau_2 = \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm kHz }} \approx 83.3\,{\rm µ s }, \hspace{0.5cm} \tau_5 = \frac{\pi/3}{2 \pi \cdot 5\,\,{\rm kHz }} \approx 33.3\,{\rm µ s }.$$
  • The lower-frequency signal is thus delayed more.