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

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[[File:P_ID913__LZI_Z_2_6_neu.png|right|frame|AM–modulator (top) and synchronous demodulator (bottom)]]
+
[[File:P_ID913__LZI_Z_2_6_neu.png|right|frame|Amplitude modulator (top), <br>synchronous demodulator]]
 
The depicted block diagram shows a transmission system  
 
The depicted block diagram shows a transmission system  
*with&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]]&nbsp; (ZSB-AM)  
+
*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]] (SD).  
+
*and&nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]]&nbsp; $\rm (SD)$.  
  
  
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\cdot {\rm sin}(\omega_5  t ) .$$
 
\cdot {\rm sin}(\omega_5  t ) .$$
  
*This signal is multiplied by the dimensionless carrier signal&nbsp; $z(t) = \cos(\omega_{\rm T} \cdot T)$&nbsp; of frequency&nbsp; $f_{\rm T} = 50 \ \rm kHz$. For ZSB&ndash;AM the dashed block is irrelevant so that the following holds for the transmit signal:
+
*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:
 
:$$s(t) = q(t) \cdot  {\rm cos}(\omega_{\rm T}  t ) .$$
 
:$$s(t) = q(t) \cdot  {\rm cos}(\omega_{\rm T}  t ) .$$
  
*In the synchronous demodulator, the receiver signal&nbsp; $r(t)$&nbsp; &ndash; dentical to the transmit signal&nbsp; $s(t)$&nbsp; in an ideal channel &ndash; is multiplied by the receive-site carrier signal&nbsp; $z_{\rm E}(t)$&nbsp; where the following applies:
+
*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:
 
:$$z_{\rm E}(t) = K \cdot  {\rm cos}(\omega_{\rm T}  t - \Delta \varphi ) .$$
 
:$$z_{\rm E}(t) = K \cdot  {\rm cos}(\omega_{\rm T}  t - \Delta \varphi ) .$$
  
*This signal should not only be frequency-synchronous with&nbsp; $z(t)$&nbsp; but also phase-synchronous &ndash; hence the name "synchronous demodulator".  
+
*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.
 
*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.
  
*The output signal&nbsp; $b(t)$&nbsp;  of the second multiplier includes components around twice the carrier frequency in addition to the desired NF component.  
+
*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&ndash; for example with cut-off frequency&nbsp; $f_{\rm T}$&nbsp; &ndash; the sink signal&nbsp; $v(t)$&nbsp; can be obtained which ideally should be equal to the source signal&nbsp; $q(t)$&nbsp;.
+
*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.
  
  
For the transmitter, multiplication by the carrier signal&nbsp; $z(t)$&nbsp; generally results in two sidebands. In the&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:
+
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}\big [(\omega_{\rm T} -
 
:$$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_2  )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} -
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+
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]].
 
 
 
 
''Please note:''
 
*The task belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]].
 
*The topic "amplitude modulation/synchronous demodulator" is discussed in detail in the book&nbsp; [[Modulation_Methods]]&nbsp;.
 
 
 
*The following trigonometric relationships are given:
 
*The following trigonometric relationships are given:
 
:$$\cos^2(\alpha) =  {1}/{2} \cdot \big [ 1 +  
 
:$$\cos^2(\alpha) =  {1}/{2} \cdot \big [ 1 +  
Line 61: Line 55:
 
  \beta)+ \sin(\alpha + \beta)
 
  \beta)+ \sin(\alpha + \beta)
 
  \big] \hspace{0.05cm}.$$
 
  \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.
 +
  
  
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<quiz display=simple>
 
<quiz display=simple>
{What is the sink signal&nbsp;$v(t)$&nbsp; for ZSB-AM and phase-synchronous 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?
+
{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% }
  
  
{The following holds: &nbsp;$K = 2$. Specify the sink signal&nbsp;$v(t)$&nbsp; considering a phase shift&nbsp;$\Delta \varphi$&nbsp; an. <br>Which of the following statements are true?
+
{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="[]"}
- $v(t) = q(t)$ holds independently of&nbsp;$\Delta \varphi$&nbsp;.
+
- $v(t) = q(t)$&nbsp; holds independently of&nbsp; $\Delta \varphi$&nbsp;.
 
+ $\Delta \varphi \ne 0$&nbsp; results in frequency-independent attenuation.
 
+ $\Delta \varphi \ne 0$&nbsp; results in frequency-independent attenuation.
- A phase shift&nbsp;$\Delta \varphi \ne 0$&nbsp; results in attenuation distortions.
+
- A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in attenuation distortions.
- A phase shift&nbsp;$\Delta \varphi \ne 0$&nbsp; results in phase distortions.
+
- A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in phase distortions.
+ $v(t) = q(t)/2$ holds with&nbsp;$\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$&nbsp;.
+
+ $v(t) = q(t)/2$&nbsp; holds with&nbsp; $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$.
  
  
{Which statements hold for synchronous demodulation of the ESB&ndash;signal if a phase shift of&nbsp;$\Delta \varphi$&nbsp; is considered?
+
{Which statements hold for synchronous demodulation of the SSB signal if a phase shift of&nbsp;$\Delta \varphi$&nbsp; is considered?
 
|type="[]"}
 
|type="[]"}
- Regardless of&nbsp;$\Delta \varphi$,&nbsp; &nbsp;$v(t) = q(t)$ holds.
+
- Regardless of&nbsp; $\Delta \varphi$,&nbsp; $v(t) = q(t)$&nbsp; holds.
 
- $\Delta \varphi \ne 0$&nbsp; results in frequency-independent attenuation.
 
- $\Delta \varphi \ne 0$&nbsp; results in frequency-independent attenuation.
- A phase shift&nbsp;$\Delta \varphi \ne 0$&nbsp; results in attenuation distortions.
+
- A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in attenuation distortions.
+ A phase shift&nbsp;$\Delta \varphi \ne 0$&nbsp; results in phase distortions.
+
+ A phase shift&nbsp; $\Delta \varphi \ne 0$&nbsp; results in phase distortions.
- $v(t) = q(t)/2$ with&nbsp;$\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$&nbsp;.
+
- $v(t) = q(t)/2$&nbsp; holds with&nbsp; $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$&nbsp;.
  
  
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*Using the trigonometric relation&nbsp; $\cos^2(\omega_{\rm T} t)  =  {1}/{2} \cdot\big[ 1  +
 
*Using the trigonometric relation&nbsp; $\cos^2(\omega_{\rm T} t)  =  {1}/{2} \cdot\big[ 1  +
  \cos(2\omega_{\rm T} t)\big]$,&nbsp;  
+
  \cos(2\omega_{\rm T} t)\big]$,&nbsp; one obtains:
 
:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot
 
:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot
  \cos(2\omega_{\rm T} t)$$ is obtained.
+
  \cos(2\omega_{\rm T} t).$$  
  
*Der zweite Anteil liegt um die doppelte Trägerfrequenz &nbsp; &rArr; &nbsp; $2 f_{\rm T}$.&nbsp;  
+
*The second component is located at around twice the carrier frequency&nbsp; &rArr; &nbsp; $2 f_{\rm T}$.&nbsp;  
*Dieser wird durch den Tiefpass&nbsp; $($mit der Grenzfrequenz&nbsp; $  f_{\rm G} = f_{\rm T})$&nbsp; entfernt.  
+
*This is removed by the low-pass filter&nbsp; $($with the cut-off frequency&nbsp; $  f_{\rm G} = f_{\rm T})$&nbsp;.  
*Damit erhält man: &nbsp; $v(t) = {K}/{2} \cdot q(t) .$  
+
*Hence, the following is obtained: &nbsp; $v(t) = {K}/{2} \cdot q(t) .$  
*Mit&nbsp; $\underline {K = 2}$&nbsp; ergibt sich eine ideale Demodulation &nbsp; &rArr; &nbsp; $v(t) =  q(t)$.
+
*An ideal demodulation&nbsp; &rArr; &nbsp; $v(t) =  q(t)$&nbsp; is obtained with&nbsp; $\underline {K = 2}$&nbsp;.
  
  
  
'''(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
 
   \big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$
 
   \big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$
  
sowie des nachgeschalteten Tiefpasses, der wieder den Anteil um die doppelte Trägerfrequenz entfernt, erhält man hier mit&nbsp; $ {K = 2}$:
+
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}$:
 
:$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$
 
:$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$
  
Richtig sind die <u>Lösungsvorschläge 2 und 5</u>:
+
<u>Proposed solutions 2 and 5</u>&nbsp; are correct:
*Ein Phasenversatz &nbsp;$\Delta \varphi$&nbsp; führt hier nur zu einer frequenzunabhängigen Dämpfung und nicht zu Dämpfungs&ndash; oder Phasenverzerrungen.  
+
*A phase shift&nbsp;$\Delta \varphi$&nbsp; only results in frequency-independent attenuation and not in attenuation distortions or phase distortions.  
*Ein Phasenversatz um &nbsp;$\varphi =\pm 60^\circ$&nbsp; hat jeweils eine Halbierung des Signals zur Folge.  
+
*A phase shift by&nbsp;$\varphi =\pm 60^\circ$&nbsp; results in halving of the signal amplitude.  
  
  
  
'''(3)'''&nbsp; Richtig ist hier der <u>Lösungsvorschlag 4</u>.  
+
'''(3)'''&nbsp; Here, <u>proposed solution 4</u>&nbsp; is correct.  
*Bei beiden Summanden tritt genau der gleiche Phasenversatz&nbsp; $\Delta \varphi$&nbsp; 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}\big[ \omega_2 \cdot (t - \tau_2) \big]+
 
:$$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],$$
 
{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 &nbsp;$\varphi =60^\circ$ entsprechend&nbsp; $\pi/3$&nbsp; 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 &micro; s }, \hspace{0.5cm}
 
83.3\,{\rm &micro; s }, \hspace{0.5cm}
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33.3\,{\rm &micro; 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}}
  

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.