Difference between revisions of "Aufgaben:Exercise 2.7Z: DSB-AM and Envelope Demodulator"

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===Solution===
 
===Solution===
 
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[[File:P_ID1035__Mod_Z_2_7_a.png|right|frame|Quellensignal im Bereich bis  $1\text{ ms}$]]
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[[File:P_ID1035__Mod_Z_2_7_a.png|right|frame|Source signal in the region up to  $1\text{ ms}$]]
'''(1)'''  Die Grafik zeigt, dass das Quellensignal alle Werte zwischen  $–4 \ \rm V$  und  $+3.667\ \rm  V$  annehmen kann.  
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'''(1)'''  The graph shows that the source signal can take on all values between   $–4 \ \rm V$  and  $+3.667\ \rm  V$  annehmen kann.  
*Der maximale Betrag tritt zum Beispiel zum Zeitpunkt  $t = t_0 =0.75\ \rm   ms$  auf:
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*For example, the maximum magnitude occurs at time   $t = t_0 =0.75\ \rm ms$ :
 
:$$q(t = t_0)  =  2 \,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t_0 ) + 2 \,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t_0 )$$
 
:$$q(t = t_0)  =  2 \,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t_0 ) + 2 \,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t_0 )$$
 
:$$\Rightarrow \hspace{0.3cm}q(t = 0.75 \,{\rm ms})  =  2 \,{\rm V} \cdot \cos(3 \pi) + 2 \,{\rm V} \cdot \sin(7.5 \pi)= -4 \,{\rm V}\hspace{0.05cm}.$$
 
:$$\Rightarrow \hspace{0.3cm}q(t = 0.75 \,{\rm ms})  =  2 \,{\rm V} \cdot \cos(3 \pi) + 2 \,{\rm V} \cdot \sin(7.5 \pi)= -4 \,{\rm V}\hspace{0.05cm}.$$
  
*Daraus folgt für den maximalen Betrag:   $q_{\rm max}\hspace{0.15cm}\underline{ = 4 \ \rm V}$.
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*From this, it follows for the maximum magnitude:   $q_{\rm max}\hspace{0.15cm}\underline{ = 4 \ \rm V}$.
  
  
  
'''(2)'''  In der Angabenseite–Grafik  gibt das Gewicht der Diraclinie bei  $f = 0$  die Amplitude des zugesetzten Trägers an.  
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'''(2)'''  In the graph on the exercise page, the weight of the Dirac line  $f = 0$  indicates the amplitude of the added carrier.
*Diese ist  $A_{\rm T}\hspace{0.15cm}\underline{ = 4\ \rm V }$.  
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*This is  $A_{\rm T}\hspace{0.15cm}\underline{ = 4\ \rm V }$.  
*Daraus erhält man den Modulationsgrad  $m = q_{\rm max}/A_{\rm T} \hspace{0.15cm}\underline{ = 1}$.
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*From this, we get the modulation depth  $m = q_{\rm max}/A_{\rm T} \hspace{0.15cm}\underline{ = 1}$.
  
  
  
'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
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'''(3)'''&nbsp;<u>Answers 2 and 3</u> are correct:
*Da der Modulationsgrad nicht größer als&nbsp; $m = 1$&nbsp; ist, führt auch der Hüllkurvendemodulator nicht zu Verzerrungen.  
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*Since the modulation depth is not greater than&nbsp; $m = 1$&nbsp;, the envelope demodulator does not cause distortion either.
*Der wesentliche Vorteil der Hüllkurvendemodulation ist, dass keine Frequenz– und Phasensynchronität notwendig ist.  
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*The main advantage of envelope demodulation is that no frequency and phase synchronization is necessary.
*Nachteilig ist, dass im Gegensatz zur Synchrondemodulation beim Sender eine deutlich höhere Leistung aufgebracht werden muss.  
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*A disadvantage ist that a significantly higher power must be applied at the transmitter relative to synchronous demodulation.  
*Bei&nbsp; $m = 1$&nbsp; ergibt sich gegenüber der ZSB–AM ohne Träger die dreifache Sendeleistung.  
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*When&nbsp; $m = 1$&nbsp; this results in three times the transmit power compared to DSB-AM without a carrier.
  
  
 
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[[File:P_ID1036__Mod_Z_2_7_d.png|right|frame|Equivalent low-pass Signal <br>in the complex plane]]
[[File:P_ID1036__Mod_Z_2_7_d.png|right|frame|Äquivalentes Tiefpass–Signal <br>in der komplexen Ebene]]
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'''(4)'''&nbsp; <u>Answers 1 and 3</u> are correct:
'''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
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*When&nbsp; $ω_2 = 2 π · 2 \ \rm kHz$&nbsp; and&nbsp; $ω_5 = 2 π · \ \rm 5 kHz$&nbsp;:
*Mit&nbsp; $ω_2 = 2 π · 2 \ \rm kHz$&nbsp; und&nbsp; $ω_5 = 2 π · \ \rm 5 kHz$&nbsp; gilt:
 
 
:$$ r_{\rm TP}(t) = 4 \,{\rm V} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t}  
 
:$$ r_{\rm TP}(t) = 4 \,{\rm V} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t}  
 
\hspace{-0.05cm}-\hspace{-0.05cm} \hspace{0.15cm}{\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} {\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{0.05cm}. \hspace{0.1cm}$$
 
\hspace{-0.05cm}-\hspace{-0.05cm} \hspace{0.15cm}{\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} {\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{0.05cm}. \hspace{0.1cm}$$
*Bei der Konstruktion der Ortskurve&nbsp; $r_{TP}(t)$&nbsp; sind somit genau fünf Zeiger zu berücksichtigen &nbsp; &rArr; &nbsp; Antwort 1 ist richtig.&nbsp; Die Grafik zeigt eine Momentaufnahme zum Zeitpunkt&nbsp; $t = 0$.
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*Thus, in constructing the locus &nbsp; $r_{TP}(t)$&nbsp;, there are exactly five pointers to consider &nbsp; &rArr; &nbsp; answer 1 is correct. The graph shows a snapshot at time &nbsp; $t = 0$.
:*Der (rote) Träger ist für alle Zeiten durch den reellen Zeiger der Länge&nbsp; $4 \ \rm V$ gegeben.&nbsp; Im Gegensatz zum Zeigerdiagramm (Darstellung des analytischen Signals) dreht dieser nicht &nbsp; &rArr; &nbsp; Antwort 2 ist falsch.
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:*The (red) carrier is given by the real pointer of length&nbsp; $4 \ \rm V$ for all time points. In contrast to the pointer diagram (showing the analytic signal), this does not rotate &nbsp; &rArr; &nbsp; Answer 2 is false.
:*Die dritte Aussage ist ebenso wie die Aussage 1 richtig:&nbsp; Die Drehzeiger bei negativen Frequenzen drehen in mathematisch negativer Richtung&nbsp; (im Uhrzeigersinn)&nbsp; im Gegensatz zu den beiden Zeigern mit&nbsp; $f > 0$.  
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:*The third statement is similarly correct as the first: The rotating pointers at negative frequencies rotate in mathematically negative direction (clockwise) in contrast to the two pointers with nbsp; $f > 0$.
:*Die letzte Aussage trifft nicht zu.&nbsp; Je größer die Frequenz&nbsp; $f$&nbsp; ist, um so schneller dreht der zugehörige Zeiger.
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:*The last statement is not true. The larger the frequency &nbsp; $f$&nbsp;, the faster the associated pointer rotates.
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[[File:P_ID1037__Mod_Z_2_7_e.png|right|frame|Ortskurve für verzerrungsfreie Hüllkurvendemodulation]]
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[[File:P_ID1037__Mod_Z_2_7_e.png|right|frame|Locus curve for distortionless envelope demodulation]]
'''(5)'''&nbsp; Richtig sind die <u>Aussagen 1 und 2</u>:
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'''(5)'''&nbsp; <u>Statements 1 and 2</u> are correct:
  
*Im betrachteten Beispiel kann für das äquivalente TP–Signal auch geschrieben werden:
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*In the example considere, the equivalent lowpass signal can be written as:
 
:$$r_{\rm TP}(t) = q(t) + A_{\rm T} \hspace{0.05cm}.$$
 
:$$r_{\rm TP}(t) = q(t) + A_{\rm T} \hspace{0.05cm}.$$
*Damit ist offensichtlich, dass&nbsp; $r_{\rm TP}(t)$&nbsp; stets reell ist.&nbsp; Aus den Teilaufgaben&nbsp; '''(1)'''&nbsp; und&nbsp; '''(2)'''&nbsp; folgt zudem &nbsp; $r_{\rm TP}(t) ≥ 0$.
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*Thus, it is obvious that &nbsp; $r_{\rm TP}(t)$&nbsp; is always real.&nbsp; Moreover, it follows from subtasks&nbsp; '''(1)'''&nbsp; and&nbsp; '''(2)'''&nbsp; tht &nbsp; $r_{\rm TP}(t) ≥ 0$.
  
  
Das bedeutet:
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This means:
:*Die Ortskurve ist hier eine horizontale Gerade auf der reellen Gerade und liegt stets in der rechten Halbebene.  
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:*Here, the locus curve is a horizontal line on the real plane and always lies in the right half-plane.
:*Dies sind die beiden notwendigen Bedingungen, dass mit einem Hüllkurvendemodulator das Nachrichtensignal verzerrungsfrei wiedergewonnen werden kann.  
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:*These are the two necessary conditions for an envelope demodulator to recover the message signal without distortion.
:*Ist eine dieser Voraussetzungen nicht erfüllt, so kommt es zu&nbsp; <u>'''nichtlinearen'''</u>&nbsp; Verzerrungen, nicht zu linearen &nbsp; &rArr; &nbsp; Antwort 3 ist falsch.  
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:*If one of these conditions is not satisfied, &nbsp; <u>'''nonlinear'''</u>&nbsp; distortions arise, not linear ones &nbsp; &rArr; &nbsp; Answer 3 is wrong.  
  
 
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Revision as of 21:03, 20 December 2021

Spectrum  $R_{\rm TP}(f)$  of the received signal in the equivalent low-pass range

Assume a source signal

$$ q(t) = 2 \,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t ) + 2 \,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t )\hspace{0.05cm}.$$

This is modulated according to the modulation method "DSB-AM with carrier" and transmitted through an ideal channel. The influence of noise can be disregarded.


The graph shows the spectrum $R_{\rm TP}(f)$  of the received signal in the equivalent low-pass region, which is composed of Dirac lines at   $f = 0$  (originating from the carrier),  at  $±2\ \rm kHz$  (originating from the cosine component)  and at  $±5\ \rm kHz$  (originating from the sine component) .


  • The locus curve is the plot of the equivalent low-pass signal  $r_{\rm TP}(t)$  in the complex plane,
  • where  $r_{\rm TP}(t)$  is the Fourier retransform of  $R_{\ \rm TP}(f)$ .





Hints:


Questions

1

Estimate the maximum magnitude  $q_{\rm max} = {\rm Max} |q(t)|$  of the source signal S.

$q_{\rm max} \ = \ $

$\ \rm V$

2

What is the amplitude  $A_{\rm T}$  of the carrier signal added at the transmitter?  What modulation depth  $m$  results from this?

$A_{\rm T} \ = \ $

$\ \rm V$
$m \ = \ $

3

Which of these are arguments for or against using an envelope demodulator? Assume the alternative would be a synchronous demodulator.

With the envelope demodulator, distortion-free demodulation is not possible in the example considered.
One can do without frequency and phase synchronization.
A smaller transmit power would be needed using a synchronous demodulator.

4

Calculate the equivalent low-pass signal  $r_{\rm TP}(t)$   ⇒   "locus curve",  using the Fourier retransform of  $R_{\rm TP}(f)$ . Which statements are true?

The locus curve  $r_{\rm TP}(t)$  is composed of five pointers.
The carrier rotates with a rotation speed  $ω_{\rm T}$.
The rotational pointers of the negative frequencies rotate clockwise.
The pointer for  $2 \ \rm kHz$  rotates twice as fast as the one for  $5 \ \rm kHz$.

5

Which statements can be made based on the locus curve? Answer the following questions by considering the application of envelope demodulation.

A distortionless demodulation is only possible when  $r_{\rm TP}(t)$  is real at all times.
A distortionless demodulation is only possible when  $r_{\rm TP}(t)$  does not become negative at any point in time.
If the first two conditions mentioned are not met, linear distortions will occur.


Solution

Source signal in the region up to  $1\text{ ms}$

(1)  The graph shows that the source signal can take on all values between   $–4 \ \rm V$  and  $+3.667\ \rm V$  annehmen kann.

  • For example, the maximum magnitude occurs at time   $t = t_0 =0.75\ \rm ms$ :
$$q(t = t_0) = 2 \,{\rm V} \cdot \cos(2 \pi \cdot 2\,{\rm kHz} \cdot t_0 ) + 2 \,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t_0 )$$
$$\Rightarrow \hspace{0.3cm}q(t = 0.75 \,{\rm ms}) = 2 \,{\rm V} \cdot \cos(3 \pi) + 2 \,{\rm V} \cdot \sin(7.5 \pi)= -4 \,{\rm V}\hspace{0.05cm}.$$
  • From this, it follows for the maximum magnitude:   $q_{\rm max}\hspace{0.15cm}\underline{ = 4 \ \rm V}$.


(2)  In the graph on the exercise page, the weight of the Dirac line  $f = 0$  indicates the amplitude of the added carrier.

  • This is  $A_{\rm T}\hspace{0.15cm}\underline{ = 4\ \rm V }$.
  • From this, we get the modulation depth  $m = q_{\rm max}/A_{\rm T} \hspace{0.15cm}\underline{ = 1}$.


(3) Answers 2 and 3 are correct:

  • Since the modulation depth is not greater than  $m = 1$ , the envelope demodulator does not cause distortion either.
  • The main advantage of envelope demodulation is that no frequency and phase synchronization is necessary.
  • A disadvantage ist that a significantly higher power must be applied at the transmitter relative to synchronous demodulation.
  • When  $m = 1$  this results in three times the transmit power compared to DSB-AM without a carrier.


Equivalent low-pass Signal
in the complex plane

(4)  Answers 1 and 3 are correct:

  • When  $ω_2 = 2 π · 2 \ \rm kHz$  and  $ω_5 = 2 π · \ \rm 5 kHz$ :
$$ r_{\rm TP}(t) = 4 \,{\rm V} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} 1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 2}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}-\hspace{-0.05cm} \hspace{0.15cm}{\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{-0.05cm}+\hspace{-0.05cm} {\rm j} \cdot1 \,{\rm V} \cdot {\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm 5}\cdot \hspace{0.03cm}t} \hspace{0.05cm}. \hspace{0.1cm}$$
  • Thus, in constructing the locus   $r_{TP}(t)$ , there are exactly five pointers to consider   ⇒   answer 1 is correct. The graph shows a snapshot at time   $t = 0$.
  • The (red) carrier is given by the real pointer of length  $4 \ \rm V$ for all time points. In contrast to the pointer diagram (showing the analytic signal), this does not rotate   ⇒   Answer 2 is false.
  • The third statement is similarly correct as the first: The rotating pointers at negative frequencies rotate in mathematically negative direction (clockwise) in contrast to the two pointers with nbsp; $f > 0$.
  • The last statement is not true. The larger the frequency   $f$ , the faster the associated pointer rotates.




Locus curve for distortionless envelope demodulation

(5)  Statements 1 and 2 are correct:

  • In the example considere, the equivalent lowpass signal can be written as:
$$r_{\rm TP}(t) = q(t) + A_{\rm T} \hspace{0.05cm}.$$
  • Thus, it is obvious that   $r_{\rm TP}(t)$  is always real.  Moreover, it follows from subtasks  (1)  and  (2)  tht   $r_{\rm TP}(t) ≥ 0$.


This means:

  • Here, the locus curve is a horizontal line on the real plane and always lies in the right half-plane.
  • These are the two necessary conditions for an envelope demodulator to recover the message signal without distortion.
  • If one of these conditions is not satisfied,   nonlinear  distortions arise, not linear ones   ⇒   Answer 3 is wrong.