Difference between revisions of "Aufgaben:Exercise 2.5: DSB-AM via a Gaussian channel"

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{{quiz-Header|Buchseite=Modulationsverfahren/Synchrondemodulation}}
 
{{quiz-Header|Buchseite=Modulationsverfahren/Synchrondemodulation}}
  
[[File:P_ID1010__Mod_A_2_5.png|right|]]
+
[[File:EN_Mod_A_2_5.png|right|frame|DSB-AM over a distorting channel]]
Das hier betrachtete Übertragungssystem setzt sich aus folgenden Blöcken zusammen:
+
The communication system considered here is composed of the following blocks:
:*ZSB–AM ohne Träger mit $f_T = 50 kHz$ bzw. $f_T = 55 kHz$:
+
*DSB-AM without carrier   $(f_{\rm T} = 50 \ \rm  kHz$  or  $f_{\rm T} = 55 \ \rm  kHz)$:
$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
+
:$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
:* Gaußförmiger Bandpass–Kanalfrequenzgang:
+
* Gaussian band-pass channel;  the magnitude  $|f|$   in the exponent causes  $H_K(–f) = H_K(f)$  to hold:
$$H_{\rm K}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{|f| - f_{\rm M}}{\Delta f_{\rm K}}\right)^2} ,\hspace{0.2cm} f_{\rm M} = 50\,{\rm kHz},\hspace{0.2cm} \Delta f_{\rm K} = 10\,{\rm kHz}\hspace{0.05cm}.$$
+
:$$H_{\rm K}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (({|f| - f_{\rm M}})/{\Delta f_{\rm K}}\right)^2} ,\hspace{0.2cm} f_{\rm M} = 50\,{\rm kHz},\hspace{0.2cm} \Delta f_{\rm K} = 10\,{\rm kHz}\hspace{0.05cm}.$$
Der Betrag $|f|$ im Exponenten berücksichtigt, dass $H_K(–f) = H_K(f)$ gilt.
+
*The synchronous demodulator has optimal parameters such that the sink signal  $v(t)$  completely coincides with the source signal  $q(t)$  when  $H_{\rm K}(f) = 1$  ("ideal channel").
:* Synchrondemodulator mit optimalen Kenngrößen, so dass das Sinkensignal $υ(t)$ vollständig mit dem Quellensignal $q(t)$ übereinstimmt, wenn $H_K(f) = 1$ ist.
 
Auf der Seite [http://en.lntwww.de/Modulationsverfahren/Synchrondemodulation#Einfluss_linearer_Kanalverzerrungen_.281.29  Einfluss linearer Kanalverzerrungen] wurde gezeigt, dass das gesamte System durch den resultierenden Frequenzgang
 
$$H_{\rm MKD}(f) = \frac{1}{2} \cdot \left[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\right]$$
 
ausreichend genau charakterisiert ist. Der Index steht hierbei für „Modulator–Kanal–Demodulator”.
 
  
Das Quellensignal q(t) setzt sich aus zwei Cosinus-Schwingungen zusammen:
 
$$q(t) = 2\,{\rm V}\cdot \cos (2 \pi \cdot 1\,{\rm kHz} \cdot t)+ 3\,{\rm V}\cdot \cos (2 \pi \cdot 5\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
 
'''Hinweis:''' Diese Aufgabe bezieht sich auf den Lehrstoff von [http://en.lntwww.de/Modulationsverfahren/Synchrondemodulation Kapitel 2.2].
 
  
===Fragebogen===
+
On the page  [[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions| Influence of linear channel distortions]]  it was shown that the entire system is is sufficiently accurately characterized by the resulting frequency response
 +
:$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]$$
 +
Here the subscript stands for  $\rm M$odulator – $\rm K$  (for German "Kanal" i.e. channel) – $\rm D$emodulator.
 +
 
 +
The source signal  $q(t)$  is composed of two cosine oscillations:
 +
:$$q(t) = 2\,{\rm V}\cdot \cos (2 \pi \cdot 1\,{\rm kHz} \cdot t)+ 3\,{\rm V}\cdot \cos (2 \pi \cdot 5\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Hints:
 +
*This exercise belongs to the chapter  [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
 +
*Particular reference is made to the page   [[Modulation_Methods/Synchronous_Demodulation#Influence_of_linear_channel_distortions|Influence of linear channel distortions]].
 +
 +
 
 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Berechnen Sie den resultierenden Frequenzgang $H_{MKD}(f)$ für $f_T = 50 kHz$. Welche Werte ergeben sich für $f = 1 kHz$ und $f = 5 kHz$?
+
{Calculate the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; for $f_{\rm T} = 50 \ \rm  kHz$.&nbsp;
 +
What are the values for $f = 1 \ \rm  kHz$&nbsp; and &nbsp;$f = 5 \ \rm  kHz$?
 
|type="{}"}
 
|type="{}"}
$f_T = 50 kHz:  |H_{MKD} (f = 1 kHz)|$ = { 0.969 3% }
+
$|H_{\rm MKD} (f = 1\ \rm  kHz)| \ = \ $ { 0.969 3% }
$f_T = 50 kHz:  |H_{MKD} (f = 5 kHz)|$ = { 0.456 3% }
+
$|H_{\rm MKD} (f = 5\ \rm  kHz)| \ = \ $ { 0.456 3% }
  
{Berechnen Sie das Sinkensignal $υ(t)$. Geben Sie die Amplituden $A_1$ und $A_5$ des 1 kHz– bzw. 5 kHz–Anteils an.
+
{Calculate the sink signal &nbsp;$v(t)$.&nbsp; Specify the amplitudes &nbsp;$A_1$&nbsp; and &nbsp;$A_5$&nbsp; of the &nbsp;$1\ \rm kHz$ component and the &nbsp;$5\ \rm kHz$&nbsp; component.
 
|type="{}"}
 
|type="{}"}
$f_T = 50 kHz:  A_1$ = { 1.938 3% } $\text{ V }$  
+
$A_1 \ = \ $ { 1.938 3% } $\ \text{ V }$  
$f_T = 50 kHz:  A_5$ = { 1.368 3% } $\text{ V }$  
+
$A_5 \ = \ $ { 1.368 3% } $\ \text{ V }$  
  
  
{Berechnen Sie den resultierenden Frequenzgang $H_{MKD}(f)$ für $f_T = 55 kHz$. Welche Werte ergeben sich nun für $f = 1 kHz$ und $f = 5 kHz$?
+
{Calculate the resulting frequency response &nbsp;$H_{\rm MKD}(f)$&nbsp; for &nbsp;$f_{\rm T} = 55 \ \rm  kHz$. &nbsp; Now, what are the values for &nbsp;$f = 1 \ \rm  kHz$&nbsp; and &nbsp;$f = 5 \ \rm  kHz$?
 
|type="{}"}
 
|type="{}"}
$f_T = 55 kHz:  |H_{MKD} (f = 1 kHz)|$ = { 0.463 3 % }  
+
$|H_{\rm MKD} (f = 1\ \rm  kHz)| \ = \ $ { 0.463 3% }  
$f_T = 55 kHz:  |H_{MKD} (f = 5 kHz)|$ = { 0.521 3 % }
+
$|H_{\rm MKD} (f = 5\ \rm  kHz)| \ = \ $ { 0.521 3% }
  
{Berechnen Sie das Sinkensignal $υ(t)$. Geben Sie hierfür die Amplituden $A_1$ und $A_5$ des 1 kHz– bzw. 5 kHz–Anteils an.  
+
{Calculate the sink signal &nbsp;$v(t)$.&nbsp; Specify the amplitudes &nbsp;$A_1$&nbsp; and &nbsp;$A_5$&nbsp; of the two components.
 
|type="{}"}
 
|type="{}"}
$f_T = 55 kHz:  A_1$ = { 0.926 3% } $\text{ V }$  
+
$A_1 \ = \ $ { 0.926 3% } $\ \text{ V }$  
$f_T = 55 kHz:  A_1$ = { 1.563 3% } $\text{ V }$  
+
$A_5 \ = \ $ { 1.563 3% } $\ \text{ V }$  
  
{Gibt es eine Trägerfrequenz $f_T$, die bei dem gegebenen Quellensignal und dem gegebenen Kanal zu keinen Verzerrungen führt? Begründen Sie Ihre Antwort.
+
{Is there a carrier frequency $f_{\rm T}$ that results in no distortion for the given source signal and channel?&nbsp;  Justify your answer.
|type="[]"}
+
|type="()"}
+ ja
+
+ Yes,
- nein
+
- No.
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
+
[[File:P_ID1011__Mod_A_2_5_a.png|right|frame|Resulting baseband frequency response for&nbsp; $f_{\rm T} = f_{\rm M}$]]
'''2.'''
+
'''(1)'''&nbsp; The equation given states that the band-pass frequency response &nbsp; $H_{\rm K}(f)$&nbsp; has to be shifted to left and to right by the carrier frequency $f_{\rm T}$,&nbsp; and the two components have to be added up. 
'''3.'''
+
*The factor&nbsp; $1/2$&nbsp; must still be taken into account&nbsp; (see plot).
'''4.'''
+
 
'''5.'''
+
*At low frequencies,&nbsp; this results in a Gaussian function around the center frequency&nbsp;  "0":
'''6.'''
+
:$$H_{\rm MKD}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left ({f}/{\Delta f_{\rm K}}\right)^2} \hspace{0.05cm}.$$
'''7.'''
+
*The two components at &nbsp; $±2f_{\rm T}$&nbsp; need not be considered further.&nbsp;  For the two frequencies we are looking for &nbsp; $f_1 = 1\ \rm  kHz$&nbsp; and &nbsp; $f_5 = 5 \ \rm kHz$,&nbsp; we obtain:
 +
:$$ H_{\rm MKD}(f = f_1) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{1\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/100}\hspace{0.15cm}\underline {\approx 0.969} \hspace{0.05cm},$$
 +
:$$H_{\rm MKD}(f = f_5) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/4} \hspace{0.3cm}\hspace{0.15cm}\underline {\approx 0.456} \hspace{0.05cm}.$$
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; With&nbsp; $ω_1 = 2π · 1\ \rm  kHz$&nbsp; and&nbsp; $ω_5 = 2π · 5 \ \rm  kHz$,&nbsp; it holds that:
 +
:$$ v(t)  =  0.969 \cdot 2\,{\rm V}\cdot \cos (\omega_1 \cdot t)+ 0.456 \cdot 3\,{\rm V}\cdot \cos (\omega_5 \cdot t) =  \underline { 1.938\,{\rm V}}\cdot \cos (\omega_1 \cdot t) + \hspace{0.15cm}\underline {1.368\,{\rm V}}\cdot \cos (\omega_5 \cdot t) \hspace{0.05cm}.$$
 +
*It can be seen that now&nbsp; &ndash; unlike the source signal &nbsp; $q(t)$ &ndash; &nbsp; the component at &nbsp; $1 \ \rm kHz$ &nbsp; &rArr; &nbsp; $A_1 = 1.938 \ \rm V$&nbsp; is larger than the &nbsp;  $5 \ \rm kHz$ component &nbsp; &rArr; &nbsp; $A_5 = 1.368 \ \rm V$, because the channel attenuates the &nbsp; $49 \ \rm kHz$&nbsp; and&nbsp; $51 \ \rm kHz$&nbsp; frequencies less than the spectral components at &nbsp; $45 \ \rm kHz$&nbsp; and&nbsp; $55 \ \rm kHz$.
 +
 
 +
 
 +
 
 +
'''(3)'''&nbsp; The two spectral functions shifted by &nbsp; $±f_{\rm T}$&nbsp; now no longer come to lie directly on top of each other,&nbsp; but there are an offset from each other by&nbsp; $10 \ \rm kHz$&nbsp;.
 +
*The resulting frequency response&nbsp; $H_{\rm MKD}(f)$&nbsp; is thus no longer Gaussian,&nbsp; but characterized according to the sketch below:
 +
:$$H_{\rm MKD}(f ) = {1}/{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f - 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f + 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right]\hspace{0.05cm}.$$
 +
 
 +
*For the frequencies&nbsp; $f_1$&nbsp; and&nbsp; $f_5$&nbsp; we get:
 +
 
 +
:$$H_{\rm MKD}(f = 1\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 56\,{\rm kHz}) + H_{\rm K}(f = -54\,{\rm kHz})\right]=$$
 +
:$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{56\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-54\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.161 + 0.302 \hspace{0.15cm}\underline {= 0.463}\hspace{0.05cm},$$
 +
:$$H_{\rm MKD}(f = 5\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 60\,{\rm kHz}) + H_{\rm K}(f = -50\,{\rm kHz})\right]= \hspace{0.75cm}$$
 +
:$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{60\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-50\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.022 + 0.500 \hspace{0.15cm}\underline {= 0.521}\hspace{0.05cm}.$$
 +
[[File:P_ID1012__Mod_A_2_5_c.png|right|frame|Resulting baseband frequency response for $f_{\rm T} \ne f_{\rm M}$]]
 +
*While the synchronous demodulator extracts information about the message signal from both sidebands in the same way at &nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$,&nbsp; the lower sideband (LSB) provides the larger contribution at&nbsp; $f_{\rm T} = 55\ \rm  kHz$.
 +
 
 +
 +
*For example, the LSB of the &nbsp; $5 \ \rm kHz$ component is now exactly at &nbsp; $f_{\rm M} = 50 \ \rm kHz$&nbsp; and is transmitted undamped, while the USB is subject to heavy attenuation at &nbsp; $60 \ \rm kHz$&nbsp;.
 +
 
 +
 
 +
 
 +
'''(4)'''&nbsp; With the result of the previous subtask,&nbsp; one obtains:
 +
:$$ A_1  = 0.463 \cdot 2\,{\rm V}\hspace{0.15cm}\underline { = 0.926\,{\rm V}}\hspace{0.05cm},$$
 +
:$$A_5  = 0.521 \cdot 3\,{\rm V} \hspace{0.15cm}\underline {= 1.563\,{\rm V}}\hspace{0.05cm}.$$
 +
*In this case,&nbsp; the linear distortions are even less strong,&nbsp; since now also the&nbsp; $1 \ \rm kHz$ portion is attenuated more.
 +
 
 +
 
 +
 
 +
'''(5)'''&nbsp; YES is correct:
 +
*With the carrier frequency&nbsp; $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$,&nbsp; the&nbsp; $5 \ \rm kHz$&nbsp; component is more attenuated than the&nbsp; $1 \ \rm kHz$&nbsp; component,&nbsp; while at&nbsp; $f_{\rm T}  = 55 \ {\rm kHz} \ne f_{\rm M}$,&nbsp; the&nbsp; $1 \ \rm kHz$ component is slightly more attenuated.
 +
*If one chooses&nbsp; $f_{\rm T}  \approx 54.5 \ \rm kHz$&nbsp; for example,&nbsp; both components are attenuated equally &nbsp; $($by about the factor&nbsp; $0.53)$&nbsp; and there is no or less distortion.
 +
*However,&nbsp; this result is only valid for the source signal considered.  &nbsp; Another&nbsp; $q(t)$&nbsp; with two spectral components would require a different&nbsp; "optimal carrier frequency".&nbsp; For a source signal with three or more spectral lines,&nbsp; linear distortions would always occur.
 +
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Modulationsverfahren|^2.2 Synchrondemodulation^]]
+
[[Category:Modulation Methods: Exercises|^2.2 Synchronous Demodulation^]]

Latest revision as of 13:26, 6 December 2021

DSB-AM over a distorting channel

The communication system considered here is composed of the following blocks:

  • DSB-AM without carrier  $(f_{\rm T} = 50 \ \rm kHz$  or  $f_{\rm T} = 55 \ \rm kHz)$:
$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
  • Gaussian band-pass channel;  the magnitude  $|f|$  in the exponent causes  $H_K(–f) = H_K(f)$  to hold:
$$H_{\rm K}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (({|f| - f_{\rm M}})/{\Delta f_{\rm K}}\right)^2} ,\hspace{0.2cm} f_{\rm M} = 50\,{\rm kHz},\hspace{0.2cm} \Delta f_{\rm K} = 10\,{\rm kHz}\hspace{0.05cm}.$$
  • The synchronous demodulator has optimal parameters such that the sink signal  $v(t)$  completely coincides with the source signal  $q(t)$  when  $H_{\rm K}(f) = 1$  ("ideal channel").


On the page   Influence of linear channel distortions  it was shown that the entire system is is sufficiently accurately characterized by the resulting frequency response

$$H_{\rm MKD}(f) = {1}/{2} \cdot \big[ H_{\rm K}(f + f_{\rm T}) + H_{\rm K}(f - f_{\rm T})\big]$$

Here the subscript stands for  $\rm M$odulator – $\rm K$  (for German "Kanal" i.e. channel) – $\rm D$emodulator.

The source signal  $q(t)$  is composed of two cosine oscillations:

$$q(t) = 2\,{\rm V}\cdot \cos (2 \pi \cdot 1\,{\rm kHz} \cdot t)+ 3\,{\rm V}\cdot \cos (2 \pi \cdot 5\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$




Hints:


Questions

1

Calculate the resulting frequency response  $H_{\rm MKD}(f)$  for $f_{\rm T} = 50 \ \rm kHz$.  What are the values for $f = 1 \ \rm kHz$  and  $f = 5 \ \rm kHz$?

$|H_{\rm MKD} (f = 1\ \rm kHz)| \ = \ $

$|H_{\rm MKD} (f = 5\ \rm kHz)| \ = \ $

2

Calculate the sink signal  $v(t)$.  Specify the amplitudes  $A_1$  and  $A_5$  of the  $1\ \rm kHz$ component and the  $5\ \rm kHz$  component.

$A_1 \ = \ $

$\ \text{ V }$
$A_5 \ = \ $

$\ \text{ V }$

3

Calculate the resulting frequency response  $H_{\rm MKD}(f)$  for  $f_{\rm T} = 55 \ \rm kHz$.   Now, what are the values for  $f = 1 \ \rm kHz$  and  $f = 5 \ \rm kHz$?

$|H_{\rm MKD} (f = 1\ \rm kHz)| \ = \ $

$|H_{\rm MKD} (f = 5\ \rm kHz)| \ = \ $

4

Calculate the sink signal  $v(t)$.  Specify the amplitudes  $A_1$  and  $A_5$  of the two components.

$A_1 \ = \ $

$\ \text{ V }$
$A_5 \ = \ $

$\ \text{ V }$

5

Is there a carrier frequency $f_{\rm T}$ that results in no distortion for the given source signal and channel?  Justify your answer.

Yes,
No.


Solution

Resulting baseband frequency response for  $f_{\rm T} = f_{\rm M}$

(1)  The equation given states that the band-pass frequency response   $H_{\rm K}(f)$  has to be shifted to left and to right by the carrier frequency $f_{\rm T}$,  and the two components have to be added up.

  • The factor  $1/2$  must still be taken into account  (see plot).
  • At low frequencies,  this results in a Gaussian function around the center frequency  "0":
$$H_{\rm MKD}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left ({f}/{\Delta f_{\rm K}}\right)^2} \hspace{0.05cm}.$$
  • The two components at   $±2f_{\rm T}$  need not be considered further.  For the two frequencies we are looking for   $f_1 = 1\ \rm kHz$  and   $f_5 = 5 \ \rm kHz$,  we obtain:
$$ H_{\rm MKD}(f = f_1) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{1\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/100}\hspace{0.15cm}\underline {\approx 0.969} \hspace{0.05cm},$$
$$H_{\rm MKD}(f = f_5) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/4} \hspace{0.3cm}\hspace{0.15cm}\underline {\approx 0.456} \hspace{0.05cm}.$$


(2)  With  $ω_1 = 2π · 1\ \rm kHz$  and  $ω_5 = 2π · 5 \ \rm kHz$,  it holds that:

$$ v(t) = 0.969 \cdot 2\,{\rm V}\cdot \cos (\omega_1 \cdot t)+ 0.456 \cdot 3\,{\rm V}\cdot \cos (\omega_5 \cdot t) = \underline { 1.938\,{\rm V}}\cdot \cos (\omega_1 \cdot t) + \hspace{0.15cm}\underline {1.368\,{\rm V}}\cdot \cos (\omega_5 \cdot t) \hspace{0.05cm}.$$
  • It can be seen that now  – unlike the source signal   $q(t)$ –   the component at   $1 \ \rm kHz$   ⇒   $A_1 = 1.938 \ \rm V$  is larger than the   $5 \ \rm kHz$ component   ⇒   $A_5 = 1.368 \ \rm V$, because the channel attenuates the   $49 \ \rm kHz$  and  $51 \ \rm kHz$  frequencies less than the spectral components at   $45 \ \rm kHz$  and  $55 \ \rm kHz$.


(3)  The two spectral functions shifted by   $±f_{\rm T}$  now no longer come to lie directly on top of each other,  but there are an offset from each other by  $10 \ \rm kHz$ .

  • The resulting frequency response  $H_{\rm MKD}(f)$  is thus no longer Gaussian,  but characterized according to the sketch below:
$$H_{\rm MKD}(f ) = {1}/{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f - 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f + 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right]\hspace{0.05cm}.$$
  • For the frequencies  $f_1$  and  $f_5$  we get:
$$H_{\rm MKD}(f = 1\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 56\,{\rm kHz}) + H_{\rm K}(f = -54\,{\rm kHz})\right]=$$
$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{56\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-54\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.161 + 0.302 \hspace{0.15cm}\underline {= 0.463}\hspace{0.05cm},$$
$$H_{\rm MKD}(f = 5\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 60\,{\rm kHz}) + H_{\rm K}(f = -50\,{\rm kHz})\right]= \hspace{0.75cm}$$
$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{60\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-50\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.022 + 0.500 \hspace{0.15cm}\underline {= 0.521}\hspace{0.05cm}.$$
Resulting baseband frequency response for $f_{\rm T} \ne f_{\rm M}$
  • While the synchronous demodulator extracts information about the message signal from both sidebands in the same way at   $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$,  the lower sideband (LSB) provides the larger contribution at  $f_{\rm T} = 55\ \rm kHz$.


  • For example, the LSB of the   $5 \ \rm kHz$ component is now exactly at   $f_{\rm M} = 50 \ \rm kHz$  and is transmitted undamped, while the USB is subject to heavy attenuation at   $60 \ \rm kHz$ .


(4)  With the result of the previous subtask,  one obtains:

$$ A_1 = 0.463 \cdot 2\,{\rm V}\hspace{0.15cm}\underline { = 0.926\,{\rm V}}\hspace{0.05cm},$$
$$A_5 = 0.521 \cdot 3\,{\rm V} \hspace{0.15cm}\underline {= 1.563\,{\rm V}}\hspace{0.05cm}.$$
  • In this case,  the linear distortions are even less strong,  since now also the  $1 \ \rm kHz$ portion is attenuated more.


(5)  YES is correct:

  • With the carrier frequency  $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$,  the  $5 \ \rm kHz$  component is more attenuated than the  $1 \ \rm kHz$  component,  while at  $f_{\rm T} = 55 \ {\rm kHz} \ne f_{\rm M}$,  the  $1 \ \rm kHz$ component is slightly more attenuated.
  • If one chooses  $f_{\rm T} \approx 54.5 \ \rm kHz$  for example,  both components are attenuated equally   $($by about the factor  $0.53)$  and there is no or less distortion.
  • However,  this result is only valid for the source signal considered.   Another  $q(t)$  with two spectral components would require a different  "optimal carrier frequency".  For a source signal with three or more spectral lines,  linear distortions would always occur.