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

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{Is there a carrier frequency  $f_{\rm T}$ that results in no distortion for the given source signal and channel?  Justify your answer.
 
{Is there a carrier frequency  $f_{\rm T}$ that results in no distortion for the given source signal and channel?  Justify your answer.
 
|type="()"}
 
|type="()"}
+ Ja,  
+
+ Yes,  
- Nein.  
+
- No.  
  
  
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===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
[[File:P_ID1011__Mod_A_2_5_a.png|right|frame|Resulierender Basisbandfrequenzgang für  $f_{\rm T} = f_{\rm M}$]]
+
[[File:P_ID1011__Mod_A_2_5_a.png|right|frame|Resulting baseband frequency response for  $f_{\rm T} = f_{\rm M}$]]
'''(1)'''  Die angegebene Gleichung besagt, dass der BP–Frequenzgang  $H_{\rm K}(f)$  jeweils um die Trägerfrequenz  $f_{\rm T}$  nach links und rechts verschoben und die beiden Anteile aufaddiert werden müssen. 
+
'''(1)'''  The equation given states that the BP frequency response   $H_{\rm K}(f)$  has to be shifted left and right by the carrier frequency   $f_{\rm T}$ , respectively, and the two components have to be added up.
*Es ist noch der Faktor  $1/2$  zu berücksichtigen (siehe Skizze).
+
*The factor   $1/2$  must still be taken into account (see plot).
  
*Bei niedrigen Frequenzen ergibt sich dann eine Gaußfunktion um die Mittenfrequenz „0”:
+
*At low frequencies, this results in a Gaussian function around the center frequency  "0":
 
:$$H_{\rm MKD}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left ({f}/{\Delta f_{\rm K}}\right)^2} \hspace{0.05cm}.$$
 
:$$H_{\rm MKD}(f) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left ({f}/{\Delta f_{\rm K}}\right)^2} \hspace{0.05cm}.$$
*Die beiden Anteile bei  $±2f_{\rm T}$  müssen nicht weiter betrachtet werden.  Für die zwei gesuchten Frequenzen  $f_1 = 1\ \rm  kHz$  und  $f_5 = 5 \ \rm kHz$  erhält man:
+
*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:
 
:$$ H_{\rm MKD}(f = f_1) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{1\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/100}\hspace{0.15cm}\underline {\approx 0.969} \hspace{0.05cm},$$  
 
:$$ H_{\rm MKD}(f = f_1) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{1\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/100}\hspace{0.15cm}\underline {\approx 0.969} \hspace{0.05cm},$$  
 
:$$H_{\rm MKD}(f = f_5) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/4} \hspace{0.3cm}\hspace{0.15cm}\underline {\approx 0.456} \hspace{0.05cm}.$$
 
:$$H_{\rm MKD}(f = f_5) = {\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2} = {\rm e}^{-\pi/4} \hspace{0.3cm}\hspace{0.15cm}\underline {\approx 0.456} \hspace{0.05cm}.$$
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'''(2)'''  Mit  $ω_1 = 2π · 1\ \rm  kHz$  und  $ω_5 = 2π · 5 \ \rm  kHz$  gilt:
+
'''(2)'''  With  $ω_1 = 2π · 1\ \rm  kHz$  and  $ω_5 = 2π · 5 \ \rm  kHz$ , it holds that:
 
:$$ v(t)  =  0.969 \cdot 2\,{\rm V}\cdot \cos (\omega_1 \cdot t)+ 0.456 \cdot 3\,{\rm V}\cdot \cos (\omega_5 \cdot t) =  \underline { 1.938\,{\rm V}}\cdot \cos (\omega_1 \cdot t) + \hspace{0.15cm}\underline {1.368\,{\rm V}}\cdot \cos (\omega_5 \cdot t) \hspace{0.05cm}.$$  
 
:$$ v(t)  =  0.969 \cdot 2\,{\rm V}\cdot \cos (\omega_1 \cdot t)+ 0.456 \cdot 3\,{\rm V}\cdot \cos (\omega_5 \cdot t) =  \underline { 1.938\,{\rm V}}\cdot \cos (\omega_1 \cdot t) + \hspace{0.15cm}\underline {1.368\,{\rm V}}\cdot \cos (\omega_5 \cdot t) \hspace{0.05cm}.$$  
*Man erkennt, dass nun –  im Gegensatz zum Quellensignal  $q(t)$  – der Anteil bei  $1 \ \rm kHz$   ⇒   $A_1 = 1.938 \ \rm V$  größer ist als der   $5 \ \rm kHz$–Anteil   ⇒   $A_5 = 1.368 \ \rm V$, da der Kanal die Frequenzen  $49 \ \rm kHz$  und  $51 \ \rm kHz$  weniger dämpft als die Spektralanteile bei  $45 \ \rm kHz$  und  $55 \ \rm kHz$.
+
*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   $45 \ \rm kHz$  und  $55 \ \rm kHz$.
  
  
  
'''(3)'''  Die beiden um  $±f_{\rm T}$  verschobenen Spektralfunktionen kommen nun nicht mehr direkt übereinander zu liegen, sondern sind um  $10 \ \rm kHz$  gegeneinander versetzt.  
+
'''(3)'''  The two spectral functions shifted by   $±f_{\rm T}$  now no longer come to lie directly on top of each other, but are offset from each other by   $10 \ \rm kHz$ .  
*Der resultierende Frequenzgang  $H_{\rm MKD}(f)$  ist somit nicht mehr gaußförmig, sondern es gilt entsprechend der unteren Skizze:
+
*The resulting frequency response   $H_{\rm MKD}(f)$  is thus no longer Gaussian, but characterized according to the sketch below:
 
:$$H_{\rm MKD}(f ) = {1}/{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f - 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f + 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right]\hspace{0.05cm}.$$
 
:$$H_{\rm MKD}(f ) = {1}/{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f - 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{f + 5\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right]\hspace{0.05cm}.$$
  
*Für die Frequenzen  $f_1$  und  $f_5$  erhält man:
+
*For the frequencies  $f_1$  and  $f_5$  we get:
  
 
:$$H_{\rm MKD}(f = 1\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 56\,{\rm kHz}) + H_{\rm K}(f = -54\,{\rm kHz})\right]=$$
 
:$$H_{\rm MKD}(f = 1\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 56\,{\rm kHz}) + H_{\rm K}(f = -54\,{\rm kHz})\right]=$$
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:$$H_{\rm MKD}(f = 5\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 60\,{\rm kHz}) + H_{\rm K}(f = -50\,{\rm kHz})\right]= \hspace{0.75cm}$$  
 
:$$H_{\rm MKD}(f = 5\,{\rm kHz}) = \frac{1}{2} \cdot \left[ H_{\rm K}(f = 60\,{\rm kHz}) + H_{\rm K}(f = -50\,{\rm kHz})\right]= \hspace{0.75cm}$$  
 
:$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{60\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-50\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.022 + 0.500 \hspace{0.15cm}\underline {= 0.521}\hspace{0.05cm}.$$
 
:$$\hspace{1.25cm}= \frac{1}{2}\cdot \left[{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{60\, {\rm kHz}- 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}+{\rm e}^{-\pi \cdot \hspace{0.05cm} \left (\frac{-50\, {\rm kHz}+ 50\,{\rm kHz}}{10\,{\rm kHz}}\right)^2}\right] = 0.022 + 0.500 \hspace{0.15cm}\underline {= 0.521}\hspace{0.05cm}.$$
[[File:P_ID1012__Mod_A_2_5_c.png|right|frame|Resulierender Basisbandfrequenzgang für $f_{\rm T} \ne f_{\rm M}$]]
+
[[File:P_ID1012__Mod_A_2_5_c.png|right|frame|Resulting baseband frequency response for $f_{\rm T} \ne f_{\rm M}$]]
*Während bei  $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$  der Synchrondemodulator die Information über das Nachrichtensignal aus beiden Seitenbändern in gleicher Weise gewinnt, liefert mit  $f_{\rm T} = 55\ \rm  kHz$  das untere Seitenband (USB) den größeren Beitrag.
+
*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  $f_{\rm T} = 55\ \rm  kHz$ .
  
 
   
 
   
*Zum Beispiel liegt das USB des  $5 \ \rm kHz$–Anteils nun genau bei  $f_{\rm M} = 50 \ \rm kHz$  und wird ungedämpft übertragen, während das OSB bei  $60 \ \rm kHz$  starken Dämpfungen unterliegt.
+
*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)'''  Mit dem Ergebnis der letzten Teilaufgabe erhält man:
+
'''(4)'''  With the result of the previous subtask, one obtains:
 
:$$ A_1  = 0.463 \cdot 2\,{\rm V}\hspace{0.15cm}\underline { = 0.926\,{\rm V}}\hspace{0.05cm},$$
 
:$$ A_1  = 0.463 \cdot 2\,{\rm V}\hspace{0.15cm}\underline { = 0.926\,{\rm V}}\hspace{0.05cm},$$
 
:$$A_5  = 0.521 \cdot 3\,{\rm V} \hspace{0.15cm}\underline {= 1.563\,{\rm V}}\hspace{0.05cm}.$$
 
:$$A_5  = 0.521 \cdot 3\,{\rm V} \hspace{0.15cm}\underline {= 1.563\,{\rm V}}\hspace{0.05cm}.$$
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'''(5)'''  Richtig ist JA:
+
'''(5)'''  YES is correct:
*Mit der Trägerfrequenz  $f_{\rm T} = f_{\rm M} = 50 \ \rm kHz$  wird der  $5 \ \rm kHz$–Anteil stärker gedämpft als der  $1 \ \rm kHz$–Anteil, während mit  $f_{\rm T}  = 55 \ {\rm kHz} \ne f_{\rm M}$  der  $1 \ \rm kHz$–Anteil etwas mehr gedämpft wird.  
+
*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.
*Wählt man nun zum Beispiel  $f_{\rm T}  \approx 54.5 \ \rm kHz$, so werden beide Anteile gleich gedämpft  $($etwa um den Faktor $0.53)$  und es gibt keine / weniger Verzerrungen.  
+
*If one then chooses  $f_{\rm T}  \approx 54.5 \ \rm kHz$ for example, both components are attenuated equally   $($etwa um den Faktor $0.53)$  and there is no or less distortion.
*Dieses Ergebnis gilt allerdings nur für das betrachtete Quellensignal.  Ein anderes  $q(t)$  mit ebenfalls zwei Spektralanteilen würde eine andere „optimale Trägerfrequenz” erfordern.  
+
*However, this result is only valid for the source signal considered.   Another   $q(t)$  also with two spectral components would require a different "optimal carrier frequency".  
*Bei einem Nachrichtensignal mit drei oder mehr Spektrallinien würde es stets zu linearen Verzerrungen kommen.
+
*For a message signal with three or more spectral lines, linear distortions would always occur.
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Revision as of 19:09, 1 December 2021

DSB-AM over a distorting channel

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

  • DSB-AM without carrier with  $f_{\rm T} = 50 \ \rm kHz$  and  $f_{\rm T} = 55 \ \rm kHz$:
$$ s(t) = q(t) \cdot \cos (2 \pi f_{\rm T} \hspace{0.05cm} t).$$
  • GGaussian bandpass 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 index stands for Modulator–K (for german "Kanal" i.e. Channel) –Demodulator.

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  $1\ \rm kHz$– and  $5\ \rm kHz$ components, respectively.

$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  $1\ \rm kHz$– and  $5\ \rm kHz$ components, respectively.

$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 BP frequency response   $H_{\rm K}(f)$  has to be shifted left and right by the carrier frequency   $f_{\rm T}$ , respectively, 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   $45 \ \rm kHz$  und  $55 \ \rm kHz$.


(3)  The two spectral functions shifted by   $±f_{\rm T}$  now no longer come to lie directly on top of each other, but are 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 diesem Fall sind die linearen Verzerrungen sogar weniger stark, da nun auch der  $1 \ \rm kHz$–Anteil stärker gedämpft wird.


(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 then chooses  $f_{\rm T} \approx 54.5 \ \rm kHz$ for example, both components are attenuated equally   $($etwa um den Faktor $0.53)$  and there is no or less distortion.
  • However, this result is only valid for the source signal considered.   Another   $q(t)$  also with two spectral components would require a different "optimal carrier frequency".
  • For a message signal with three or more spectral lines, linear distortions would always occur.