Aufgaben:Exercise 2.5: DSB-AM via a Gaussian channel

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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:

• This exercise belongs to the chapter  Synchronous Demodulation.
• Particular reference is made to the page  Influence of linear channel distortions.

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

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.