Aufgaben:Exercise 2.8: Asymmetrical Channel

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Equivalent low-pass signal
in the complex plane

A cosine-shaped source signal  $q(t)$  with amplitude  $A_{\rm N}$  and frequency  $f_{\rm N}$  is DSB amplitude modulated, such that the modulated signal is given by:

$$ s(t) = \big[ q(t) + A_{\rm T}\big] \cdot \cos(2 \pi \cdot f_{\rm T} \cdot t ) \hspace{0.05cm}.$$

The transmission channel exhibits linear distortions:

• While the lower sideband $($LSB frequency:    $f_{\rm T} - f_{\rm N})$  and the carrier are transmitted undistorted,
• the upper sideband $($USB-Frequenz:    $f_{\rm T} + f_{\rm N})$  is weighted with the attenuation factor  $α_{\rm O} = 0.25$ .

The graph shows the locus curve, i.e., the representation of the equivalent low-pass signal  $r_{\rm TP}(t)$  in the complex plane.

Evaluating the signal  $r(t)$  with an ideal envelope demodulator, we obtain a sink signal $v(t)$, which can be approximated as follows:

$$v(t) = 2.424 \,{\rm V} \cdot \cos(\omega_{\rm N} \cdot t ) -0.148 \,{\rm V} \cdot \cos(2\omega_{\rm N} \cdot t )+ 0.056 \,{\rm V} \cdot \cos(3\omega_{\rm N} \cdot t )-\text{ ...}$$

For this measurement, the message frequency  $f_{\rm N} = 2 \ \rm kHz$  was used.

In subtask  (7)  the signal-to-noise power ratio   $\rm (SNR)$  should be calculated as follows:

$$ \rho_{v } = \frac{P_{v 1}}{P_{\varepsilon }} \hspace{0.05cm}.$$

Here,  $P_{v1} = α^2 · P_q$  and  $P_ε$  denote the "powers" of both signals:

$$ v_1(t) = 2.424 \,{\rm V} \cdot \cos(\omega_{\rm N} \cdot t )\hspace{0.05cm},$$

$$ \varepsilon(t) = v(t) - v_1(t) \approx -0.148 \,{\rm V} \cdot \cos(2\omega_{\rm N} \cdot t )+ 0.056 \,{\rm V} \cdot \cos(3\omega_{\rm N} \cdot t ) \hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  Envelope Demodulation.
• Particular reference is made to the page  Description using the equivalent low-pass signal.

Questions

1

Give the low-pass signal  $r_{\rm TP}(t)$  in its analytical form. What value results for time  $t = 0$?

$r_{\rm TP}(t=0) \ = \ $ $\ \rm V$

2

What are the amplitude values  $A_{\rm T}$  and  $A_{\rm N}$?

$A_{\rm T} \ = \ $ $\ \rm V$

$A_{\rm N} \ = \ $ $\ \rm V$

3

Let  $f_{\rm N} \hspace{0.15cm}\underline{= 2 \ \rm kHz}$.  At which time  $t_1$  is the starting point   (1)  first reached again after  $t = 0$ ?

$t_1 \ = \ $ $\ \rm ms$

4

At which point in time  $t_2$  is the elliptical point   (2)  with value  $\rm j · 3\ V$  first reached?

$t_2 \ = \ $ $\ \rm ms$

5

Calculate the magnitude function (envelope)  $a(t)$  and the phase function  $ϕ(t)$  for this time point  $t_2$.

$a(t = t_2) \ = \ $ $\ \rm V$

$ϕ(t = t_2)\ = \ $ $\ \rm degrees$

6

Calculate the distortion factor  $K$  for  $f_{\rm N} \hspace{0.15cm}\underline{= 2 \ \rm kHz}$.

$K \ = \ $ $\ \text{%}$

7

Calculate the signal-to-noise power ratio  $\rm (SNR)$ for  $f_{\rm N}\hspace{0.15cm}\underline{ = 2 \ \rm kHz}$  according to the given definition.

$ρ_v \ = \ $

8

What distortion factor results from otherwise equal conditions for the message frequency  $f_{\rm N} \hspace{0.15cm}\underline{= 4 \ \rm kHz}$?

$K \ = \ $ $\ \text{%}$

Solution

Solution

(1)  For a cosine-shaped source signal and attenuation of the upper sideband, it holds that:

$$ r_{\rm TP}(t) = A_{\rm T} + \frac{A_{\rm N}}{2} \cdot \alpha_{\rm O} \cdot{\rm e}^{{\rm j} \cdot \hspace{0.03cm}\omega_{\rm N}\cdot t} + \frac{A_{\rm N}}{2} \cdot{\rm e}^{-{\rm j} \cdot \hspace{0.03cm}\omega_{\rm N}\cdot t}\hspace{0.05cm}.$$

• At time   $t = 0$  all vectors point in the direction of the real axis.
• Thus   $r_{\rm TP}(t = 0)\hspace{0.15cm}\underline { = 15 \ \rm V}$  can be read from the graph on the exercise page.

(2)  The carrier amplitude is defined by the center of the ellipse:   $A_{\rm T}\hspace{0.15cm}\underline { = 10 \ \rm V}$.

• From the equation given in the first subtask, the amplitude  $A_{\rm N}$  can thus also be calculated:

$$ \frac{A_{\rm N}}{2} \cdot ( 1+ \alpha_0) = r_{\rm TP}(t= 0) - A_{\rm T} = 5 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm N} \hspace{0.15cm}\underline {= 8 \,{\rm V}} \hspace{0.05cm}.$$

• The point marked   (2)  can be used as a check:

$$\frac{A_{\rm N}}{2} \cdot ( 1- \alpha_0) = 3 \,{\rm V}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A_{\rm N} = 8 \,{\rm V} \hspace{0.05cm}.$$

(3)  The necessary time for one cycle t  $t_1$  is equal to the period of the source signal, i.e.,

$$t_1= 1/f_{\rm N} \hspace{0.15cm}\underline {=0.5 \ \rm ms}.$$

(4)  Since the lower sideband is larger than the upper sideband, the peak of the pointer composite moves clockwise around the ellipse.

• Point  (2)  is first reached at time  $t_2 = 3/4 · t_1\hspace{0.15cm}\underline { = 0.375 \ \rm ms}$  .

Calculation of   $t_2$  and  $t_3$

(5)  The pointer length at time   $t_2$  can be determined with the   Pythagorean Theorem :

$$ a(t = t_2) = \sqrt{(10 \,{\rm V})^2 + (3 \,{\rm V})^2}\hspace{0.15cm}\underline { = 10.44 \,{\rm V}}\hspace{0.05cm}.$$

• The phase function is:

$$\phi(t = t_2) = {\rm arctan} \frac{3 \,{\rm V}}{10 \,{\rm V}} \hspace{0.15cm}\underline {= 16.7^{\circ}}\hspace{0.05cm}.$$

• The maximum phase  $ϕ_{\rm max}$  is slightly larger.  It occurs (with a positive sign) at time   $t_3 < t_2$  when a straight line from the origin is tangent to the ellipse.
• By setting up the ellipse equation, this point   $(x_3$,  $y_3)$  can be accurately calculated analytically.
• From this, the following would hold for the maximum phase:

  $\phi_{\rm max} = {\rm arctan} \ {y_3}/{x_3} \hspace{0.05cm}.$

(6)  The distortion factors of second and third ordern can be obtained from the given equation for nbsp; $v(t)$  $($valid for $f_{\rm N} = 2 \ \rm kHz)$ , and are:

$$ K_2 = \frac{0.148 \,{\rm V}}{2.424 \,{\rm V}} = 0.061, \hspace{0.3cm} K_3 = \frac{0.056 \,{\rm V}}{2.424 \,{\rm V}} = 0.023 \hspace{0.05cm}.$$

• Thus for the total distortion factor we get:

$$K = \sqrt{K_2^2 + K_3^2 }\hspace{0.15cm}\underline { \approx 6.6 \%}.$$

(7)  Frot he power of the useful signal and the interference signal, we obtain:

$$ P_{v 1} = \frac{(2.424 \,{\rm V})^2}{2} = 2.94 \,{\rm V}^2\hspace{0.05cm},\hspace{0.3cm} P_{\varepsilon} = \frac{(-0.148 \,{\rm V})^2}{2} + \frac{(0.056 \,{\rm V})^2}{2}= 0.0125 \,{\rm V}^2\hspace{0.05cm}$$

• This gives the signal-to-noise power ratio   $\rm (SNR)$:

$$\rho_{v} = \frac{P_{v 1}}{P_{\varepsilon }}= \frac{(2.94 \,{\rm V})^2}{0.0125 \,{\rm V}^2} \hspace{0.15cm}\underline {\approx 230} = \frac{1}{K^2} \hspace{0.05cm}.$$

• If, on the other hand, the amplitude distortion were also assigned to the error signal, we would arrive at a much smaller  $\rm SNR$.
• When $P_q = A_{\rm N}^2/2 = 8 \ \rm V^2$  and  $P_{\varepsilon}\hspace{0.02cm}' = \overline{(v(t)-q(t))^2} = {1}/{2}\cdot ( 4 \,{\rm V} - 2.424 \,{\rm V})^2 + P_{\varepsilon}= 1.254 \,{\rm V}^2$  one would get:

$$\rho_{v }\hspace{0.02cm}' = \frac{8 \,{\rm V}^2}{1.254 \,{\rm V}^2} \approx 6.4\hspace{0.05cm}.$$

(8)  All calculations are valid regardless of the message frequency   $f_{\rm N}$ if the attenuation factor of the upper sideband remains at   $α_{\rm O} = 0.25$ .

• Thus, the same distortion factor   $K\hspace{0.15cm}\underline { \approx 6.6 \%}$ is obtained for   $f_{\rm N} = 4 \ \rm kHz$ .