Aufgaben:Exercise 2.3Z: Asymmetrical Characteristic Operation

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System and signal examples

The cosine signal

$$x(t) = A \cdot \cos(\omega_0 t)$$

is applied to the input of a system  $S$  where  $A = 0.5$  shall always hold for the amplitude.  The system  $S$  consists of

• the addition of a direct (DC) component  $C$,
• a nonlinearity with the characteristic curve

$$g(x) = \sin(x) \hspace{0.05cm} \approx x -{x^3}\hspace{-0.1cm}/{6} = g_3(x),$$

• as well as an ideal high-pass filter that allows all frequencies to pass unaltered except for a direct (DC) signal  $(f = 0)$.

The output signal of the overall system can generally be depicted as follows:

$$y(t) = A_0 + A_1 \cdot \cos(\omega_0 t) + A_2 \cdot \cos(2\omega_0 t) + A_3 \cdot \cos(3\omega_0 t) + \hspace{0.05cm}\text{...}$$

The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.

This would result in exactly the same constellation as in  Exercise 2.3  for  $C = 0$  in whose subtask  (2)  the distortion factor was calculated:

• $K = K_{g3} \approx 1.08 \%$  für  $A = 0.5$,
• $K = K_{g3} \approx 4.76 \%$  für  $A = 1.0$.

Considering the constants  $A = C = 0.5$  the following holds for the input signal of the nonlinearity:

$$x_{\rm C}(t) = C + A \cdot \cos(\omega_0 t) = {1}/{2} + {1}/{2}\cdot \cos(\omega_0 t).$$

• So,  the characteristic curve is operated asymmetrically with values between  $0$  and  $1$.
• In the above graph,  the signals  $x_{\rm C}(t)$  and  $y_{\rm C}(t)$  are plotted additionally directly before and after the characteristic curve  $g(x)$ .

Please note:

• The exercise belongs to the chapter  Nonlinear Distortions.

• The following trigonometric relations are assumed to be known:

$$\cos^2(\alpha) = {1}/{2} + {1}/{2} \cdot \cos(2\alpha)\hspace{0.05cm}, \hspace{0.3cm} \cos^3(\alpha) = {3}/{4} \cdot \cos(\alpha) + {1}/{4} \cdot \cos(3\alpha) \hspace{0.05cm}.$$

Questions

1

Compute the output signal  $y(t)$  considering the high-pass filter.  What is the direct (DC) signal component  $A_0$?

$A_0 \ = \ $

2

State the other Fourier coefficients of the signal  $y(t)$ .

$A_1 \ = \ $

$A_2 \ = \ $

$A_3 \ = \ $

$A_4 \ = \ $

3

Compute the distortion factor of the overall system.

$K \ = \ $ $\ \%$

4

Compute the maximum and the minimum value of the signal  $y(t)$.

$y_\text{max} \ = \ $

$y_\text{min} \ = \ $

Solution

Solution

(1)  Considering the cubic approximation $g_3(x)$  the following is obtained before the high-pass filter:

$$y_{\rm C}(t) = g_3\big[x_{\rm C}(t)\big] = \big[ C + A \cdot \cos(\omega_0 t)\big] - {1}/{6} \cdot \big[ C + A \cdot \cos(\omega_0 t)\big]^3 $$

$$\Rightarrow \; y_{\rm C}(t) = C + A \cdot \cos(\omega_0 t) - {1}/{6} \cdot \big[ C^3 + 3 \cdot C^2 \cdot A \cdot \cos(\omega_0 t) + \hspace{0.09cm}3 \cdot C \cdot A^2 \cdot \cos^2(\omega_0 t) + A^3 \cdot \cos^3(\omega_0 t)\big].$$

• The signal  $y_{\rm C}(t)$  contains a direct (DC) component  $C - C^3/6$  which is no longer included in the signal  $y(t)$  due to the high-pass filter:

$$\underline{ A_0 = 0}.$$

(2)  Applying the given trigonometric relations the following coefficients with  $A= C = 0.5$  are obtained:

$$A_1 = A - {1}/{6}\cdot 3 \cdot C^2 \cdot A - {1}/{6} \cdot {3}/{4}\cdot A^3 = {1}/{2} - {1}/{16} - {1}/{64} = {27}/{64} \hspace{0.15cm}\underline{ \approx 0.422},$$

$$A_2 = - {1}/{6}\cdot 3 \cdot {1}/{2}\cdot C \cdot A^2 = - \frac{1}{32} \hspace{0.15cm}\underline{\approx -0.031},$$

$$A_3 = - {1}/{6}\cdot \frac{1}{4}\cdot A^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$

• Higher order terms do not occur.  Thus,   $\underline{A_4 = 0}$  holds.

(3)  In this task,  the higher order distortion factors are  $K_2 = 2/27 \approx 7.41\%$  and  $K_3 = 1/81 \approx 1.23\%$.

• Thereby, the following is obtained for the overall distortion factor:

$$K = \sqrt{K_2^2 + K_3^2} \hspace{0.15cm}\underline{\approx7.51 \%}.$$

(4)  The maximum value occurs at time  $t = 0$  and at multiples of  $T$ :

$$y_{\rm max}= y(t=0) = A_1 + A_2 + A_3 = 0.422 -0.031 -0.005 \hspace{0.15cm}\underline{= 0.386}.$$

• The minimum values are located exactly in the middle between two maxima and it holds that:

$$y_{\rm min}= - A_1 + A_2 - A_3 = -0.422 -0.031 +0.005\hspace{0.15cm}\underline{ = -0.448}.$$

• The signal  $y(t)$  is shifted downward by  $0.448$  compared to the signal  drawn in the sketch on the information page.
• This signal value is obtained from the following equation considering  $A = C = 1/2$:

$$C - \frac{C \cdot A^2}{4}- \frac{C^3}{6} = {1}/{2} - {1}/{32}- {1}/{48} = 0.448.$$