Aufgaben:Exercise 2.3: Sinusoidal Characteristic

• Return to book

Sinusoidal characteristic curve

We consider a system with input  $x(t)$  and output  $y(t)$. For simplicity of description, the signals are considered to be dimensionless.

The relationship between the input signal  $x(t)$  and the output signal  $y(t)$  is given in the range between  $-\pi/2$  and  $+\pi/2$  by the following characteristic curve.

$$g(x) = \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \hspace{0.05cm}\text{...}$$

The second part of this equation describes the series expansion of the sine function.

As approximations for the nonlinear characteristic curve the following is used in this task:

$$g_1(x) = x\hspace{0.05cm},$$

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

$$g_5(x) = x- x^3\hspace{-0.1cm}/{6}+x^5\hspace{-0.1cm}/{120}\hspace{0.05cm}.$$

The input signal  $x(t) = A \cdot \cos(\omega_0 \cdot t)$  is always assumed wherat the values  $A = 0.5$,  $A = 1.0$  and  $A = 1.5$  are to be considered for the (dimensionless) signal amplitude.

Please note:

• The task belongs to the chapter  Nonlinear Distortion.
• The resulting signal curves for  $x(t)$  and  $y(t)$  are shown graphically on the page  Description of nonlinear systems .

• All powers required here refer to the resistance  $R = 1 \ \rm \Omega$  and thus have the unit  ${\rm V}^2$.
• The following trigonometric relations are assumed to be known:

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

$$ \cos^5(\alpha) = {10}/{16} \cdot \cos(\alpha) + {5}/{16} \cdot \cos(3\alpha) + {1}/{16} \cdot \cos(5\alpha)\hspace{0.05cm}.$$

Questions

1

What distortion factor  $K$  is obtained with the approximation  $\underline{g_1(x)}$  of the characteristic curve independent of the amplitude  $A$  of the input signal?

$K \ = \ $ $\ \%$

2

Compute the distortion factor  $K$  for the input signal  $x(t) = A \cdot \cos(\omega_0 \cdot t)$  and the approximation  $\underline{g_3(x)}$.
What values arise as a result for  $A = 0.5$  and  $A = 1.0$?

$A = 0.5\hspace{-0.08cm}:\ \ K \ = \ $ $\ \%$

$A = 1.0\hspace{-0.08cm}:\ \ K \ = \ $ $\ \%$

3

What is the distortion factor for  $\underline{A = 1.0}$  considering the approximation  $\underline{g_5(x)}$?

$K \ = \ $ $\ \%$

4

Which of the following statements are true? Here,  $K$  denotes the distortion factor of the sine function  $g(x)$.
$K_{\rm g3}$  and  $K_{\rm g5}$  are based on the approximations  $g_3(x)$  and  $g_5(x)$, respectively.

	$K_{\rm g5}$  generally represents a better approximation for  $K$  than  $K_{\rm g3}$.
	$K_{\rm g3} < K_{\rm g5}$ holds for  $A = 1.0$  .
	$K_{\rm g3} \approx K_{\rm g5}$ will hold for  $A = 0.5$  .

Solution

Solution

(1)  The very inaccurate approximation  $g_1(x) = x$  is linear in  $x$  and therefore does not result in nonlinear distortions. Hence, the distortion factor is $\underline{K = 0}$.

(2)  The analytical spectrum (positive frequencies only) of the input signal is:

$$X_+(f) = A \cdot {\rm \delta}(f- f_0) .$$

• Then, the following signal is applied to the output of the nonlinear characteristic curve  $g_3(x)$ :

$$y(t) = A \cdot {\rm cos}(\omega_0 t ) - \frac{A^3}{6} \cdot {\rm cos}^3(\omega_0 t )= A \cdot {\rm cos}(\omega_0 t ) - \frac{3}{4} \cdot \frac{A^3}{6} \cdot {\rm cos}(\omega_0 t )- \frac{1}{4} \cdot \frac{A^3}{6} \cdot {\rm cos}(3\omega_0 t ) = A_1 \cdot {\rm cos}(\omega_0 t ) + A_3 \cdot {\rm cos}(3\omega_0 t ).$$

• For the coefficients  $A_1$  and $A_3$ the following is obtained by comparison of coefficients:

$$A_1 = A - {A^3}\hspace{-0.1cm}/{8}, \hspace{0.5cm}A_3 = - {A^3}\hspace{-0.1cm}/{24}.$$

• Using $A = 0.5$ the following is obtained: $A_1 \approx 0.484$ and  $A_3 \approx 0.005$. Thus, the distortion factor is:

$$K = K_3 ={|A_3|}/{A_1}= {0.005}/{0.484} \hspace{0.15cm}\underline{ = 1.08\%}.$$

Note that for the approximation  $g_3(x)$  only the cubic part  $K_3$  of the distortion factor is effective.

• For  $A = 1.0$  and  $A = 1.5$  the following numerical values arise as aresult:

$$A = 1.0: A_1 \approx 0.875, \hspace{0.2cm} A_3 \approx -0.041\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \hspace{0.15cm}\underline{K \approx 4.76\%}\; \; \Rightarrow \; \; K_{g3},$$

$$A = 1.5: A_1 \approx 1.078, \hspace{0.2cm} A_3 \approx -0.140\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \hspace{0.15cm}{K \approx 13 \%}.$$

(3)  Similarly as in subtask  (2) 

$$y(t) = A_1 \cdot {\rm cos}(\omega_0 t ) + A_3 \cdot {\rm cos}(3\omega_0 t )+ A_5 \cdot {\rm cos}(5\omega_0 t )$$

with the following coefficients:

$$A_1 = A - {A^3}\hspace{-0.1cm}/{8} + {A^5}\hspace{-0.1cm}/{192},\hspace{0.3cm} A_3 = - {A^3}\hspace{-0.1cm}/{24} + {A^5}\hspace{-0.1cm}/{384},\hspace{0.3cm} A_5 = {A^5}\hspace{-0.1cm}/{1920}.$$

• Daraus ergeben sich mit  $A=1$  die Zahlenwerte:

$$A_1 \approx 1 -0.125 +0.005 = 0.880,\hspace{0.3cm} A_3 \approx -0.042 +0.003 = -0.039,\hspace{0.3cm} A_5 \approx 0.0005$$

$$\Rightarrow \hspace{0.3cm}K_3 = {|A_3|}/{A_1}= 0.0443,\hspace{0.3cm}K_5 = {|A_5|}/{A_1}= 0.0006 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} K = \sqrt{K_3^2 + K_5^2} \hspace{0.15cm}\underline{\approx 4.45\%} \; \; \Rightarrow \; \; K_{g5}.$$

(4)  Richtig sind die Lösungsvorschläge 1 und 3:

• Der Ansatz  $g_5(x)$  ist im gesamten Bereich eine bessere Näherung für die Sinusfunktion  $g(x)$  als die Näherung  $g_3(x)$.
• Deshalb ist der in der Teilaufgabe  (3)  berechnete Wert  $K_{g5}$  eine bessere Näherung für den tatsächlichen Klirrfaktor als  $K_{g3}$.
  die erste Aussage ist somit richtig.
• Die zweite Aussage ist falsch, wie schon die Berechnung für  $A=1$  gezeigt hat:   $K_{g3} \approx 4.76 \%$  ist größer als  $K_{g5} \approx 4.45 \%$.
• Der Grund hierfür ist, dass $g_3(x)$ unterhalb von $g_5(x)$ liegt und damit auch eine größere Abweichung vom linearen Verlauf vorliegt.
• Für  $A=0.5$  wird  $K_{g5} \approx K_{g3} = 1.08 \%$  gelten.
• Die Kennlinie auf der Angabenseite zeigt, dass für  $|x| \le 0.5$  die Funktionen  $g_3(x)$  und  $g_5(x)$  innerhalb der Zeichengenauigkeit nicht zu unterscheiden sind.
• Damit ergeben sich auch gleiche Klirrfaktoren.