Exercise 2.4Z: Characteristics Measurement

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Given characteristic  $y = g(x)$

It is known that the characteristic curve can be represented as follows for a nonlinear system:

$$y(t) = c_1 \cdot x(t) + c_2 \cdot x^2(t).$$

Since the distortions are nonlinear no frequency response  $H(f)$  can be given.

To determine the dimensionless coefficient  $c_1$  as well as the quadratic coefficient  $c_2$  different rectangular pulses  $x(t)$  – characterized by the amplitude  $A_x$  and the width  $T_x$  – are now applied to the input and the pulse amplitude  $A_y$  at the output is measured in each case.

The first three trials generate the following values:

• $A_x = 1 \ {\rm V}, \; \; T_x = 8 \ {\rm ms}$ :     $A_y = 0.55 \ {\rm V}$,
• $A_x = 2 \ {\rm V}, \; \; T_x = 4 \ {\rm ms}$ :     $A_y = 1.20 \ {\rm V}$,
• $A_x = 3 \ {\rm V}, \; \; T_x = 2 \ {\rm ms}$ :     $A_y = 1.95 \ {\rm V}$.

For the subtasks  (3)  and  (4)  let the input signal  $x(t)$  be a harmonic oscillation because only for such an oscillation a distortion factor can be specified.

In contrast, a triangular pulse with amplitude  $A_x = 3 \ {\rm V}$  and the one-sided pulse duration  $T_x = 2 \ {\rm ms}$  is considered for the subtask  (5) :

$$x(t) = A_x \cdot ( 1 - {|t|}/{T_x})$$

Please note:

• The exercise belongs to the chapter  Nonlinear Distortions.

• The following abbreviations are used in the formulation of the questions:

$$y_1(t) = c_1 \cdot x(t), \hspace{0.5cm} y_2(t) = c_2 \cdot x^2(t).$$

Questions

1

A rectangular pulse  $x(t)$  with amplitude  $A_x$  and duration  $T_x$  is applied to the input. 
Which statements hold for the output pulse  $y(t)$?

	The output pulse  $y(t)$  is triangular in shape.
	The amplitudes at the input and output are the same  ⇒   $A_y = A_x$.
	The pulse duration is not changed by the system  ⇒   $T_y = T_x$.

2

Compute the first two coefficients of the Taylor series.

$c_1 \ = \ $

$c_2 \ = \ $ $\ \rm 1/V$

3

Which distortion factor  $K$  is measured with the test signal  $x(t) = 1 \hspace{0.08cm} {\rm V} \cdot \cos(\omega_0 \cdot t)$ ?  That is:   $\underline{A_x = 1\hspace{0.08cm} \rm V}$.

$K \ = \ $ $\ \%$

4

Which distortion factor  $K$  is measured with the test signal  $x(t) = 3 \hspace{0.08cm} {\rm V} \cdot \cos(\omega_0 \cdot t)$ ?  That is:   $\underline{A_x = 3\hspace{0.08cm} \rm V}$.

$K \ = \ $ $\ \%$

5

Which output pulse  $y(t)$  arises as a result when the input pulse is triangular?  What are the signal values at  $t = 0$  and  $t = T_x/2$?

$y(t = 0) \ = \ $ $\ \rm V$

$y(t = T_x/2) \ = \ $ $\ \rm V$

Solution

Solution

(1)  Proposed solution 3 is the only correct one:

• If the input pulse $x(t)$  is rectangular, then $x^2(t)$  is also a rectangle with height $A_x^2$  between  $0$  and  $T_x$,  outside zero.
• The overall output signal $y(t)$  is thus also rectangular with the amplitude

$$A_y= c_1 \cdot A_x + c_2 \cdot A_x^2 .$$

• The following holds for the pulse duration:   $T_y = T_x$.

(2)  The following system of linear equations can be specified with the first two sets of parameters:

$$c_1 \cdot 1\,{\rm V} + c_2 \cdot (1\,{\rm V})^2 = 0.55\,{\rm V},$$

$$c_1 \cdot 2\,{\rm V} + c_2 \cdot (2\,{\rm V})^2 = 1.20\,{\rm V}.\hspace{0.05cm}$$

• The following is obtained by multiplying the first equation by  $-2$  and adding the two equations:

$$c_2 \cdot 2\,{\rm V}^2 = 0.1\,{\rm V} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} c_2 \hspace{0.15cm}\underline{= 0.05\cdot{1/\rm V}}.$$

• The linear coefficient is thus  $c_1 \hspace{0.15cm}\underline{= 0.5}.$

• The third set of parameters can be used to verify the result:

$$c_1 \cdot 3\,{\rm V} + c_2 \cdot (3\,{\rm V})^2 = 0.5 \cdot 3\,{\rm V}+ 0.05 \ {1}/{\rm V}\cdot 9\,{\rm V}^2 = 1.95\,{\rm V}.$$

(3)  The specification of a distortion factor requires the use of a harmonic oscillation at the input.

• If  $X_+(f) = 1 \ {\rm V} \cdot \delta (f - f_0)$  holds,  then the spectrum of the analytic signal at the output is:

$$Y_{+}(f)={c_2}/{2}\cdot A_x^2 \cdot \delta(f) + c_1\cdot A_x \cdot \delta(f- f_0)+ {c_2}/{2}\cdot A_x^2 \cdot \delta(f- 2 f_0).$$

• The Dirac function at  $f = 0$  follows from the trigonometric transformation  $\cos^2(\alpha) = 1/2 + 1/2 \cdot \cos(\alpha).$

• With  $A_1 = c_1 \cdot A_x = 0.5 \ {\rm V}$  and  $A_2 = (c_2/2) \cdot A_x^2 = 0.025 \ {\rm V}^2$  the following is thus obtained for the distortion factor:

$$K= \frac{A_2}{A_1}= \frac{c_2/2 \cdot A_x}{c_1 }= \frac{0.025}{0.5} \hspace{0.15cm}\underline{= 5 \%}.$$

(4)  According to the solution of the last subtask  $K$  is proportional to  $A_x$. Therefore, one now obtains  $K \hspace{0.15cm}\underline{= 15 \%}.$

(5)  Now the output signal is:

$$y(t)= c_1\cdot A_x \cdot \left( 1 - {|\hspace{0.05cm}t\hspace{0.05cm}|}/{T_x}\right) +\hspace{0.1cm} {c_2}\cdot A_x^2 \cdot \left( 1 - {|\hspace{0.05cm}t\hspace{0.05cm}|}/{T_x}\right)^2.$$

• The following values occur at times $t = 0$  and  $t = T_x/2$:

$$y(t=0) = c_1\cdot A_x + {c_2}\cdot A_x^2 \hspace{0.15cm}\underline{= 1.95\,{\rm V}},$$

$$y(t=T_x/2) = c_1\cdot A_x \cdot {1}/{2} + \hspace{0.1cm}{c_2}\cdot A_x^2 \cdot {1}/{4}= 0.75\,{\rm V}+ 0.1125\,{\rm V} \hspace{0.15cm}\underline{ = 0.8625\,{\rm V}}.$$