Aufgaben:Exercise 2.4: Distortion Factor and Distortion Power

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On the meaning of the distortion factor

A cosine signal

$$x_1(t) = A_x \cdot \cos(\omega_0 t)$$

with the amplitude  $A_x = 1 \ \rm V$  is applied to the input of a message transmission system to test it. Then, the following signal occurs at the system output:

$$y_1(t) = {0.992 \,\rm V} \cdot \cos(\omega_0 t) - {0.062 \,\rm V} \cdot \cos(2\omega_0 t)+ \hspace{0.05cm}\text{...}$$

In the upper graph the signals  $x_1(t)$  and  $y_1(t)$  are shown. Harmonics with amplitudes smaller than  $10 \ \rm mV$  are not considered here.

The bottom image shows the input signal  $x_2(t)$  with the ampiltude  $A_x = 2 \ \rm V$  and the corresponding output signal, again without harmonics smaller than  $10 \ \rm mV$:

$$y_2(t) \hspace{-0.05cm}=\hspace{-0.05cm}{1.938 \,\rm V} \cdot \cos(\omega_0 t)\hspace{-0.05cm} -\hspace{-0.05cm} {0.234 \,\rm V} \cdot \cos(2\omega_0 t) \hspace{-0.05cm}+\hspace{-0.05cm} {0.058 \,\rm V} \cdot \cos(3\omega_0 t)\hspace{-0.05cm} -\hspace{-0.05cm}{0.018 \,\rm V} \cdot \cos(4\omega_0 t) \hspace{-0.05cm}+\hspace{-0.05cm} \hspace{0.05cm}\text{...}$$

It is obvious that the indices "1" and "2" respectively denote the normalised amplitude of the input signal.

The system is supposed to be analysed based on the signal–to–distortion–power ratio

$$\rho_{\rm V} = { P_{x}}/{P_{\rm V}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot \lg \hspace{0.1cm}\rho_{\rm V} = 10 \cdot \lg \hspace{0.1cm}{ P_{x}}/{P_{\rm V}}\hspace{0.3cm} \left( {\rm in \hspace{0.15cm} dB} \right)$$

defined in the section  Quantitative measure for the signal distortions  and the distortion factor  $K$ :

• $P_x$  denotes the power of the input signal.
• The distortion power  $P_{\rm V}$  respectively represents the power  (the root mean square)  of the difference signal  $\varepsilon(t) = y(t) - x(t)$ .

To determine the powers  $P_{x}$  and  $P_{\rm V}$  it is necessary to take the average of the squared signals in each case. However, it is easier to calculate the powers in the frequency domain in this task.

Please note:

• The task belongs to the chapter  Nonlinear Distortion.

• All powers required here refer to the resistance $R = 1 \ \rm \Omega$  and thus have the unit ${\rm V}^2$.

Questions

1

Compute the distortion factor  $K$  for the input amplitude  $\underline{ A_x = 1\ \rm V}$.

$K \ = \ $ $\%$

2

What is the distortion factor with the input amplitude  $\underline{ A_x = 2\ \rm V}$?

$K \ = \ $ $\%$

3

Which statements are true for the signals $x_2(t)$  and  $y_2(t)$ ?

	The lower half-wave is more peaked than the upper half-wave.
	The maximum– and Mminimum values of $y_2(t)$  are asymmetrically zero.
	A different frequency would result in a different distortion factor.

4

Wie groß ist die Leistung  $P_x$  des Eingangssignals  $x_2(t)$  in  ${\rm V}^2$, also umgerechnet auf den Bezugswiderstand  $R = 1 \ \rm \Omega$?

$P_x \ = \ $ $\ {\rm V}^2$

5

What is the power $P_{\rm V}$  of the difference signal $\varepsilon_2(t)$   ⇒   "distortion power"?

$P_{\rm V} \ = \ $ $\ {\rm V}^2$

6

What is the signal–to–distortion–power ratio in  ${\rm dB}$?

$10 \cdot {\rm lg} \ \rho_{\rm V} \ = \ $ $\ {\rm dB}$

7

Which of the following statements are true for cosine-shaped input signals?

	The distortion factor can be computed using the coefficients $A_1$,  $A_2$,  $A_3$,  ...  of the output variable alone.
	The signal–to–distortion–power ratio  $10 \cdot {\rm lg} \ \rho_{\rm V}$  is computable using the coefficients $A_1$,  $A_2$,  $A_3$,  ...  alone.
	For the special case  [$A_1 = A_x$   ⇒   no change of the fundamental wave],   $\rho_{\rm V}$  and  $K$  can be converted directly into each othe.

Solution

Solution

(1)  Considering the input amplitude  $A_x = 1 \ \rm V$  corresponding to the upper sketch only the second order distortion factor provides a relevant contribution. Therefore, the following holds:

$$K \approx K_2 = \frac{0.062 \,\,{\rm V}}{0.992 \,\,{\rm V}} \hspace{0.15cm}\underline{\approx 6.25 \%}.$$

(2)  For the input amplitude  $A_x = 2 \ \rm V$  (bottom sketch) the various distortion factors are:

$$K_2 = \frac{0.234 \,\,{\rm V}}{1.938 \,\,{\rm V}} \approx 0.121, \hspace{0.5cm} K_3 = \frac{0.058 \,\,{\rm V}}{1.938 \,\,{\rm V}} \approx 0.030, \hspace{0.5cm}K_4 = \frac{0.018 \,\,{\rm V}}{1.938 \,\,{\rm V}} \approx 0.009.$$

• Thus, the overall distortion factor is:

$$K = \sqrt{K_2^2 + K_3^2 + K_4^2 +\text{ ...} }\hspace{0.15cm}\underline{ \approx 12.5 \%}.$$

(3)  The first two proposed solutions are correct:

• Here, the nonlinear distortions cause the lower half-wave to be more peaked than the upper half-wave.
• In addition,   $y_{\rm max} = 1.75 \ \rm V$  and  $y_{\rm min} = -2.25 \ \rm V$ holds since  $y(t)$  does not contain any direct (DC) signals. Hence, there is no symmetry with respect to the zero line anymore.
• For a nonlinear system, the distortion factor  $K$  is independent of the frequency of the cosine input signal but strongly dependent on its amplitude.

(4)  The root mean square (rms) value of a cosine signal is known to be  $\sqrt{0.5}$–times the amplitude. The square of this is the "power":

$$P_x = \frac{A_x^2}{2} = \frac{(2 \,{\rm V})^2}{2}\hspace{0.15cm}\underline{ = 2\,{\rm V^2}}.$$

• Actually, the power also depends on the reference resistance $R$  and has the unit "watt".
• $P_x = 2 \ \rm W$ is obtained with $R = 1 \ \rm \Omega$ , so exactly the same numerical value as in this simpler calculation.

(5)  Bezeichnet man

• mit  $A_1$  die Amplitude der Grundwelle von  $y_2(t)$, und
• mit  $A_2$,  $A_3$  und  $A_4$  die so genannten Oberwellen,

so erhält man für die Verzerrungsleistung durch Berechnung im Frequenzbereich:

$$P_{\rm V} = \frac{1}{2} \cdot \big[ (A_1 - A_x)^2 + A_2^2+ A_3^2+ A_4^2\big] = \frac{1}{2} \cdot \big[ (-2 \,{\rm V} \hspace{-0.05cm}+ \hspace{-0.05cm}1.938 \,{\rm V} )^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (0.234 \,{\rm V})^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (0.058 \,{\rm V})^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (0.018 \,{\rm V})^2 \big] \hspace{0.15cm}\underline{\approx 0.031 \,{\rm V}^2}.$$

Hierbei bezeichnet  $A_x$  die Amplitude des Eingangssignals. Die Vorzeichen der Oberwellen spielen bei dieser Berechnung keine Rolle.

(6)  Mit den Ergebnissen der Unterpunkte  (4)  und  (5)  erhält man:

$$10 \cdot \lg \rho_{V} = 10 \cdot \lg \frac{P_x}{P_{\rm V}}= 10 \cdot \lg \frac{2.000\,{\rm V^2}}{0.031 \,{\rm V}^2} \hspace{0.15cm}\underline{\approx 18.10 \,{\rm dB}}.$$

(7)  Richtig sind die Lösungsvorschläge 1 und 3:

• Die erste Aussage ist richtig, denn es gilt

$$K^2 = \frac{A_2^2 + A_3^2 + A_4^2 + ... }{A_1^2}.$$

• Dagegen gilt für den Kehrwert des Signal–zu–Verzerrungs–Leistungsverhältnisses:

$${1}/{\rho_{\rm V}} = \frac{(A_1 - A_x)^2+A_2^2 + A_3^2 + A_4^2 + \text{...} }{A_x^2}.$$

• Bei der Berechnung der Verzerrungsleistung  $P_{\rm V}$  wird auch eine Verfälschung der Grundwellenamplitude  $($diese ist nun  $A_1$  anstelle von  $A_x)$  berücksichtigt. Außerdem wird die Verzerrungsleistung nicht auf  $A_1^2$, sondern auf  $A_x^2$  bezogen.

• Allgemein gilt zwischen dem Signal–zu–Verzerrungs–Leistungsverhältnis und dem Klirrfaktor folgender Zusammenhang:

$${\rho_{\rm V}} = \frac{A_x^2}{(A_1 - A_x)^2 + K^2 \cdot A_1^2}.$$

• Mit  $A_1 = A_x$  vereinfacht sich diese Gleichung wie folgt:

$${\rho_{\rm V}} = {1}/{ K^2 }.$$

Anmerkungen:

• Ein Klirrfaktor von  $1\%$  entspricht in diesem Fall dem Ergebnis  $10 \cdot \lg \rho_{\rm V} = 40 \,{\rm dB}$.
• Mit dem Klirrfaktor  $K = 0.125$  aus Teilaufgabe  (2)  hätte man mit der Näherung  $A_1 \approx A_x$  sofort  $10 \cdot \lg \rho_{\rm V} = 18.06 \,{\rm dB}$  erhalten.
• Der unter Punkt  (7)  errechnete tatsächliche Wert  $(18.10 \ \rm dB)$  weicht hiervon nur unwesentlich ab.