View source for Aufgaben:Exercise 2.4: Distortion Factor and Distortion Power


{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Nonlinear_Distortion
}}

[[File:P_ID897__LZI_A_2_4.png|right|frame|On the meaning of the distortion factor]]
A cosine signal
:$$x_1(t) =  A_x  \cdot \cos(\omega_0 t)$$

with the amplitude&nbsp; $A_x = 1 \ \rm V$&nbsp; 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&nbsp; $x_1(t)$&nbsp; and&nbsp; $y_1(t)$&nbsp; are shown. Harmonics with amplitudes smaller than&nbsp; $10 \ \rm mV$&nbsp; are not considered here.


The bottom image shows the input signal&nbsp; $x_2(t)$&nbsp; with the ampiltude&nbsp; $A_x = 2 \ \rm V$&nbsp; and the corresponding output signal, again without harmonics smaller than&nbsp; $10 \ \rm mV$:
:$$y_2(t) \hspace{-0.05cm}=\hspace{-0.05cm}{1.938 \,\rm V}  \cdot \cos(\omega_0 t)\hspace{-0.05cm} -\hspace{-0.05cm} {0.234
\,\rm V} \cdot \cos(2\omega_0 t) \hspace{-0.05cm}+\hspace{-0.05cm}  {0.058 \,\rm V} \cdot
\cos(3\omega_0 t)\hspace{-0.05cm} -\hspace{-0.05cm}{0.018 \,\rm V} \cdot \cos(4\omega_0 t) \hspace{-0.05cm}+\hspace{-0.05cm}
\hspace{0.05cm}\text{...}$$

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&ndash;to&ndash;distortion&ndash;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&nbsp;[[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions#Quantitative_measure_for_the_signal_distortions| Quantitative measure for the signal distortions]]&nbsp; and the distortion factor&nbsp; $K$&nbsp;:
* $P_x$&nbsp; denotes the power of the input signal.
* The distortion power&nbsp; $P_{\rm V}$&nbsp; respectively represents the power&nbsp; (the root mean square)&nbsp; of the difference signal&nbsp; $\varepsilon(t) = y(t) - x(t)$&nbsp;.


To determine the powers&nbsp; $P_{x}$&nbsp; and&nbsp; $P_{\rm V}$&nbsp; 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&nbsp; [[Linear_and_Time_Invariant_Systems/Nonlinear_Distortion|Nonlinear Distortion]].
 
*All powers required here refer to the resistance&nbsp;$R = 1 \ \rm \Omega$&nbsp; and thus have the unit&nbsp;${\rm V}^2$.


===Questions===

<quiz display=simple>
{Compute the distortion factor&nbsp; $K$&nbsp; for the input amplitude&nbsp;  $\underline{ A_x = 1\ \rm V}$.
|type="{}"}
$K \ = \ $  { 6.25 3% } $\%$


{What is the distortion factor with the input amplitude&nbsp; $\underline{ A_x = 2\ \rm V}$?
|type="{}"}
$K \ = \ $ { 12.5 3% } $\%$


{Which statements are true for the signals&nbsp;$x_2(t)$&nbsp; and &nbsp;$y_2(t)$&nbsp;?
|type="[]"}
+ The lower half-wave is more peaked than the upper half-wave.
+ The maximum&ndash; and minimum values of&nbsp;$y_2(t)$&nbsp; are asymmetrically zero.
- A different frequency would result in a different distortion factor.


{What is the power&nbsp;$P_x$&nbsp; of the input signal&nbsp;$x_2(t)$&nbsp; in &nbsp;${\rm V}^2$, i.e. in terms of the reference resistance &nbsp;$R = 1 \ \rm \Omega$?
|type="{}"}
$P_x \ = \ $  { 2 1% } $\ {\rm V}^2$


{What is the power&nbsp;$P_{\rm V}$&nbsp; of the difference signal&nbsp;$\varepsilon_2(t)$ &nbsp; &rArr; &nbsp; "distortion power"? 
|type="{}"}
$P_{\rm V} \ = \ $  { 0.031 3% } $\ {\rm V}^2$


{What is the signal&ndash;to&ndash;distortion&ndash;power ratio in&nbsp; ${\rm dB}$?
|type="{}"}
$10 \cdot {\rm lg} \ \rho_{\rm V} \ = \ $ { 18.1 3% } $\ {\rm dB}$


{Which of the following statements are true for cosine-shaped input signals?
|type="[]"}
+ The distortion factor can be computed using the coefficients&nbsp;$A_1$,&nbsp; $A_2$,&nbsp; $A_3$,&nbsp; ...&nbsp; of the output variable alone.
- The signal&ndash;to&ndash;distortion&ndash;power ratio&nbsp; $10 \cdot {\rm lg} \ \rho_{\rm V}$&nbsp; is computable using the coefficients&nbsp;$A_1$,&nbsp; $A_2$,&nbsp; $A_3$,&nbsp; ...&nbsp; alone.
+ For the special case&nbsp; [$A_1 = A_x$ &nbsp; &rArr; &nbsp; no change of the fundamental wave],&nbsp; &nbsp;$\rho_{\rm V}$&nbsp; and &nbsp;$K$&nbsp; can be converted directly into each other.



</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; Considering the input amplitude&nbsp; $A_x = 1 \ \rm V$&nbsp; 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)'''&nbsp; For the input amplitude&nbsp; $A_x = 2 \ \rm V$&nbsp; (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)'''&nbsp; <u>The first two proposed solutions</u> are correct:
*Here, the nonlinear distortions cause the lower half-wave to be more peaked than the upper half-wave. 
*In addition, &nbsp; $y_{\rm max} = 1.75 \ \rm V$&nbsp; and&nbsp; $y_{\rm min} = -2.25 \ \rm V$ hold since&nbsp; $y(t)$&nbsp; 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&nbsp; $K$&nbsp; is independent of the frequency of the cosine input signal but strongly dependent on its amplitude. 



'''(4)'''&nbsp; The root mean square (rms) value of a cosine signal is known to be&nbsp; $\sqrt{0.5}$&ndash;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&nbsp;$R$&nbsp; and has the unit "watt". 
*$P_x =  2 \ \rm W$ is obtained with&nbsp;$R = 1 \ \rm \Omega$&nbsp;, so exactly the same numerical value as in this simpler calculation.



'''(5)'''&nbsp; Denoting 
*the amplitude of the fundamental wave of&nbsp; $y_2(t)$ by&nbsp; $A_1$&nbsp; and 
*the so-called harmonics by&nbsp; $A_2$,&nbsp; $A_3$&nbsp; and&nbsp; $A_4$&nbsp;, 


the distortion power is thus by computing it in the frequency domain:
:$$P_{\rm V} = \frac{1}{2} \cdot \big[ (A_1 - A_x)^2 + A_2^2+
A_3^2+ A_4^2\big] =  \frac{1}{2} \cdot \big[ (-2
\,{\rm V} \hspace{-0.05cm}+ \hspace{-0.05cm}1.938 \,{\rm V} )^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (0.234 \,{\rm V})^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (0.058 \,{\rm V})^2 \hspace{-0.05cm}+ \hspace{-0.05cm} (0.018
\,{\rm V})^2 \big] \hspace{0.15cm}\underline{\approx 0.031 \,{\rm V}^2}.$$

Here,&nbsp; $A_x$&nbsp; denotes the amplitude of the input signal. The signs of the harmonics do not matter in this calculation.



'''(6)'''&nbsp; Using the results of subtasks&nbsp; '''(4)'''&nbsp; and&nbsp; '''(5)'''&nbsp; the following is obtained:
:$$10 \cdot \lg \rho_{V} =  10 \cdot \lg \frac{P_x}{P_{\rm V}}=  10
\cdot \lg \frac{2.000\,{\rm V^2}}{0.031 \,{\rm V}^2} \hspace{0.15cm}\underline{\approx 18.10
\,{\rm dB}}.$$


'''(7)'''&nbsp; <u>Proposed solutions 1 and 3</u> are correct:
*The first statement is correct because it holds that 
:$$K^2 = \frac{A_2^2 + A_3^2 + A_4^2 + ... }{A_1^2}.$$

*In contrast, the following holds for the reciprocal of the signal&ndash;to&ndash;distortion&ndash;power ratio:
:$${1}/{\rho_{\rm V}} = \frac{(A_1 - A_x)^2+A_2^2 + A_3^2 + A_4^2
+ \text{...} }{A_x^2}.$$

*When calculating the distortion power&nbsp; $P_{\rm V}$&nbsp; a falsification of the amplitude of the fundamental wave&nbsp; $($this is now&nbsp; $A_1$&nbsp; instead of&nbsp; $A_x)$&nbsp; is also taken into account. Moreover, the distortion power is not in terms of&nbsp; $A_1^2$ but in terms of&nbsp; $A_x^2$&nbsp;. 

*Generally, the following relationship holds between the signal&ndash;to&ndash;distortion&ndash;power ratio and the distortion factor:
:$${\rho_{\rm V}} = \frac{A_x^2}{(A_1 - A_x)^2 + K^2 \cdot A_1^2}.$$

*With&nbsp; $A_1 = A_x$&nbsp; this equation is simplified as follows:
:$${\rho_{\rm V}} = {1}/{ K^2 }.$$


''Remarks:''
*A distortion factor of&nbsp; $1\%$&nbsp; in this case corresponds to the result&nbsp; $10 \cdot \lg \rho_{\rm V} = 40 \,{\rm dB}$.
*Using the distortion factor&nbsp; $K = 0.125$&nbsp; from subtask&nbsp; '''(2)'''&nbsp; &nbsp; $10 \cdot \lg \rho_{\rm V} = 18.06 \,{\rm dB}$&nbsp; would have been obtained immediately with the approximation&nbsp; $A_1 \approx A_x$&nbsp;. 
*The actual value &nbsp; $(18.10 \ \rm dB)$&nbsp; calculated in subtask&nbsp; '''(7)'''&nbsp; differs from this only insignificantly. 


{{ML-Fuß}}



[[Category:Linear and Time-Invariant Systems: Exercises|^2.2 Nichtlineare Verzerrungen^]]