Aufgaben:Exercise 2.4: Distortion Factor and Distortion Power - Revision history

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	Then, the following signal occurs at the system output:		Then, the following signal occurs at the system output:
	:$$y_1(t) = {0.992 \,\rm V} \cdot \cos(\omega_0 t) - {0.062 \,\~~rmV~~} \cdot \cos(2\omega_0 t)+ \hspace{0.05cm}\text{...}$$		:$$y_1(t) = {0.992 \,\rm V} \cdot \cos(\omega_0 t) - {0.062 \,\rm V} \cdot \cos(2\omega_0 t)+ \hspace{0.05cm}\text{...}$$

	In the upper graph the signals  $x_1(t)$  and  $y_1(t)$  are shown. Harmonics with amplitudes  $\lt 10 \ \rm mV$  are not considered here.		In the upper graph the signals  $x_1(t)$  and  $y_1(t)$  are shown. Harmonics with amplitudes  $\lt 10 \ \rm mV$  are not considered here.
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.2 Nonlinear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.2 Nonlinear Distortions^]]
	[[de:Aufgaben:2.~~4_Klirrfaktor_und_Verzerrungsleistung~~]]		[[de:Aufgaben:Aufgabe 2.4: Klirrfaktor und Verzerrungsleistung]]

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	Then, the following signal occurs at the system output:		Then, the following signal occurs at the system output:
	:$$y_1(t) = {0.992 \,\rm V} \cdot \cos(\omega_0 t) - {0.062 \,\rm		:$$y_1(t) = {0.992 \,\rm V} \cdot \cos(\omega_0 t) - {0.062 \,\rmV} \cdot \cos(2\omega_0 t)+ \hspace{0.05cm}\text{...}$$
	V} \cdot \cos(2\omega_0 t)+ \hspace{0.05cm}\text{...}$$

	In the upper graph the signals  $x_1(t)$  and  $y_1(t)$  are shown. Harmonics with amplitudes  $\lt 10 \ \rm mV$  are not considered here.		In the upper graph the signals  $x_1(t)$  and  $y_1(t)$  are shown. Harmonics with amplitudes  $\lt 10 \ \rm mV$  are not considered here.
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	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  $10 \ \rm mV$:		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  $10 \ \rm mV$:
	:$$y_2(t) \hspace{-0.05cm}=\hspace{-0.05cm}{1.938 \,\rm V} \cdot \cos(\omega_0 t)\hspace{-0.05cm} -\hspace{-0.05cm} {0.234		:$$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{...}$$
	\,\rm V} \cdot \cos(2\omega_0 t) \hspace{-0.05cm}+\hspace{-0.05cm} {0.058 \,\rm V} \cdot
	\cos(3\omega_0 t)\hspace{-0.05cm} -\hspace{-0.05cm}{0.018 \,\rm V} \cdot \cos(4\omega_0 t) \hspace{-0.05cm}+\hspace{-0.05cm}
	\hspace{0.05cm}\text{...}$$

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

	The system is supposed to be analyzed based on the signal–to–distortion–power ratio		The system is supposed to be analyzed based on the signal–to–distortion–power ratio
	:$$\rho_{\rm V} = { P_{x}}/{P_{\rm V}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot \lg \hspace{0.1cm}\rho_{\rm V} =		:$$\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)$$
	10 \cdot \lg \hspace{0.1cm}{ P_{x}}/{P_{\rm V}}\hspace{0.3cm} \left( {\rm in \hspace{0.15cm} dB} \right)$$
	defined in the section [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions#Quantitative_measure_for_the_signal_distortions\| Quantitative measure for the signal distortions]]  and the distortion factor  $K$ :		defined in the section [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions#Quantitative_measure_for_the_signal_distortions\| Quantitative measure for the signal distortions]]  and the distortion factor  $K$ :
	* $P_x$  denotes the power of the input signal.		* $P_x$  denotes the power of the input signal.
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(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:		'''(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:
	:$$K \approx K_2 = \frac{0.062 \,\,{\rm V}}{0.992 \,\,{\rm V}}		:$$K \approx K_2 = \frac{0.062 \,\,{\rm V}}{0.992 \,\,{\rm V}}\hspace{0.15cm}\underline{\approx 6.25 \%}.$$
	\hspace{0.15cm}\underline{\approx 6.25 \%}.$$



	'''(2)'''  For the input amplitude  $A_x = 2 \ \rm V$  (bottom sketch) the various distortion factors are:		'''(2)'''  For the input amplitude  $A_x = 2 \ \rm V$  (bottom sketch) the various distortion factors are:
	:$$K_2 = \frac{0.234 \,\,{\rm V}}{1.938 \,\,{\rm V}} \approx 0.121,		:$$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.$$
	\hspace{0.5cm} K_3 = \frac{0.058 \,\,{\rm V}}{1.938 \,\,{\rm V}}
	\approx 0.030, \hspace{0.5cm}K_4 = \frac{0.018 \,\,{\rm V}}{1.938
	\,\,{\rm V}} \approx 0.009.$$

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

	Here,  $A_x$  denotes the amplitude of the input signal.  The signs of the harmonics do not matter in this calculation.		Here,  $A_x$  denotes the amplitude of the input signal.  The signs of the harmonics do not matter in this calculation.
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	'''(6)'''  Using the results of subtasks  '''(4)'''  and  '''(5)'''  the following is obtained:		'''(6)'''  Using the results of subtasks  '''(4)'''  and  '''(5)'''  the following is obtained:
	:$$10 \cdot \lg \rho_{V} = 10 \cdot \lg \frac{P_x}{P_{\rm V}}= 10		:$$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}}.$$
	\cdot \lg \frac{2.000\,{\rm V^2}}{0.031 \,{\rm V}^2} \hspace{0.15cm}\underline{\approx 18.10
	\,{\rm dB}}.$$


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

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

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.2 Nonlinear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.2 Nonlinear Distortions^]]
			[[de:Aufgaben:2.4_Klirrfaktor_und_Verzerrungsleistung]]

Guenter at 12:56, 1 October 2021

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Guenter at 12:45, 1 October 2021

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	''Please note:''		''Please note:''
	*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Nonlinear_Distortion\|Nonlinear ~~Distortion~~]].		*The task belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Nonlinear_Distortion\|Nonlinear Distortions]].

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

Guenter at 10:25, 1 October 2021

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	}}		}}

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

	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:		with the amplitude  $A_x = 1 \ \rm V$  is applied to the input of a communication system to test it.

			Then, the following signal occurs at the system output:
	:$$y_1(t) = {0.992 \,\rm V} \cdot \cos(\omega_0 t) - {0.062 \,\rm		:$$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{...}$$		V} \cdot \cos(2\omega_0 t)+ \hspace{0.05cm}\text{...}$$

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


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	\hspace{0.05cm}\text{...}$$		\hspace{0.05cm}\text{...}$$

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

	The system is supposed to be analyzed based on the signal–to–distortion–power ratio		The system is supposed to be analyzed based on the signal–to–distortion–power ratio
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	defined in the section [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions#Quantitative_measure_for_the_signal_distortions\| Quantitative measure for the signal distortions]]  and the distortion factor  $K$ :		defined in the section [[Linear_and_Time_Invariant_Systems/Classification_of_the_Distortions#Quantitative_measure_for_the_signal_distortions\| Quantitative measure for the signal distortions]]  and the distortion factor  $K$ :
	* $P_x$  denotes the power of the input signal.		* $P_x$  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  $\varepsilon(t) = y(t) - x(t)$ .		* The distortion power  (German:  "Verrzerrungsleistung"   ⇒   "V")  $P_{\rm V}$  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.

	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.

Javier: Text replacement - "analyse" to "analyze"

2021-09-22T11:47:37Z

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	It is obvious that the indices "1" and "2" respectively denote the normalised amplitude of the input signal.		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		The system is supposed to be analyzed based on the signal–to–distortion–power ratio
	:$$\rho_{\rm V} = { P_{x}}/{P_{\rm V}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 10 \cdot \lg \hspace{0.1cm}\rho_{\rm V} =		:$$\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)$$		10 \cdot \lg \hspace{0.1cm}{ P_{x}}/{P_{\rm V}}\hspace{0.3cm} \left( {\rm in \hspace{0.15cm} dB} \right)$$

Guenter at 10:10, 20 September 2021

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

Guenter: Guenter moved page Aufgaben:Aufgabe 2.4: Klirrfaktor und Verzerrungsleistung to Aufgaben:Exercise 2.4: Distortion Factor and Distortion Power

2021-09-20T10:07:17Z

Guenter moved page Aufgaben:Aufgabe 2.4: Klirrfaktor und Verzerrungsleistung to Aufgaben:Exercise 2.4: Distortion Factor and Distortion Power

← Older revision	Revision as of 12:07, 20 September 2021
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Oezer at 15:00, 15 September 2021

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	+ \text{...} }{A_x^2}.$$		+ \text{...} }{A_x^2}.$$

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

	*Generally, the following relationship holds between the signal–to–distortion–power ratio and the distortion factor:		*Generally, the following relationship holds between the signal–to–distortion–power ratio and the distortion factor:

← Older revision		Revision as of 14:56, 1 October 2021
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	'''(3)'''  <u>The first two proposed solutions</u> are correct:		'''(3)'''  <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.		*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  $y_{\rm min} = -2.25 \ \rm V$ hold since  $y(t)$  does not contain any direct (DC) signals. Hence, there is no symmetry with respect to the zero line anymore.		*In addition,   $y_{\rm max} = 1.75 \ \rm V$  and  $y_{\rm min} = -2.25 \ \rm V$ hold since  $y(t)$  does not contain any direct (DC) signals. <br>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.		*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":		'''(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":
	:$$P_x = \frac{A_x^2}{2} = \frac{(2 \,{\rm V})^2}{2}\hspace{0.15cm}\underline{ = 2\,{\rm V^2}}.$$		:$$P_x = \frac{A_x^2}{2} = \frac{(2 \,{\rm V})^2}{2}\hspace{0.15cm}\underline{ = 2\,{\rm V^2}}.$$

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

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


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	\,{\rm V})^2 \big] \hspace{0.15cm}\underline{\approx 0.031 \,{\rm V}^2}.$$		\,{\rm V})^2 \big] \hspace{0.15cm}\underline{\approx 0.031 \,{\rm V}^2}.$$

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


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	+ \text{...} }{A_x^2}.$$		+ \text{...} }{A_x^2}.$$

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

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