Aufgaben:Exercise 2.3Z: Asymmetrical Characteristic Operation - Revision history

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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.~~3Z_Kennlinienbetrieb_asymmetrisch~~]]		[[de:Aufgaben:Aufgabe 2.3Z: Kennlinienbetrieb asymmetrisch]]

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	The output signal of the overall system can generally be depicted as follows:		The output signal of the overall system can generally be depicted as follows:
	:$$y(t) = A_0 + A_1 \cdot \cos(\omega_0 t) + A_2 \cdot \cos(2\omega_0 t) +		:$$y(t) = A_0 + A_1 \cdot \cos(\omega_0 t) + A_2 \cdot \cos(2\omega_0 t) +A_3 \cdot \cos(3\omega_0 t) + \hspace{0.05cm}\text{...}$$
	A_3 \cdot \cos(3\omega_0 t) + \hspace{0.05cm}\text{...}$$

	The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.		The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.
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	*The following trigonometric relations are assumed to be known:		*The following trigonometric relations are assumed to be known:
	:$$\cos^2(\alpha) = {1}/{2} + {1}/{2}		:$$\cos^2(\alpha) = {1}/{2} + {1}/{2}\cdot \cos(2\alpha)\hspace{0.05cm}, \hspace{0.3cm}\cos^3(\alpha) = {3}/{4} \cdot \cos(\alpha) + {1}/{4} \cdot \cos(3\alpha)\hspace{0.05cm}.$$
	\cdot \cos(2\alpha)\hspace{0.05cm}, \hspace{0.3cm}
	\cos^3(\alpha) = {3}/{4} \cdot \cos(\alpha) + {1}/{4} \cdot \cos(3\alpha)
	\hspace{0.05cm}.$$


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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Considering the cubic approximation $g_3(x)$  the following is obtained before the high-pass filter:		'''(1)'''  Considering the cubic approximation $g_3(x)$  the following is obtained before the high-pass filter:
	:$$y_{\rm C}(t) = g_3\big[x_{\rm C}(t)\big] = \big[ C + A \cdot \cos(\~~omega_0~~		:$$y_{\rm C}(t) = g_3\big[x_{\rm C}(t)\big] = \big[ C + A \cdot \cos(\omega_0t)\big] - {1}/{6} \cdot \big[ C + A \cdot \cos(\omega_0t)\big]^3 $$
	t)\big] - {1}/{6} \cdot \big[ C + A \cdot \cos(\~~omega_0~~		:$$\Rightarrow \; y_{\rm C}(t) =C + A \cdot \cos(\omega_0t) - {1}/{6} \cdot \big[ C^3 + 3 \cdot C^2 \cdot A \cdot \cos(\omega_0t) + \hspace{0.09cm}3 \cdot C \cdot A^2 \cdot \cos^2(\omega_0t) + A^3 \cdot \cos^3(\omega_0 t)\big].$$
	t)\big]^3 $$
	:$$\Rightarrow \; y_{\rm C}(t) =
	C + A \cdot \cos(\~~omega_0~~
	t) - {1}/{6} \cdot \big[ C^3 + 3 \cdot C^2 \cdot A \cdot \cos(\~~omega_0~~
	t) + \hspace{0.09cm}3 \cdot C \cdot A^2 \cdot \cos^2(\~~omega_0~~
	t) + A^3 \cdot \cos^3(\omega_0 t)\big].$$

	*The signal  $y_{\rm C}(t)$  contains a direct (DC) component  $C - C^3/6$  which is no longer included in the signal  $y(t)$  due to the high-pass filter:		*The signal  $y_{\rm C}(t)$  contains a direct (DC) component  $C - C^3/6$  which is no longer included in the signal  $y(t)$  due to the high-pass filter:
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	'''(2)'''  Applying the given trigonometric relations the following coefficients with  $A= C = 0.5$  are obtained:		'''(2)'''  Applying the given trigonometric relations the following coefficients with  $A= C = 0.5$  are obtained:
	:$$A_1 = A - {1}/{6}\cdot 3 \cdot C^2 \cdot A - {1}/{6} \cdot {3}/{4}\~~cdot~~		:$$A_1 = A - {1}/{6}\cdot 3 \cdot C^2 \cdot A - {1}/{6} \cdot {3}/{4}\cdotA^3 = {1}/{2} - {1}/{16} - {1}/{64} = {27}/{64}\hspace{0.15cm}\underline{ \approx 0.422},$$
	A^3 = {1}/{2} - {1}/{16} - {1}/{64} = {27}/{64}		:$$A_2 = - {1}/{6}\cdot 3 \cdot {1}/{2}\cdotC \cdot A^2 = - \frac{1}{32} \hspace{0.15cm}\underline{\approx -0.031},$$
	\hspace{0.15cm}\underline{ \approx 0.422},$$		:$$A_3 = - {1}/{6}\cdot \frac{1}{4}\cdotA^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$
	:$$A_2 = - {1}/{6}\cdot 3 \cdot {1}/{2}\~~cdot~~
	C \cdot A^2 = - \frac{1}{32} \hspace{0.15cm}\underline{\approx -0.031},$$
	:$$A_3 = - {1}/{6}\cdot \frac{1}{4}\~~cdot~~
	A^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$

	*Higher order terms do not occur.  Thus,   $\underline{A_4 = 0}$  holds.		*Higher order terms do not occur.  Thus,   $\underline{A_4 = 0}$  holds.
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	'''(4)'''  The maximum value occurs at time  $t = 0$  and at multiples of  $T$ :		'''(4)'''  The maximum value occurs at time  $t = 0$  and at multiples of  $T$ :
	:$$y_{\rm max}= y(t=0) = A_1 + A_2 + A_3 = 0.422 -0.031 -0.005 \hspace{0.15cm}\underline{=		:$$y_{\rm max}= y(t=0) = A_1 + A_2 + A_3 = 0.422 -0.031 -0.005 \hspace{0.15cm}\underline{=0.386}.$$
	0.386}.$$

	*The minimum values are located exactly in the middle between two maxima and it holds that:		*The minimum values are located exactly in the middle between two maxima and it holds that:
	:$$y_{\rm min}= - A_1 + A_2 - A_3 = -0.422 -0.031 +0.005\hspace{0.15cm}\underline{ =		:$$y_{\rm min}= - A_1 + A_2 - A_3 = -0.422 -0.031 +0.005\hspace{0.15cm}\underline{ =-0.448}.$$
	-0.448}.$$

	*The signal  $y(t)$  is shifted downward by  $0.448$  compared to the signal  drawn in the sketch on the information page.		*The signal  $y(t)$  is shifted downward by  $0.448$  compared to the signal  drawn in the sketch on the information page.

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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:2.3Z_Kennlinienbetrieb_asymmetrisch]]

Guenter at 13:50, 29 September 2021

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Guenter at 13:31, 29 September 2021

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

	*The following trigonometric relations are assumed to be known:		*The following trigonometric relations are assumed to be known:

Guenter at 13:30, 29 September 2021

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	The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.		The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.

	This would result in exactly the same constellation as in  [[Aufgaben:2.~~3_Sinusförmige_Kennlinie~~\|Exercise 2.3]]  for  $C = 0$  in whose subtask  '''(2)'''  the distortion factor was calculated:		This would result in exactly the same constellation as in  [[Aufgaben:Exercise_2.3:_Sinusoidal_Characteristic\|Exercise 2.3]]  for  $C = 0$  in whose subtask  '''(2)'''  the distortion factor was calculated:
	*$K = K_{g3} \approx 1.08 \%$  für  $A = 0.5$,		*$K = K_{g3} \approx 1.08 \%$  für  $A = 0.5$,
	*$K = K_{g3} \approx 4.76 \%$  für  $A = 1.0$.		*$K = K_{g3} \approx 4.76 \%$  für  $A = 1.0$.

Guenter at 13:29, 29 September 2021

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Guenter at 10:09, 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 2.3Z: Kennlinienbetrieb asymmetrisch to Aufgaben:Exercise 2.3Z: Asymmetrical Characteristic Operation

2021-09-20T10:05:13Z

Guenter moved page Aufgaben:Aufgabe 2.3Z: Kennlinienbetrieb asymmetrisch to Aufgaben:Exercise 2.3Z: Asymmetrical Characteristic Operation

← Older revision	Revision as of 12:05, 20 September 2021
(No difference)

Oezer at 22:26, 14 September 2021

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	A^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$		A^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$

	*Higher order terms do not occur. Thus, h  $\underline{A_4 = 0}$ holds.		*Higher order terms do not occur. Thus,   $\underline{A_4 = 0}$ holds.

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	t) + A^3 \cdot \cos^3(\omega_0 t)\big].$$		t) + A^3 \cdot \cos^3(\omega_0 t)\big].$$

	*The signal  $y_{\rm C}(t)$  contains a direct (DC) component  $C - C^3/6$ which is no longer included in the signal  $y(t)$  due to the high-pass filter:		*The signal  $y_{\rm C}(t)$  contains a direct (DC) component  $C - C^3/6$  which is no longer included in the signal  $y(t)$  due to the high-pass filter:
	:$$\underline{ A_0 = 0}.$$		:$$\underline{ A_0 = 0}.$$


	'''(2)'''  Applying the given trigonometric relations the following coefficients with  $A= C = 0.5$ are obtained:		'''(2)'''  Applying the given trigonometric relations the following coefficients with  $A= C = 0.5$  are obtained:
	:$$A_1 = A - {1}/{6}\cdot 3 \cdot C^2 \cdot A - {1}/{6} \cdot {3}/{4}\cdot		:$$A_1 = A - {1}/{6}\cdot 3 \cdot C^2 \cdot A - {1}/{6} \cdot {3}/{4}\cdot
	A^3 = {1}/{2} - {1}/{16} - {1}/{64} = {27}/{64}		A^3 = {1}/{2} - {1}/{16} - {1}/{64} = {27}/{64}
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	A^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$		A^3 = - {1}/{192} \hspace{0.15cm}\underline{\approx -0.005}.$$

	*Higher order terms do not occur. Thus,   $\underline{A_4 = 0}$ holds.		*Higher order terms do not occur.  Thus,   $\underline{A_4 = 0}$  holds.



	'''(3)'''  In this task, the higher order distortion factors are  $K_2 = 2/27 \approx 7.41\%$  ~~und~~  $K_3 = 1/81 \approx 1.23\%$.		'''(3)'''  In this task,  the higher order distortion factors are  $K_2 = 2/27 \approx 7.41\%$  and  $K_3 = 1/81 \approx 1.23\%$.
	*Thereby, the following is obtained for the overall distortion factor:		*Thereby, the following is obtained for the overall distortion factor:
	:$$K = \sqrt{K_2^2 + K_3^2} \hspace{0.15cm}\underline{\approx7.51 \%}.$$		:$$K = \sqrt{K_2^2 + K_3^2} \hspace{0.15cm}\underline{\approx7.51 \%}.$$
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	-0.448}.$$		-0.448}.$$

	*The signal  $y(t)$  is shifted downward by $0.448$  compared to the signal  drawn in the sketch on the information page.		*The signal  $y(t)$  is shifted downward by  $0.448$  compared to the signal  drawn in the sketch on the information page.
	*This signal value is obtained from the following equation considering  $A = C = 1/2$:		*This signal value is obtained from the following equation considering  $A = C = 1/2$:
	:$$C - \frac{C \cdot A^2}{4}- \frac{C^3}{6} = {1}/{2} - {1}/{32}- {1}/{48} = 0.448.$$		:$$C - \frac{C \cdot A^2}{4}- \frac{C^3}{6} = {1}/{2} - {1}/{32}- {1}/{48} = 0.448.$$

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

	[[File:P_ID895__LZI_Z_2_3.png\|right\|frame\|~~Influence of nonlinear distortions~~]]		[[File:P_ID895__LZI_Z_2_3.png\|right\|frame\|System and signal examples]]
	The cosine signal		The cosine signal
	:$$x(t) = A \cdot \cos(\omega_0 t)$$		:$$x(t) = A \cdot \cos(\omega_0 t)$$

	is applied to the input of a system  $S$  where  $A = 0.5$  shall always hold for the amplitude. The system  $S$  consists of		is applied to the input of a system  $S$  where  $A = 0.5$  shall always hold for the amplitude.  The system  $S$  consists of
	*the addition of a direct (DC) component  $C$,		*the addition of a direct (DC) component  $C$,
	*a nonlinearity with the characteristic curve		*a nonlinearity with the characteristic curve
	:$$g(x) = \sin(x) \hspace{0.05cm} \approx x -{x^3}\hspace{-0.1cm}/{6} = g_3(x),$$		:$$g(x) = \sin(x) \hspace{0.05cm} \approx x -{x^3}\hspace{-0.1cm}/{6} = g_3(x),$$
	*as well as an ideal high-pass filter that allows all frequencies to pass unaltered except for a direct (DC) signal  $(f = 0)$~~ ~~.		*as well as an ideal high-pass filter that allows all frequencies to pass unaltered except for a direct (DC) signal  $(f = 0)$.


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	The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.		The sinusoidal characteristic curve  $g(x)$  is to be approximated by the cubic approximation $g_3(x)$  throughout the whole problem according to the above equation.

	This would result in exactly the same constellation as in  [[Aufgaben:2.3_Sinusförmige_Kennlinie\|Exercise 2.3]] for  $C = 0$  in whose subtask  '''(2)'''  the distortion factor was calculated:		This would result in exactly the same constellation as in  [[Aufgaben:2.3_Sinusförmige_Kennlinie\|Exercise 2.3]]  for  $C = 0$  in whose subtask  '''(2)'''  the distortion factor was calculated:
	*$K = K_{g3} \approx 1.08 \%$  für  $A = 0.5$,		*$K = K_{g3} \approx 1.08 \%$  für  $A = 0.5$,
	*$K = K_{g3} \approx 4.76 \%$  für  $A = 1.0$.		*$K = K_{g3} \approx 4.76 \%$  für  $A = 1.0$.
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	:$$x_{\rm C}(t) = C + A \cdot \cos(\omega_0 t) = {1}/{2} + {1}/{2}\cdot \cos(\omega_0 t).$$		:$$x_{\rm C}(t) = C + A \cdot \cos(\omega_0 t) = {1}/{2} + {1}/{2}\cdot \cos(\omega_0 t).$$

	*So, the characteristic curve is operated asymmetrically with values between  $0$  and  $1$.		*So,  the characteristic curve is operated asymmetrically with values between  $0$  and  $1$.
	*In the above graph, the signals  $x_{\rm C}(t)$  and  $y_{\rm C}(t)$  are plotted additionally directly before and after the characteristic curve  $g(x)$ .		*In the above graph,  the signals  $x_{\rm C}(t)$  and  $y_{\rm C}(t)$  are plotted additionally directly before and after the characteristic curve  $g(x)$ .


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

	*The following trigonometric relations are assumed to be known:		*The following trigonometric relations are assumed to be known:
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	<quiz display=simple>		<quiz display=simple>
	{Compute the output signal  $y(t)$  considering the high-pass filter. What is the direct (DC) signal component  $A_0$?		{Compute the output signal  $y(t)$  considering the high-pass filter.  What is the direct (DC) signal component  $A_0$?
	\|type="{}"}		\|type="{}"}
	$A_0 \ = \ $ { 0. }		$A_0 \ = \ $ { 0. }