Aufgaben:Exercise 2.2Z: Non-Linearities - Revision history

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2026-03-16T15:53:24Z

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	[[Category:Signal Representation: Exercises\|^2.2 Direct Current Signal^]]		[[Category:Signal Representation: Exercises\|^2.2 Direct Current Signal^]]
	[[de:Aufgaben:2.~~2Z_Nichtlinearitäten~~]]		[[de:Aufgaben:Aufgabe 2.2Z: Nichtlinearitäten]]

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	'''(3)'''  Due to the periodicity and symmetry, averaging in the range from  $0$  bis  $T_0/2$ is sufficient.		'''(3)'''  Due to the periodicity and symmetry, averaging in the range from  $0$  bis  $T_0/2$ is sufficient.
	* With the corresponding characteristic curve, the following then applies:		* With the corresponding characteristic curve, the following then applies:
	:$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2		:$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$
	\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Signal Representation: Exercises\|^2.2 Direct Current Signal^]]		[[Category:Signal Representation: Exercises\|^2.2 Direct Current Signal^]]
			[[de:Aufgaben:2.2Z_Nichtlinearitäten]]

2021-04-14T11:24:51Z

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	'''(3)'''  Due to the periodicity and symmetry, averaging in the range from  $0$  bis  $T_0/2$ is sufficient.		'''(3)'''  Due to the periodicity and symmetry, averaging in the range from  $0$  bis  $T_0/2$ is sufficient.
	* With the corresponding characteristic curve, the following then applies::		* With the corresponding characteristic curve, the following then applies:
	:$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2		:$$z_0=\frac{1}{T_0/2}\int^{T_0/2}_0 x^2(t)\,{\rm d}t=\frac{4\rm V^2}{T_0/2}\int^{T_0/2}_0 ({2t}/{T_0})^2\, {\rm d}t={4}/{3}\rm \;V^2
	\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$		\hspace{0.15cm}\underline{\approx1.333\rm \;V^2}.$$

2021-04-12T15:02:05Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The DC signal  $x_0$  is the mean ~~value~~ of ~~the~~ signal  ${x(t)}$. Averaging over a period duration  $T_0 = 1 \, \text{ms}$ is sufficient. One obtains:		'''(1)'''  The DC signal  $x_0$  is the mean of signal  ${x(t)}$.  Averaging over a period duration  $T_0 = 1 \, \text{ms}$ is sufficient.  One obtains:
	:$$x_0=\frac{1}{T_0}\int^{T_0}_0 x(t)\,{\rm d} t \hspace{0.15cm}\underline{=1\,\rm V}.$$		:$$x_0=\frac{1}{T_0}\int^{T_0}_0 x(t)\,{\rm d} t \hspace{0.15cm}\underline{=1\,\rm V}.$$



	'''(2)'''  In half the time  ${y(t)} = 1\, \text{V}$, in the other half is is between  $0$  and  $1\, \text{V}$  with the mean ~~value~~ at  $0.5 \,\text{V}$  ⇒   $y_0 \hspace{0.15cm}\underline{= 0.75 \,\text{V}}$.		'''(2)'''  In half the time  ${y(t)} = 1\, \text{V}$, in the other half is is between  $0$  and  $1\, \text{V}$  with the mean at  $0.5 \,\text{V}$  ⇒   $y_0 \hspace{0.15cm}\underline{= 0.75 \,\text{V}}$.

2021-04-12T15:00:25Z

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	We start from the triangular signal  ${x(t)}$  according to the figure above.		We start from the triangular signal  ${x(t)}$  according to the figure above.

	If we apply this signal to an amplitude limiter, we get the signal		*If we apply this signal to an amplitude limiter, we get the signal
	:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm for}\quad x(t)\le \rm 1V \atop {\rm else}}\right..$$		:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm for}\quad x(t)\le \rm 1V \atop {\rm else}}\right..$$
	~~A second~~ non-linearity provides the signal		*Another non-linearity provides the signal
	:$$z(t)=x^2(t).$$		:$$z(t)=x^2(t).$$
	The DC signal components are designated  $x_0$,  $y_0$  and  $z_0$  in the following.		The DC signal components are designated  $x_0$,  $y_0$  and  $z_0$  in the following.




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	''Hint:''		''Hint:''
	*This ~~exercise~~ belongs to ~~the~~ chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal\|Direct Current Signal - Limit Case of a Periodic Signal]].		*This task belongs to chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal\|Direct Current Signal - Limit Case of a Periodic Signal]].

2021-04-12T14:57:19Z

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	[[Category:Signal Representation: Exercises\|^2.2 Direct Current ~~Signal - Limit Case of a Periodic~~ Signal^]]		[[Category:Signal Representation: Exercises\|^2.2 Direct Current Signal^]]

2021-03-26T09:18:18Z

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	}}		}}

	[[File:P_ID322__Sig_Z_2_2.png\|right\|frame\|~~Gleichanteil nach Nichtlinearitäten~~]]		[[File:P_ID322__Sig_Z_2_2.png\|right\|frame\|DC component after non-linearities]]
	We start from the triangular signal  ${x(t)}$  according to the figure above.		We start from the triangular signal  ${x(t)}$  according to the figure above.

	If we apply this signal to an amplitude limiter, we get the signal		If we apply this signal to an amplitude limiter, we get the signal
	:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm ~~f\ddot{u}r~~}\quad x(t)\le \rm 1V \atop {\rm ~~sonst~~}}\right..$$		:$$y(t)=\left\{ {x(t)\atop \rm 1V}{\hspace{0.5cm} {\rm for}\quad x(t)\le \rm 1V \atop {\rm else}}\right..$$
	A second non-linearity provides the signal		A second non-linearity provides the signal
	:$$z(t)=x^2(t).$$		:$$z(t)=x^2(t).$$
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	''Hint:''		''Hint:''
	*This exercise belongs to the chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal\|Direct Current Signal - Limit Case of a Periodic Signal.]].		*This exercise belongs to the chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal\|Direct Current Signal - Limit Case of a Periodic Signal]].

2021-03-23T12:39:08Z

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	[[Category:~~Exercises for~~ Signal Representation\|^2.2 Direct Current Signal - Limit Case of a Periodic Signal^]]		[[Category:Signal Representation: Exercises\|^2.2 Direct Current Signal - Limit Case of a Periodic Signal^]]