Aufgaben:Exercise 2.1: Rectification - Revision history

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	[[de:Aufgaben:2.~~1_Gleichrichtung~~]]		[[de:Aufgaben:Aufgabe 2.1: Gleichrichtung]]

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	'''(2)'''  The period duration  $x(t)$  is  $T_0 = 2\,\text{ms}$. The inverse ~~amounts~~ to the base frequency  $f_0 \hspace{0.1cm}\underline{ = 500\,\text{Hz}}$.		'''(2)'''  The period duration  $x(t)$  is  $T_0 = 2\,\text{ms}$. The inverse magnitudes to the base frequency  $f_0 \hspace{0.1cm}\underline{ = 500\,\text{Hz}}$.

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	{What is the circular frequency  $\omega_0$  of the signal  $z(t)$?		{What is the basic circular frequency  $\omega_0$  of the signal  $z(t)$?
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	$\omega_0 \ = \ $ { 6283 3% }   $\text{1/s}$		$\omega_0 \ = \ $ { 6283 3% }   $\text{1/s}$

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	'''(3)'''  The half-wave rectification does not change the duration of the period, see the left graph~~. Thus the following still applies~~  $T_0 \hspace{0.1cm}\underline{= 2\,\text{ms}}$.		'''(3)'''  The half-wave rectification does not change the duration of the period, see the left graph:  $T_0 \hspace{0.1cm}\underline{= 2\,\text{ms}}$.

	[[File:P_ID262__Sig_A_2_1_a.png\|center\|frame\|Periodic triangular signals]]		[[File:P_ID262__Sig_A_2_1_a.png\|center\|frame\|Periodic triangular signals]]

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	{What is the ~~base angular~~ frequency  $\omega_0$  of the signal  $z(t)$?		{What is the circular frequency  $\omega_0$  of the signal  $z(t)$?
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	$\omega_0 \ = \ $ { 6283 3% }   $\text{1/s}$		$\omega_0 \ = \ $ { 6283 3% }   $\text{1/s}$

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	[[File:P_ID239__Sig_A_2_1.png\|250px\|right\|frame\|Periodic triangular signal]]		[[File:P_ID239__Sig_A_2_1.png\|250px\|right\|frame\|Periodic triangular signal]]
	The graph shows the periodic signal  $x(t)$. If  $x(t)$ is applied to the input of a non-linearity with the characteristic curve		The graph shows the periodic signal  $x(t)$.  If  $x(t)$ is applied to the input of a non-linearity with the characteristic curve

	:$$y=g(x)=\left\{ {x \; \rm for\; \it x \geq \rm 0, \atop {\rm 0 \;\;\; \rm else,}}\right.$$		:$$y=g(x)=\left\{ {x \; \rm for\; \it x \geq \rm 0, \atop {\rm 0 \;\;\; \rm else,}}\right.$$

	the ~~signall~~  $y(t)$ is obtained at the output. A second non-linear characteristic		the signal  $y(t)$ is obtained at the output.  A second non-linear characteristic

	:$$z=h(x)=\|x\|$$		:$$z=h(x)=\|x\|$$
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	''~~Hint~~:''		''Note:''
	*This exercise belongs to the chapter  [[Signal_Representation/General_Description\|General description of periodic signals]].		*This exercise belongs to the chapter  [[Signal_Representation/General_Description\|General description of periodic signals]].

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	[[Category:Signal Representation: Exercises\|^2.1 ~~General~~ Description of Periodic Signals^]]		[[Category:Signal Representation: Exercises\|^2.1 Description of Periodic Signals^]]

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	[[Category:Signal Representation: Exercises\|^2.1 General Description ~~about~~ Periodic Signals^]]		[[Category:Signal Representation: Exercises\|^2.1 General Description of Periodic Signals^]]

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