Aufgaben:Exercise 2.2: DC Component of Signals - Revision history

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	'''(3)'''  Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component  $A_0$ .		'''(3)'''  Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component  $A_0$ .
	*For signal  $x_3(t)$  the period duration is  $T_0 = 3\,\text{ms}$.  This results in the required DC component:$$A_0=\rm \frac{1}{3\,ms}\cdot \big[1\,V\cdot 1\,ms+(-1\,V)\cdot 2\,ms \big]		*For signal  $x_3(t)$  the period duration is  $T_0 = 3\,\text{ms}$.  This results in the required DC component:$$A_0=\rm \frac{1}{3\,ms}\cdot \big[1\,V\cdot 1\,ms+(-1\,V)\cdot 2\,ms \big]\hspace{0.15cm}\underline{=-0.333\,V}.$$
	\hspace{0.15cm}\underline{=-0.333\,V}.$$

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Guenter at 16:30, 17 May 2021

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	[[File:P_ID273__Sig_A_2_2.png\|right\|frame\|Square wave signal with ~~and~~ without DC component]]		[[File:P_ID273__Sig_A_2_2.png\|right\|frame\|Square wave signal with/ without DC component]]

	The graph shows six time signals defined for all times $($from  $-\infty$  to  $+\infty)$.  For all sample signals  $x_i(t)$  the associated spectral function can be written as:		The graph shows six time signals defined for all times $($from  $-\infty$  to  $+\infty)$.  For all sample signals  $x_i(t)$  the associated spectral function can be written as:

Guenter at 14:49, 12 April 2021

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Guenter at 14:42, 12 April 2021

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	Here:		Here:
	*$A_0$  is the DC component of the signal.		*$A_0$  is the DC component of the signal.
	*$\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component		*$\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component:
	:$$\Delta x_i(t) = x_i(t) - A_0.$$		:$$\Delta x_i(t) = x_i(t) - A_0.$$





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Guenter at 14:40, 12 April 2021

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	[[File:P_ID273__Sig_A_2_2.png\|right\|frame\|Square wave signal with and without DC component]]		[[File:P_ID273__Sig_A_2_2.png\|right\|frame\|Square wave signal with and without DC component]]

	The graph shows ~~some~~ time signals defined for all times $($from  $-\infty$  to  $+\infty)$ . For all ~~six~~ sample signals  $x_i(t)$  the associated spectral function can be written as:		The graph shows six time signals defined for all times $($from  $-\infty$  to  $+\infty)$.  For all sample signals  $x_i(t)$  the associated spectral function can be written as:

	:$$X_i(f)=A_0\cdot{\rm \delta}(f)+\Delta X_i(f).$$		:$$X_i(f)=A_0\cdot{\rm \delta}(f)+\Delta X_i(f).$$

	Here		Here:
	*$A_0$  is the DC component~~, and~~		*$A_0$  is the DC component of the signal.
	*$\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component  $\Delta x_i(t) = x_i(t) - A_0$.		*$\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component
			:$$\Delta x_i(t) = x_i(t) - A_0.$$

Guenter at 14:36, 12 April 2021

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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^]]

Guenter: Guenter moved page Aufgaben:Exercise 2.2: Direct Current Component of Signals to Aufgaben:Exercise 2.2: DC Component of Signals

2021-04-12T14:33:46Z

Guenter moved page Aufgaben:Exercise 2.2: Direct Current Component of Signals to Aufgaben:Exercise 2.2: DC Component of Signals

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Noah at 09:21, 26 March 2021

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	[[File:P_ID273__Sig_A_2_2.png\|right\|frame\|~~Rechtecksignale mit und ohne Gleichanteil~~]]		[[File:P_ID273__Sig_A_2_2.png\|right\|frame\|Square wave signal with and without DC component]]

	The graph shows some time signals defined for all times $($from  $-\infty$  to  $+\infty)$ . For all six sample signals  $x_i(t)$  the associated spectral function can be written as:		The graph shows some time signals defined for all times $($from  $-\infty$  to  $+\infty)$ . For all six sample signals  $x_i(t)$  the associated spectral function can be written as:

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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  <br>[[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal\|Direct Current Signal - Limit Case of a Periodic Signal]].


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	'''(2)'''  Only <u>solution 5 is correct</u>:		'''(2)'''  Only <u>solution 5 is correct</u>:
	*If the DC component   $1\text{V}$ is subtracted from the signal   $x_5(t)$ , the residual signal  $\Delta x_5(t) = x5(t) - 1\text{V}$  is ~~equal to~~ zero.		*If the DC component   $1\text{V}$ is subtracted from the signal   $x_5(t)$,  the residual signal  $\Delta x_5(t) = x5(t) - 1\text{V}$  is zero.
	*Accordignly, the spectral function is  $\Delta X_5(f) = 0$.		*Accordignly, the spectral function is  $\Delta X_5(f) = 0$.
	*For all other time courses  $\Delta x_i(t)$  ~~is not equal to zero~~ and thus the associated spectral function   $\Delta X_i(f)$ ~~is also not equal to zero~~.		*For all other time courses  $\Delta x_i(t)ßne 0$  and thus the associated spectral function   $\Delta X_i(f)\ne 0$,  too.



	'''(3)'''  Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component  $A_0$ .		'''(3)'''  Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component  $A_0$ .
	*For ~~the sample~~ signal  $x_3(t)$  the period duration is  $T_0 = 3\,\text{ms}$. This results in the required DC component:$$A_0=\rm \frac{1}{3\,ms}\cdot \big[1\,V\cdot 1\,ms+(-1\,V)\cdot 2\,ms \big]		*For signal  $x_3(t)$  the period duration is  $T_0 = 3\,\text{ms}$.  This results in the required DC component:$$A_0=\rm \frac{1}{3\,ms}\cdot \big[1\,V\cdot 1\,ms+(-1\,V)\cdot 2\,ms \big]
	\hspace{0.15cm}\underline{=-0.333\,V}.$$		\hspace{0.15cm}\underline{=-0.333\,V}.$$

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	'''(4)'''  The signal  $x_4(t)$  can be written as:  $x_4(t) = 0.5 \,{\rm V} + Δx_4(t)$.		'''(4)'''  The signal  $x_4(t)$  can be written as:  $x_4(t) = 0.5 \,{\rm V} + Δx_4(t)$.
	*Here  $Δx_4(t)$  denotes a ~~square wave~~ pulse with amplitude  $0.5 \,{\rm V} $  and duration  $4 \,{\rm ms} $, which due to its finite duration does not contribute to the DC signal component..		*Here  $Δx_4(t)$  denotes a rectangular pulse with amplitude  $0.5 \,{\rm V} $  and duration  $4 \,{\rm ms} $,
			*which due to its finite duration does not contribute to the DC signal component.
	*Therefore  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$ applies here.		*Therefore  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$ applies here.

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	:$$A_0=\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{-T_{\rm M}/2}^{0}0 {\rm V} \cdot\, {\rm d } {\it t }+\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{0}^{+T_{\rm M}/2}1 \rm V \ {\rm d }{\it t }.$$		:$$A_0=\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{-T_{\rm M}/2}^{0}0 {\rm V} \cdot\, {\rm d } {\it t }+\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{0}^{+T_{\rm M}/2}1 \rm V \ {\rm d }{\it t }.$$

	*Only the second term makes a contribution. From this follows again :  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$.		*Only the second term makes a contribution.  From this follows again :  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$.
	{{ML-Fuß}}		{{ML-Fuß}}

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