Aufgaben:Exercise 3.2: From the Spectrum to the Signal - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:04Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
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	:$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$		:$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$

	The associated time function  $x(t)$  can be determined with the help of  [[Signal_Representation/~~Fourier_Transform_and_Its_Inverse~~#The_second_Fourier_integral\|The second Fourier integral]] :		The associated time function  $x(t)$  can be determined with the help of  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral\|The second Fourier integral]] :

	:$$x(t) = \int_{ - \infty }^{ + \infty } {X(f)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f = x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$		:$$x(t) = \int_{ - \infty }^{ + \infty } {X(f)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f = x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$
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	[[Category:Signal Representation: Exercises\|^3.1 Fourier Transform and Its Inverse^]]		[[Category:Signal Representation: Exercises\|^3.1 Fourier Transform and Its Inverse^]]
	[[de:Aufgaben:3.~~2_Vom_Spektrum_zum_Signal~~]]		[[de:Aufgaben:Aufgabe 3.2: Vom Spektrum zum Signal]]

Maintenance script: Add German interlanguage link

2026-03-16T12:32:15Z

Add German interlanguage link

← Older revision		Revision as of 14:32, 16 March 2026
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	[[Category:Signal Representation: Exercises\|^3.1 Fourier Transform and Its Inverse^]]		[[Category:Signal Representation: Exercises\|^3.1 Fourier Transform and Its Inverse^]]
			[[de:Aufgaben:3.2_Vom_Spektrum_zum_Signal]]

Guenter at 14:25, 21 April 2021

2021-04-21T14:25:32Z

← Older revision		Revision as of 16:25, 21 April 2021
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct is the <u>proposed solution 2</u>   ⇒  $x(t)$  is <u>purely real</u>:		'''(1)'''  Correct is the <u>proposed solution 2</u>   ⇒  $x(t)$  is <u>purely real</u>:
	*For the imaginary signal component  ⇒   $x_{\rm I}(t)$  the integrand is an odd function (even numerator, odd denominator)..		*For the imaginary signal component  ⇒   $x_{\rm I}(t)$  the integrand is an odd function  (even numerator, odd denominator).
	*Thus the integral from  $-\infty$  bis  $+\infty$  is zero.		*Thus the integral from  $-\infty$  bis  $+\infty$  is zero.
	*In contrast, for the real component  $x_{\rm R}(t)$   ⇒   even integrand (odd numerator, odd denominator) yields a non-zero value.		*In contrast, for the real component  $x_{\rm R}(t)$   ⇒   even integrand  (odd numerator, odd denominator)  yields a non-zero value.



	'''(2)'''  With the abbreviation  $a = 2\pi t$  can be written for the time signal:		'''(2)'''  With the abbreviation  $a = 2\pi t$  can be written for the time signal:


	:$$x(t) = x_{\rm R} \left( t \right) = \frac{{4\,{\rm V}}}{\pi }\int_0^\infty {\frac{{\sin( {af} )}}{f}}\hspace{0.1cm} {\rm d}f.$$		:$$x(t) = x_{\rm R} \left( t \right) = \frac{{4\,{\rm V}}}{\pi }\int_0^\infty {\frac{{\sin( {af} )}}{f}}\hspace{0.1cm} {\rm d}f.$$
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	'''(3)'''  ~~At ~~ $t ~~= 0~~$    $x(t)$  ~~has a jump point.~~ The right-hand limit value for  $t \rightarrow 0$  ~~lautet~~  $x_+ = +2\,\text{V}$.		'''(3)'''  $x(t)$  has a jumping point at  $t = 0$.  The right-hand limit value for  $t \rightarrow 0$  is  $x_+ = +2\,\text{V}$.
	*If one approaches the ~~jump~~ point of negative times as close as desired, one obtains  $x_– = -\hspace{-0.05cm}2\,\text{V}$. The following then applies to the actual signal value at  $t = 0$~~ ~~:		*If one approaches the jumping point of negative times as close as desired, one obtains  $x_– = -\hspace{-0.05cm}2\,\text{V}$.
			*The following then applies to the actual signal value at  $t = 0$:

	:$$x( {t = 0} ) = {1}/{2}\cdot ( x_{+} + x_{-} ) \hspace{0.15 cm}\underline{= 0}.$$		:$$x( {t = 0} ) = {1}/{2}\cdot ( x_{+} + x_{-} ) \hspace{0.15 cm}\underline{= 0}.$$
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	Here is a second solution:		Here is a second solution:
	*The right-hand limit for  $f → 0$  is  $X_+ = -\text{j} \cdot \infty$, the left-hand limit  $X_- = \text{j} \cdot \infty$.		*The right–hand limit for  $f → 0$  is  $X_+ = -\text{j} \cdot \infty$,– and the left–hand limit  $X_- = \text{j} \cdot \infty$.
	*So the relationship also applies with regard to the spectral value at   $f = 0$~~ ~~:		*So the relationship also applies with regard to the spectral value at   $f = 0$:
	:$$X( {f = 0}) = {1}/{2}\cdot \left( {X_{ +} + X_{-} } \right) = 0.$$		:$$X( {f = 0}) = {1}/{2}\cdot \left( {X_{ +} + X_{-} } \right) = 0.$$

Guenter at 14:19, 21 April 2021

2021-04-21T14:19:54Z

← Older revision		Revision as of 16:19, 21 April 2021
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	:$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$		:$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$

	The associated time function  $x(t)$  can be determined with the help of  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#~~The_Second_Fourier_Integral~~\|The ~~Second~~ Fourier ~~Integral~~]] :		The associated time function  $x(t)$  can be determined with the help of  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral\|The second Fourier integral]] :

	:$$x(t) = \int_{ - \infty }^{ + \infty } {X(f)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f = x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$		:$$x(t) = \int_{ - \infty }^{ + \infty } {X(f)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f = x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$

Guenter at 14:18, 21 April 2021

2021-04-21T14:18:22Z

← Older revision		Revision as of 16:18, 21 April 2021
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	:$$x_{\rm R} (t) = 2\,{\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{{\sin ( {2\pi ft} )}}{ {\pi f}}}\hspace{0.1cm} {\rm d}f, $$		:$$x_{\rm R} (t) = 2\,{\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{{\sin ( {2\pi ft} )}}{ {\pi f}}}\hspace{0.1cm} {\rm d}f, $$
	:$$x_{\rm I} (t) = -2\, {\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{ {\cos ( {2\pi ft} )}}{ {\pi f}}} \hspace{0.1cm}{\rm d}f.$$		:$$x_{\rm I} (t) = -2\, {\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{ {\cos ( {2\pi ft} )}}{ {\pi f}}} \hspace{0.1cm}{\rm d}f.$$





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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Transform_and_Its_Inverse\|Fourier Transform and ~~Its~~ Inverse]].		*This exercise belongs to the chapter  [[Signal_Representation/Fourier_Transform_and_Its_Inverse\|Fourier Transform and its Inverse]].

	*If necessary, use the following information for the solution:		*If necessary, use the following information for the solution:
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	<quiz display=simple>		<quiz display=simple>
	{Which of the following statements are true for the time signal  $x(t)$ ?		{Which of the following statements are true for the time signal  $x(t)$ ?
	\|type="[]"}		\|type="()"}
	- $x(t)$  is a complex function.		- $x(t)$  is a complex function.
	+ $x(t)$  is purely real.		+ $x(t)$  is purely real.
	- $x(t)$  is purely imaginary.		- $x(t)$  is purely imaginary.

	{Calculate the signal curve  $x(t)$  in the entire definition area. Which values occur at the times  $t = 1\, \text{ms}$  and  $t = -\hspace{-0.05cm}1\, \text{ ms}$~~  auf~~?		{Calculate the signal curve  $x(t)$  in the entire definition area.  Which values occur at the times  $t = 1\, \text{ms}$  and  $t = -\hspace{-0.05cm}1\, \text{ ms}$?
	\|type="{}"}		\|type="{}"}
	$x(t=+1\, \text{ms}) \ = \ $ { 2 3% } $\ \text{V}$		$x(t=+1\, \text{ms}) \ = \ $ { 2 3% } $\ \text{V}$

Noah at 13:41, 26 March 2021

2021-03-26T13:41:58Z

← Older revision		Revision as of 15:41, 26 March 2021
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	<quiz display=simple>		<quiz display=simple>
	{Which of the following statements are true for the time ~~signa~~  $x(t)$ ?		{Which of the following statements are true for the time signal  $x(t)$ ?
	\|type="[]"}		\|type="[]"}
	- $x(t)$  is a complex function.		- $x(t)$  is a complex function.

Noah at 13:36, 26 March 2021

2021-03-26T13:36:16Z

← Older revision		Revision as of 15:36, 26 March 2021
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	}}		}}

	[[File:P_ID495__Sig_A_3_2.png\|right\|frame\|~~Spektraldarstellung der Sprungfunktion~~]]		[[File:P_ID495__Sig_A_3_2.png\|right\|frame\|Spectral representation of the unit step function]]

	Given the spectral function		Given the spectral function
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct is the <u>proposed solution 2</u>   ⇒  $x(t)$  is <u>purely real:</u>:		'''(1)'''  Correct is the <u>proposed solution 2</u>   ⇒  $x(t)$  is <u>purely real</u>:
	*For the imaginary signal component  ⇒   $x_{\rm I}(t)$  the integrand is an odd function (~~gerader Zähler~~, ~~ungerader Nenner~~).		*For the imaginary signal component  ⇒   $x_{\rm I}(t)$  the integrand is an odd function (even numerator, odd denominator)..
	*Thus the integral from  $-\infty$  bis  $+\infty$  is zero.		*Thus the integral from  $-\infty$  bis  $+\infty$  is zero.
	*In contrast, for the real component  $x_{\rm R}(t)$   ⇒   even integrand (odd numerator, odd denominator) yields a non-zero value.		*In contrast, for the real component  $x_{\rm R}(t)$   ⇒   even integrand (odd numerator, odd denominator) yields a non-zero value.

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:51Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

← Older revision		Revision as of 14:39, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:~~Exercises for~~ Signal Representation\|^3.1 Fourier Transform and Its Inverse^]]		[[Category:Signal Representation: Exercises\|^3.1 Fourier Transform and Its Inverse^]]

Noah at 21:47, 17 January 2021

2021-01-17T21:47:49Z

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Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.2: Vom Spektrum zum Signal to Aufgaben:Exercise 3.2: From the Spectrum to the Signal

2020-11-02T14:28:25Z

Oezdemir moved page Aufgaben:Aufgabe 3.2: Vom Spektrum zum Signal to Aufgaben:Exercise 3.2: From the Spectrum to the Signal

← Older revision	Revision as of 16:28, 2 November 2020
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