Aufgaben:Exercise 3.8Z: Convolution of Two Rectangles - Revision history

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	[[de:Aufgaben:3.~~8Z_Faltung_zweier_Rechtecke~~]]		[[de:Aufgaben:Aufgabe 3.8Z: Faltung zweier Rechtecke]]

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	[[File:P_ID536__Sig_Z_3_8_a_neu.png\|right\|frame\|To illustrate the convolution  $x(t) \star h(t)$;<br>the abscissas have been renamed:  $\tau$]]		[[File:P_ID536__Sig_Z_3_8_a_neu.png\|right\|frame\|To illustrate the convolution  $x(t) \star h(t)$;<br>the abscissas have been renamed:  $\tau$]]
	'''(1)'''  In general, the following applies to the convolution integral:		'''(1)'''  In general, the following applies to the convolution integral:
	:$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$		:$$y(t) = \int_{ - \infty }^{+ \infty} {x( \tau ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$

	The signal value at time  $t = 1 \,\text{ms}$  can be calculated as follows:		The signal value at time  $t = 1 \,\text{ms}$  can be calculated as follows:

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Guenter: Guenter moved page Aufgaben:Exercise 3.8Z:Convolution of Two Rectangles to Aufgaben:Exercise 3.8Z: Convolution of Two Rectangles

2021-05-24T12:56:52Z

Guenter moved page Aufgaben:Exercise 3.8Z:Convolution of Two Rectangles to Aufgaben:Exercise 3.8Z: Convolution of Two Rectangles

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Guenter at 12:57, 29 April 2021

2021-04-29T12:57:17Z

Guenter at 12:46, 29 April 2021

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	+ The output signal  ${y(t)}$  has a trapezoidal shape.		+ The output signal  ${y(t)}$  has a trapezoidal shape.
	- The spectrum is   ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.		- The spectrum is   ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
	+ With  $T = 2 \,\text{ms}$ , a triangular shape would result.		+ With  $T = 2 \,\text{ms}$,  a triangular shape would result.


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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID536__Sig_Z_3_8_a_neu.png\|right\|frame\|To illustrate the ~~graphical~~ convolution $x(t) \star h(t)$]]		[[File:P_ID536__Sig_Z_3_8_a_neu.png\|right\|frame\|To illustrate the convolution  $x(t) \star h(t)$;<br>the abscissas have been renamed to  $"\tau"$]]
	'''(1)'''  In general, the following applies to the convolution integral:		'''(1)'''  In general, the following applies to the convolution integral:
	:$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$		:$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$
	~~''Hints:''  The abscissas in the graph opposite have been renamed  $\tau$ .~~

	The signal value at time  $t = 1 \,\text{ms}$  ms can be calculated as follows:		The signal value at time  $t = 1 \,\text{ms}$  ms can be calculated as follows:

Guenter at 12:39, 29 April 2021

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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|"The Convolution Theorem and Operation"]].		*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|"The Convolution Theorem and Operation"]].
	*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#~~Graphical_Convolution~~\|"Graphical ~~Convolution~~"]].		*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_convolution\|"Graphical convolution"]].
	*The topic of this section is also illustrated in the interactive applet  [[Applets:Graphical_Convolution\| "Graphical Convolution"]].		*The topic of this section is also illustrated in the interactive applet  [[Applets:Graphical_Convolution\| "Graphical Convolution"]].

Guenter at 12:38, 29 April 2021

2021-04-29T12:38:34Z

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	}}		}}

	[[File:P_ID535__Sig_Z_3_8.png\|right\|frame\|~~For the~~ convolution of two rectangles		[[File:P_ID535__Sig_Z_3_8.png\|right\|frame\|The convolution of two rectangles  $x(t)$  and  $h(t)$
	]]		]]
	At the input of a causal LTI system (i.e. linear and time-invariant) with a rectangular impulse response  ${h(t)}$  of duration  $2 \,\text{ms}$ , a rectangular ~~impulse~~  ${x(t)}$  of duration  $T = 3 \,\text{ms}$  and amplitude  $A = 2\,\text{ V}$  is applied. ~~The square-wave functions each start at the time~~  ~~$t = 0$.~~		At the input of a causal LTI system (i.e. linear and time-invariant)
			*with a rectangular impulse response  ${h(t)}$  of duration  $2 \,\text{ms}$ ,
			*a rectangular pulse  ${x(t)}$  of duration  $T = 3 \,\text{ms}$  and amplitude  $A = 2\,\text{ V}$  is applied.

	~~In this task you are to calculate the output signal  ${y(t)}$  with the help of the graphic convolution. As you can easily check, the output signal  ${y(t)}$~~
	*differs from zero only in the range from  $0$  to  $5 \, \text{ms}$  and
	*is symmetrical at the time  $t = 2.5 \, \text{ms}$.

			The rectangular functions each start at the time  $t = 0$.

			In this task you are to calculate the output signal  ${y(t)}$  with the help of the "Graphical Convolution".

			As you can easily check, the output signal  ${y(t)}$
			*differs from zero only in the range between  $0$  and  $5 \, \text{ms}$,  and
			*is symmetrical at the time  $t = 2.5 \, \text{ms}$.




			''Hints:''
			*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|"The Convolution Theorem and Operation"]].
			*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution\|"Graphical Convolution"]].
			*The topic of this section is also illustrated in the interactive applet  [[Applets:Graphical_Convolution\| "Graphical Convolution"]].


	~~''Hinweise:''~~
	*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].
	*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution\|Graphical Convolution]]
	*The topic of this section is also illustrated in the interactive applet  [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung\|To illustrate the graphical convolution]].




	===Questions===		===Questions===
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	{Determine the signal values for the ~~time points~~  $t = 3 \,\text{ms}$  and  $t = 4 \,\text{ms}$  by exploiting the symmetry properties.		{Determine the signal values for the times  $t = 3 \,\text{ms}$  and  $t = 4 \,\text{ms}$  by exploiting the symmetry properties.
	\|type="{}"}		\|type="{}"}
	$y(t = 3 \,\text{ms})\ = \ $ { 1.2 3% }  $\text{V}$		$y(t = 3 \,\text{ms})\ = \ $ { 1.2 3% }  $\text{V}$
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	\|type="[]"}		\|type="[]"}
	+ The output signal  ${y(t)}$  has a trapezoidal shape.		+ The output signal  ${y(t)}$  has a trapezoidal shape.
	- The spectrum is:   ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.		- The spectrum is   ${Y(f)} = Y_0 \cdot \text{si}^{2}(\pi f T)$.
	+ With  $T = 2 \,\text{ms}$ , a triangular shape would result.		+ With  $T = 2 \,\text{ms}$ , a triangular shape would result.

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

2021-03-23T12:38:46Z

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	[[Category:~~Exercises for~~ Signal Representation\|^3.4 The Convolution Theorem^]]		[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]

Noah at 21:45, 15 February 2021

2021-02-15T21:45:00Z

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	*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].		*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].
	*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution\|Graphical Convolution]]		*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution\|Graphical Convolution]]
	*The topic of this section is also illustrated in the interactive applet  [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung\|To illustrate the graphical convolution]] ~~veranschaulicht~~.		*The topic of this section is also illustrated in the interactive applet  [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung\|To illustrate the graphical convolution]].

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	*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].		*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|The Convolution Theorem and Operation]].
	*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution\|Graphical Convolution]]		*It mainly refers to the page  [[Signal_Representation/The_Convolution_Theorem_and_Operation#Graphical_Convolution\|Graphical Convolution]]
	*The topic of this section is also illustrated in the interactive applet  [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung\|To illustrate the graphical convolution]] ~~veranschaulicht~~.		*The topic of this section is also illustrated in the interactive applet  [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung\|To illustrate the graphical convolution]].

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID536__Sig_Z_3_8_a_neu.png\|right\|frame\|To illustrate the convolution  $x(t) \star h(t)$;<br>the abscissas have been renamed to  $"\tau"$]]		[[File:P_ID536__Sig_Z_3_8_a_neu.png\|right\|frame\|To illustrate the convolution  $x(t) \star h(t)$;<br>the abscissas have been renamed:  $\tau$]]
	'''(1)'''  In general, the following applies to the convolution integral:		'''(1)'''  In general, the following applies to the convolution integral:
	:$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$		:$$y(t) = \int_{ - \infty }^{ + \infty } {x( \tau ) \cdot h( {t - \tau } )}\hspace{0.1cm} {\rm d}\tau.$$

	The signal value at time  $t = 1 \,\text{ms}$  ms can be calculated as follows:		The signal value at time  $t = 1 \,\text{ms}$  can be calculated as follows:
	*~~Reflection~~ of the impulse response  ${h(\tau)}$,		*Mirroring of the impulse response  ${h(\tau)}$,
	*~~shift~~ by  $t = 1 \text{ ms}$  to the right (violet curve in the sketch),		*shifting by  $t = 1 \text{ ms}$  to the right (violet curve in the sketch),
	*multiplication of the two functions and integration.		*multiplication of the two functions and integration.


	The product is also rectangular with the height  $2 \text{ V} \cdot 300 \; \text{1/s}$  and width  $1 \,\text{ms}$. This results in for the area:		The product is also rectangular with the height  $2 \text{ V} \cdot 300 \; \text{1/s}$  and width  $1 \,\text{ms}$. This results for the area:
	:$$y( {t = 1\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= 0.6\;{\rm{V}}}{\rm{.}}$$		:$$y( {t = 1\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= 0.6\;{\rm{V}}}{\rm{.}}$$
	The green rectangle illustrates the calculation of the second signal value. Now the resulting rectangle is twice as wide after the multiplication and we get:		The green rectangle illustrates the calculation of the second signal value.  Now the resulting rectangle is twice as wide after the multiplication and we get:
	:$$y( {t = 2\;{\rm{ms}}} ) = 2\;{\rm{V}} \cdot {\rm{300}}\;{1}/{{\rm{s}}} \cdot 2\;{\rm{ms}}\hspace{0.15 cm}\underline{={\rm{1.2}}\;{\rm{V}}}{\rm{.}}$$		:$$y( {t = 2\;{\rm{ms}}} ) = 2\;{\rm{V}} \cdot {\rm{300}}\;{1}/{{\rm{s}}} \cdot 2\;{\rm{ms}}\hspace{0.15 cm}\underline{={\rm{1.2}}\;{\rm{V}}}{\rm{.}}$$


	'''(2)'''  Because of the symmetry of ${y(t)}$ with respect to the time  $t = 2.5\, \text {ms}$  holds:		'''(2)'''  Because of the symmetry of  ${y(t)}$  with respect to the time  $t = 2.5\, \text {ms}$  holds:
	:$$y( {t = 3\;{\rm{ms}}} ) = y( {t = 2\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= {\rm{1}}{\rm{.2}}\;{\rm{V}}}{\rm{,}}$$		:$$y( {t = 3\;{\rm{ms}}} ) = y( {t = 2\;{\rm{ms}}} ) \hspace{0.15 cm}\underline{= {\rm{1}}{\rm{.2}}\;{\rm{V}}}{\rm{,}}$$
	:$$y( {t = 4\;{\rm{ms}}} ) = y( {t = 1\;{\rm{ms}}} )\hspace{0.15 cm}\underline{ = 0.6\;{\rm{V}}}{\rm{.}}$$		:$$y( {t = 4\;{\rm{ms}}} ) = y( {t = 1\;{\rm{ms}}} )\hspace{0.15 cm}\underline{ = 0.6\;{\rm{V}}}{\rm{.}}$$


	[[File:P_ID537__Sig_Z_3_8_c.png\|right\|frame\|~~Convolution~~ result  $y(t)$]]		[[File:P_ID537__Sig_Z_3_8_c.png\|right\|frame\|Overall result  $y(t)$]]
	'''(3)'''  In subtasks  '''(1)'''  and  '''(2)'''  the signal values were calculated at discrete time points.		'''(3)'''  Proposed solutions <u>1 and 3</u> are correct:
			*In subtasks  '''(1)'''  and  '''(2)'''  the signal values were calculated at discrete time points.
	*All points are to be connected by straight line segments, since the integration over rectangular functions of increasing width results in a linear course.		*All points are to be connected by straight line segments, since the integration over rectangular functions of increasing width results in a linear course.
	*This means:  The output signal  ${y(t)}$  is trapezoidal.		*This means:  The output signal  ${y(t)}$  is trapezoidal.
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	*The associated spectrum is complex and reads:		*The associated spectrum is complex and reads:
	:$$Y(f) = 6 \cdot 10^{ - 3} \;{{\rm{V}}}/{{{\rm{Hz}}}} \cdot {\mathop{\rm si}\nolimits} ( {2\;{\rm{ms}}\cdot{\rm{\pi }}f} ) \cdot {\mathop{\rm si}\nolimits} ( {3\;{\rm{ms}}\cdot{\rm{\pi }}f}) \cdot {\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \hspace{0.05cm}\cdot \hspace{0.05cm}2.5\;{\rm{ms}}\hspace{0.05cm}\cdot \hspace{0.05cm} \pi }}f} .$$		:$$Y(f) = 6 \cdot 10^{ - 3} \;{{\rm{V}}}/{{{\rm{Hz}}}} \cdot {\mathop{\rm si}\nolimits} ( {2\;{\rm{ms}}\cdot{\rm{\pi }}f} ) \cdot {\mathop{\rm si}\nolimits} ( {3\;{\rm{ms}}\cdot{\rm{\pi }}f}) \cdot {\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \hspace{0.05cm}\cdot \hspace{0.05cm}2.5\;{\rm{ms}}\hspace{0.05cm}\cdot \hspace{0.05cm} \pi }}f} .$$
	*If the input pulse  ${x(t)}$  ~~had~~ the duration  $T = 2\, \text {ms}$, ~~had~~ the duration  ${y(t)}$  would show a triangular waveform between  ${t = 0}$  and  $t = 4 \text { ms}$ .		*If the input pulse  ${x(t)}$  would have the duration  $T = 2\, \text {ms}$, the duration  ${y(t)}$  would show a triangular waveform between  ${t = 0}$  and  $t = 4 \text { ms}$.  The maximum  $1.2 \, \text {V}$  would then only result at the time  $t = 2 \, \text {ms}$.
	*The maximum  $1.2 \, \text {V}$  would then only result at the time  $t = 2 \, \text {ms}$.


	~~Proposed solutions <u>1 and 3</u> are therefore correct.~~
	{{ML-Fuß}}		{{ML-Fuß}}