Aufgaben:Exercise 3.9: Convolution of Rectangle and Gaussian Pulse - Revision history

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	[[de:Aufgaben:3.~~9_Faltung_von_Rechteck_und_Gauß~~]]		[[de:Aufgaben:Aufgabe 3.9: Faltung von Rechteck und Gauß]]

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	'''(3)'''  With the Dirac delta  $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the Dirac function:		'''(3)'''  With the Dirac delta  $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the Dirac function:

	:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$		:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{- {\rm{\pi}}( {\Delta f \cdot t})^2 }.$$

	*At time  $t = 0$ , one also obtains here with a good approximation  $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.		*At time  $t = 0$ , one also obtains here with a good approximation  $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.

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			[[de:Aufgaben:3.9_Faltung_von_Rechteck_und_Gauß]]

Javier: Text replacement - "Dirac impulse" to "Dirac delta"

2021-05-26T07:05:07Z

Text replacement - "Dirac impulse" to "Dirac delta"

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	* a rectangular pulse  $x_1(t)$  with amplitude  $A_1 =1\,\text{V}$  and duration  $T_1 = 20\,\text{ns}$  (red curve),		* a rectangular pulse  $x_1(t)$  with amplitude  $A_1 =1\,\text{V}$  and duration  $T_1 = 20\,\text{ns}$  (red curve),
	* a rectangular pulse  $x_2(t)$  with amplitude  $A_2 =10\,\text{V}$  and duration  $T_2 = 2\,\text{ns}$  (violet curve),		* a rectangular pulse  $x_2(t)$  with amplitude  $A_2 =10\,\text{V}$  and duration  $T_2 = 2\,\text{ns}$  (violet curve),
	* a Dirac ~~impulse~~  $x_3(t)$  with impulse weight  $2 \cdot 10^{–8}\text{ Vs}$  (green arrow).		* a Dirac delta  $x_3(t)$  with impulse weight  $2 \cdot 10^{–8}\text{ Vs}$  (green arrow).


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	'''(3)'''  With the Dirac ~~impulse~~  $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the Dirac function:		'''(3)'''  With the Dirac delta  $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the Dirac function:

	:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$		:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
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	One can see from comparing the results from  '''(2)'''  and   '''(3)''', that  $y_3(t) \approx y_2(t)$  gilt.		One can see from comparing the results from  '''(2)'''  and   '''(3)''', that  $y_3(t) \approx y_2(t)$  gilt.
	*The reason for this is that the Dirac ~~impulse~~ is a good approximation for a rectangular input impulse of the same area if the rectangular duration  $T$  is significantly smaller than the equivalent impulse duration  $\Delta t$  of the impulse response.		*The reason for this is that the Dirac delta is a good approximation for a rectangular input impulse of the same area if the rectangular duration  $T$  is significantly smaller than the equivalent impulse duration  $\Delta t$  of the impulse response.
	*This means for our example:   If the duration  $T$  of the rectangular input impulse  $x(t)$  is clearly smaller than the  "equivalent pulse duration"  $\Delta t$  of the Gaussian impulse response  $h(t)$, then  $y(t)$  is also almost Gaussian.  But:   '''Gaussian (once) folded with non–Gaussian never results in (exactly) Gaussian!'''		*This means for our example:   If the duration  $T$  of the rectangular input impulse  $x(t)$  is clearly smaller than the  "equivalent pulse duration"  $\Delta t$  of the Gaussian impulse response  $h(t)$, then  $y(t)$  is also almost Gaussian.  But:   '''Gaussian (once) folded with non–Gaussian never results in (exactly) Gaussian!'''
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:16, 29 April 2021

2021-04-29T14:16:38Z

Guenter at 13:45, 29 April 2021

2021-04-29T13:45:18Z

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

2021-03-23T12:38:49Z

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

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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 13:08, 1 February 2021

2021-02-01T13:08:50Z

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	<quiz display=simple>		<quiz display=simple>
	{Calculate the ~~signa~~  $y_1(t) = x_1(t) \ast h(t)$. <br>What values result at times $t = 0$ und $t = 20\,\text{ns}$ with the approximation $(2\pi )^{1/2} \approx 2.5$?		{Calculate the signal  $y_1(t) = x_1(t) \ast h(t)$. <br>What values result at times $t = 0$ und $t = 20\,\text{ns}$ with the approximation $(2\pi )^{1/2} \approx 2.5$?
	\|type="{}"}		\|type="{}"}
	$y_1(t=0)\ = \ $ { 0.682 3% }  $\text{V}$		$y_1(t=0)\ = \ $ { 0.682 3% }  $\text{V}$

Noah at 13:08, 1 February 2021

2021-02-01T13:08:24Z

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	}}		}}

	[[File:P_ID540__Sig_A_3_9_neu.png\|250px\|right\|frame\|~~Rechteckförmiges~~  $x(t)$  ~~und gaußförmiges~~  $h(t)$]]		[[File:P_ID540__Sig_A_3_9_neu.png\|250px\|right\|frame\|Rectangular  $x(t)$  and gaussian  $h(t)$]]

	We consider a Gaussian low pass with the equivalent bandwidth  $\Delta f = 40 \,\text{MHz}$:		We consider a Gaussian low pass with the equivalent bandwidth  $\Delta f = 40 \,\text{MHz}$:

Noah at 13:07, 1 February 2021

2021-02-01T13:07:04Z

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	'''(3)'''  ~~Beim diracförmigen Eingangssignal~~  $x_3(t)$  ~~ist das Ausgangssignal~~  $y_3(t)$  ~~gleich der Impulsantwort~~  $h(t)$, ~~gewichtet mit dem Gewicht der Diracfunktion~~:		'''(3)'''  With the dirac-shaped input signal   $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the dirac function:

	:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$		:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$

	*~~Zum Zeitpunkt~~  $t = 0$  ~~erhält man auch hier mit guter Näherung~~  $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.		*At time  $t = 0$ , one also obtains here with a good approximation  $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.
	*~~Nach~~  $20\, \rm ns$  ~~ist der Ausgangsimpuls um den Faktor~~  ${\rm e}^{–0.64π} \hspace{0.15 cm}\underline{\approx 0.136}$  ~~kleiner und man erhält~~  $y_3( t = 20 \,\text{ns}) ≈ 0.11 \,\text{V}$.		*After  $20\, \rm ns$ , the output pulse is smaller by a factor of  ${\rm e}^{–0.64π} \hspace{0.15 cm}\underline{\approx 0.136}$  and one obtains  $y_3( t = 20 \,\text{ns}) ≈ 0.11 \,\text{V}$.


	~~Man erkennt aus dem Vergleich der Resultate aus~~  '''(2)'''  ~~und~~   '''(3)''', ~~dass~~  $y_3(t)$ ≈ $y_2(t)$  gilt.		One can see from comparing the results from  '''(2)'''  and   '''(3)''', that  $y_3(t)$ ≈ $y_2(t)$  gilt.
	*~~Der Grund hierfür ist, dass der Diracimpuls eine gute Näherung für einen rechteckförmigen Eingangsimpuls gleicher Fläche ist, wenn die Rechteckdauer~~  $T$  ~~deutlich kleiner als die äquivalente Impulsdauer~~  $\Delta t$  ~~der Impulsantwort ist~~.		*The reason for this is that the Dirac impulse is a good approximation for a rectangular input impulse of the same area if the rectangular duration  $T$  is significantly smaller than the equivalent impulse duration  $\Delta t$  of the impulse response.
	*Das heißt für unser Beispiel:  ~~Ist die Dauer~~  $T$  ~~des rechteckförmigen Eingangsimpulses~~  $x(t)$  ~~deutlich kleiner als die äquivalente Dauer~~  $\Delta t$  ~~der gaußförmigen Impulsantwort~~  $h(t)$, ~~dann ist auch der Ausgangsimpuls~~  $y(t)$  ~~nahezu gaußförmig~~. ~~Aber~~:   '''~~Gauß~~ (~~einmal~~) ~~gefaltet mit Nicht~~–~~Gauß ergibt nie~~ (~~exakt~~) ~~Gauß~~!'''		This means for our example:   If the duration  $T$  of the rectangular input impulse  $x(t)$  is clearly smaller than the equivalent duration  $\Delta t$  of the Gaussian impulse response  $h(t)$, then the output impulse  $y(t)$  is also almost Gaussian. But:   '''Gaussian (once) folded with non–Gaussian never results in (exactly) Gaussian!'''
	{{ML-Fuß}}		{{ML-Fuß}}

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

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	:$$y_1( t ) = A_1 \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm} {\rm{d}}u.}$$		:$$y_1( t ) = A_1 \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm} {\rm{d}}u.}$$

	*Here the substitution  $u = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \tau$  was used. The integration limits are at:		*Here the substitution  $u = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \tau$  was used.  The integration limits are at:

	:$$u_1 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \big( {t - T_1 /2} \big),\hspace{0.5cm}u_2 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \big( {t + T_1 /2} \big).$$		:$$u_1 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \big( {t - T_1 /2} \big),\hspace{0.5cm}u_2 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \big( {t + T_1 /2} \big).$$
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	'''(3)'''  With the ~~dirac-shaped input signal~~   $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the ~~dirac~~ function:		'''(3)'''  With the Dirac impulse  $x_3(t)$ , the output signal  $y_3(t)$  is equal to the impulse response  $h(t)$, weighted by the weight of the Dirac function:

	:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$		:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
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	One can see from comparing the results from  '''(2)'''  and   '''(3)''', that  $y_3(t)~~$ ≈ $~~y_2(t)$  gilt.		One can see from comparing the results from  '''(2)'''  and   '''(3)''', that  $y_3(t) \approx y_2(t)$  gilt.
	*The reason for this is that the Dirac impulse is a good approximation for a rectangular input impulse of the same area if the rectangular duration  $T$  is significantly smaller than the equivalent impulse duration  $\Delta t$  of the impulse response.		*The reason for this is that the Dirac impulse is a good approximation for a rectangular input impulse of the same area if the rectangular duration  $T$  is significantly smaller than the equivalent impulse duration  $\Delta t$  of the impulse response.
	This means for our example:   If the duration  $T$  of the rectangular input impulse  $x(t)$  is clearly smaller than the equivalent duration  $\Delta t$  of the Gaussian impulse response  $h(t)$, then ~~the output impulse~~  $y(t)$  is also almost Gaussian. But:   '''Gaussian (once) folded with non–Gaussian never results in (exactly) Gaussian!'''		*This means for our example:   If the duration  $T$  of the rectangular input impulse  $x(t)$  is clearly smaller than the  "equivalent pulse duration"  $\Delta t$  of the Gaussian impulse response  $h(t)$, then  $y(t)$  is also almost Gaussian.  But:   '''Gaussian (once) folded with non–Gaussian never results in (exactly) Gaussian!'''
	{{ML-Fuß}}		{{ML-Fuß}}

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

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	}}		}}

	[[File:P_ID540__Sig_A_3_9_neu.png\|250px\|right\|frame\|~~Rectangular~~  $x(t)$  and ~~gaussian~~  $h(t)$]]		[[File:P_ID540__Sig_A_3_9_neu.png\|250px\|right\|frame\|Convolution of rectangle  $x(t)$  and Gaussian pulse  $h(t)$]]

	We consider a Gaussian low pass with the equivalent bandwidth  $\Delta f = 40 \,\text{MHz}$:		We consider a Gaussian low–pass with the equivalent bandwidth  $\Delta f = 40 \,\text{MHz}$:

	:$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$		:$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$
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	:$$h( t ) = \Delta f \cdot {\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm} \cdot \hspace{0.05cm} t} )^2 } .$$		:$$h( t ) = \Delta f \cdot {\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm} \cdot \hspace{0.05cm} t} )^2 } .$$

	From the sketch it can be seen that the equivalent time duration   ⇒   $\Delta t = 1/\Delta f = 25\,\text{ns}$  the impulse response  $h(t)$  can be read at the two inflection points of the Gaussian function.		From the sketch it can be seen that the  "equivalent time duration"   ⇒   $\Delta t = 1/\Delta f = 25\,\text{ns}$  of the impulse response  $h(t)$  can be read at the two inflection points of the Gaussian function.

	Three different pulse-like signals are now applied to the input of the low-pass filter:		Three different pulse-like signals are now applied to the input of the low-pass filter:
	* a ~~square-wave~~ pulse  $x_1(t)$  with amplitude  $A_1 =1\,\text{V}$  and duration  $T_1 = 20\,\text{ns}$  (red curve),		* a rectangular pulse  $x_1(t)$  with amplitude  $A_1 =1\,\text{V}$  and duration  $T_1 = 20\,\text{ns}$  (red curve),
	* a rectangular pulse  $x_2(t)$  with amplitude  $A_2 =10\,\text{V}$  and duration  $T_2 = 2\,\text{ns}$  (violet curve),		* a rectangular pulse  $x_2(t)$  with amplitude  $A_2 =10\,\text{V}$  and duration  $T_2 = 2\,\text{ns}$  (violet curve),
	* a Dirac ~~pulse~~  $x_3(t)$  with ~~pulse~~ weight  $2 \cdot 10^{–8}\text{ Vs}$  (green arrow).		* a Dirac impulse  $x_3(t)$  with impulse weight  $2 \cdot 10^{–8}\text{ Vs}$  (green arrow).


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	*To answer the questions, you can use the complementary Gaussian error integral, which is defined as follows:		*To answer the questions, you can use the complementary Gaussian error integral, which is defined as follows:
	[[File:P_ID541__Sig_A_3_9Tab_neu.png\|right\|frame\|Some values of the Q-function]]		[[File:P_ID541__Sig_A_3_9Tab_neu.png\|right\|frame\|Some values of the  $\rm Q$–function]]
	:$${\rm Q}( x ) = \frac{1}{ {\sqrt {2{\rm{\pi }}} }}\int_{\it x}^\infty {{\rm{e}}^{{{ - {\it u}}}^{\rm{2}} {\rm{/2}}} }\hspace{0.1cm}{\rm{d}}{\it u}.$$		:$${\rm Q}( x ) = \frac{1}{ {\sqrt {2{\rm{\pi }}} }}\int_{\it x}^\infty {{\rm{e}}^{{{ - {\it u}}}^{\rm{2}} {\rm{/2}}} }\hspace{0.1cm}{\rm{d}}{\it u}.$$

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	<quiz display=simple>		<quiz display=simple>
	{Calculate the signal  $y_1(t) = x_1(t) \ast h(t)$. ~~<br>~~What values result at times $t = 0$ ~~und~~ $t = 20\,\text{ns}$ with the approximation $(2\pi )^{1/2} \approx 2.5$?		{Calculate the signal  $y_1(t) = x_1(t) \ast h(t)$.  What values result at times  $t = 0$  and  $t = 20\,\text{ns}$  with the approximation  $(2\pi )^{1/2} \approx 2.5$?
	\|type="{}"}		\|type="{}"}
	$y_1(t=0)\ = \ $ { 0.682 3% }  $\text{V}$		$y_1(t=0)\ = \ $ { 0.682 3% }  $\text{V}$
	$y_1(t=20\,\text{ns})\ = \ $ { 0.158 3% }  $\text{V}$		$y_1(t=20\,\text{ns})\ = \ $ { 0.158 3% }  $\text{V}$

	{What are the signal values of the output signal  $y_2(t) = x_2(t) \ast h(t)$ at the points ~~in time considered~~?		{What are the signal values of the output signal  $y_2(t) = x_2(t) \ast h(t)$  at the considered time points?
	\|type="{}"}		\|type="{}"}
	$y_2(t=0)\ = \ $ { 0.8 3% }  $\text{V}$		$y_2(t=0)\ = \ $ { 0.8 3% }  $\text{V}$
	$y_2(t=20 \,\text{ns})\ = \ $ { 0.11 3% }  $\text{V}$		$y_2(t=20 \,\text{ns})\ = \ $ { 0.11 3% }  $\text{V}$

	{What is the ~~magnitude~~ of the output signa  $y_3(t) = x_3(t) \ast h(t)$  at the points ~~in time considered~~? Interpret the result.		{What is the value of the output signa  $y_3(t) = x_3(t) \ast h(t)$  at the considered time points?  Interpret the result.
	\|type="{}"}		\|type="{}"}
	$y_3(t=0)\ = \ $ { 0.8 3% }  $\text{V}$		$y_3(t=0)\ = \ $ { 0.8 3% }  $\text{V}$