Aufgaben:Exercise 3.9Z: Convolution of Gaussian Pulses - Revision history

Guenter at 14:45, 29 April 2021

2021-04-29T14:45:06Z

Guenter at 14:34, 29 April 2021

2021-04-29T14:34:57Z

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

2021-03-23T12:38:57Z

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

Noah at 13:30, 1 February 2021

2021-02-01T13:30:15Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.9Z: Gauß gefaltet mit Gauß to Aufgaben:Exercise 3.9Z: Convolution of Gaussian Pulses

2020-11-18T22:11:17Z

Oezdemir moved page Aufgabe 3.9Z: Gauß gefaltet mit Gauß to Exercise 3.9Z: Convolution of Gaussian Pulses

Oezdemir at 22:04, 18 November 2020

2020-11-18T22:04:00Z

Javier: Text replacement - "Signaldarstellung/Faltungssatz und Faltungsoperation" to "Signal Representation/The Convolution Theorem and Operation"

2020-09-01T09:55:34Z

Text replacement - "Signaldarstellung/Faltungssatz und Faltungsoperation" to "Signal Representation/The Convolution Theorem and Operation"

Javier: Text replacement - "Signal_Representation/Faltungssatz_und_Faltungsoperation" to "Signal_Representation/The_Convolution_Theorem_and_Operation"

2020-09-01T09:53:12Z

Text replacement - "Signal_Representation/Faltungssatz_und_Faltungsoperation" to "Signal_Representation/The_Convolution_Theorem_and_Operation"

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

2020-08-24T08:28:13Z

Text replacement - "Category:Aufgaben zu Signaldarstellung" to "Category:Exercises for Signal Representation"

Javier: Text replacement - "[[Signaldarstellung/" to "[[Signal_Representation/"

2020-07-09T09:44:02Z

Text replacement - "[[Signaldarstellung/" to "[[Signal_Representation/"

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID544__Sig_Z_3_9.png|right|frame|Gaussian pulses&nbsp;  $x(t)$&nbsp; and&nbsp; $h(t)$]]
+[[File:P_ID544__Sig_Z_3_9.png|right|frame|Gaussian pulses&nbsp;  $x(t)$, &nbsp; $h(t)$]]
 The convolution result of two Gaussian functions is to be determined.&nbsp; We consider
 *a Gaussian input pulse&nbsp; ${x(t)}$&nbsp; with amplitude $x_0 = 1\,\text{V}$ and&nbsp; "equivalent pulse duration"&nbsp; $\Delta t_x = 4 \,\text{ms}$,&nbsp; as well as
@@ Line 50: / Line 50: @@
 :$$X( f ) = x_0  \cdot \Delta t_x  \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_x  \hspace{0.05cm}\cdot \hspace{0.05cm} f} \right)^2 } , \hspace{0.5cm}H(f) = {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_h  \hspace{0.05cm}\cdot \hspace{0.05cm}f} \right)^2 } .$$
 *The values we are looking for are
-:$$X(f = 0)\;\underline{ = 4 \,\text{mV/Hz}},$$
+:$$X(f = 0)\;\underline{ = 4 \,\text{mV/Hz}}, \hspace{0.5cm}
-:$$H(f = 0)\; \underline{= 1}.$$
+H(f = 0)\; \underline{= 1}.$$
-[[File:P_ID589__Sig_Z_3_9_b_neu.png|right|frame|Convolution result for &bdquo;$\rm Gauss  \ast  Gauss$&rdquo;]]
+[[File:P_ID589__Sig_Z_3_9_b_neu.png|right|frame|Gaussian spektra&nbsp; $X(f)$, &nbsp; &nbsp; $Y(f)$ &nbsp; &nbsp; &ndash; &nbsp; &nbsp; Gaussian pulses&nbsp; $x(t)$, &nbsp; &nbsp; $y(t)$]]
-'''(2)'''&nbsp;  Convolution in the time domain corresponds to multiplication in the frequency domain:
+'''(2)'''&nbsp;  Convolution in time domain corresponds to multiplication in frequency domain:
 :$$Y(f) = X(f) \cdot H(f) = x_0  \cdot \Delta t_x  \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_x^2  + \Delta t_h^2 } \right)f^2 } .$$
 *With the abbreviation&nbsp; $\Delta t_y = (\Delta t_x^2 + \Delta t_h^2)^{1/2} = 5\, \text{ms}$&nbsp; one can write for this:
@@ Line 69: / Line 69: @@
 *This gives:
 :$$y(t) = x(t) * h(t) = x_0  \cdot \frac{\Delta t_x }{\Delta t_y } \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {t/\Delta t_y } \right)^2 } .$$
-*The maximum value of the signal&nbsp; ${y(t)}$&nbsp; is also at&nbsp; $t = 0$&nbsp; and is&nbsp; $y_0 \hspace{0.15cm}\underline{= 0.8 \text{ V} }$.
+*The maximum value of the signal&nbsp; ${y(t)}$&nbsp; is also at &nbsp; $t = 0$&nbsp; and is &nbsp; $y_0 \hspace{0.15cm}\underline{= 0.8 \text{ V} }$.
-*The equivalent pulse duration results in&nbsp; $\Delta t_y \hspace{0.15cm}\underline{= 5 \text{ ms}}$&nbsp; (see above picture, right sketch).
+*The equivalent pulse duration results in&nbsp; $\Delta t_y \hspace{0.15cm}\underline{= 5 \text{ ms}}$&nbsp; (see above graphic, right sketch).
 *This means:&nbsp; The Gaussian&nbsp; ${H(f)}$&nbsp; causes the output pulse&nbsp; ${y(t)}$&nbsp; to be smaller and wider than the input pulse&nbsp; ${x(t)}$&nbsp;.
-*The pulse shape remains Gaussian because: &nbsp; '''Gaussian convoluted with Gaussian always results in Gaussian!'''
+*The pulse shape remains Gaussian. &nbsp; Because: &nbsp; '''Gaussian convoluted with Gaussian always results in Gaussian!'''
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID544__Sig_Z_3_9.png|right|frame|Gaussian $x(t)$ and $h(t)$]]
+[[File:P_ID544__Sig_Z_3_9.png|right|frame|Gaussian pulses&nbsp;  $x(t)$&nbsp; and&nbsp; $h(t)$]]
-The convolution result of two Gaussian functions is to be determined. We consider a Gaussian input impulse&nbsp; ${x(t)}$&nbsp; with amplitude $x_0 = 1\,\text{V}$ and equivalent duration&nbsp; $\Delta t_x = 4 \,\text{ms}$&nbsp; as well as a likewise Gaussian impulse response&nbsp; ${h(t)}$, which has the equivalent duration&nbsp; $\Delta t_h = 3 \,\text{ms}$&nbsp;:
+The convolution result of two Gaussian functions is to be determined.&nbsp; We consider
 :$$x( t ) = x_0  \cdot {\rm{e}}^{ - {\rm{\pi }}( {t/\Delta t_x } )^2 } ,$$
 :$$h( t ) = \frac{1}{\Delta t_h } \cdot {\rm{e}}^{ - {\rm{\pi }}( {t/\Delta t_h } )^2 } .$$
-The output signal&nbsp; ${y(t)} = {x(t)} ∗{h(t)}$ is sought, whereby the diversions via the spectral functions is to be taken.
+The output signal&nbsp; ${y(t)} = {x(t)} ∗{h(t)}$&nbsp; is sought, whereby the diversions via the spectral functions is to be taken.
@@ Line 23: / Line 25: @@
 <quiz display=simple>
-{Give the spectral functions&nbsp; ${X(f)}$&nbsp; and&nbsp; ${H(f)}$&nbsp; an. Which values result for&nbsp; $f = 0$?
+{Give the spectral functions&nbsp; ${X(f)}$&nbsp; and&nbsp; ${H(f)}$&nbsp; an.&nbsp; Which values result for&nbsp; $f = 0$?
 |type="{}"}
 $X(f = 0)\ = \ $ { 4 3% } &nbsp;$\text{mV/Hz}$
@@ Line 29: / Line 31: @@
-{Calculate the spectral function&nbsp; ${Y(f)}$&nbsp; of the output signal. What is the spectral value at&nbsp; $f = 0$?
+{Calculate the spectral function&nbsp; ${Y(f)}$&nbsp; of the output signal.&nbsp; What is the spectral value at&nbsp; $f = 0$?
 |type="{}"}
 $Y(f = 0)\ = \ $ { 4 3% } &nbsp;$\text{mV/Hz}$
-{Calculate the output pulse&nbsp; ${y(t)}$. What values result for the amplitude&nbsp; $y_0 = y(t = 0)$&nbsp; and the equivalent pulse duration&nbsp; $\Delta t_y$?
+{Calculate the output pulse&nbsp; ${y(t)}$.&nbsp; What values result for the amplitude&nbsp; $y_0 = y(t = 0)$&nbsp; and the equivalent pulse duration&nbsp; $\Delta t_y$?
 |type="{}"}
 $y_0\ = \ $ { 0.8 3% } &nbsp;$\text{V}$

← Older revision		Revision as of 12:38, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^3.4 The Convolution Theorem^]]	+	[[Category:Signal Representation: Exercises\|^3.4 The Convolution Theorem^]]

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID544__Sig_Z_3_9.png|right|frame|Gaußförmige $x(t)$ und $h(t)$]]
+[[File:P_ID544__Sig_Z_3_9.png|right|frame|Gaussian $x(t)$ and $h(t)$]]
-Es soll das Faltungsergebnis zweier Gaußfunktionen ermittelt werden. Wir betrachten einen gaußförmigen Eingangsimpuls&nbsp; ${x(t)}$&nbsp; mit Amplitude $x_0 = 1\,\text{V}$ und äquivalenter Dauer&nbsp; $\Delta t_x = 4 \,\text{ms}$&nbsp; sowie eine ebenfalls gaußförmige Impulsantwort&nbsp; ${h(t)}$, welche die äquivalente Dauer&nbsp; $\Delta t_h = 3 \,\text{ms}$&nbsp; aufweist:
+The convolution result of two Gaussian functions is to be determined. We consider a Gaussian input impulse&nbsp; ${x(t)}$&nbsp; with amplitude $x_0 = 1\,\text{V}$ and equivalent duration&nbsp; $\Delta t_x = 4 \,\text{ms}$&nbsp; as well as a likewise Gaussian impulse response&nbsp; ${h(t)}$, which has the equivalent duration&nbsp; $\Delta t_h = 3 \,\text{ms}$&nbsp;:
 :$$x( t ) = x_0  \cdot {\rm{e}}^{ - {\rm{\pi }}( {t/\Delta t_x } )^2 } ,$$
 :$$h( t ) = \frac{1}{\Delta t_h } \cdot {\rm{e}}^{ - {\rm{\pi }}( {t/\Delta t_h } )^2 } .$$
-Gesucht ist das Ausgangssignal&nbsp; ${y(t)} = {x(t)} ∗{h(t)}$, wobei der Umweg über die Spektralfunktionen gegangen werden soll.
+The output signal&nbsp; ${y(t)} = {x(t)} ∗{h(t)}$ is sought, whereby the diversions via the spectral functions is to be taken.
@@ Line 13: / Line 13: @@
-''Hinweis:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|Faltungssatz und Faltungsoperation]].
+''Hint:''
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Geben Sie die Spektralfunktionen&nbsp; ${X(f)}$&nbsp; und&nbsp; ${H(f)}$&nbsp; an. Welche Werte ergeben sich für&nbsp; $f = 0$?
+{Give the spectral functions&nbsp; ${X(f)}$&nbsp; and&nbsp; ${H(f)}$&nbsp; an. Which values result for&nbsp; $f = 0$?
 |type="{}"}
 $X(f = 0)\ = \ $ { 4 3% } &nbsp;$\text{mV/Hz}$
@@ Line 28: / Line 29: @@
-{Berechnen Sie die Spektralfunktion&nbsp; ${Y(f)}$&nbsp; des Ausgangssignals. Wie groß ist der Spektralwert bei&nbsp; $f = 0$?
+{Calculate the spectral function&nbsp; ${Y(f)}$&nbsp; of the output signal. What is the spectral value at&nbsp; $f = 0$?
 |type="{}"}
 $Y(f = 0)\ = \ $ { 4 3% } &nbsp;$\text{mV/Hz}$
-{Berechnen Sie den Ausgangsimpuls&nbsp; ${y(t)}$. Welche Werte ergeben sich für die Amplitude&nbsp; $y_0 = y(t = 0)$&nbsp; und die äquivalente Impulsdauer&nbsp; $\Delta t_y$?
+{Calculate the output pulse&nbsp; ${y(t)}$. What values result for the amplitude&nbsp; $y_0 = y(t = 0)$&nbsp; and the equivalent pulse duration&nbsp; $\Delta t_y$?
 |type="{}"}
 $y_0\ = \ $ { 0.8 3% } &nbsp;$\text{V}$
@@ Line 42: / Line 43: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Durch Fouriertransformation erhält man:
+'''(1)'''&nbsp;  By Fourier transformation one obtains:
 :$$X( f ) = x_0  \cdot \Delta t_x  \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_x  \hspace{0.05cm}\cdot \hspace{0.05cm} f} \right)^2 } , \hspace{0.5cm}H(f) = {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_h  \hspace{0.05cm}\cdot \hspace{0.05cm}f} \right)^2 } .$$
-*Die gesuchten Werte sind
+*The values we are looking for are
 :$$X(f = 0)\;\underline{ = 4 \,\text{mV/Hz}},$$
 :$$H(f = 0)\; \underline{= 1}.$$
-[[File:P_ID589__Sig_Z_3_9_b_neu.png|right|frame|Faltungsergebnis für &bdquo;$\rm Gauß  \ast  Gauß$&rdquo;]]
+[[File:P_ID589__Sig_Z_3_9_b_neu.png|right|frame|Convolution result for &bdquo;$\rm Gauss  \ast  Gauss$&rdquo;]]
-'''(2)'''&nbsp;  Der Faltung im Zeitbereich entspricht die Multiplikation im Frequenzbereich:
+'''(2)'''&nbsp;  Convolution in the time domain corresponds to multiplication in the frequency domain:
 :$$Y(f) = X(f) \cdot H(f) = x_0  \cdot \Delta t_x  \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_x^2  + \Delta t_h^2 } \right)f^2 } .$$
-*Mit der Abkürzung&nbsp; $\Delta t_y = (\Delta t_x^2 + \Delta t_h^2)^{1/2} = 5\, \text{ms}$&nbsp; kann man hierfür schreiben:
+*With the abbreviation&nbsp; $\Delta t_y = (\Delta t_x^2 + \Delta t_h^2)^{1/2} = 5\, \text{ms}$&nbsp; one can write for this:
 :$$Y(f) = x_0  \cdot \Delta t_x  \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_y  \hspace{0.05cm}\cdot \hspace{0.05cm} f} \right)^2 } .$$
-*Bei der Frequenz&nbsp; $f = 0$&nbsp; sind die Spektralwerte am Eingang und Ausgang des Gaußfilters gleich, also gilt:
+*At frequency&nbsp; $f = 0$&nbsp;, the spectral values at the input and output of the Gaussian filter are equal, so:
 :$$Y(f = 0) \;\underline{= 4 \text{ mV/Hz}}.$$
-*Der Funktionsverlauf von&nbsp; ${Y(f)}$&nbsp; ist schmaler als&nbsp; ${X(f)}$&nbsp; und schmaler als&nbsp; ${H(f)}$.
+*The function curve of&nbsp; ${Y(f)}$&nbsp; is narrower than&nbsp; ${X(f)}$&nbsp; and narrower than&nbsp; ${H(f)}$.
-'''(3)'''&nbsp;  Es gilt die folgende Fourierkorrespondenz:
+'''(3)'''&nbsp;  The following Fourier correspondence holds:
 :$${\rm{e}}^{ - {\rm{\pi }}\left( {\Delta t_y  \hspace{0.05cm}\cdot \hspace{0.05cm} f} \right)^2 }\bullet\!\!\!-\!\!\!-\!\!\!-\!\!\circ\, \frac{1}{\Delta t_y } \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {t/\Delta t_y } \right)^2 } .$$
-*Damit erhält man:
+*This gives:
 :$$y(t) = x(t) * h(t) = x_0  \cdot \frac{\Delta t_x }{\Delta t_y } \cdot {\rm{e}}^{ - {\rm{\pi }}\left( {t/\Delta t_y } \right)^2 } .$$
-*Der Maximalwert des Signals&nbsp; ${y(t)}$&nbsp; liegt ebenfalls bei&nbsp; $t = 0$&nbsp; und beträgt&nbsp; $y_0 \hspace{0.15cm}\underline{= 0.8 \text{ V} }$.
+*The maximum value of the signal&nbsp; ${y(t)}$&nbsp; is also at&nbsp; $t = 0$&nbsp; and is&nbsp; $y_0 \hspace{0.15cm}\underline{= 0.8 \text{ V} }$.
-*Die äquivalente Impulsdauer ergibt sich zu&nbsp; $\Delta t_y \hspace{0.15cm}\underline{= 5 \text{ ms}}$&nbsp; (siehe obiges Bild, rechte Skizze).
+*The equivalent pulse duration results in&nbsp; $\Delta t_y \hspace{0.15cm}\underline{= 5 \text{ ms}}$&nbsp; (see above picture, right sketch).
-*Das bedeutet:&nbsp; Das Gaußfilter&nbsp; ${H(f)}$&nbsp; bewirkt, dass der Ausgangsimpuls&nbsp; ${y(t)}$&nbsp; kleiner und breiter als der Eingangsimpuls&nbsp; ${x(t)}$&nbsp; ist.
+*This means:&nbsp; The Gaussian&nbsp; ${H(f)}$&nbsp; causes the output pulse&nbsp; ${y(t)}$&nbsp; to be smaller and wider than the input pulse&nbsp; ${x(t)}$&nbsp;.
-*Die Impulsform bleibt weiterhin gaußförmig, weil: &nbsp; '''Gauß gefaltet mit Gauß ergibt immer Gauß!'''
+*The pulse shape remains Gaussian because: &nbsp; '''Gaussian convoluted with Gaussian always results in Gaussian!'''
 {{ML-Fuß}}

← Older revision		Revision as of 22:04, 18 November 2020
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−	[[Category:Exercises for Signal Representation\|^3. ~~Aperiodische Signale - Impulse~~^]]	+	[[Category:Exercises for Signal Representation\|^3.4 The Convolution Theorem^]]

← Older revision	Revision as of 22:11, 18 November 2020
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−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Faltungssatz und Faltungsoperation~~	+	{{quiz-Header\|Buchseite=Signal Representation/The Convolution Theorem and Operation
	}}		}}

← Older revision		Revision as of 09:53, 1 September 2020
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	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/~~Faltungssatz_und_Faltungsoperation~~\|Faltungssatz und Faltungsoperation]].	+	*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|Faltungssatz und Faltungsoperation]].

← Older revision		Revision as of 08:28, 24 August 2020
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−	[[Category:~~Aufgaben zu Signaldarstellung~~\|^3. Aperiodische Signale - Impulse^]]	+	[[Category:Exercises for Signal Representation\|^3. Aperiodische Signale - Impulse^]]

← Older revision		Revision as of 09:44, 9 July 2020
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	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[~~Signaldarstellung~~/Faltungssatz_und_Faltungsoperation\|Faltungssatz und Faltungsoperation]].	+	*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/Faltungssatz_und_Faltungsoperation\|Faltungssatz und Faltungsoperation]].