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

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

2021-05-26T07:05:07Z

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

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"

Noah at 13:08, 1 February 2021

2021-02-01T13:08:50Z

Noah at 13:08, 1 February 2021

2021-02-01T13:08:24Z

Noah at 13:07, 1 February 2021

2021-02-01T13:07:04Z

Noah at 21:39, 28 January 2021

2021-01-28T21:39:28Z

Noah at 21:28, 28 January 2021

2021-01-28T21:28:17Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 3.9: Faltung von Rechteck und Gauß to Aufgaben:Exercise 3.9: Convolution of Rectangle and Gaussian Pulse

2020-11-18T22:10:31Z

Oezdemir moved page Aufgabe 3.9: Faltung von Rechteck und Gauß to Exercise 3.9: Convolution of Rectangle and Gaussian Pulse

@@ Line 18: / Line 18: @@
 * a rectangular pulse&nbsp; $x_1(t)$&nbsp; with amplitude&nbsp; $A_1 =1\,\text{V}$&nbsp; and duration&nbsp; $T_1 = 20\,\text{ns}$&nbsp;  (red curve),
 * a rectangular pulse&nbsp; $x_2(t)$&nbsp; with amplitude&nbsp; $A_2 =10\,\text{V}$&nbsp; and duration&nbsp; $T_2 = 2\,\text{ns}$&nbsp; (violet curve),
-* a Dirac impulse&nbsp; $x_3(t)$&nbsp; with impulse weight&nbsp; $2 \cdot 10^{–8}\text{ Vs}$&nbsp;  (green arrow).
+* a Dirac delta&nbsp; $x_3(t)$&nbsp; with impulse weight&nbsp; $2 \cdot 10^{–8}\text{ Vs}$&nbsp;  (green arrow).
@@ Line 92: / Line 92: @@
-'''(3)'''&nbsp; With the Dirac impulse&nbsp; $x_3(t)$&nbsp;, the output signal&nbsp; $y_3(t)$&nbsp; is equal to the impulse response&nbsp; $h(t)$, weighted by the weight of the Dirac function:
+'''(3)'''&nbsp; With the Dirac delta&nbsp; $x_3(t)$&nbsp;, the output signal&nbsp; $y_3(t)$&nbsp; is equal to the impulse response&nbsp; $h(t)$, weighted by the weight of the Dirac function:
 :$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
@@ Line 101: / Line 101: @@
 One can see from comparing the results from&nbsp; '''(2)'''&nbsp; and &nbsp; '''(3)''', that&nbsp; $y_3(t) \approx y_2(t)$&nbsp; 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&nbsp; $T$&nbsp; is significantly smaller than the equivalent impulse duration&nbsp; $\Delta t$&nbsp; 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&nbsp; $T$&nbsp; is significantly smaller than the equivalent impulse duration&nbsp; $\Delta t$&nbsp; of the impulse response.
 *This means for our example: &nbsp; If the duration&nbsp; $T$&nbsp; of the rectangular input impulse&nbsp; $x(t)$&nbsp; is clearly smaller than the&nbsp; "equivalent pulse duration"&nbsp; $\Delta t$&nbsp; of the Gaussian impulse response&nbsp; $h(t)$, then&nbsp; $y(t)$&nbsp; is also almost Gaussian.&nbsp; But:  &nbsp; '''Gaussian (once) folded with non&ndash;Gaussian never results in (exactly) Gaussian!'''
 {{ML-Fuß}}

@@ Line 67: / Line 67: @@
 :$$y_1( t ) = A_1  \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau  = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm}  {\rm{d}}u.}$$
-*Here the substitution&nbsp; $u = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \tau$&nbsp; was used. The integration limits are at:
+*Here the substitution&nbsp; $u = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \tau$&nbsp; was used.&nbsp; The integration limits are at:
 :$$u_1  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \big( {t - T_1 /2} \big),\hspace{0.5cm}u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \big( {t + T_1 /2} \big).$$
@@ Line 92: / Line 92: @@
-'''(3)'''&nbsp; With the dirac-shaped input signal &nbsp; $x_3(t)$&nbsp;, the output signal&nbsp; $y_3(t)$&nbsp; is equal to the impulse response&nbsp; $h(t)$, weighted by the weight of the dirac function:
+'''(3)'''&nbsp; With the Dirac impulse&nbsp; $x_3(t)$&nbsp;, the output signal&nbsp; $y_3(t)$&nbsp; is equal to the impulse response&nbsp; $h(t)$, weighted by the weight of the Dirac function:
 :$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
@@ Line 100: / Line 100: @@
-One can see from comparing the results from&nbsp; '''(2)'''&nbsp; and &nbsp; '''(3)''', that&nbsp; $y_3(t)$ ≈ $y_2(t)$&nbsp; gilt.
+One can see from comparing the results from&nbsp; '''(2)'''&nbsp; and &nbsp; '''(3)''', that&nbsp; $y_3(t) \approx y_2(t)$&nbsp; 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&nbsp; $T$&nbsp; is significantly smaller than the equivalent impulse duration&nbsp; $\Delta t$&nbsp; of the impulse response.
-This means for our example: &nbsp; If the duration&nbsp; $T$&nbsp; of the rectangular input impulse&nbsp; $x(t)$&nbsp; is clearly smaller than the equivalent duration&nbsp; $\Delta t$&nbsp; of the Gaussian impulse response&nbsp; $h(t)$, then the output impulse&nbsp; $y(t)$&nbsp; is also almost Gaussian. But:  &nbsp; '''Gaussian (once) folded with non&ndash;Gaussian never results in (exactly) Gaussian!'''
+*This means for our example: &nbsp; If the duration&nbsp; $T$&nbsp; of the rectangular input impulse&nbsp; $x(t)$&nbsp; is clearly smaller than the&nbsp; "equivalent pulse duration"&nbsp; $\Delta t$&nbsp; of the Gaussian impulse response&nbsp; $h(t)$, then&nbsp; $y(t)$&nbsp; is also almost Gaussian.&nbsp; But:  &nbsp; '''Gaussian (once) folded with non&ndash;Gaussian never results in (exactly) Gaussian!'''
 {{ML-Fuß}}
 __NOEDITSECTION__
 [[Category:Signal Representation: Exercises|^3.4 The Convolution Theorem^]]

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID540__Sig_A_3_9_neu.png|250px|right|frame|Rectangular&nbsp; $x(t)$&nbsp;  and gaussian&nbsp;  $h(t)$]]
+[[File:P_ID540__Sig_A_3_9_neu.png|250px|right|frame|Convolution of rectangle&nbsp; $x(t)$ &nbsp;and Gaussian pulse&nbsp;  $h(t)$]]
-We consider a Gaussian low pass with the equivalent bandwidth&nbsp; $\Delta f = 40 \,\text{MHz}$:
+We consider a Gaussian low&ndash;pass with the equivalent bandwidth&nbsp; $\Delta f = 40 \,\text{MHz}$:
 :$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$
@@ Line 13: / Line 13: @@
 :$$h( t ) = \Delta f \cdot {\rm{e}}^{{\rm{ - \pi }}( {\Delta f  \hspace{0.05cm} \cdot \hspace{0.05cm} t} )^2 } .$$
-From the sketch it can be seen that the equivalent time duration &nbsp; ⇒  &nbsp; $\Delta t = 1/\Delta f = 25\,\text{ns}$&nbsp; the impulse response&nbsp; $h(t)$&nbsp;  can be read at the two inflection points of the Gaussian function.
+From the sketch it can be seen that the&nbsp; "equivalent time duration" &nbsp; ⇒  &nbsp; $\Delta t = 1/\Delta f = 25\,\text{ns}$&nbsp; of the impulse response&nbsp; $h(t)$&nbsp;  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:
-* a square-wave pulse&nbsp; $x_1(t)$&nbsp; with amplitude&nbsp; $A_1 =1\,\text{V}$&nbsp; and duration&nbsp; $T_1 = 20\,\text{ns}$&nbsp;  (red curve),
+* a rectangular pulse&nbsp; $x_1(t)$&nbsp; with amplitude&nbsp; $A_1 =1\,\text{V}$&nbsp; and duration&nbsp; $T_1 = 20\,\text{ns}$&nbsp;  (red curve),
 * a rectangular pulse&nbsp; $x_2(t)$&nbsp; with amplitude&nbsp; $A_2 =10\,\text{V}$&nbsp; and duration&nbsp; $T_2 = 2\,\text{ns}$&nbsp; (violet curve),
-* a Dirac pulse&nbsp; $x_3(t)$&nbsp; with pulse weight&nbsp; $2 \cdot 10^{–8}\text{ Vs}$&nbsp;  (green arrow).
+* a Dirac impulse&nbsp; $x_3(t)$&nbsp; with impulse weight&nbsp; $2 \cdot 10^{–8}\text{ Vs}$&nbsp;  (green arrow).
@@ Line 29: / Line 29: @@
 *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&nbsp; $\rm Q$&ndash;function]]
 :$${\rm Q}( x ) = \frac{1}{ {\sqrt {2{\rm{\pi }}} }}\int_{\it x}^\infty  {{\rm{e}}^{{{ - {\it u}}}^{\rm{2}} {\rm{/2}}} }\hspace{0.1cm}{\rm{d}}{\it u}.$$
@@ Line 42: / Line 42: @@
 <quiz display=simple>
-{Calculate the signal&nbsp; $y_1(t) = x_1(t) \ast h(t)$. <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&nbsp; $y_1(t) = x_1(t) \ast h(t)$.&nbsp; What values result at times&nbsp; $t = 0$&nbsp; and&nbsp; $t = 20\,\text{ns}$&nbsp;  with the approximation&nbsp; $(2\pi )^{1/2} \approx 2.5$?
 |type="{}"}
 $y_1(t=0)\ = \ $ { 0.682 3% } &nbsp;$\text{V}$
 $y_1(t=20\,\text{ns})\ = \ $  { 0.158 3% } &nbsp;$\text{V}$
-{What are the signal values of the output signal&nbsp; $y_2(t) = x_2(t) \ast h(t)$ at the points in time considered?
+{What are the signal values of the output signal&nbsp; $y_2(t) = x_2(t) \ast h(t)$&nbsp; at the  considered time points?
 |type="{}"}
 $y_2(t=0)\ = \ $ { 0.8 3% } &nbsp;$\text{V}$
 $y_2(t=20 \,\text{ns})\ = \ $ { 0.11 3% } &nbsp;$\text{V}$
-{What is the magnitude of the output signa&nbsp; $y_3(t) = x_3(t) \ast h(t)$&nbsp; at the points in time considered? Interpret the result.
+{What is the value of the output signa&nbsp; $y_3(t) = x_3(t) \ast h(t)$&nbsp; at the  considered time points?&nbsp; Interpret the result.
 |type="{}"}
 $y_3(t=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 42: / Line 42: @@
 <quiz display=simple>
-{Calculate the signa&nbsp; $y_1(t) = x_1(t) \ast h(t)$. <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&nbsp; $y_1(t) = x_1(t) \ast h(t)$. <br>What values result at times $t = 0$ und $t = 20\,\text{ns}$  with the approximation $(2\pi )^{1/2} \approx 2.5$?
 |type="{}"}
 $y_1(t=0)\ = \ $ { 0.682 3% } &nbsp;$\text{V}$

@@ Line 92: / Line 92: @@
-'''(3)'''&nbsp; Beim diracförmigen Eingangssignal&nbsp; $x_3(t)$&nbsp; ist das Ausgangssignal&nbsp; $y_3(t)$&nbsp; gleich der Impulsantwort&nbsp; $h(t)$, gewichtet mit dem Gewicht der Diracfunktion:
+'''(3)'''&nbsp; With the dirac-shaped input signal &nbsp; $x_3(t)$&nbsp;, the output signal&nbsp; $y_3(t)$&nbsp; is equal to the impulse response&nbsp; $h(t)$, weighted by the weight of the dirac function:
 :$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
-*Zum Zeitpunkt&nbsp; $t = 0$&nbsp; erhält man auch hier mit guter Näherung&nbsp; $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.
+*At time&nbsp; $t = 0$&nbsp;, one also obtains here with a good approximation&nbsp; $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.
-*Nach&nbsp; $20\, \rm ns$&nbsp; ist der Ausgangsimpuls um den Faktor&nbsp; ${\rm e}^{–0.64π} \hspace{0.15 cm}\underline{\approx 0.136}$&nbsp; kleiner und man erhält&nbsp; $y_3( t = 20 \,\text{ns}) ≈ 0.11  \,\text{V}$.
+*After&nbsp; $20\, \rm ns$&nbsp;, the output pulse is smaller by a factor of&nbsp; ${\rm e}^{–0.64π} \hspace{0.15 cm}\underline{\approx 0.136}$&nbsp; and one obtains&nbsp; $y_3( t = 20 \,\text{ns}) ≈ 0.11  \,\text{V}$.
-Man erkennt aus dem Vergleich der Resultate aus&nbsp; '''(2)'''&nbsp; und &nbsp; '''(3)''', dass&nbsp; $y_3(t)$ ≈ $y_2(t)$&nbsp; gilt.
+One can see from comparing the results from&nbsp; '''(2)'''&nbsp; and &nbsp; '''(3)''', that&nbsp; $y_3(t)$ ≈ $y_2(t)$&nbsp; gilt.
-*Der Grund hierfür ist, dass der Diracimpuls eine gute Näherung für einen rechteckförmigen Eingangsimpuls gleicher Fläche ist, wenn die Rechteckdauer&nbsp; $T$&nbsp; deutlich kleiner als die äquivalente Impulsdauer&nbsp; $\Delta t$&nbsp; 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&nbsp; $T$&nbsp; is significantly smaller than the equivalent impulse duration&nbsp; $\Delta t$&nbsp; of the impulse response.
-*Das heißt für unser Beispiel:&nbsp; Ist die Dauer&nbsp; $T$&nbsp; des rechteckförmigen Eingangsimpulses&nbsp; $x(t)$&nbsp; deutlich kleiner als die äquivalente Dauer&nbsp; $\Delta t$&nbsp; der gaußförmigen Impulsantwort&nbsp; $h(t)$, dann ist auch der Ausgangsimpuls&nbsp; $y(t)$&nbsp; nahezu gaußförmig. Aber: &nbsp; '''Gauß (einmal) gefaltet mit Nicht&ndash;Gauß ergibt nie (exakt) Gauß!'''
+This means for our example: &nbsp; If the duration&nbsp; $T$&nbsp; of the rectangular input impulse&nbsp; $x(t)$&nbsp; is clearly smaller than the equivalent duration&nbsp; $\Delta t$&nbsp; of the Gaussian impulse response&nbsp; $h(t)$, then the output impulse&nbsp; $y(t)$&nbsp; is also almost Gaussian. But:  &nbsp; '''Gaussian (once) folded with non&ndash;Gaussian never results in (exactly) Gaussian!'''
 {{ML-Fuß}}
 __NOEDITSECTION__
 [[Category:Exercises for Signal Representation|^3.4 The Convolution Theorem^]]

@@ Line 42: / Line 42: @@
 <quiz display=simple>
-{Berechnen Sie das Signal&nbsp; $y_1(t) = x_1(t) \ast h(t)$. <br>Welche Werte ergeben sich zu den Zeiten $t = 0$ und $t = 20\,\text{ns}$  mit der Näherung $(2\pi )^{1/2} \approx 2.5$?
+{Calculate the signa&nbsp; $y_1(t) = x_1(t) \ast h(t)$. <br>What values result at times $t = 0$ und $t = 20\,\text{ns}$  with the approximation $(2\pi )^{1/2} \approx 2.5$?
 |type="{}"}
 $y_1(t=0)\ = \ $ { 0.682 3% } &nbsp;$\text{V}$
 $y_1(t=20\,\text{ns})\ = \ $  { 0.158 3% } &nbsp;$\text{V}$
-{Welche Signalwerte ergeben sich beim Ausgangssignal&nbsp; $y_2(t) = x_2(t) \ast h(t)$ zu den betrachteten Zeitpunkten?
+{What are the signal values of the output signal&nbsp; $y_2(t) = x_2(t) \ast h(t)$ at the points in time considered?
 |type="{}"}
 $y_2(t=0)\ = \ $ { 0.8 3% } &nbsp;$\text{V}$
 $y_2(t=20 \,\text{ns})\ = \ $ { 0.11 3% } &nbsp;$\text{V}$
-{Wie groß ist das Ausgangssignal&nbsp; $y_3(t) = x_3(t) \ast h(t)$&nbsp; zu den betrachteten Zeitpunkten? Interpretieren Sie das Ergebnis.
+{What is the magnitude of the output signa&nbsp; $y_3(t) = x_3(t) \ast h(t)$&nbsp; at the points in time considered? Interpret the result.
 |type="{}"}
 $y_3(t=0)\ = \ $ { 0.8 3% } &nbsp;$\text{V}$
@@ Line 61: / Line 61: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Das Faltungsintegral lautet hier:
+'''(1)'''&nbsp;  The convolution integral here is:
 :$$y_1( t ) = A_1  \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau  = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm}  {\rm{d}}u.}$$
-*Hierbei wurde die Substitution&nbsp; $u = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \tau$&nbsp; verwendet. Die Integrationsgrenzen liegen bei:
+*Here the substitution&nbsp; $u = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \tau$&nbsp; was used. The integration limits are at:
 :$$u_1  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \big( {t - T_1 /2} \big),\hspace{0.5cm}u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \big( {t + T_1 /2} \big).$$
-*Mit dem komplementären Gaußschen Fehlerintegral kann hierfür auch geschrieben werden:
+*Using the complementary Gaussian error integral, it is also possible to write for this:
 :$$y_1 (t) = A_1  \cdot \big[ {{\rm Q} ( {u_1 } ) - {\rm Q}( {u_2 } )} \big].$$
-*Für den Zeitpunkt&nbsp; $t = 0$&nbsp; erhält man mit&nbsp; $(2\pi )^{1/2} \approx 2.5$:
+*For time&nbsp; $t = 0$&nbsp; one obtains with&nbsp; $(2\pi )^{1/2} \approx 2.5$:
 :$$u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \frac{ {T_1 }}{2} \approx 2.5 \cdot 4 \cdot 10^{7} \;{\rm{1/s}} \cdot 10^{-8} \;{\rm{s}} = 1.$$
-*Mit&nbsp; $u_1 = -u_2 = -1$&nbsp; folgt für die beiden gesuchten Signalwerte:
+*With&nbsp; $u_1 = -u_2 = -1$&nbsp;, it follows for the two signal values we are looking for:
 :$$y_1 ( {t = 0} ) \approx A_1  \cdot \big[ {{\rm Q}( { - 1} ) - {\rm Q}(+ 1 )} \big] = 1\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.841 - 0}}{\rm{.159}}} \big] \hspace{0.15 cm}\underline{= 0.682\;{\rm{V}}}{\rm{,}}$$
@@ Line 85: / Line 85: @@
-'''(2)'''&nbsp; Analog zur ersten Musterlösung erhält man für den schmaleren Eingangsimpuls&nbsp; $x_2(t)$:
+'''(2)'''&nbsp; Analogous to the first sample solution, one obtains&nbsp; $x_2(t)$ for the narrower input pulse:
 :$$y_2 ( {t = 0} ) \approx A_2  \cdot \big[ {{\rm Q}( { - 0.1} ) - {\rm Q}( {0.1} )} \big] = 10\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.540 - 0}}{\rm{.460}}} \big] \hspace{0.15 cm}\underline{= 0.80\;{\rm{V}}}{\rm{,}}$$

@@ Line 5: / Line 5: @@
 [[File:P_ID540__Sig_A_3_9_neu.png|250px|right|frame|Rechteckförmiges&nbsp; $x(t)$&nbsp;  und gaußförmiges&nbsp;  $h(t)$]]
-Wir betrachten einen gaußförmigen Tiefpass mit der äquivalenten Bandbreite&nbsp; $\Delta f = 40 \,\text{MHz}$:
+We consider a Gaussian low pass with the equivalent bandwidth&nbsp; $\Delta f = 40 \,\text{MHz}$:
 :$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$
-Die dazugehörige Impulsantwort lautet:
+The corresponding impulse response is:
 :$$h( t ) = \Delta f \cdot {\rm{e}}^{{\rm{ - \pi }}( {\Delta f  \hspace{0.05cm} \cdot \hspace{0.05cm} t} )^2 } .$$
-Aus der Skizze ist zu ersehen, dass die äquivalente Zeitdauer &nbsp; ⇒  &nbsp; $\Delta t = 1/\Delta f = 25\,\text{ns}$&nbsp; der Impulsantwort&nbsp; $h(t)$&nbsp;  an den beiden Wendepunkten der Gaußfunktion abgelesen werden kann.
+From the sketch it can be seen that the equivalent time duration &nbsp; ⇒  &nbsp; $\Delta t = 1/\Delta f = 25\,\text{ns}$&nbsp; the impulse response&nbsp; $h(t)$&nbsp;  can be read at the two inflection points of the Gaussian function.
-An den Eingang des Tiefpasses werden nun drei verschiedene impulsartige Signale angelegt:
+Three different pulse-like signals are now applied to the input of the low-pass filter:
-* ein Rechteckimpuls&nbsp; $x_1(t)$&nbsp; mit Amplitude&nbsp; $A_1 =1\,\text{V}$&nbsp; und Dauer&nbsp; $T_1 = 20\,\text{ns}$&nbsp;  (roter Verlauf),
+* a square-wave pulse&nbsp; $x_1(t)$&nbsp; with amplitude&nbsp; $A_1 =1\,\text{V}$&nbsp; and duration&nbsp; $T_1 = 20\,\text{ns}$&nbsp;  (red curve),
-* ein Rechteckimpuls&nbsp; $x_2(t)$&nbsp; mit Amplitude&nbsp; $A_2 =10\,\text{V}$&nbsp; und Dauer&nbsp; $T_2 = 2\,\text{ns}$&nbsp; (violetter Verlauf),
+* a rectangular pulse&nbsp; $x_2(t)$&nbsp; with amplitude&nbsp; $A_2 =10\,\text{V}$&nbsp; and duration&nbsp; $T_2 = 2\,\text{ns}$&nbsp; (violet curve),
-* ein Diracimpuls&nbsp; $x_3(t)$&nbsp; mit dem Impulsgewicht&nbsp; $2 \cdot 10^{–8}\text{ Vs}$&nbsp;  (grüner Pfeil).
+* a Dirac pulse&nbsp; $x_3(t)$&nbsp; with pulse weight&nbsp; $2 \cdot 10^{–8}\text{ Vs}$&nbsp;  (green arrow).
@@ Line 25: / Line 25: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|Faltungssatz und Faltungsoperation]].
+*This exercise belongs to the chapter&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
-*Zur Beantwortung der Fragen können Sie das komplementäre Gaußsche Fehlerintegral benutzen, das wie folgt definiert ist:
+*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|Einige Werte der Q-Funktion]]
+[[File:P_ID541__Sig_A_3_9Tab_neu.png|right|frame|Some values of the Q-function]]
 :$${\rm Q}( x ) = \frac{1}{ {\sqrt {2{\rm{\pi }}} }}\int_{\it x}^\infty  {{\rm{e}}^{{{ - {\it u}}}^{\rm{2}} {\rm{/2}}} }\hspace{0.1cm}{\rm{d}}{\it u}.$$
-Nebenstehende Tabelle gibt einige Funktionswerte wieder.
+This table gives some function values.
 <br clear=all>
@@ Line 39: / Line 39: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>

← Older revision	Revision as of 22:10, 18 November 2020
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