Difference between revisions of "Aufgaben:Exercise 3.9: Convolution of Rectangle and Gaussian Pulse"

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{{quiz-Header|Buchseite=Signaldarstellung/Faltungssatz und Faltungsoperation
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{{quiz-Header|Buchseite=Signal Representation/The Convolution Theorem and Operation
 
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[[File:P_ID540__Sig_A_3_9_neu.png|250px|right|Faltung von Rechteck und Gauß]]
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[[File:P_ID540__Sig_A_3_9_neu.png|250px|right|frame|Convolution of rectangle  $x(t)$  and Gaussian pulse   $h(t)$]]
  
Wir betrachten in der Aufgabe einen gaußförmigen Tiefpass mit der äquivalenten Bandbreite $\Delta f = 40 \,\text{MHz}$:
+
We consider a Gaussian low–pass with the equivalent bandwidth  $\Delta f = 40 \,\text{MHz}$:
 
   
 
   
$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$
+
:$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$
  
Die dazugehörige Impulsantwort lautet:
+
The corresponding impulse response is:
 
   
 
   
$$h( t ) = \Delta f \cdot {\rm{e}}^{{\rm{ - \pi }}( {\Delta f  \hspace{0.05cm} \cdot \hspace{0.05cm} t} )^2 } .$$
+
:$$h( t ) = \Delta f \cdot {\rm{e}}^{{\rm{ - \pi }}( {\Delta f  \hspace{0.05cm} \cdot \hspace{0.05cm} t} )^2 } .$$
  
Aus der Skizze ist zu ersehen, dass die äquivalente Zeitdauer   ⇒    $\Delta t = 1/\Delta f = 25\,\text{ns}$ der Impulsantwort $h(t)$  an den beiden Wendepunkten der Gaußfunktion abgelesen werden kann.
+
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.
  
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 $x_1(t)$ mit der Amplitude $A_1 =1\,\text{V}$ und der Dauer $T_1 = 20\,\text{ns}$  (roter Kurvenverlauf),
+
* a rectangular pulse  $x_1(t)$  with amplitude  $A_1 =1\,\text{V}$  and duration  $T_1 = 20\,\text{ns}$  (red curve),
* ein Rechteckimpuls $x_2(t)$ mit der Amplitude $A_2 =10\,\text{V}$ und der Dauer $T_2 = 2\,\text{ns}$ (violetter Kurvenverlauf),
+
* a rectangular pulse  $x_2(t)$  with amplitude  $A_2 =10\,\text{V}$  and duration  $T_2 = 2\,\text{ns}$  (violet curve),
* ein Diracimpuls $x_3(t)$ mit dem Impulsgewicht $2 \cdot 10^{–8},\text{Vs}$  (grüner Pfeil).
+
* a Dirac delta  $x_3(t)$  with impulse weight  $2 \cdot 10^{–8}\text{ Vs}$  (green arrow).
  
  
''Hinweise:''  
+
 
*Die Aufgabe gehört zum  Kapitel [[Signaldarstellung/Faltungssatz_und_Faltungsoperation|Faltungssatz und Faltungsoperation]].
+
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
+
 
*Zur Lösung der nachfolgenden Fragen können Sie das komplementäre Gaußsche Fehlerintegral benutzen, das wie folgt definiert ist:
+
 
 +
''Hints:''  
 +
*This exercise belongs to the chapter  [[Signal_Representation/The_Convolution_Theorem_and_Operation|The Convolution Theorem and Operation]].
 
   
 
   
$${\rm Q}( x ) = \frac{1}{ {\sqrt {2{\rm{\pi }}} }}\int_{\it x}^\infty  {{\rm{e}}^{{{ - {\it u}}}^{\rm{2}} {\rm{/2}}} }\hspace{0.1cm}{\rm{d}}{\it u}.$$
+
*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  $\rm Q$–function]]
 +
:$${\rm Q}( x ) = \frac{1}{ {\sqrt {2{\rm{\pi }}} }}\int_{\it x}^\infty  {{\rm{e}}^{{{ - {\it u}}}^{\rm{2}} {\rm{/2}}} }\hspace{0.1cm}{\rm{d}}{\it u}.$$  
  
Die nachfolgende Tabelle gibt einige Funktionswerte wieder:
 
  
[[File:P_ID541__Sig_A_3_9Tab_neu.png|Einige Werte der Q-Funktion]]
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This table gives some function values.
 +
<br clear=all>
  
  
===Fragebogen===
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 +
 
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie das Signal $y_1(t) = x_1(t) \ast h(t)$. 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 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="{}"}
 
|type="{}"}
$y_1(t=0)$ &nbsp;= { 0.682 3% } &nbsp;$\text{V}$
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$y_1(t=0)\ = \ $ { 0.682 3% } &nbsp;$\text{V}$
$y_1(t=20\,\text{ns})$ &nbsp;= { 0.158 3% } &nbsp;$\text{V}$
+
$y_1(t=20\,\text{ns})\ = \ $  { 0.158 3% } &nbsp;$\text{V}$
  
{Welche Signalwerte ergeben sich beim Ausgangssignal $y_2(t) = x_2(t) \ast h(t)$ zu den Zeitpunkten $t$ = 0 und $t$ = 20 ns?
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{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="{}"}
 
|type="{}"}
$y_2(t=0)$ &nbsp;= { 0.8 3% } &nbsp;$\text{V}$
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$y_2(t=0)\ = \ $ { 0.8 3% } &nbsp;$\text{V}$
$y_2(t=20 \,\text{ns})$ &nbsp;= { 0.11 3% } &nbsp;$\text{V}$
+
$y_2(t=20 \,\text{ns})\ = \ $ { 0.11 3% } &nbsp;$\text{V}$
  
{Wie groß ist das Ausgangssignal $y_3(t) = x_3(t) \ast h(t)$ zu den betrachteten Zeitpunkten? Interpretieren Sie das Ergebnis.
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{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="{}"}
 
|type="{}"}
$y_3(t=0)$ &nbsp;= { 0.8 3% } &nbsp;$\text{V}$
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$y_3(t=0)\ = \ $ { 0.8 3% } &nbsp;$\text{V}$
$y_3(t=20\, \text{ns})$ &nbsp;= { 0.11 3% } &nbsp;$\text{V}$
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$y_3(t=20\, \text{ns})\ = \ $ { 0.11 3% } &nbsp;$\text{V}$
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
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'''1.'''  Das Faltungsintegral lautet hier:
+
'''(1)'''&nbsp; The convolution integral here is:
  
$$y_1( t ) = A_1  \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau  = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm}  {\rm{d}}u.}$$
+
:$$y_1( t ) = A_1  \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau  = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm}  {\rm{d}}u.}$$
 
   
 
   
Hierbei wurde die Substitution $u = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \tau$ verwendet. Die Integrationsgrenzen liegen bei:
+
*Here the substitution&nbsp; $u = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \tau$&nbsp; was used.&nbsp; The integration limits are at:
 
   
 
   
$$u_1  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \left( {t - T_1 /2} \right),\hspace{0.5cm}u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \left( {t + T_1 /2} \right).$$
+
:$$u_1  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \big( {t - T_1 /2} \big),\hspace{0.5cm}u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \big( {t + T_1 /2} \big).$$
  
Mit dem komplementären Gaußschen Fehlerintegral kann hierfür auch geschrieben werden:
+
*Using the complementary Gaussian error integral, it is also possible to write for this:
 
   
 
   
$$y_1 (t) = A_1  \cdot \left[ {{\rm Q} ( {u_1 } ) - {\rm Q}( {u_2 } )} \right].$$
+
:$$y_1 (t) = A_1  \cdot \big[ {{\rm Q} ( {u_1 } ) - {\rm Q}( {u_2 } )} \big].$$
  
Für den Zeitpunkt $t = 0$ erhält man mit $(2\pi )^{1/2} \approx 2.5$:
+
*For time&nbsp; $t = 0$&nbsp; one obtains with&nbsp; $(2\pi )^{1/2} \approx 2.5$:
 
   
 
   
$$u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \frac{ {T_1 }}{2} \approx 2.5 \cdot 4 \cdot 10^{7} \;{\rm{1/s}} \cdot 10^{-8} \;{\rm{s}} = 1.$$
+
:$$u_2  = \sqrt {2{\rm{\pi }}}  \cdot \Delta f \cdot \frac{ {T_1 }}{2} \approx 2.5 \cdot 4 \cdot 10^{7} \;{\rm{1/s}} \cdot 10^{-8} \;{\rm{s}} = 1.$$
  
Mit $u_1 = –u_2 = –1$ 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:
 
   
 
   
$$y_1 ( {t = 0} ) \approx A_1  \cdot \left[ {{\rm Q}( { - 1} ) - {\rm Q}(+ 1 )} \right] = 1\;{\rm{V}} \cdot \left[ {{\rm{0}}{\rm{.841 - 0}}{\rm{.159}}} \right] \hspace{0.15 cm}\underline{= 0.682\;{\rm{V}}}{\rm{,}}$$
+
:$$y_1 ( {t = 0} ) \approx A_1  \cdot \big[ {{\rm Q}( { - 1} ) - {\rm Q}(+ 1 )} \big] = 1\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.841 - 0}}{\rm{.159}}} \big] \hspace{0.15 cm}\underline{= 0.682\;{\rm{V}}}{\rm{,}}$$
$$y_1 ( {t = 20\;{\rm{ns}}} ) \approx A_1  \cdot \left[ {{\rm Q}( 1 ) - {\rm Q}( 3 )} \right] = 1\;{\rm{V}} \cdot \left[ {{\rm{0}}{\rm{.159 - 0}}{\rm{.001}}} \right] \hspace{0.15 cm}\underline{= 0.158\;{\rm{V}}}{\rm{.}}$$
+
:$$y_1 ( {t = 20\;{\rm{ns}}} ) \approx A_1  \cdot \big[ {{\rm Q}( 1 ) - {\rm Q}( 3 )} \big] = 1\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.159 - 0}}{\rm{.001}}} \big] \hspace{0.15 cm}\underline{= 0.158\;{\rm{V}}}{\rm{.}}$$
  
  
'''2.''' Analog zur erstgen Musterlösung erhält man für den schmaleren Eingangsimpuls $x_2(t)$:
+
'''(2)'''&nbsp; Analogous to the first sample solution, one obtains&nbsp; $x_2(t)$ for the narrower input pulse:
 
   
 
   
$$y_2 ( {t = 0} ) \approx A_2  \cdot \left[ {{\rm Q}( { - 0.1} ) - {\rm Q}( {0.1} )} \right] = 10\;{\rm{V}} \cdot \left[ {{\rm{0}}{\rm{.540 - 0}}{\rm{.460}}} \right] \hspace{0.15 cm}\underline{= 0.80\;{\rm{V}}}{\rm{,}}$$
+
:$$y_2 ( {t = 0} ) \approx A_2  \cdot \big[ {{\rm Q}( { - 0.1} ) - {\rm Q}( {0.1} )} \big] = 10\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.540 - 0}}{\rm{.460}}} \big] \hspace{0.15 cm}\underline{= 0.80\;{\rm{V}}}{\rm{,}}$$
 
   
 
   
$$y_2 ( {t = 20\,{\rm ns}} ) \approx A_2  \cdot \left[ {{\rm Q}( {1.9} ) - {\rm Q}( {2.1} )} \right] = 10\;{\rm{V}} \cdot \left[ {{\rm{0}}{\rm{.029 - 0}}{\rm{.018}}} \right] \hspace{0.15 cm}\underline{= 0.11\;{\rm{V}}}{\rm{.}}$$
+
:$$y_2 ( {t = 20\,{\rm ns}} ) \approx A_2  \cdot \big[ {{\rm Q}( {1.9} ) - {\rm Q}( {2.1} )} \big] = 10\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.029 - 0}}{\rm{.018}}} \big] \hspace{0.15 cm}\underline{= 0.11\;{\rm{V}}}{\rm{.}}$$
  
  
'''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)'''&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:
 
   
 
   
$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
+
:$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
 +
 
 +
*At time&nbsp; $t = 0$&nbsp;, one also obtains here with a good approximation&nbsp; $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.
 +
*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; $y_3( t = 20 \,\text{ns}) ≈ 0.11  \,\text{V}$.
  
*Zum Zeitpunkt $t = 0$ erhält man auch hier mit guter Näherung $0.8\, {\rm V}$.
 
*Nach $20$ Nanosekunden ist der Ausgangsimpuls um den Faktor ${\rm e}^{–0.64π} \approx 0.136$ kleiner und man erhält das Ergebnis $y_3( t = 20 \,\text{ns}) ≈ 0.11  \,\text{V}$.
 
  
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&nbsp; '''(2)'''&nbsp; and &nbsp; '''(3)''', that&nbsp; $y_3(t) \approx y_2(t)$&nbsp; gilt.  
*Der Grund hierfür ist, dass der Diracimpuls eine gute Näherung für einen rechteckförmigen Eingangsimpuls gleicher Fläche ist, wenn die Rechteckdauer $T$ deutlich kleiner ist als die äquivalente Impulsdauer $\Delta t$ der Impulsantwort.  
+
*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.
*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.
+
*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}}
 
{{ML-Fuß}}
  
 
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[[Category:Aufgaben zu Signaldarstellung|^3. Aperiodische Signale - Impulse^]]
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[[Category:Signal Representation: Exercises|^3.4 The Convolution Theorem^]]

Latest revision as of 08:05, 26 May 2021

Convolution of rectangle  $x(t)$  and Gaussian pulse  $h(t)$

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

$$H( f ) = {\rm{e}}^{{\rm{ - \pi }}( {f/\Delta f} )^2 } .$$

The corresponding impulse response is:

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

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:

  • 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 Dirac delta  $x_3(t)$  with impulse weight  $2 \cdot 10^{–8}\text{ Vs}$  (green arrow).




Hints:

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


This table gives some function values.



Questions

1

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  $(2\pi )^{1/2} \approx 2.5$?

$y_1(t=0)\ = \ $

 $\text{V}$
$y_1(t=20\,\text{ns})\ = \ $

 $\text{V}$

2

What are the signal values of the output signal  $y_2(t) = x_2(t) \ast h(t)$  at the considered time points?

$y_2(t=0)\ = \ $

 $\text{V}$
$y_2(t=20 \,\text{ns})\ = \ $

 $\text{V}$

3

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.

$y_3(t=0)\ = \ $

 $\text{V}$
$y_3(t=20\, \text{ns})\ = \ $

 $\text{V}$


Solution

(1)  The convolution integral here is:

$$y_1( t ) = A_1 \cdot \Delta f \cdot \int_{t - T_1 /2}^{t + T_1 /2} {{\rm{e}}^{{\rm{ - \pi }}( {\Delta f \hspace{0.05cm}\cdot \hspace{0.05cm} \tau } )^2 } }\hspace{0.1cm} {\rm{d}}\tau = \frac{A_1 }{\sqrt{2\pi }} \cdot\int_{u_1 }^{u_2 } {{\rm{e}}^{ - u^2 /2}\hspace{0.1cm} {\rm{d}}u.}$$
  • Here the substitution  $u = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \tau$  was used.  The integration limits are at:
$$u_1 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \big( {t - T_1 /2} \big),\hspace{0.5cm}u_2 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \big( {t + T_1 /2} \big).$$
  • Using the complementary Gaussian error integral, it is also possible to write for this:
$$y_1 (t) = A_1 \cdot \big[ {{\rm Q} ( {u_1 } ) - {\rm Q}( {u_2 } )} \big].$$
  • For time  $t = 0$  one obtains with  $(2\pi )^{1/2} \approx 2.5$:
$$u_2 = \sqrt {2{\rm{\pi }}} \cdot \Delta f \cdot \frac{ {T_1 }}{2} \approx 2.5 \cdot 4 \cdot 10^{7} \;{\rm{1/s}} \cdot 10^{-8} \;{\rm{s}} = 1.$$
  • With  $u_1 = -u_2 = -1$ , it follows for the two signal values we are looking for:
$$y_1 ( {t = 0} ) \approx A_1 \cdot \big[ {{\rm Q}( { - 1} ) - {\rm Q}(+ 1 )} \big] = 1\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.841 - 0}}{\rm{.159}}} \big] \hspace{0.15 cm}\underline{= 0.682\;{\rm{V}}}{\rm{,}}$$
$$y_1 ( {t = 20\;{\rm{ns}}} ) \approx A_1 \cdot \big[ {{\rm Q}( 1 ) - {\rm Q}( 3 )} \big] = 1\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.159 - 0}}{\rm{.001}}} \big] \hspace{0.15 cm}\underline{= 0.158\;{\rm{V}}}{\rm{.}}$$


(2)  Analogous to the first sample solution, one obtains  $x_2(t)$ for the narrower input pulse:

$$y_2 ( {t = 0} ) \approx A_2 \cdot \big[ {{\rm Q}( { - 0.1} ) - {\rm Q}( {0.1} )} \big] = 10\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.540 - 0}}{\rm{.460}}} \big] \hspace{0.15 cm}\underline{= 0.80\;{\rm{V}}}{\rm{,}}$$
$$y_2 ( {t = 20\,{\rm ns}} ) \approx A_2 \cdot \big[ {{\rm Q}( {1.9} ) - {\rm Q}( {2.1} )} \big] = 10\;{\rm{V}} \cdot \big[ {{\rm{0}}{\rm{.029 - 0}}{\rm{.018}}} \big] \hspace{0.15 cm}\underline{= 0.11\;{\rm{V}}}{\rm{.}}$$


(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:

$$y_3 (t) = 2 \cdot 10^{ - 8} \,{\rm{Vs}} \cdot 4 \cdot 10^7 \;{\rm{1/s}} \cdot {\rm{e}}^{ - {\rm{\pi }}( {\Delta f \cdot t})^2 }.$$
  • At time  $t = 0$ , one also obtains here with a good approximation  $y_3( t=0)\hspace{0.15 cm}\underline{ =0.8\, {\rm V}}$.
  • 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  $y_3( t = 20 \,\text{ns}) ≈ 0.11 \,\text{V}$.


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 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!