Difference between revisions of "Aufgaben:Exercise 1.4Z: Everything Rectangular"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Systembeschreibung im Zeitbereich}}
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{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain}}
  
[[File:P_ID834__LZI_Z_1_4.png |right|Periodisches Rechtecksignal und Rechteckfilter (Aufgabe Z1.4)]]
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[[File:P_ID834__LZI_Z_1_4.png |right|frame|Periodic rectangular signal and <br>filter with rectangular impulse response]]
Wir betrachten das periodische Rechtecksignal $x(t)$ gemäß obiger Skizze, dessen Periodendauer $T_0 = 2T$ ist. Dieses Signal besitzt Spektralanteile bei der Grundfrequenz $f_0 = 1/T_0 = 1/(2T)$ und allen ungeradzahligen Vielfachen davon, d.h. bei $3f_0, 5f_0,$ usw. Zusätzlich gibt es einen Gleichanteil.  
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We consider the periodic rectangular signal&nbsp; $x(t)$&nbsp;, whose periodic duration is&nbsp; $T_0 = 2T$&nbsp;, according to the sketch above.  
  
Dazu betrachten wir zwei Filter A und B mit jeweils rechteckförmiger Impulsantwort $h_{\rm A}(t)$ mit Dauer $6T$ bzw. $h_{\rm B}(t)$ mit der Dauer $5T$. Die Höhen der beiden Impulsantworten sind so gewählt, dass die Flächen der Rechtecke jeweils 1 ergeben.  
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*This signal has spectral components at the fundamental frequency&nbsp; $f_0 = 1/T_0 = 1/(2T)$&nbsp; and at all odd multiples thereof, that is, at&nbsp; $3f_0$,&nbsp; $5f_0,$&nbsp; and so on. In addition, there is a direct component.  
  
'''Hinweis:''' Die Aufgabe bezieht sich auf den Theorieteil von [[Lineare_zeitinvariante_Systeme/Systembeschreibung_im_Zeitbereich|Kapitel 1.2]]. Informationen zur Faltung finden Sie im [[Signaldarstellung/Faltungssatz_und_Faltungsoperation|Kapitel 3.4]] des Buches „Signaldarstellung”.  
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*For this purpose, we consider two filters&nbsp; $\rm A$&nbsp; and&nbsp; $\rm B$&nbsp; each with rectangular impulse response&nbsp; $h_{\rm A}(t)$&nbsp; with duration&nbsp; $6T$&nbsp; and&nbsp; $h_{\rm B}(t)$&nbsp; with duration&nbsp; $5T$, respectively.
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*The heights of the two impulse responses are such that the areas of the rectangles each add up to&nbsp; $1$&nbsp;.  
  
  
===Fragebogen===
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''Please note:''
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*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]].
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*For information on convolution, see the chapter&nbsp;  [[Signal_Representation/The_Convolution_Theorem_and_Operation|convolution theorem and operation]]&nbsp; in the book "Signal Representation”.
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*We also refer you to the interactive applet&nbsp; [[Applets:Zur_Verdeutlichung_der_grafischen_Faltung|Zur Verdeutlichung der graphischen Faltung]].
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 +
 +
 
 +
 
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie das Ausgangssignal $y_{\rm A}(t)$ von Filter A, insbesondere die Werte bei $t = 0$ und $t = T$.  
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{Compute the output signal&nbsp; $y_{\rm A}(t)$&nbsp; of the filter&nbsp; $\rm A$, in particular the values at&nbsp; $t = 0$&nbsp; and&nbsp; $t = T$.  
 
|type="{}"}
 
|type="{}"}
$y_{\rm A}(t = 0) =$ { 1 } V
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$y_{\rm A}(t = 0) \ =\ $ { 1 3% } &nbsp;$\rm V$
$y_{\rm A}(t = T) =$ { 1 } V
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$y_{\rm A}(t = T) \ =\ $ { 1 3% } &nbsp;$\rm V$
  
  
{Geben Sie die Betragsfunktion $|H_{\rm A}(f)|$ an. Welcher Wert ergibt sich bei der Frequenz $f = f_0$? Interpretieren Sie das Ergebnis der Teilaufgabe 1).
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{Give the absolute value function&nbsp; $|H_{\rm A}(f)|$&nbsp;. &nbsp; What value is obtained at frequency&nbsp; $f = f_0$? <br>Interpret the result of the subtask&nbsp; '''(1)'''.
 
|type="{}"}
 
|type="{}"}
$|H_{\rm A}(f = f_0)| =$ { 0 }
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$|H_{\rm A}(f = f_0)| \ =\ $ { 0. }
  
  
{Berechnen Sie das Ausgangssignal $y_{\rm B}(t)$ von Filter B, insbesondere die Werte bei $t = 0$ und $t = T$.  
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{Compute the output signal&nbsp; $y_{\rm B}(t)$&nbsp; of the filter&nbsp; $\rm B$, in particular the values at&nbsp; $t = 0$&nbsp; and&nbsp; $t = T$.  
 
|type="{}"}
 
|type="{}"}
$y_{\rm B}(t = 0) =$ { 0.8 } V
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$y_{\rm B}(t = 0) \ =\ $ { 0.8 3% } &nbsp;$\rm V$
$y_{\rm B}(t = T) =$ { 1.2 } V
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$y_{\rm B}(t = T) \ =\ $ { 1.2 3% } &nbsp;$\rm V$
  
  
{Wie lautet die Betragsfunktion $|H_{\rm B}(f)|$, insbesondere bei den Frequenzen $f = f_0$ und $f = 3 · f_0$? Interpretieren Sie damit das Ergebnis von 3).  
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{What is the absolute value function&nbsp; $|H_{\rm B}(f)|$, especially at frequencies&nbsp; $f = f_0$&nbsp; and&nbsp; $f = 3 · f_0$? <br>Use this to interpret the result of the subtask&nbsp; '''(3)'''.  
 
|type="{}"}
 
|type="{}"}
$|H_{\rm B}(f = f_0)| =$ { 0.127 5%  }
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$|H_{\rm B}(f = f_0)| \ =\ $ { 0.127 5%  }
$|H_{\rm B}(f = 3f_0)| =$ { 0.042 5%  }
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$|H_{\rm B}(f = 3f_0)| \ =\ $ { 0.042 5%  }
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.'''
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'''(1)'''&nbsp; The output signal is the result of the convolution operation between&nbsp; $x(t)$&nbsp; and&nbsp; $h_{\rm A}(t)$:
'''2.'''
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:$$y_{\rm A}(t) = x (t) * h_{\rm A} (t) = \int_{ - \infty }^{ + \infty } {x ( \tau  )}  \cdot h_{\rm A} ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
'''3.'''
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*Because of the rectangular function and the duration&nbsp; $6T$&nbsp; this can also be written as follows:
'''4.'''
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:$$y_{\rm A}(t) = \frac{1}{6T}\cdot \int_{t-6T}^{t}x(\tau)\hspace{0.15cm} {\rm d}\tau.$$
'''5.'''
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*It can be seen that this equation gives the same result&nbsp; $y_{\rm A}(t) \rm \underline{\: = 1V}$&nbsp; for all&nbsp; $t$&nbsp;.
'''6.'''
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'''7.'''
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'''(2)'''&nbsp; The magnitude of the frequency response is&nbsp; $|H_{\rm A}(f)| = |{\rm si}(\pi \cdot f \cdot 6T)|.$ &nbsp; This has zeros at an interval of&nbsp; $1/(6T)$&nbsp;.
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*So, there are also zeros at&nbsp; $f_0$,&nbsp; $3f_0$,&nbsp; $5f_0$&nbsp; etc., respectively.
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*In particular,&nbsp; $|H_{\rm A}(f = f_0)| \underline{\: = 0}$ holds, too.
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*From the spectrum&nbsp; $X(f)$&nbsp; only the direct component&nbsp; $1 \hspace{0.05cm} \rm V$&nbsp; remains unchanged.
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*In contrast to this, all other spectral lines in&nbsp; $Y_{\rm A}(f)$&nbsp; are no longer included.
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[[File:P_ID836__LZI_Z_1_4_c.png | Graphical illustration of the convolution operation| rechts|frame]]
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'''(3)'''&nbsp; Analogous to the subtask&nbsp; '''(1)'''&nbsp; for the output signal the following can be recorded:
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:$$y_{\rm B}(t) = \frac{1}{5T}\cdot \int_{t-5T}^{t}x(\tau)\hspace{0.15cm} {\rm d}\tau.$$
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*This results in a triangular shape fluctuating around the mean&nbsp; $1 \ \rm V$&nbsp; &nbsp; &rArr; &nbsp; see bottom graph.
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*Since two rectangles and three gaps each are covered by the integration interval, the following holds for&nbsp; $t = 0,&nbsp; t = 2T,$&nbsp; etc.:
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:$$y_{\rm B}(t) = \frac{2\,{\rm V} \cdot 2T }{5T}  \hspace{0.15cm}\underline{= 0.8\,{\rm V} =y_{\rm B}(t=0) }.$$
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*For&nbsp; $t = T,\ 3T, \ 5T, $&nbsp; etc., there are three rectangles and two gaps each to be considered: One obtains:
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:$$y_{\rm B}(t)  \underline{\: = 1.2 \: {\rm V}=y_{\rm B}(t=T)}.$$
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 +
 
 +
 
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'''(4)'''&nbsp;  The magnitude function is generally or at frequencies&nbsp; $f = f_0 = 1/(2T)$&nbsp; and&nbsp; $f = 3f_0$:
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:$$\begin{align*} |H_{\rm B}(f)| & = |{\rm si}(\pi \cdot f \cdot 5T)|, \\ |H_{\rm B}(f = f_0)| & = |{\rm si}(\pi \frac{5T}{2T})| = |{\rm si}(2.5\pi )| = \frac{1}{2.5 \pi}  \hspace{0.15cm}\underline{= 0.127}, \\ |H_{\rm B}(f = 3f_0)| & =  |{\rm si}(7.5\pi )| = \frac{1}{7.5 \pi}  \hspace{0.15cm}\underline{=0.042}.\end{align*}$$
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Interpretation:
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*The spectral components of the rectangular signal at&nbsp; $f_0,&nbsp; 3f_0,$&nbsp; etc., although now no longer suppressed, are increasingly attenuated as the frequency increases, in such a way that the rectangular curve is converted into a periodic triangular signal. &nbsp; The direct component&nbsp; $(1 \hspace{0.05cm} \rm V)$&nbsp; remains unchanged here, too.
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*Thus, both filters provide the average value of the input signal. &nbsp; For the signal&nbsp; $x(t)$&nbsp; at hand the filter&nbsp; $\rm A$&nbsp; is more suitable than the filter&nbsp; $\rm B$ for the determination of the mean value, because for the former the length of the impulse response is a multiple of the period&nbsp; $T_0 = 2T$&nbsp;.
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*If this condition – as with the filter&nbsp; $\rm B$ – is not fulfilled, an error signal (triangular in this example) is still superimposed on the mean value.
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{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^Kapitelx^]]
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[[Category:Linear and Time-Invariant Systems: Exercises|^1.2 System Description in Time Domain^]]

Latest revision as of 18:49, 5 September 2021

Periodic rectangular signal and
filter with rectangular impulse response

We consider the periodic rectangular signal  $x(t)$ , whose periodic duration is  $T_0 = 2T$ , according to the sketch above.

  • This signal has spectral components at the fundamental frequency  $f_0 = 1/T_0 = 1/(2T)$  and at all odd multiples thereof, that is, at  $3f_0$,  $5f_0,$  and so on. In addition, there is a direct component.
  • For this purpose, we consider two filters  $\rm A$  and  $\rm B$  each with rectangular impulse response  $h_{\rm A}(t)$  with duration  $6T$  and  $h_{\rm B}(t)$  with duration  $5T$, respectively.
  • The heights of the two impulse responses are such that the areas of the rectangles each add up to  $1$ .




Please note:



Questions

1

Compute the output signal  $y_{\rm A}(t)$  of the filter  $\rm A$, in particular the values at  $t = 0$  and  $t = T$.

$y_{\rm A}(t = 0) \ =\ $

 $\rm V$
$y_{\rm A}(t = T) \ =\ $

 $\rm V$

2

Give the absolute value function  $|H_{\rm A}(f)|$ .   What value is obtained at frequency  $f = f_0$?
Interpret the result of the subtask  (1).

$|H_{\rm A}(f = f_0)| \ =\ $

3

Compute the output signal  $y_{\rm B}(t)$  of the filter  $\rm B$, in particular the values at  $t = 0$  and  $t = T$.

$y_{\rm B}(t = 0) \ =\ $

 $\rm V$
$y_{\rm B}(t = T) \ =\ $

 $\rm V$

4

What is the absolute value function  $|H_{\rm B}(f)|$, especially at frequencies  $f = f_0$  and  $f = 3 · f_0$?
Use this to interpret the result of the subtask  (3).

$|H_{\rm B}(f = f_0)| \ =\ $

$|H_{\rm B}(f = 3f_0)| \ =\ $


Solution

(1)  The output signal is the result of the convolution operation between  $x(t)$  and  $h_{\rm A}(t)$:

$$y_{\rm A}(t) = x (t) * h_{\rm A} (t) = \int_{ - \infty }^{ + \infty } {x ( \tau )} \cdot h_{\rm A} ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
  • Because of the rectangular function and the duration  $6T$  this can also be written as follows:
$$y_{\rm A}(t) = \frac{1}{6T}\cdot \int_{t-6T}^{t}x(\tau)\hspace{0.15cm} {\rm d}\tau.$$
  • It can be seen that this equation gives the same result  $y_{\rm A}(t) \rm \underline{\: = 1V}$  for all  $t$ .


(2)  The magnitude of the frequency response is  $|H_{\rm A}(f)| = |{\rm si}(\pi \cdot f \cdot 6T)|.$   This has zeros at an interval of  $1/(6T)$ .

  • So, there are also zeros at  $f_0$,  $3f_0$,  $5f_0$  etc., respectively.
  • In particular,  $|H_{\rm A}(f = f_0)| \underline{\: = 0}$ holds, too.
  • From the spectrum  $X(f)$  only the direct component  $1 \hspace{0.05cm} \rm V$  remains unchanged.
  • In contrast to this, all other spectral lines in  $Y_{\rm A}(f)$  are no longer included.


rechts

(3)  Analogous to the subtask  (1)  for the output signal the following can be recorded:

$$y_{\rm B}(t) = \frac{1}{5T}\cdot \int_{t-5T}^{t}x(\tau)\hspace{0.15cm} {\rm d}\tau.$$
  • This results in a triangular shape fluctuating around the mean  $1 \ \rm V$    ⇒   see bottom graph.
  • Since two rectangles and three gaps each are covered by the integration interval, the following holds for  $t = 0,  t = 2T,$  etc.:
$$y_{\rm B}(t) = \frac{2\,{\rm V} \cdot 2T }{5T} \hspace{0.15cm}\underline{= 0.8\,{\rm V} =y_{\rm B}(t=0) }.$$
  • For  $t = T,\ 3T, \ 5T, $  etc., there are three rectangles and two gaps each to be considered: One obtains:
$$y_{\rm B}(t) \underline{\: = 1.2 \: {\rm V}=y_{\rm B}(t=T)}.$$


(4)  The magnitude function is generally or at frequencies  $f = f_0 = 1/(2T)$  and  $f = 3f_0$:

$$\begin{align*} |H_{\rm B}(f)| & = |{\rm si}(\pi \cdot f \cdot 5T)|, \\ |H_{\rm B}(f = f_0)| & = |{\rm si}(\pi \frac{5T}{2T})| = |{\rm si}(2.5\pi )| = \frac{1}{2.5 \pi} \hspace{0.15cm}\underline{= 0.127}, \\ |H_{\rm B}(f = 3f_0)| & = |{\rm si}(7.5\pi )| = \frac{1}{7.5 \pi} \hspace{0.15cm}\underline{=0.042}.\end{align*}$$

Interpretation:

  • The spectral components of the rectangular signal at  $f_0,  3f_0,$  etc., although now no longer suppressed, are increasingly attenuated as the frequency increases, in such a way that the rectangular curve is converted into a periodic triangular signal.   The direct component  $(1 \hspace{0.05cm} \rm V)$  remains unchanged here, too.
  • Thus, both filters provide the average value of the input signal.   For the signal  $x(t)$  at hand the filter  $\rm A$  is more suitable than the filter  $\rm B$ for the determination of the mean value, because for the former the length of the impulse response is a multiple of the period  $T_0 = 2T$ .
  • If this condition – as with the filter  $\rm B$ – is not fulfilled, an error signal (triangular in this example) is still superimposed on the mean value.