Difference between revisions of "Aufgaben:Exercise 1.3Z: Exponentially Decreasing Impulse Response"

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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_ID819__LZI_Z_1_3.png |right|frame|Abfallende Impulsantwort]]
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[[File:P_ID819__LZI_Z_1_3.png |right|frame|Decreasing impulse response]]
Gemessen wurde die Impulsantwort  $h(t)$  eines LZI–Systems,  
+
The impulse response  $h(t)$  of an LTI system, which
*die für alle Zeiten&nbsp; $t < 0$&nbsp; identisch Null ist,
+
*is identically zero for all times&nbsp; $t < 0$,
*sich zur Zeit&nbsp; $t > 0$&nbsp; sprungartig verändert, und
+
*changes abruptly at time&nbsp; $t > 0$,&nbsp; and
*für&nbsp; $t > 0$&nbsp; entsprechend einer Exponentialfunktion abfällt:
+
*decreases for&nbsp; $t > 0$&nbsp; according to an exponential function:
:$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}.$$
+
:$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T},$$ was measured.
Der Parameter sei&nbsp; $T = 1 \hspace{0.05cm} \rm ms$. In der Teilaufgabe&nbsp; '''(3)''' ist nach der&nbsp; 3dB–Grenzfrequenz&nbsp; $f_{\rm G}$&nbsp; gefragt, die wie folgt (implizit) definiert ist:  
+
Let the parameter be&nbsp; $T = 1 \hspace{0.15cm} \rm ms$.&nbsp; In the subtask&nbsp; '''(3)''' &nbsp; the 3dB cut-off frequency&nbsp; $f_{\rm G}$&nbsp; is to be determined, which is (implicitly) defined as follows:  
 
:$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$
 
:$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$
  
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''Hinweise:''  
+
''Please note:''  
*Die Aufgabe gehört zum  Kapitel&nbsp; [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Zeitbereich|Systembeschreibung im Zeitbereich]]  
+
*The exercise belongs to the chapter&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain|System Description in Time Domain]].
*Gegeben ist das folgende bestimmte Integral:  
+
*The following definite integral is given:  
 
:$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2}  \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$
 
:$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2}  \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$
 
   
 
   
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===Fragebogen===
+
===Questions===
  
  
 
<quiz display=simple>
 
<quiz display=simple>
{Berechnen Sie den Frequenzgang&nbsp; $H(f)$. Welcher Wert ergibt sich für&nbsp; $f = 0$?
+
{Compute the frequency response&nbsp; $H(f)$.&nbsp; What value is obtained for&nbsp; $f = 0$?
 
|type="{}"}
 
|type="{}"}
 
$H(f = 0) \ = \ $ { 1 3% }
 
$H(f = 0) \ = \ $ { 1 3% }
  
  
{Welchen Wert besitzt die Impulsantwort zur Zeit&nbsp; $t = 0$?  
+
{What is the value of the impulse response at time&nbsp; $t = 0$?  
 
|type="{}"}
 
|type="{}"}
 
$h(t = 0)  \ = \ $ { 500 3% } &nbsp;$\rm 1/s$
 
$h(t = 0)  \ = \ $ { 500 3% } &nbsp;$\rm 1/s$
  
  
{Berechnen Sie die 3dB–Grenzfrequenz&nbsp; $f_{\rm G}$.
+
{Compute the 3dB cut-off frequency&nbsp; $f_{\rm G}$.
 
|type="{}"}
 
|type="{}"}
 
$f_{\rm G}  \ =\ $ { 159 3% } &nbsp;$\rm Hz$
 
$f_{\rm G}  \ =\ $ { 159 3% } &nbsp;$\rm Hz$
  
  
{Welche der folgenden Aussagen sind zutreffend?
+
{Which of the following statements are true?
 
|type="[]"}
 
|type="[]"}
+ Das betrachtete System ist kausal.
+
+ The considered system is causal.
- Das betrachtete System hat Hochpass–Charakter.  
+
- The considered system has high-pass filter characteristics.  
- Liegt am Systemeingang ein Cosinussignal der Frequenz&nbsp; $f_{\rm G}$&nbsp; an, so ist das Ausgangssignal ebenfalls cosinusförmig.  
+
- If a cosine signal of frequency&nbsp; $f_{\rm G}$&nbsp; is applied to the system input, the output signal is also cosine-shaped.  
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Der Frequenzgang&nbsp; $H(f)$&nbsp; ist die Fouriertransformierte von&nbsp; $h(t)$:
+
'''(1)'''&nbsp; The frequency response&nbsp; $H(f)$&nbsp; is the Fourier transform of&nbsp; $h(t)$:
 
:$$H(f) = \int_{-\infty}^{+\infty}h(t) \cdot {\rm e}^{\hspace{0.05cm}{-\rm j}2\pi ft}\hspace{0.15cm} {\rm d}t = \frac{1}{T} \cdot \int_{0}^{+\infty} {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\hspace{0.15cm}
 
:$$H(f) = \int_{-\infty}^{+\infty}h(t) \cdot {\rm e}^{\hspace{0.05cm}{-\rm j}2\pi ft}\hspace{0.15cm} {\rm d}t = \frac{1}{T} \cdot \int_{0}^{+\infty} {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\hspace{0.15cm}
 
{\rm d}t.$$  
 
{\rm d}t.$$  
*Die Integration führt zum Ergebnis:
+
*Integration leads to the result:
 
:$$H(f)  = \left[ \frac{-1/T}{{\rm j}2\pi f+{1}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T})
 
:$$H(f)  = \left[ \frac{-1/T}{{\rm j}2\pi f+{1}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T})
 
t}\right]_{0}^{\infty}= \frac{1}{1+{\rm j} \cdot 2\pi fT}.$$
 
t}\right]_{0}^{\infty}= \frac{1}{1+{\rm j} \cdot 2\pi fT}.$$
*Bei der Frequenz&nbsp; $f = 0$&nbsp; hat der Frequenzgang den Wert&nbsp; $H(f = 0) \; \underline{= 1}$.  
+
*At frequency&nbsp; $f = 0$&nbsp; the frequency response has the value&nbsp; $H(f = 0) \; \underline{= 1}$.  
  
  
  
'''(2)'''&nbsp; Dieser Frequenzgang kann mit Real– und Imaginärteil auch wie folgt geschrieben werden:  
+
'''(2)'''&nbsp; This frequency response can also be written with real and imaginary parts as follows:  
 
:$$H(f)  =  \frac{1}{1+(2\pi fT)^2} -{\rm j} \cdot  \frac{2\pi fT}{1+(2\pi fT)^2}.$$
 
:$$H(f)  =  \frac{1}{1+(2\pi fT)^2} -{\rm j} \cdot  \frac{2\pi fT}{1+(2\pi fT)^2}.$$
*Die Impulsantwort an der Stelle&nbsp; $t = 0$&nbsp; ist gleich dem Integral über&nbsp; $H(f)$.  
+
*The impulse response at time&nbsp; $t = 0$&nbsp; is equal to the integral over&nbsp; $H(f)$.  
*Da der Imaginärteil ungerade ist, muss nur über den Realteil integriert werden.  
+
*Since the imaginary part is odd only the real part has to be integrated over.  
*Unter Ausnutzung der Symmetrieeigenschaft erhält man:  
+
*Using the symmetry property one obtains:  
 
:$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2}  \hspace{0.1cm}{\rm
 
:$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2}  \hspace{0.1cm}{\rm
 
  d}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2}  \hspace{0.1cm}{\rm d}x .$$
 
  d}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2}  \hspace{0.1cm}{\rm d}x .$$
*Unter Benutzung des angegebenen bestimmten Integrals mit dem Resultat&nbsp; $π/2$&nbsp; ergibt sich:  
+
*Using the given definite integral with the result&nbsp; $π/2$&nbsp; the following is obtained:  
 
:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
 
:$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
*Das Ergebnis zeigt , dass die Impulsantwort bei&nbsp; $t = 0$&nbsp; gleich dem Mittelwert aus links&ndash; und rechtsseitigem Grenzwert ist.  
+
*The result shows that the impulse response at&nbsp; $t = 0$&nbsp; is equal to the mean value of the left-hand and right-hand limits.  
  
  
  
'''(3)'''&nbsp; Der Amplitudengang lautet bei dieser Aufgabe bzw. allgemein mit der 3dB-Grenzfrequenz&nbsp; $f_{\rm G}$:  
+
'''(3)'''&nbsp; The amplitude response in this task or in general with the 3dB cut-off frequency&nbsp; $f_{\rm G}$ is:  
 
:$$|H(f)|  =  \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$
 
:$$|H(f)|  =  \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$
*Durch Koeffizientenvergleich erhält man:  
+
*By comparing the coefficients one obtains:  
:$$f_{\rm G} =   \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$
+
:$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$
  
  
  
'''(4)'''&nbsp; Richtig ist  <u> der erste Lösungsvorschlag}</u>.:
+
'''(4)'''&nbsp; <u>The first statement</u> is correct:
*Wegen&nbsp; $h(t) = 0$&nbsp; &nbsp;für&nbsp; $t < 0$&nbsp; ist das System tatsächlich kausal. Es handelt sich um einen <u>Tiefpass erster Ordnung</u>.  
+
*Due to&nbsp; $h(t) = 0$&nbsp; &nbsp;for&nbsp; $t < 0$:&nbsp; The system is indeed causal.&nbsp; It is a <u>low-pass filter of first order</u>.  
*Dagegen müsste ein Hochpass folgende Bedingung erfüllen:  
+
*In contrast, a high-pass filter would have to satisfy the following condition:  
 
:$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
 
:$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
*$H(f)$&nbsp; ist eine komplexe Funktion. Der Phasengang lautet&nbsp; (siehe&nbsp; [[Aufgaben:Aufgabe_1.1Z:_Tiefpass_1._und_2._Ordnung|Aufgabe 1.1Z]]):  
+
*$H(f)$&nbsp; is a complex function. The phase response is&nbsp; (see&nbsp; [[Aufgaben:Exercise_1.1Z:_Low-Pass_Filter_of_1st_and_2nd_Order|Exercise 1.1Z]]):  
 
:$$b(f) = \arctan {f}/{f_{\rm G}}.$$
 
:$$b(f) = \arctan {f}/{f_{\rm G}}.$$
*Für die Frequenz&nbsp; $f = f_{\rm G}$&nbsp; erhält man&nbsp; $b(f = f_{\rm G}) = π/4 = 45^\circ$.  
+
*For the frequency&nbsp; $f = f_{\rm G}$&nbsp; one obtains&nbsp; $b(f = f_{\rm G}) = π/4 = 45^\circ$.  
*Liegt am Eingang ein Cosinussignal der Frequenz&nbsp; $f = f_{\rm G}$&nbsp; an, so ergibt sich für das Ausgangssignal:
+
*If a cosine signal of frequency&nbsp; $f = f_{\rm G}$&nbsp; is applied ot the input, the output signal is given by:
 
:$$y(t) = K \cdot \cos( 2 \pi f_{\rm G} t - 45^{\circ}).$$
 
:$$y(t) = K \cdot \cos( 2 \pi f_{\rm G} t - 45^{\circ}).$$
*Dieses Signal ist zwar eine harmonische Schwingung, aber kein Cosinussignal.  
+
*This signal is a harmonic oscillation but not a cosine signal.  
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Exercises for Linear and Time-Invariant Systems|^1.2 Systembeschreibung im Zeitbereich^]]
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[[Category:Linear and Time-Invariant Systems: Exercises|^1.2 System Description in Time Domain^]]

Latest revision as of 18:17, 18 July 2021

Decreasing impulse response

The impulse response  $h(t)$  of an LTI system, which

  • is identically zero for all times  $t < 0$,
  • changes abruptly at time  $t > 0$,  and
  • decreases for  $t > 0$  according to an exponential function:
$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T},$$ was measured.

Let the parameter be  $T = 1 \hspace{0.15cm} \rm ms$.  In the subtask  (3)   the 3dB cut-off frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:

$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$





Please note:

$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$



Questions

1

Compute the frequency response  $H(f)$.  What value is obtained for  $f = 0$?

$H(f = 0) \ = \ $

2

What is the value of the impulse response at time  $t = 0$?

$h(t = 0) \ = \ $

 $\rm 1/s$

3

Compute the 3dB cut-off frequency  $f_{\rm G}$.

$f_{\rm G} \ =\ $

 $\rm Hz$

4

Which of the following statements are true?

The considered system is causal.
The considered system has high-pass filter characteristics.
If a cosine signal of frequency  $f_{\rm G}$  is applied to the system input, the output signal is also cosine-shaped.


Solution

(1)  The frequency response  $H(f)$  is the Fourier transform of  $h(t)$:

$$H(f) = \int_{-\infty}^{+\infty}h(t) \cdot {\rm e}^{\hspace{0.05cm}{-\rm j}2\pi ft}\hspace{0.15cm} {\rm d}t = \frac{1}{T} \cdot \int_{0}^{+\infty} {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\hspace{0.15cm} {\rm d}t.$$
  • Integration leads to the result:
$$H(f) = \left[ \frac{-1/T}{{\rm j}2\pi f+{1}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\right]_{0}^{\infty}= \frac{1}{1+{\rm j} \cdot 2\pi fT}.$$
  • At frequency  $f = 0$  the frequency response has the value  $H(f = 0) \; \underline{= 1}$.


(2)  This frequency response can also be written with real and imaginary parts as follows:

$$H(f) = \frac{1}{1+(2\pi fT)^2} -{\rm j} \cdot \frac{2\pi fT}{1+(2\pi fT)^2}.$$
  • The impulse response at time  $t = 0$  is equal to the integral over  $H(f)$.
  • Since the imaginary part is odd only the real part has to be integrated over.
  • Using the symmetry property one obtains:
$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2} \hspace{0.1cm}{\rm d}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x .$$
  • Using the given definite integral with the result  $π/2$  the following is obtained:
$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
  • The result shows that the impulse response at  $t = 0$  is equal to the mean value of the left-hand and right-hand limits.


(3)  The amplitude response in this task or in general with the 3dB cut-off frequency  $f_{\rm G}$ is:

$$|H(f)| = \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$
  • By comparing the coefficients one obtains:
$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$


(4)  The first statement is correct:

  • Due to  $h(t) = 0$   for  $t < 0$:  The system is indeed causal.  It is a low-pass filter of first order.
  • In contrast, a high-pass filter would have to satisfy the following condition:
$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
  • $H(f)$  is a complex function. The phase response is  (see  Exercise 1.1Z):
$$b(f) = \arctan {f}/{f_{\rm G}}.$$
  • For the frequency  $f = f_{\rm G}$  one obtains  $b(f = f_{\rm G}) = π/4 = 45^\circ$.
  • If a cosine signal of frequency  $f = f_{\rm G}$  is applied ot the input, the output signal is given by:
$$y(t) = K \cdot \cos( 2 \pi f_{\rm G} t - 45^{\circ}).$$
  • This signal is a harmonic oscillation but not a cosine signal.