Aufgaben:Exercise 3.1: Causality Considerations - Revision history

Guenter at 15:10, 9 October 2021

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Guenter at 14:35, 9 October 2021

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Guenter at 14:33, 9 October 2021

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Oezer at 07:22, 24 September 2021

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Oezer at 07:20, 24 September 2021

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Oezer at 07:18, 24 September 2021

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Oezer at 07:16, 24 September 2021

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Oezer at 06:59, 24 September 2021

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Oezer at 00:46, 24 September 2021

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Oezer at 00:42, 24 September 2021

2021-09-24T00:42:17Z

@@ Line 72: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Proposed solution 2</u> is correct:
+'''(1)'''&nbsp; <u>Proposed solution 2</u>&nbsp; is correct:
 *The given transfer function can be computed according to the voltage divider principle. &nbsp; The following holds:
 :$$H_1(f = 0) = 0, \hspace{0.2cm}H_1(f \rightarrow \infty) = 1$$
@@ Line 80: / Line 80: @@
-'''(2)'''&nbsp; <u>Yes</u> is correct:
+'''(2)'''&nbsp; <u>Yes</u>&nbsp; is correct:
-*Every real network is causal.&nbsp; The impulse response&nbsp; $h(t)$&nbsp; is equal to the output signal&nbsp; $y(t)$ if at time&nbsp; $t= 0$&nbsp; an extremely short impulse &ndash; a so-called Dirac impulse &ndash; is applied to the input.
+*Every real network is causal.&nbsp; The impulse response&nbsp; $h(t)$&nbsp; is equal to the output signal&nbsp; $y(t)$&nbsp; if at time&nbsp; $t= 0$&nbsp; an extremely short impulse &ndash; a so-called Dirac delta impulse &ndash; is applied to the input.
-*Then, a signal of course cannot occur at the output already for times&nbsp; $t< 0$&nbsp; for causality reasons:
+*Then,&nbsp; a signal cannot occur at the output already for times&nbsp; $t< 0$&nbsp; for causality reasons:
 :$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
   t<0 \hspace{0.05cm}.$$
@@ Line 90: / Line 90: @@
   \hspace{0.05cm}.$$
 *The second transfer function describes the low-pass function equivalent to&nbsp; $H_1(f)$,&nbsp; which results in the exponential function in the time domain.
-*The "$1$" becomes a Dirac function.&nbsp; Considering&nbsp; $T = 2\pi \cdot f_{\rm G}$&nbsp; the following thus holds for&nbsp; $t \ge 0$:
+*The&nbsp; "$1$"&nbsp; becomes a Dirac delta function.&nbsp; Considering&nbsp; $T = 2\pi \cdot f_{\rm G}$&nbsp; the following thus holds for&nbsp; $t \ge 0$:
 :$$h_1(t) = \delta(t) - {1}/{T} \cdot {\rm e}^{-t/T} \hspace{0.05cm}.$$
-*In contrast, &nbsp; $h_1(t)= 0$ holds for&nbsp; $t< 0$&nbsp;, which would prove causality.
+*In contrast,&nbsp; $h_1(t)= 0$ holds for&nbsp; $t< 0$,&nbsp; which would prove causality.
@@ Line 110: / Line 110: @@
-'''(4)'''&nbsp; <u>The first two proposed solutions</u> are correct:
+'''(4)'''&nbsp; <u>The first two proposed solutions</u>&nbsp; are correct:
-*Since the impulse response is &nbsp;$h_1(t) = 0$&nbsp; for &nbsp;$t < 0$&nbsp;, the convolution operation &nbsp;$h_2(t) = h_1(t) \star h_1(t)$&nbsp; also satisfies the causality condition.&nbsp; Similarly, the&nbsp; $n$&ndash;fold convolution yields a causal impulse response: &nbsp;  $h_n(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm}
+*Since the impulse response is &nbsp;$h_1(t) = 0$&nbsp; for &nbsp;$t < 0$,&nbsp; the convolution operation &nbsp;$h_2(t) = h_1(t) \star h_1(t)$&nbsp; also satisfies the causality condition.&nbsp;
   t<0 \hspace{0.05cm}.$
-*However, the real and imaginary parts of the spectral function &nbsp;$H_2(f)$&nbsp; are related via the Hilbert transformation for a causal impulse response &nbsp;$h_2(t)$&nbsp;.&nbsp; Thus, considering the shortcut &nbsp;$x = f/f_{\rm G}$&nbsp; and the result of the subtask&nbsp; '''(3)'''&nbsp; the following holds:
+*However,&nbsp; the real and imaginary parts of the spectral function &nbsp;$H_2(f)$&nbsp; are related via the Hilbert transformation for a causal impulse response &nbsp;$h_2(t)$&nbsp;.&nbsp;
 :$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad
 \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad

@@ Line 35: / Line 35: @@
 Please note:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem|Conclusions from the Allocation Theorem]].
-*Reference is also made to the theory pages &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]]&nbsp; and &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition|Partial fraction decomposition]].
+*Reference is also made to the theory pages &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert_transform|Hilbert transformation]]&nbsp; and &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition|Partial fraction decomposition]].

@@ Line 8: / Line 8: @@
   \hspace{0.05cm},$$
-where&nbsp; $f_{\rm G}$&nbsp; represents the 3dB cut-off frequency:
+where&nbsp; $f_{\rm G}$&nbsp; represents the&nbsp; $\rm 3dB$&nbsp; cut-off frequency:
 :$$f_{\rm G} = \frac{R}{2 \pi \cdot L}
   \hspace{0.05cm}.$$
-By cascading &nbsp;$n$&nbsp; two-port networks &nbsp;$H_1(f)$&nbsp; built the same way, the transfer function
+By cascading &nbsp;$n$&nbsp; two-port networks &nbsp;$H_1(f)$&nbsp; built in the same way, the following transfer function is obtained:
 :$$H_n(f) = \big [H_1(f)\big ]^n =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^n}{\big [1+{\rm j}\cdot f/f_{\rm G}\big ]^n}
-  \hspace{0.05cm}$$ is obtained.
+  \hspace{0.05cm}.$$
-*Here, a suitable resistor decoupling is presumed, but this is not important for solving this exercise.
+*Here, a suitable resistor decoupling is presumed,&nbsp; but this is not important for solving this exercise.
 *The lower graph shows for example the realization of the transfer function &nbsp;$H_2(f)$.
-In this exercise, such a two-port network is considered with respect to its causality properties.
+In this exercise,&nbsp; such a two-port network is considered with respect to its causality properties.
-For any causal system, the real and imaginary parts of the spectral function &nbsp;$H(f)$&nbsp; satisfy the&nbsp; [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]], which is expressed by the following abbreviation:
+For any causal system,&nbsp; the real and imaginary parts of the spectral function &nbsp;$H(f)$&nbsp; satisfy the&nbsp; [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]],&nbsp; which is expressed by the following abbreviation:
 :$${\rm Im} \left\{ H(f) \right \}  \quad
 \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad
 {\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$
-Since the Hilbert transformation provides important information not only for transfer functions but also for time signals, the correspondence is often expressed by the general variable &nbsp;$x$&nbsp;, which is to be interpreted - depending on the application - as normalized frequency or normalized time.
+Since the Hilbert transformation provides important information not only for transfer functions but also for time signals,&nbsp; the correspondence is often expressed by the general variable &nbsp;$x$,&nbsp; which is to be interpreted - depending on the application - as normalized frequency or as normalized time.
@@ Line 33: / Line 33: @@
 Please note:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem|Conclusions from the Allocation Theorem]].
 *Reference is also made to the theory pages &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]]&nbsp; and &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition|Partial fraction decomposition]].
@@ Line 57: / Line 54: @@
-{Compute the transfer function &nbsp;$H_2(f)$.&nbsp; What is the complex value for &nbsp;$f = f_{\rm G}$?
+{Compute the transfer function &nbsp;$H_2(f)$. &nbsp; What is the complex value for &nbsp;$f = f_{\rm G}$?
 |type="{}"}
 ${\rm Re}\big[H_2(f = f_{\rm G})\big] \ = \ $  { 0. }

@@ Line 114: / Line 114: @@
 '''(4)'''&nbsp; <u>The first two proposed solutions</u> are correct:
-*Since the impulse response is &nbsp;$h_1(t) = 0$&nbsp; for &nbsp;$t < 0$&nbsp;, the convolution operation &nbsp;$h_2(t) = h_1(t) \star h_1(t)$&nbsp; also satisfies the causality condition.&nbsp; Similarly, the&nbsp; $n$&ndash;fold convolution yields a causal impulse response: &nbsp;  $h_n(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
+*Since the impulse response is &nbsp;$h_1(t) = 0$&nbsp; for &nbsp;$t < 0$&nbsp;, the convolution operation &nbsp;$h_2(t) = h_1(t) \star h_1(t)$&nbsp; also satisfies the causality condition.&nbsp; Similarly, the&nbsp; $n$&ndash;fold convolution yields a causal impulse response: &nbsp;  $h_n(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm}
   t<0 \hspace{0.05cm}.$
 *However, the real and imaginary parts of the spectral function &nbsp;$H_2(f)$&nbsp; are related via the Hilbert transformation for a causal impulse response &nbsp;$h_2(t)$&nbsp;.&nbsp; Thus, considering the shortcut &nbsp;$x = f/f_{\rm G}$&nbsp; and the result of the subtask&nbsp; '''(3)'''&nbsp; the following holds:

@@ Line 88: / Line 88: @@
 :$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
   t<0 \hspace{0.05cm}.$$
-*Formally, this can be shown as follows: &nbsp; The high-pass&ndash;transfer function&nbsp; $H_1(f)$&nbsp; can be rearranged as follows:
+*Formally, this can be shown as follows: &nbsp; The high-pass transfer function&nbsp; $H_1(f)$&nbsp; can be rearranged as follows:
 :$$H_1(f) = \frac{{\rm j}\cdot f/f_{\rm G}}{1+{\rm j}\cdot f/f_{\rm G}}
   = 1- \frac{1}{1+{\rm j}\cdot f/f_{\rm G}}

@@ Line 99: / Line 99: @@
-'''(3)'''&nbsp; Die Hintereinanderschaltung zweier Hochpässe führt zu folgender Übertragungsfunktion:
+'''(3)'''&nbsp; The series connection of two high-pass filters results in the following transfer function:
 :$$H_2(f) = \big [H_1(f)\big ]^2  =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^2}{\big [1+{\rm j}\cdot f/f_{\rm G}\big ]^2}
   =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^2 \cdot \big [(1-{\rm j}\cdot f/f_{\rm G})\big ]^2}
@@ Line 106: / Line 106: @@
   {\big [1+(f/f_{\rm G})^2 \big ]^2}\hspace{0.05cm}.$$
-*Mit&nbsp; $f = f_{\rm G}$&nbsp; folgt daraus:
+*With&nbsp; $f = f_{\rm G}$&nbsp; from this it follows that:
 :$$H_2(f = f_{\rm G})  = \frac{1 - 1 +{\rm j}\cdot 2}
   {4}= {\rm j} /{2} \hspace{0.5cm}\Rightarrow \hspace{0.5cm}{\rm Re} \left\{ H_2(f = f_{\rm G}) \right \}  \hspace{0.15cm}\underline{  = 0}, \hspace{0.4cm}
@@ Line 113: / Line 113: @@
-'''(4)'''&nbsp; Richtig sind <u>die beiden ersten Lösungsvorschläge</u>:
+'''(4)'''&nbsp; <u>The first two proposed solutions</u> are correct:
-* Da  für &nbsp;$t < 0$&nbsp; die Impulsantwort &nbsp;$h_1(t) = 0$&nbsp;  ist, erfüllt auch die Faltungsoperation &nbsp;$h_2(t) = h_1(t) \star h_1(t)$&nbsp; die Kausalitätsbedingung.&nbsp; Ebenso ergibt die&nbsp; $n$&ndash;fache Faltung eine kausale Impulsantwort: &nbsp;  $h_n(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
+*Since the impulse response is &nbsp;$h_1(t) = 0$&nbsp; for &nbsp;$t < 0$&nbsp;, the convolution operation &nbsp;$h_2(t) = h_1(t) \star h_1(t)$&nbsp; also satisfies the causality condition.&nbsp; Similarly, the&nbsp; $n$&ndash;fold convolution yields a causal impulse response: &nbsp;  $h_n(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
   t<0 \hspace{0.05cm}.$
-*Bei kausaler Impulsantwort &nbsp;$h_2(t)$&nbsp; hängen aber der Real&ndash; und der Imaginärteil der Spektralfunktion &nbsp;$H_2(f)$&nbsp; über die Hilbert&ndash;Transformation zusammen.&nbsp; Mit der Abkürzung &nbsp;$x = f/f_{\rm G}$&nbsp;  und dem Ergebnis der Teilaufgabe&nbsp; '''(3)'''&nbsp; gilt somit:
+*However, the real and imaginary parts of the spectral function &nbsp;$H_2(f)$&nbsp; are related via the Hilbert transformation for a causal impulse response &nbsp;$h_2(t)$&nbsp;.&nbsp; Thus, considering the shortcut &nbsp;$x = f/f_{\rm G}$&nbsp; and the result of the subtask&nbsp; '''(3)'''&nbsp; the following holds:
 :$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad
 \bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad

@@ Line 45: / Line 45: @@
 <quiz display=simple>
-{Wie kann &nbsp;$H_1(f)$&nbsp; charakterisiert werden?
+{How can &nbsp;$H_1(f)$&nbsp; be characterized?
 |type="()"}
-- $H_1(f)$&nbsp; beschreibt einen Tiefpass.
+- $H_1(f)$&nbsp; describes a low-pass filter.
-+ $H_1(f)$&nbsp; beschreibt einen Hochpass.
++ $H_1(f)$&nbsp; describes a high-pass filter.
-{Beschreibt &nbsp;$H_1(f)$&nbsp; ein kausales Netzwerk?
+{Does &nbsp;$H_1(f)$&nbsp; describe a causal network?
 |type="()"}
-+ Ja.
++ Yes.
-- Nein.
+- No.
-{Berechnen Sie die Übertragungsfunktion &nbsp;$H_2(f)$.&nbsp; Welcher komplexe Wert ergibt sich für &nbsp;$f = f_{\rm G}$?
+{Compute the transfer function &nbsp;$H_2(f)$.&nbsp; What is the complex value for &nbsp;$f = f_{\rm G}$?
 |type="{}"}
 ${\rm Re}\big[H_2(f = f_{\rm G})\big] \ = \ $  { 0. }
@@ Line 63: / Line 63: @@
-{Welche der folgenden Aussagen treffen zu?
+{Which of the following statements are true?
 |type="[]"}
-+ $H_2(f)$&nbsp; beschreibt ein kausales System.
++ $H_2(f)$&nbsp; describes a causal system.
-+ Die Ausdrücke&nbsp; $(x^4 - x^2)/(x^4 +2 x^2 + 1)$&nbsp; und &nbsp;$2x^3/(x^4 +2 x^2 + 1)$&nbsp; sind ein Hilbert&ndash;Paar.
++ The expressions&nbsp; $(x^4 - x^2)/(x^4 +2 x^2 + 1)$&nbsp; and &nbsp;$2x^3/(x^4 +2 x^2 + 1)$&nbsp; are a Hilbert pair.
-- Für &nbsp;$n > 2$&nbsp; ist die Kausalitätsbedingung nicht erfüllt.
+- The causality condition is not satisfied for &nbsp;$n > 2$&nbsp;.

@@ Line 38: / Line 38: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem|Conclusions from the Allocation Theorem]].
-*Bezug genommen wird auch auf die Therieseiten &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]]&nbsp; sowie &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition|Partial fraction decomposition]].
+*Reference is also made to the theory pages &nbsp;[[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation|Hilbert transformation]]&nbsp; and &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition|Partial fraction decomposition]].