Aufgaben:Exercise 3.1: Causality Considerations - Revision history

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.1 Conclusions from the Assignment Theorem^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.1 Conclusions from the Assignment Theorem^]]
	[[de:Aufgaben:3.~~1_Kausalitätsbetrachtungen~~]]		[[de:Aufgaben:Aufgabe 3.1: Kausalitätsbetrachtungen]]

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	[[File:EN_LZI_A_3_1.png\|right\|frame\|Two two-port networks]]		[[File:EN_LZI_A_3_1.png\|right\|frame\|Two two-port networks]]
	The graph shows above the two-port network with the transfer function		The graph shows above the two-port network with the transfer function
	:$$H_1(f) = \frac{{\rm j}\cdot f/f_{\rm G}}{1+{\rm j}\cdot f/f_{\rm G}}		:$$H_1(f) = \frac{{\rm j}\cdot f/f_{\rm G}}{1+{\rm j}\cdot f/f_{\rm G}}\hspace{0.05cm},$$
	\hspace{0.05cm},$$

	where  $f_{\rm G}$  represents the  $\rm 3dB$  cut-off frequency:		where  $f_{\rm G}$  represents the  $\rm 3dB$  cut-off frequency:
	:$$f_{\rm G} = \frac{R}{2 \pi \cdot L}		:$$f_{\rm G} = \frac{R}{2 \pi \cdot L}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	By cascading  $n$  two-port networks  $H_1(f)$  built in the same way, the following transfer function is obtained:		By cascading  $n$  two-port networks  $H_1(f)$  built in the same way, the following transfer function is obtained:
	:$$H_n(f) = \big [H_1(f)\big ]^n =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^n}{\big [1+{\rm j}\cdot f/f_{\rm G}\big ]^n}		:$$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}.$$
	\hspace{0.05cm}.$$

	*Here, a suitable resistor decoupling is presumed,  but this is not important for solving this exercise.		*Here, a suitable resistor decoupling is presumed,  but this is not important for solving this exercise.
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	For any causal system,  the real and imaginary parts of the spectral function  $H(f)$  satisfy the  [[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,  the real and imaginary parts of the spectral function  $H(f)$  satisfy the  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation\|Hilbert transformation]],  which is expressed by the following abbreviation:
	:$${\rm Im} \left\{ H(f) \right \} \quad		:$${\rm Im} \left\{ H(f) \right \} \quad\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad{\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$
	\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad
	{\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$

	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  $x$,  which is to be interpreted - depending on the application - as normalized frequency or as normalized time.		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  $x$,  which is to be interpreted - depending on the application - as normalized frequency or as normalized time.
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	*Every real network is causal.  The impulse response  $h(t)$  is equal to the output signal  $y(t)$  if at time  $t= 0$  an extremely short impulse – a so-called Dirac delta impulse – is applied to the input.		*Every real network is causal.  The impulse response  $h(t)$  is equal to the output signal  $y(t)$  if at time  $t= 0$  an extremely short impulse – a so-called Dirac delta impulse – is applied to the input.
	*Then,  a signal cannot occur at the output already for times  $t< 0$  for causality reasons:		*Then,  a signal cannot occur at the output already for times  $t< 0$  for causality reasons:
	:$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}		:$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}t<0 \hspace{0.05cm}.$$
	t<0 \hspace{0.05cm}.$$
	*Formally, this can be shown as follows:   The high-pass transfer function  $H_1(f)$  can be rearranged as follows:		*Formally, this can be shown as follows:   The high-pass transfer function  $H_1(f)$  can be rearranged as follows:
	:$$H_1(f) = \frac{{\rm j}\cdot f/f_{\rm G}}{1+{\rm j}\cdot f/f_{\rm G}}		:$$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}}\hspace{0.05cm}.$$
	= 1- \frac{1}{1+{\rm j}\cdot f/f_{\rm G}}
	\hspace{0.05cm}.$$
	*The second transfer function describes the low-pass function equivalent to  $H_1(f)$,  which results in the exponential function in the time domain.		*The second transfer function describes the low-pass function equivalent to  $H_1(f)$,  which results in the exponential function in the time domain.
	*The  "$1$"  becomes a Dirac delta function.  Considering  $T = 2\pi \cdot f_{\rm G}$  the following thus holds for  $t \ge 0$:		*The  "$1$"  becomes a Dirac delta function.  Considering  $T = 2\pi \cdot f_{\rm G}$  the following thus holds for  $t \ge 0$:
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	'''(3)'''  The series connection of two high-pass filters results in the following transfer function:		'''(3)'''  The series connection of two high-pass filters results in the following transfer function:
	:$$H_2(f) = \big [H_1(f)\big ]^2 =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^2}{\big [1+{\rm j}\cdot f/f_{\rm G}\big ]^2}		:$$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}{\big [(1+{\rm j}\cdot f/f_{\rm G}) \cdot (1-{\rm j}\cdot f/f_{\rm G})\big ]^2}= \frac{(f/f_{\rm G})^4 - (f/f_{\rm G})^2 +{\rm j}\cdot 2\cdot (f/f_{\rm G})^3)}{\big [1+(f/f_{\rm G})^2 \big ]^2}\hspace{0.05cm}.$$
	=\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^2 \cdot \big [(1-{\rm j}\cdot f/f_{\rm G})\big ]^2}
	{\big [(1+{\rm j}\cdot f/f_{\rm G}) \cdot (1-{\rm j}\cdot f/f_{\rm G})\big ]^2}= \frac{(f/f_{\rm G})^4 - (f/f_{\rm G})^2 +{\rm j}\cdot 2
	\cdot (f/f_{\rm G})^3)}
	{\big [1+(f/f_{\rm G})^2 \big ]^2}\hspace{0.05cm}.$$

	*With  $f = f_{\rm G}$  from this it follows that:		*With  $f = f_{\rm G}$  from this it follows that:
	:$$H_2(f = f_{\rm G}) = \frac{1 - 1 +{\rm j}\cdot 2}		:$$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}{\rm Im} \left\{ H_2(f = f_{\rm G}) \right \} \hspace{0.15cm}\underline{ = 0.5}\hspace{0.05cm}.$$
	{4}= {\rm j} /{2} \hspace{0.5cm}\Rightarrow \hspace{0.5cm}{\rm Re} \left\{ H_2(f = f_{\rm G}) \right \} \hspace{0.15cm}\underline{ = 0}, \hspace{0.4cm}
	{\rm Im} \left\{ H_2(f = f_{\rm G}) \right \} \hspace{0.15cm}\underline{ = 0.5}\hspace{0.05cm}.$$


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	*However,  the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .		*However,  the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .
	*Thus,  considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:		*Thus,  considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:
	:$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad		:$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad\frac{2x^3}{x^4+2 x^2+1}\hspace{0.05cm}.$$
	\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad
	\frac{2x^3}{x^4+2 x^2+1}\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.1 Conclusions from the Assignment Theorem^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.1 Conclusions from the Assignment Theorem^]]
			[[de:Aufgaben:3.1_Kausalitätsbetrachtungen]]

Guenter at 15:10, 9 October 2021

2021-10-09T15:10:13Z

← Older revision		Revision as of 17:10, 9 October 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Proposed solution 2</u> is correct:		'''(1)'''  <u>Proposed solution 2</u>  is correct:
	*The given transfer function can be computed according to the voltage divider principle.   The following holds:		*The given transfer function can be computed according to the voltage divider principle.   The following holds:
	:$$H_1(f = 0) = 0, \hspace{0.2cm}H_1(f \rightarrow \infty) = 1$$		:$$H_1(f = 0) = 0, \hspace{0.2cm}H_1(f \rightarrow \infty) = 1$$
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	'''(2)'''  <u>Yes</u> is correct:		'''(2)'''  <u>Yes</u>  is correct:
	*Every real network is causal.  The impulse response  $h(t)$  is equal to the output signal  $y(t)$ if at time  $t= 0$  an extremely short impulse – a so-called Dirac impulse – is applied to the input.		*Every real network is causal.  The impulse response  $h(t)$  is equal to the output signal  $y(t)$  if at time  $t= 0$  an extremely short impulse – a so-called Dirac delta impulse – is applied to the input.
	*Then, a signal ~~of course~~ cannot occur at the output already for times  $t< 0$  for causality reasons:		*Then,  a signal cannot occur at the output already for times  $t< 0$  for causality reasons:
	:$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}		:$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
	t<0 \hspace{0.05cm}.$$		t<0 \hspace{0.05cm}.$$
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*The second transfer function describes the low-pass function equivalent to  $H_1(f)$,  which results in the exponential function in the time domain.		*The second transfer function describes the low-pass function equivalent to  $H_1(f)$,  which results in the exponential function in the time domain.
	*The "$1$" becomes a Dirac function.  Considering  $T = 2\pi \cdot f_{\rm G}$  the following thus holds for  $t \ge 0$:		*The  "$1$"  becomes a Dirac delta function.  Considering  $T = 2\pi \cdot f_{\rm G}$  the following thus holds for  $t \ge 0$:
	:$$h_1(t) = \delta(t) - {1}/{T} \cdot {\rm e}^{-t/T} \hspace{0.05cm}.$$		:$$h_1(t) = \delta(t) - {1}/{T} \cdot {\rm e}^{-t/T} \hspace{0.05cm}.$$
	*In contrast,   $h_1(t)= 0$ holds for  $t< 0$ , which would prove causality.		*In contrast,  $h_1(t)= 0$ holds for  $t< 0$,  which would prove causality.


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	'''(4)'''  <u>The first two proposed solutions</u> are correct:		'''(4)'''  <u>The first two proposed solutions</u>  are correct:
	*Since the impulse response is  $h_1(t) = 0$  for  $t < 0$ , the convolution operation  $h_2(t) = h_1(t) \star h_1(t)$  also satisfies the causality condition.  Similarly, the  $n$–fold convolution yields a causal impulse response:   $h_n(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm}		*Since the impulse response is  $h_1(t) = 0$  for  $t < 0$,  the convolution operation  $h_2(t) = h_1(t) \star h_1(t)$  also satisfies the causality condition.
			*Similarly, the  $n$–fold convolution yields a causal impulse response:   $h_n(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm}
	t<0 \hspace{0.05cm}.$		t<0 \hspace{0.05cm}.$
	*However, the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .  Thus, considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:		*However,  the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .
			*Thus,  considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:
	:$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad		:$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad
	\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad		\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad

Guenter at 14:35, 9 October 2021

2021-10-09T14:35:04Z

← Older revision		Revision as of 16:35, 9 October 2021
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	Please note:		Please note:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem\|Conclusions from the Allocation Theorem]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem\|Conclusions from the Allocation Theorem]].
	*Reference is also made to the theory pages  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#~~Hilbert transformation~~\|Hilbert transformation]]  and  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition\|Partial fraction decomposition]].		*Reference is also made to the theory pages  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert_transform\|Hilbert transformation]]  and  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition\|Partial fraction decomposition]].

Guenter at 14:33, 9 October 2021

2021-10-09T14:33:59Z

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	\hspace{0.05cm},$$		\hspace{0.05cm},$$

	where  $f_{\rm G}$  represents the 3dB cut-off frequency:		where  $f_{\rm G}$  represents the  $\rm 3dB$  cut-off frequency:
	:$$f_{\rm G} = \frac{R}{2 \pi \cdot L}		:$$f_{\rm G} = \frac{R}{2 \pi \cdot L}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	By cascading  $n$  two-port networks  $H_1(f)$  built the same way, the transfer function		By cascading  $n$  two-port networks  $H_1(f)$  built in the same way, the following transfer function is obtained:
	:$$H_n(f) = \big [H_1(f)\big ]^n =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^n}{\big [1+{\rm j}\cdot f/f_{\rm G}\big ]^n}		:$$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,  but this is not important for solving this exercise.
	*The lower graph shows for example the realization of the transfer function  $H_2(f)$.		*The lower graph shows for example the realization of the transfer function  $H_2(f)$.


	In this exercise, such a two-port network is considered with respect to its causality properties.		In this exercise,  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  $H(f)$  satisfy the  [[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,  the real and imaginary parts of the spectral function  $H(f)$  satisfy the  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation\|Hilbert transformation]],  which is expressed by the following abbreviation:
	:$${\rm Im} \left\{ H(f) \right \} \quad		:$${\rm Im} \left\{ H(f) \right \} \quad
	\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad		\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad
	{\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$		{\rm Re} \left\{ H(f) \right \}\hspace{0.05cm}.$$

	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  $x$ , 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,  the correspondence is often expressed by the general variable  $x$,  which is to be interpreted - depending on the application - as normalized frequency or as normalized time.


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			Please note:


	''Please note:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem\|Conclusions from the Allocation Theorem]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem\|Conclusions from the Allocation Theorem]].
	*Reference is also made to the theory pages  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation\|Hilbert transformation]]  and  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition\|Partial fraction decomposition]].		*Reference is also made to the theory pages  [[Linear_and_Time_Invariant_Systems/Conclusions_from_the_Allocation_Theorem#Hilbert transformation\|Hilbert transformation]]  and  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial fraction decomposition\|Partial fraction decomposition]].
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	{Compute the transfer function  $H_2(f)$.  What is the complex value for  $f = f_{\rm G}$?		{Compute the transfer function  $H_2(f)$.   What is the complex value for  $f = f_{\rm G}$?
	\|type="{}"}		\|type="{}"}
	${\rm Re}\big[H_2(f = f_{\rm G})\big] \ = \ $ { 0. }		${\rm Re}\big[H_2(f = f_{\rm G})\big] \ = \ $ { 0. }

Oezer at 07:22, 24 September 2021

2021-09-24T07:22:33Z

← Older revision		Revision as of 09:22, 24 September 2021
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	'''(4)'''  <u>The first two proposed solutions</u> are correct:		'''(4)'''  <u>The first two proposed solutions</u> are correct:
	*Since the impulse response is  $h_1(t) = 0$  for  $t < 0$ , the convolution operation  $h_2(t) = h_1(t) \star h_1(t)$  also satisfies the causality condition.  Similarly, the  $n$–fold convolution yields a causal impulse response:   $h_n(t) = 0 \hspace{0.2cm}{\rm{~~f\ddot{u}r~~}} \hspace{0.2cm}		*Since the impulse response is  $h_1(t) = 0$  for  $t < 0$ , the convolution operation  $h_2(t) = h_1(t) \star h_1(t)$  also satisfies the causality condition.  Similarly, the  $n$–fold convolution yields a causal impulse response:   $h_n(t) = 0 \hspace{0.2cm}{\rm{for}} \hspace{0.2cm}
	t<0 \hspace{0.05cm}.$		t<0 \hspace{0.05cm}.$
	*However, the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .  Thus, considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:		*However, the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .  Thus, considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:

Oezer at 07:20, 24 September 2021

2021-09-24T07:20:17Z

← Older revision		Revision as of 09:20, 24 September 2021
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	:$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}		:$$y(t) = h(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
	t<0 \hspace{0.05cm}.$$		t<0 \hspace{0.05cm}.$$
	*Formally, this can be shown as follows:   The high-pass~~–~~transfer function  $H_1(f)$  can be rearranged as follows:		*Formally, this can be shown as follows:   The high-pass transfer function  $H_1(f)$  can be rearranged as follows:
	:$$H_1(f) = \frac{{\rm j}\cdot f/f_{\rm G}}{1+{\rm j}\cdot f/f_{\rm G}}		:$$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}}		= 1- \frac{1}{1+{\rm j}\cdot f/f_{\rm G}}

Oezer at 07:18, 24 September 2021

2021-09-24T07:18:35Z

← Older revision		Revision as of 09:18, 24 September 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Proposed solution 2</u> is ~~cocrrect~~:		'''(1)'''  <u>Proposed solution 2</u> is correct:
	*The given transfer function can be computed according to the voltage divider principle.   The following holds:		*The given transfer function can be computed according to the voltage divider principle.   The following holds:
	:$$H_1(f = 0) = 0, \hspace{0.2cm}H_1(f \rightarrow \infty) = 1$$		:$$H_1(f = 0) = 0, \hspace{0.2cm}H_1(f \rightarrow \infty) = 1$$
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	'''(2)'''  <u>Yes</u> is ~~cocrrect~~:		'''(2)'''  <u>Yes</u> is correct:
	*Every real network is causal.  The impulse response  $h(t)$  is equal to the output signal  $y(t)$ if at time  $t= 0$  an extremely short impulse – a so-called Dirac impulse – is applied to the input.		*Every real network is causal.  The impulse response  $h(t)$  is equal to the output signal  $y(t)$ if at time  $t= 0$  an extremely short impulse – a so-called Dirac impulse – is applied to the input.
	*Then, a signal of course cannot occur at the output already for times  $t< 0$  for causality reasons:		*Then, a signal of course cannot occur at the output already for times  $t< 0$  for causality reasons:

Oezer at 07:16, 24 September 2021

2021-09-24T07:16:24Z

← Older revision		Revision as of 09:16, 24 September 2021
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	'''(3)'''  ~~Die Hintereinanderschaltung zweier Hochpässe führt zu folgender Übertragungsfunktion~~:		'''(3)'''  The series connection of two high-pass filters results in the following transfer function:
	:$$H_2(f) = \big [H_1(f)\big ]^2 =\frac{\big [{\rm j}\cdot f/f_{\rm G}\big ]^2}{\big [1+{\rm j}\cdot f/f_{\rm G}\big ]^2}		:$$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}		=\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}.$$		{\big [1+(f/f_{\rm G})^2 \big ]^2}\hspace{0.05cm}.$$

	*~~Mit~~  $f = f_{\rm G}$  ~~folgt daraus~~:		*With  $f = f_{\rm G}$  from this it follows that:
	:$$H_2(f = f_{\rm G}) = \frac{1 - 1 +{\rm j}\cdot 2}		:$$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}		{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)'''  ~~Richtig sind~~ <u>~~die beiden ersten Lösungsvorschläge~~</u>:		'''(4)'''  <u>The first two proposed solutions</u> are correct:
	* ~~Da für~~  $t < 0$  ~~die Impulsantwort~~  $~~h_1(~~t~~) =~~ 0$  ~~ist~~, ~~erfüllt auch die Faltungsoperation~~  $h_2(t) = h_1(t) \star h_1(t)$  ~~die Kausalitätsbedingung~~.  ~~Ebenso ergibt die~~  $n$–~~fache Faltung eine kausale Impulsantwort~~:   $h_n(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}		*Since the impulse response is  $h_1(t) = 0$  for  $t < 0$ , the convolution operation  $h_2(t) = h_1(t) \star h_1(t)$  also satisfies the causality condition.  Similarly, the  $n$–fold convolution yields a causal impulse response:   $h_n(t) = 0 \hspace{0.2cm}{\rm{f\ddot{u}r}} \hspace{0.2cm}
	t<0 \hspace{0.05cm}.$		t<0 \hspace{0.05cm}.$
	*~~Bei kausaler Impulsantwort~~  $~~h_2~~(t)$  ~~hängen aber der Real– und der Imaginärteil der Spektralfunktion~~  $~~H_2~~(f)$  ~~über die Hilbert–Transformation zusammen~~.  ~~Mit der Abkürzung~~  $x = f/f_{\rm G}$  ~~und dem Ergebnis der Teilaufgabe~~  '''(3)'''  ~~gilt somit~~:		*However, the real and imaginary parts of the spectral function  $H_2(f)$  are related via the Hilbert transformation for a causal impulse response  $h_2(t)$ .  Thus, considering the shortcut  $x = f/f_{\rm G}$  and the result of the subtask  '''(3)'''  the following holds:
	:$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad		:$$\frac{x^4- x^2}{x^4+2 x^2+1} \quad
	\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad		\bullet\!\!-\!\!\!-\!\!\!-\!\!\hspace{-0.05cm}\rightarrow\quad