Aufgaben:Exercise 3.7Z: Partial Fraction Decomposition - Revision history

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]
	[[de:~~Zusatzaufgaben~~:3.~~7_Partialbruchzerlegung~~]]		[[de:Aufgaben:Aufgabe 3.7Z: Partialbruchzerlegung]]

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	Rather,  a  "partial fraction decomposition"  corresponding to		Rather,  a  "partial fraction decomposition"  corresponding to
	:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)		:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)\hspace{0.05cm}$$
	\hspace{0.05cm}$$
	must be made beforehand. Then,		must be made beforehand. Then,
	:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)		:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)\hspace{0.05cm}$$
	\hspace{0.05cm}$$
	holds for the impulse response.		holds for the impulse response.
	$h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$,  where the condition  $Z' < N'$  is satisfied.		$h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$,  where the condition  $Z' < N'$  is satisfied.
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	Furthermore,  in this exercise the  $p$–transfer function		Furthermore,  in this exercise the  $p$–transfer function
	:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}		:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}\hspace{0.05cm}$$
	\hspace{0.05cm}$$
	⇒   "configuration $(5)$" will be examined in more detail,  which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.		⇒   "configuration $(5)$" will be examined in more detail,  which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.

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	'''(2)'''  The  <u> suggested solution 4</u>  is correct:		'''(2)'''  The  <u> suggested solution 4</u>  is correct:
	*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:		*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
	:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}		:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}=\frac{p/A}{{p/A}+2+ {A/p}}= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2}= H_{\rm L}^{(4)}(p)\hspace{0.05cm}.$$
	=\frac{p/A}{{p/A}+2+ {A/p}}
	= \hspace{0.2cm}\frac{p^2}{p^2 + 2A \cdot p + A^2} = \frac{p^2}{(p+A)^2
	}= H_{\rm L}^{(4)}(p)
	\hspace{0.05cm}.$$
	*The double zero is at  $p_{\rm o} = 0$  and the double pole at  $p_{\rm x} = -A = -2$.		*The double zero is at  $p_{\rm o} = 0$  and the double pole at  $p_{\rm x} = -A = -2$.

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	'''(3)'''  The following holds for configuration  $(1)$:		'''(3)'''  The following holds for configuration  $(1)$:
	:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)		:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)=2}\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\hspace{0.15cm}\underline{H_{\rm L}\hspace{-0.05cm}'(p =0)
	=2}
	\hspace{0.05cm}.$$



	'''(4)'''  Similarly,  the following is obtained for configuration  $(2)$:		'''(4)'''  Similarly,  the following is obtained for configuration  $(2)$:
	:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=		:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
	\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=		:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}\hspace{0.05cm}.$$
	\hspace{0.2cm}\frac{p^2 +4\cdot p +8 -8\cdot p}{p^2 +4\cdot p
	+8} =1- \frac{8\cdot p}{p^2 +4\cdot p +8}=1- H_{\rm L}\hspace{-0.05cm}'(p)$$
	:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 8
	\cdot \frac{p}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
	\hspace{0.05cm}.$$

	Thus,  the  <u> suggested solutions 2 and 3</u>  are correct in contrast to statement 1:		Thus,  the  <u> suggested solutions 2 and 3</u>  are correct in contrast to statement 1:
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	'''(5)'''  The following applies for configuration  $(3)$ :		'''(5)'''  The following applies for configuration  $(3)$ :
	:$$H_{\rm L}(p) =		:$$H_{\rm L}(p) =\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
	\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}		:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}\hspace{0.05cm}.$$
	= 1- H_{\rm L}\hspace{-0.05cm}'(p)$$
	:$$\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = 4
	\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
	\hspace{0.05cm}.$$
	*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.		*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
	*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only   <u> suggested solution 2</u>   is correct here.		*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only   <u> suggested solution 2</u>   is correct here.
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	'''(6)'''  Finally,  the following holds for configuration  $(4)$:		'''(6)'''  Finally,  the following holds for configuration  $(4)$:
	:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}		:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4\cdot \frac{p+1}{(p+2)^2}\hspace{0.05cm}.$$
	= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{0.05cm}'(p) = 4
	\cdot \frac{p+1}{(p+2)^2}
	\hspace{0.05cm}.$$
	<u>Suggested solution 2</u>  is correct here.  In general,  it can be said that:		<u>Suggested solution 2</u>  is correct here.  In general,  it can be said that:
	*The partial fraction decomposition changes the number and position of the zeros.		*The partial fraction decomposition changes the number and position of the zeros.

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^3.3 Inverse Laplace Transform^]]
			[[de:Zusatzaufgaben:3.7_Partialbruchzerlegung]]

Guenter at 13:55, 25 January 2022

2022-01-25T13:55:58Z

← Older revision		Revision as of 15:55, 25 January 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The <u> suggested solutions 1 and 2</u> are correct:		'''(1)'''  The  <u> suggested solutions 1 and 2</u>  are correct:
	*According to the criteria given in exercise 3.4Z, there is always an all-pass filter at hand if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.		*According to the criteria given in exercise 3.4Z,  there is always an all-pass filter at hand <br>if there is a corresponding zero  $p_{\rm o} = + A + {\rm j} \cdot B$  in the right $p$–half-plane for each pole  $p_{\rm x} = - A + {\rm j} \cdot B$  in the left half-plane.
	*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $\|H(f)\| = 1$.		*Considering  $K = 1$  the attenuation function is then  $a(f) = 0 \ \rm Np$   ⇒   $\|H(f)\| = 1$.
	*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.		*The following can be seen from the graph on the information page:   The configurations  $(1)$ and  $(2)$ satisfy exactly these symmetry properties.
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	'''(2)'''  The <u> suggested solution 4</u> is correct:		'''(2)'''  The  <u> suggested solution 4</u>  is correct:
	*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:		*The transfer function  $H_{\rm L}^{(5)}(p)$  is also described by configuration  $(4)$  as the following calculation shows:
	:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}		:$$H_{\rm L}^{(5)}(p) \hspace{0.25cm} = \hspace{0.2cm} \frac{p/A}{(\sqrt{p/A}+\sqrt{A/p})^2}
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	}= H_{\rm L}^{(4)}(p)		}= H_{\rm L}^{(4)}(p)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*The double zero is at  $p_{\rm o} = 0$ and the double pole at  $p_{\rm x} = -A = -2$.		*The double zero is at  $p_{\rm o} = 0$  and the double pole at  $p_{\rm x} = -A = -2$.



	'''(3)'''  The following holds for configuration  $(1)$~~ ~~:		'''(3)'''  The following holds for configuration  $(1)$:
	:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)		:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}
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	'''(4)'''  Similarly, the following is obtained for configuration  $(2)$:		'''(4)'''  Similarly,  the following is obtained for configuration  $(2)$:
	:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=		:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
	\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=		\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Thus, the <u> suggested solutions 2 and 3</u> are correct in contrast to statement 1:		Thus,  the  <u> suggested solutions 2 and 3</u>  are correct in contrast to statement 1:
	* While  $H_{\rm L}(p)$  has two conjugate complex zeros,		* While  $H_{\rm L}(p)$  has two conjugate complex zeros,
	* ~~ ~~$H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.		* $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.


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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.		*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
	*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only <u> suggested solution 2</u> is correct here.		*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only   <u> suggested solution 2</u>   is correct here.



	'''(6)'''  Finally, the following holds for configuration  $(4)$:		'''(6)'''  Finally,  the following holds for configuration  $(4)$:
	:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}		:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
	= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}		= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
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	\cdot \frac{p+1}{(p+2)^2}		\cdot \frac{p+1}{(p+2)^2}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	<u>Suggested solution 2</u> is ~~also~~ correct here. In general, it can be said that:		<u>Suggested solution 2</u>  is correct here.  In general,  it can be said that:
	*The partial fraction decomposition changes the number and position of the zeros.		*The partial fraction decomposition changes the number and position of the zeros.
	* On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.		* On the contrary,  the poles of  $H_{\rm L}\hspace{0.01cm}'(p)$  are always identical to those of  $H_{\rm L}(p)$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 13:45, 25 January 2022

2022-01-25T13:45:35Z

Oezer at 10:02, 26 October 2021

2021-10-26T10:02:24Z

← Older revision		Revision as of 12:02, 26 October 2021
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	'''(5)'''  The following applies for ~~Konfiguration~~  $(3)$ :		'''(5)'''  The following applies for configuration  $(3)$ :
	:$$H_{\rm L}(p) =		:$$H_{\rm L}(p) =
	\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}		\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}

Oezer at 10:00, 26 October 2021

2021-10-26T10:00:35Z

← Older revision		Revision as of 12:00, 26 October 2021
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	{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for ~~the~~ configuration  $(2)$.  Which statements are true here?		{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(2)$.  Which statements are true here?
	\|type="[]"}		\|type="[]"}
	- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.		- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
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	{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for ~~the~~ configuration  $(3)$.  Which statements are true here?		{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(3)$.  Which statements are true here?
	\|type="[]"}		\|type="[]"}
	- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.		- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.
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	{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for ~~the~~ configuration  $(4)$.  Which statements are true here?		{Compute  $H_{\rm L}\hspace{0.01cm}'(p)$  for configuration  $(4)$.  Which statements are true here?
	\|type="[]"}		\|type="[]"}
	- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.		- $H_{\rm L}\hspace{0.01cm}'(p)$  has the same zeros as  $H_{\rm L}(p)$.

Oezer at 09:59, 26 October 2021

2021-10-26T09:59:33Z

← Older revision		Revision as of 11:59, 26 October 2021
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	{Which ~~quadripole~~ has the transfer function  $H_{\rm L}^{(5)}(p)$?		{Which two-port network has the transfer function  $H_{\rm L}^{(5)}(p)$?
	\|type="()"}		\|type="()"}
	- Configuration  $(1)$,		- Configuration  $(1)$,

Oezer at 09:57, 26 October 2021

2021-10-26T09:57:00Z

← Older revision		Revision as of 11:57, 26 October 2021
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	'''(4)'''  ~~In gleicher Weise ergibt sich für die Konfiguration~~  $(2)$:		'''(4)'''  Similarly, the following is obtained for configuration  $(2)$:
	:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=		:$$H_{\rm L}(p) =\frac{(p-2 - {\rm j} \cdot 2)(p-2 + {\rm j} \cdot 2)}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}=
	\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=		\frac{p^2 -4\cdot p +8 }{p^2 +4\cdot p +8}=
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Richtig sind also die~~ <u> ~~Lösungsvorschläge~~ 2 ~~und~~ 3</u> ~~im Gegensatz zur Aussage~~ 1:		Thus, the <u> suggested solutions 2 and 3</u> are correct in contrast to statement 1:
	* ~~Während~~  $H_{\rm L}(p)$  ~~zwei konjugiert–komplexe Nullstellen aufweist~~,		* While  $H_{\rm L}(p)$  has two conjugate complex zeros,
	*~~besitzt~~  $H_{\rm L}\hspace{0.01cm}'(p)$  ~~nur eine einzige Nullstelle bei~~  $p_{\rm o}\hspace{0.01cm}' = 0$.		*  $H_{\rm L}\hspace{0.01cm}'(p)$  only has a single zero at  $p_{\rm o}\hspace{0.01cm}' = 0$.




	'''(5)'''  ~~Für die~~ Konfiguration  $(3)$  ~~gilt~~:		'''(5)'''  The following applies for Konfiguration  $(3)$ :
	:$$H_{\rm L}(p) =		:$$H_{\rm L}(p) =
	\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}		\frac{p^2 }{p^2 +4\cdot p +8}=\frac{p^2 +4\cdot p +8 -4\cdot p -8 }{p^2 +4\cdot p +8}
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	\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}		\cdot \frac{p+2}{(p+2 - {\rm j} \cdot 2)(p+2 + {\rm j} \cdot 2)}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*~~Die Nullstelle von~~  $H_{\rm L}\hspace{0.01cm}'(p)$  ~~liegt nun bei~~  $p_{\rm o}\hspace{0.01cm}' = -2$.		*The zero of  $H_{\rm L}\hspace{0.01cm}'(p)$  is now at  $p_{\rm o}\hspace{0.01cm}' = -2$.
	*~~Die Konstante ist~~  $K\hspace{0.01cm}' = 4$   ⇒   ~~richtig ist hier nur der~~ <u> ~~Lösungsvorschlag~~ 2</u>.		*The constant is  $K\hspace{0.01cm}' = 4$   ⇒   only <u> suggested solution 2</u> is correct here.



	'''(6)'''  ~~Schließlich gilt für die Konfiguration~~  $(4)$:		'''(6)'''  Finally, the following holds for configuration  $(4)$:
	:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}		:$$H_{\rm L}(p) = \frac{p^2 }{(p+2)^2}=\frac{p^2 +4\cdot p +4 -4\cdot p -4 }{p^2 +4\cdot p +4}
	= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}		= 1- \frac{4\cdot p +4 }{p^2 +4\cdot p +4}
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	\cdot \frac{p+1}{(p+2)^2}		\cdot \frac{p+1}{(p+2)^2}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	~~Richtig ist auch hier~~ <u>~~der Lösungsvorschlag~~ 2</u>. ~~Allgemein lässt sich sagen~~:		<u>Suggested solution 2</u> is also correct here. In general, it can be said that:
	*~~Durch die Partialbruchzerlegung wird die Anzahl und die Lage der Nullstellen verändert~~.		*The partial fraction decomposition changes the number and position of the zeros.
	*~~Die Pole von~~ $H_{\rm L}\hspace{0.01cm}'(p)$ ~~sind dagegen stets identisch mit denen von~~ $H_{\rm L}(p)$.		* On the contrary, the poles of $H_{\rm L}\hspace{0.01cm}'(p)$ are always identical to those of $H_{\rm L}(p)$.
	{{ML-Fuß}}		{{ML-Fuß}}

Oezer at 07:30, 26 October 2021

2021-10-26T07:30:18Z

← Older revision		Revision as of 09:30, 26 October 2021
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	'''(3)'''  The following holds for ~~the~~ configuration  $(1)$ :		'''(3)'''  The following holds for configuration  $(1)$ :
	:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)		:$$H_{\rm L}(p) =\frac{p-2}{p+2}=\frac{p+2-4}{p+2}= 1 - \frac{4}{p+2}=1- H_{\rm L}\hspace{-0.05cm}'(p)
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm L}\hspace{-0.05cm}'(p) = \frac{4}{p+2}

← Older revision		Revision as of 15:45, 25 January 2022
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	[[File:P_ID1789__LZI_Z_3_7.png\|right\|frame\|Pole-zero diagrams]]		[[File:P_ID1789__LZI_Z_3_7.png\|right\|frame\|Pole-zero diagrams]]
	In the graph, four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$~~ ~~.		In the graph,  four two-port networks are given by their pole–zero diagrams  $H_{\rm L}(p)$.
	* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.		* They all have in common that the number  $Z$  of zeros is equal to the number  $N$  of poles.
	*The constant factor in each case is  $K=1$.		*The constant factor in each case is  $K=1$.


	In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$~~ ~~.		In the special case  $Z = N$  the residue theorem cannot be applied directly to compute the impulse response  $h(t)$.

	Rather, a partial fraction decomposition corresponding to		Rather,  a  "partial fraction decomposition"  corresponding to
	:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)		:$$H_{\rm L}(p) =1- H_{\rm L}\hspace{0.05cm}'(p)
	\hspace{0.05cm}$$		\hspace{0.05cm}$$
	must be made beforehand. Then,		must be made beforehand. Then,
	:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)		:$$h(t) = \delta(t)- h\hspace{0.03cm}'(t)
	\hspace{0.05cm}$$ holds for the impulse response.		\hspace{0.05cm}$$
	$h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$ , where the condition  $Z' < N'$  is satisfied.		holds for the impulse response.
			$h\hspace{0.03cm}'(t)$  is the inverse Laplace transform of  $H_{\rm L}\hspace{0.05cm}'(p)$,  where the condition  $Z' < N'$  is satisfied.

	Two of the four configurations given are so-called ''all-pass filters''.		Two of the four configurations given are so-called  "all-pass filters".
	*This refers to two-port networks for which the Fourier ~~spectral function~~ satisfies the condition  $\|H(f)\| = 1$   ⇒   $a(f) = 0$ .		*This refers to two-port networks for which the Fourier spectrum satisfies the condition  $\|H(f)\| = 1$   ⇒   $a(f) = 0$ .
	*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters\|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.		*In [[Aufgaben:Exercise_3.4Z:_Various_All-Pass_Filters\|Exercise 3.4Z]] it is given how the poles and zeros of such an all-pass filter must be positioned.


	Furthermore, in this exercise the  $p$–transfer function		Furthermore,  in this exercise the  $p$–transfer function
	:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}		:$$H_{\rm L}^{(5)}(p) =\frac{p/A}{\left (\sqrt{p/A}+\sqrt{A/p} \right )^2}
	\hspace{0.05cm}$$		\hspace{0.05cm}$$
	⇒   "configuration $(5)$" will be examined in more detail, which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.		⇒   "configuration $(5)$" will be examined in more detail,  which can be represented by one of the four pole–zero diagrams given in the graph if the parameter  $A$  is chosen correctly.


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	''Please note:''		Please note:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform\|Inverse Laplace Transform]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform\|Inverse Laplace Transform]].
	*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition\|Partial fraction decomposition]].		*In particular, reference is made to the page  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Partial_fraction_decomposition\|Partial fraction decomposition]].
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	{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''. ~~<br>~~Enter the function value for  $p = 0$~~ ~~.		{Compute the function  $H_{\rm L}\hspace{0.01cm}'(p)$  after a partial fraction decomposition for configuration  '''(1)'''.  Enter the function value for  $p = 0$.
	\|type="{}"}		\|type="{}"}
	$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }		$H_{\rm L}\hspace{0.01cm}'(p = 0) \ = \ $ { 2 3% }