Aufgaben:Exercise 3.5Z: Application of the Residue Theorem - Revision history

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	'''(2)'''  Considering  $Y_{\rm L}(p) = 2/(p+1)$  it follows from the residue theorem with  $I=1$:		'''(2)'''  Considering  $Y_{\rm L}(p) = 2/(p+1)$  it follows from the residue theorem with  $I=1$:
	:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= 2 \cdot {\~~rme~~}^{- \hspace{0.05cm}t}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y(t=1)=\frac{2}{\rm e} \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (purely\hspace{0.15cm}real)}}\hspace{0.05cm} .$$		:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= 2 \cdot {\rm e}^{- \hspace{0.05cm}t}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y(t=1)=\frac{2}{\rm e} \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (purely\hspace{0.15cm}real)}}\hspace{0.05cm} .$$


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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.~~5_Anwendung_des_Residuensatzes~~]]		[[de:Aufgaben:Aufgabe 3.5Z: Anwendung des Residuensatzes]]

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 [[File:P_ID1783__LZI_Z_3_5_d.png|right|frame|Signal curve of configuration&nbsp; $\rm E$]]
 '''(4)'''&nbsp; Now &nbsp;$I=2$ holds.&nbsp;  The residuals of &nbsp;$p_{x1}$&nbsp; and &nbsp;$p_{x2}$&nbsp; yield:
-:$$y_1(t) =
+:$$y_1(t) =\frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot  {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot\hspace{0.05cm}t}\bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}=\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot  {\rm e}^{\hspace{0.05cm}p_{{\rm x}1}\hspace{0.05cm}\cdot\hspace{0.05cm}t}\hspace{0.05cm} ,$$:$$ y_2(t) =\frac {K } { p_{{\rm x}2}-p_{{\rm x}1}} \cdot  {\rm e}^{\hspace{0.05cm}p_{{\rm x}2}\hspace{0.05cm}\cdot\hspace{0.05cm}t}=-\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot  {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot\hspace{0.05cm}t}$$:$$y(t)= y_1(t)+y_2(t) =\frac {2 \cdot {\rm e}^{\hspace{0.05cm}-0.2\hspace{0.08cm}\cdot\hspace{0.05cm}t}}{{\rm j} \cdot 3 \pi} \cdot \big [ \cos(.) + {\rm j} \cdot \sin(.)- \cos(.) + {\rm j} \cdot \sin(.)\big ]$$:$$\Rightarrow\hspace{0.3cm}y(t)=\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2\hspace{0.08cm}\cdot\hspace{0.05cm}t}\cdot  \sin(1.5\pi \cdot t)$$:$$\Rightarrow\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2\hspace{0.08cm}\cdot\hspace{0.05cm}t}  \hspace{0.15cm}\underline{= -0.347}\hspace{0.05cm} .$$
 The graph shows the&nbsp; (purely real)&nbsp; signal curve&nbsp; $y(t)$&nbsp; for this configuration.

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	[[File:P_ID1782__LZI_Z_3_5_c.png\|right\|frame\|Complex signals at a single complex pole]]		[[File:P_ID1782__LZI_Z_3_5_c.png\|right\|frame\|Complex signals at a single complex pole]]
	'''(3)'''  Using the same procedure as in subtask  '''(2)'''  the following is obtained:		'''(3)'''  Using the same procedure as in subtask  '''(2)'''  the following is obtained:
	:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+		:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}= 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2 \hspace{0.08cm}\cdot\hspace{0.05cm}t}\cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.08cm}\cdot \hspace{0.05cm}1.5\pi\hspace{0.05cm}\cdot\hspace{0.05cm}t}\hspace{0.05cm} .$$Due to the second term,  it is a complex signal whose phase rotates in the mathematically positive direction  (counterclockwise) .For time  $t=1$,  the following holds::$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [\cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi)\big ]= - {\rm j} \cdot 1.638$$:$$\Rightarrow\hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}\hspace{0.05cm} .$$
	\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}
	= 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2 \hspace{0.08cm}\cdot
	\hspace{0.05cm}t}\cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.08cm}\cdot \hspace{0.05cm}1.5
	\pi\hspace{0.05cm}\cdot
	\hspace{0.05cm}t}
	\hspace{0.05cm} .$$Due to the second term,  it is a complex signal whose phase rotates in the mathematically positive direction  (counterclockwise) .For time  $t=1$,  the following holds::$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [
	\cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi)
	\big ]= - {\rm j} \cdot 1.638$$:$$\Rightarrow
	\hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}
	\hspace{0.05cm} .$$
	*The left graph shows the signal for a pole at  $p_x = -2 + {\rm j} \cdot 1.5 \pi$.		*The left graph shows the signal for a pole at  $p_x = -2 + {\rm j} \cdot 1.5 \pi$.
	*The right graph shows the conjugate complex signal to it can be seen for  $p_x = -2 - {\rm j} \cdot 1.5 \pi$.		*The right graph shows the conjugate complex signal to it can be seen for  $p_x = -2 - {\rm j} \cdot 1.5 \pi$.

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	In this case:		In this case:
	:$$y(t) = \sum_{i=1}^{I} \left \{		:$$y(t) = \sum_{i=1}^{I} \left \{Y_{\rm L}(p)\cdot (p - p_{{\rm x}i})\cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right\} \hspace{0.05cm}.$$
	Y_{\rm L}(p)\cdot (p - p_{{\rm x}i})\cdot {\rm e}^{\hspace{0.05cm}p
	\hspace{0.05cm}t}
	\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right
	\} \hspace{0.05cm}.$$
	$I$  indicates the number of distinguishable poles;    $I = N$ holds  for all given constellations.		$I$  indicates the number of distinguishable poles;    $I = N$ holds  for all given constellations.

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	*This condition is not met for the configurations  $\rm B$,  $\rm D$ and  $\rm F$.		*This condition is not met for the configurations  $\rm B$,  $\rm D$ and  $\rm F$.
	*First,  a partial fraction decomposition must be made here,  for example for configuration  $\rm B$  with  $p_x = -1$:		*First,  a partial fraction decomposition must be made here,  for example for configuration  $\rm B$  with  $p_x = -1$:
	:$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1}		:$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1}\hspace{0.05cm} .$$
	\hspace{0.05cm} .$$



	'''(2)'''  Considering  $Y_{\rm L}(p) = 2/(p+1)$  it follows from the residue theorem with  $I=1$:		'''(2)'''  Considering  $Y_{\rm L}(p) = 2/(p+1)$  it follows from the residue theorem with  $I=1$:
	:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}		:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t}\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= 2 \cdot {\rme}^{- \hspace{0.05cm}t}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y(t=1)=\frac{2}{\rm e} \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (purely\hspace{0.15cm}real)}}\hspace{0.05cm} .$$
	\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= 2 \cdot {\rm
	e}^{- \hspace{0.05cm}t}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y(t=1)
	=\frac{2}{\rm e} \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (purely\hspace{0.15cm}real)}}
	\hspace{0.05cm} .$$


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	\pi\hspace{0.05cm}\cdot		\pi\hspace{0.05cm}\cdot
	\hspace{0.05cm}t}		\hspace{0.05cm}t}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$Due to the second term,  it is a complex signal whose phase rotates in the mathematically positive direction  (counterclockwise) .For time  $t=1$,  the following holds::$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [
	*Due to the second term,  it is a complex signal whose phase rotates in the mathematically positive direction  (counterclockwise) .
	*For time  $t=1$,  the following holds:
	:$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [
	\cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi)		\cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi)
	\big ]= - {\rm j} \cdot 1.638$$		\big ]= - {\rm j} \cdot 1.638$$:$$\Rightarrow
	:$$\Rightarrow
	\hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}		\hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
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	\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}1}\hspace{0.05cm}\cdot		\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}1}\hspace{0.05cm}\cdot
	\hspace{0.05cm}t}		\hspace{0.05cm}t}
	\hspace{0.05cm} ,$$		\hspace{0.05cm} ,$$:$$ y_2(t) =
	:$$ y_2(t) =
	\frac {K } { p_{{\rm x}2}-p_{{\rm x}1}} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}2}\hspace{0.05cm}\cdot		\frac {K } { p_{{\rm x}2}-p_{{\rm x}1}} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}2}\hspace{0.05cm}\cdot
	\hspace{0.05cm}t}=		\hspace{0.05cm}t}=
	-\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot		-\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot
	\hspace{0.05cm}t}$$		\hspace{0.05cm}t}$$:$$y(t)= y_1(t)+y_2(t) =
	:$$y(t)= y_1(t)+y_2(t) =
	\frac {2 \cdot {\rm e}^{\hspace{0.05cm}-0.2		\frac {2 \cdot {\rm e}^{\hspace{0.05cm}-0.2
	\hspace{0.08cm}\cdot		\hspace{0.08cm}\cdot
	\hspace{0.05cm}t}}{{\rm j} \cdot 3 \pi} \cdot \big [ \cos(.) + {\rm j} \cdot \sin(.)		\hspace{0.05cm}t}}{{\rm j} \cdot 3 \pi} \cdot \big [ \cos(.) + {\rm j} \cdot \sin(.)
	- \cos(.) + {\rm j} \cdot \sin(.)\big ]$$		- \cos(.) + {\rm j} \cdot \sin(.)\big ]$$:$$\Rightarrow
	:$$\Rightarrow
	\hspace{0.3cm}y(t)=		\hspace{0.3cm}y(t)=
	\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2		\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
	\hspace{0.08cm}\cdot		\hspace{0.08cm}\cdot
	\hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$		\hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$:$$\Rightarrow
	:$$\Rightarrow
	\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2		\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
	\hspace{0.08cm}\cdot		\hspace{0.08cm}\cdot

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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:3.5_Anwendung_des_Residuensatzes]]

Guenter at 10:12, 10 November 2021

2021-11-10T10:12:56Z

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	In the case that the number  $Z$  of zeros is less than the number  $N$  of poles,  the corresponding time signal  $y(t)$  can be determined directly by applying the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]] .		In the case that the number  $Z$  of zeros is less than the number  $N$  of poles,  the corresponding time signal  $y(t)$  can be determined directly by applying the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]] .

	In this case,		In this case:
	:$$y(t) = \sum_{i=1}^{I} \left \{		:$$y(t) = \sum_{i=1}^{I} \left \{
	Y_{\rm L}(p)\cdot (p - p_{{\rm x}i})\cdot {\rm e}^{\hspace{0.05cm}p		Y_{\rm L}(p)\cdot (p - p_{{\rm x}i})\cdot {\rm e}^{\hspace{0.05cm}p
	\hspace{0.05cm}t}		\hspace{0.05cm}t}
	\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right		\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right
	\} \hspace{0.05cm}$$ ~~apllies.~~		\} \hspace{0.05cm}.$$
	$I$  indicates the number of distinguishable poles;    $I = N$ holds  for all given constellations.		$I$  indicates the number of distinguishable poles;    $I = N$ holds  for all given constellations.

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Suggested solutions 2, 4 and 6</u> are correct:		'''(1)'''  <u>Suggested solutions 2,  4  and  6</u>  are correct:
	*The prerequisite for the application of the residue theorem is that there are fewer zeros than poles, that is,  $Z < N$  must hold.		*The prerequisite for the application of the residue theorem is that there are fewer zeros than poles,  that is,  $Z < N$  must hold.
	*This condition is not met for the configurations  $\rm B$,  $\rm D$ and  $\rm F$.		*This condition is not met for the configurations  $\rm B$,  $\rm D$ and  $\rm F$.
	*First, a partial fraction decomposition must be made here, for example for configuration  $\rm B$  with  $p_x = -1$:		*First,  a partial fraction decomposition must be made here,  for example for configuration  $\rm B$  with  $p_x = -1$:
	:$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1}		:$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
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	'''(3)'''  Using the same procedure as in subtask  '''(2)'''  the following is obtained ~~now~~:		[[File:P_ID1782__LZI_Z_3_5_c.png\|right\|frame\|Complex signals at a single complex pole]]
			'''(3)'''  Using the same procedure as in subtask  '''(2)'''  the following is obtained:
	:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+		:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+
	\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}		\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}
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	\hspace{0.05cm}t}		\hspace{0.05cm}t}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
	*Due to the second term, it is a complex signal whose phase rotates in the mathematically positive direction  (counterclockwise) .		*Due to the second term,  it is a complex signal whose phase rotates in the mathematically positive direction  (counterclockwise) .
	*For time  $t=1$,  the following holds:		*For time  $t=1$,  the following holds:
	:$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [		:$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [
	\cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi)		\cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi)
	\big ]= - {\rm j} \cdot 1.638~~\hspace{0.3cm}~~\Rightarrow		\big ]= - {\rm j} \cdot 1.638$$
			:$$\Rightarrow
	\hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}		\hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$
	*The left graph shows the ~~complex~~ signal for a pole at  $p_x = -2 + {\rm j} \cdot 1.5 \pi$.  ~~On the~~ right the conjugate complex signal to it can be seen for  $p_x = -2 - {\rm j} \cdot 1.5 \pi$.		*The left graph shows the signal for a pole at  $p_x = -2 + {\rm j} \cdot 1.5 \pi$.
			*The right graph shows the conjugate complex signal to it can be seen for  $p_x = -2 - {\rm j} \cdot 1.5 \pi$.

	~~[[File:P_ID1782__LZI_Z_3_5_c.png\|center\|frame\|Complex signals at a pole]]~~


	'''(4)'''  Now  $I=2$ holds. The residuals of  $p_{x1}$  and  $p_{x2}$  yield:		[[File:P_ID1783__LZI_Z_3_5_d.png\|right\|frame\|Signal curve of configuration  $\rm E$]]
			'''(4)'''  Now  $I=2$ holds.  The residuals of  $p_{x1}$  and  $p_{x2}$  yield:
	:$$y_1(t) =		:$$y_1(t) =
	\frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot		\frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot
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	-\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot		-\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot
	\hspace{0.05cm}t}$$		\hspace{0.05cm}t}$$
	:$$~~\Rightarrow~~		:$$y(t)= y_1(t)+y_2(t) =
	~~\hspace{0.3cm}~~y(t)= y_1(t)+y_2(t) =
	\frac {2 \cdot {\rm e}^{\hspace{0.05cm}-0.2		\frac {2 \cdot {\rm e}^{\hspace{0.05cm}-0.2
	\hspace{0.08cm}\cdot		\hspace{0.08cm}\cdot
	\hspace{0.05cm}t}}{{\rm j} \cdot 3 \pi} \cdot \big [ \cos(.) + {\rm j} \cdot \sin(.)		\hspace{0.05cm}t}}{{\rm j} \cdot 3 \pi} \cdot \big [ \cos(.) + {\rm j} \cdot \sin(.)
	- \cos(.) + {\rm j} \cdot \sin(.)\big ]=		- \cos(.) + {\rm j} \cdot \sin(.)\big ]$$
			:$$\Rightarrow
			\hspace{0.3cm}y(t)=
	\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2		\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
	\hspace{0.08cm}\cdot		\hspace{0.08cm}\cdot
	\hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$		\hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$
	~~[[File:P_ID1783__LZI_Z_3_5_d.png\|right\|frame\|Signal curve of configuration $\rm E$]]~~
	:$$\Rightarrow		:$$\Rightarrow
	\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2		\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	The graph shows the (purely real) signal curve  $y(t)$  for this configuration.		The graph shows the  (purely real)  signal curve  $y(t)$  for this configuration.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 09:57, 10 November 2021

2021-11-10T09:57:46Z

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	}}		}}

	[[File:P_ID1781__LZI_Z_3_5.png\|right\|frame\|Six ~~different <br>~~pole–zero configurations]]		[[File:P_ID1781__LZI_Z_3_5.png\|right\|frame\|Six pole–zero configurations]]
	Let the spectral function  $Y_{\rm L}(p)$  be given in pole–zero notation characterized by		Let the spectral function  $Y_{\rm L}(p)$  be given in pole–zero notation characterized by
	*$Z$  zeros  $p_{{\rm o}i}$,		*$Z$  zeros  $p_{{\rm o}i}$,
Line 10:		Line 10:


	In the following, the configurations shown in the diagram are considered. Let always  $K= 2$ hold.		In the following,  the configurations shown in the diagram are considered.  Let always  $K= 2$ hold.

	In the case that the number  $Z$  of zeros is less than the number  $N$  of poles, the corresponding time signal  $y(t)$  can be determined directly by applying the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]] .		In the case that the number  $Z$  of zeros is less than the number  $N$  of poles,  the corresponding time signal  $y(t)$  can be determined directly by applying the  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem\|residue theorem]] .

	In this case,		In this case,
Line 20:		Line 20:
	\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right		\bigg \|_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right
	\} \hspace{0.05cm}$$ apllies.		\} \hspace{0.05cm}$$ apllies.
	$I$  indicates the number of distinguishable poles;    $I = N$ holds for all given constellations.		$I$  indicates the number of distinguishable poles;    $I = N$ holds  for all given constellations.


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			Please note:
			*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform\|Inverse Laplace Transform]].
			*If the time signal  $y(t)$  is complex,  then  $Y_{\rm L}(p)$  cannot be realized as a circuit.  However, the application of the residue theorem is still possible.
	''Please note:''		*The complex frequency  $p$,  the zeros  $p_{{\rm o}i}$  as well as the poles  $p_{{\rm x}i}$  each describe normalized quantities without units in this exercise.
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform\|Inverse Laplace Transform]].		*Thus,  time  $t$  is dimensionless, too.

	*If the time signal  $y(t)$  is complex, then  $Y_{\rm L}(p)$  cannot be realized as a circuit.  However, the application of the residue theorem is still possible.
	*The complex frequency  $p$, the zeros  $p_{{\rm o}i}$  as well as the poles  $p_{{\rm x}i}$  each describe normalized quantities without units in this exercise.
	*Thus, time  $t$  is dimensionless, too.


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	<quiz display=simple>		<quiz display=simple>
	{For which configurations can the residue theorem <u>not be applied directly</u>?		{For which configurations can the residue theorem  <u>not be applied directly</u>?
	\|type="[]"}		\|type="[]"}
	- Configuration  $\rm A$,		- Configuration  $\rm A$,
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	{Compute  $y(t)$  for configuration  $\rm A$ with  $K= 2$  and  $p_{\rm x} = -1$.  What is the numerical value for time  $t = 1$?		{Compute  $y(t)$  for configuration  $\rm A$  with  $K= 2$  and  $p_{\rm x} = -1$.  What is the numerical value for time  $t = 1$?
	\|type="{}"}		\|type="{}"}
	$\ {\rm Re}\{y(t = 1)\} \ = \ $ { 0.736 3% }		$\ {\rm Re}\{y(t = 1)\} \ = \ $ { 0.736 3% }
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	{Compute  $y(t)$  for configuration  $\rm C$ with  $K= 2$  and  $p_{\rm x} = -0.2 + {\rm j} \cdot 1.5\pi$.  What numerical value is obtained for time  $t = 1$?		{Compute  $y(t)$  for configuration  $\rm C$  with  $K= 2$  and  $p_{\rm x} = -0.2 + {\rm j} \cdot 1.5\pi$.  What numerical value is obtained for time  $t = 1$?
	\|type="{}"}		\|type="{}"}
	$\ {\rm Re}\{y(t = 1)\} \ = \ $ { 0. }		$\ {\rm Re}\{y(t = 1)\} \ = \ $ { 0. }
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	{What signal value  $y(t = 1)$  is obtained for the constellation  $\rm E$ with  $K= 2$  and two poles at  $p_{\rm x} = -0.2 \pm {\rm j} \cdot 1.5\pi$?		{What signal value  $y(t = 1)$  is obtained for the constellation  $\rm E$  with  $K= 2$  and two poles at  $p_{\rm x} = -0.2 \pm {\rm j} \cdot 1.5\pi$?
	\|type="{}"}		\|type="{}"}
	$\ {\rm Re}\{y(t = 1)\} \ = \ $ { -0.357--0.337 }		$\ {\rm Re}\{y(t = 1)\} \ = \ $ { -0.357--0.337 }

Oezer at 21:22, 19 October 2021

2021-10-19T21:22:34Z

← Older revision		Revision as of 23:22, 19 October 2021
Line 92:		Line 92:


	'''(3)'''  Using the same procedure as in ~~the~~ subtask  '''(2)'''  the following is obtained now:		'''(3)'''  Using the same procedure as in subtask  '''(2)'''  the following is obtained now:
	:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+		:$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+
	\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}		\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}

Oezer at 21:20, 19 October 2021

2021-10-19T21:20:06Z

← Older revision		Revision as of 23:20, 19 October 2021
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	*The prerequisite for the application of the residue theorem is that there are fewer zeros than poles, that is,  $Z < N$  must hold.		*The prerequisite for the application of the residue theorem is that there are fewer zeros than poles, that is,  $Z < N$  must hold.
	*This condition is not met for the configurations  $\rm B$,  $\rm D$ and  $\rm F$.		*This condition is not met for the configurations  $\rm B$,  $\rm D$ and  $\rm F$.
	*First, a partial fraction decomposition must be made here, for example for ~~the~~ configuration  $\rm B$  with  $p_x = -1$:		*First, a partial fraction decomposition must be made here, for example for configuration  $\rm B$  with  $p_x = -1$:
	:$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1}		:$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1}
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

Oezer at 20:49, 19 October 2021

2021-10-19T20:49:41Z

← Older revision		Revision as of 22:49, 19 October 2021
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	'''(4)'''  ~~Nun gilt~~  $I=2$. ~~Die Residien von~~  $p_{x1}$  ~~bzw.~~  $p_{x2}$  ~~liefern~~:		'''(4)'''  Now  $I=2$ holds. The residuals of  $p_{x1}$  and  $p_{x2}$  yield:
	:$$y_1(t) =		:$$y_1(t) =
	\frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot		\frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot
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	\hspace{0.08cm}\cdot		\hspace{0.08cm}\cdot
	\hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$		\hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$
	[[File:P_ID1783__LZI_Z_3_5_d.png\|right\|frame\|~~Signalverlauf der Konfiguration~~ $\rm E$]]		[[File:P_ID1783__LZI_Z_3_5_d.png\|right\|frame\|Signal curve of configuration $\rm E$]]
	:$$\Rightarrow		:$$\Rightarrow
	\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2		\hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
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	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

	~~Die Grafik zeigt den~~ (~~rein reellen~~) ~~Signalverlauf~~  $y(t)$  ~~für diese Konfiguration~~.		The graph shows the (purely real) signal curve  $y(t)$  for this configuration.

	{{ML-Fuß}}		{{ML-Fuß}}