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

Guenter at 10:12, 10 November 2021

2021-11-10T10:12:56Z

Guenter at 09:57, 10 November 2021

2021-11-10T09:57:46Z

Oezer at 21:22, 19 October 2021

2021-10-19T21:22:34Z

Oezer at 21:20, 19 October 2021

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Oezer at 20:49, 19 October 2021

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Oezer at 20:46, 19 October 2021

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Oezer at 20:45, 19 October 2021

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Oezer at 11:44, 19 October 2021

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Oezer at 22:05, 18 October 2021

2021-10-18T22:05:47Z

Oezer at 18:50, 18 October 2021

2021-10-18T18:50:23Z

@@ Line 14: / Line 14: @@
 In the case that the number&nbsp; $Z$&nbsp; of zeros is less than the number&nbsp; $N$&nbsp; of poles,&nbsp; the corresponding time signal &nbsp;$y(t)$&nbsp; can be determined directly by applying the&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp;.
-In this case,
+In this case:
 :$$y(t) = \sum_{i=1}^{I} \left \{
   Y_{\rm L}(p)\cdot (p - p_{{\rm x}i})\cdot  {\rm e}^{\hspace{0.05cm}p
   \hspace{0.05cm}t}
   \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right
-  \} \hspace{0.05cm}$$ apllies.
+  \} \hspace{0.05cm}.$$
 $I$&nbsp; indicates the number of distinguishable poles;&nbsp; &nbsp; $I = N$ holds&nbsp; for all given constellations.
@@ Line 70: / Line 70: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Suggested solutions 2, 4 and 6</u> are correct:
+'''(1)'''&nbsp; <u>Suggested solutions 2,&nbsp; 4&nbsp; and&nbsp; 6</u>&nbsp; are correct:
-*The prerequisite for the application of the residue theorem is that there are fewer zeros than poles, that is, &nbsp;$Z < N$&nbsp; must hold.
+*The prerequisite for the application of the residue theorem is that there are fewer zeros than poles,&nbsp; that is, &nbsp;$Z < N$&nbsp; must hold.
 *This condition is not met for the configurations &nbsp;$\rm B$, &nbsp;$\rm D$ and &nbsp;$\rm F$.
-*First, a partial fraction decomposition must be made here, for example for configuration &nbsp;$\rm B$&nbsp; with &nbsp;$p_x =  -1$:
+*First,&nbsp; a partial fraction decomposition must be made here,&nbsp; for example for configuration &nbsp;$\rm B$&nbsp; with &nbsp;$p_x =  -1$:
 :$$Y_{\rm L}(p)=   \frac {p} {p +1}= 1-\frac {1} {p +1}
   \hspace{0.05cm} .$$
@@ Line 88: / Line 88: @@
-'''(3)'''&nbsp; Using the same procedure as in subtask&nbsp; '''(2)'''&nbsp; the following is obtained now:
+[[File:P_ID1782__LZI_Z_3_5_c.png|right|frame|Complex signals at a single complex pole]]
 :$$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}
@@ Line 96: / Line 97: @@
   \hspace{0.05cm}t}
   \hspace{0.05cm} .$$
-*Due to the second term, it is a complex signal whose phase rotates in the mathematically positive direction&nbsp; (counterclockwise)&nbsp;.
+*Due to the second term,&nbsp; it is a complex signal whose phase rotates in the mathematically positive direction&nbsp; (counterclockwise)&nbsp;.
 *For time &nbsp;$t=1$,&nbsp; the following holds:
 :$$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\hspace{0.3cm}\Rightarrow
+  \big ]= - {\rm j} \cdot 1.638$$
   \hspace{0.3cm}{\rm Re}\{y(t = 1)\}  \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm}  {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638}
   \hspace{0.05cm} .$$
-*The left graph shows the complex signal for a pole at &nbsp;$p_x =  -2 + {\rm j} \cdot 1.5 \pi$.&nbsp; On the right the conjugate complex signal to it can be seen for &nbsp;$p_x =  -2 - {\rm j} \cdot 1.5 \pi$.
+*The left graph shows the signal for a pole at &nbsp;$p_x =  -2 + {\rm j} \cdot 1.5 \pi$.&nbsp;
+*The right graph shows the conjugate complex signal to it can be seen for &nbsp;$p_x =  -2 - {\rm j} \cdot 1.5 \pi$.
-'''(4)'''&nbsp; Now &nbsp;$I=2$ holds.  The residuals of &nbsp;$p_{x1}$&nbsp; and &nbsp;$p_{x2}$&nbsp; yield:
+[[File:P_ID1783__LZI_Z_3_5_d.png|right|frame|Signal curve of configuration&nbsp; $\rm E$]]
 :$$y_1(t) =
   \frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot  {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot
@@ Line 122: / Line 124: @@
   -\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot  {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot
   \hspace{0.05cm}t}$$
-:$$\Rightarrow
+:$$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 ]=
+  - \cos(.) + {\rm j} \cdot \sin(.)\big ]$$
   \frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
 \hspace{0.08cm}\cdot
   \hspace{0.05cm}t}\cdot  \sin(1.5\pi \cdot t)$$
 :$$\Rightarrow
 \hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
@@ Line 138: / Line 140: @@
   \hspace{0.05cm} .$$
-The graph shows the (purely real) signal curve&nbsp; $y(t)$&nbsp; for this configuration.
+The graph shows the&nbsp; (purely real)&nbsp; signal curve&nbsp; $y(t)$&nbsp; for this configuration.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[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 &nbsp;$Y_{\rm L}(p)$&nbsp; be given in pole&ndash;zero notation characterized by
 *$Z$&nbsp; zeros&nbsp; $p_{{\rm o}i}$,
@@ Line 10: / Line 10: @@
-In the following, the configurations shown in the diagram are considered. Let always &nbsp;$K= 2$ hold.
+In the following,&nbsp; the configurations shown in the diagram are considered.&nbsp; Let always &nbsp;$K= 2$ hold.
-In the case that the number&nbsp; $Z$&nbsp; of zeros is less than the number&nbsp; $N$&nbsp; of poles, the corresponding time signal &nbsp;$y(t)$&nbsp; can be determined directly by applying the&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp;.
+In the case that the number&nbsp; $Z$&nbsp; of zeros is less than the number&nbsp; $N$&nbsp; of poles,&nbsp; the corresponding time signal &nbsp;$y(t)$&nbsp; can be determined directly by applying the&nbsp; [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp;.
 In this case,
@@ Line 20: / Line 20: @@
   \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}i}} \right
   \} \hspace{0.05cm}$$ apllies.
-$I$&nbsp; indicates the number of distinguishable poles;&nbsp; &nbsp; $I = N$ holds for all given constellations.
+$I$&nbsp; indicates the number of distinguishable poles;&nbsp; &nbsp; $I = N$ holds&nbsp; for all given constellations.
@@ Line 27: / Line 27: @@
 Please note:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
+*If the time signal &nbsp;$y(t)$&nbsp; is complex,&nbsp; then &nbsp;$Y_{\rm L}(p)$&nbsp; cannot be realized as a circuit.&nbsp; However, the application of the residue theorem is still possible.
-''Please note:''
+*The complex frequency &nbsp;$p$,&nbsp; the zeros &nbsp;$p_{{\rm o}i}$&nbsp; as well as the poles &nbsp;$p_{{\rm x}i}$&nbsp; each describe normalized quantities without units in this exercise.
-*The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
+*Thus,&nbsp; time&nbsp; $t$&nbsp; is dimensionless, too.
@@ Line 41: / Line 37: @@
 <quiz display=simple>
-{For which configurations can the residue theorem <u>not be applied directly</u>?
+{For which configurations can the residue theorem&nbsp; <u>not be applied directly</u>?
 |type="[]"}
 - Configuration &nbsp;$\rm A$,
@@ Line 51: / Line 47: @@
-{Compute &nbsp;$y(t)$&nbsp; for configuration &nbsp;$\rm A$ with &nbsp;$K= 2$&nbsp; and &nbsp;$p_{\rm x} = -1$.&nbsp; What is the numerical value for time &nbsp;$t = 1$?
+{Compute &nbsp;$y(t)$&nbsp; for configuration &nbsp;$\rm A$&nbsp; with &nbsp;$K= 2$&nbsp; and &nbsp;$p_{\rm x} = -1$.&nbsp; What is the numerical value for time &nbsp;$t = 1$?
 |type="{}"}
 $\ {\rm Re}\{y(t = 1)\}  \ = \ $  { 0.736 3% }
@@ Line 57: / Line 53: @@
-{Compute &nbsp;$y(t)$&nbsp; for configuration &nbsp;$\rm C$ with &nbsp;$K= 2$&nbsp; and &nbsp;$p_{\rm x} = -0.2 + {\rm j} \cdot 1.5\pi$.&nbsp; What numerical value is obtained for time &nbsp;$t = 1$?
+{Compute &nbsp;$y(t)$&nbsp; for configuration &nbsp;$\rm C$&nbsp; with &nbsp;$K= 2$&nbsp; and &nbsp;$p_{\rm x} = -0.2 + {\rm j} \cdot 1.5\pi$.&nbsp; What numerical value is obtained for time &nbsp;$t = 1$?
 |type="{}"}
 $\ {\rm Re}\{y(t = 1)\}  \ = \ $ { 0. }
@@ Line 63: / Line 59: @@
-{What signal value &nbsp;$y(t = 1)$&nbsp; is obtained for the constellation &nbsp;$\rm E$ with &nbsp;$K= 2$&nbsp; and two poles at &nbsp;$p_{\rm x} = -0.2 \pm {\rm j} \cdot 1.5\pi$?
+{What signal value &nbsp;$y(t = 1)$&nbsp; is obtained for the constellation &nbsp;$\rm E$&nbsp; with &nbsp;$K= 2$&nbsp; and two poles at &nbsp;$p_{\rm x} = -0.2 \pm {\rm j} \cdot 1.5\pi$?
 |type="{}"}
   $\ {\rm Re}\{y(t = 1)\}  \ = \ $ { -0.357--0.337 }

@@ Line 92: / Line 92: @@
-'''(3)'''&nbsp; Using the same procedure as in the subtask&nbsp; '''(2)'''&nbsp; the following is obtained now:
+'''(3)'''&nbsp; Using the same procedure as in subtask&nbsp; '''(2)'''&nbsp; the following is obtained now:
 :$$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}

@@ Line 77: / Line 77: @@
 *The prerequisite for the application of the residue theorem is that there are fewer zeros than poles, that is, &nbsp;$Z < N$&nbsp; must hold.
 *This condition is not met for the configurations &nbsp;$\rm B$, &nbsp;$\rm D$ and &nbsp;$\rm F$.
-*First, a partial fraction decomposition must be made here, for example for the configuration &nbsp;$\rm B$&nbsp; with &nbsp;$p_x =  -1$:
+*First, a partial fraction decomposition must be made here, for example for configuration &nbsp;$\rm B$&nbsp; with &nbsp;$p_x =  -1$:
 :$$Y_{\rm L}(p)=   \frac {p} {p +1}= 1-\frac {1} {p +1}
   \hspace{0.05cm} .$$

@@ Line 113: / Line 113: @@
-'''(4)'''&nbsp; Nun gilt &nbsp;$I=2$. Die Residien von &nbsp;$p_{x1}$&nbsp; bzw. &nbsp;$p_{x2}$&nbsp; liefern:
+'''(4)'''&nbsp; Now &nbsp;$I=2$ holds.  The residuals of &nbsp;$p_{x1}$&nbsp; and &nbsp;$p_{x2}$&nbsp; yield:
 :$$y_1(t) =
   \frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot  {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot
@@ Line 135: / Line 135: @@
 \hspace{0.08cm}\cdot
   \hspace{0.05cm}t}\cdot  \sin(1.5\pi \cdot t)$$
-[[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 $\rm E$]]
 :$$\Rightarrow
 \hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2
@@ Line 142: / Line 142: @@
   \hspace{0.05cm} .$$
-Die Grafik zeigt den (rein reellen) Signalverlauf&nbsp; $y(t)$&nbsp; für diese Konfiguration.
+The graph shows the (purely real) signal curve&nbsp; $y(t)$&nbsp; for this configuration.
 {{ML-Fuß}}

@@ Line 107: / Line 107: @@
   \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} .$$
-*Die linke Grafik zeigt das komplexe Signal für einen Pol bei &nbsp;$p_x =  -2 + {\rm j} \cdot 1.5 \pi$.&nbsp; Rechts sieht man das dazu konjugiert&ndash;komplexe Signal für &nbsp;$p_x =  -2 - {\rm j} \cdot 1.5 \pi$.
+*The left graph shows the complex signal for a pole at &nbsp;$p_x =  -2 + {\rm j} \cdot 1.5 \pi$.&nbsp; On the right the conjugate komplex signal to it can be seen for &nbsp;$p_x =  -2 - {\rm j} \cdot 1.5 \pi$.

@@ Line 83: / Line 83: @@
-'''(2)'''&nbsp; Mit  &nbsp;$Y_{\rm L}(p) = 2/(p+1)$&nbsp; ergibt sich aus dem Residuensatz mit &nbsp;$I=1$:
+'''(2)'''&nbsp; Considering &nbsp;$Y_{\rm L}(p) = 2/(p+1)$&nbsp; it follows from the residue theorem with &nbsp;$I=1$:
 :$$y(t) = 2 \cdot  {\rm e}^{\hspace{0.05cm}p  \hspace{0.05cm}t}
   \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= 2 \cdot  {\rm
   e}^{-  \hspace{0.05cm}t}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y(t=1)
-  =\frac{2}{\rm e}  \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (rein\hspace{0.15cm}reell)}}
+  =\frac{2}{\rm e}  \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (purely\hspace{0.15cm}real)}}
   \hspace{0.05cm} .$$
-'''(3)'''&nbsp; Bei gleicher Vorgehensweise wie in der Teilaufgabe&nbsp; '''(2)'''&nbsp; erhält man nun:
+'''(3)'''&nbsp; Using the same procedure as in the subtask&nbsp; '''(2)'''&nbsp; the following is obtained now:
 :$$y(t) = 2 \cdot  {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+
   \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t}
@@ Line 100: / Line 100: @@
   \hspace{0.05cm}t}
   \hspace{0.05cm} .$$
-*Aufgrund des zweiten Terms handelt es sich um ein komplexes Signal, dessen Phase in mathematisch positiver Richtung&nbsp; (entgegen dem Uhrzeigersinn)&nbsp; dreht.
+*Due to the second term, it is a complex signal whose phase rotates in the mathematically positive direction&nbsp; (counterclockwise)&nbsp;.
-*Für den Zeitpunkt &nbsp;$t=1$&nbsp; gilt:
+*For time &nbsp;$t=1$,&nbsp; 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)
@@ Line 110: / Line 110: @@
-[[File:P_ID1782__LZI_Z_3_5_c.png|center|frame|Komplexe Signale bei einem Pol]]
+[[File:P_ID1782__LZI_Z_3_5_c.png|center|frame|Complex signals at a pole]]

@@ Line 33: / Line 33: @@
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
-*Ist das Zeitsignal &nbsp;$y(t)$&nbsp; komplex, so kann &nbsp;$Y_{\rm L}(p)$&nbsp; nicht als Schaltung realisiert werden.&nbsp; Die Anwendung des Residuensatzes ist aber trotzdem möglich.
+*If the time signal &nbsp;$y(t)$&nbsp; is complex, then &nbsp;$Y_{\rm L}(p)$&nbsp; cannot be realized as a circuit.&nbsp; However, the application of the residue theorem is still possible.
-*Die komplexe Frequenz &nbsp;$p$, die Nullstellen &nbsp;$p_{{\rm o}i}$&nbsp; sowie die Pole &nbsp;$p_{{\rm x}i}$&nbsp; beschreiben in dieser Aufgabe jeweils normierte Größen ohne Einheit.
+*The complex frequency &nbsp;$p$, the zeros &nbsp;$p_{{\rm o}i}$&nbsp; as well as the poles &nbsp;$p_{{\rm x}i}$&nbsp; each describe normalized quantities without units in this exercise.
-*Damit ist auch die Zeit&nbsp; $t$&nbsp; dimensionslos.
+*Thus, time&nbsp; $t$&nbsp; is dimensionless, too.
@@ Line 41: / Line 41: @@
 <quiz display=simple>
-{Bei welchen Konfigurationen lässt sich der Residuensatz <u>nicht direkt</u> anwenden?
+{For which configurations can the residue theorem <u>not be applied directly</u>?
 |type="[]"}
-- Konfiguration &nbsp;$\rm A$,
+- Configuration &nbsp;$\rm A$,
-+ Konfiguration &nbsp;$\rm B$,
++ Configuration &nbsp;$\rm B$,
-- Konfiguration &nbsp;$\rm C$,
+- Configuration &nbsp;$\rm C$,
-+ Konfiguration &nbsp;$\rm D$,
++ Configuration &nbsp;$\rm D$,
-- Konfiguration &nbsp;$\rm E$,
+- Configuration &nbsp;$\rm E$,
-+ Konfiguration &nbsp;$\rm F$.
++ Configuration &nbsp;$\rm F$.
-{Berechnen Sie &nbsp;$y(t)$&nbsp; für die Konfiguration &nbsp;$\rm A$ mit &nbsp;$K= 2$&nbsp; und &nbsp;$p_{\rm x} = -1$.&nbsp; Welcher Zahlenwert ergibt sich für den Zeitpunkt &nbsp;$t = 1$?
+{Compute &nbsp;$y(t)$&nbsp; for configuration &nbsp;$\rm A$ with &nbsp;$K= 2$&nbsp; and &nbsp;$p_{\rm x} = -1$.&nbsp; What is the numerical value for time &nbsp;$t = 1$?
 |type="{}"}
 $\ {\rm Re}\{y(t = 1)\}  \ = \ $  { 0.736 3% }