Aufgaben:Exercise 3.6Z: Two Imaginary Poles - Revision history

Guenter at 16:33, 9 December 2021

2021-12-09T16:33:43Z

Guenter at 16:29, 9 December 2021

2021-12-09T16:29:01Z

Oezer at 08:15, 25 October 2021

2021-10-25T08:15:01Z

Oezer at 08:14, 25 October 2021

2021-10-25T08:14:17Z

Oezer at 08:12, 25 October 2021

2021-10-25T08:12:44Z

Oezer at 08:11, 25 October 2021

2021-10-25T08:11:12Z

Oezer at 08:10, 25 October 2021

2021-10-25T08:10:12Z

Oezer at 08:02, 25 October 2021

2021-10-25T08:02:43Z

Oezer at 07:54, 25 October 2021

2021-10-25T07:54:32Z

Oezer at 00:21, 25 October 2021

2021-10-25T00:21:33Z

@@ Line 71: / Line 71: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The <u>suggested solutions 1, 3 and 4</u> are correct:
+'''(1)'''&nbsp; The&nbsp; <u>suggested solutions 1, 3 and 4</u>&nbsp; are correct:
 *The following is obtained for signal&nbsp; $x(t)$&nbsp; for positive times by applying the residue theorem:
 :$$x_1(t)\hspace{0.25cm} =  \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
@@ Line 94: / Line 94: @@
-'''(2)'''&nbsp; The <u>suggested solutions 2 and 4</u> are correct:
+'''(2)'''&nbsp; The&nbsp; <u>suggested solutions 2 and 4</u>&nbsp; are correct:
-*In principle, this subtask could be solved in the same way as subtask&nbsp; '''(1)'''.
+*In principle,&nbsp; this subtask could be solved in the same way as subtask&nbsp; '''(1)'''.
-*However, the integration theorem can also be used.
+*However,&nbsp; the integration theorem can also be used.
 *This says among other things that multiplication by&nbsp; $1/p$&nbsp; in the spectral domain corresponds to integration in the time domain:
 :$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
@@ Line 102: / Line 102: @@
   \hspace{0.05cm} .$$
-''Please note:'' &nbsp; &nbsp; The causal cosine signal&nbsp; $x(t)$&nbsp; and the causal sine signal&nbsp; $y(t)$&nbsp; are shown on the information page of&nbsp; [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]&nbsp; as&nbsp; $c_{\rm K}(t)$&nbsp; and&nbsp; $s_{\rm K}(t)$,&nbsp; respectively.
+Please note:&nbsp; The causal cosine signal&nbsp; $x(t)$&nbsp; and the causal sine signal&nbsp; $y(t)$&nbsp; are shown on the information page of&nbsp; [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]&nbsp; as&nbsp; $c_{\rm K}(t)$&nbsp; and&nbsp; $s_{\rm K}(t)$,&nbsp; respectively.
-'''(3)'''&nbsp; The <u>suggested solutions 1 and 3</u> are correct:
+'''(3)'''&nbsp; The&nbsp; <u>suggested solutions 1 and 3</u>&nbsp; are correct:
-*A comparison with the computation of &nbsp;$x(t)$&nbsp; shows that &nbsp;$z(t) = \cos (\beta \cdot t)$&nbsp; holds for &nbsp;$t \ge 0$&nbsp; and &nbsp;$z(t) = 0$&nbsp; for &nbsp;$t < 0$&nbsp;.
+*A comparison with the computation of &nbsp;$x(t)$&nbsp; shows that &nbsp;$z(t) = \cos (\beta \cdot t)$&nbsp; holds for &nbsp;$t \ge 0$&nbsp; and &nbsp;$z(t) = 0$&nbsp; for &nbsp;$t < 0$.
 *The limit process for &nbsp;$\beta &#8594; 0$&nbsp; thus results in the step function &nbsp;$\gamma(t)$.
 *The same result is obtained by consideration in the spectral domain:

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
+[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles <br>and one zero ]]
-In this exercise, we consider a causal signal &nbsp;$x(t)$&nbsp; with the Laplace transform
+In this exercise,&nbsp; we consider a causal signal &nbsp;$x(t)$&nbsp; with the Laplace transform
 :$$X_{\rm L}(p) =
   \frac { p} { p^2 + 4 \pi^2}=
@@ Line 11: / Line 11: @@
 corresponding to the graph&nbsp; (one red zero and two green poles).
-In contrast, the signal &nbsp;$y(t)$&nbsp; has the Laplace spectral function
+*In contrast,&nbsp; the signal &nbsp;$y(t)$&nbsp; has the Laplace spectral function
 :$$Y_{\rm L}(p) =
   \frac { 1} { p^2 + 4 \pi^2}
   \hspace{0.05cm}.$$
-Thus, the red zero does not belong to &nbsp;$Y_{\rm L}(p)$.
+:Thus,&nbsp; the red zero does not belong to &nbsp;$Y_{\rm L}(p)$.
-Finally, the signal &nbsp;$z(t)$&nbsp; with the Laplace tansform
+*Finally, the signal &nbsp;$z(t)$&nbsp; with the Laplace tansform
 :$$Z_{\rm L}(p) =
   \frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
   \hspace{0.05cm}$$
-is considered, in particular the limiting case for &nbsp;$\beta &#8594; 0$.
+:is considered, in particular the limiting case for &nbsp;$\beta &#8594; 0$.
@@ Line 27: / Line 27: @@
 Please note:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
-*The frequency variable&nbsp; $p$&nbsp; is normalized such that time&nbsp; $t$&nbsp; is in microseconds after applying the residue theorem.
+*The frequency variable&nbsp; $p$ &nbsp; is normalized such that time&nbsp; $t$&nbsp; is in microseconds after applying the residue theorem.
 *A result &nbsp;$t = 1$&nbsp; is thus to be interpreted as &nbsp;$t = T$&nbsp; with &nbsp;$T = 1 \ \rm &micro; s$&nbsp;.
 *The &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp; is as follows using the example of the function &nbsp;$X_{\rm L}(p)$&nbsp; with two simple poles at &nbsp;$ \pm {\rm j} \cdot \beta$:
@@ Line 46: / Line 43: @@
 <quiz display=simple>
-{Compute the signal &nbsp;$x(t)$. Which of the following statements are correct?
+{Compute the signal &nbsp;$x(t)$.&nbsp; Which of the following statements are correct?
 |type="[]"}
 + $x(t)$&nbsp; is a causal cosine signal.
 - $x(t)$&nbsp; is a causal sinusoidal signal.
 + The amplitude of&nbsp; $x(t)$&nbsp; is&nbsp; $1$.
-+ The period of $x(t)$ is&nbsp; $T = 1 \ \rm &micro; s$.
++ The period of&nbsp; $x(t)$&nbsp; is&nbsp; $T = 1 \ \rm &micro; s$.

← Older revision		Revision as of 08:15, 25 October 2021
Line 105:		Line 105:
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

−	''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  ~~als~~  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.	+	''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  as  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

← Older revision		Revision as of 08:14, 25 October 2021
Line 105:		Line 105:
	\hspace{0.05cm} .$$		\hspace{0.05cm} .$$

−	''Please note'':     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.	+	''Please note:''     The causal cosine signal  $x(t)$  and the causal sine signal  $y(t)$  are shown on the information page of  [[Aufgaben:Exercise_3.6:_Transient_Behavior\|Exercise 3.6]]  als  $c_{\rm K}(t)$  and  $s_{\rm K}(t)$,  respectively.

@@ Line 50: / Line 50: @@
 + $x(t)$&nbsp; is a causal cosine signal.
 - $x(t)$&nbsp; is a causal sinusoidal signal.
-+ The amplitude ofn&nbsp; $x(t)$&nbsp; is&nbsp; $1$.
++ The amplitude of&nbsp; $x(t)$&nbsp; is&nbsp; $1$.
 + The period of $x(t)$ is&nbsp; $T = 1 \ \rm &micro; s$.

@@ Line 105: / Line 105: @@
   \hspace{0.05cm} .$$
-''Please note'': &nbsp; &nbsp; The causal cosine signal&nbsp; $x(t)$&nbsp; and the causal sine signal&nbsp; $y(t)$&nbsp; sind auf dem Angabenblatt zu&nbsp; [[Aufgaben:3.6_Einschwingverhalten|Exercise 3.6]]&nbsp; als&nbsp; $c_{\rm K}(t)$&nbsp; bzw.&nbsp; $s_{\rm K}(t)$&nbsp; dargestellt.
+''Please note'': &nbsp; &nbsp; The causal cosine signal&nbsp; $x(t)$&nbsp; and the causal sine signal&nbsp; $y(t)$&nbsp; are shown on the information page of&nbsp; [[Aufgaben:Exercise_3.6:_Transient_Behavior|Exercise 3.6]]&nbsp; als&nbsp; $c_{\rm K}(t)$&nbsp; and&nbsp; $s_{\rm K}(t)$,&nbsp; respectively.
-'''(3)'''&nbsp; Richtig sind <u>die Lösungsvorschläge 1 und 3</u>:
+'''(3)'''&nbsp; The <u>suggested solutions 1 and 3</u> are correct:
-*Ein Vergleich mit der Berechnung von &nbsp;$x(t)$&nbsp; zeigt, dass &nbsp;$z(t) = \cos (\beta \cdot t)$&nbsp; für &nbsp;$t \ge 0$&nbsp; und &nbsp;$z(t) = 0$&nbsp; für &nbsp;$t < 0$&nbsp; gilt.
+*A comparison with the computation of &nbsp;$x(t)$&nbsp; shows that &nbsp;$z(t) = \cos (\beta \cdot t)$&nbsp; holds for &nbsp;$t \ge 0$&nbsp; and &nbsp;$z(t) = 0$&nbsp; holds for &nbsp;$t < 0$&nbsp;.
-*Der Grenzübergang für &nbsp;$\beta &#8594; 0$&nbsp; führt damit zur Sprungfunktion &nbsp;$\gamma(t)$.
+*The limit process for &nbsp;$\beta &#8594; 0$&nbsp; thus results in the step function &nbsp;$\gamma(t)$.
-*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
+*The same result is obtained by consideration in the spectral domain:
 :$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
   \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}

@@ Line 17: / Line 17: @@
 Thus, the red zero does not belong to &nbsp;$Y_{\rm L}(p)$.
-Abschließend wird noch das Signal &nbsp;$z(t)$&nbsp; mit der Laplace&ndash;Transformierten
+Finally, the signal &nbsp;$z(t)$&nbsp; with the Laplace tansform
 :$$Z_{\rm L}(p) =
   \frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)}
   \hspace{0.05cm}$$
-betrachtet, insbesondere der Grenzfall für &nbsp;$\beta &#8594; 0$.
+is considered, in particular the limiting case for &nbsp;$\beta &#8594; 0$.
@@ Line 32: / Line 32: @@
 ''Please note:''
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform|Inverse Laplace Transform]].
-*Die Frequenzvariable&nbsp; $p$&nbsp; ist so normiert, dass nach Anwendung des Residuensatzes die Zeit&nbsp; $t$&nbsp; in Mikrosekunden angegeben ist.
+*The frequency variable&nbsp; $p$&nbsp; is normalized such that time&nbsp; $t$&nbsp; is in microseconds after applying the residue theorem.
-*Ein Ergebnis &nbsp;$t = 1$&nbsp; ist somit als &nbsp;$t = T$&nbsp; mit &nbsp;$T = 1 \ \rm &micro; s$&nbsp; zu interpretieren.
+*A result &nbsp;$t = 1$&nbsp; is thus to be interpreted as &nbsp;$t = T$&nbsp; with &nbsp;$T = 1 \ \rm &micro; s$&nbsp;.
-*Der &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp; lautet am Beispiel der Funktion &nbsp;$X_{\rm L}(p)$&nbsp; mit zwei einfachen Polstellen bei &nbsp;$ \pm {\rm j} \cdot \beta$:
+*The &nbsp;[[Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform#Formulation_of_the_residue_theorem|residue theorem]]&nbsp; is as follows using the example of the function &nbsp;$X_{\rm L}(p)$&nbsp; with two simple poles at &nbsp;$ \pm {\rm j} \cdot \beta$:
 :$$x(t)  =  X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot  {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot
   \hspace{0.05cm}t}  \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
@@ Line 46: / Line 46: @@
 <quiz display=simple>
-{Berechnen Sie das Signal &nbsp;$x(t)$. Welche der folgenden Aussagen sind richtig?
+{Compute the signal &nbsp;$x(t)$. Which of the following statements are correct?
 |type="[]"}
-+ $x(t)$&nbsp; ist ein kausales Cosinussignal.
++ $x(t)$&nbsp; is a causal cosine signal.
-- $x(t)$&nbsp; ist ein kausales Sinussignal.
+- $x(t)$&nbsp; is a causal sinusoidal signal.
-+ Die Amplitude von&nbsp; $x(t)$&nbsp; ist&nbsp; $1$.
++ The amplitude ofn&nbsp; $x(t)$&nbsp; is&nbsp; $1$.
-+ Die Periodendauer von $x(t)$ ist&nbsp; $T = 1 \ \rm &micro; s$.
++ The period of $x(t)$ is&nbsp; $T = 1 \ \rm &micro; s$.
-{Berechnen Sie das Signal &nbsp;$y(t)$. Welche der folgenden Aussagen sind richtig?
+{Compute the signal &nbsp;$y(t)$. Which of the following statements are correct?
 |type="[]"}
-- $y(t)$&nbsp; ist ein kausales Cosinussignal.
+- $y(t)$&nbsp; is a causal cosine signal.
-+ $y(t)$&nbsp; ist ein kausales Sinussignal.
++ $y(t)$&nbsp; is a causal sinusoidal signal.
-- Die Amplitude von&nbsp; $y(t)$&nbsp; ist&nbsp; $1$.
+- The amplitude of&nbsp; $y(t)$&nbsp; is&nbsp; $1$.
-+ Die Periodendauer von&nbsp; $y(t)$&nbsp; ist&nbsp; $T = 1 \ \rm &micro; s$.
++ The period of&nbsp; $y(t)$&nbsp; is&nbsp; $T = 1 \ \rm &micro; s$.
-{Welche Aussagen treffen für das Signal &nbsp;$z(t)$&nbsp; zu?
+{Which statements are true for the signal &nbsp;$z(t)$&nbsp;?
 |type="[]"}
-+ Für &nbsp;$ \beta > 0$&nbsp; verläuft &nbsp;$z(t)$&nbsp; cosinusförmig.
++ For &nbsp;$ \beta > 0$,&nbsp; &nbsp;$z(t)$&nbsp; is cosine-shaped.
-- Für &nbsp;$ \beta > 0$&nbsp; verläuft &nbsp;$z(t)$&nbsp; sinusförmig.
+- For &nbsp;$ \beta > 0$,&nbsp; &nbsp;$z(t)$&nbsp; is sinusoidal.
-+ Der Grenzfall &nbsp;$\beta &#8594; 0$&nbsp; führt zur Sprungfunktion &nbsp;$\gamma(t)$.
++ The limiting case &nbsp;$\beta &#8594; 0$&nbsp; results in the step function &nbsp;$\gamma(t)$.