View source for Aufgaben:Exercise 3.6Z: Two Imaginary Poles


{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Inverse_Laplace_Transform
}}

[[File:P_ID1786__LZI_Z_3_6.png|right|frame|Two imaginary poles and one zero ]]
In this exercise, we consider a causal signal &nbsp;$x(t)$&nbsp; with the Laplace transform
:$$X_{\rm L}(p) =
 \frac { p} { p^2 + 4 \pi^2}=
 \frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)}
 \hspace{0.05cm}$$
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
:$$Y_{\rm L}(p) =
 \frac { 1} { p^2 + 4 \pi^2}
 \hspace{0.05cm}.$$
Thus, the red zero does not belong to &nbsp;$Y_{\rm L}(p)$.

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;$\beta &#8594; 0$.








''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. 
*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$:
:$$x(t)  =  X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot  {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot 
 \hspace{0.05cm}t}  \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
 \beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot  {\rm e}^{\hspace{0.03cm}p
 \hspace{0.05cm} \cdot\hspace{0.05cm}t}  \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it
 \beta}}
 \hspace{0.05cm}.$$


===Questions===

<quiz display=simple>
{Compute the signal &nbsp;$x(t)$. 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 ofn&nbsp; $x(t)$&nbsp; is&nbsp; $1$.
+ The period of $x(t)$ is&nbsp; $T = 1 \ \rm &micro; s$.


{Compute the signal &nbsp;$y(t)$. Which of the following statements are correct?
|type="[]"}
- $y(t)$&nbsp; is a causal cosine signal.
+ $y(t)$&nbsp; is a causal sinusoidal signal.
- The amplitude of&nbsp; $y(t)$&nbsp; is&nbsp; $1$.
+ The period of&nbsp; $y(t)$&nbsp; is&nbsp; $T = 1 \ \rm &micro; s$.


{Which statements are true for the signal &nbsp;$z(t)$&nbsp;?
|type="[]"}
+ For &nbsp;$ \beta > 0$,&nbsp; &nbsp;$z(t)$&nbsp; is cosine-shaped.
- For &nbsp;$ \beta > 0$,&nbsp; &nbsp;$z(t)$&nbsp; is sinusoidal.
+ The limiting case &nbsp;$\beta &#8594; 0$&nbsp; results in the step function &nbsp;$\gamma(t)$.



</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; The <u>suggested solutions 1, 3 and 4</u> are correct:
*The following is obtained for signal&nbsp; $x(t)$&nbsp; for positive times by applying the residue theorem:
:$$x_1(t)\hspace{0.25cm} =  \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}
 \hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}=
 \frac {p} { p+{\rm j} \cdot 2\pi}\cdot  {\rm e}^{\hspace{0.05cm}p
 \hspace{0.05cm}\cdot \hspace{0.05cm}t}
 \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
 \frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$
:$$ x_2(t)\hspace{0.25cm} =  \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}}
 \hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}=
 \frac {p} { p-{\rm j} \cdot 2\pi}\cdot  {\rm e}^{\hspace{0.05cm}p
 \hspace{0.05cm}\cdot \hspace{0.05cm}t}
 \bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}=
 \frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}
 \hspace{0.05cm} .$$
:$$\Rightarrow  \hspace{0.3cm} x(t) = x_1(t) + x_2(t) =
 {1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
 t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi
 t}\right ] = \cos(2\pi t)
 \hspace{0.05cm} .$$



'''(2)'''&nbsp; Richtig sind <u>die Lösungsvorschläge 2 und 4</u>:
*Prinzipiell könnte diese Teilaufgabe in gleicher Weise gelöst werden wie die Teilaufgabe&nbsp; '''(1)'''. 
*Man kann aber auch den Integrationssatz heranziehen. 
*Dieser besagt unter anderem, dass die Multiplikation mit&nbsp; $1/p$&nbsp; im Spektralbereich der Integration im Zeitbereich entspricht:
:$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi
 \tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t)
 \hspace{0.05cm} .$$

''Hinweis'': &nbsp; &nbsp; Das kausale Cosinussignal&nbsp; $x(t)$&nbsp; sowie das kausale Sinussignal&nbsp; $y(t)$&nbsp; sind auf dem Angabenblatt zu&nbsp; [[Aufgaben:3.6_Einschwingverhalten|Aufgabe 3.6]]&nbsp; als&nbsp; $c_{\rm K}(t)$&nbsp; bzw.&nbsp; $s_{\rm K}(t)$&nbsp; dargestellt.



'''(3)'''&nbsp; Richtig sind <u>die Lösungsvorschläge 1 und 3</u>:
*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. 
*Der Grenzübergang für &nbsp;$\beta &#8594; 0$&nbsp; führt damit zur Sprungfunktion &nbsp;$\gamma(t)$. 
*Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
:$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p}
 \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}
  z(t) = \gamma(t)
 \hspace{0.05cm} .$$
{{ML-Fuß}}



[[Category:Linear and Time-Invariant Systems: Exercises|^3.3 Inverse Laplace Transform^]]