Aufgaben:Exercise 3.5Z: Application of the Residue Theorem

• Return to book

Six different
pole–zero configurations

Let the spectral function  $Y_{\rm L}(p)$  be given in pole–zero notation characterized by

• $Z$  zeros  $p_{{\rm o}i}$,
• $N$  poles  $p_{{\rm x}i}$, and
• the constant  $K$.

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  residue theorem .

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.

$I$  indicates the number of distinguishable poles;    $I = N$ holds for all given constellations.

Please note:

• The exercise belongs to the chapter  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.
• 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.

Questions

1

For which configurations can the residue theorem not be applied directly?

	Configuration  $\rm A$,
	Configuration  $\rm B$,
	Configuration  $\rm C$,
	Configuration  $\rm D$,
	Configuration  $\rm E$,
	Configuration  $\rm F$.

2

Compute  $y(t)$  for configuration  $\rm A$ with  $K= 2$  and  $p_{\rm x} = -1$.  What is the numerical value for time  $t = 1$?

$\ {\rm Re}\{y(t = 1)\} \ = \ $

$\ {\rm Im}\{y(t = 1)\} \ = \ $

3

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  $t = 1$?

$\ {\rm Re}\{y(t = 1)\} \ = \ $

$\ {\rm Im}\{y(t = 1)\} \ = \ $

4

What signal value  $y(t = 1)$  is obtained for the constellation  $\rm E$ with  $K= 2$  and two poles at  $p_{\rm x} = -0.2 \pm {\rm j} \cdot 1.5\pi$?

$\ {\rm Re}\{y(t = 1)\} \ = \ $

$\ {\rm Im}\{y(t = 1)\} \ = \ $

Solution

Solution

(1)  Suggested solutions 2, 4 and 6 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.
• 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  $p_x = -1$:

$$Y_{\rm L}(p)= \frac {p} {p +1}= 1-\frac {1} {p +1} \hspace{0.05cm} .$$

(2)  Mit  $Y_{\rm L}(p) = 2/(p+1)$  ergibt sich aus dem Residuensatz mit  $I=1$:

$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}-1}= 2 \cdot {\rm e}^{- \hspace{0.05cm}t}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y(t=1) =\frac{2}{\rm e} \hspace{0.15cm}\underline{ \approx 0.736 \hspace{0.15cm}{\rm (rein\hspace{0.15cm}reell)}} \hspace{0.05cm} .$$

(3)  Bei gleicher Vorgehensweise wie in der Teilaufgabe  (2)  erhält man nun:

$$y(t) = 2 \cdot {\rm e}^{\hspace{0.05cm}-(0.2 \hspace{0.05cm}+ \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}1.5 \pi) \hspace{0.05cm} \cdot \hspace{0.05cm}t} = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2 \hspace{0.08cm}\cdot \hspace{0.05cm}t}\cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.08cm}\cdot \hspace{0.05cm}1.5 \pi\hspace{0.05cm}\cdot \hspace{0.05cm}t} \hspace{0.05cm} .$$

• Aufgrund des zweiten Terms handelt es sich um ein komplexes Signal, dessen Phase in mathematisch positiver Richtung  (entgegen dem Uhrzeigersinn)  dreht.
• Für den Zeitpunkt  $t=1$  gilt:

$$y(t = 1) = 2 \cdot {\rm e}^{\hspace{0.05cm}-0.2} \cdot \big [ \cos(1.5 \pi) + {\rm j} \cdot \sin(1.5 \pi) \big ]= - {\rm j} \cdot 1.638\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}\{y(t = 1)\} \hspace{0.15cm}\underline{ = 0},\hspace{0.2cm} {\rm Im}\{y(t = 1)\} \hspace{0.15cm}\underline{=- 1.638} \hspace{0.05cm} .$$

• Die linke Grafik zeigt das komplexe Signal für einen Pol bei  $p_x = -2 + {\rm j} \cdot 1.5 \pi$.  Rechts sieht man das dazu konjugiert–komplexe Signal für  $p_x = -2 - {\rm j} \cdot 1.5 \pi$.

Komplexe Signale bei einem Pol

(4)  Nun gilt  $I=2$. Die Residien von  $p_{x1}$  bzw.  $p_{x2}$  liefern:

$$y_1(t) = \frac {K \cdot (p-p_{{\rm x}1})} { (p-p_{{\rm x}1})(p-p_{{\rm x}2})} \cdot {\rm e}^{\hspace{0.05cm}p\hspace{0.05cm}\cdot \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}}= \frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}1}\hspace{0.05cm}\cdot \hspace{0.05cm}t} \hspace{0.05cm} ,$$

$$ y_2(t) = \frac {K } { p_{{\rm x}2}-p_{{\rm x}1}} \cdot {\rm e}^{\hspace{0.05cm}p_{{\rm x}2}\hspace{0.05cm}\cdot \hspace{0.05cm}t}= -\frac {K } { p_{{\rm x}1}-p_{{\rm x}2}} \cdot {\rm e}^{-p_{{\rm x}1}\hspace{0.05cm}\cdot \hspace{0.05cm}t}$$

$$\Rightarrow \hspace{0.3cm}y(t)= y_1(t)+y_2(t) = \frac {2 \cdot {\rm e}^{\hspace{0.05cm}-0.2 \hspace{0.08cm}\cdot \hspace{0.05cm}t}}{{\rm j} \cdot 3 \pi} \cdot \big [ \cos(.) + {\rm j} \cdot \sin(.) - \cos(.) + {\rm j} \cdot \sin(.)\big ]= \frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2 \hspace{0.08cm}\cdot \hspace{0.05cm}t}\cdot \sin(1.5\pi \cdot t)$$

Signalverlauf der Konfiguration $\rm E$

$$\Rightarrow \hspace{0.3cm}y(t=1)= -\frac {4 }{ 3 \pi} \cdot {\rm e}^{\hspace{0.05cm}-0.2 \hspace{0.08cm}\cdot \hspace{0.05cm}t} \hspace{0.15cm}\underline{= -0.347} \hspace{0.05cm} .$$

Die Grafik zeigt den (rein reellen) Signalverlauf  $y(t)$  für diese Konfiguration.