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Exercise 3.6Z: Two Imaginary Poles

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Two imaginary poles and one zero

In this exercise, we consider a causal signal  x(t)  with the Laplace transform

XL(p)=pp2+4π2=p(pj2π)(p+j2π)

corresponding to the graph  (one red zero and two green poles).

In contrast, the signal  y(t)  has the Laplace spectral function

YL(p)=1p2+4π2.

Thus, the red zero does not belong to  YL(p).

Finally, the signal  z(t)  with the Laplace tansform

ZL(p)=p(pjβ)(p+jβ)

is considered, in particular the limiting case for  β0.





Please note:

  • The exercise belongs to the chapter  Inverse Laplace Transform.
  • The frequency variable  p  is normalized such that time  t  is in microseconds after applying the residue theorem.
  • A result  t=1  is thus to be interpreted as  t=T  with  T=1 µs .
  • The  residue theorem  is as follows using the example of the function  XL(p)  with two simple poles at  ±jβ:
x(t)=XL(p)(pjβ)ept|p=jβ+XL(p)(p+jβ)ept|p=jβ.


Questions

1

Compute the signal  x(t). Which of the following statements are correct?

x(t)  is a causal cosine signal.
x(t)  is a causal sinusoidal signal.
The amplitude ofn  x(t)  is  1.
The period of x(t) is  T=1 µs.

2

Compute the signal  y(t). Which of the following statements are correct?

y(t)  is a causal cosine signal.
y(t)  is a causal sinusoidal signal.
The amplitude of  y(t)  is  1.
The period of  y(t)  is  T=1 µs.

3

Which statements are true for the signal  z(t) ?

For  β>0,   z(t)  is cosine-shaped.
For  β>0,   z(t)  is sinusoidal.
The limiting case  β0  results in the step function  γ(t).


Solution

(1)  The suggested solutions 1, 3 and 4 are correct:

  • The following is obtained for signal  x(t)  for positive times by applying the residue theorem:
x1(t)=Res|p=px1{XL(p)ept}=pp+j2πept|p=j2π=12ej2πt,
x2(t)=Res|p=px2{XL(p)ept}=ppj2πept|p=j2π=12ej2πt.
x(t)=x1(t)+x2(t)=1/2[ej2πt+ej2πt]=cos(2πt).


(2)  The suggested solutions 2 and 4 are correct:

  • In principle, this subtask could be solved in the same way as subtask  (1).
  • However, the integration theorem can also be used.
  • This says among other things that multiplication by  1/p  in the spectral domain corresponds to integration in the time domain:
YL(p)=1/pXL(p)t0:y(t)=tcos(2πτ)dτ=1/(2π)sin(2πt).

Please note:     The causal cosine signal  x(t)  and the causal sine signal  y(t)  sind auf dem Angabenblatt zu  Exercise 3.6  als  cK(t)  bzw.  sK(t)  dargestellt.


(3)  Richtig sind die Lösungsvorschläge 1 und 3:

  • Ein Vergleich mit der Berechnung von  x(t)  zeigt, dass  z(t)=cos(βt)  für  t0  und  z(t)=0  für  t<0  gilt.
  • Der Grenzübergang für  β0  führt damit zur Sprungfunktion  γ(t).
  • Zum gleichen Ergebnis kommt man durch die Betrachtung im Spektralbereich:
ZL(p)=limβ0pp2+β2=1/pz(t)=γ(t).