Question

Five facts about real signal x(t) with Laplace transform X(s) are given below :

1. X(s) has exactly two poles.
2.X(s) has no zeros in the finite s-plane.
3. X(s) has a pole at s = -1 + j.
4. $e^{2t}x\left(t\right)$ is not absolutely integrable.
5. X(0) = 8.

• A. $\frac{8}{s^2+2s+1},Re\left[s\right]>-1$
• B. $\frac{16}{s^2+2s+2},Re\left[s\right]>-1$
• C. $\frac{16}{s^2+2s+2},Re\left[s\right]<-1$
• D. $\frac{8}{s^2+2s+1},Re\left[s\right]<-1$

Answer 1

The correct option is B $\frac{16}{s^2+2s+2},Re\left[s\right]>-1$

X(s) is of the form,

$X\left(s\right)=\frac{A}{(s+a)(s+b)}$

Given that one pole is at s = -1 + j. Since x(t) is real that poles of X(s) must occure in conjugate reciprocal pairs.

$s=-1-j$ is another pole of X(s).

$\therefore X\left(s\right)=\frac{A}{(s+1-j)(s+1+j)}=\frac{A}{(s+1{)}^2+1}$

$=\frac{A}{s^2+2s+2}$

Given,

$X\left(0\right)=8$

$X\left(0\right)=\frac{A}{2}$

$8=\frac{A}{2}$

$A=16$

$\therefore X\left(s\right)=\frac{16}{s^2+2s+2}$

Let R denote the ROC of X(s). From the pole locations we know that there are two possible choices of R

R may either be Re[s] < -1 or Re[s]> -1
Given that,

$e^{2t}x\left(t\right)\stackrel{L.T}{⟶}X(s-2)$

The ROC of Y(s) is R shifted by 2 to the right. Since it is given that y(t) is not absolutely integrable.
The ROC of Y(s) should not include the $j\omega$ axis. This is possible only if R is Re[s]>-1.