Question

The Laplace transform of a continuous-time signal $x\left(t\right)$ is $X\left(s\right)=\frac{5-s}{s^2-s-2}\cdot$ If the Fourier transform of this signal exists, then $x\left(t\right)$ is

Answer 1

$X\left(s\right)=\frac{5-s}{s^2-s-2}=\frac{5-s}{(s+1)(s-2)}$

$\frac{5-s}{(s+1)(s-2)}=\frac{A}{(s+1)}\frac{B}{(s-2)}$

$5-s=A(s-2)+B(s+1)$

$s=2,3=3B$

$\Rightarrow B=1$

$s=-1,6=-3A$

$\Rightarrow A=-2$

$\therefore x\left(t\right)=-2e^{-t}u\left(t\right)-e^{2t}u(-t)$

$ROC:-1<\sigma <2$

$\text{Note : ROC of Laplace transform should include}\sigma =0\text{line.}$