$ \text { Find inverse Laplace Transform of } X(S)=\frac{S^2+9 S+1}{S\left[S^2+6 S+8\right]} $ $ \text { Find ROC for } i) \operatorname{Re}(s)\gt0 $ $ \text { ii) } \operatorname{Re}(s)\lt-4 \quad \text { iii })-2\gt\operatorname{Re}(s)\gt-4 $

Solution:

$ \mathrm{X}(S)=\frac{S^2+9 S+1}{S\left[S^2+6 S+8\right]}\\ $

$ =\frac{S^2+9 S+1}{S(S+4)(S+2)}\\ $

$ =\frac{A}{S}+\frac{B}{(S+4)}+\frac{C}{(S+2)}\\ $

$ S^2+9 S+1=A(S+4)(S+2)+\mathrm{B} S(S+2)+\mathrm{C} S(S+4)\\ $

$ \therefore X(S)=\frac{\frac{1}{8}}{S}+\frac{-\frac{19}{8}}{(S+4)}+\frac{\frac{13}{4}}{(S+2)}\\ $

Applying inverse Laplace transform,

$ x(t)=\frac{1}{8} u(t)-\frac{19}{8} e^{-4 t} u(t)+\frac{13}{4} e^{-2 t} u(t)\\ $

i) $\boldsymbol{R e}(\boldsymbol{s})\gt0$

ROC lies a right side of all poles,

$ \therefore x(t)=\frac{1}{8} u(t)-\frac{19}{8} e^{-4 t} u(t)+\frac{13}{4} e^{-2 t} u(t)\\ $

ii) $\operatorname{Re}(s)\lt-4$

ROC lies on the left side of all poles,

$ \therefore x(t)=-\frac{1}{8} u(-t)+\frac{19}{8} e^{-4 t} u(-t)-\frac{13}{4} e^{-2 t} u(-t)\\ $

iii) $-\mathbf{2}\gt\operatorname{Re}(\boldsymbol{s})\gt-\mathbf{4}$

ROC lies left side of poles $s=-2, s=0$ and right side of pole $s=-4$

$ \therefore x(t)=-\frac{1}{8} u(-t)-\frac{19}{8} e^{-4 t} u(t)-\frac{13}{4} e^{-2 t} u(-t)\\ $