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$\text { Find inverse Laplace Transform of } X(S)=\frac{S^2+9 S+1}{S\left[S^2+6 S+8\right]}$ $\text { Find ROC for } i) \operatorname{Re}(s)>0$

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

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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 $s=-4$

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