Answer to Question #261614 in Electrical Engineering for Mafizur Alam

Question

Find the Inverse Laplace transform of X(s)= -s^2-2s+1/(s+2)(s+3)^2. ROC: Re{s}>-2

Answer 1

$L^{-1}\left\{\frac{-s^2-2s+1}{\left(s+2\right)\left(s+3\right)^2}\right\}\\ =L^{-1}\left\{\frac{1}{s+2}-\frac{2}{s+3}+\frac{2}{\left(s+3\right)^2}\right\}\\ \mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}\\ \mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:\quad L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}\\ =L^{-1}\left\{\frac{1}{s+2}\right\}-L^{-1}\left\{\frac{2}{s+3}\right\}+L^{-1}\left\{\frac{2}{\left(s+3\right)^2}\right\}\\ =e^{-2t}-2e^{-3t}+e^{-3t}\cdot \:2t$