Inverse Laplace transform of X(s)=(s-3+e^(-2s))/(s^2+4)

Question : Find Inverse Laplace transform of X(s)=(s-3+e^(-2s))/(s^2+4) Solution : To Find Inverse Laplace transform of X(s)=(s-3+e^(-2s))/(s^2+4) \[X\left(s\right)=\frac{s-3+e^{-2s}}{s^2+4}\] We can write function X(s) as \[X\left(s\right)=\frac{s}{s^2+4\ }-\frac{3}{s^2+4\ }+\frac{e^{-2s}}{s^2+4}\] First we shall write it as follows \[X\left(s\right)=\frac{s}{s^2+4\ }-\frac{3}{s^2+4\ }+\frac{1}{2}\frac{2e^{-2s}}{s^2+4}\] Now we shall…

This Question has been answered. 

Please Subscribe to See Answer or To Get Homework Help