Question : Find Inverse Laplace transform of F(s)=(s-2)/(s^2+ω^2 )
Solution :
To Find Inverse Laplace transform of F(s)=(s-2)/(s^2+ω^2 ) Partial fraction of F(s) is given as \[F\left(s\right)=\frac{s-2}{s^2+\omega^2}\] It can be written as \[F\left(s\right)=\frac{s}{s^2+\omega^2}-\frac{2}{\omega}\frac{\omega}{s^2+\omega^2}\] Applying inverse Laplace transeform \[L^{-1}\left(F\left(s\right)\right)=L^{-1}\left[\frac{s}{s^2+\omega^2}-\frac{2}{\omega}\frac{\omega}{s^2+\omega^2}\right]\] \[L^{-1}\left(F\left(s\right)\right)=L^{-1}\left[\frac{s}{s^2+\omega^2}\right]-\frac{2}{\omega}L^{-1}\left[\frac{\omega}{s^2+\omega^2}\right]\] We know that inverse Laplace transeform formulas \[L^{-1}\left[\frac{s}{s^2+a^2}\right]=\cos(at)\ \ \ ,\ L^{-1}\left[\frac{a}{s^2+a^2}\right]=\sin(at)\] \[\Rightarrow\ \ f\left(t\right)=\cos{\left(\omega t\right)}-\frac{2}{\omega}\sin(\omega t)\] Is the required Inverse Laplace transform of F(s)=(s-2)/(s^2+ω^2 )