Question

Find the inverse laplace transform of the function by using the convolution theorem. f(s) = 1 s5(s2 + 1)

Answer 1

By the convolution theorem,


`f(t)=(g*h)(t)=\displaystyle\int_(\tau=0)^(\tau=t)g(\tau)h(t-\tau)\,\mathrm d\tau\implies F(s)=G(s)H(s)`

where
`F(s),G(s),H(s)` are the respective Laplace transforms of
`f(t),g(t),h(t)`.

We have


`F(s)=\frac1{s^5(s^2+1)}=\frac1{4!}(4!)/(s^5)*\frac1{s^2+1}`

If we let
`G(s)=(4!)/(s^5)` and
`H(s)=\frac1{s^2+1}`, then clearly
`g(t)=t^4` and
`h(t)=\sin t`, so by the convolution theorem,


`f(t)=\frac1{4!}t^4*\sin t=\displaystyle\frac1{24}\int_(\tau=0)^(\tau=t)\tau^4\sin(t-\tau)\,\mathrm d\tau=(t^4)/(24)-\frac{t^2}2+1-\cos t`