Question

Inverse laplace transform of H(s) = 1/(s+4)^2

Answer 1

Inverse Laplace of
`(1)/((S+4)^2)` will be
`te^(-4t)`

Explanation:

We have to find the inverse Laplace transform of
`H(S)=(1)/((S+4)^2)`

We know that of
`(1)/(s+4)` is
`e^(-4t)`

As in H(s) there is square of
`s+4`

So i inverse Laplace there will be multiplication of t

So the inverse Laplace of
`(1)/((s+4)^2)` will be
`te^(-4t)`

`L^(-1)(1)/((S+4)^2)=te^(-4t)`