Question

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

Answer 1

Answer:

Inverse Laplace of $\frac{1}{(S+4)^2}$ will be $te^{-4t}$

Step-by-step explanation:

We have to find the inverse Laplace transform of $H(S)=\frac{1}{(S+4)^2}$

We know that of $\frac{1}{s+4}$ 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 $\frac{1}{(s+4)^2}$  will be $te^{-4t}$

$L^{-1}\frac{1}{(S+4)^2}=te^{-4t}$