Answer:
Inverse Laplace of [tex]\frac{1}{(S+4)^2}[/tex] will be [tex]te^{-4t}[/tex]
Step-by-step explanation:
We have to find the inverse Laplace transform of [tex]H(S)=\frac{1}{(S+4)^2}[/tex]
We know that of [tex]\frac{1}{s+4}[/tex] is [tex]e^{-4t}[/tex]
As in H(s) there is square of [tex]s+4[/tex]
So i inverse Laplace there will be multiplication of t
So the inverse Laplace of [tex]\frac{1}{(s+4)^2}[/tex] will be [tex]te^{-4t}[/tex]
[tex]L^{-1}\frac{1}{(S+4)^2}=te^{-4t}[/tex]
