Inverse laplace

Question

Find the inverse Laplace transform of 𝐹(𝑠), where 𝐹(𝑠)=

Answer 1

Hi Armed,

This problem can be made easy if you complete the square of (s^s + 2s + 2) and then apply the shift rule.

F(s) = 1 / [s (s^s + 2s + 2)] = 1 / [s (s^s + 2s + 1 + 1)] = 1 / [s ((s + 1)² + 1 )]

Taking the inverse Laplace of both sides, we get:

f(t) = L^-1 {1 / [s ((s + 1)² + 1 )]} = e^-t L^-1 {1 / [(s - 1) (s² + 1 )]}

Now, we can use partial fraction on 1 / [(s - 1) (s² + 1 )], or do some algebraic manipulations (easier in this case):

1 / [(s - 1) (s² + 1 )] = (1 / 2) [ [1 / (s - 1)]- [ (s + 1) / (s² + 1)] ] = (1 / 2) [ [1 / (s - 1)]- [ (s / (s² + 1)] - [1 / (s² + 1)] ]

f(t) = e^-t L^-1 {(1 / 2) [ [1 / (s - 1)]- [ (s / (s² + 1)] - [1 / (s² + 1)] ] } = (1/2) e^-t [e^t - cos t - sin t]

f(t) = (1/2) [1 - e^-t (cos t + sin t)]