What is the easiest way to find the inverse Laplace of F(s)?

Question

$$ F(s)= \frac{1}{(s-1)^2(1-1/s^2)} $$

Do I have to multiply by $s^2/s^2$ and then use partial fractions or is there a way to use the convolution theorem?

Answer 1

This can be rewritten as (another form is also possible):

$$F(s) = \dfrac{s^2}{(s-1)^3 (s+1)}$$

Using Partial fractions, this is:

$$F(s) = \dfrac{s^2}{(s-1)^3 (s+1)}= -\dfrac{1}{8 (s+1)} + \dfrac{1}{8 (s-1)}+\dfrac{3}{4 (s-1)^2}+\dfrac{1}{2 (s-1)^3}$$

Now, you can find the inverse laplace transform using a table or using the definitions.

We arrive at:

$$F(t) = \dfrac{1}{8} e^{-t} (2 e^{2 t} t^2+6 e^{2 t} t+e^{2 t}-1)$$

A second form is possible (of course the ILT is identical):

$$F(s) = = \dfrac{s^2}{(s-1)^3 (s+1)}= \dfrac{s^2}{(s-1)^2 (s^2-1)}$$