# Inverse Laplace Transform

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• Apr 15th 2008, 05:32 PM
thedoge
Inverse Laplace Transform
Find the inverse Laplace Transform f(t)

$F \left( s \right) ={\frac {s \left( 1-{e}^{-2\,s} \right) }{{s}^{2}+{
\pi }^{2}}}
$

The top half looks like the inverse staircase transform, [[a/t]], but that doesnt explain the denominator looking like the bottom of a sin transform ... not sure how to tackle this.
• Apr 15th 2008, 05:48 PM
TheEmptySet
Quote:

Originally Posted by thedoge
Find the inverse Laplace Transform f(t)

$F \left( s \right) ={\frac {s \left( 1-{e}^{-2\,s} \right) }{{s}^{2}+{
\pi }^{2}}}
$

the inverse transform with unit step is given by the formula

$\mathcal{L}^{-1}(e^{-as}F(s))=f(t-a)\mathcal{U}(t-a) \mbox{ with } a >0$

lets expand to get

$\frac{s(1-e^{-2s})}{s^+\pi^2}=\frac{s}{s^2+\pi^2}+e^{-2s}\frac{s}{s^2+\pi^2}$

The inverse transform of

$\frac{s}{s^2+\pi^2}=\cos(\pi t)$

so we get

$\cos(\pi t) -\mathcal{U}(t-\pi)\cos(\pi (t-\pi))$
• Apr 15th 2008, 05:50 PM
thedoge
Thanks again empty! You're saving me here, and I'm learning to boot=]

That was much easier than I thought it'd be.
• Apr 21st 2008, 10:56 PM
Danshader
shouldn't the answer be:
cos(pi.t) + H(t-2)[cos(pi.(t-2))]

since a = 2 and we need to find f(t-2) instead of f(t- pi).