What is the Laplace transform of a convolution?

[wildman]

Homework Statement

Find R(\tau) if a) S(\omega) = \frac{1}{(4+\omega^2)^2}

Homework Equations

I have given \frac{4}{4+\omega^2} <==> e^{-2|\tau|}

The Attempt at a Solution

So S(\omega) = \frac{1}{(4+\omega^2)^2}= <br /> \frac{1}{16}\frac{4}{(4+\omega^2)}\frac{4}{(4+\omega^2)}R(\tau)= \frac{1}{16} e^{-2|\tau|} * e^{-2|\tau|}

Where * is convolutionSo

R(\Tau) = \frac {1}{8}\int_{0}^{\infty} e^{-2(\tau-\alpha)} e^{-2\alpha} d\alpha

But that turns out to be infinite. Does anyone have any idea where I went wrong?

 

[benorin]

The Laplace transform is <br /> \frac{a}{a^2+\omega^2} \Leftrightarrow \sin a\tau<br />