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 $$\frac{a}{a^2+\omega^2} \Leftrightarrow \sin a\tau$$