Convolution math homework

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Homework Statement



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

Homework Equations



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

The Attempt at a Solution


So [tex] S(\omega) = \frac{1}{(4+\omega^2)^2}=
\frac{1}{16}\frac{4}{(4+\omega^2)}\frac{4}{(4+\omega^2)} [/tex]


[tex] R(\tau)= \frac{1}{16} e^{-2|\tau|} * e^{-2|\tau|} [/tex]

Where * is convolution


So

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

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

Answers and Replies

  • #2
benorin
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The Laplace transform is [tex]
\frac{a}{a^2+\omega^2} \Leftrightarrow \sin a\tau
[/tex]
 
Last edited:

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