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What's the inverse laplace transform of s/((s-.5)^2+1)?

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Wolfram alpha to the rescue! (:… – Nick Alger Apr 15 '11 at 7:26

1 Answer 1

Hint: $$\mathcal{L}(e^{-at}\cos\omega t) = \frac{s + a}{(s+a)^2 + \omega^2}$$ $$\mathcal{L}(e^{-at}\sin\omega t) = \frac{\omega}{(s+a)^2 + \omega^2}.$$

So maybe you can rewrite $$\frac{s}{(s-0.5)^2 + 1}$$ as a sum of two such functions and then take the inverse Laplace transform.

Alternatively: If you don't mind dealing with complex numbers, you might consider decomposing your function via partial fractions and using $\mathcal{L}(e^{at}) = 1/(s-a)$, though this might be annoying as a calculation (I haven't worked out the details).

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Blah, seems so obvious now. Must be because it's 4 am. Thanks a bunch =) – user9616 Apr 15 '11 at 7:34
@Anthony: Yeah, late night calculations are like that. No problem at all. :-) – Jesse Madnick Apr 15 '11 at 7:35

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