What's the inverse laplace transform of s/((s-.5)^2+1)?
Inverse Laplace Transform
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integral-transforms
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0Wolfram alpha to the rescue! (: http://www.wolframalpha.com/input/?i=What%27s%20the%20inverse%20laplace%20transform%20of%20s/%28%28s-.5%29%5E2%2b1%29?%20 – 2011-04-15
1 Answers
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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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0@Anthony: Yeah, late night calculations are like that. No problem at all. :-) – 2011-04-15