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I have the following formula, which I need to get the inverse laplace transform of:

$\frac{2s}{s^2 + 6s + 13}$

I've managed to get $2\mathrm{e}^{-3t}\cos(2t)$, that's rather simple - but according to matlab there's one more term: $-3 \mathrm{e}^{-3t} \sin(2t)$

I don't really understand why.

I used completion of square on the denominator and got $\frac{2s}{(s+3)^2+4}$, which seemed to fit the formula for damped cosine.

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We have (through the residue theorem or partial fraction decomposition): $$\mathcal{L}^{-1}\left(\frac{2s}{(s+3)^2+4}\right)=\mathcal{L}^{-1}\left(\frac{1-\frac{3}{2}i}{s-(-3-2i)}+\frac{1+\frac{3}{2}i}{s-(-3+2i)}\right) $$ and since: $$\mathcal{L}^{-1}\left(\frac{1}{s-\alpha}\right) = e^{\alpha t}$$ it follows that: $$\begin{eqnarray*}\mathcal{L}^{-1}\left(\frac{2s}{(s+3)^2+4}\right)&=&\left(1-\frac{3}{2}i\right)e^{-(3+2i)t}+\left(1+\frac{3}{2}i\right)e^{-(3-2i)t}\\&=&\color{red}{e^{-3t}\left(2\cos(2t)-3\sin(2t)\right)}.\end{eqnarray*}$$

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  • $\begingroup$ hm, I don't understand this method... but I've meanwhile figured it out the original way, so it's OK $\endgroup$
    – MightyPork
    Feb 15, 2015 at 19:05

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