Inverse Laplace Transform of a polynomial fraction

Question

How do I find the inverse Laplace transform of $\;\;\large\frac{4s}{(s^2+4)^2}\;\;$?

Answer 1

Hints: Let $F(s)$ be the Laplace transform of $f(x)$, then

1) $ \mathcal{L}\left\{\sin(ax)\right\}=\frac{a}{a^2+s^2} $

2) $ \mathcal{L}\left\{x f(x)\right\} = \frac{d}{ds}F(s). $

See here for a related problem.

Answer 2

Unless you are working out of tables, these can be a little harder because of the squared polynomial in the denominator. To find the ILT, you use the definition:

$$f(t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds\: e^{s t} \frac{4 s}{(s^2+4)^2}$$

This is an integral in the complex plane; if your are doing these then you are familiar with the technique: define the so-called Bromwich contour and use the residue theorem. That is,

$$f(t) = \sum \text{Res}_{s=\pm 2 i}\: \frac{4 s \, e^{s t}}{(s^2+4)^2}$$

To evaluate the residues at the poles $s=\pm2 i$, we must evaluate

$$\text{Res}_{s=\pm 2 i}\: \frac{4 s \, e^{s t}}{(s^2+4)^2} = \lim_{s \rightarrow \pm 2 i} \frac{d}{ds} \left [ (s \mp 2 i)^2 \frac{4 s \, e^{s t}}{(s^2+4)^2} \right ]$$

Doing this is an exercise in taking derivatives and evaluating them at the poles. Note that we are excluding the pieces where the integrand blows up at these poles. Leaving the work to you, I get

$$f(t) = t\,\sin{2 t} $$

Answer 3

Rule:

$$ \mathcal{L}\left\{tsin(wt)\right\} = \dfrac{2 \omega s}{(s^2 + \omega^2)^2} \iff \mathcal{L}^{-1}\left\{ \dfrac{2 \omega s}{(s^2 + \omega^2)^2} \right\} = tsin(wt) $$

If you consider $\omega=2$ in your question the corresponding time domain expression will be $2\sin(2t)$.