2

I can't seem to get this Laplace working using the convolution method.

$H(s) = \frac{1}{s^2(s+2)}$

Which I can't get to work using convolution. So I am separating it into $\frac{1}{s^2} * \frac{1}{s+2}$ and then I think I go from here with convolution?

How do I do this?

This takes forms $t$ and $e^{-2t}$

So I plugged it into the theorem and I got $\int_o^\tau (t-\tau)e^{-2\tau} \mathrm{d\tau}$ which gives me $2t$, but this is not the answer in the book? What is wrong?

Nevermind I got it, I will post a self answer soon(Integral above forgot to do int by parts).

Michael
  • 92
  • 5

2 Answers2

1

We can find the inverse Laplace transform by taking Mellin inversion formula. In the problem, we have a simple pole at $s=-2$ and a pole of order two at $s=0$. \begin{align} h(t) &= \mathcal{L}^{-1}\{F(s)\}\\ &=\frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{s^2(s+2)}ds\\ &= \lim_{s\to 0}\frac{d}{ds}s^2\frac{e^{st}}{s^2(s+2)}+\lim_{s\to -2}(s+2)\frac{e^{st}}{s^2(s+2)}\\ &= \frac{2t - 1 + e^{-2t}}{4} \end{align}

dustin
  • 8,559
0

To find the inverse Laplace you need to take the convolution of the functions $x$ and $e^{-2x}$. See a related problem.

Community
  • 1
Mhenni Benghorbal
  • 48,099