inverse laplace transform

**TshingY** · May 3rd 2009, 11:17 AM

how would i transform this inverse laplace?

${\frac{s}{(s^2+2s+5)^2}}$

**TheEmptySet** · May 3rd 2009, 12:31 PM

Originally Posted by TshingY

how would i transform this inverse laplace?

${\frac{s}{(s^2+2s+5)^2}}$

There are a few different ways.....

First complete the square in the denominator to get

$F(s)=\frac{s}{[(s+1)^2+4]^2}$

Note that
$F(s)=G(s+1)-H(s+1)$

where G and H are below

$G(s)= \frac{s}{[s^2+4]}$ and

$H(s) = \frac{1}{[s^2+4]}$

notice that each of these is a translation on the s axis

$L^{-1}F(s)=L^{-1}G(s+1) - L^{-1}H(s+1)=e^{-t}L^{-1}G(s)-L^{-1}H(s)$

$G(s)=\frac{1}{2} (-1)\frac{d}{ds}\left( \frac{1}{s^2+4}\right)$ so the inverse transform is

$L^{-1} G(s) = \frac{t}{4}\sin(2t)$

$H(s)=\frac{1}{s^2+4}\cdot \frac{1}{s^2+4}$

This is the convolution of $\frac{1}{2}\sin(2t)$ with its self so its inverse transform is

$L^{-1} H(s)=\frac{1}{4}\int_{0}^{t}\sin(2t)\sin(2t-2\tau)d\tau=\frac{\sin(2t)}{4}\left( \frac{-\cos(2t-2\tau)}{-2}\right)\bigg|_{0}^{t}$

$=\frac{\sin(2t)}{8}(1-\cos(2t)$

Just plug this in above and we are done

YAY

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