Inverse Laplace Transform

**tremor** · October 10th 2010, 12:12 PM

Could someone help with the inverse transform of this:

$\frac{2e^{-2s}}{s((s+1)^{2}+1)}$

It's a little messier than I'm used to - I'm not quite sure how to handle the heaviside function as I can't find any examples like this. Thanks in advance for any help!!

**chisigma** · October 11th 2010, 05:34 AM

We can proceed 'step by step' as follows...

$\displaystyle \mathcal{L}^{-1} \{\frac{1}{1+s^{2}}\} = \sin t \implies$

$\displaystyle \implies \mathcal{L}^{-1} \{\frac{1}{1+(1+s^{2})}\}= e^{-t}\ \sin t \implies$

$\displaystyle \implies \mathcal{L}^{-1} \{\frac{1}{s\ \{1+(1+s^{2})\}}\}= \int_{0}^{t} e^{-\tau}\ \sin \tau\ d \tau= \frac{1}{2} \ \{1-e^{-t}\ (\sin t + \cos t)\}\implies$

$\displaystyle \implies \mathcal{L}^{-1} \{\frac{2\ e^{-2 s}}{s\ \{1+(1+s^{2})\}}\}= \{1-e^{-(t-2)}\ [\sin (t-2) + \cos (t-2)] \}\ \mathcal{U} (t-2)$

Kind regards

$\chi$ $\sigma$

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