Inverse Laplace Transforms

**kpizle** · April 28th 2010, 03:52 AM

Hey

Having trouble with a couple of inverse Laplace transforms:

$L^{-1} {\frac{s}{(s-2)^2 + 9}}$

I know this uses $cos{kt}$ and $e^{at}$ , but can't figure out how to get that -2 out of there...

Also,

$L^{-1} {\frac{s+4}{s^2 + 4s + 8}}$

I'm just lost on this one.

Please help!

**chisigma** · April 28th 2010, 04:23 AM

From the tables...

$\mathcal {L}^{-1} \{\frac{1}{(s-b)^{2} + a^{2}}\} = \frac{e^{b t} \cdot \sin a t}{a}$ (1)

... so that is...

$\mathcal {L}^{-1} \{\frac{1}{(s-2)^{2} + 9}\} = \frac{e^{2 t} \cdot \sin 3 t}{3}$ (2)

A basic properties of the LT is...

$\mathcal {L} \{f^{'} (t)\}= s\cdot \mathcal {L} \{f(t)\} - f(0)$ (3)

... so that...

$\mathcal {L}^{-1} \{\frac{s}{(s-2)^{2} + 9}\} = e^{2 t}\cdot (\frac{2}{3}\cdot \sin 3 t + \cos 3t)$ (4)

Kind regards

$\chi$ $\sigma$

**chisigma** · April 28th 2010, 01:35 PM

Is...

$\frac{s+4}{s^{2} + 4 s + 8} = \frac{s+4}{(s +2)^{2} + 4}$ (1)

... and because [see previous post...] is...

$\mathcal {L}^{-1} \{\frac{1}{(s+2)^{2} + 4}\} = \frac{e^{-2t}\cdot \sin 2t}{2}$ (2)

... and ...

$\mathcal {L}^{-1} \{\frac{s}{(s+2)^{2} + 4}\} = e^{-2t}\cdot (\cos 2t - \sin 2t)$ (3)

... we conclude that...

$\mathcal {L}^{-1} \{\frac{s+4}{(s+2)^{2} + 4}\} = e^{-2t}\cdot (\sin 2t + \cos 2t)$ (4)

Kind regards

$\chi$ $\sigma$

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