## Inverting Laplace Transforms

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factorialite
Posts: 16
Joined: Thu Nov 01, 2007 2:12 am UTC

### Inverting Laplace Transforms

I'm supposed to find the inverse Laplace transform of the given equation:

F(s) = 2*e-2s/s2-4.

Things I know:uc(t) is the L-1 of e-cs/s.
The inverse Laplace of a/s2-a2 and s/s2-a2 are sinh and cosh, respectively.

Things I don't know: How to solve this equation.
How to solve Laplace transforms in general (especially when it seems like there are many different transforms in the same equation.)

Can anybody help me?

aguacate
Posts: 209
Joined: Fri Feb 16, 2007 10:29 pm UTC

### Re: Inverting Laplace Transforms

All I know is that laplace transforms kicked ass when I was analyzing circuits. I vaguely remember that multiplication in the laplace domain was addition in the time domain. So the inverse of F(s) would work out to be

L-1(e-2s)+L-1(2/s2-4)

ATCG
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### Re: Inverting Laplace Transforms

You might investigate the convolution theorem.
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Quanglement Entree
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### Re: Inverting Laplace Transforms

Convolution isn't needed here, which is a good thing, because I haven't looked into it.

Try calculating L[uc(t) * f(t - c)](s). You should find it comes out to exp(-cs) * L[f(t)](s). My lecturers called this the "shifting theorem". You can then use this backwards to solve the sort of problem you have given.
Because it's homocide when you start killing time.

massivefoot
Posts: 39
Joined: Mon Nov 05, 2007 8:19 am UTC

### Re: Inverting Laplace Transforms

factorialite, the general way to invert a Laplace transform F is:

with the contour to the right of all the poles of F in the complex plane. Usually you close the contour in either the left of right hand plane depending on the value of t.

Hope this helps!

Herman
Posts: 559
Joined: Wed May 02, 2007 2:46 am UTC

### Re: Inverting Laplace Transforms

You could figure it out directly, with massivefoot's formula. But this looks like it'll be pretty easy to convolute, and that's how I learned it. The method goes like this:

You have two functions, G(s) and H(s) that you recognize from your Laplace transform tables.

g(t) = L-1G(s), and h(t) = L-1H(s)

Then, L-1(G(s)*H(s)) = int(0-->t)g(t-x)h(x)dx

Where int(0-->t) is the integral from 0 to t. x is just a dummy variable used for integration; your final answer should be in terms of t, as expected.

Note that g and h are interchangable, so you have a choice of which one gets t-x and which one gets x in the integral above. Pick the one that looks algebraically easier.

factorialite
Posts: 16
Joined: Thu Nov 01, 2007 2:12 am UTC

### Re: Inverting Laplace Transforms

Thanks a lot guys! I got it, using massivefoot's.

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