Inverting Laplace Transforms

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Inverting Laplace Transforms

Postby factorialite » Thu Nov 01, 2007 2:26 am UTC

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?

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Re: Inverting Laplace Transforms

Postby aguacate » Thu Nov 01, 2007 4:58 am UTC

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


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Re: Inverting Laplace Transforms

Postby ATCG » Thu Nov 01, 2007 7:45 am UTC

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

Postby Quanglement Entree » Sat Nov 03, 2007 9:28 am UTC

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.
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Re: Inverting Laplace Transforms

Postby massivefoot » Mon Nov 05, 2007 8:24 am UTC

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!

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Re: Inverting Laplace Transforms

Postby Herman » Mon Nov 05, 2007 4:10 pm UTC

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.

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Re: Inverting Laplace Transforms

Postby factorialite » Mon Nov 12, 2007 3:10 pm UTC

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

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