Laplace transformation

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boks

Joined Oct 10, 2008
218
How can I transform

\(F(s) = \frac{1}{(s-2)^2}\)

?


I cannot see that any of the functions in my table is useful...

 
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silvrstring

Joined Mar 27, 2008
159
if L{e^(at)} = 1/(s-a) and L{t} = 1/ s^2

then inverse L{ 1/(s-2)^2 } = f(t) = te^(at)

There's a rule for it (a better way to put it). I have to go, though. If nobody posts it, put it up later.
 

steveb

Joined Jul 3, 2008
2,436
You can use the partial fraction expansion

Oops, I see your problem. You already thought of the partial fraction expansion, but that doesn't work for the degeneracy.

The above from silvrstring is correct: you can put the following in your table. It's needed for degeneracy cases with partial fraction expansions.

\( {{1}\over{(s+\alpha)^n}} \) transforms to \( {{t^{n-1}\over{(n-1)!}} e^{-\alpha t}\; u(t) \)

with region of convergence Re{s} > \( - \alpha \)

\( {{1}\over{(s+\alpha)^n}} \) transforms to \( -{{t^{n-1}\over{(n-1)!}} e^{-\alpha t}\; u(-t) \)

with region of convergence Re{s} < \( - \alpha \)
 
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