# Laplace transformation

Discussion in 'Math' started by boks, Dec 11, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
How can I transform

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

?

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

Last edited: Dec 11, 2008
2. ### steveb Senior Member

Jul 3, 2008
2,432
469
You can use the partial fraction expansion

3. ### silvrstring Active Member

Mar 27, 2008
159
0
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.

4. ### steveb Senior Member

Jul 3, 2008
2,432
469

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} < $- \alpha$

Last edited: Dec 11, 2008
5. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
I solved it, thanks.