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...

    [​IMG]
     
    Last edited: Dec 11, 2008
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    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
    2,433
    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.
     
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