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Differential equations - inverse laplace transforms

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    We are given the following LaPlace Transformations and we are to find g(t) and h(t).

    L(g)(t) = 1/[(1-e^(-2s))(s-1)]

    L(h)(t) = (e^-s)/s(1-e^-s)

    2. Relevant equations

    By looking at the Table of LaPlace Transformations, the only one I found that could possibly work for both of these is

    f(t+T) = f(t) ---> (integral from 0 to T of e^(-st)f(t) dt)/ 1-e^(-sT)

    3. The attempt at a solution

    For the first one, I broke it into two separate equations, 1/(1-e^-2s) * 1/(s-1) and for the second part it would be e^t. However, I am very confused on the equation I found in the Table and I'm not sure how to use it on either of these.
     
  2. jcsd
  3. Mar 8, 2008 #2

    Tom Mattson

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    Gold Member

    That formula makes it easy to compute the Laplace transform of a periodic function, but I don't see how it's going to make it easy to do the inverse transform. I think the following is a better approach.

    * Use the summation formula for convergent geometric series:

    [tex]\sum_{n=0}^\infty x^n = \frac{1}{1-x}[/tex], [tex]|x|<1[/tex].

    Here, [itex]x=e^-Ts[/itex]. Since [itex]T>0[/itex], you can apply the above sum formula.

    Once you have done that, then...

    * Make repeated use the time shift property of Laplace transforms:

    [tex]e^{-Ts}F(s) \longrightarrow f(t-T)U(t-T)[/tex]

    I think you'll have more success taking that route.
     
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