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Math Help - Inverse Laplace Transform

  1. #1
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    Inverse Laplace Transform

    Find the inverse Laplace Transform f(t)

     F \left( s \right) ={\frac {s \left( 1-{e}^{-2\,s} \right) }{{s}^{2}+{<br />
\pi }^{2}}}<br />

    The top half looks like the inverse staircase transform, [[a/t]], but that doesnt explain the denominator looking like the bottom of a sin transform ... not sure how to tackle this.
    Last edited by thedoge; April 15th 2008 at 05:49 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by thedoge View Post
    Find the inverse Laplace Transform f(t)

     F \left( s \right) ={\frac {s \left( 1-{e}^{-2\,s} \right) }{{s}^{2}+{<br />
\pi }^{2}}}<br />
    the inverse transform with unit step is given by the formula

    \mathcal{L}^{-1}(e^{-as}F(s))=f(t-a)\mathcal{U}(t-a) \mbox{ with } a >0

    lets expand to get

    \frac{s(1-e^{-2s})}{s^+\pi^2}=\frac{s}{s^2+\pi^2}+e^{-2s}\frac{s}{s^2+\pi^2}

    The inverse transform of

    \frac{s}{s^2+\pi^2}=\cos(\pi t)

    so we get

    \cos(\pi t) -\mathcal{U}(t-\pi)\cos(\pi (t-\pi))
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  3. #3
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    Thanks again empty! You're saving me here, and I'm learning to boot=]

    [edit]

    That was much easier than I thought it'd be.
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  4. #4
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    shouldn't the answer be:
    cos(pi.t) + H(t-2)[cos(pi.(t-2))]

    since a = 2 and we need to find f(t-2) instead of f(t- pi).
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