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Math Help - Piecewise laplace transformation

  1. #1
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    Piecewise laplace transformation

    Here is the question:

    F(t) = {

    0, t<4
    5 sin(pi * t) 4<=t<5
    0, t>=5

    How do I find the Laplace transformation of this function?

    My answer is (5pi/(s^2+pi^2))(e^(-4s)-e^(-5s)) but it's wrong..

    help plz!!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Another more useful expression of the f(t) is...

    f(t) = 5\cdot \{\sin \pi t\cdot \mathcal{U}(t-4) - \sin \pi t\cdot \mathcal{U}(t-5)\}

    ... where \mathcal{U} (*) is the 'Haeviside step function'. Now You have to compute the LT of (1) 'term by term' in appropriate way...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 22nd 2009 at 09:48 PM.
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  3. #3
    Senior Member Pinkk's Avatar
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    The definition of the Laplace transform of a function f(t) is L[f(t)]=\int_{0}^{\infty}f(t)e^{-st}\,\,dt

    For this particular function we have:

    L[f(t)]=\int_{0}^{4}0e^{-st}\,\,dt + \int_{4}^{5}5sin(\pi t)e^{-st}\,\,dt + \int_{5}^{\infty}0e^{-st}\,\,dt

    The first and third integral terms above are equal to 0, so the Laplace transform would be \int_{4}^{5}5sin(\pi t)e^{-st}\,\,dt
    Now, using integration by parts, I got the answer:

    L[f(t)]=\frac{5\pi (e^{4s} + e^{5s})}{e^{9s}(s^{2} + \pi^{2})}

    I kinda rushed through the integration, so check for yourself what the integral equates to.
    Last edited by Pinkk; November 23rd 2009 at 12:02 PM.
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