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Math Help - inverse laplace transform

  1. #1
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    inverse laplace transform

    I have the laplace transform:

    x(s) = 1 / ((s + a)^2 + (w^2 - a^2))

    what is the inverse laplace if w^2 < a^2 ????

    I have tried attempting the question without any luck. Step by step solution showing how it is solved would be very helpful for me to understand
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  2. #2
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    Re: inverse laplace transform

    x(t)=-\frac{e^{t \left(-a-\sqrt{a^2-w^2}\right)}-e^{t \left(-a+\sqrt{a^2-w^2}\right)}}{2 \sqrt{a^2-w^2}}
    Last edited by MaxJasper; August 20th 2012 at 05:16 PM.
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  3. #3
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    Re: inverse laplace transform

    Quote Originally Posted by MaxJasper View Post
    x(t)=-\frac{e^{t \left(-a-\sqrt{a^2-w^2}\right)}-e^{t \left(-a+\sqrt{a^2-w^2}\right)}}{2 \sqrt{a^2-w^2}}
    How exactly did you get to that solution? I thought you would use partial fractions in some way?
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  4. #4
    Senior Member MaxJasper's Avatar
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    Re: inverse laplace transform

    You can find InvLaplace for the following f(s) in any math handbook:
    f(s)=\mathcal{L}_s^{-1}\left[\frac{1}{(a+s)^2+b^2}\right](t)

    F(t)=\frac{e^{-a t} \text{Sin}[b t]}{b}
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  5. #5
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    Re: inverse laplace transform

    Or you can do it the hard way using complex analysis. If you know some complex analysis, read on. The inverse Laplace transform is the following

    \mathcal{L}^{-1} \{x(s)\} = x(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{st}x(s)\,ds

    where \gamma is real and greater than the real part of any singularity of the function x(s). The integral itself is unwieldy but we can use the Cauchy Residue theorem to calculate it. Assuming that the values a,w are real, we see that the singularities of x(s) are just

    s_1 = -a + \sqrt{a^2-w^2},\quad s_2 = -a - \sqrt{a^2-w^2}.

    Using the formulas for residues from simple poles we have

     x(t) = \frac{e^{s_2 t}}{(s_2+a)-\sqrt{a^2-w^2}} + \frac{e^{s_1 t}}{(s_1+a)+\sqrt{a^2-w^2}},

    which after simplifying leads to the expression given by MaxJasper. If you want I can elaborate on any of the parts.
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  6. #6
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    Re: inverse laplace transform

    Is this for w^2 < a^2?
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  7. #7
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    Re: inverse laplace transform

    Yes
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  8. #8
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    Re: inverse laplace transform

    And what about the answer in the case w^2 = a^2??
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