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Math Help - inverse laplace transform

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    inverse laplace transform

    how would i transform this inverse laplace?

    {\frac{s}{(s^2+2s+5)^2}}
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    Quote Originally Posted by TshingY View Post
    how would i transform this inverse laplace?

    {\frac{s}{(s^2+2s+5)^2}}
    There are a few different ways.....

    First complete the square in the denominator to get

    F(s)=\frac{s}{[(s+1)^2+4]^2}

    Note that
    F(s)=G(s+1)-H(s+1)

    where G and H are below

    G(s)= \frac{s}{[s^2+4]} and

    H(s) = \frac{1}{[s^2+4]}

    notice that each of these is a translation on the s axis

    L^{-1}F(s)=L^{-1}G(s+1) - L^{-1}H(s+1)=e^{-t}L^{-1}G(s)-L^{-1}H(s)

    G(s)=\frac{1}{2} (-1)\frac{d}{ds}\left( \frac{1}{s^2+4}\right) so the inverse transform is

    L^{-1} G(s) = \frac{t}{4}\sin(2t)

    H(s)=\frac{1}{s^2+4}\cdot \frac{1}{s^2+4}

    This is the convolution of \frac{1}{2}\sin(2t) with its self so its inverse transform is

    L^{-1} H(s)=\frac{1}{4}\int_{0}^{t}\sin(2t)\sin(2t-2\tau)d\tau=\frac{\sin(2t)}{4}\left( \frac{-\cos(2t-2\tau)}{-2}\right)\bigg|_{0}^{t}

    =\frac{\sin(2t)}{8}(1-\cos(2t)

    Just plug this in above and we are done

    YAY
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