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Math Help - Inverse Laplace Transform

  1. #1
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    Inverse Laplace Transform

    Could someone help with the inverse transform of this:

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

    It's a little messier than I'm used to - I'm not quite sure how to handle the heaviside function as I can't find any examples like this. Thanks in advance for any help!!
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  2. #2
    MHF Contributor chisigma's Avatar
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    We can proceed 'step by step' as follows...

    \displaystyle \mathcal{L}^{-1} \{\frac{1}{1+s^{2}}\} = \sin t \implies

    \displaystyle \implies \mathcal{L}^{-1} \{\frac{1}{1+(1+s^{2})}\}= e^{-t}\ \sin t \implies

    \displaystyle \implies \mathcal{L}^{-1} \{\frac{1}{s\ \{1+(1+s^{2})\}}\}= \int_{0}^{t} e^{-\tau}\ \sin \tau\ d \tau=  \frac{1}{2} \ \{1-e^{-t}\ (\sin t + \cos t)\}\implies

    \displaystyle \implies \mathcal{L}^{-1} \{\frac{2\ e^{-2 s}}{s\ \{1+(1+s^{2})\}}\}= \{1-e^{-(t-2)}\ [\sin (t-2) + \cos (t-2)] \}\ \mathcal{U} (t-2)

    Kind regards

    \chi \sigma
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