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Math Help - inverse laplace transform

  1. #1
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    inverse laplace transform

    How to show that
    L^{-1} [\frac{6}{(s^2+2s+2)^2}]=3\exp(-t)sin(t)-3t\exp(-t)cos(t)
    where L^{-1}[F(s)] = f(t) is the inverse laplace ?
    Ty
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: inverse laplace transform

    \mathcal{L}\{3e^{-t}\sin t-3e^{-t}\cos t\}=3\mathcal{L}\{e^{-t}\sin t\}-3\mathcal{L}\{e^{-t}\cos t\} . Now use

    \mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a)^2+b^2},\quad \mathcal{L}\{e^{-at}\cos bt\}=\frac{s+a}{(s+a)^2+b^2} .
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  3. #3
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    Re: inverse laplace transform

    Quote Originally Posted by FernandoRevilla View Post
    \mathcal{L}\{3e^{-t}\sin t-3e^{-t}\cos t\}=3\mathcal{L}\{e^{-t}\sin t\}-3\mathcal{L}\{e^{-t}\cos t\} . Now use

    \mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a)^2+b^2},\quad \mathcal{L}\{e^{-at}\cos bt\}=\frac{s+a}{(s+a)^2+b^2} .
    Done, thanks!
    Now i have to find
    L^{-1} [\frac{6}{(s^2+2s+2)^2}]
    without knowing the result previously.
    Thank
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  4. #4
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    Re: inverse laplace transform

    use partial fractions
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  5. #5
    Super Member General's Avatar
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    Re: inverse laplace transform

    Partial Fraction Decomposition will not do anything here.

    Use the Convolution Theorem.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: inverse laplace transform

    Quote Originally Posted by General View Post
    Partial Fraction Decomposition will not do anything here.

    Use the Convolution Theorem.
    The way indicated by General is the 'winning key'!...

    Setting F(s)= \frac{1}{s^{2}+2\ s + 2}= \frac{1}{1+(s+1)^2} is \mathcal{L}^{-1} \{F(s)\}= e^{-t}\ \sin t, so that is...

    \mathcal{L}^{-1} \{F^{2}(s)\} = \int_{0}^{t} e^{-\tau}\ e^{\tau - t}\ \sin \tau\ \sin (t-\tau)\ d \tau =

    = e^{-t}\ (\sin t\ \int_{0}^{t} \sin \tau\ \cos \tau\ d \tau - \cos t\ \int_{0}^{t} \sin^{2} \tau\ d \tau) =

    = \frac{e^{-t}}{2}\ (\sin t - \sin t\ \cos^{2} t - t\ \cos t + \sin t\ \cos^{2} t) =

    = \frac{e^{-t}}{2}\ \{\sin t - t\ \cos t \}

    Kind regards

    \chi \sigma
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