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Thread: Laplace Transform : Convolution

  1. #1
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    Laplace Transform : Convolution

    L { Integral (t- T)^2 Cos(2T) dT }

    T = tau

    The integral goes from 0 to t


    I am unsure how to get to the answer, which is 2/ (s^2)(s^2+4)
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  2. #2
    MHF Contributor chisigma's Avatar
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    Given $\displaystyle f(t)$ and $\displaystyle g(t)$ two functions and $\displaystyle F(s)$ and $\displaystyle G(s)$ their L-trasforms. The convolution is defined as...

    $\displaystyle f*g = \int_{0}^{t} f(t-\tau)\cdot g(\tau)\cdot d\tau$ (1)

    Is...

    $\displaystyle \mathcal{L} \{f*g\} = F(s)\cdot G(s)= \mathcal{L}\{t^{2}\}\cdot \mathcal{L} \{\cos 2t\} = \frac{2}{s^{3}} \cdot \frac{s}{s^{2} + 4} = \frac{2}{s^{2}\cdot (s^{2} +4)}$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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