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Math Help - Laplace Transform

  1. #1
    Member Maccaman's Avatar
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    Laplace Transform

    Could someone please help with the following:

    Find the laplace transform of the following:

     f(t) = e^{4t} \ \sin^2(3t)

    I dont know if my method is correct, but here is what I started with.

    f(t) becomes

     f(t) = \frac{1}{2} e^{4t} - \frac{1}{2} e^{4t} \cos(6t)

    Let  g(t) = e^{4t}, ... h(t) = e^{4t} \cos(6t)

    and
     A = \frac{1}{2}, B = -\frac{1}{2}

    Then  F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} (A(g(t)) + B(h(t) )

    skipping a few lines of calculation....

     = \frac{1}{2} \int_0^{\infty} e^{(4-s)t} dt + (\frac{-1}{2} ) \ \mathcal{L} (\cos(6t) e^{4t})

     = \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}

    but I am not really sure. Im lost
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Maccaman View Post
    Find the Laplace transform of the following:

     f(t) = e^{4t} \ \sin^2(3t)

    .....

     = \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}

    but I am not really sure. Im lost
    Your answer looks correct to me.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Maccaman View Post
    Could someone please help with the following:

    Find the laplace transform of the following:

     f(t) = e^{4t} \ \sin^2(3t)

    I dont know if my method is correct, but here is what I started with.

    f(t) becomes

     f(t) = \frac{1}{2} e^{4t} - \frac{1}{2} e^{4t} \cos(6t)

    Let  g(t) = e^{4t}, ... h(t) = e^{4t} \cos(6t)

    and
     A = \frac{1}{2}, B = -\frac{1}{2}

    Then  F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} (A(g(t)) + B(h(t) )

    skipping a few lines of calculation....

     = \frac{1}{2} \int_0^{\infty} e^{(4-s)t} dt + (\frac{-1}{2} ) \ \mathcal{L} (\cos(6t) e^{4t})

     = \frac{1}{2s-8} + \frac{-1}{2} \times \frac{s-4}{(s-4)^2 + 6^2}

    but I am not really sure. Im lost
    You can just use Frequency shifting \mathcal{L}\left\{ {{e^{\alpha t}}f\left( t \right)} \right\} = F\left( {s - \alpha } \right)

    So g\left( t \right) = {e^{4t}}{\sin ^2}3t

    Then f\left( t \right) = {\sin ^2}3t

    \mathcal{L}\left\{ {{{\sin }^2}3t} \right\} = \mathcal{L}\left\{ {\frac{1}{2}\left( {1 - \cos 6t} \right)} \right\} = \frac{1}{2}\left( {\frac{1}{s} - \frac{s}{{{s^2} + 36}}} \right)

    Finally \mathcal{L}\left\{ {{e^{4t}}{{\sin }^2}3t} \right\} = \frac{1}{2}\left( {\frac{1}{{s - 4}} - \frac{{s - 4}}{{{{\left( {s - 4} \right)}^2} + 36}}} \right)
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