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Math Help - Laplace Transform - Convolution Theorem

  1. #1
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    Post Laplace Transform - Convolution Theorem

    Q)The output y(t) of a linear system, with impulse response of h(t), subjected to an input function x(t) is given by y(t) = h(t) * x(t)

    If x(t) = u(t) and H(s) = 4/s^2 +2s + 5 ,

    Solve for y(t) by using the Convolution Theorem.

    My approach:
    I've tried to apply the partial fraction to find the coefficient of
    (4/s^2 + 2s +5) X1/s = (As + B/s^2 +2s + 5) + (C/s)

    A=-4/5; B=-8/5; C=4/5

    Then apply,
    inverse L{Y(s)} = y(t)
    = inverse L{[(-4/5)s + (-8/5)/s^2 + 2s +5]+ 4/5s}
    .....

    but can't get the final answer (4/5e^-t cos2t - 2/5e^-t sin2t + 4/5) u(t)

    Please help to enlighten me on this
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Octopus View Post
    Q)The output y(t) of a linear system, with impulse response of h(t), subjected to an input function x(t) is given by y(t) = h(t) * x(t)

    If x(t) = u(t) and H(s) = 4/s^2 +2s + 5 ,

    Solve for y(t) by using the Convolution Theorem.

    My approach:
    I've tried to apply the partial fraction to find the coefficient of
    (4/s^2 + 2s +5) X1/s = (As + B/s^2 +2s + 5) + (C/s)

    A=-4/5; B=-8/5; C=4/5

    Then apply,
    inverse L{Y(s)} = y(t)
    = inverse L{[(-4/5)s + (-8/5)/s^2 + 2s +5]+ 4/5s}
    .....

    but can't get the final answer (4/5e^-t cos2t - 2/5e^-t sin2t + 4/5) u(t)

    Please help to enlighten me on this
    (\mathcal{L}u)(s)=1/s

    So:

     <br />
\mathcal{L}y=\mathcal{L}(h*u)=H(s)(\mathcal{L}u)(s  )=\frac{4}{s^2 + 2s +5} \times \frac{1}{s}=\frac{4}{s^3 + 2s^2 +5s}<br />

    Now use partial fractions ...

    CB
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