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Math Help - Convolution

  1. #1
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    Convolution

    Given:

    y'' +4y = sin3t
    y'(0) = y(0) = 0

    From this, I can pretty easily find Y(s) to be:

    Y(s) = 3/[(s^2 + 3^2) (s^2 + 2^2)]

    In the above, I can see sin3t and (1/2) sin2t, but I'm having trouble seeing how to get back to just plain y(t). Any help will be appreciated.
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  2. #2
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    Quote Originally Posted by dsprice View Post
    Given:

    y'' +4y = sin3t
    y'(0) = y(0) = 0

    From this, I can pretty easily find Y(s) to be:

    Y(s) = 3/[(s^2 + 3^2) (s^2 + 2^2)]

    In the above, I can see sin3t and (1/2) sin2t, but I'm having trouble seeing how to get back to just plain y(t). Any help will be appreciated.
    Use partial fractions!

     \frac{3}{(s^2 + 3^2)(s^2 + 2^2)} = \frac{As+B}{s^2 + 3^2} + \frac{Cs+D}{s^2+2^2}

    Multiply through by the denominator of the LHS:

     3 = (As+B)(s^2 + 4) + (Cs+D)(s^2 + 9)

    Expand it out:

     3 = As^3 + 4As + Bs^2 4B + Cs^3 + 9Cs + Ds^2 + 9D

    Not collect like terms in s:

     3 = (A+C) s^3 + (B+D)s^2 + (4A+9C)s + (4B+9D)

    Now, if you compare coefficients of both sides of the equation, you get the following:

     A+C = 0
     B+D = 0
     4A+9C = 0
     4B + 9D = 3

    This is two groups of two equations in two unknowns. Easily solveable!
    Last edited by Mush; October 18th 2009 at 10:18 AM.
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