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Math Help - Tricky inverse Laplace

  1. #1
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    Tricky inverse Laplace

    G'day,

    I struggle with Laplace transforms and calculating their inverses. Could someone please help me with the following?

     F(s) = \frac{6s^2 + 9s +1}{(s+1)^2 (s-1)}

    Thanks
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  2. #2
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    Quote Originally Posted by Jimmy_W View Post
    G'day,

    I struggle with Laplace transforms and calculating their inverses. Could someone please help me with the following?

     F(s) = \frac{6s^2 + 9s +1}{(s+1)^2 (s-1)}
    Split it into partial fractions F(s) = \frac A{s-1} + \frac B{s+1} + \frac C{(s+1)^2}. Then use your table of inverse Laplace transforms to find the inverse transform of each fraction.
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  3. #3
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    I havent gotten far, even with your help.

    So from your post I have

     6s^2 + 9s+1 = A(s+1)^2 + B(s-1)(s+1)^2 + C (s-1)(s+1)^2

    So for s = 1,

     16 = 4A and so A = 4

    but when i cant get B and C.
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  4. #4
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    Quote Originally Posted by Jimmy_W View Post
    I havent gotten far, even with your help.

    So from your post I have

     6s^2 + 9s+1 = A(s+1)^2 + B(s-1)(s+1)^2 + C (s-1)(s+1)^2

    So for s = 1,

     16 = 4A and so A = 4

    but when i cant get B and C.
    You should have had 6s^2 + 9s+1 \equiv A(s+1)^2 + B(s-1)(s+1) + C(s-1). Try it now.
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  5. #5
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    I get A = 4, C=-2, B = 1. Is this correct?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    6s^2 + 9s +1 = 6(s+1)^2 - 3(s+1) -2

    F(s) = 6/(s-1) -3/[(s+1)(s-1)] -2/[(s+1)^2(s-1)]

    The first 2 transforms are standards the last could be done fairly simple with convolution
    f(t) = -2te^(-t) g(t) = e^t
    Last edited by Calculus26; August 27th 2009 at 08:18 PM.
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