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Math Help - laplace transforms

  1. #1
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    Exclamation laplace transforms

    I need to find the laplace transforms of these functions:
    a) (t-1)^4 <---dont know which property to use
    b) t*exp(-t)*sin(2t)
    c) cos(kt)^2 k = constant

    Any help would be great.
    My attempts are:
    b) 1/s^2 * F(s+1)*4/(s^2+4)
    Answer should be: 4(s+1)/[(s+1)^2+4]^2

    c) Answer should be: 4(3s^2-4)/(s^2+4)^3
    Last edited by ben.mahoney@tesco.net; March 22nd 2009 at 12:59 PM.
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  2. #2
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    I need to find the laplace transforms of these functions:
    a) (t-1)^4 <---dont know which property to use
    b) t*exp(-t)*sin(2t)
    c) cos(kt)^2 k = constant

    Any help would be great.
    My attempts are:
    b) 1/s^2 * F(s+1)*4/(s^2+4)
    Answer should be: 4(s+1)/[(s+1)^2+4]^2

    c) Answer should be: 4(3s^2-4)/(s^2+4)^3
    a) Expand (t - 1)^4 and take the Laplace transform of each term.

    -------------------------------------------------------------------------------------

    b) Here is the chain of logic you need to use:


    1. LT \left[e^{at} F(t)\right] = f(s - a) where f(s) = LT[F(t)].


    So LT \left[e^{-t} t \sin (2t) \right] = f(s + 1) where f(s) = LT[t \sin (2t)].


    2. LT[t G(t)] = - g'(s) where g(s) = LT[G(t)].


    So LT[t \sin (2t)] = - g'(s) where g(s) = LT[\sin (2t)].


    And I assume you can look up the relevant tables and get LT[\sin (2t)].

    --------------------------------------------------------------------------------

    c) Note that \cos^2 (kt) = \frac{\cos (2kt) + 1}{2} using the usual double angle formula.
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