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Laplace Transform - Step Function

  1. Aug 5, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the Laplace transform of f(t) = t^2 - 18 for 0 < t < 3 and f(t) = (t-3)^2 for t>3?


    2. Relevant equations

    Laplace Transforms

    3. Work

    Using Heaviside/step function, made equation into:

    f(t) = t^2 - 18 + u(t-3)( (t-3)^2 - (t^2-18) )

    Then, using Laplace transforms, found:

    L{f(t)} = ( 2 / s^3 ) - ( 18 / s ) + e^(-3s)*L{27-6t}

    I know that I have to put t in the L{27-6t} into either t-3 or t+3, but not sure which.

    I went under the assumption that it t+3 would substitute for the t, so applying Laplace transforms I found:

    L{27-6t} = L{27-6(t+3)} = L{27-6t-18} = L{9-6t} = 9/s - 6/s^2

    Leaving me with the final answer:

    L{f(t)} = 2/s^3 - 18/s + e^-3s*(9/s - 6/s^2)

    ...any chance this is close to right?
     
    Last edited: Aug 5, 2007
  2. jcsd
  3. Aug 6, 2007 #2
    em,you're quite right
    Let
    u(t-3)( (t-3)^2 - (t^2-18) )=u(t-3)g(t-3)
    You're using the property of time shifting. Put t into t+3 and you get u(t)g(t).
    Then
    L{u(t-3)g(t-3)}=e^(-3s) L{u(t)g(t))}

    ps:
    g(t-3)=27-6t
    g(t)=9-6t
     
    Last edited: Aug 6, 2007
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