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Discontinuous Forcing terms

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Find Laplace transformation
    H(t-1)t^2


    2. Relevant equations




    3. The attempt at a solution

    [2(e^-s)f(t^2)]/s^3

    I'm not sure how to get t^2 in terms of t-1

    I know the answer is (2+2s+s^2)/s^3
     
  2. jcsd
  3. Mar 24, 2012 #2

    LCKurtz

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    Just express ##t^2## as a finite Taylor series about ##t=1##. Alternatively you can use this formula$$
    \mathcal L(u(t-a)f(t)) = e^{-as}\mathcal L(f(t+a)$$which would give, in your case$$
    \mathcal L(u(t-1)t^2)=e^{-s}\mathcal L((t+1)^2)
    = e^{-s}\mathcal L(t^2+2t+1)$$
     
  4. Mar 25, 2012 #3
    can you show me the proof on that formula because it is no where in my book
     
  5. Mar 25, 2012 #4

    LCKurtz

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    $$\mathcal L(u(t-a)f(t)=\int_a^\infty e^{-st}f(t)dt$$Let ##v = t-a##$$
    =\int_0^\infty e^{-s(v+a)}f(v+a)dv =e^{-as}\int_0^\infty e^{-sv}f(v+a)dv
    =e^{-as}\mathcal L(f(t+a))$$This formula is handy because, as in your problem, the function f(t) isn't usually given in terms of the argument of the step function.
     
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