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Laplace Transform of this function

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  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    We want to find the Laplace transform for

    f(t): 0 for t≤2 and (t-2)2 for t≥2

    2. Relevant equations
    I know that Lap{uc f(t-c)} = e-csLap{f(t)}=e-csF(s)
    I rewrite f(t)=0+g(t) where g(t) = 0 for 0≤t<2 and (t-2)2 for t≥2
    so we can write f(t)=g(t)= u2(t)*(t-2)2
    Lap{f(t)}= (e-2s/s) * Lap{(t-2)2}


    3. The attempt at a solution

    The answer in my bok is 2*(e-2s)/(s3)
    But Lap{(t-2)2} is (2/s3) - (4/s2)+(4/s)

    why ?!
     
  2. jcsd
  3. Oct 10, 2016 #2

    BvU

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    (t-2)^2 is not the same as your f(t) (especially in the range 0<t<2 :smile:) !
     
  4. Oct 10, 2016 #3
    TNX !

    Now I've got it!:smile:
     
  5. Oct 10, 2016 #4

    Ray Vickson

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    Start with your formula ##{\cal L}[u(t-c) f(t-c)](s) = e^{-cs}{\cal L}[f(t)](s)##. So if ##F(s) = {\cal L}[t^2](s)##, then the answer you want is ##\text{answer} = e^{-2s} F(s)##.
     
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