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Use the second shifting theorem to find the Laplace transform

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the second shifting theorem to find the Laplace transform of

    f(t) = {t2, t<4
    {t, t[itex]\geq[/itex]4

    2. Relevant equations

    L{f(t - a)u(t - a)} = e-as F(s)

    3. The attempt at a solution
    Okay so I applied the unit step function to get the equation in to the form
    f(t) = t2 - t2(u)(t - 4) + t(u)(t - 4)
    = t2(1 -u(t - 4) + t(u)(t - 4))

    However now I get lost as I know it needs to be in the form f(t - a)u(t - a) and I see that the second part of equation is but the first part is confusing me I think maybe I am suppose to complete the square though I just can't see how?
     
  2. jcsd
  3. Aug 23, 2011 #2
    You've missplaced a parenthesis. There should be no t^3 power in your expression for f(t).

    Anyway, you don't need to complete the square; that would not be appropriate. Think about what g(t1) needs to be so that g(t-a) = t2.
     
  4. Aug 24, 2011 #3
    Sorry but I am still not sure what to do with the t2 but with my second term, t2(u)(t - 4) is right to do this..

    L{t2(u)(t - 4)
    = t2 8t +16 - 8t - 16
    = (t - 4)2 + 8t - 16)(u)(t - 4)
    = ((t - 4)2 _ 8(t - 4) + 16)(u)(t - 4)
    = (2/s3 + 8/s2 and here I am not sure what 16 becomes?)e-4s
     
  5. Aug 24, 2011 #4

    LCKurtz

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  6. Aug 24, 2011 #5
    Sorry I posted this before than saw you reply in my other thread so replied with the example I was working with.. should I have linked you to this thread instead is that still okay?
     
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