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Laplace transformation

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data
    fing the laplace transformation of f(t) where
    -1 if 0<t<2
    f(t)= e^3t if 2<t<4
    2t if 4<t

    2. Relevant equations

    3. The attempt at a solution

    I can get to the part where f(t) = -1 + (e^3t +1)u(t-2) + (2t-e^3t)u(t-4)
    then I get lost ...Any help is appreciated. Thank you all in advance
  2. jcsd
  3. Nov 28, 2008 #2
    Hi Karmel,

    First note that the -1 you have there should be -u(t); also, a minor point, but I'll assume u(t) is the Heaviside function restricted to t>0 (since f is only defined on t>0).

    Your task now is to work with the formula [tex]g(t-a)u(t-a)\mapsto e^{-as}F(s)[/tex] to take the Laplace transform of f(t). For example, we may rewrite the second term as

    [tex](e^{3t} + 1)u(t-2) = (e^6e^{3(t-2)} + 1)u(t-2)[/tex],

    and so by the above formula this transforms to [tex]e^{-2s}\left(e^6\frac{1}{s-3} + \frac{1}{s}\right)[/tex].

    Your turn. Show us your work if you get stuck.
    Last edited: Nov 28, 2008
  4. Nov 28, 2008 #3
    Okay so by using the examples from the book and the above example for the last part of the above problem I get and if Im right it will be luck!

    (2t-e^3t)u(t-4) = (2t-(e^12)(e^3(t-4))u(t-4) which transforms into e^-4s(-e^12((1/s-3)+(2/s^2))

    so assuming that this part is right. Do I leave the answer as is or do I need to multiply throughout and see if I can add anything and simplyfy it..

  5. Nov 29, 2008 #4
    Almost, but you haven't completely made the factor multiplied by u(t-4) look like a function g(t-4) like in formula I gave above. We need to write it as

    [tex](2t - e^{3t})u(t - 4) = (2(t - 4) + 8 - e^{12}e^{3(t - 4)})u(t - 4)[/tex].

    Once you have transformed each term of f(t), then you have found the Laplace transform of f(t), as required.
  6. Nov 30, 2008 #5
    Thank you so much for the help and assuming I havent made any errors I do believe I have finally arrivied at the right answer. Once again thank you karmel
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