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Math Help - second shifting theorem

  1. #1
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    second shifting theorem

    Hi all,

    I am having trouble gettin the answer to the attached question. I need to find f(t) and then find L[f(t)] using the t -shifting theorem.

    I tried working it out...

    f(t) = e^t - e^t u(t-2)

    But when i take the Laplace transform i can't get the answer. The answer is

    (1 - e^(2 - 2s))/ (s -1)

    Thanks in advance,
    ArTiCk
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  2. #2
    MHF Contributor chisigma's Avatar
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    Applying the defintion of LT we have...

    \mathcal{L}\{f(t)\} = \int_{0}^{2} e^{-(s-1)t}\cdot dt = - \frac{1}{s-1} |e^{-(s-1)t}|_{0}^{2} = \frac{1-e^{-2(s-1)}}{s-1}

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    I am having trouble gettin the answer to the attached question. I need to find f(t) and then find L[f(t)] using the t -shifting theorem.

    I tried working it out...

    f(t) = e^t - e^t u(t-2)

    But when i take the Laplace transform i can't get the answer. The answer is

    (1 - e^(2 - 2s))/ (s -1)

    Thanks in advance,
    ArTiCk
    To use the theorem you mention, you should note that f(t) = e^t [H(t) - H(t - 2)].

    Then LT[f(t)] = G(s - 1) where G(s) = LT[g(t)] and g(t) = H(t) - H(t - 2).

    The answer so obtained is the same as the answer found by the previous poster.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    To use the theorem you mention, you should note that f(t) = e^t [H(t) - H(t - 2)].

    Then LT[f(t)] = G(s - 1) where G(s) = LT[g(t)] and g(t) = H(t) - H(t - 2).

    The answer so obtained is the same as the answer found by the previous poster.
    Hi,

    I don't quite understand your notation. I found f(t) using step functions and that gave f(t) = e^t - e^t u (t -2). From my lectures the t shifiting theorem is:

    If LT[f(t)] = F (s), then

    LT[f(t-a) u(t-a)] = e^(-as)F(s)

    Could you please explain what H(t) is?

    Thanks once again,
    ArTiCk
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  5. #5
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    Quote Originally Posted by ArTiCK View Post
    Hi,

    I don't quite understand your notation. I found f(t) using step functions and that gave f(t) = e^t - e^t u (t -2). Mr F says: It should be clear from my previous reply (go back and read it) that this is wrong. It does not give a 'pulse' between t = 0 and t = 2.

    From my lectures the t shifiting theorem is:

    If LT[f(t)] = F (s), then

    LT[f(t-a) u(t-a)] = e^(-as)F(s)

    Could you please explain what H(t) is?

    Thanks once again,
    ArTiCk
    The notation H(t) (the Heaviside step function) means the same as u(t) (the unit step function).
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