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Math Help - laplace transform..

  1. #1
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    laplace transform..

    [FONT=monospace]

    \mathcal{L}^{-1}\left\{e^{-as}F\left(s\right)\right\}=u\left(t-a\right)f\left(t-a\right)<br />

    i have
    <br />
\frac{2e^{-s}}{s}<br />
    i know that 1/s is 1
    but i cant do a shift of t by 1 here
    there is not t in the end
    ??
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by transgalactic View Post
    [font=monospace]

    \mathcal{L}^{-1}\left\{e^{-as}F\left(s\right)\right\}=u\left(t-a\right)f\left(t-a\right)<br />

    i have
    <br />
\frac{2e^{-s}}{s}<br />
    i know that 1/s is 1
    but i cant do a shift of t by 1 here
    there is not t in the end
    ??
     <br />
2u(t-1)<br />

    CB
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  3. #3
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by transgalactic View Post
    [font=monospace]

    \mathcal{L}^{-1}\left\{e^{-as}F\left(s\right)\right\}=u\left(t-a\right)f\left(t-a\right)<br />

    i have
    <br />
\frac{2e^{-s}}{s}<br />
    i know that 1/s is 1
    but i cant do a shift of t by 1 here
    there is not t in the end
    ??
    If f(t) = 1 then f(t - 1) = 1.

    Alternatively, you can use the well known result that \mathcal{L}^{-1} \left[ \frac{F(s)}{s}\right] = \int_0^t f(u) \, du where f(t) = \mathcal{L}^{-1} [F(s)]. Note that \mathcal{L}^{-1} [e^{-s}] = \delta (t - 1) (the Dirac delta-function).
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