laplace transform..

**transgalactic** · August 17th 2009, 10:17 PM

[FONT=monospace]

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

i have
$ \frac{2e^{-s}}{s} $
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
??

**CaptainBlack** · August 17th 2009, 10:55 PM

Originally Posted by transgalactic

[font=monospace]

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

i have
$ \frac{2e^{-s}}{s} $
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
??

$ 2u(t-1) $

CB

**mr fantastic** · August 17th 2009, 11:07 PM

Originally Posted by transgalactic

[font=monospace]

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

i have
$ \frac{2e^{-s}}{s} $
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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