# laplace transform..

• Aug 17th 2009, 10:17 PM
transgalactic
laplace transform..
[FONT=monospace]

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

i have
$\displaystyle \frac{2e^{-s}}{s}$
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
??
• Aug 17th 2009, 10:55 PM
CaptainBlack
Quote:

Originally Posted by transgalactic
[font=monospace]

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

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

$\displaystyle 2u(t-1)$

CB
• Aug 17th 2009, 11:07 PM
mr fantastic
Quote:

Originally Posted by transgalactic
[font=monospace]

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

i have
$\displaystyle \frac{2e^{-s}}{s}$
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 $\displaystyle \mathcal{L}^{-1} \left[ \frac{F(s)}{s}\right] = \int_0^t f(u) \, du$ where $\displaystyle f(t) = \mathcal{L}^{-1} [F(s)]$. Note that $\displaystyle \mathcal{L}^{-1} [e^{-s}] = \delta (t - 1)$ (the Dirac delta-function).