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<math>= \int_{-\infty}^{\infty} x(t)e^{-at}e^{-j\omega t} dt\!</math> | <math>= \int_{-\infty}^{\infty} x(t)e^{-at}e^{-j\omega t} dt\!</math> | ||
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| + | <math>= F(x(t)e^{-at})\!</math> | ||
Revision as of 16:50, 24 November 2008
The Laplace Transform
The Laplace Transform is a generalization of the Fourier Transform. Instead of considering only the imaginary axis, $ j\omega\! $, (as the Fourier Transform does) the Laplace Transform considers all complex values represented by the general complex variable $ s\! $. Take the following simple picture:
Fourier Transform: $ x(t) --> X(\omega)\! $ where $ \omega\! $ is a frequency.
Laplace Transform: $ x(t) --> X(s)\! $ where $ s\! $ is a complex variable.
Mathematically, the Laplace Transform is represented as follows:
Let's consider the case where $ s = j\omega\! $.
$ X(s)|_{s=j\omega} = X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt = F(x(t)) = X (\omega)\! $
So, the Laplace Transform, $ X(s)\! $, evaluated on the imaginary axis, $ X(j\omega)\! $, is equal to the Fourier Transform, $ F(x(t))\! $, at $ \omega\! $. Said another way, the Fourier Transform, $ X(\omega)\! $, is the restriction of the Laplace Transform, $ X(s)\! $, on the imaginary axis, $ s = j\omega\! $.
It is actually possible to obtain the Laplace Transform, $ X(s)\! $, from the Fourier Transform, $ X(\omega)\! $.
$ X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt\! $, $ let\! $ $ s = a + j\omega \! $
$ = \int_{-\infty}^{\infty} x(t)e^{-(a+j\omega)t} dt\! $
$ = \int_{-\infty}^{\infty} x(t)e^{-at}e^{-j\omega t} dt\! $
$ = F(x(t)e^{-at})\! $
