(New page: ==A Laplace Transform Example== <math> x(t)=-e^{-2t}u(-t)</math> Therefore the Laplace Transform is: <math>X(s) = \int_{-\infty}^{\infty} x(t)e^{-st}dt</math> <math> = \int_{-\infty}^{...) |
(No difference)
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Latest revision as of 18:14, 17 November 2008
A Laplace Transform Example
$ x(t)=-e^{-2t}u(-t) $
Therefore the Laplace Transform is:
$ X(s) = \int_{-\infty}^{\infty} x(t)e^{-st}dt $
$ = \int_{-\infty}^{\infty} -e^{-2t}u(-t)e^{-st}dt $
$ = \int_{-\infty}^{0}-e^{-2t}e^{-st}dt $
$ = \int_{-\infty}^{0}-e^{-2t}e^{-(a+j\omega )t}dt $
$ = \int_{-\infty}^{0}-e^{-(2+a)t}e^{-j\omega t}dt $
$ = \frac{-e^{-(2+a)t}e^{-j\omega t}}{-(2+a+j\omega )}|_{-\infty}^{0} $
If 2+a<=0 then integral diverges
else:
$ X(s) = \frac{-1}{-(2+a+j\omega)} - 0 $
$ = \frac{1}{2+s} $
With a ROC: a<-2