Revision as of 03:18, 8 October 2008 by Sparksj (Talk)

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For the signal:

$ x(t) = e^{-8(t + 1)} u(t + 1) $

$ X(\omega) = \int_{-\infty}^\infty e^{-8(t + 1)} u(t + 1) e^{-j\omega t} \mathrm{d}t $

when $ t + 1 < 0 $ or $ t < -1 $, $ u(t + 1) = 0 $ so,

$ X(\omega) = \int_{-1}^\infty e^{-8(t + 1)} e^{-j\omega t}\mathrm{d}t $

$ = e^{-8} \int_{-1}^\infty e^{-8t} e^{-j \omega t}\mathrm{d}t $ $ = e^{-8} \int_{-1}^\infty e^{-(8+j\omega)t} \mathrm{d}t $

$ = e^{-8} \frac{ -e^{-(8+j\omega)t}}{(8+j\omega)t}\bigg|_{t=-1}^\infty $ $ = e^{-8} \frac{ e^{(8+j\omega)}}{(8+j\omega)} $

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