(New page: == The Chosen Signal == <math> x(t) = e^{3t} u(t+2) + e^{4t} u(t-2) \!</math> == The Fourier Transform == <math> X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \!</math> <m...) |
(No difference)
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Revision as of 17:50, 8 October 2008
The Chosen Signal
$ x(t) = e^{3t} u(t+2) + e^{4t} u(t-2) \! $
The Fourier Transform
$ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \! $
$ X(\omega) = \int_{-2}^{\infty} e^{3t} e^{-j\omega t} dt + \int_{2}^{\infty} e^{4t} e^{-j\omega t} dt \! $
$ X(\omega) = \int_{-2}^{\infty} e^{(3-j\omega )t} dt + \int_{2}^{\infty} e^{(4-j\omega )t} dt \! $
$ X(\omega) = {\left. \frac{e^{(j\omega -3)t}}{(j\omega - 3 )} \right]^{\infty}_{-2} } + {\left. \frac{e^{( j\omega -4)t}}{(j\omega -4 )} \right]^{\infty}_2 }\, $
$ X(\omega) = \frac{e^{-2*(3-j\omega)}}{(j\omega - 3 )} + \frac{e^{( 4- j\omega )2}}{(j\omega -4 )} $
