(New page: == Chosen Signal to Transform == The signal we will transform here will be <math>x(t)=e^{2jt}*(u(t+4)-u(t-4))</math> ==Transform by integral== <math> = \int_{-\infty}^{\infty}e^{2jt}*(u...) |
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Revision as of 10:00, 8 October 2008
Chosen Signal to Transform
The signal we will transform here will be $ x(t)=e^{2jt}*(u(t+4)-u(t-4)) $
Transform by integral
$ = \int_{-\infty}^{\infty}e^{2jt}*(u(t+4)-u(t-4))e^{-j\omega t}dt\, $
$ = \int_{-4}^{4}e^{2jt}e^{-j\omega t}dt \, $
$ = \int_{-4}^{4}e^{2jt -j\omega t}dt\, $
$ = \int_{-4}^{4}e^{t*(2j -j\omega )}dt \, $
$ = \frac{e^{2jt - j\omega t}}{2j-j\omega}]_{-4}^{4} \, $
$ = \frac{e^{8j - 4j\omega} - e^{-8j + 4j\omega}}{2j-j\omega} \, $
$ = \frac{e^{j(8 - 4\omega )} - e^{-j(8 - 4\omega )}}{j(2-\omega )} \, $
$ = \frac{2sin(8 - 4\omega )}{2-\omega }\, $
