Revision as of 12:25, 7 October 2008 by Anders89 (Talk)

Signal

$ x(t) = e^{3jt}*(u(t+5) - u(t-5)) + e^{-2t}*u(t)\, $


Transformed

$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\, $

$ = \int_{-\infty}^{\infty}e^{3jt}*(u(t+5) - u(t-5))e^{-j\omega t}dt + \int_{-\infty}^{\infty}e^{-2t}u(t)e^{-j\omega t}dt\, $

$ = \int_{-5}^{5}e^{3jt}e^{-j\omega t}dt + \int_{0}^{\infty}e^{-2t}e^{-j\omega t}dt\, $

$ = \int_{-5}^{5}e^{3jt -j\omega t}dt + \int_{0}^{\infty}e^{-2t -j\omega t}dt\, $

$ = \int_{-5}^{5}e^{t*(3j -j\omega )}dt + \int_{0}^{\infty}e^{t*(-2 -j\omega )}dt\, $

$ = \frac{e^{3jt - j\omega t}{3j-j\omega t}}]_{-5}^{5} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang