Revision as of 17:48, 8 October 2008 by Verman (Talk)

$ x(t) = e^{-3t} , t>3 \, $, $ x(t)= e^{-6t} , 0 \le t \le 3 $, $ x(t)= 0 , t < 0 \, $

$ x(t)= e^{-3t} u(t-3) + e^{-6t}( u(t-3)-u(t))\, $

$ X(\omega) = \int^\infty_\infty e^{-3t}e^{-j\omega t} dt + \int^2_0 e^{-6t}e^{-j\omega t} dt\, $

$ X(\omega) = \int^\infty_\infty e^{-(3+j\omega)t} dt + \int^3_0 e^{-(6+j\omega) t} dt\, $

$ X(\omega) = {\left. \frac{e^{-(j\omega + 3)t}}{-(j\omega +3)} \right]^{\infty}_0 } + {\left. \frac{e^{-(j\omega + 6)t}}{-(j\omega +6)} \right]^3_0 }\, $

$ X(\omega) = \frac{e^{-(3j\omega + 9)}}{j\omega +3} - \frac{e^{-(3j\omega + 18)t}}{-j\omega +6} + \frac{1}{6+j\omega} \, $

$ X(\omega) = \frac{e^{-(3j\omega + 9)}}{j\omega +3} + \frac{1 - e^{-(3j\omega + 18)t}}{-j\omega +6} \, $

Alumni Liaison

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

Buyue Zhang