Revision as of 09:41, 3 October 2008 by Jpfister (Talk)

Fourier Transform

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

$ x(t)=t^2 u(t) $

$ X(\omega)=\int_{-\infty}^{\infty}t^2 u(t) e^{-j\omega t}dt \; = \int_{0}^{\infty}t^2 e^{-j\omega t}dt $

Integration by Parts

$ u=t^2 \; \; \; \; \; \; \; \; \; \; dv = e^(-j \omega t) $

$ du=2t \; \; \; \; \; \; \; \; \; dv = \frac{1}{-j\omega}e^(-j \omega t) $

Alumni Liaison

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

Buyue Zhang