Revision as of 16:06, 7 October 2008 by Longja (Talk)

The Signal

$ X(j \omega) = \cos(4 \omega + \frac{\pi}{3}) $

Taken from 4.22.b from the course book, it looks interesting and I want to try it.


The Inverse Fourier Transform

$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(j \omega)e^{j\omega t}d\omega $


$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\cos(4 \omega + \frac{\pi}{3})e^{j\omega t}d\omega $


$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{ e^{j(4 \omega + \frac{\pi}{3})} + e^{j(4 \omega + \frac{\pi}{3})}}{2}e^{j\omega t}d\omega $

Alumni Liaison

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

Buyue Zhang