Line 3: Line 3:
 
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{-j\omega t}d\omega</math>
 
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{-j\omega t}d\omega</math>
  
<math>X(\omega} = \dirac{\omega - 4\pi}</math>
+
<math>X(\omega} = \delta(\omega - 4\pi)</math>

Revision as of 10:18, 3 October 2008

Inverse Fourier Transform

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

$ X(\omega} = \delta(\omega - 4\pi) $

Alumni Liaison

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

Buyue Zhang