(Inverse Fourier Transform)
Line 2: Line 2:
  
  
<math> \chi (\omega) = 2 \pi \sigma (\omega - \pi) </math>
+
<math> \chi(\omega) = 2 \pi \sigma (\omega - \pi) </math>
  
 
<math> x[n] = frac{1}{2\pi}\int_{-\infty}^{\infty} \delta (\omega - \pi)e^{j\omega t} dw </math>
 
<math> x[n] = frac{1}{2\pi}\int_{-\infty}^{\infty} \delta (\omega - \pi)e^{j\omega t} dw </math>
  
 
<math> x[n] = \int_{-\infty}^\infty \delta (\omega - \pi)e^{j\omega t} dw </math>
 
<math> x[n] = \int_{-\infty}^\infty \delta (\omega - \pi)e^{j\omega t} dw </math>

Revision as of 19:02, 8 October 2008

Inverse Fourier Transform

$ \chi(\omega) = 2 \pi \sigma (\omega - \pi) $

$ x[n] = frac{1}{2\pi}\int_{-\infty}^{\infty} \delta (\omega - \pi)e^{j\omega t} dw $

$ x[n] = \int_{-\infty}^\infty \delta (\omega - \pi)e^{j\omega t} dw $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood