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<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} 2\pi \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 19:03, 8 October 2008

Inverse Fourier Transform

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

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

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

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn