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<math>a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n}</math><br>
 
<math>a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n}</math><br>
 +
<math>a_k = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-jk\pi N n}</math><br>

Revision as of 15:46, 22 September 2008

DT signal:

$ X[n] = 6\cos(3 \pi n + \pi)\, $

$ N = \frac{2\pi}{3\pi} K \, $

Where K is the smallest integer that makes N an integer. K would be 3.

$ N = \frac{2\pi}{3\pi} 3 \, $

$ N = 2 \, $


$ X[0] = -6 \, $
$ X[1] = 6 \, $
$ X[2] = -6 \, $
$ X[-1] = 6 \, $


$ a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n} $
$ a_k = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-jk\pi N n} $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin