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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}</math><br> | + | <math>a_k = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-jk\pi n}</math><br><br> |
<math>a_0 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{0}</math><br> | <math>a_0 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{0}</math><br> | ||
| + | <math>a_0 = \frac{1}{2} (-6 + 6) = 0</math><br><br> | ||
| + | <math>a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n}</math><br> | ||
| + | <math>a_1 = \frac{1}{2} (-6 * e^0 + 6 * e^{-j\pi})</math><br> | ||
| + | <math>a_1 = \frac{1}{2} (-6 * 1 + 6 * -1) = -6</math><br><br> | ||
Latest revision as of 15:56, 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} $
$ a_0 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{0} $
$ a_0 = \frac{1}{2} (-6 + 6) = 0 $
$ a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n} $
$ a_1 = \frac{1}{2} (-6 * e^0 + 6 * e^{-j\pi}) $
$ a_1 = \frac{1}{2} (-6 * 1 + 6 * -1) = -6 $
