(New page: DT signal: <br> <math>X(t) = 6\cos(3 \pi n + \pi)\,</math><br> <math>N = \frac{2/pi}{3\pi} K \,</math><br> Where K is the smallest integer that makes N an integer.)
 
 
(13 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
DT signal:
 
DT signal:
<br>
+
<br><br>
<math>X(t) = 6\cos(3 \pi n + \pi)\,</math><br>
+
<math>X[n] = 6\cos(3 \pi n + \pi)\,</math><br><br>
<math>N = \frac{2/pi}{3\pi} K \,</math><br>
+
<math>N = \frac{2\pi}{3\pi} K \,</math><br><br>
 
Where K is the smallest integer that makes N an integer.
 
Where K is the smallest integer that makes N an integer.
 +
K would be 3.<br><br>
 +
<math>N = \frac{2\pi}{3\pi} 3 \,</math><br><br>
 +
<math>N = 2 \,</math><br>
 +
<br>
 +
<br>
 +
<math>X[0] = -6 \,</math><br>
 +
<math>X[1] = 6 \,</math><br>
 +
<math>X[2] = -6 \,</math><br>
 +
<math>X[-1] = 6 \,</math><br>
 +
<br>
 +
<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><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 14:56, 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 $

Alumni Liaison

EISL lab graduate

Mu Qiao