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<br><math>N = \frac{2\pi}{3\pi} 3 \,</math>
 
<br><math>N = \frac{2\pi}{3\pi} 3 \,</math>
 
<br><br><math>N = 2 \,</math>
 
<br><br><math>N = 2 \,</math>
 +
Prove that it is periodic:
 +
x(0)= -3 ; x(1)=3 ; x(2)=-3 ; x(3) = 3 etc.
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<br>
 +
<br>
 +
<math>a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n}</math>
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<br><br>
 +
<math>a_k = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-jk\pi n}</math>
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<br><br>
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<math>a_0 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{0}</math>
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<br><br>
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<math>a_0 = \frac{1}{2}  (-3 + 3) = 0</math>
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<br><br>
 +
<math>a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n}</math>
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<br><br>
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<math>a_1 = \frac{1}{2} (-3 * e^0 + 3 * e^{-j\pi})</math>
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<br><br>
 +
<math>a_1 = \frac{1}{2} (-3 * 1 + 3 * -1) = -3</math>

Latest revision as of 08:36, 25 September 2008

DT signal


$ X[n] = 3\cos(3 \pi n + \pi)\, $
Find a value for k that makes N an integer:
$ N = \frac{2\pi}{3\pi} K \, $
K equals 3 results in a integer.
$ N = \frac{2\pi}{3\pi} 3 \, $

$ N = 2 \, $ Prove that it is periodic: x(0)= -3 ; x(1)=3 ; x(2)=-3 ; x(3) = 3 etc.

$ 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} (-3 + 3) = 0 $

$ a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n} $

$ a_1 = \frac{1}{2} (-3 * e^0 + 3 * e^{-j\pi}) $

$ a_1 = \frac{1}{2} (-3 * 1 + 3 * -1) = -3 $

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