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<math>a_0 = \frac{1}{4}\sum_{n=0}^{3} x[n]e^{-j0\pi n}
 
<math>a_0 = \frac{1}{4}\sum_{n=0}^{3} x[n]e^{-j0\pi n}
</math>
+
= \frac{1}{4} (0+2+0+2) = 1</math>

Revision as of 18:47, 25 September 2008

Discrete time Fourier series

$ x[n] = \sum_{k=N} a_ke^{-jkw_0n} $

$ a_k = \frac{1}{N}\sum_{n=N} x[n]e^{-jkw_0n} $

assume that

x[0] = 0

x[1] = 2

x[2] = 0

x[3] = 2

x[4] = 0

.

.

.

.


N = 4

let's solve for coefficients

$ w_0 = \frac{2\pi}{4} $

$ a_0 = \frac{1}{4}\sum_{n=0}^{3} x[n]e^{-j0\pi n} = \frac{1}{4} (0+2+0+2) = 1 $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood