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<math>w_0 = \frac{2\pi}{4}</math> | <math>w_0 = \frac{2\pi}{4}</math> | ||
| − | <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/2) n} |
| − | </math> | + | = \frac{1}{4} (0+2+0+2) = 1</math> |
| + | |||
| + | <math>a_1 = \frac{1}{4}\sum_{n=0}^{3} x[n]e^{-j(\pi/2) n} | ||
| + | = \frac{1}{4} ((0*1)+(2*-j)+0+(2*-j)) = \frac {-2j}{2}</math> | ||
Latest revision as of 18:03, 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/2) n} = \frac{1}{4} (0+2+0+2) = 1 $
$ a_1 = \frac{1}{4}\sum_{n=0}^{3} x[n]e^{-j(\pi/2) n} = \frac{1}{4} ((0*1)+(2*-j)+0+(2*-j)) = \frac {-2j}{2} $
