| Line 9: | Line 9: | ||
Let the signal be | Let the signal be | ||
| + | x[n] = 2cos(5πn) | ||
| + | |||
| + | N = 2 | ||
| + | |||
| + | <math>a_k = \frac{1}{2} \sum_{n=0}^{N-1}x[n] e^{-jk\frac{2\pi}{2} n}</math> | ||
| + | |||
| + | <math>a0 = \frac{1}{2} \sum_{n=0}^{1}x[n] e^{0}</math> | ||
| + | |||
| + | <math>a_0 = \frac{1}{2} *-2</math> | ||
| + | |||
| + | a0 = −1 similarly a1 = -2 | ||
Latest revision as of 18:22, 26 September 2008
Define a periodic DT signal and compute its Fourier series coefficients.
For DT,
$ x[n]=\sum_{k=0}^{N-1} a_k e^{jk\frac{2\pi}{N} n} $
where
$ a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} $.
Let the signal be
x[n] = 2cos(5πn)
N = 2
$ a_k = \frac{1}{2} \sum_{n=0}^{N-1}x[n] e^{-jk\frac{2\pi}{2} n} $
$ a0 = \frac{1}{2} \sum_{n=0}^{1}x[n] e^{0} $
$ a_0 = \frac{1}{2} *-2 $
a0 = −1 similarly a1 = -2
