| Line 10: | Line 10: | ||
<math>a_{k} = \frac{1}{4} \sum_{n=0}^{3} x[n] e^{-jk \frac{\pi}{2} n}</math> | <math>a_{k} = \frac{1}{4} \sum_{n=0}^{3} x[n] e^{-jk \frac{\pi}{2} n}</math> | ||
| + | |||
| + | and: | ||
| + | |||
| + | <math>a[0] = 0 | ||
| + | a[1] = 1 | ||
| + | a[2] = 2 | ||
| + | a[3] = 1 | ||
| + | a[4] = 0</math> | ||
| + | |||
| + | Therefore: | ||
| + | |||
| + | <math>a_{0} = \frac{1}{4} \sum_{n=0}^{3} x[n]</math> | ||
Revision as of 17:48, 23 September 2008
Periodic DT Signal
The following plot shows two periods of the periodic DT signal $ x[n] $, a sawtooth:
Fourier Series Coefficients
$ a_{k} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-jk \frac{2 \pi}{N} n} $
From the plot above, N = 4:
$ a_{k} = \frac{1}{4} \sum_{n=0}^{3} x[n] e^{-jk \frac{\pi}{2} n} $
and:
$ a[0] = 0 a[1] = 1 a[2] = 2 a[3] = 1 a[4] = 0 $
Therefore:
$ a_{0} = \frac{1}{4} \sum_{n=0}^{3} x[n] $
