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Therefore: | Therefore: | ||
| − | <math>a_{0} = \frac{1}{4} \sum_{n=0}^{3} x[n]</math> | + | <math>a_{0} = \frac{1}{4} \sum_{n=0}^{3} x[n] = \frac{1}{4}(0+1+2+1)</math> |
| + | |||
| + | '''<math>a_{0} = 1</math>''' | ||
Revision as of 16:52, 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:
$ x[0] = 0 $
$ x[1] = 1 $
$ x[2] = 2 $
$ x[3] = 1 $
$ x[4] = 0 $
Therefore:
$ a_{0} = \frac{1}{4} \sum_{n=0}^{3} x[n] = \frac{1}{4}(0+1+2+1) $
$ a_{0} = 1 $
