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Coupling 1. and 2. together, we see that the interval is twice the fundamental period. The coefficient <math>a_{0}</math> is equal to the average value of the signal over one period, so <math>a_{0} = \frac{0}{2} = 0</math>. | Coupling 1. and 2. together, we see that the interval is twice the fundamental period. The coefficient <math>a_{0}</math> is equal to the average value of the signal over one period, so <math>a_{0} = \frac{0}{2} = 0</math>. | ||
| − | Recall that <math>a_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jk \frac{2\pi}{N}n | + | Recall that <math>a_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jk \frac{2\pi}{N}n} = \frac{1}{4} \sum_{n=0}^{3} x[n]e^{-jk \frac{\pi}{2}n}</math> |
Revision as of 19:54, 24 September 2008
Determine the Periodic Signal
A periodic signal x[n] has the following characteristics:
1. N = 4
2. The average value of the signal over the interval $ 0 \leq n \leq 7 $ is 0.
3. $ \sum_{n=0}^{3}x[n](-j)^n = -20j $
4. $ a_{-k} \; = \; a_{k} $
Solution
Coupling 1. and 2. together, we see that the interval is twice the fundamental period. The coefficient $ a_{0} $ is equal to the average value of the signal over one period, so $ a_{0} = \frac{0}{2} = 0 $.
Recall that $ a_{k}=\frac{1}{N} \sum_{n=0}^{N-1} x[n]e^{-jk \frac{2\pi}{N}n} = \frac{1}{4} \sum_{n=0}^{3} x[n]e^{-jk \frac{\pi}{2}n} $
