| Line 9: | Line 9: | ||
4. <font "size"=4> <math>a_{-k} \; = \; a_{k}</math> </font> | 4. <font "size"=4> <math>a_{-k} \; = \; a_{k}</math> </font> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | 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 1 period, so <math>a_{0} = \frac{0}{2} = 0</math>. | ||
Revision as of 19:50, 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 1 period, so $ a_{0} = \frac{0}{2} = 0 $.
