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<center><math>= \frac{4}{6} = \frac{2}{3} = a_0\,</math></center> | <center><math>= \frac{4}{6} = \frac{2}{3} = a_0\,</math></center> | ||
| − | Now that we know <math>a_0\,</math>, we know that <math>x[n] = \frac{2}{3} + \sum_{1}^{5}a_k e^{jk\frac{\pi}{3}n}\,</math> | + | Now that we know <math>a_0\,</math>, we know that <math>x[n] = \frac{2}{3} + \sum_{n=1}^{5}a_k e^{jk\frac{\pi}{3}n}\,</math> |
Revision as of 13:28, 25 September 2008
Guess the Periodic Signal
A certain periodic signal has the following properties:
1. N = 6
2. $ \sum_{n=0}^{5}x[n] = 4 $
3. $ \sum_{n=1}^{6}(-1)^nx[n] = 2 $
4. $ a_k = a_{k+3}\, $
Answer
From 1. we know that $ x[n] = \sum_{n=0}^{5}a_k e^{jk\frac{\pi}{3}n}\, $
Using 2., it is apparent that this is the formula for $ a_k\, $. Specifically, for $ a_0\, $, since the only thing under the sum is $ x[n]\, $. So,
Now that we know $ a_0\, $, we know that $ x[n] = \frac{2}{3} + \sum_{n=1}^{5}a_k e^{jk\frac{\pi}{3}n}\, $
