| Line 29: | Line 29: | ||
<math>a_3 = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-3j\frac{\pi}{3}n} = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-j\pi n}\,</math> | <math>a_3 = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-3j\frac{\pi}{3}n} = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-j\pi n}\,</math> | ||
| − | Using the property that <math>e^{-j\pi n} = (e^{-j\pi})^{n} | + | Using the property that <math>e^{-j\pi n} = (e^{-j\pi})^{n} = (-1)^{n} \,</math> |
Revision as of 12:36, 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}\, $
Since $ \omega_0 = \frac{\pi}{3}\, $, let's try and find $ a_3\, $,
$ a_3 = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-3j\frac{\pi}{3}n} = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-j\pi n}\, $
Using the property that $ e^{-j\pi n} = (e^{-j\pi})^{n} = (-1)^{n} \, $
