Revision as of 13:32, 25 September 2008 by Zcurosh (Talk)

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,

$ \frac{1}{6}\sum_{n=0}^{5}x[n] = 4\, $, and


$ = \frac{4}{6} = \frac{2}{3} = a_0\, $

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}\, $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva