Revision as of 09:59, 26 September 2008 by Rwijaya (Talk)

Informations

1. $ N = 2\, $

2. $ a_k = 0\, $ for all |k|>1

3. $ \sum_{n=0}^{3}x[n]=2 $

4. $ \sum_{n=0}^{3}(-1)^nx[n]=5 $

Inspections

From first information, we can directly subtitute N into:

$ x[n]=\sum_{n=0}^{3}a_ke^{jk(2\pi/2)n}\, $

$ x[n]=\sum_{n=0}^{3}a_ke^{jk\pi n}\, $

From third information, we can find $ a_0\, $:

$ a_0=\frac{1}{4}\sum_{n=0}^{3}x[n]=\frac{1}{4}x2 = \frac{1}{2} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood