Periodic Signal

1. $ N = 4\, $

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

3. $ \sum_{n=0}^{6}x[n]=4 $

4. $ \sum_{n=0}^{3}(-1)^nx[n]=10 $

Inspection :

Using Info 1 :

$ x[n]=\sum_{n=0}^{6}a_ke^{jk(2\pi/4)n}\, $

$ x[n]=\sum_{n=0}^{6}a_ke^{jk(\pi/2)n}\, $

Using third info to find $ a_0\, $:

$ a_0=\frac{1}{16}\sum_{n=0}^{6}x[n]=\frac{1}{16} 4 = \frac{1}{4} $

Using fourth info, we can find $ a_1\, $:

$ a_0=\frac{1}{16}\sum_{n=0}^{6}(-1)^nx[n]\, $

$ a_0=\frac{1}{16}\sum_{n=0}^{6}x[n]e^{-jn\pi}\, $

$ a_0=\frac{5}{8} $

Source:

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin