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:
