Suppose a DT signal x[n] satisfies
1. x[n] is periodic and period N=8.
2. $ \sum_{n=0}^{7}x[n]=5 $
3.$ a_{k+3} = a_k $
Find x[n].
Answer:
from 1 we deduce that x[n]= $ \sum_{n=0}^{8}a_k e^{-jk \frac {\pi}{4} n}=5 $
from 2 we have $ a_0 = avg = \frac {5}{8} $
from 3 we have $ a_3 = a_6 = a_0 = \frac {5}{8} $
