Problem
We have a DT signal x[n] such that:
1. x[n] is periodic and period N=8.
2. $ \sum_{n=0}^{7}x[n]=12 $
3.$ a_{k+3} = a_k $
4. x[n] has minimum power among all signals that satisfy 1,2,3.
Find x[n].
Solution
From 1 we know:
$ x[n] = \sum_{n=0}^{7}a_k e^{-jk \frac {\pi}{4} n} $
From 2 we know:
$ a_0 = avg = 12/8 = 3/2 $
From 3 we know:
$ a_0 = a_3 = a_6 = 3/2 $
From 4 we know:
We want to minimize the power, so:
Power = $ \frac {1}{8} \sum_{n=0}^{7} |x[n]|^2 = \sum_{n=0}^{7} |{a_k}|^2 $
To minimize this, $ a_1=a_2=a_4=a_5=a_7=0 $
So:
$ x[n] = 3/2 * (1 + e^{-j \frac {3\pi}{4} n} + e^{-j \frac {6\pi}{4} n}) $
