Suppose a DT signal x[n] satisfies

1. x[n] is periodic and period N=6.

2. $ \sum_{n=0}^{5}x[n]=5 $

3.$ a_{k+2} = a_k $

4. x[n] has minimum power among all signals that satisfy 1,2,3.

Find x[n].



Answer:

from 1 we have that x[n]= $ \sum_{n=0}^{5}a_k e^{-jk \frac {\pi}{3} n} $

from 2 we have $ a_0 = avg = \frac {5}{6} $

from 3 we have $ a_2 = a_4= a_0 = \frac {5}{6} $

from 4, power of x[n] = $ \frac {1}{6} \sum_{n=0}^{5} |x[n]|^2 = \sum_{n=0}^{5} |{a_k}|^2 $

to get a minimum value, $ a_1=a_3=a_5=0 $

Thus $ x[n]=\frac{5}{6} (1+e^{-j \frac {2\pi}{3}n}+e^{-j \frac {4\pi}{3}n}) $

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Mu Qiao