Guessing the periodic signal:
Hint
1. Period of x[n] is N=4.
2. $ a_0=3 $.
3.$ \sum_{n=4}^{7}(-1)^{n}x[n]=4 $.
4.x[n] has minimum power among all the signals that satisfy 1,2,3.
Solution
From (1), we can get $ \omega_{0}=2\pi/N=\frac{1}{2}\pi $,so x[n]=$ \sum_{k=0}^{N-1}a_ke^{jk\frac{1}{2}\pi n} $.
From (2),x[n]=3+$ \sum_{k=1}^{3}a_ke^{jk\frac{1}{2}\pi n} $.
Since N=4, we have x[1]=x[5],x[2]=x[6],x[3]=x[7], and $ (-1)^n $ is periodic with 4,too
From (3),we can get $ a_2=\frac{1}{4}sum_{n=4}^{7}e^{-j\pi n}x[n]=\frac{1}{4}sum_{n=4}^{7}(-1)^nx[n]=1 $
So,
$ x[n]= 3+a_1e^{j\frac{\pi}{2}n}+e^{j\frac{2\pi}{2}n}+a_3e^{j\frac{3\pi}{2}n} $
Power of x[n] is
$ P=\frac{1}{4}\sum{n=0}^{3}|x[n]|^2 $
$ =\sum{k=0}^{3}|a_k|^2 $
$ =3^2+|a_1|^2+1^2+|a_3|^2\, $
the minimum happens when $ |a_1|^2=0\, $ and $ |a_3|^2=0\, $
Answer:
$ x[n]=3+e^{j\pi n}\, $
