Problem
Suppose a DT signal satisfies the following properties:
1)x[n] is periodic with period 12.
2)$ \sum_{k=0}^{11}x[n]=24 $
3)$ \sum_{k=5}^{16}(-1)^nx[n]=6 $
4)x[n] has minimum power among all signals that satisfy the above properties.
Find x[n].
Solution
Since x[n] is periodic with period 12, it can be wriiten as
$ x[n]=\sum_{k=0}^{11}a_ke^{jk\frac{2\pi}{12}n} $
$ a_0=\frac{1}{N}\sum_{k=0}^{N-1}x[n] $, so from property (2), $ a_0=\frac{1}{12}\sum_{k=0}^{11}x[n]=\frac{1}{12}(24)=2 $
Now we must find $ a_1 $, $ a_2 $, $ a_3 $, $ a_4 $, $ a_5 $, $ a_6 $, $ a_7 $, $ a_8 $, $ a_9 $, $ a_{10} $,& $ a_{11} $ from the other two properties.
Since $ x[n]=\sum_{k=0}^{11}a_ke^{jk\frac{2\pi}{12}n}=\sum_{k=0}^{11}a_ke^{jk\frac{\pi}{6}n} $, we can choose to find $ a_6 $ first because $ a_6=\frac{1}{12}\sum_{k=0}^{11}x[n]e^{-j6\frac{\pi}{6}n}=\frac{1}{12}\sum_{k=0}^{11}x[n]e^{-j\pi n} $
$ e^{-j\pi}=cos[-\pi n]+jsin[-\pi n] $, but n is an integer.
This becomes $ (-1)^n $ since the cosine function alternates between 1 and -1 for multiples of $ \pi $, so $ a_6=\frac{1}{12}\sum_{k=0}^{11}x[n](-1)^n $
Since x[n] is periodic, x[1]=x[N+1], x[2]=x[N+2], etc and using this fact, the $ \sum_{k=5}^{16}x[n](-1)^n=\sum_{k=0}^{11}x[n](-1)^n $
Now we can solve for $ a_6 $, $ a_6=\frac{1}{12}\sum_{k=0}^{11}x[n]e^{-j\pi n}=\frac{1}{12}\sum_{k=5}^{16}x[n](-1)^n=\frac{1}{12}6=\frac{1}{2} $
Right now, we have $ x[n]=2+a_1e^{j\frac{\pi}{6}n}+a_2e^{j\frac{\pi}{3}n}+a_3e^{j\frac{\pi}{2}n}+a_4e^{j\frac{2\pi}{3}n}+a_5e^{j\frac{5\pi}{6}n}+\frac{1}{2}e^{j\pi n}+a_7e^{j\frac{7\pi}{6}n}+a_8e^{j\frac{4\pi}{3}n}+a_9e^{j\frac{3\pi}{2}n}+a_{10}e^{j\frac{5\pi}{3}n}+a_{11}e^{j\frac{11\pi}{6}n} $
The only property left is property 4.
$ P=\frac{1}{N}\sum_{k=0}^{N-1}(x[n])^2=\frac{1}{12}(2^2+(a_1)^2+(a_2)^2+(a_3)^2+(a_4)^2+(a_5)^2+(\frac{1}{6})^2+(a_7)^2+(a_8)^2+(a_9)^2+(a_{10})^2+(a_{11})^2) $
To minimize the power take $ a_1=a_2=a_3=a_4=a_5=a_7=a_8=a_9=a_{10}=a_{11}=0 $
Therefore,
$ x[n]=2+\frac{1}{6}e^{j\pi n}=2+\frac{1}{6}(-1)^n $
