Guessing The Periodic Signal
Given
1. x[n] is periodic with period N=6
2.$ \sum_{k=0}^{5}x[n]=2 $
3.$ \sum_{k=2}^{7}(-1)^nx[n]=1 $
4. x[n] has minimum power among all the signals that satisfy 1,2,3
Solution
1. $ \Rightarrow x[n]=\sum_{n=0}^{5}a_k e^{jk \frac {\pi}{3} n} $
2. $ \Rightarrow a_0 = \frac{2}{6} = \frac{1}{3} $
3. $ \Rightarrow a_3 = \frac{1}{6}\sum_{k=0}^{5}(-1)^nx[n] $
- $ \sum_{k=2}^{7}(-1)^nx[n]= \sum_{k=0}^{5}(-1)^nx[n] $
- $ \begin{align}\therefore a_3 &= \frac{1}{6}\sum_{k=2}^{7}(-1)^nx[n] \\ &=\left (\frac{1}{6}\right )\cdot 1= \frac{1}{6} \end{align} $
- $ x[n]=\frac{1}{3}+a_1e^{j\frac{\pi}{3}n}+a_2e^{j\frac{2\pi}{3}n}+\frac{1}{6}e^{j\pi n}+a_4e^{j\frac{4\pi}{3}n}+a_5e^{j\frac{5\pi}{3}n}+a_6e^{j2\pi n} $
4. $ \Rightarrow $ To minimum the power, we set the rest of $ a_k $ to zero
$ \therefore x[n]=\frac{1}{3}+\frac{1}{6}e^{j\pi n} $
