Problem
1. x[n] is periodic with period N = 7
2. $ \sum_{n = 0}^{6} x[n] = 2\! $
3. $ a_1 = a_4\! $
4. $ \sum_{n = 3}^{9} (-1)^n x[n] = 3 \! $
Solution
First Property gives us:
$ x[n] = \frac{1}{7} \sum_{n = 0}^{6} a_k e^{jk \frac{2\pi}{7}n} \! $
Second Property gives us:
$ \frac{1}{7}\sum_{n = 0}^{6} x[n] = \frac{2}{7} \! $ which is $ a_0 \! $
Fourth Property gives us:
$ a_4 = \frac{1}{7} \sum_{n = 0}^{6} x[n] e^{-j\pi n} \! $ and $ e^{-j\pi n} = (-1)^n \! $
$ a_4 = \frac{1}{7} \sum_{n = 3}^{9} x[n] (-1)^n \! $ and $ \sum_{n = 3}^{9} x[n] (-1)^n = 3 \! $ by Property 4
$ a_4 = (\frac{1}{7})(3) = \frac{3}{7} \! $
Third Property gives us:
$ a_4 = a_1 = \frac{3}{7} \! $
Signal
$ x[n] = \frac{2}{7} + \frac{3}{7} e^{j\frac{2\pi}{7}n} + \frac{3}{7} e^{j\frac{8\pi}{7}n} \! $
