Periodic Signal Properties
1. Period is equal to 2
2. $ a_k = a_{k-2}\, $
3. $ \sum^{5}_{2} x[n] = 0\, $
4. $ \sum^{5}_{2} x[n](-1)^n = 10\, $
5. Signal is Even
Signal Determination
From the first property, we can clearly say that N = 2.
$ a_0 = \frac{1}{2}\sum^{1}_{0} x[n] e^{-jn\pi 0} = \frac{1}{2}\sum^{1}_{0} x[n]\, $
Notice that according to property 3 and 2, we can shift the sum to start from 0, and decrease the total sum values as the signal is periodic, and repeat itself every $ n = 2\, $
$ \sum^{5}_{2} x[n] = \sum^{3}_{0} x[n] = \sum^{1}_{0} x[n] = 0\, $
Thus , $ a_0 = \frac{1}{2}\sum^{1}_{0} x[n] = 0\, $
$ a_1 = \frac{1}{2}\sum^{1}_{0} x[n] e^{-jn\pi 1} = \frac{1}{2}\sum^{1}_{0} x[n]e^{-jn\pi}\, $
Notice that $ e^{-j\pi}\, $ is equivilent to $ -1\, $
Thus $ a_1 = \frac{1}{2}\sum^{1}_{0} x[n] (-1)^n\, $
From property 4, we can modify the properties to make it look similar to our formula.
$ \sum^{5}_{2} x[n](-1)^n = \sum^{3}_{0} x[n](-1)^n = \sum^{1}_{0} x[n](-1)^n = 10\, $
Subtituting the formula, we can see that $ a_1 = \frac{1}{2}(10) = 5\, $
We can proceed to calculate the formula for $ x[n]\, $
$ x[n] = \sum^{1}_{k=0} a_k e^{j\pi k n} = 0 + 5e^{j\pi n} = 5(-1)^n $
