Guess the signal
1. DT signal x[n] is even.
2. X[n] has a period of 2.
3. $ \sum_{n=0}^{1}x[n]=3 $
4. $ \sum_{n=0}^{1}(-1)^nx[n]=5 $
Answer
From 2. we know the period = 2, therefore:
$ x[n]=\sum_{n=0}^{1}a_ke^{jk(2\pi/2)n} $
From 3. we know that:
$ a_0=\frac{1}{2}\sum_{n=0}^{1}x[n]=\frac{1}{2}3 = \frac{3}{2} $
From 4. we know:
$ a_1=\frac{1}{2}\sum_{n=0}^{1}x[n](-1)^n= \frac{1}{2}\sum_{n=0}^{1}x[n]e^{-jn\pi} = \frac{5}{2} $
Therefore our function x[n]:
$ x[n]=\frac{3}{2}+\frac{5}{2}e^{jn\pi} $
