DT Signal:
1. Signal is periodic with N = 4
2. $ \sum_{k = 0}^{3}x[n] = (2 + j) $
3. for the given value of k, $ e^{jk2\pi} = 1\, $, then that $ a_{k} $
Solution
$ a_{0} = \frac{2 + j}{4} $
DT Signal:
1. Signal is periodic with N = 4
2. $ \sum_{k = 0}^{3}x[n] = (2 + j) $
3. for the given value of k, $ e^{jk2\pi} = 1\, $, then that $ a_{k} $
$ a_{0} = \frac{2 + j}{4} $