DT Signal:

1. Signal is periodic with N = 4

2. $ \sum_{k = 0}^{3}x[n] = (2 + j) $

3. $ a_{1} = a{2}\, $

4. for the given value of k, $ e^{jk\frac{2\pi}{N}} = -1\, $, then that $ a_{k} = \frac{1}{2}\, $

5. All other $ a_{k} = 0\, $


Solution

$ a_{0} = \frac{2 + j}{4} $

$ a_{1} = \frac{1}{2} $

$ a_{2} = \frac{1}{2} $

$ a_{3} = 0 $

$ x[n] = \frac{2 + j}{4} + \frac{1}{2}e^{j\frac{\pi}{2}n} + \frac{1}{2}e^{j\pi n} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett