Guess the Periodic Signal
- $ \sum_{n = 0}^{1} x[n] = 8 $
- $ \sum_{n = 0}^{1} (-1)^n x[n] = 3 $
- $ a_k = 0 $ when |k| > 1
- The period of the DT signal x[n] = 2
Answer
x[n] with Fourier series representation (pg. 230):
$ x[n] = \sum_{k = N}^{1}a_k * e $$ (jk(2\pi /N)n $
$ \,\ a_0 = 0.5 * 8 = 4 $
$ a_1 = \frac{1}{2} \sum_{n = 0}^{1} x[n] * e $$ (-j\pi n) $ $ = \frac{3}{2} $
Final Answer:
$ \,\ x[n] = 4 + \frac{3}{2} * e $$ (-j\pi n) $
