Periodic DT Signal

$ x[n] = \sum_{k=0}^{N-1}a_k e^{jk\frac{2\pi}{N} n} $


Where $ a_k = \frac{1}{N} \sum_{n=0}^{N-1}x[n] e^{-jk\frac{2\pi}{N} n} $


$ \omega = \frac{2\pi *k}{N} $

if

 $ x[n] = 2cos(5\pi n) $

then

   $ N = \frac{2\pi}{5\pi}*k = \frac{2}{5}*5 = 2 $

So

$ a_k = \frac{1}{2} \sum_{n=0}^{N-1}x[n] e^{-jk\frac{2\pi}{2} n} $

$ a_0 = \frac{1}{2} \sum_{n=0}^{1}x[n] e^{0} $

$ a_0 = \frac{1}{2} *-2 $

$ a_0 = -1 $

$ a_1 = \frac{1}{2} \sum_{n=0}^{1}x[n] e^{-j\pi n} $

$ a_1 = \frac{1}{2} (-2*e^{0} + 2*e^{-j\pi}) $

$ a_1 = \frac{1}{2} (-2 + -2) = -2 $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin