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

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

$ \frac{2\pi}{N} =\omega_0 $


Signal

$ x[n]=3+sin\bigg(\frac{\pi}{2}n\bigg) $

$ \frac{w_0}{2\pi}=\frac{1}{4} $, which is rational, so it is periodic.


Fourier Series

$ x[n]=3+sin\bigg(\frac{\pi}{2}n\bigg) $

$ x[n]=3+\frac{e^{i\frac{\pi}{2}n}-e^{-i\frac{\pi}{2}n}}{2i} $

$ x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n} $

$ x[0]=3,x[1]=4,x[2]=3,x[3]=2,x[4]=3, x[5]=4\! $

So, the fundamental period is 4.


Fourier Series Coefficients

$ a_0=3\! $

$ a_1=\frac{1}{2i} $

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

$ a_k,even=3\! $

$ a_k=\frac{1}{2i} $ for all $ k=1,5,9,13,..., 4n+1\! $

$ a_k=-\frac{1}{2i} $ for all $ k=3,7,11,15,..., 4n+3\! $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood