Let $ x[n]\ $ be a real periodic sequence with fundamental period $ N_0\ $ and Fourier coefficients $ c_k = a_k+jb_k\ $, where $ a_k\ $ and $ b_k\ $ are both real.

Show that $ a_{-k} = {a}_{k}\ $ and $ b_{-k}=-b_k\ $.


If $ x[n]\ $ is real we have (equation for Fourier coefficients):

$ c_{-k} = \frac{1}{N_0} \sum_{n=0}^{N_0-1} x[n]{e}^{jk \omega_0 n} $

and further:

$ = \left ( \frac{1}{N_0} \sum_{n=0}^{N_0-1} x[n]e^{-jk \omega_0 n} \right ) ^{*} = c^{*}_k $

Therefore:

$ c_{-k} = a_{-k} + jb_{-k} =(a_k + b_k)^{*} = a_k - jb_k\ $

So now we can see that:

$ a_{-k} = a_k\ $ and $ b_{-k} = -b_k\ $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva