(New page: Category: ECE Category: ECE 301 Category: Fall Category: 2007 Category: mboutin Category: Defintions Category: Fourier Category: Fourier Series Let <math> ...) |
|||

Line 11: | Line 11: | ||

Show that <math> a_{-k} = {a}_{k}\ </math> and <math> b_{-k}=-b_k\ </math>. | Show that <math> a_{-k} = {a}_{k}\ </math> and <math> b_{-k}=-b_k\ </math>. | ||

+ | |||

+ | |||

If <math> x[n]\ </math> is real we have (equation for Fourier coefficients): | If <math> x[n]\ </math> is real we have (equation for Fourier coefficients): |

## Latest revision as of 12:11, 12 December 2008

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\ $