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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>. | ||
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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\ $
