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Let <math>x[n]</math> be a real periodic sequence with fundamental period <math>{N}_{0}</math> and Fourier coefficients <math>{c}_{k}={a}_{k}+j{b}_{k}</math> where ak and bk are both real. | Let <math>x[n]</math> be a real periodic sequence with fundamental period <math>{N}_{0}</math> and Fourier coefficients <math>{c}_{k}={a}_{k}+j{b}_{k}</math> where ak and bk are both real. | ||
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<math> {a}_{-k}={a}_{k}</math> and | <math> {a}_{-k}={a}_{k}</math> and | ||
<math> {b}_{-k}=-{b}_{k}</math> | <math> {b}_{-k}=-{b}_{k}</math> | ||
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| + | [[Frequency Response Example_OldKiwi]] | ||
Latest revision as of 16:36, 30 March 2008
Even odd Fourrier Series Coefficients
Let $ x[n] $ be a real periodic sequence with fundamental period $ {N}_{0} $ and Fourier coefficients $ {c}_{k}={a}_{k}+j{b}_{k} $ where ak and bk 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}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k} $
So now we can see that:
$ {a}_{-k}={a}_{k} $ and $ {b}_{-k}=-{b}_{k} $
