| (5 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
[[Image:Lecture5_OldKiwi.pdf]] | [[Image:Lecture5_OldKiwi.pdf]] | ||
| − | + | Even odd Fourrier Series Coefficients | |
| − | <math> | + | 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. |
| − | |||
| − | |||
| − | |||
| − | + | Show that <math>{a}_{-k}={a}_{k}</math> and <math>{b}_{-k}=-{b}_{k}</math>. | |
| − | + | ||
| − | |||
| − | |||
| − | If | + | If <math>x[n]</math> is real we have (equation for Fourier coefficients): |
| − | |||
| − | |||
| − | + | <math>{c}_{-k}=\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{jk{\omega}_{0}n}</math> | |
| − | + | ||
and further: | and further: | ||
| − | + | <math>={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k}</math> | |
| − | + | ||
Therefore: | Therefore: | ||
| − | + | <math>{c}_{-k}={a}_{-k}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k}</math> | |
| − | + | ||
So now we can see that: | So now we can see that: | ||
| − | + | <math> {a}_{-k}={a}_{k}</math> and | |
| + | <math> {b}_{-k}=-{b}_{k}</math> | ||
| − | + | [[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} $
