(New page: [[[[Media:Lec5]]]])
 
 
(10 intermediate revisions by one other user not shown)
Line 1: Line 1:
[[[[Media:Lec5]]]]
+
[[Image:Lecture5_OldKiwi.pdf]]
 +
 
 +
Even odd Fourrier Series Coefficients
 +
 
 +
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 <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:
 +
 
 +
<math>={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k}</math>
 +
 
 +
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:
 +
 
 +
  <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

File:Lecture5 OldKiwi.pdf

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} $

Frequency Response Example_OldKiwi

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010