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[[Image:Lecture5_Old Kiwi.pdf]]
 
[[Image:Lecture5_Old Kiwi.pdf]]
  
Let |four| be a real periodic sequence with fundamental period |zero| and Fourier coefficients |one|, where ak and bk are both real.
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Even odd Fourrier Series Coefficients
  
.. |zero| image:: tex
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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.
          :alt: tex:{N}_{0}
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.. |one| image:: tex
 
          :alt: tex:{c}_{k}={a}_{k}+j{b}_{k}
 
  
Show that |two| and |three|.
 
  
.. |two| image:: tex
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Show that <math>{a}_{-k}={a}_{k}</math> and <math>{b}_{-k}=-{b}_{k}</math>.
          :alt: tex:{a}_{-k}={a}_{k}
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.. |three| image:: tex
 
          :alt: tex:{b}_{-k}=-{b}_{k}
 
  
If |four| is real we have (equation for Fourier coefficients):  
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If <math>x[n]</math> is real we have (equation for Fourier coefficients):  
  
.. |four| image:: tex
 
          :alt: tex:x[n]
 
  
.. image:: tex
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<math>{c}_{-k}=\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{jk{\omega}_{0}n}</math>
  :alt: tex:{c}_{-k}=\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{jk{\omega}_{0}n}
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and further:
 
and further:
  
.. image:: tex
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<math>={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k}</math>
  :alt: tex:={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k}
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Therefore:
 
Therefore:
  
.. image:: tex
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<math>{c}_{-k}={a}_{-k}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k}</math>
  :alt: tex:{c}_{-k}={a}_{-k}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k}
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So now we can see that:
 
So now we can see that:
  
                    |five| and |six|
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  <math> {a}_{-k}={a}_{k}</math> and
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  <math> {b}_{-k}=-{b}_{k}</math>
  
.. |five| image:: tex
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[[Frequency Response Example_Old Kiwi]]
  :alt: tex:{a}_{-k}={a}_{k}
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.. |six| image:: tex
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  :alt: tex:{b}_{-k}=-{b}_{k}
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Latest revision as of 16:36, 30 March 2008

File:Lecture5 Old Kiwi.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_Old Kiwi

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