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(Fourier Series for DT signals)
 
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Now consider the signal <math>x[n]=sin(3 \pi)\,</math>
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Now consider the signal <math>x[n]=sin(3 \pi n)\,</math>
  
 
It's periodic because <math>\frac{\omega_0}{2\pi} = \frac{3\pi}{2\pi} =1.5 </math>  is a rational number.
 
It's periodic because <math>\frac{\omega_0}{2\pi} = \frac{3\pi}{2\pi} =1.5 </math>  is a rational number.
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<math>a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j\pi n} </math>
 
<math>a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j\pi n} </math>
  
<math>a_1=\frac{1}{2}(x[0] e^{0} + x[1]e^{-j\pi}) </math>
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<math>a_1=\frac{1}{2}(x[0] e^{0} + x[1]e^{-j\pi})=1 </math>
  
<math>a_1=1 \,</math>
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Finally,
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<math>a_{k_{even}}=0\,</math>
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<math>a_{k_{odd}}=1\,</math>

Latest revision as of 05:49, 25 September 2008

Fourier Series for DT signals

Let $ x[n]\, $ be a periodic DT signal with fundamental period N.

Then $ x[n]=\sum_{k=0}^{N-1} a_k e^{jk\frac{2\pi}{N} n} $


where $ a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} $


note that $ \frac{2\pi}{N} =\omega_0 $



Now consider the signal $ x[n]=sin(3 \pi n)\, $

It's periodic because $ \frac{\omega_0}{2\pi} = \frac{3\pi}{2\pi} =1.5 $ is a rational number.


Notice that $ x[0]= 1, x[1]=-1, x[2]=1, x[3]=-1 \, $ etc


Thus the fundamental period is 2.

Then...

$ a_0=\frac{1+(-1)}{2}=0 $

$ a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j1\frac{2\pi}{2} n} $

$ a_1=\frac{1}{2}\sum_{n=0}^{1} x[n] e^{-j\pi n} $

$ a_1=\frac{1}{2}(x[0] e^{0} + x[1]e^{-j\pi})=1 $


Finally,

$ a_{k_{even}}=0\, $

$ a_{k_{odd}}=1\, $

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