(New page: <math>x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n}</math> <math>a_k=\frac{1}{N}\sum_{n=0}^{N-1} x[n] e^{-jk\frac{2\pi}{N} n} </math> <math>\frac{2\pi}{N} =\omega_0</math> ...)
 
 
Line 23: Line 23:
 
<math>x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n}</math>
 
<math>x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n}</math>
  
 
+
<math>x[0]=3,x[1]=4,x[2]=3,x[3]=2,x[4]=3, x[5]=4\!</math>
<font size = 3.5><math>x[0]=3,x[1]=4,x[2]=3,x[3]=2,x[4]=3, x[5]=4</math></font>
+
  
 
So, the fundamental period is 4.
 
So, the fundamental period is 4.
Line 31: Line 30:
 
== Fourier Series Coefficients ==
 
== Fourier Series Coefficients ==
  
<font size = 3.5><math>a_0=3</math></font>
+
<math>a_0=3\!</math>
  
 
<math>a_1=\frac{1}{2i}</math>
 
<math>a_1=\frac{1}{2i}</math>
Line 37: Line 36:
 
<math>a_{-1}=-\frac{1}{2i}</math>
 
<math>a_{-1}=-\frac{1}{2i}</math>
  
<font size = 3.5><math>a_k,even=3</math></font>
+
<math>a_k,even=3\!</math>
  
<math>a_k=\frac{1}{2i}</math> for all <math>k=1,5,9,13,..., 4n+1</math>
+
<math>a_k=\frac{1}{2i}</math> for all <math>k=1,5,9,13,..., 4n+1\!</math>
  
<math>a_k=-\frac{1}{2i}</math> for all <math>k=3,7,11,15,..., 4n+3</math>
+
<math>a_k=-\frac{1}{2i}</math> for all <math>k=3,7,11,15,..., 4n+3\!</math>

Latest revision as of 13:35, 26 September 2008

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

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

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


Signal

$ x[n]=3+sin\bigg(\frac{\pi}{2}n\bigg) $

$ \frac{w_0}{2\pi}=\frac{1}{4} $, which is rational, so it is periodic.


Fourier Series

$ x[n]=3+sin\bigg(\frac{\pi}{2}n\bigg) $

$ x[n]=3+\frac{e^{i\frac{\pi}{2}n}-e^{-i\frac{\pi}{2}n}}{2i} $

$ x[n]=3+ \frac{1}{2i} e^{i\frac{\pi}{2}n}-\frac{1}{2i}e^{-i\frac{\pi}{2}n} $

$ x[0]=3,x[1]=4,x[2]=3,x[3]=2,x[4]=3, x[5]=4\! $

So, the fundamental period is 4.


Fourier Series Coefficients

$ a_0=3\! $

$ a_1=\frac{1}{2i} $

$ a_{-1}=-\frac{1}{2i} $

$ a_k,even=3\! $

$ a_k=\frac{1}{2i} $ for all $ k=1,5,9,13,..., 4n+1\! $

$ a_k=-\frac{1}{2i} $ for all $ k=3,7,11,15,..., 4n+3\! $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn