Line 4: Line 4:
 
8 + sin<math>( \frac{2  pi  n}{N} )</math> + 8cos<math>( \frac{4  pi  n}{N} )</math>
 
8 + sin<math>( \frac{2  pi  n}{N} )</math> + 8cos<math>( \frac{4  pi  n}{N} )</math>
  
= 8 + <math>( \frac{1}{2j})</math>  {<math>( e^( \frac{j2  pi  n}{N} ) </math>- <math>e^( \frac{-j2 pi n}{N} )</math>} + 4 {<math>e^( \frac{j4 pi n}{N} </math> + <math>e^( \frac{-j4 pi n}{N} </math>}
+
= 8 + <math>( \frac{1}{2j})</math>  {<math>( e^( \frac{j2  pi  n}{N} ) </math> - <math>e^( \frac{-j2 pi n}{N} )</math>} + 4 {<math>e^( \frac{j4 pi n}{N} </math> + <math>e^( \frac{-j4 pi n}{N} </math>}
  
= 8 + <math>( \frac{-1j}{2})</math> { e^(<math>( \frac{j2  pi  n}{N} )</math>)} + <math>( \frac{1j}{2})</math> { e^(<math>( \frac{-j2  pi  n}{N} )</math>)} +4 { e^(<math>( \frac{j4  pi   n}{N} )</math>)} +4 { e^(<math>( \frac{-j4  pi  n}{N} )</math>)}
+
= 8 + <math>( \frac{-1j}{2})</math> <math>( e^( \frac{j2  pi  n}{N} ) </math> + <math>( \frac{1j}{2})</math> <math>( e^( \frac{-j2  pi  n}{N} ) </math> +4 <math>( e^( \frac{j4  pi n}{N} ) </math> +4 <math>( e^( \frac{-j4  pi  n}{N} ) </math>  
  
 
Therfore, we have the coefficients as  
 
Therfore, we have the coefficients as  

Latest revision as of 18:11, 26 September 2008

Let the DT siganl be


8 + sin$ ( \frac{2 pi n}{N} ) $ + 8cos$ ( \frac{4 pi n}{N} ) $

= 8 + $ ( \frac{1}{2j}) $ {$ ( e^( \frac{j2 pi n}{N} ) $ - $ e^( \frac{-j2 pi n}{N} ) $} + 4 {$ e^( \frac{j4 pi n}{N} $ + $ e^( \frac{-j4 pi n}{N} $}

= 8 + $ ( \frac{-1j}{2}) $ $ ( e^( \frac{j2 pi n}{N} ) $ + $ ( \frac{1j}{2}) $ $ ( e^( \frac{-j2 pi n}{N} ) $ +4 $ ( e^( \frac{j4 pi n}{N} ) $ +4 $ ( e^( \frac{-j4 pi n}{N} ) $

Therfore, we have the coefficients as

$ a_0 $ = 8

$ a_1 $ = $ ( \frac{-1 j }{2} ) $

$ a_-1 $ = $ ( \frac{1 j }{2} ) $


$ a_2 $ = 4


$ a_-2 $ = 4

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett