(New page: Let the DT siganl be 8 + sin<math>( \frac{2*pi*n}{N} )</math> + 8cos<math>( \frac{4*pi*n}{N} )</math>)
 
 
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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{-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
 +
 
 +
<math>a_0</math> = 8
 +
 
 +
<math>a_1</math> = <math>( \frac{-1 j }{2} )</math>
 +
 
 +
<math>a_-1</math> = <math>( \frac{1 j }{2} )</math>
 +
 
 +
 
 +
<math>a_2</math> = 4
 +
 
 +
 
 +
<math>a_-2</math> = 4

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

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Sean Hu, ECE PhD 2009