| 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>- | + | = 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> { 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>)} | ||
Revision as of 17:07, 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
