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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> | + | = 8 + <math>( \frac{1}{2j})</math> <math>( e^( \frac{j2 pi n}{N} ) </math>- e^(<math>( \frac{-j2 pi n}{N} )</math> } + 8 { e^(<math>( \frac{j4 pi n}{N} )</math>) - 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>)} | = 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>)} | ||
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<math>a_-2</math> = 4 | <math>a_-2</math> = 4 | ||
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Revision as of 17:01, 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} ) $ } + 8 { 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
