Line 8: Line 8:
 
= 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>)}  
  
Therfore, we have te coeffients as  
+
Therfore, we have the coefficients as  
  
 
<math>a_0</math> = 8
 
<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

Revision as of 17:57, 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

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang