Line 2: Line 2:
  
  
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> { e^(<math>( \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{1}{2j})</math> { e^(<math>( \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>)}  
  
 
Therfore, we have te coeffients as  
 
Therfore, we have te coeffients as  
  
 
<math>a_0</math> = 8
 
<math>a_0</math> = 8

Revision as of 17:50, 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 te coeffients as

$ a_0 $ = 8

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood