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

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood