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Unformatted text preview: For x4 = [s; z4] : X4 = [X[0] X[2] X[4] X[6] X[8] X[10] X[12] X[14]].' (since x has exactly twice as many entries as [s;z4]) For x5 = [z4 ; s] : X4 = [X[0] X[2] X[4] X[6] X[8] X[10] X[12] X[14]].' (circular shift of x4 by 4=8/2 units, so multiplication by (1)^k in the frequency domain) For x6 = [s ; z4 ; s ; z4] : X6 = [2*X[0] 0 2*X[2] 0 2*X[4] 0 2*X[6] 0 . .. 2*X[8] 0 2*X[10] 0 2*X[12] 0 2*X[14] 0 ].' (periodic extension of [s;z4] to twice its length) For x7 = [s ; s ; z4 ; z4] : x7 = x + (P^4)*x and thus X7 = X + F^(4)*X Since 4*(2*pi/16) = pi/2, we have X7[k] = (1+(j)^k)*X[k] for every k (k=0:15) ....
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 Spring '08
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