solved16 - For x4 = [s; z4] : X4 = [X[0] X[2] X[4] X[6]...

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S 16.1 (P 3.26) _______________ (i) x is obtained from y by zero-padding to double its length. Thus Y = [ X0 X2 X4 X6 X8 X10 ].' (ii) x is also obtained from [a b c].' by zero-padding to four times its length. Thus [ a b c ].' <--> [ X0 X4 X8 ].' S is a periodic extension of [a b c].' to four times its length. Thus S = 4*[ X0 0 0 0 X4 0 0 0 X8 0 0 0 ].' S 16.2 (P 3.27) _______________ For x1=s : X1 = [X[0] X[4] X[8] X[12]].' (since the zero-padded vector has length = 4*length of s). For x2 = [s ; s] : X2 = [2*X[0] 0 2*X[4] 0 2*X[8] 0 2*X[12] 0].' (periodic extension of s to twice its length) For x3 = [s ; s ; s ; s ; s] : X3 = [5*X[0]; z4 ; 5*X[4] ; z4 ; 5*X[8] ; z4 ; 5*X[12] ; z4]
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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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solved16 - For x4 = [s; z4] : X4 = [X[0] X[2] X[4] X[6]...

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