K
COS cos(xt + y) =
t= 1
cos(y + 0.5x)sin(Kx) - sin(y + 0.5x)(1 - cos(Kx))
4 sin(0.5 x)
(A.57)
Substituting K = [(n-τ)/s], x = 2kπs/n, y = 2kπ(τ-0.5)/n it follows that
K ~ n, (A.58)
s
1 - cos(Kx) ~ 2k2π2τ2, (A.59)
n2
sin(Kx) ~ -2kπτ , (A.60)
n
sin(0.5x) ~ kπ s , cos(0.5 x) ~ 1, (A.61)
n
sin(y + 0.5x) ~ 2kπ (τ + 0.5s___°J2, cos(y + 0.5x) ~ 1, (A.62)
n
hence
[( n-τ )/s ]
∑ cos(2kπ (js + τ - 0.5)/n)
j=1 (A.63)
-2kπτ /n - (2kπ (τ +0.5s-0.5)/n2kk2π 2τ2/n2) _ τ
4 k π s / n 2 s
and consequently
lim cos[2kπ (t - 0.5)/n] = -ɪ. (A.64)
‘ 2 s
Next, observe that
49
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