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# HW7sol - ENEE 630 Homework#7 Solution to 1 Recall that m =...

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ENEE 630 Homework #7 Solution to 1 Recall that Γ m = - Δ m 1 P m 1 where Δ m 1 = m 1 summationdisplay k =0 a m 1 ,k r ( k - m ) We have r ( m ) = - Γ m P m 1 - m 1 summationdisplay k =1 a m 1 ,k r ( m - k ) or, Γ m = - r ( m ) P m 1 - 1 P m 1 m 1 summationdisplay k =1 a m 1 ,k r ( m - k ) We also need, a m,k = a m 1 ,k + Γ m a m 1 ,m k , k = 0 , 1 , . . . , m P m = P m 1 (1 - | Γ m | 2 ) (a) P 0 = r (0) = 1 Γ 1 = - r (1) P 0 = - 0 . 8 P 1 = P 0 (1 - | Γ 1 | 2 ) = 0 . 36 Γ 2 = - r (2) P 1 - r (1) P 1 Γ 1 = 1 9 P 2 = P 1 (1 - | Γ 2 | 2 ) = 0 . 356 Also, a 22 = Γ 2 a 21 = Γ 1 + Γ 2 Γ 1 = - 0 . 889 Δ 2 = - 0 . 0444 Γ 3 = - Δ 2 P 2 = 0 . 125 1

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(b) See Figure 1. -1 z -1 z -1 z -0.8 -0.8 0.111 0.281 0.281 0.111 u[n] Figure 1: P-1 (c) P 3 = P 2 (1 - | Γ 3 | 2 ) = 0 . 35. See Figure 2 P m m 1 1 2 3 Figure 2: P-1 Solution to 2 (a) (i) The tap weights of a prediction error filter is a m = parenleftbigg 1 - w 0 parenrightbigg where w 0 = R 1 M r M and r M = E [ u M ( n - 1) u 1 ( n )] = σ 2 α e jw e j 2 w . . . e jMw = σ 2 α e jw s M ( w ) 2
with s M ( w ) = 1 e jw .

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