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630_hw_4_sol

# 630_hw_4_sol - Advanced DSP HW#4 Solution Set ENEE 630...

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Advanced DSP HW#4 Solution Set ENEE 630 Homework Solution #4 Solution to 1 Recall that Γ m = - Δ m - 1 P m - 1 Δ m - 1 = m - 1 X k =0 a m - 1 ,k r ( k - m ) 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 , Δ 0 = r ( - 1) = 0 . 8 Γ 1 = - Δ 0 /P 0 = - 0 . 8 , a 1 , 0 = 1 , a 1 , 1 = - 0 . 8 P 1 = P 0 (1 - Γ 2 1 ) = 0 . 36 Δ 1 = r (2) a 1 , 0 + r (1) a 1 , 1 = - 0 . 04 Γ 2 = - Δ 1 /P 1 = 1 / 9 , a 2 , 0 = 1 , a 2 , 1 = - 8 / 9 , a 2 , 2 = 1 / 9 P 2 = P 1 (1 - Γ 2 2 ) = 16 / 45 Δ 2 = r (3) a 2 , 0 + r (2) a 2 , 1 + r (1) a 2 , 2 = - 2 / 45 Γ 3 = - Δ 2 /P 2 = 1 / 8 P 3 = P 2 (1 - Γ 2 3 ) = 7 / 20 (b) See Figure 1. Figure 1: (c) See Figure 2 1

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Advanced DSP HW#4 Solution Set 0 1 2 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 2: Solution to 2 (a) (i) The tap weights of a prediction error filter is a m = ˆ 1 - w 0 ! 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 ) with s M ( w ) 4 = 1 e - jw . . . e - jw ( M - 1) Similarly, we can show that R M = E [ u M ( n - 1) u H M ( n - 1)] = σ 2 α s M ( w ) s H M ( w ) + σ 2 v I M Recall the matrix inversion lemma , which says: A = B - 1 + CD - 1 C H A - 1 = ( A + BCD ) - 1 + A - 1 B ( C - 1 + DA - 1 B ) - 1 DA - 1 2
Advanced DSP HW#4 Solution Set Let B - 1 = σ 2 v I M , D - 1 = σ 2 α , C = s M ( w ) then, R - 1 M = σ - 2 v I M - σ - 2 v σ 2 v σ - 2 α + s H M ( w ) s M ( w ) | {z } M s M ( w ) s H M ( w ) Thus, w 0 = R - 1 M r M = e - jw σ 2 v σ - 2 α + M s M ( w ) Recall that P M = r (0) - r

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630_hw_4_sol - Advanced DSP HW#4 Solution Set ENEE 630...

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