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Unformatted text preview: 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 = r (0) = 1 , Δ = r ( 1) = 0 . 8 Γ 1 = Δ /P = . 8 , a 1 , = 1 , a 1 , 1 = . 8 P 1 = P (1 Γ 2 1 ) = 0 . 36 Δ 1 = r (2) a 1 , + r (1) a 1 , 1 = . 04 Γ 2 = Δ 1 /P 1 = 1 / 9 , a 2 , = 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 , + 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 Advanced DSP HW#4 Solution Set 1 2 3 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 ! where w = 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...
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This note was uploaded on 10/31/2010 for the course EE 630 taught by Professor Wu during the Spring '10 term at Aarhus Universitet, Aarhus.
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