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630_rec_9_sol - ENEE630 ADSP 1 RECITATION 9 w solution...

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ENEE630 ADSP RECITATION 9 w/ solution November 12/13, 2009 1. [Problem 8.3 continued] Assume v ( n ) and w ( n ) are white Gaussian random processes with zero mean and variance 1. The two filters in Fig. R9.1 are G ( z ) = 1 1 - 0 . 4 z - 1 and H ( z ) = 2 1 - 0 . 5 z - 1 . Figure R9.1: (a) Design a 1-order Wiener filter such that the desired output is u ( n ). What is the MSE? (b) Design a 2-order Wiener filter. What is the MSE? Solution: (a) R x = " r u (0) + 1 r u (1) r u (1) r u (0) + 1 # , and p xd = " r u (0) r u (1) # . The filter is w = R - 1 x p with MSE r u (0) - p H xd R - 1 x p xd . (b) Similar to (a), except R x = r u (0) + 1 r u (1) r u (2) r u (1) r u (0) + 1 r u (1) r u (2) r u (1) r u (0) + 1 , and p xd = r u (0) r u (1) r u (2) . MSE is still the same expression, i.e. r u (0) - p H xd R - 1 x p xd . Note: 1. In general, for a p -order AR model, given { σ 2 v , a 1 , a 2 , . . . , a p } , we can find { r (0) , r (1) , r (2) , . . . } ; and vice versa. They are related by Yule-Walker Equations. 2. r ( - k ) = r * ( k ) in general (and hence matrix R is Hermitian), and r ( - k ) = r ( k ) for real-valued signals. r (0) is the power of sequence u ( n ), and hence r (0) > 0 from physical point of view.
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