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630_rec_10f09_sol

# 630_rec_10f09_sol - ENEE630 ADSP 1 RECITATION 10 w solution...

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ENEE630 ADSP RECITATION 10 w/ solution November 19/20, 2009 1. Given an real-valued AR(3) model with parameters Γ 1 = - 4 / 5, Γ 2 = 1 / 9, Γ 3 = 1 / 8, and r (0) = 1. Find r (1) , r (2), and r (3). Solution: Since Γ 1 = - r (1) /r (0), r (1) = - Γ 1 r (0) = 4 / 5. P 0 = r (0) = 1. Γ 1 = - 4 / 5. Then, a 1 , 0 = 1 , a 1 , 1 = - 4 / 5. P 1 = (1 - | Γ 1 | 2 ) P 0 = 9 / 25. Δ 1 = - P 1 Γ 2 = - 1 / 25. Also, Δ 1 = r ( - 2) a 1 , 0 + r ( - 1) a 1 , 1 . Hence, r (2) = r ( - 2) = Δ 1 - r ( - 1) a 1 , 1 = 3 / 5. a 2 , 0 = 1 , a 2 , 1 = - 4 / 5 + 1 / 9( - 4 / 5) = - 8 / 9 , a 2 , 2 = Γ 2 = 1 / 9. P 2 = (1 - | Γ 2 | 2 ) P 1 = 16 / 45. Δ 2 = - P 2 Γ 3 = - 2 / 45 = r ( - 3) a 2 , 0 + r ( - 2) a 2 , 1 + r ( - 1) a 2 , 2 , from which we solve r (3) = 2 / 5. 2. Consider the MA(1) process x ( n ) = v ( n ) + bv ( n - 1) with v ( n ) being a zero-mean white sequence with variance 1. If we use Γ k to represent this system, prove that Γ m +1 = Γ 2 m Γ m - 1 (1 - | Γ m | 2 ) . Solution: Note that r ( k ) = 0 for | k | ≥ 2. Γ m +1 = - Δ m P m . Δ m = m k =0 r ( k - ( m + 1)) a m,k = r ( - 1) a m,m = r ( - 1)Γ m . Therefore, Γ m +1 Γ m = Δ m P m P m - 1 Δ m - 1 = Γ m Γ m - 1 (1 - | Γ m | 2 ) . 3. Given a p -order AR random process { x ( n ) } , it can be equivalently represented by any of the three following sets of values: • { r (0) , r (1) , . . . , r ( p ) } • { a 1 , a 2 , . . . , a p } and r (0) • { Γ 1 , Γ 2 , . . . ,

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630_rec_10f09_sol - ENEE630 ADSP 1 RECITATION 10 w solution...

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