630_hw_3_sol

# 630_hw_3_sol - ENEE 630 Homework Solution#3 Solution to 1(a...

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Unformatted text preview: ENEE 630 Homework Solution #3 Solution to 1 (a) u ( n ) = v ( n ) + w 1 u ( n- 1) = n- 1 X k =0 w k 1 v ( n- k ) where w 1 =- a 1 and (assuming) u(0)=0. Thus, E [ u ( n )] =      m v ( 1- w n 1 1- w 1 ) w 1 6 = 1 nm v w 1 = 1 where m v = E [ v ( n )]. Therefore, { u(n) } is nonstationary . However, if | a 1 | < 1, then E[u(n)] → m v 1- w 1 as n → ∞ . (b) Since r (0) + a 1 r (1) = σ 2 v and Yule-Walker equation r (1) + a 1 r (0) = 0 , we get r (0) = σ 2 v 1- a 2 1 . (c) Further, r ( k + 1) + a 1 r ( k ) = 0 . Therefore, r ( k ) = (- a 1 ) | k | σ 2 v 1- a 2 1 . Figure 1 shows the sketches for two different cases. Solution to 2 (a) ˆ r (0) r (1) r (1) r (0) !ˆ 1- . 5 ! = ˆ r (1) r (2) ! (b) r (1) = 2 3 r (0) r (2) = 1 6 r (0) 1 r(k) k Case 1: 0 < a < 1 1 r(k) k Case 2: -1 < a < 0 1 Figure 1: 2 (c) Var[u(n)]=r(0). Note that, σ 2 v = 2 X n =0 a n r ( n ) = r (0) + a 1 r (1) + a 2 r (2) Thus, r(0)= σ 2 u =1.2 Solution to 3 X ( z ) = V ( z )(1 + 0 . 75 z- 1 + 0 . 25 z- 2 ) H ( z ) = X ( z ) V ( z ) = 1 + 0 . 75 z- 1 + 0 . 25 z- 2 Use the division, we have 1 1 + 0 . 75 z- 1 + 0 . 25 z- 2 = 1- 3 4 z- 1 + 5 16 z- 2- 3 64...
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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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630_hw_3_sol - ENEE 630 Homework Solution#3 Solution to 1(a...

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