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630F09_h5_sol - ENEE 630 Homework#5 Solution to 1(a u(n =...

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ENEE 630 Homework #5 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 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) Note that, E [ v ( n ) v ( k )] = ( σ 2 v n = k 0 n 6 = k Use u ( n ) = n - 1 k =0 w k 1 v ( n - k ), we get V ar [ u ( n )] = σ 2 v n - 1 X k =0 w 2 k 1 = σ 2 v ( 1 - w 2 n 1 1 - w 2 1 ) w 1 6 = 1 2 v w 1 = 1 Thus, Var[u(n)] σ 2 v 1 - a 2 1 as n → ∞ . (c) Observing for the real-valued sequence r ( k ) = E [ u ( n ) u ( n - k )] = σ 2 v n - 1 X l =0 w k +2 l 1 = σ 2 v w k 1 ( 1 - w 2 n 1 1 - w 2 1 ) w 1 6 = 1 2 v w 1 = 1 we have r(k) σ 2 v w k 1 1 - w 2 1 as n → ∞ for w 1 6 = 1. Figure 1 shows the sketches for two different cases. 1
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r(k) k Case 1: 0 < a < 1 1 r(k) k Case 2: -1 < a < 0 1 Figure 1: Solution to 2 (a) r (0) r (1) r (1) r (0) 1 - 0 . 5 = r (1) r (2) (b) r (1) = 2 3 r (0) r (2) = 1 6 r (0) (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 Since 1 1 + 0 . 75 z - 1 + 0 . 25 z - 2 1 - 3 4 z - 1 + 5 16 z - 2 - 3 64 z - 3 - 11 256 z - 4 + 45 1024 z - 5 + . . . 2
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(a) An AR(2) approximation of the MA process is: H ( z ) = 1 1 - 0 . 75 z - 1 + 0 . 25 z - 2 (b) An AR(5) approximation of the MA process is: H ( z ) = 1 1 - 0 . 75 z - 1 + 0 . 3125 z - 2 - 0 . 0469 z - 3 - 0 . 043 z - 4 + 0 . 0489 z - 5 Solution to 4
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