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Unformatted text preview: 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 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 nσ 2 v w 1 = 1 Thus, Var[u(n)] → σ 2 v 1 a 2 1 as n → ∞ . (c) Observing for the realvalued 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 nσ 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....
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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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