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Unformatted text preview: ECE 302: Homework 7 Prof. Saul Gelfand Spring 2010 Due April 30, 2010 Problem 1 a. Figure 1 shows the only two possible outcomes of X n , each which have probability 1 2 . b. We have that for all n , X n is either 1 or 1, with probability 1 2 for each case. Therefore p X n ( x ) = braceleftbigg 1 2 , x = 1 1 2 , x = 1 c. We first notice that no matter whether heads or tails were tossed, that X n = X n + k when k is even, and X n = X n + k when k is odd. Since X n always has equal probability of being 1 or 1 for all n , then: k even ⇒ p X n ,X n + k ( x 1 , x 2 ) = braceleftbigg 1 2 , x 1 = x 2 = 1 1 2 , x 1 = x 2 = 1 k odd ⇒ p X n ,X n + k ( x 1 , x 2 ) = braceleftbigg 1 2 , x 1 = x 2 = 1 1 2 , x 1 = x 2 = 1 d. The mean function m X ( n ) = E [ X n ] = summationdisplay x xp X n ( x ) = ( 1) 1 2 + (1) 1 2 = 0 . 15 510.5 0.5 1 n X n5 510.5 0.5 1 n X n (a) (b) Figure 1: Problem 9.3(a): Realizations of X n when (a) heads is tossed; (b) tails is tossed.1 1 2 0.2 0.4 0.6 0.8 1 t X(t)1 1 210.80.60.40.2 t X(t) (a) (b) Figure 2: Problem 9.5(a): Realizations of X ( t ) when (a) A = 1; (b) A = 1. The autocovariance C X ( n, n + k ) = E [ X n X n + k ] m X ( n ) m X ( n + k ) = parenleftBigg summationdisplay x 1 ,x 2 x 1 x 2 p X n X n + k ( x 1 , x 2 ) parenrightBigg (0)(0) = braceleftbigg 1 , k even 1 , k odd Problem 2 We have that X ( t ) has two possible outcomes, shown in Figure 2, which are equallylikely. a. We see from Figure 2 that for t ∈ [0 , 1], X ( t ) = 1 with probability 2 . 5 and X ( t ) = 1 with probability 0 . 5; for t / ∈ [0 , 1], X ( t ) = 0 with probability 1. Therefore: p X ( t ) ( x ) = braceleftbigg 1 2 δ ( x 1) + 1 2 δ ( x + 1) , t ∈ [0 , 1] δ ( x ) , else. b. m X ( t ) = E [ X ( t )] = summationdisplay x xp X ( t ) ( x ) = braceleftbigg 1 2 1 2 = 0 , t ∈ [0 , 1] , else Thus, m x ( t ) = 0 for all t . c. To compute the joint pmf, p X ( t ) X ( t + d ) ( x 1 , x 2 ), we explore several cases for t and t + d . If t and t + d are both in the range [0 , 1] (i.e. t ∈ [0 , 1], d ∈ [ t, 1 t ]), then X ( t ) = X ( t + d ) = ± 1. If t ∈ [0 , 1] and d / ∈ [ t, 1 t ], then X ( t ) = ± 1 and X ( t + d ) = 0. Similarly, if t / ∈ [0 , 1] and d ∈ [ t, 1 t ], then X ( t ) = 0 and X ( t + d ) = ± 1. Finally, if t / ∈ [0 , 1] and d / ∈ [ t, 1 t ], then X ( t ) = X ( t + d ) = 0....
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This note was uploaded on 02/07/2011 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.
 Fall '08
 GELFAND

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