part37

part37 - ECE 863 Analysis of Stochastic Systems Fall 2001...

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ECE863 1 Prof. Hayder Radha ECE 863 – Analysis of Stochastic Systems Fall 2001 Solutions for Homework Set #8 Michigan State University Department of Electrical & Computer Engineering
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ECE863 2 Prof. Hayder Radha 6.5 a) Within the interval [] t0 , 1 , X(t) can take only two possible values: () () Xt gt 1 =+ or 1 =− Both of these values are equally likely since A takes on 1 ± with equal probability. 1 PXt 1 1 t 0 ,1 2 = = ∀ ∈   () [] 0 1 t 0 == b) For , 1 () () () () X mt 1 PX t 1 1PX t 1 + −=   X mt 0 t 0 , 1 =∀ Also, X , 1 () X t 6.5 c) Here, we have two random variables: Xt and Xt d + The joint pmf of + depends on the time instances (t) and (t + d).
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ECE863 3 Prof. Hayder Radha () [] When t and t d 0,1 +∈ , then both X(t) and X(t + d) have the same value (either both of them = +1 if A = 1 OR both of them = -1 if A = -1). 1 PXt 1 ,Xt d 1 2 1 1 1 2  =+ + =  =− + = t0 , 1 and t d 0,1 ∀∈ [] () When t 0,1 and t d ∈+ , then 1 1 ,Xt d 0 2 1 1 2 + = = + = = , 1 td 0 , 1 +∉ When both t and t d 0,1 , then 0 0 1 = =
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ECE863 4 Prof. Hayder Radha 6.5 d) () XX X C t ,t d EXt Xt d m tm t d  += + +  EXtXt d =+ There is a couple of ways that can be used to evaluate C X (t, t + d). We can use the joint pmf we derived above to compute C X (t, t + d). When (t) and (t + d) are in the interval [0,1], then the product {X(t)X(t + d)} takes on only one possible value: X(t)X(t + d) =1 {} 11 111 1 1 22 X PXtXt d PXt Xt d + = == + = + = + = = . ., , X Cttd 1 [] when t 0,1 a n d td 0 , 1 +∈ X Ct , otherwise Another (easier) approach for computing C X (t, t + d) is as follows: 2 2 X EA gtA gt d EAgt gt d EA gtgt d + It can be easily shown that E[A 2 ]=1. Therefore, ()[] 10 1 0 & X tt d , td g tg td otherwise + =
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ECE863 5 Prof. Hayder Radha 6.9 a) For a given value of t, At and B are independent (since A and B are independent). Therefore, for: () Zt A t B =+ the pdf () () B At fZ fa f b =∗ where At A t = () ( ) B fz fu f z u d u −∞ =− Since A t = u A 1u ff tt  =   A B f z u d u
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ECE863 6 Prof. Hayder Radha 6.9 b) () [] Z EZt m t = [] [] tEA EB =+ A B tm m ( ) ( ) zz 12 Z 1 Z 2 C t,t R t,t m t m t =− ( ) ( ) Z12 1 2 Rt , t EZ tZ t  =  EA tBA t B =++ 22 A B 1 2 EA tt mm t t + + AA 1 B 2 B m m + + () () B1 2 1 1 2 , t mmtt m t t m t t + + + 1 2 C t ,t VAR A t t VAR B
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ECE863 7 Prof. Hayder Radha 6.15 a) () [] ZX Y EAt EXt Y m t mt m =+   ()( ) ( ) zz Z12 1 2 C t,t R t,t m t m t =− ( ) ( ) z 12 1 2 Rt , t EZtZt = 12 EX t YX t Y =++ 22 1 2 EX tt EXY t t EY + + 1 2 EXtt XY t XY + + () () Z1 Z2 X 1 Y X 2 Y mtmt m tm m t m + X11 X Y 1 2 Y mtt mm t t m + + () [] X 1
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part37 - ECE 863 Analysis of Stochastic Systems Fall 2001...

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