EE562a_Final2005_Solutions

EE562a_Final2005_Solutions - EE 562a Final Exam Solutions...

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EE 562a Final Exam Solutions August 8, 2005 Inst: Dr. C.W. Walker Problem Points Score 1 10 2 12 3 10 4 10 5 10 6 12 7 10 8 16 9 10 Total 100

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Provided Data. Some of the following may be useful on this test. Laplace Transforms: 1 ←→ 1 s ,e at 1 s + a ( t ) 1 ,t 1 s 2 , if y ( t ) Y ( s )then y 0 ( t ) sY ( s ) Y (0) . N X k = M r k = r M r N +1 1 r ,r 6 =1 . 1
Problem 1. A mean zero normal random vector X is observed so that X N (0 , K X ), where the covariance matrix K X is unknown. This data is generated by the operation X = HY + n where the transformation H is linear and Y N (0 , K Y )and n N (0 , K n ). Both K Y and K n are known. Find ˆ Y in terms of H , K Y , K n and X ,where ˆ Y is the optimal estimator of Y in the mean-square sense. Note: Y is un- correlated with n . Solution. ˆ Y = K Y H h HK Y H + K n i 1 X . Problem 2. Consider a real Gaussian random sequence x ( n ), n anon - negative integer, with E [ x ( n )] = 0 ,E h x ( n ) 2 i =1 [ x ( n ) x ( m )] = ρ | n m | where 0 <ρ< 1. Let y ( n )= 1 n n 1 X k =0 x ( k ) ,n> 0 0 ,n =0 . a. Is x ( n ) wide sense stationary? Solution. Yes. b. Find the covariance of y ( n ) and state whether or not it is wide-sense stationary. Solution. First consider m>n . E [ y ( n )] = 0 so K Y ( n, m E [ y ( n ) y ( m )] = E " 1 n n 1 X k =0 x ( k ) 1 m m 1 X l =0 x ( l ) # = 1 nm n 1 X k =0 m 1 X l =0 E [ x ( k ) x ( l )] = 1 nm n 1 X k =0 m 1 X l =0 ρ | k l | 2

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= 1 nm n 1 X k =0 k X l =0 ρ k l + m 1 X l = k +1 ρ l k = 1 nm n 1 X k =0 ρ k · 1 ρ k 1 1 ρ 1 + ρ k · ρ k +1 ρ m 1 ρ ! = 1 nm n 1 X k =0 ρ k ρ 1 1 ρ 1 + ρ ρ m k 1 ρ ! = 1 nm " 1 1 ρ 1 1 ρ n 1 ρ 1 ! + 1 1 ρ ρ m · 1 ρ n 1 ρ 1 !# or K Y ( n, m )= 1 nm " 1 ρ n ρ m + ρ m n + n ( ρ ρ 1 ) (1 ρ )(1 ρ 1 ) # . If m<n since K Y ( n, m K Y ( m, n ) we can interchange the roles of m and n in the last expression to get K Y ( n, m 1 nm " 1 ρ n ρ m + ρ n m + m ( ρ ρ 1 ) (1 ρ )(1 ρ 1 ) # . If m = n we get K Y ( n, m 1 nm " 1 ρ n + n (1 ρ 1 ) (1 ρ )(1 ρ 1 ) # . Putting the above together we get K Y ( n, m 1 nm " 1 ρ n ρ m + ρ | n m | +max { n, m } ( ρ ρ 1 ) (1 ρ )(1 ρ 1 ) # ,m 6 = n 1 nm " 1 ρ n + n (1 ρ 1 ) (1 ρ )(1 ρ 1 ) # = n.
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This note was uploaded on 09/13/2008 for the course EE 562a taught by Professor Toddbrun during the Summer '07 term at USC.

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EE562a_Final2005_Solutions - EE 562a Final Exam Solutions...

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