Final Solution 2008 new - Detection & Estimation...

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Unformatted text preview: Detection & Estimation Spring, 2008 Final Exam Solution 1. (a) MVU = x[n](-1) n =0 N -1 n , N N -1 n =0 2 variance = 2 N n (b) ML estimate = x[n](-1) N N -1 n =0 (c) ^ A= N 1 + 2 2 A 2 2 x[n](-1) N -1 n=0 n (d) MMSE = N 1 + 2 2 A 2 2 x[n](-1) n , Bmse = 1 N 2 1 + 2 2 A 2. (a) x[n]e -n n =0 N -1 , x[n]e n =0 N -1 - 2 n N -1 - 2 n e n=0 2 (b) I ( ) = N -1 - 3 n e n =0 2 n =0 2 N -1 e - 4 n n =0 2 e N -1 - 3 n ^ (c) Var ( A) 2 e - 4 n N -1 n=0 e - 2 n e - 4 n - ( e - 3 n ) 2 n=0 n=0 n=0 N -1 N -1 N -1 n=0 n=0 n=0 N -1 N -1 N -1 - 2 n - 4 n - 3 n 2 n=0 n=0 N -1 N -1 N -1 ^ , Var ( B) 2 e - 2 n N -1 n=0 e - 2 n e - 4 n - ( e - 3 n ) 2 n=0 n=0 n=0 N -1 N -1 N -1 ^ (d) A = e- 4n x[n]e- n - e-3n x[n]e- 2n n=0 N -1 e e N -1 n=0 N -1 - ( e n=0 N -1 ) ^ B= - e - 3n x[n]e - n + e - 2 n x[n]e - 2 n N -1 e e n=0 n=0 n=0 n=0 n=0 N -1 N -1 N -1 - 2 n - 4 n - 3 n 2 - ( e n=0 ) 3. (a) L(x[0]) 0.5exp{x[0]} 1/ 3/ x[0] (b) The NP test decides H1 if ln(0.2 + e -3 ) 3 - < x[0] < PD = (c) PD 1 1.5+0.5 ln(PFA+e-3) 3 1 + ln(0.2 + e -3 ) 2 2 e-1-e-3 1 PFA (d) We decide H1 if ln 4 3 < x[0] < We decide H0 otherwise. 4. (a) NP detector N -1 - n x[ n]e 0 n =N -1 > r ' for A > 0 - 2 n e n =0 , N -1 - n x[n]e n =0 N -1 - 2 n < r" for A < 0 e n=0 (b) UMP test Decide H1 if x[n]e n =0 N -1 n =0 N -1 - n e - 2 n >r (c) From midterm, we know that ^ A= x[n]e n =0 N -1 n =0 N -1 - n e ( - 2 n 1 2 ( 2 ) e N - n =0 ^ ( x [ n ] - Ae - n ) 2 2 2 N -1 N -1 LG ( x ) = 1 2 2 N -1 n =0 )N e - n =0 2 2 x2 [n] ln( LG ( x )) = e - 2 n 2 2 0 ( n =N -1 n =0 x[n]e e - 2 n N -1 - n )2 We decide H 1 if x[n]e n =0 N -1 - n > r ' or x[n]e n =0 N -1 n =0 N -1 - n e - 2 n > r" (d) ^ A= x[n]e n =0 N -1 n =0 N -1 - n e - 2 n ^2 0 = ^ 12 = 1 N 1 N x [ n] 2 n =0 N -1 ^ ( x[n] - Ae ) n =0 N -1 - n 2 ^ ^ ^2 p ( x; A, 12 , H1 ) 0 N LG ( x ) = =( 2)2 ^2 ^ p ( x ; 0 , H 0 ) 1 ...
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