SampleMidSol

SampleMidSol - EE 562a Midterm Solution October 15, 2008 1...

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Unformatted text preview: EE 562a Midterm Solution October 15, 2008 1 1. (a) Substitution into well developed results gives Say H 1 is true x ( u ) + A 1 1 < x ( u )- A 1 1 x t ( u ) 1 1 < . P { error } 1 2 A 2 (b) When H 1 is true, z ( u ) , x t ( u ) 1 1 =- A 1 1 + n ( u, 1) n ( u, 2) t 1 1 =- 2 A + n ( u, 1) + n ( u, 2) and this quantity is a Gaussian random variable with probability density p z ( u ) ( r ) = N 1 ( r + 2 A, 2). P (error |H 1 ) = P ( z ( u ) > 0) = Z e- ( r +2 A ) 2 / 4 2 2 dr = Z 2 A/ 2 e- s 2 / 2 2 ds = Q (2 A/ 2) Similarly (or by symmetry), P (error |H 2 ) = Q (2 A/ 2), so P (error) = Q (2 A/ 2). (c) The correlation matrix of the switched random vector is given by K n = 0 . 5 E { w 2 ( u ) } 1 t 1 + 0 . 5 E { w 2 ( u ) } 1 t 1 = I In this case (as in (b)), when H 1 is true, z ( u ) , x t ( u ) 1 1 =- A 1 1 + n ( u, 1) n ( u, 2) t 1 1 =- 2 A + w ( u ) In this case, z ( u ) again has probability density p z ( u ) ( r ) = N 1 ( r + 2 A, 2) and therefore P (error |H 1 ) = Q (2 A/ 2). and it is easily shown that P (error) = Q (2 A/ 2). (d) The switched vector n ( u ) of part (c) is not Gaussian. There are at least three ways to view this: (1) The joint probability density p x ( z ) is non-zero only on the z 1 and z 2 axes and does not have the form associated with jointly Gaussian random variables, (2) the marginal probability density is p n ( u, 1) ( z 1 ) = 0 . 5 N 1 ( z 1 , 2) + 0 . 5 D ( z 1 ) which is not Gaussian, and (3) the transformation from w ( u ) to n ( u ) is not a simple linear transformation. When H 1 is true, one of the coordinates is guaranteed to- A (which coordinate is unknown); similarly when H 2 is true, one of the coordinates is guaranteed to be A . Since the probability that one coordinate is exactly A and the other is- A in any given observation x ( u ) is zero, an ad hoc perfect decision procedure is possible: Say H 1 is true x ( u, 1) =- A or x ( u, 2) =- A. 2 EE 562a Midterm Solution October 15, 2008 Obviously P (error |H 1 ) = P ( { x ( u, 1) 6 =- A } { x ( u, 2) 6 =- A }|H 1 ) = 0 . Similarly it can be shown that P (error |H 2 ) = 0 = P (error) . Note that the second-moment description of the vector n ( u ) in all four parts of this problem, This solution illustrates the way a decision procedure or performance evaluation can change when you are given more information about the observation....
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SampleMidSol - EE 562a Midterm Solution October 15, 2008 1...

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