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Unformatted text preview: EE 562a Homework Solutions 4 October 13, 2009 1 1. Solution: For each estimation problem you should draw the following block diagram and identify the desirable and observable quantities. Estimator Estimate Error Observable Desirable Model only For the linear estimator case we have the solution estimate = R (desirable)(observable) R 1 observable observable (a) Use the notation r ( u ) = x ( u, 1) x ( u, 2) s ( u ) = x ( u, 3) x ( u, 4) , so that we view x ( u ) as a partitioned vector x ( u ) = r ( u ) s ( u ) which means we can also view the correlation matrix of x ( u ) as a partitioned matrix R x = R r R rs R sr R s . The desirable (quantity to be estimated) in this case is s ( u ) and the observable is r ( u ). Hence the desired estimator is s ( u ) = R sr R r 1 r ( u ) . But since m x = , R x = K x = I = I O O I , the resulting estimate is s ( u ) = . This should not be too surprising because the components of x ( u ) are uncorrelated, so there is no second moment information in r ( u ) about s ( u ). (b) In this part, the desirable is y ( u ) and the observable is again r ( u ), so y ( u ) = R yr R r 1 r ( u ) . 2 EE 562a Homework Solutions 4 October 13, 2009 We must now get the crosscorrelation term, so we partition G G = G r G s , where G r = 6 2 1 6 0 1 0 0 , G s = 3 0 2 3 6 2 1 6 . Then the desirable is y ( u ) = G r r ( u ) + G s s ( u ) , and the crosscorrelation matrix is R yr = E { y ( u ) r ( u ) } = E { G r r ( u ) r ( u ) } + E { G s s ( u ) r ( u ) } = G r R r + G s R sr = G r I + O = G r . The LMMSE estimate is then y ( u ) = G r r ( u ) = 6 2 1 6 0 1 0 0 x ( u, 1) x ( u, 2) . (c) First we write down the solution to the two estimation problems w ( u ) = R wv R 1 v v ( u ) , z ( u ) = R zv R 1 v v ( u ) . The only thing to do is find R zv . R zv = E { z ( u ) v ( u ) } = E { Hw ( u ) v ( u ) } = HR wv . So we have z ( u ) = HR wv R 1 v v ( u ) = H w ( u ) . What happens if z ( u ) = g ( w ( u )), where g ( ) is a nonlinear vector function? (d) The first thing to realize is that the result in (a) also gave us an estimate of x ( u ) based on the observation r ( u ): x ( u ) = r ( u ) , where we have used the simple fact the the best linear estimate of r ( u ) based on the observation r ( u ) is r ( u ) = r ( u )! Thus we identify x ( u ) w ( u ) , y ( u ) z ( u ) , r ( u ) v ( u ) , G H . In other words we could have gotten y ( u ) by y ( u ) = G x ( u ) . 2. Solution: This problem deals with the relationship between LMMSE Estimation and Mini mum Distance Hypothesis Testing. As mentioned in class, the fundamental difference between hypothesis testing and estimation is that in hypothesis testing the signal (desirable) is known to be an element of a finite set (just one of two values in the binary hypothesis test). In this problem we consider estimating a signal that takes on only two values (using LMMSE).this problem we consider estimating a signal that takes on only two values (using LMMSE)....
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This note was uploaded on 04/03/2010 for the course EE 562a taught by Professor Toddbrun during the Fall '07 term at USC.
 Fall '07
 ToddBrun

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