soln5

soln5 - EE 562a Homework Solutions 5 22 March 2006 1 1. (a)...

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Unformatted text preview: EE 562a Homework Solutions 5 22 March 2006 1 1. (a) Since the observation and desired signal are jointly Gaussian and mean zero we know that optimal unconstrained solution is the LMMSE estimator. Therefore we have that v i ( u ) , E { v ( u ) | x i ( u ) } = R vx ( i ) R- 1 x ( i ) x i ( u ) i = 1 , 2 , 3 ... In particular, we have for i = k + 1 v k +1 ( u ) = R vx ( k + 1) R- 1 x ( k + 1) x k +1 ( u ) . The key to this problem is to partition the observations as x k +1 ( u ) = x ( u,k + 1) x k ( u ) . With this we obtain the following updates R vx ( k + 1) = E v ( u ) x ( u,k + 1) x t k ( u ) = r x v ( k + 1) R vx ( k ) R x ( k + 1) = E x ( u,k + 1) x k ( u ) x ( u,k + 1) x t k ( u ) = 2 x ( k + 1) r t x v ( k + 1) r x v ( k + 1) R x ( k ) . We use the matrix inversion lemma to get = R- 1 vx ( k + 1) = 1 - 1 r t x v ( k + 1) R- 1 x ( k )- 1 R- 1 x ( k ) r x v ( k + 1) R- 1 x ( k )- 1 R- 1 x ( k ) r x v ( k + 1) r t x v ( k + 1) R- 1 x ( k ) , where we have added lines to show the partitioning more clearly and we have introduced the term for simplicity = 2 x ( k + 1)- r t x v ( k + 1) R- 1 x ( k ) r x v ( k + 1) . We can then multiply the two partitioned matrices together yielding R vx ( k + 1) R- 1 x ( k + 1) = M P , where the matrices M and P are M = 1 r x v ( k + 1)- R vx ( k ) R- 1 x ( k ) r x v ( k + 1) ( n 1) 2 EE 562a Homework Solutions 5 22 March 2006 P = R vx ( k ) R- 1 x ( k )- 1 r x v ( k + 1)- R vx ( k ) R- 1 x ( k ) r x v ( k + 1) r t x v ( k + 1) R- 1 x ( k ) ( n k ) . Using this partitioned form, along with the partitioned version of x k +1 ( u ) = v k +1 ( u ) = M x ( u,k + 1) + Px k ( u ) = R vx ( k ) R- 1 x ( k ) x k ( u ) + 1 r x v ( k + 1)- R vx ( k ) R- 1 x ( k ) r x v ( k + 1) x ( u,k + 1)- r t x v ( k + 1) R- 1 x ( k ) x k ( u ) = v k ( u ) + g ( k + 1) x ( u,k + 1)- r t x v ( k + 1) R- 1 x ( k ) x k ( u ) . This is in the desired form with the Kalman Gain vector defined by g ( k + 1) = r x v ( k + 1)- R vx ( k ) R- 1 x ( k ) r x v ( k + 1) 2 x ( k + 1)- r t x v ( k + 1) R- 1 x ( k ) r x v ( k + 1) . (b) The innovation provided by x ( u,k + 1) is x ( u,k + 1)- r t x v ( k + 1) R- 1 x ( k ) x k ( u ) , which can identified as the MMSE estimate of x ( u,k +1) based on the observation x k ( u ). So the from of this recursive estimator is intuitively satisfying: we update the estimate by a linear function of the error in predicting x ( u,k + 1) from x k ( u ). This may not seem all that useful if you consider the fact that we still need to estimate x ( u,k +1) from x k ( u ), which means that we need to compute R- 1 x ( k ) in order to update the estimate. A recursion on this inverse matrix can be found when we have a fixed model for the relation between the observation and the desirable. For example, the state variable model would have the linear observation equation:...
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soln5 - EE 562a Homework Solutions 5 22 March 2006 1 1. (a)...

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