The complication with problem 22 is that the cost is

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Unformatted text preview: t r b = {ra , ra+1 , · · · , rb }. The symbol-by-symbol MAP estimate of s(k) is the a symbol Am which maximizes P (s(k) = Am /r k+D ) = 1 This is equivalent to the problem: s (k ) = A m p(r k+D /s(k) = Am ) P (s(k) = Am ) 1 p (r k + D ) 1 . (21) max p(rk+D /s(k) = Am ) P (s(k) = Am ) , 1 p(r k+D /s(k) = Am ) 1 (22) which reduces to s (k ) = A m max (23) (i.e. the ML symbol estimation) for equally probably symbols. The complication with problem (22) is that the cost is implicitly a function of ”nuisance” symbols s(n) ; n = 1, 2, · · · , k − 1, k + 1, k + 2, · · · , k + D . This is overcome, as detailed in the Course Text, by marginalizing over the nuisance symbols. A time-recursive algorithm for generating the costs in problem (22) is derived in the Course Text (based on a paper by Abend and Fritchman (1970)). The resulting algorithm, given data rk ; k = 1, 2, · · · , k + D , is: s (k ) = A max m p(r k+D /s(k) = Am ) P (s(k) = Am ) 1 = s(k+1) s(k+2) ··· s (k + D ) pk−1 (s(k) , s(k+1) , s(k+2) , · · · , s(k+D) ) (24) where pk (s(k) , s(k+1) , · · · , s(k+D) ) = (k +D −L) (k +D −L+1) pk (rk+D...
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This document was uploaded on 10/12/2009.

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