32 for k 4 receiver observations r1 r2 r3 r4

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Unformatted text preview: p · M L−1 paths at stage n to stage n + 1 by adding the ˆ new branch cost to the previous path cost; 3. at stage n + 1, for each state, compare all path cost into the state and keep only the one with lowest cost (i.e. prune the rest of the paths). Example 4 - Coherent DPSK: Consider the coherent DPSK MLSE problem discussed last week in Subsection 2.3.2. For K = 4 receiver observations, {r1 , r2 , r3 , r4 } = {1.3, −.7, 1.1, −0.2}, the MLSE cost was K min 4 S1 D (r 4 , S 4 ) = 1 1 n=1 ||rn − Sn ||2 , (18) where Sn is the state (and also the symbol) at time n for the hypothesized sequence S 4 . For this problem, the state can be either “1” or “-1”. The trellis diagram, 1 first shown in Figure 2 from Lecture 4, is redrawn here in Figure 2 to show the stage-by-stage implementation of the Viterbi algorithm. As before, each branch label with its cost ||rn − Sn ||2. Figure 2(a) shows processing up to stage 2, where the first Viterbi pruning takes place. At each state, costs for all incoming paths...
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This document was uploaded on 10/12/2009.

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