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Unformatted text preview: are computed and compared, and all but the lowest-cost path is discarded. The state is then labeled with the cost of the survivor path. Figure 2(b), which illustrates the stage-3 processing, Kevin Buckley - 2007 7 shows only the survivor paths up to stage 2, which are extended to the stage-3 states by adding the branch costs to the path costs stored at the previous states. The paths are then pruned as in the previous stage. Figure 2(c) illustrates that this process continues for each future stage. Figure 2(d) shows the results after pruning at stage 4. As shown, two paths have survived. Note that looking back from stage 4, the survivor paths happen to merge at stage 3. Thus, regardless of what happens in the future, the optimum path will transverse states 0,1,0 at stages 1,2,3. So the MLSE for this DPSK problem will have, for its 1-st three bits, 0,1,1. If stage 4 is the last stage, then the ﬁnal MLSE will correspond to the highlighted path, i.e. 0,1,1,1.
2 2 n=1 r 1 1.3 =
.09 n=2 r 2 −.7 =
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- Spring '09