# 3 n2 r 2 7 289 n3 r 3 11 1 0 1 0 2 8 12 1

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Unformatted text preview: ) 2 1 1 0 1 0 | r 1− 1 | =(.3) 2 2 1 0 1 0 2 (.3) (−1.7) 1 0 1 0 2 2.89 8.18 1 0 1 0 | r + 1 | =(2.3) 1 −1 1 0 1 0 .18 5.29 (.3) 2 1 5.38 0 1 0 (b) n=0 n=1 r 1 1.3 = n=2 r 2 −.7 = 2.89 n=3 r 3 1.1 = (.1) 2 1 1 0 1 0 .09 1 0 1 0 1 0 1 0 2 (2.1) (.1) 1 0 1 0 2 2.98 .27 7.3 1 0 1 0 −1 1 0 1 0 5.29 1 0 1 0 .18 (2.1) 2 1 4.59 0 1 0 (c) n=0 n=1 r 1 1.3 = n=2 r 2 −.7 = 2.89 n=3 r 3 1.1 = 1 0 1 0 2 (.8) (−1.2) 1 0 1 0 2 n=4 r 4 −.2 = (−1.2) 2 1 1 0 1 0 .09 1 0 1 0 1 0 1 0 .27 1.71 6.03 1 0 1 0 −1 1 0 1 0 5.29 1 0 1 0 .18 1 0 1 0 .91 4.59 (.8) 2 1 5.23 0 1 0 (d) n=0 n=1 r 1 1.3 = n=2 r 2 −.7 = 2.89 n=3 r 3 1.1 = 1 0 1 0 .27 n=4 r 4 −.2 = 1.71 1 1 0 1 0 .09 1 0 1 0 1 0 1 0 1 0 1 0 −1 1 0 1 0 5.29 1 0 1 0 .18 1 0 1 0 4.59 1 0 1 0 .91 1 0 1 0 Figure 2: Illustration of the Viterbi algorithm for coherent CPSK MLSE. Kevin Buckley - 2007 8 Example 5 - PRS: Consider the PRS MLSE problem discussed last week in Subsection 2.3.3. Figure 3 illustrates the application of the Viterbi algori...
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## This document was uploaded on 10/12/2009.

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