20072ee132A_1_ca5_sol

20072ee132A_1_ca5_sol - EE132A, Spring 2007 Prof. John...

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EE132A, Spring 2007 Communication Systems Prof. John Villasenor Handout# 25 TA: Choo Chin (Jeffrey) Tan Computer Assignment 5 Solutions Part 1 The following table shows the encoded outputs for each one of the inputs. Input Output 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 1 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 1 0 1 1 0 1 1 0 0 1 1 0 1 0 1 1 1 1 1 1 1 1 0 1 1 0 0 1 1 1 Part 4 (a) When a single error is added to the sequences, all of them are decoded correctly. (b) When two errors are added to the sequences, all of them are decoded correctly. Note that we must traceback from the 00 state (that is, start and end the Viterbi algorithm at the 00 state) to be able to recover from two errors. (c) When three errors are added to the sequences, not all of them are decoded correctly. (d) From the results of parts a, b, and c, we can conclude that the code can reliably correct up to a maximum of two errors. For this code, the minimum (Hamming) distance is 5. We can see this by looking at the table from part 1, and noting that all of the possible outputs differ in at least 5 bits. Since the decoder selects the one that
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20072ee132A_1_ca5_sol - EE132A, Spring 2007 Prof. John...

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