solution6

solution6 - UNIVERSITY OF CALIFORNIA SAN DIEGO Electrical...

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UNIVERSITY OF CALIFORNIA, SAN DIEGO Electrical & Computer Engineering Department ECE 259B - Winter Quarter 2011 Probabilistic Coding Solutions to Problem Set #6 1. (a) As discussed in class, there are no stopping sets of weight less than 3 in the Tanner graph G corresponding to the “Venn diagram” parity check matrix H : H = 1110100 1011010 1101001 There are 7 stopping sets of size 3 corresponding to the weight-3 codewords with support sets: { 4 , 6 , 7 }{ 3 , 5 , 6 2 , 5 , 7 2 , 3 , 4 1 , 4 , 5 1 , 3 , 7 1 , 2 , 6 } . There are 3 other weight-3 stopping sets: { 1 , 2 , 3 1 , 2 , 4 1 , 3 , 4 } . We denote this set of 10 weight-3 stopping sets by S . (b) If we append to H the redundant parity-check equation obtained by adding the 3 rows of the parity-check matrix H , the resulting parity-check matrix H 1 is: H 1 = 1000111 The corresponding Tanner graph G 1 consists of G with an additional check node and its connecting edges:
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c 0 c 1 c 2 c 3 c 4 c 5 c 6 If we denote the set of weight-3 stopping sets in G 1 by S 1 , then necessarily S 1 S ,s ince adding redundant checks cannot create new stopping sets. Also S 1 must contain the 7 stopping sets corresponding to the supports of the 7 weight-3 codewords. By inspection, we see that the new check node has exactly one edge connecting to the other 3 stopping sets in S , namely the edge connecting to variable node 1. Therefore, these sets are not stopping sets in G 1 . 2. (a) The rate- 1 5 repetition code C has 2 length-5 codewords, namely the all-zeros and all-ones codewords ,[00000]and[11111] .
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This note was uploaded on 04/02/2011 for the course ECE 264 taught by Professor Song during the Spring '11 term at UCSB.

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solution6 - UNIVERSITY OF CALIFORNIA SAN DIEGO Electrical...

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