Reed-Muller Codes Problem Set 4 - University of Illinois, Fall 2005, Assignments of Electrical and Electronics Engineering

Download Reed-Muller Codes Problem Set 4 - University of Illinois, Fall 2005 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! University of Illinois Fall 2005 ECE 556/CS 577/MATH 579: Problem Set 4 Due: September 22, 8:30 a.m. Reading: Blahut, Algebraic Codes for Data Transmission, Chapters 2 and 4. This Problem Set contains four problems 1. Consider the 2nd-order Reed-Muller code of length 16, which is also an extended Hamming code and thus has parity-check matrix H =  1111 1111 1111 1111 0101 0101 0101 0101 0011 0011 0011 0011 0000 1111 0000 1111 0000 0000 1111 1111  . Note that the rows of H are the truth tables of the polynomials 1, x1, x2, x3, x4 re- spectively and the grouping into sets of four columns is strictly for human visual convenience. (a) Decode the received word 0001 0110 0110 0011 using the Reed decoding algorithm to obtain the data bits a1,2, a1,3.a1,4.a2,3, a2,4, a3,4, a1, a2, a3, a4 and a0. (b) Decode the same received word using the parity-check matrix to obtain the most likely transmitted codeword. (c) Now consider the use of an r-th order Reed-Muller code on an erasures-only channel. A code of minimum distance d can be used to correct up to d − 1 erasures, and in Problem 3 of Problem Set 3, we saw how, by setting the erasures to 0s and to 1s, and using two decodings, the erasures can be corrected. Describe how the Reed decoding algorithm can be modified to produce the data bits (coefficients of the degree-r polynomial) with only one decoding, provided that no more than 2m−r − 1 erasures have occurred. Then use this modified algorithm to decode ???1 0110 0110 0011 for the 2nd-order Reed-Muller code of length 16. 2. Let A(m)(z) denote the weight enumerator polynomial of the first-order Reed-Muller code of length 2m. (a) What is A(1)(z)? (b) Use the Plotkin representation of Reed-Muller codes to prove that A(m)(z) = A(m−1)(z2) + 2mz2 m−1 . (c) Prove that A(m)(z) = 1 + (2m+1 − 2)z2m−1 + z2m .