Question

7) a) Le l be the ternary \left [ 8,3 \right ] narrow-senseBCH code of designed distance \delta =5, which has defining set T=\left \{ 1,2,3,4,5,6 \right \}.Use the primitive root 8th root of unity you chose in 4a) to  avoid recomputing the the table of powers. If 

g(x)=x^{5}-x^{4}+x^{3}+x^{2}-1

 

Image ignouassignments-ignouacademy-com--p-ternarynbspnbspnarrowsensenbspcode-63666

                   Figure 1: Encoder for convolutional code.

is the generator polynomial of  Image ignouassignments-ignouacademy-com--p-lnbspbe-35400 and

y(x)=x^{7}-x^{6}-x^{4}-x^{3}

is the received word, find the transmitted codeword.

10 Mar 2024
Answer :
Word Count : 1405
Let’s break this problem into steps. --- ## 1. Understand the problem This is an error-correction coding problem where: - We have a ternary narrow-sense BCH code (over \( GF(3) \)) with a certain designed distance. - Defining set \( T \) is given, probably in the original full problem statement but missing here. - They mention a primitive 8th root of unity chosen in problem 4a (not provided here), so we need to recall or reconstruct that. - We have a generator polynomial \( g(x) \) for a convolutional code ? — wait, reading carefully, it says: “is the generator polynomial of % and … is the received word, find the transmitted codeword.” The convolutional code diagram is just Figure 1, maybe not directly relevant — maybe the problem is actually about a cyclic block code not convolutional. The received word is given in polynomial form but cut off in the text you provided. --- My guess from standard problems: We have a primitive 8th root of unity \( \alpha \) in \( GF(3^m) \) for \( m \) such that \( 8 \mid 3^m - 1 \). For \( GF(3) \), smallest \( m \) such that \( 8 \mid 3^m - 1 \) is \( m=2 \)? Check: \( 3^2 - 1 = 8 \), yes. So \( GF(9) \) contains primitive 8th roots of unity. Let’s say \( \alpha \) is a primitive 8th root of unity in \( GF(9) \) with primitive polynomial \( x^2 + x + 2 \) or \( x^2 + 1 \) over \( GF(3) \) — but \( x^2 + 1 \) is irreducible over \( GF(3) \) since \( -1 \equiv 2 \), no, check: \( x^2 + 1 \) mod 3: \( 0^2+1=1\), \( 1^2+1=2\), \( 2^2+1=5\equiv 2\), never 0, so irreducible, so \( GF(9) \cong GF(3)[x]/(x^2+1) \). Let \( \beta \) be a root: \( \beta^2 + 1 = 0 \), so \( \beta^2 = 2 \). Now \( \beta^4 = 4 \equiv 1 \) mod 3? Wait \( 2^2 = 4 \equiv 1 \) mod 3, yes, ___ ___ ______ ______ _________ _____ __________ ________.
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