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Construct a (5, 2) code that corrects single errors.

Short Answer

Expert verified

The 5,2 codes that correct single errors are:00000,01010,10111,11101

Step by step solution

01

Theorem 16.6 and Corollary 16.4

Theorem 16.6

If G is a k×n standard generator matrix, then uG|uBk is a systematic n,k code

Corollary 16.4

A linear code detects 2t errors and corrects terrors if and only if the Hamming weight of every nonzero codeword is at least 2t+1.

02

Constructing a (5, 2) code that corrects single errors

According to Theorem 16.6 to construct a 5,2code, the dimension of the generator matrix should be 2×5

Let the generator matrix,

G=1011101010

The message words of the code 5,2are:

00,01,10,11

Therefore, to get the code words we must multiply them with the generator

001011101010=00000011011101010=01010101011101010=10111111011101010=11101

Therefore, the code words for the given matrix generator are:

00000,01010,10111,11101

Now we see the humming weight of nonzero codewords is at least 3=2×1+1

Therefore, from corollary 16.4, with t=1,code generated by the matrix G corrects single error.

Hence 5,2 code that corrects single errors are: 00000,01010,10111,11101

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Most popular questions from this chapter

Question:- If fxZ2xandα is an element in some extension field of ,Z2 provek1,fα2k=fαk2 [Hint Lemma 11.24 ]

Question 15: Let C be the BCH code of Examples 1 and 2, with codewords written as polynomials of degree . Suppose the codeword c (x)transmitted with errors in the coefficients of xiand xj and role="math" localid="1659179965727" r(x) is received. Then D(x)=(x+αi)(x+αj)K[x], whose roots are role="math" localid="1659179995612" aiand role="math" localid="1659179986054" aj , is the error-locator polynomial. Express the coeffecients ofD(x)in terms of r(α),r(α2),r(α3)as follows.

  1. Show that r(x)-c(x)=xi+xj.
  2. Show that r(αk)=αki+αkjfork=1,2,3 .[ See the bold face statement on page 495.]
  3. Show that D(x)=x2+(αi+αj)x+αi+j=x2+r(α)x+αi+j .
  4. Show that αi+j=r(α2)+r(α3)r(α) . [Hint: Show that r(α)3=(αi+αj)3=α3i+α3j+αi+j(αi+αj)=r(α3)+r(α)αi+jand solve for ,αi+jnote thatr(α)2=r(α2) .]

Question 11: Let K=2(α)be a finite field of order2r , whose multiplicative group is generated byα . For each i, let mixbe the polynomial of αiover2 . Ifn=2r-1 , prove that each mixdivides data-custom-editor="chemistry" xn-1. [Hint:αn=1 (Why?) Use theorem 11.6]

Prove that an element ofB6is a codeword in the6,5 parity-check code(Example 2) if the sum of its digits is0. [Hint: Compare the sum of the firstfive digits with the sixth digit.]

Suppose only three messages are needed (for instance, "go," "slow down," "stop"). Find the smallest possible nso that these messages may be transmitted in an (n,k)code that corrects single errors.

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