**1 Answer**

written 6.5 years ago by | • modified 6.5 years ago |

The **generator matrix** will be a kn matrix with rank k. Since the choice of the basic vectors is not
unique, the generator matrix is not unique for a given linear code.
The generator matrix converts the vector of length k to a vector of length n. Let the input vector be
represented by i. The coded symbol will be given by

C = iG

Where c is called the codeword and I is called the information word.

The generator matrix provides a concise and efficient way of representing a linear block code. The
nÃ— k matrix can generate q^{k} codewords.

**For length 7 binary cyclic codes we have the factorization
into irreducible polynomials:**

x^{7} âˆ’ 1 = (x âˆ’ 1)(x^{3} + x + 1)(x^{3} + x^{2 }+ 1).

Since we are looking at binary codes, all the minus signs can be replaced by plus signs:

x^{7 }+ 1 = (x + 1)(x^{3} + x + 1)(x^{3} + x^{2} + 1).

As there are 3 irreducible factors, there are 2^{3} = 8 cyclic codes.The 8 generator polynomials are:

(i) 1 = 1

(ii) x + 1 = x + 1

(iii) x^{3}+ x + 1 = x^{3} + x + 1

(iv) x^{3 }+ x^{2} + 1 = x^{3} + x^{2} + 1

(v) (x + 1)(x^{3} + x + 1) = x^{4 }+ x^{3 }+ x^{2} + 1

(vi) (x + 1)(x^{3} + x^{2} + 1) = x^{4 }+ x^{2} + x + 1

(vii) (x^{3} + x + 1)(x^{3 }+ x^{2} + 1) = x^{6 }+ x^{5} + x^{4 }+ x^{3} + x^{2} + x + 1

(viii) (x + 1)(x^{3 }+ x + 1)(x^{3 }+ x^{2 }+ 1) = x^{7 }+ 1

Here in (i) the polynomial 1 generates all F^{7}2. In (ii) we find the parity
check code and in (vii) the repetition code. As mentioned before, in (viii)
we view the 0-code as being generated by x^{7} + 1.

The polynomials of (iii) and (iv) have degree 3 and so generate [7, 4] codes, which we shall later see are Hamming codes. The [7,3] codes of (v) and (vi) are the duals of the Hamming codes.

For (7, 4) cyclic code, the polynomial 1+x^{7} can be factorized as 1+x^{7}=(1+x)(1+x+x^{3})(1+x^{2}+x^{3}),
G(x) =1+x+x^{3}, the minimum distance is 3 of single-error.

Dividing x^{3}, x^{4}, x^{5}& x^{6} by g(x), we have

x^{3} =g(x) +1+x

x^{4} =xg(x)+x+x^{2}

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

x^{6}=(1+x+x^{3})g(x)+1+x^{2}

Rearranging the above, we have

V_{0}(x)=1+x+x^{3}

V_{1}(x)=x+x^{2}+x^{4}

V_{2}(x)=1+x+x^{2}+x^{5}

V_{3}(x)=1+x^{2}+x^{6}

Considering above equation in matrix form, we obtain the generator matrix of order of (4*7) in systematic form in cyclic code.

1 | 1 | 0 | 1 | 0 | 0 | 0 |
---|---|---|---|---|---|---|

0 | 1 | 1 | 0 | 1 | 0 | 0 |

0 | 0 | 1 | 1 | 0 | 1 | 0 |

0 | 0 | 0 | 1 | 1 | 0 | 1 |