Example 3.1
Consider a
code with block length n = 7. The polynomial x 7 + 1
has the following factors:
x 7 +1 = (x +1)(x 3
+ x 2 + 1)(x 3 + x + 1)
To generate
a (7, 4) cyclic code, we may take as a generator polynomial one of the
following two polynomials:
g1(x) = x 3 + x 2+
1
g2(x) = x 3 + x+ 1
The code words
in the (7, 4) code generated by g1(x) = x 3 + x 2
+ 1
are given in
Table 3.1
|
Information
bits
|
Code words
|
|
x
3 x 2 x 1 x 0
|
x 6 x 5 x 4 x 3 x 2
x 1 x 0
|
|
0
0 0 0
0
0 0 1
0
0 1 0
0
0 1 1
0
1 0 0
0
1 0 1
0
1 1 0
0
1 1 1
1
0 0
0
1
0 0 1
1
0 1 0
1
0 1 1
1
1 0 0
1
1 0
1
1
1 1 0
1 1 1
1
|
0 0 0 0
0 0 0
0 0
0 1
1 0 1
0 0 1 1
0 1 0
0 0 1 0
1 1 1
0 1
1 0
1 0 0
0
1 1 1
0 0 1
0 1
0 1
1 1 0
0 1
0 0
0 1 1
1 1
0 1
0 0 0
1
1 0
0 1
0 1
1
1 1
0 0
1 0
1
1 1 1
1 1 1
1 0
1
1 1
0 0
1
0 1 0
0 0 1
1
0 0 0
1 1 0
1
0 0 1
0 1 1
|
Table 3.1 (7, 4) Cyclic Code Generated By
g1(x) = x 3 +x 2 +1
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