B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

14.4 Cyclic Codes 815
Regardless of the data values {di}, d(x) g(x) still has degree n - 1 or less while being a
multiple of g(x). Hence, a codeword is formed by using Eq. (14.16). There are a total of 2 k
such distinct polynomials (codewords) of the data polynomial d (x), corresponding to 2 k data
vectors. Thus, we have a linear (n, k) code generated by Eq. (14.14). To prove that this code is
cyclic, let
c(x) = CJX n -l + c 2 x 11 - 2 + · · · + C n
be a code polynomial in this code [Eq. (14.16)1. Then,
xc(x) = CJX n + c 2 x n - l + · · · + C n X
= c1 (x n + 1 ) + (c2x"- 1 + c3x n - 2 + · · · + C n X + c1 )
= c1 (x n + 1) + c < l l
(x)
Because xc(x) is xd (x)g(x), and g(x) is a factor of x n + 1,c O> (x) must also be a multiple
of g(x) and can also be expressed as d(x)g(x) for some data vector d. Therefore, c ( l l (x)
is also a code polynomial. Continuing this way, we see that c C 2>
(x), c < 3> (x), ... are all code
polynomials generated by Eq. (14. 16). Thus, the linear (n, k) code generated by d(x)g(x) is
indeed cyclic.
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Exam pie 1 4. 3 Find a generator polynomial g(x) for a (7, 4) cyclic code, and find code vectors for the following
data vectors: 1010, 1111, 0001, and 1000.
;,,
In this case n = 7 and n - k = 3, and
x 7 + l = (x + l)(x 3 + x + l ) (x 3 + x 2 + 1)
For a (7, 4) code, the generator polynomial must be of the order n - k = 3. In this case,
there are two possible choices for g(x) : x 3 + x + 1 or x 3 + x 2 + I . Let us pick the latter,
that is,
g(x) = x 3 + x 2 + 1
as a possible generator polynomial. For d = [l 0 1 0],
and the code polynomial is
Hence,
d(x) = x 3 +x
c(x) = d(x)g(x)
= (x 3 +x)(x 3 +x 2 + 1)
= x 6 + x 5 + x 4 + X
C = 1110010