U. Glaeser

We leave as an exercise proving that Algorithm 2 produces indeed a codeword in C.
Example 7 Consider the [7,4] cyclic code over GF(2) of Example 6. If we want to encode systematically
the information vector = (1, 0, 0, 1) (or u(x) = 1 + x 3 ), we have to obtain first the residue of dividing
x 3 u(x) = x 3 + x 6 by g(x). This residue is r(x) = x + x 2 . Hence, the output of the encoder is c(x) = u(x) −
x 4 r(x) = 1 + x 3 + x 5 + x 6 u
. In vector form, this gives c = (1 0 0 1 0 1 1). �
Reed Solomon Codes
Throughout this subsection, the codes considered are over the field GF(2 ν ). Let α be a primitive element
in GF(2 ν ), i.e., = 1, α i ≠ 1 for i � 0 mod 2 ν − 1. A Reed–Solomon (RS) code of length n = 2 ν α − 1
and dimension k is the cyclic code generated by
2ν −1
Each α i is a root of unity, x − α i divides x n − 1, hence, g divides x n − 1 and the code is cyclic.
An equivalent way of describing a RS code is as the set of polynomials over GF(2 ν ) of degree ≤ n − 1
with roots α, α 2 ,…,α n−k , i.e., F is in the code if and only if deg(F) ≤ n − 1 and F(α) = F(α 2 ) = … =
F(α n−k ) = 0.
This property allows us to find a parity check matrix for a RS code. Say that F(x) = F 0 + F 1x +…+
F n−1x n−1 is in the code. Let 1 ≤ i ≤ n − k, then
© 2002 by CRC Press LLC
g( x)
= x – α
F α i
( ) x α 2
( – )…xα n−k−1
( – ) x α n−k
–
i( n−1)
( )
( ) F0 F1α i = + + … + Fn−1α = 0
(34.58)
In other words, Eq. (34.58) tells us that codeword (F 0, F 1,…, F n−1) is orthogonal to the vectors (1, α i ,
α 2i ,…, α i(n−1) ), 1 ≤ i ≤ n − k. Hence, these vectors are the rows of a parity check matrix for the RS code.
A parity check matrix of an [n, k] RS code over GF(2 ν ) is then
H
1 α α 2 K α n−1
1 α 2
α 4 2( n−1 )
K α
M M M O M
1 α n−k α n−k ( )2 K α n−k
⎛ ⎞
⎜ ⎟
⎜ ⎟
= ⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ ( ) ( n−1 ) ⎟
⎝ ⎠
(34.59)
In order to show that H is in fact a parity check matrix, we need to prove that the rows of H are linearly
independent. The next lemma provides an even stronger result.
Lemma 6 Any set of n − k columns in matrix H defined by Eq. (34.59) is linearly independent.
Proof: Take a set 0 ≤ i1 < i2 < … < in−k ≤ n − 1 of columns of H. Denote by αj, 1 ≤ j ≤ n − k. Columns
i1, i2, … , in−k are linearly independent if and only if their determinant is nonzero, i.e., if and only if
α 1 α 2 K α n−k
α i j
( α1) det
2
( α2) 2
K ( αn−k ) 2
M M O M
( α1) n−k
( α2) n−k
K ( αn−k ) n−k
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟ ≠
0
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ ⎠
(34.60)