This implies the error has magnitude e j and occurred

This implies the error has magnitude e
j
and occurred in the j-th position.
Because c ( x )= g ( x ) i ( x ) for some polynomial i ( x ) it is clear that
S
1
= r ( 2 ) = g ( 2 ) i ( 2 )+ e ( 2 )= e ( 2 )= e
j
2
j
S
2
= r ( 4 ) = g ( 4 ) i ( 4 )+ e ( 4 )= e ( 4 )= e
j
2
2 j
.
Thus the location of the error can be determined from
j = log
2
(
r ( 4 )
r ( 2 )
)= log
2
( 2
j
) .
The magnitude of the error can be determined by
e
j
= S
1
/ X
1
= S
2
1
/ S
2
.
Suppose we receive
r ( x )= 2 x
3
+ 4 x
2
+ 2 x
1
+ 3 .
Then r ( 2 )= 4 and r ( 4 )= 3. The error locator is X
1
= 2
j
=
r ( 4 )
r ( 2 )
= 2
1
thus j = 1. The error magnitude is r ( 2 ) / X
1
=
4 / 2 = 4 ∗ 3 = 2. Thus the error polynomial is e ( x )= 2 x
1
. Thus
c ( x )= r ( x ) − e ( x )= 2 x
3
+ 4 x
2
+ 2 x
1
+ 3 − 2 x
1
= 2 x
3
+ 4 x
2
+ 0 x
1
+ 3
which is indeed a codeword.
November 13, 2019 322

CHAPTER 8. LINEAR BLOCK CODES 8.14
8.14.4 Example: (10,6) Reed-Solomon code over GF(11)
The (10,6) code over GF(11) has distance 5 which means it can correct two errors. Let’s see how this is done.
First note that 2 is a primitive element in this field. That is, the powers of 2 generate all the nonzero elements
of the field.
l 2
l
l 2
l
0 1 5 10
1 2 6 9
2 4 7 7
3 8 8 3
4 5 9 6
Second note that
x
10
− 1 =( x − 1 )( x − 2 )( x − 3 )( x − 4 )( x − 5 )( x − 6 )( x − 7 )( x − 8 )( x − 9 )( x − 10 )
so that any combination of these factors will yield a cyclic code.
Consider the generator polynomial for this code is
g ( x )=( x − 2 )( x − 2
2
)( x − 2
3
)( x − 2
4
) .
A codeword c ( x ) is a multiple of the generator polynomial c ( x )= g ( x ) ∗ m ( x ) . A received polynomial
is r ( x )= c ( x )+ e ( x ) . We evaluate the received polynomial at the zeros of the generator polynomial.
Suppose only two errors occurred at locations i and l . Let Y
1
and Y
2
be the errors that were made. Let
X
1
= 2
i
and X
2
= 2
l
be indicators of the locations of the errors.
S
1
= r ( 2
1
)= Y
1
2
i
+ Y
2
2
l
= Y
1
X
1
+ Y
2
X
2
S
2
= r ( 2
2
)= Y
1
2
2 i
+ Y
2
2
2 l
= Y
1
X
2
1
+ Y
2
X
2
2
S
3
= r ( 2
3
)= Y
1
2
3 i
+ Y
2
2
3 l
= Y
1
X
3
1
+ Y
2
X
3
2
S
4
= r ( 2
4
)= Y
1
2
4 i
+ Y
2
2
4 l
= Y
1
X
4
1
+ Y
2
X
4
2
The goal is to solve these four equations in four unknowns to determine the error magnitudes and the
error locations. Note that if we determine X
1
and X
2
then we can easily solve for Y
1
and Y
2
because the
equations become linear. As such,
bracketleftbigg
X
1
X
2
X
2
1
X
2
2
bracketrightbiggbracketleftbigg
Y
1
Y
2
bracketrightbigg
=
bracketleftbigg
S
1
S
2
bracketrightbigg
bracketleftbigg
Y
1
Y
2
bracketrightbigg
=
bracketleftbigg
X
1
X
2
X
2
1
X
2
2
bracketrightbigg
− 1
bracketleftbigg
S
1
S
2
bracketrightbigg
bracketleftbigg
Y
1
Y
2
bracketrightbigg
= ( X
1
X
2
2
− X
2
1
X
2
)
− 1
bracketleftbigg
X
2
2
− X
2
− X
2
1
X
1
bracketrightbiggbracketleftbigg
S
1
S
2
bracketrightbigg
.
Consider the polynomial Λ ( x ) given by
Λ ( x )=( 1 − xX
1
)( 1 − xX
2
)= 1 + Λ
1
x + Λ
2
x
2
Clearly Λ ( X
− 1
1
)= Λ ( X
− 1
2
)= 0. So
1 + Λ
1
X
− 1
1
+ Λ
2
X
− 2
1
= 0
Y
1
X
j + 2
1
( 1 + Λ
1
X
− 1
1
+ Λ
2
X
− 2
1
) = 0
Y
1
X
j + 2
1
+ Λ
1
Y
1
X
j + 1
1
+ Λ
2
Y
1
X
j
1
= 0 .
November 13, 2019 323

8.14 CHAPTER 8. LINEAR BLOCK CODES
Similarly
1 + Λ
1
X
− 1
2
+ Λ
2
X
− 2
2
= 0
Y
2
X
j + 2
2
( 1 + Λ
1
X
− 1
2
+ Λ
2
X
− 2
2
) = 0
Y
2
X
j + 2
2
+ Λ
1
Y
1
X
j + 1
2
+ Λ
2
Y
2
X
j
2
= 0 .