This implies the error has magnitude e j and occurred in the j th position

This implies the error has magnitude e j and occurred

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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
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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
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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 .
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