This implies the error has magnitudeejand occurred in the j-th position.Becausec(x)=g(x)i(x)for some polynomiali(x)it is clear thatS1=r(2) =g(2)i(2)+e(2)=e(2)=ej2jS2=r(4) =g(4)i(4)+e(4)=e(4)=ej22j.Thus the location of the error can be determined fromj=log2(r(4)r(2))=log2(2j).The magnitude of the error can be determined byej=S1/X1=S21/S2.Suppose we receiver(x)=2x3+4x2+2x1+3.Thenr(2)=4 andr(4)=3. The error locator isX1=2j=r(4)r(2)=21thusj=1. The error magnitude isr(2)/X1=4/2=4∗3=2. Thus the error polynomial ise(x)=2x1. Thusc(x)=r(x)−e(x)=2x3+4x2+2x1+3−2x1=2x3+4x2+0x1+3which is indeed a codeword.November 13, 2019322
CHAPTER 8. LINEAR BLOCK CODES22.214.171.124Example: (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 elementsof the field.l2ll2l015101269247738834596Second note thatx10−1=(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 isg(x)=(x−2)(x−22)(x−23)(x−24).A codewordc(x)is a multiple of the generator polynomialc(x)=g(x)∗m(x). A received polynomialisr(x)=c(x)+e(x). We evaluate the received polynomial at the zeros of the generator polynomial.Suppose only two errors occurred at locationsiandl. LetY1andY2be the errors that were made. LetX1=2iandX2=2lbe indicators of the locations of the errors.S1=r(21)=Y12i+Y22l=Y1X1+Y2X2S2=r(22)=Y122i+Y222l=Y1X21+Y2X22S3=r(23)=Y123i+Y223l=Y1X31+Y2X32S4=r(24)=Y124i+Y224l=Y1X41+Y2X42The goal is to solve these four equations in four unknowns to determine the error magnitudes and theerror locations. Note that if we determineX1andX2then we can easily solve forY1andY2because theequations become linear. As such,bracketleftbiggX1X2X21X22bracketrightbiggbracketleftbiggY1Y2bracketrightbigg=bracketleftbiggS1S2bracketrightbiggbracketleftbiggY1Y2bracketrightbigg=bracketleftbiggX1X2X21X22bracketrightbigg−1bracketleftbiggS1S2bracketrightbiggbracketleftbiggY1Y2bracketrightbigg= (X1X22−X21X2)−1bracketleftbiggX22−X2−X21X1bracketrightbiggbracketleftbiggS1S2bracketrightbigg.Consider the polynomialΛ(x)given byΛ(x)=(1−xX1)(1−xX2)=1+Λ1x+Λ2x2ClearlyΛ(X−11)=Λ(X−12)=0. So1+Λ1X−11+Λ2X−21=0Y1Xj+21(1+Λ1X−11+Λ2X−21) =0Y1Xj+21+Λ1Y1Xj+11+Λ2Y1Xj1=0.November 13, 2019323
8.14CHAPTER 8. LINEAR BLOCK CODESSimilarly1+Λ1X−12+Λ2X−22=0Y2Xj+22(1+Λ1X−12+Λ2X−22) =0Y2Xj+22+Λ1Y1Xj+12+Λ2Y2Xj2=0.