纠错编码习题解答第一章1.1 Solution: p=0.05(1)The correct decoding Pc is Pc= P0 =C40p0(1-p)4=0.8145(2)The decoding error Pe is Pe = P2+P4 = C42p2(1-p)2 + C44p4(1-p)0 = 0.0135(3)The decoding failure Pf is Pf= C41p(1-p)3 + C43p3(1-p) = 0.17201.2 Solution: Because the success rate does not fall below 99%,then the decoding failure Pf 1% .And p1, Pf = P1 = n*0.001*0.999n-1 0.01 So n=10 .then the maximum blocklength n such that the success ratedoes not fall below 99% is 10.1.3 Solution: p=0.01 Pf = P2 = C42p2(1-p)2 = C42 * 0.012 * 0.992 = 0.000588 So the decoding failure rate is 0.000588.1.4 Solution: (a)Error: There is one error (b)Correct (c)Failure (d)Error: There is two error1.5 Solution: S1 = v1+v2+v3+v4+v6+v8+v9+v12 S2 = v2+v3+v4+v5+v7+v9+v10+v13 S3 = v3+v4+v5+v6+v8+v10+v11+v14 S4 = v1+v2+v3+v5+v7+v8+v11+v15Error pattern e Error Syndrome s(e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,e11,e12,e13,e14,e15) (s1,s2,s3,s4)0000 0000 0000 000 0000 0000 0000 0000 001 0001 0000 0000 0000 010 0010 0000 0000 0000 100 0100 0000 0000 0001 000 1000 0000 0000 0010 000 0011 0000 0000 0100 000 01100000 0000 1000 000 1100 0000 0001 0000 000 1011 0000 0010 0000 000 0101 0000 0100 0000 000 1010 0000 1000 0000 000 0111 0001 0000 0000 000 11100010 0000 0000 000 1111 0100 0000 0000 000 1101 1000 0000 0000 000 10011.6 Solution (1)s=(0000) e = 0000 0000 0000 000 c= e+v1 = (000000000000000)+(100010011001001)= (100010011001001) (2)s=(1011) e = 0000 0001 0000 000 c= e+v2 = (000000010000000)+(001001110100110)= (001001100100110)1.7 Solution (1) v=(1011 110) s=(110)e = (001 0000) c=(1001 110) (2) v=(1100 110) s=(100) e = (0000 100) c=(1100 010) (3) v=(0001 011) s=(000) e=(0000 000) c=(0001 011)第二章2.1 Solution:cisins0000000000000111001011010010100101001110011101110011001101001011001101011010111101111110From the table ,we can know every Ci has only Cj .2.2 Solution 1 1 1 1 1 1 1 r1-r2 1 0 0 0 1 0 1G2= 0 1 1 1 0 1 0 r2-r3 0 1 0 0 1 1 1 = G1 0 0 1 1 1 0 1 0 0 1 0 1 1 0 0 0 0 1 0 1 1 r3-r4 0 0 0 1 0 1 1G1 is systematic form.And every liner coder is equivalent to a systematic linear codeSo the (7,4) linear codes generated by G1 and G2 equivalent. 2.3 SolutionInformation wordCodeword000000000001001110010010101011011011100100011101101101110110110111111000 (a) C0 = (000) G = (000000)C1 = (001) G = (001110)C2 = (010) G = (010101)C3 = (011) G = (011011)C4 = (100) G = (100011)C5 = (101) G = (101101)C6 = (110) G = (110110)C7 = (111) G = (111000)(b)If p=0 thenInformation wordCodeword000000000001001001010010010011011011100100100101101101110110110111111111If p=1 thenInformation wordCodeword0000001110010011100100101010110111001001000111011010101101100011111110002.4 SolutionBecause the (4,3) even-parity code is a linear code ,The minimum distance d(Ci,Cj)= Wmin = 2The error detection limit is L=2-1=1The error correction is t=(2-1)/2=0.5.Namely 0.2.5 Solution 1 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 0 1 0 0 r4-r3-r2-r1 0 1 1 1 0 1 0 0H= 1 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 1So the systematic forms is 1 1 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 0 1 Because H = PT | In-k and G=Ik | PThen G = 1 0 0 0 1 0 1 1 0 1 0 0 1 1 1 0 0 0 1 0 1 1 0 1 0 0 0 1 0 1 1 12.6 Solution 1 1 1 0 1 0 0 0 1 0 1 1 0 1 1 1 0 1 0 0 HT = 1 1 1 1H= 1 1 0 1 0 0 1 0 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1So eivi=c+eisi=viHTeici=vi+eiconclusion000001000110011101010000010001100011Correct codeword01010000001100111000absentDecoding failure000110100111100111010010000001011001Incorrect codeword第三章3.1 solution Because x3+1 x4+x+1 x7+x3+1