Chapter 16

Chapter 16 - , .. Chapter 16 M 2...

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Unformatted text preview: , .. Chapter 16 M 2 >’:<33)=(%3)+<P)+(%3)+(%3) '=0 2048; 1+23+23x11+23x77=2048 16.1-2 (a) There are ) ways in which j positions can be chosen fiom n . But for a ternery code, a digit can be mistaken for two other digits. Hence the number of possible errors in j places is (7)04)" or 3" 23" i(;)21 ->3""‘ 2 ' (;)2i j=0 jsO (b) (11,6) code fort = 2 35 2(5‘)+({‘)2+(5‘)22 = 1+22+220=243 This is satisfied exactly. 16.1-3 For (18,7) code to correct up to 3 errors 2 in“) or 2(s')+(:*)+(a=)+(a°) 1-0 1 I I =1+li+ “- + 13' =1+18+153+816=988 17! 2! 16! 3115! 2”=2048 Hence 3 raw) .=o Clearly, there exists a possibility of 3 error correcting (18,7) code. Since the Hamming bound is oversatisfied, this code could correct some 4 error patterns in addition to all patterns with up to 3 errors. [6.2-1 GHT=[Ik HUM] =P$P =0 16.2-2 c = r16 wheredisasingle digit (Oor 1). Ford==0 c=o[iii]=[ooo] Ford=1 c=1[1 1 l]=[l l 1] 140 16.2-3 c=dG wheredisasingie digit (0 or 1). Ford=0 c=0[illil]=[00000] Ford=1 c=1[1 11 1 1]=[11111] Hence in this code a digit repeats 5 times. Such a code can correct up to two errors USing majority rule for detection. 16.2.4 0 is transmitted by [0 o o] and I is transmitted by [1 1 1] (a) This is clearly a systematic code with aspifl 16.2-5 (a) 100- 0] l 010- 01 l G= . . . . . . .. P= Notethatm=l ooo. . 1 lg P (b) tutu—un‘cQQQ “OOflGH—G (c) This is a parity check code. if a single error occurs anywhere in the code word, the parity is violated. Therefore this code can detect a single error. (d) Equation (16.93) in the text shows that cur = 0. Now r=c0e and rHT =(cee)HT =cHTeeHT "HT lfthere is no error e==0 and rHT = cur = 0 Also HT =[f Butsince m=l,I,,,=[l] M and 141 c has a single 1 element with all other elements being 0. If there is a single error in the received word r, Hence r]! T = all T = 1 (for single error) From this code we see that the distance between any two code words is at least 3. Renee dmin = 3. 16.2-7 Observe that dmin = 3 16.2-8 II T is a 15 x 4 matrix with all distinct rows. One possible HT is: 1111 1110 1101 1100 1011 1010 1001 HT: 0011 =[P] 0111 0110 0101 1000 0100 0010 0001 142 100000000001111 010000000001110 001000000001101 000100000001100 000010000001011 G=[lkP]= 000001000001010 000000100001001 000000010000011 000000001000111 000000000100110 000000000010101 Ford=10111010101 c=dG=[lOlllOlOlOl]G=lOlllOlOlOlll10 1&}9 a) 111 1 100111 :0? G= 010110 & HT=100 00 101 7,717: 010 001 (b) (e) The minimum distance between any two code words is 3. Hence, this is a single error correcting code. Since there are 6 single errors and 7 syndromes, we can correct all single errors and one double error. (d) s=eHT S l l 0 0 l 0 1 I43 (e) r s c «I 101100 110 010000 111100 111 000110 110 010000 010110 010 101010 000 000000 101010 101 16.2-10 (I) done in Prob. 162-7 011 101 110 100 010 001 (0) HT: 100000 010000 001000 000100 000010 000001 100100 six single errors 1 double error 16.2-11 1000101 0100111 G=[I*P]= 0010011 0001110 c=dG 101 110 100 010 001 111 144 Q. C 0000 0000000 0001 0001110 0010 0010011 0011 0011101 0100 0100111 101 0101 0101001 111 0110 0110100 011 0111 0111010 HT=110 1000 1000101 100 1001 1001011 010 1010 1010110 001 1011 1011000 1100 1100010 1101 1101100 1110 1110001 1111 1111111 s=eHT ; 1 0000001 001 0000010 010 0000100 100 0001000 110 0010000 011 0100000 111 1000000 101 s = r” T where r = received code c=r$e c = corrected code 16.2-12 We observe that the syndrome for all the three 2-ermr patterns 100010, 010100, or 001001 have the same syndrome namely 111. Since the decoding table specifies s = 111 for e = 100010 whenever e = 100010 occurs, it will be corrected. Ihe other two patterns will not be corrected. If for example e = 010100 occurs, 3 = 111 and we shall read from the decoding table e = 100010 and the error is not corrected. 11' we wish to correct the 2-error pattern 010100 (along with six single error patterns), the new decoding table is identical to that in Table 16.3 except for the last entry which should be 2 s 010100 111 16.2-13 From Eq. on P.737, for a simple error correcting code 2'” 2 n+1 or 2"‘8 211+] —) n-82 log2(r_1+ 1) This is satisfied for n 212 . Choose n =12. This gives a (12, 8) code. HT is chosen to have 12 distinct rows of four elements with the last 4 rows forming an identity matrix. Hence, 145 4-. Tom-- - G=U4P] 0011 0101 100000000011 0110 010000000101 0111 P 001000000110 1001 000100000111 Hr: 1010 6‘ 000010001001 1011 000001001010 1100 000000101011 1000 000000011100 0100 0010 I 0001 The number of non-zero syndromes = 16 -l = 15. There are 12 single error patterns. Hence we may be able to correct 3 double-error patterns. 5 C 0000 000000000000 0011 100000000000 0101 010000000000 0110 001000000000 0111 000100000000 1001 000010000000 1010 000001000000 1011 000000100000 1100 000000010000 1000 000000001000 0100 000000000100 0010 000000000010 0001 000000000001 1111 100000010000 1110 001000001000 1101 000000010001 16.2-14 Data word Code word 00 000000 01 011011 10 101110 11 110101 The minimum distance between any two code words is dmin =4 . Therefore, it can correct all l-error patterns. Since the code oversatisfies Hamming bound it can also correct some 2-error and possibly some 3-error patterns. 146 G=U4P] 0011 0101 100000000011 0110 010000000101 0111 P 001000000110 1001 000100000111 Hr: 1010 6‘ 000010001001 1011 000001001010 1100 000000101011 1000 000000011100 0100 0010 I 0001 The number of non-zero syndromes = 16 -l = 15. There are 12 single error patterns. Hence we may be able to correct 3 double-error patterns. 5 C 0000 000000000000 0011 100000000000 0101 010000000000 0110 001000000000 0111 000100000000 1001 000010000000 1010 000001000000 1011 000000100000 1100 000000010000 1000 000000001000 0100 000000000100 0010 000000000010 0001 000000000001 1111 100000010000 1110 001000001000 1101 000000010001 16.2-14 Data word Code word 00 000000 01 011011 10 101110 11 110101 The minimum distance between any two code words is dmin =4 . Therefore, it can correct all l-error patterns. Since the code oversatisfies Hamming bound it can also correct some 2-error and possibly some 3-error patterns. 146 (b) 100000 010000 1011 . 001000 1000 6smyeunxpumms 000100 0100 000010 0010 000001 0001 110000 0101 101000 0110 100100 1010 7dmmk¢nmpmums 100010 1100 100001 1111 011000 0011 010010 1001 2 triple-error pattems{ g: (I) i ‘1) i a : 16.3-1 Systematic (7, 4) cyclic code g(x) = x3 + x +1 Fordatallll d(x)=x3+x2+x+l 2 +::+l)=x6 +x5 w:4 +1:3 x3+x2+1 x3+x+l J:‘5+-::5+x4+x3 13(33 +X X5 +x3+xz x3+x2 x3+x+l x2+x+1 c(x)=(x3 +x+1Xx3+x+l)=x6+xs+x4+x3 +22 +x+1 The code word is 11111111 M7 Fordata 1110 d(x)=x3+x2+x x3+x2 1:3«0»J:+li1:6+x5+x4 :6 +Jc4 +x3 _____.___. x5 +x 1:5 +1:3 +1:2 x2 The code word is 1110100 A similar procedure is used to find die remaining codes (see Table l). (1)) From Table 1 it can be seen that the minimum distance between any two codes is 3. Hence this is a single error correcting code. Table 1 I (c) There are seven possible non~zero syndromes. x3+x+l for e=1000000 x3+x+1§x6 - x6+x‘ +X3 x‘ +X3 x4 +x2+x _______._ x3+x2+x x3 +x+l x2 +1 s=101 148 The remaining syndromes are shown in Table 2. 1000000 0100000 111 0010000 110 0001000 011 “"“2 0000100 100 0000010 010 0000001 001 (d) The received data is 1101100 r(::)=x‘wr"«Maw:2 41:34»x+1x6+xs +x3+x2 R +£+R x5+x‘ +x2 x5 413+):2 x4+x3 x4 +x2+x x3+x2 +x s(x)=x2+l ' :3 AH} 3:10] x2 +1 FromTabch (=1000000 c=r$e=110110061000000- 0101100 Hence d=0101 16.3-2 g(x)=xll +x9 +x7 +x6+x5+x+l 9(1) = d(x)8(x) 1. °+x5+x‘ 9 +37 +15 +x3 +x d1=000011110000,dfifi=x7+x 42=IOIGIOIOIBIm dflflaxn+x cl(x) = d1(x)g(x) = x” +1” +x13 +x'2 +x” +x9 +xs +x7 444 and c,=oooolloooi1101110010000 ¢2(x)=dz(x)g(X)=x22+x|8+x'7“‘15 l3 8 s 4 3 2 +x +3 +2 +3 +x +3 +1 and c2=10001101010000100111110 149 x2+l 2 16.3-3 x+1x3+x +x+1 X3+X2 x+1 Lil 0 Hence x3+2r2+1r+1=(x+1)(x2 +1)=(x+l)(x+l)(x+l)=(x+l)3 x4+x+1 16-3-4 x‘Hxs‘hl'4+x2+l Hence x5+x4+x2+1=(x+1)(x‘+x+1) x5+x4 ___..———- Now try dividing it“ +x+1 by 2: +1 , we get a remainder 1. Hence (1: +1) is not a factor of (1:4 +2: +1) . The ZM-order prime factors not divisible by .1: +1 are x2 and :2 +2: +1 . Since (1:4 + x + 1) is not divisible . This also yields a remainder 1. Hence it4 + x +1 does not have by x2 , we try dividing by (x2 + x +1) it cannot have a third order factor either. Hence either a first or a second order factor. This means x5+x4 +1:2 +1=(x+1)(x‘+x+1) 16.3-4 Try dividing 1:7 +1 by x+1 x6+x5+x‘+x3+x2+x+1 x+1x7+1 174-16 __.____. x6+1 16+Xs xs+l x5+x4 x4+l x4 +x3 x3+1 150 Therefore (1:7 + l) = (x+1)(x6 +x5 +1:4 +21:3 +2:2 +x+ 1) Now try dividing (1:6 +9 + x‘ + x3 + x2 + x+ 1) by (1: +1). 1: does not divide. So try dividing by (x2 +1) . 11 does not divide. Try dividing by (x2 + x + 1) . 11 does not divide. Next uy dividing by (x3 + 1) . 11 does not divide either. Now try dividing by (x3 + x + 1) . It divides. We find (1:6 +x5 +x4 +x3 +x2 +x+l)=(x3 +x+1)(1:3 +x2 +1) Since (1:3 + x2 + I) is not divisible by x or x +1 (the only two first-order prime factors), it must be a third-order prime factor. Hence 1:7 +1 s==(x+l)(x2 +1r+l)(x2 +1:2 +1) 16.3.6 For a single error correcting (7, 4) cyclic code with a generator polynomial g(x) = x3 + 1:2 +1 k=4 n=7 xk—l g(x)q xk—z g(x) x3 g(x) x6+x5+x3 x g(x) x5+x4+x2 x fix) x4+x3+x 8(x) x3+x2+l 80) Hence 1101000 ,_ 0110100 ' 0011010 0001101 Each code word is found by matrix multiplication c s (16" 1101000 0110100 0011010 0001101 1101000 0110100 ooono} 0011010 , 0001101 c=[0000] =ooooooo c=[0001] 151 The remaining codes are found in a similar manner. See table below. c 0000000 0001101 0011010 0010111 0110100 0111001 0101110 0100001 1101000 1100101 1110010 1111111 1011100 1010001 1000110 1001011 16.3-7 g(x) = 1:3 +x2 +1 The desired form is 1°°°"'°h11h21h31""5n 0100-..[email protected]--4n 0010---0h,3hnh,3---h,,,3 G’ = . . . . . . . . 0000 H1” h“ h” hm ’k P (/11: k) (kx m) The code is found by using c = 116 Proceeding with matrix multiplication, and noting that 0+0-0, 0+1=1+0=1, 1+1=01nd0x0=0, 0xl=1x0=0, 1x1=1 we get 1000110 0100011 q5=p111] nulnill =p11111q 0001101 q4=p110]6=p110010] and so on. 152 1001 1001011 1000 1000110 0111 0111001 0110 0110100 0101 0101110 0100 0100011 0011 0011010 0010 0010111 0001 0001101 0000 0000000 These results agree with those of Table 16.5 16.3-8 (8) 1011000 _0101100 ‘0010110 0001011 67 (b) The code is found by matrix multiplication. c = 46' In general g(x) = glx'H‘ + gzx'hk'l +"'gn-k+1 Forthiscase g,=l. g2=l. 33:0, g4=1 Since in” = g2, bu = 33, by = 34, the fourth row is immediately found. Thus, so far we have 0001101 Next, to get row 3, use row 4 with one lefi shifi. 0011010 0001101 The i is eliminated by adding row 4 to row 3. 0010111 0001101 Next for row 2, use row 3 with l lefi shifi. 153 0101110 0010111 0001101 The l is eliminated by adding row 4 to row 2. 0100011 \ 0010111 0001101 Next for row 1, use row 2 with 1 lefi shifi. 1000110 0100011 0010111 0001101 This is the desired form. 0000000 0001011 0010110 0011101 0101100 0100111 0111010 0110001 1011000 1010011 1001110 1000101 1110100 1111111 1100010 1101001 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 (c) All code words are at a minimum distance of 3 units. Hence this is a single error correcting code. 15.3.9 g(x)=x3+x+1.Hencerow4150001011. 1011000 6, 0101100 0010110 0001011 Row4isok. 0001011 4—mw4 Row3131eflshiflofrow4. 0010110 e—row3 Forrow2,lefishifirow3. 0101100 AndaddrowltoobtainrowZ. 0100111 e—row2 Forrow1,1efishiflrow2. 1001110 Andaddrowltoobtainrow]. 1000101 «row! 154 1000101 _0100111 "0010110 0001011 G 16.4-1 The burst (of length 5) detection ability is obvious. The single error correcting ability can be demonstrated as follows. if in any segment of b digits a single error occurs, it will violate the parity in that segment Hence we locate the segment where the error exists. This error will also cause parity violation in the augmented segment. By checking which hit in the augmented segment violates the parity, we can locate the wrong bit position exactly. 16.5-1 The code can correct any 3 bursts of length 10 or less. It can also correct any 3 random errors in each code word. 16.7.1 P5,, = kQ(./2E,,/.N)=129(J2 x 9.12) = 9325 ><10’6 ‘ 4 pa = wHJZfizfl =(33)[Q(J9—533)]‘ = 9.372 x 10'9 To achieve a value 9.872 x10"9 for P5,, , we need new value Eb [w say £5 /e\l. Then 9.872x10'9=kQ[ 2—‘-5-"’—)=129[ ii 011 J 011 Hence Q[ 355-] = 0.3227 x10—9 ‘J 011 and 15.2 =5.03=> 5% 18.18 011 w This means Eb /Jl must be increased from 9.12 to 18.18 (nearly doubled). 155 ...
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This note was uploaded on 05/07/2008 for the course ECE S-306 taught by Professor Dandakar during the Spring '08 term at Drexel.

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Chapter 16 - , .. Chapter 16 M 2...

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