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Channel Coding

Channel Coding. Channel Coding

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Channel Coding

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  1. Channel Coding www.assignmentpoint.com

  2. Channel Coding • Channel Coding is kind of signal transformation to improve performance of communication systems. Encoded signal can resist the impacts of noise interference and fading of channel hence improving error detection and correction at receiving end with same SNR. The channel encoder adds some redundant bits with original message bits according to the same prescribed value. The channel decoder at receiving end exploits the redundancy to determine the original message bits from coded data of high rate. • In this chapter following three Channel Coding techniques will be discussed along with their decoding. • Block code (linear) • Cyclic Redundancy Code (CRC) • Convolutional Code • R-S code www.assignmentpoint.com

  3. Single-Parity Check Code A single-parity check code is constructed by adding a single-parity bit to a block of data bits. The parity bit is taken on the value of 1 or 0 as needed to ensure that the summation of all the bits in the codeword yield an even or odd result. 0 1 1 0 0 0 0 1 0 0 0 1 1 1 0 0 1 1 Parity bit At receiving end, the modulo-2 sum of code word is 0 for even parity or 1 for odd parity www.assignmentpoint.com

  4. If the number of erroneous bits are even the receiver can not detect the error called undetected error. This coding can detect error but can not correct it. Let all bit errors are equally likely and occurs independently. We can write the probability of j errors occurring in a block of n symbols as, Where p is the probability that a symbol is received in error. The probability of undetected error, www.assignmentpoint.com

  5. Example-1: (4,3) even-parity error detection. Here n = 4 and k = 3. Now n/2 = 2. Let p = 10-3 = 6×10-6 www.assignmentpoint.com

  6. Convolutional code In telecommunication, a convolutional code is a type of error-correcting code in which each m-bit information symbol (each m-bit-string) to be encoded is transformed into an n-bit symbol, where m/n is the code rate (n ≥ m). www.assignmentpoint.com

  7. C1 + Information frame S C M9 M2 M1 M3 M4 M5 M6 M7 M8 + C2 Fig. below shows a (2,1,8) coder with g1 = (111101011) and g2 = (101110001) Let the message sequence, S = 1010000100000101  The channel frame would be input sequence with m = 8 tail bits of ‘0’ www.assignmentpoint.com

  8. M = 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 Information 16 bits Tail bits www.assignmentpoint.com

  9. C = 11 10 00 01 10 01 01 00 00 00 00 00 11 10 01 11 00 10 11 10 00 11 10 11 Length of C is 48 bits  Length of code sequence is determined using the relation, r = n(S+M) = 2(16+8) = 48 ; where n is the number of output; M is the number of shift register and S is the length of message string. www.assignmentpoint.com

  10. Q: Design a convolution coder with g1 = 111 and g2 = 101 with m = 2. Determine output for message sequence, S = 10011 Ans: Here is S = 1 0 0 1 1 The channel frame would be input sequence with m = 2 tail bits of ‘0’ frame, M = 1 0 0 1 1 0 0 C1 = M1+M2+M3 C2 = M1+M3 Here sum of C1 and C2 are X-OR sum The circuit diagram of the coder is shown in fig.below. C1 + m2 m1 S M1 M2 M1 + C2 www.assignmentpoint.com

  11. C = 11 10 11 11 01 01 11 i.e. the length of codeword is14 bits. Length of code sequence r = n (S+M) = 2(5+2) = 14 www.assignmentpoint.com

  12. C1 + m2 m1 S M1 M2 M1 + C2 Q: A convolution coder with g1 = 1 1 1 and g2 = 1 0 1 with m = 2. Determine output code sequence for S = 10011  Ans:  Here, g1(x)= 1+x+x2 g2(x)= 1+x2 S(x)= 1+x3+x4 C1(x)= S(x)g(1)(x) = (1+x3+x4)( 1+x+x2) = 1+x3+x4+x+x4+x5+x2+x5+x6 = 1+x+x2+x3+x6 = 1 1 1 1 0 0 1 C2(x)= S(x)g(2)(x) = (1+x3+x4)( 1+x2) = 1+x2+x3+x5+x4+x6 = 1+x2+x3+ x4+x5+x6 = 1 0 1 1 1 1 1 C = C1+ C2 = (11 10 11 11 01 01 11) www.assignmentpoint.com

  13. Graphical Presentation of Convoluational Coder • A convolutional coder can be expressed in three graphical form • code tree • state diagram • trellis diagram www.assignmentpoint.com

  14. code tree 00 00 11 00 11 01 00 11 10 0 00 11 01 01 C1 10 + 00 11 11 10 1 00 10 m2 m1 S M1 M2 M1 01 11 11 01 00 01 + 10 01 10 C2 For input message sequence 10011, the encoder sequence could be 11 10 11 11 according to the dotted line of tree diagram. At receiving end reverse operation is performed on coded data to retrieve the message. Fig. 1 Tree diagram of fig.1 www.assignmentpoint.com Fig.1

  15. b) state diagram Let us consider the convolutional coder of the fig.5.6; where content of the two memory units m1 and m2 can take any one of the four state 00, 01, 10 and 11. C1 + m2 m1 S M1 M2 M1 + 00 C2 11 M1 M2 0 0 M1 M2 0 1 1 0 11 10 01 01 M1 M2 1 0 M1 M2 1 1 10 01 www.assignmentpoint.com

  16. c) trellis diagram Above state diagram could be presented more explicitly like fig. 5.7(b) know as trellis diagram for transition between kth and (k+1)th clock. 00 11 M1 M2 0 0 M1 M2 0 1 1 0 11 10 01 01 M1 M2 1 0 M1 M2 1 1 10 01 clock(k) clock(k+1) clock(k+2) clock(k+3) 00 00 00 m1 m2 0 0 0 1 1 0 1 1 11 11 11 11 11 11 01 01 01 01 01 01 10 10 10 10 10 10 01 01 01 www.assignmentpoint.com

  17. Let us consider a trellis between k and (k+1)th clock shown in fig. 5.8 00 01 10 11 00 11 11 00 10 01 01 10 00 11 M1 M2 0 0 M1 M2 0 1 1 0 11 10 01 01 M1 M2 1 0 M1 M2 1 1 10 01 0 1 www.assignmentpoint.com

  18. 00 11 11 00 10 01 01 10 00 11 11 00 10 01 01 10 00 11 11 00 10 01 01 10 00 11 11 00 10 01 01 10 00 11 11 00 10 01 01 10 00 01 10 11 www.assignmentpoint.com

  19. 00 11 11 00 10 01 01 10 • 00 • 11 • 00 • 10 • 01 • 01 • 10 • 00 • 11 • 00 • 10 • 01 • 01 • 10 00 11 00 11 10 01 00 01 10 11 Q. Determine probable transmitted code for received code word, R=01 00 11 10 01 www.assignmentpoint.com

  20. + M1 M2 m C + MATLAB Determine encoded data of the convolutional encoder of fig. 8.1, where the message sequence is, m = [1 0 0 1 1 0 0]. Here the last two zeros of m are for two FF (memory cell) of the encoder; not the part of the message sequence. m = [1 0 0 1 1 0 0]; t = poly2trellis( 3, [7,5] ); %Formation of the sequential circuit of fig %length of generator sequence is 3 %generator g1 = 111 = 7 and g2 = 101 = 5 c = convenc(m, t) The result of above code is, c = 1 1 1 0 1 1 1 1 0 1 0 1 1 1 www.assignmentpoint.com

  21. Let us decode c by Viterbi algorithm. tb=1% traceback depth v=vitdec(c, t, tb, 'cont', 'hard') Above code gives one bit delayed version of message vector, m. v = 0 1 0 0 1 1 0 Determine bit error of above encoder and decoder system. [b,r]=biterr(v(tb+1:end), m(1:end-tb)) b = 0 r = 0 www.assignmentpoint.com

  22. b(0) b(1) b(2)…………b(n-k-1) │ m(0) m(1)……m(k-1) n-k k Linear Block Code The linear block code is one type of parity check code expressed by notation (n, k); where k is the length of message bit stream, n is the length of the code word and length of parity bits is n-k. Structure of the (n, k) code word is, www.assignmentpoint.com

  23. Let us express message block, parity bits and code word in row matrix form. Message, m = [m(0) m(1) m(2)………m(k-1) ] Parity bits,b = [b(0) b(1) b(2)……….b(n-k-1) ] Code word, c = [b(0) b(1)………b(n-k-1) | m(0) m(1) m(2)………m(k-1)] = [ c(0) c(1) c(2)………c(n-1) ] Here parity matrix b is determined using frame encoding rule. b = m.P = [ m(0) m(1)……… m(k-1) ] www.assignmentpoint.com

  24. Here P is a k×n-k matrix of 1 or 0, known as co-efficient matrix. Each row of P is chosen such that they are different from one another. b = = Now code word, C = [ m(0) m(1)…………m(k-1)] = m [ P │ Ik] ; where, Ik is a k×k identity matrix = m G ;where, G is known as generator matrix www.assignmentpoint.com

  25. Let us introduce another matrix H of (n-k)×k known as parity check matrix, H = [ In-k │ PT] Now, H G T= [ In-k │ PT] = In-kPT + PTIk = PT+PT = 0 Similarly, G HT = 0 At receiving end we receive, c HT= m G H T; c = m G = m × 0 = 0 www.assignmentpoint.com

  26. Example - 1 Design a (7, 4 ) linear block code. www.assignmentpoint.com

  27. Error Syndrome Vector: Let the code vector c is transmitted through a noisy channel and at receiving end another vector r is received with the relation, r = c+e ; where e is error vector. Any ith element of e is 1 when an error has occurred at ith position. For no error case all elements of e will be zero. Let us introduce a vector called error syndrome vector, S = r H T = (c + e )H T   = c H T+ e H T = 0 + e H T = e H T = 1 × n-k matrix For single bit error S= e H T gives ith column of H given error has occurred at ith position. www.assignmentpoint.com

  28. Example-3 Determine error syndrome vector of example-2 for r = [ 0 1 1 1 1 0 0 ] and comment on the result. Here, c = [ 0 1 1 0 1 0 0 ] and r = [ 0 1 1 1 1 0 0 ] e = [ 0 0 0 1 0 0 0 ] S = eH T = [ 0 0 0 1 0 0 0 ] = 1 1 0 This is the fourth column of the H matrix, error has occurs at 4th position of c. www.assignmentpoint.com

  29. Q. (6, 3) linear block code. Let m = [1 1 0] C = mG = [1 0 1 1 1 0] If r = [0 0 1 1 1 0] The syndrome, S = rHT= [1 0 0] Therefore error on 1st bit. www.assignmentpoint.com

  30. S = rHT =[r1 r2 r3 r4 r5 r6] = [(r1 +r4+r6) (r2 +r4+r5) (r3+r5+r6)] S1= (r1 +r4+r6) S2= (r2 +r4+r5) S3= (r3+r5+r6) www.assignmentpoint.com

  31. Again , S = rHT= (c+e)HT= eHT =[e1 e2 e3 e4 e5 e6] = [ S1 S2 S3] If e = [1 0 0 0 0 0 ] then S = eHT = [1 0 0] i.e 1st row of HT If e = [0 1 0 0 0 0 ] then S = eHT = [0 1 0] i.e 2nd row of HT If e = [0 0 1 0 0 0 ] then S = eHT = [0 0 1] i.e 3rd row of HT If e = [0 0 0 1 0 0 ] then S = eHT = [1 1 0] i.e 4tht row of HT If e = [0 0 0 0 1 0 ] then S = eHT = [0 1 1] i.e 5th row of HT If e = [0 0 0 0 0 1 ] then S = eHT = [1 0 1] i.e 6th row of HT www.assignmentpoint.com

  32. If e = [1 0 0 0 0 0 ] then S = eHT= [1 0 0] i.e 1st row of HT If e = [0 1 0 0 0 0 ] then S = eHT= [0 1 0] i.e 2nd row of HT If e = [0 0 1 0 0 0 ] then S = eHT= [0 0 1] i.e 3rd row of HT If e = [0 0 0 1 0 0 ] then S = eHT= [1 1 0] i.e 4tht row of HT If e = [0 0 0 0 1 0 ] then S = eHT= [0 1 1] i.e 5th row of HT If e = [0 0 0 0 0 1 ] then S = eHT= [1 0 1] i.e 6th row of HT www.assignmentpoint.com

  33. S1 = (r1 +r4+r6) S2 =( r2 +r4+r5) S3 = (r3+r5+r6) r1 r2 r3 r4 r5 r6 + + + S1 S2 S3 e1 e2 e3 e4 e5 e6 + + + + + + r1 r2 r3 r4 r5 r6 c1 c2 c3 c4 c5 c6 www.assignmentpoint.com

  34. MATLAB Code Determine the parity check matrix, H and the generator matrix, G for a (7, 4) linear block encoder. n = 7; k = 4; gen = cyclpoly(n, k); %generator polynomial [H,G] = cyclgen(7, gen) %H is the parity check matrix and G is the generator matrix.  The results of above code are, H = 1 0 0 1 1 1 0 0 1 0 0 1 1 1 0 0 1 1 1 0 1  G = 1 0 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 1 www.assignmentpoint.com

  35. For the message sequence, m=1 0 0 1 determine code word. m = [1 0 0 1]; C = encode(m, 7, 4, 'Linear', G); C = transpose(C) The result of above code is, C = 1 1 0 1 0 0 1  If the received sequence R=[1 1 0 1 0 0 1], determine syndrome and comment on the result. R = [1 1 0 1 0 0 1]; S = rem(R*transpose(H), 2) %Modulo-2 operation The result of above code is, S = 0 0 0 No error has occurred. If the received sequence R = [1 1 1 1 0 0 1], determine syndrome and comment on the result. www.assignmentpoint.com

  36. R = [1 1 1 1 0 0 1]; S = rem(R*transpose(H), 2) %Modulo-2 operation The result of above code is, S = 0 0 1 Which is the third column of H, error has taken place on third bit of the codeword. www.assignmentpoint.com

  37. The Cyclic Redundancy Check (CRC) www.assignmentpoint.com

  38. Error Detection Coding Cyclic Redundancy Check It should be clear by now that a major goal in designing error detection algorithms is to maximize the probability of detecting errors using only a small number of redundant bits. Cyclic redundancy checks use some fairly powerful mathematics to achieve this goal. For example, an 8-bit message consisting of the bits 10011010 corresponds to the polynomial, M(x) = 1×x7 +0×x6 + 0×x5+1×x4 + 1×x3 + 0×x2+1×x1 + 0×x0 = x7 +x4 + x3 +x1 www.assignmentpoint.com

  39. For the purposes of calculating a CRC, a sender and receiver have to agree on a divisor or generator polynomial, G(x). G(x) is a polynomial of degree k. For example, G(x) = x3 + x2 + 1. In this case, k = 3. Common CRC polynomials CRC C(x) CRC-8 x8 +x2 +x + 1 CRC-10 x10 + x9 + x5 +x4 +x1 +1 CRC-12 x12 + x11 +x3 +x2 +1 CRC-16 x16 + x15 +x2 +1 CRC-CCITT x16 + x12 +x5 +1 CRC-32 x32 + x26 +x23 + x22 +x16 +x12 + x11+ x10 +x8 +x7 +x5 +x4 +x2 + x + 1 www.assignmentpoint.com

  40. Steps of determining transmitted polynomial We wanted to create a polynomial T(x) for transmission that is derived from the original message M(x), is k bits longer than M(x), and is exactly divisible by G(x). We can do this in the following way: Multiply M(x) by xk, that is, add k zeros at the end of the message. Call this zero-extended message xk M(x). Divide xk M(x) by G(x) and find the remainder R(x). Subtract the remainder from xk M(x) to get T(x). It should be obvious that what is left at this point is a message that is exactly divisible by G(x). www.assignmentpoint.com 40

  41. Example-1 www.assignmentpoint.com

  42. Example-2 Using polynomial www.assignmentpoint.com

  43. Example-3:  Given, Message bit string M = 1 0 1 0 0 0 1 1 0 1 (10 bits) Generator bit string G = 1 1 0 1 0 1 (6 bits) Determine R(x) and T(x). Given, Message bit string, M = 1 0 1 0 0 0 1 1 0 1 (10 bits) Generator bit string, G = 1 1 0 1 0 1 (6 bits) Determine R(x) and T(x). Degree of G(x), r = 5.   Message after appending 5 zeros, 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 The corresponding polynomial, www.assignmentpoint.com

  44. www.assignmentpoint.com

  45. Message string after appending 5 zeros, 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 1 1 0 1 0 1 ) 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 ( 1 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1 _________________ 1 1 1 0 1 1 1 1 0 1 0 1 ________________ 1 1 1 0 1 0 1 1 0 1 0 1 ______________ 1 1 1 1 1 0 1 1 0 1 0 1 _________________ 1 0 1 1 0 0 1 1 0 1 0 1 ________________ 1 1 0 0 1 0 1 1 0 1 0 1 __________________ 0 1 1 1 0 (R) 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 + 0 1 1 1 0 ____________________________ T = 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 www.assignmentpoint.com

  46. Transmitted bit sequences and polynomial T = 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 The received polynomial www.assignmentpoint.com

  47. If there is some error at detecting end then polynomial of the received string will be, T(x) + E(x) where E(x) is arises from error. Now Choice of G(x): Here if E(x) is divisible by G(x) then error can’t be detected. To combat the situation, choosing of G(x) has some criteria like below.  1. One common type of error is a single-bit error, which can be expressed as E(x) = xi when it affects bit position i. If we select G(x) such that the first and the last term are nonzero, then we already have a two-term polynomial that cannot divide evenly into the one term E(x). www.assignmentpoint.com

  48. 2. For two bit error, ; m > r In this case G(x) has to be chosen such that it does not divide . For example if then G(x) is unable to divide for k ≤ 32,768. 3. For odd number of bits in error then E(x) will contains odd number of terms. For 3 bit error, . If any polynomial has odd number of terms then it does not have any factor like (x+1). If G(x) has a factor of (x+1) it would be unable to divide E(x). www.assignmentpoint.com

  49. 4. For burst error of length k ≤ r then If the degree of G(x) is r and r ≥ k-1 then G(x) is simply unable to divide E(x). 5. If then length of the burst error k = r+1 and the burst is identical to G(x) then the CRC code is failed to detect error. The probability of such case, th term = (1/2)r+1 www.assignmentpoint.com

  50. Dividing Circuit Design •  Division operation is done by a sequential digital circuit, consists of X-OR gates and a shift resister. Design procedure includes three steps. • Number flip-flop of shift resister is equal to highest order of G(x) i.e. r. • Maximum number X-OR gates could be r. • Presence or absence of X-OR gate in the circuit after each flip-flop depends upon presence / absence of terms of G(x) excluding xr. www.assignmentpoint.com

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