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1. EEE436 DIGITAL COMMUNICATION Coding EE436 Lecture Notes

2. Error Detection and Correction • Syndrome Decoding • Decoding involves parity-check information derived from the code’s coefficient matrix, P. • Associated with any systematic linear (n,k) block code is a (n-k)-by-n matrix, H called the parity-check matrix. • H is defined as • H = [In-kPT] • Where PT is the transpose of the coefficient matrix, P and is an • (n-k)-by-k matrix. • In-k is the (n-k)-by-(n-k) identity matrix. • For error detection purposes, the parity check matrix, H has the following property • c.HT = (0 0 ….. 0) (ie Null matrix) EE436 Lecture Notes

3. Syndrome Decoding • c.HT = (0 0 ….. 0) (ie Null matrix) • Since c=m.G, therefore • m.G.HT = (0 0 …. 0) • This property is satisfied only when c is correctly received. • Errors are indicated by the presence of non-zero elements in the matrix. • Let r denotes the 1-by-n received vector that results from sending the code vector c over a noisy channel. • When there is an error, the decoding operation will give a syndrome vector, s whose elements contain at least 1 non-zero element. EE436 Lecture Notes

4. Syndrome Decoding – Example for the (7,4) Hamming Code • A (7,4) Hamming code with the following parameters • n=7; k=4, m=7-4=3 • The k-by-(n-k) (4-by-3) coefficient matrix, P = • The generator matrix, G is, G = P = G = EE436 Lecture Notes

5. Syndrome Decoding –Example for (7,4) Hamming Code • Associated with the (7,4) Hamming Code is a 3-by-7 matrix, H called the parity-check matrix. • H is defined as • H = [In-kPT] • When a codeword is correctly received, the c.HT will result in a null matrix, otherwise it will result in a syndrome vector, s. EE436 Lecture Notes

6. Syndrome Decoding –Example for (7,4) Hamming Code • Example: The received code vector is [1110010], check whether this is a correct codeword • c.HT = [1110010] EE436 Lecture Notes

7. Syndrome Decoding –Example for (7,4) Hamming Code • Example: The received code vector is [1100010], check whether this is a correct codeword • c.HT = [1100010] = [0 0 1] – this is called the error syndrome EE436 Lecture Notes

8. Error pattern Error pattern is an error vector E whose nonzero element mark the position of the transmission errors in the received codeword We can work out all syndromes and find the corresponding error patterns and store them in a look up table for decoding purposes For example the (7,4) Hamming code EE436 Lecture Notes

9. Error detection & correction The error pattern, E is essentially the modulo-2 sum of the correct code vector and the erroneous received code vector. For example , c = 1110010 and r=1100010 (ie error in the 3rd bit) c + r =E 1110010 + 1100010 = 0010000 This error pattern corresponds to a syndrome vector in the look up table, 001 Recall that the syndrome vector, s = rHT s = (c + E)HT = cHT + EHT = EHT EE436 Lecture Notes

10. Error detection and correction Therefore, the decoding procedure involves working out the syndrome for the received code vector and look up for the corresponding error pattern. Then, modulo-2 sum the error pattern, E and the received vector, r , so that c = r + E, and the correct codeword can be recovered. EE436 Lecture Notes

11. Error detection and correctionExample For message word 0010, the correctly encoded codeword is c = 1110010. Due to channel noise, the received code vector is r = [1100010]. Show how the decoder recover the correct codeword. • The decoder uses r and the HT to find the error syndrome, s S=r.HT = 001 2) Using the resulting syndrome, refer the look up table for the corresponding assumed error vector, E. S=001 corresponds to assumed error vector, E = 0010000 3) Then ex-OR E and r to recover the correct codeword E+r = 0010000 + 1100010 = 1110010 EE436 Lecture Notes

12. Error detection and correctionExercise • For message word 0110, the correctly encoded codeword is c = 1000110. Due to channel noise, the received code vector is r = [1100110]. Show how the decoder recover the correct codeword. • For message word 0110, the correctly encoded codeword is c = 1000110. Due to channel noise, the received code vector is r = [1100100]. Show how the decoder performs its decoding operation. What is your observation and explain it. EE436 Lecture Notes

13. BCH Codes • A class of cyclic codes discovered in 1959 by Hocquenghem and in 1960 by Bose and Ray-Chaudhuri. • Include both binary and multilevel codes • Identified in the form of (n,k) BCH code for example (15,7) BCH code • A t-error Binary BCH codes consist of binary sequences of length n= 2m – 1 ; m indicates the corresponding Galois Field • Specified by its generator polynomial, g • The generator polynomial is specified in terms of its roots from the Galois Field, GF(2m) EE436 Lecture Notes

14. BCH Codes • To work out the corresponding generator polynomial for example (15,7) BCH code • First, we need to find m; since n=2m -1, therefore, for n=15; m = 4 • Then we need to find the primitive polynomial for m=4 from a specified reference table • Then, based on the primitive polynomial, construct the elements of GF(24) • Then find the minimal polynomials of the elements of GF(24) from a specified reference table • Then based on these minimal polynomials , we can work out the generator polynomial. EE436 Lecture Notes

15. Binary BCH Codes • For our example, (15,7) BCH code consider a Galois Field with m=4 GF(24) • A polynomial p(X) over GF(24) of degree 4 that is primitive is taken from the following Table of primitive polynomials EE436 Lecture Notes

16. Binary BCH Codes • Then, based on the primitive polynomial, 1 + X + X4 construct the elements of GF(24) • Let alpha (ά) be a primitive element in GF(2m) • Set p(ά)=1+ ά+ ά4 = 0 , then ά4 = 1+ ά • Using this relation, we can construct GF(24) elements as below EE436 Lecture Notes

17. Binary BCH Codes Then find the minimal polynomials of the elements of GF(24) from a specified reference table EE436 Lecture Notes

18. Binary BCH Codes Then the generator polynomial of a t-error correcting BCH code of length 2m – 1 is given by g(x) = LCM { ǿ1(X), ǿ2(X) , ….., ǿ2t (X) } Since every even power of ά in the elements sequence has the same minimal polynomial as some preceding odd power of ά in the elements sequence g(x) = LCM { ǿ1(X), ǿ3(X) , ….., ǿ2t-1 (X) } EE436 Lecture Notes

19. Binary BCH Codes generator polynomial – Example A(15,7) BCH code (ie n=15) m=2m-1; m=4 Therefore , refer to Galois Field with m=4 GF(24) A polynomial p(X) over GF(24) of degree 4 that is primitive is taken the table p(X)= 1 + X + X4 Then, based on the primitive polynomial, 1 + X + X4 construct the elements of GF(24) Then find the minimal polynomials of the elements of GF(24) from the table Then the generator polynomial of a t-error correcting BCH code of length 2m – 1 is given by g(x) = LCM { ǿ1(X), ǿ3(X) , ….., ǿ2t-1 (X) } EE436 Lecture Notes

20. Binary BCH Codes generator polynomial – Example Then the generator polynomial of a t-error correcting BCH code of length 2m – 1 is given by g(x) = LCM { ǿ1(X), ǿ3(X) , ….., ǿ2t-1 (X) } For 2-error correcting; t=2 Therefore, g(x) = LCM { ǿ1(X), ǿ3(X)} ǿ1(X) = 1 + X + X4 and ǿ3(X)= 1 + X + X2 + X3 + X4 g(x) = ǿ1(X). ǿ3(X) = 1 + X4 + X6 + X7 + X8 Exercise : Try out for 3-error correcting BCH code of the same length EE436 Lecture Notes

21. Binary BCH Codes generator polynomials – Example EE436 Lecture Notes

22. Non-Binary or M-ary BCH Codes • Unlike binary these codes are multilevel codes • Operates on multiple bits rather than individual bits • The general (n,k) encoder encodes k m-bit symbols into blocks consisting of n=2m-1 symbols of total m(2m-1) bits • Thus the encoding expands a block of k symbols to n symbols by adding n-k redundant symbols • An example of non-binary BCH code is the Reed-Solomon Code EE436 Lecture Notes

23. RS Codes • A t-error-correcting RS code has the following parameters • Block length n= 2m-1 • Message size k symbols • Parity-check size n-k=2t symbols • Minimum distance = 2t + 1 • Example RS(7,4) with m=3 bits EE436 Lecture Notes