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**DIGITAL COMMUNICATION Coding**

EEE436 DIGITAL COMMUNICATION Coding En. Mohd Nazri Mahmud MPhil (Cambridge, UK) BEng (Essex, UK) Room 2.14 EE436 Lecture Notes

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**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-k PT] 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

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**c.HT = (0 0 ….. 0) (ie Null matrix) Since c=m.G, therefore **

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

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**1 1 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 = 1 P = 1 G = EE436 Lecture Notes

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**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-k PT] When a codeword is correctly received, the c.HT will result in a null matrix, otherwise it will result in a syndrome vector, s. 1 EE436 Lecture Notes

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**Syndrome Decoding –Example for (7,4) Hamming Code **

Example: The received code vector is [ ], check whether this is a correct codeword c.HT = [ ] 1 EE436 Lecture Notes

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**Syndrome Decoding –Example for (7,4) Hamming Code **

Example: The received code vector is [ ], check whether this is a correct codeword c.HT = [ ] 1 = [0 0 1] – this is called the error syndrome EE436 Lecture Notes

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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

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**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 = and r= (ie error in the 3rd bit) c + r =E = 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

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**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

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**Error detection and correction Example**

For message word 0010, the correctly encoded codeword is c = Due to channel noise, the received code vector is r = [ ]. 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 = 3) Then ex-OR E and r to recover the correct codeword E+r = = EE436 Lecture Notes

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**Error detection and correction Exercise**

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

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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

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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

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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

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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

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

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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

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**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

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**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

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**Binary BCH Codes generator polynomials – Example**

EE436 Lecture Notes

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**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

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**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

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