ee436 lecture notes1 eee436 digital communication coding en. mohd nazri mahmud mphil (cambridge, uk)...

EE436 Lecture Notes 1

EEE436DIGITAL COMMUNICATION

Coding

En. Mohd Nazri Mahmud

MPhil (Cambridge, UK)

BEng (Essex, UK)

[email protected]

Room 2.14

mailto:[email protected]


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)


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.


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 1 0

0 1 1

1 1 1

1 0 1

P =

1 1 0 1 0 0 0

0 1 1 0 1 0 0

1 1 1 0 0 1 0

1 0 1 0 0 0 1

G =


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 0 0 1 0 1 1

0 1 0 1 1 1 0

0 0 1 0 1 1 1



Example: The received code vector is [1110010], check whether this is a correct codeword

c.HT = [1110010] 1 0 0

0 1 0

0 0 1

1 1 0

0 1 1

1 1 1

0 0 1



Example: The received code vector is [1100010], check whether this is a correct codeword

c.HT = [1100010] 1 0 0

0 1 0

0 0 1

1 1 0

0 1 1

1 1 1

0 0 1

= [0 0 1] – this is called the error syndrome


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


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


Error detection and correctionTherefore, 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.


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.

1) The decoder uses r and the HT to find the error syndrome, s

S=r.HT = 0012) Using the resulting syndrome, refer the look up table for

the corresponding assumed error vector, E.S=001 corresponds to assumed error vector, E = 00100003) Then ex-OR E and r to recover the correct codeword E+r = 0010000 + 1100010 = 1110010


Error detection and correctionExercise

i) 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.

ii) 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.


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)


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.


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


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


Binary BCH CodesThen find the minimal polynomials of the elements of GF(24) from a

specified reference table


Binary BCH CodesThen 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) }


Binary BCH Codes generator polynomial – Example

A(15,7) BCH code (ie n=15)

m=2m-1; m=4Therefore , 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) }


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


Binary BCH Codes generator polynomials – Example


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


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 notes1 eee436 digital communication coding en. mohd nazri mahmud mphil (cambridge, uk)...

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