EEE436

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EEE436

• DIGITAL COMMUNICATION
• Coding

En. Mohd Nazri Mahmud MPhil (Cambridge, UK) BEng (Essex, UK) nazriee_at_eng.usm.my Room 2.14
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Error Detection and Correction Syndrome Decoding
Decoding involves parity-check information
derived from the codes 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)
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Syndrome Decoding c.HT (0 0 .. 0)
(ie Null matrix) Since cm.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.
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Syndrome Decoding Example for the (7,4) Hamming
Code A (7,4) Hamming code with the following
parameters n7 k4, m7-43 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
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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 0 0 1 0 1 1
0 1 0 1 1 1 0
0 0 1 0 1 1 1
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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
1 0 0
0 1 0
0 0 1
1 1 0
0 1 1
1 1 1
0 0 1
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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
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
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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

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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 1110010
  and r1100010 (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

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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.

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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
• Sr.HT 001
• 2) Using the resulting syndrome, refer the look
  up table for the corresponding assumed error
  vector, E.
• S001 corresponds to assumed error vector, E
  0010000
• 3) Then ex-OR E and r to recover the correct
  codeword
• Er 0010000 1100010 1110010

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Error detection and correctionExercise

1. 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.
2. 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.

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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)

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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 n2m -1,
  therefore, for n15 m 4
• Then we need to find the primitive polynomial for
  m4 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.

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Binary BCH Codes

• For our example, (15,7) BCH code consider a
  Galois Field with m4 GF(24)
• A polynomial p(X) over GF(24) of degree 4 that is
  primitive is taken from the following Table of
  primitive polynomials

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Binary BCH Codes

• Then, based on the primitive polynomial, 1 X
  X4 construct the elements of GF(24)
• Let alpha (?) be a primitive element in GF(2m)
• Set p(?)1 ? ?4 0 , then ?4 1 ?
• Using this relation, we can construct GF(24)
  elements as below

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Binary BCH Codes

• Then find the minimal polynomials of the elements
  of GF(24) from a specified reference table

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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 ?1(X), ?2(X) , .., ?2t (X)
• Since every even power of ? in the elements
  sequence has the same minimal polynomial as some
  preceding odd power of ? in the elements sequence
  
• g(x) LCM ?1(X), ?3(X) , .., ?2t-1 (X)

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Binary BCH Codes generator polynomial Example
A(15,7) BCH code (ie n15) m2m-1
m4 Therefore , refer to Galois Field with m4
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 ?1(X), ?3(X) , .., ?2t-1 (X)

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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 ?1(X), ?3(X) , .., ?2t-1 (X) For
2-error correcting t2 Therefore, g(x) LCM
?1(X), ?3(X) ?1(X) 1 X X4 and ?3(X)
1 X X2 X3 X4 g(x) ?1(X). ?3(X) 1
X4 X6 X7 X8 Exercise Try out for
3-error correcting BCH code of the same length
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Binary BCH Codes generator polynomials Example


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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 n2m-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

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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-k2t symbols
• Minimum distance 2t 1
• Example RS(7,4) with m3 bits