Transcript EEE436 - School Of Electrical & Electronic Engineering

EEE436
DIGITAL COMMUNICATION
Coding
En. Mohd Nazri Mahmud
MPhil (Cambridge, UK)
BEng (Essex, UK)
[email protected]
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)-byn 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
EE436 Lecture Notes
c.HT = (0 0 ….. 0)
(ie Null matrix)
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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.
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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 =
P=
1
1
0
0
1
1
1
1
1
1
0
1
The generator matrix, G is, G =
G=
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
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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]
1
0
0
1
0
1
1
0
1
0
1
1
1
0
0
0
1
0
1
1
1
When a codeword is correctly received, the c.HT will result in a
null matrix, otherwise it will result in a syndrome vector, s.
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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 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
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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 correction
Example
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
1)
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Error detection and correction
Exercise
i)
ii)
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.
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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 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.
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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
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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
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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 { 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) }
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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) }
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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
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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 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
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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
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