Digital communication coding

1. EE436 Lecture Notes 1
EEE436
DIGITAL COMMUNICATION
Coding
En. Mohd Nazri Mahmud
MPhil (Cambridge, UK)
BEng (Essex, UK)
nazriee@eng.usm.my
Room 2.14

2. 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-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)

3. 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.

4. 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 =
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 =

5. 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-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

6. 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] 1 0 0
0 1 0
0 0 1
1 1 0
0 1 1
1 1 1
0 0 1

7. 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] 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

8. 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

9. 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

10. 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.

11. EE436 Lecture Notes 11
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.
1) 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

12. EE436 Lecture Notes 12
Error detection and correction
Exercise
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.

13. 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)

14. 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.

15. 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

16. 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

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

18. 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) }

19. 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) }

20. 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

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

22. 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

23. 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