cyclic codes error control code, Study notes of Digital Communication Systems

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MODULE 2

ERROR CONTROL CODE

Contents

• Introduction

• Linear Block codes-G and H matrix, Generation

and Decoding

• Cyclic codes-g(x) and h(x), Encoding using an (n-

k) bit shift register, syndrome calculation

• Convolution Codes: Code generation-Time

domain and Transfer domain approach and Code

Tree

Types of codes

i) Block Codes:

Block code consists of (n-k) number of check

bits(redundant bits) being added to k number of

information bits to form ‘n’ bit code-words.

i) Convolutional code:

In this code, input databits are fed as streams of

data bits which convolve to output bits based upon

the logic function of the encoder.

Linear Block codes

• Let C

and C

be any two code words(n-bits)

belonging to a set of (n, k) block code

• If C

C

, is also a n-bit code word

belonging to the same set of(n,k) block code,

such a block code is called (n,k)linear block

code.

Matrix description of linear block code

• Let the message block of k-bits(code-words) be

represented as a “row-vector” or “k-tuple” called

“message vector” is given by

[D]={ d 1 , d 2 , .....dk}

• 2

k

code-vectors can be represented by

C={c 1 ,c 2 ,.......cn}

• Also ci=di for all i= 1 , 2 ,......k
• [C]= {c 1 ,c 2 ,........,ck, ck+ 1 , ck+ 2 ,.......cn }

• (n-k) number of check bits ck+1, ck+2,.......cn are derived from ‘k’ message bits using a predetermined rule as below ck+1 = p 11 d 1 + p 21 d 2 +................+ pk1dk ck+2 = p 12 d 1 + p 22 d 2 +................+ pk2dk : : ck+1 = p1. n-kd 1 + p (^) 2.n-kd 2 +................+ pk.n-kdk
• In matrix form, [c 1 ,c 2 ,..,ck, ck+1, ck+2,...cn ]=[d 1 , d 2 , .....dk ] 1 0 0...0 p 11 p 12 .... p1. n-k 0 1 0...0 p 21 p 22 .... p2. n-k 0 0 0... pk1 pk2 .... pk. n-k [C]= [D] [G]
• [G] is called as generator matrix of order (k x n)
• [G] = [Ik ¦ P] (^) (k x n) where Ik unit matrix of order ‘k’ [P]= Parity matrix of order k x (n-k)
• Also [G] = [P ¦ Ik]

For a systematic (6,3) linear block code, the parity matrix P is given by [P] = 1 0 1 0 1 1 1 1 0 Find all possible code-vectors

• Solution: Given n=6, k=3, Since k=3, 2 k =8 message vectors given by (000)

(001), (010),^ (001),^ (011),^ (100),^ (101),^ (111)

[C]= [D] [G]

where

[G] = [Ik ¦ P]

[I 3 ¦ P] [G]= 1 0 0 : 1 0 1 0 1 0 : 0 1 1 0 0 1 : 1 1 0

• [C]=[D] [G] =[d

d

d

] 1 0 0 1 0 1 0 1 0 0 1 1 0 0 1 1 1 0 = [ d

, d

,d

, (d

+d

), (d

+d

),(d

+d

)]

• C

c

• C

g

= ( 010011) + (110110) = (100101) = c

e

For a systematic (7,4) linear block code, generated by G= Find all possible code vectors. Solution: n=7, k=4; (n-k)= 3 ∴ 2 k = 2 4 = 16 message vectors will be present. [C]= [D] [G] = [d 1 d 2 d 3 d 4 ] =[d 1 , d 2 , d 3 , d 4 , (d 1 +d 2 + d 3 ), (d 1 +d 2 + d 4 ), (d 1 +d 3 + d 4 ) ]

If C is a valid code vector, namely C=[ D G]. Then prove that CH T = where H T is the transpose of the parity check matrix H. Wkt 1 0 0.......0 p

p

.... p

1. n-k

0 1 0......0 p

p

.... p

2. n-k

0 0 0......1 p

k

p

k

.... p

k. n-k

i

th

row of [G] matrix is given by

g

i

= [ 0 0 0......1.....0 p

i

p

i

.... p

ij ........

p

k. n-k

] ith^ element (k+j)th^ element J th row of [H] matrix is given by hj = [p1j p2j ....... pij .... pkj 0 0 0 .....1.......0 ]

gi.hj T = [ 0 0 0......1.....0 pi1 pi2 .... pij ........pk. n-k ] .[p1j p2j ....... pij .... pkj 0 0 0 .....1.......0 ] T = [ 0 0 0......1.....0 pi1 pi2 .... pij ........pk. n-k ] ith element (k+j)th element Modulo-2 multiplication yields ; gi.hj T = pij + pij = pij (1+1) = pij. 0 = 0 hence proved. p1j p2j : pij : pkj 0 0 : 1 : 0

1.For a systematic (6,3) code, find all the transmitted code vector , draw the encoding circuit. If received vector R=[110010], detect and correct the error that has occurred due to noise. Given P= 1 0 1 0 1 1 1 1 0 Solution: (Refer notes) (problem has been solved during the class also)

Video link

• https://youtu.be/qI1M6UzdyQw

• ( The above video can be watched for linear

block code concept)