Systematic Cyclic Code• In a systematic code, the first k digits are

bits and the last m = n – k digits are the parity check bits.

• For systematic code, the codeword polynomial c(x) corresponding to the data polynomial d(x) is given by

• Where ρ(x) is the remainder from dividing xn-kd(x) by g(x).

( ) ( ) ( )n kc x x d x xρ−= +

( )( ) Re( )

n kx d xx mg x

ρ−

=

• To prove this we observe that

• Where q(x) is of degree k – 1 or less. We add to both sides, and because of f(x) + f(x) = 0 under modulo-2 operation, we have

Or

( ) ( )( )( ) ( )

n kx d x xq xg x g x

ρ−

= +

( ) / ( )x g xρ

( ) ( ) ( )( )

n kx d x x q xg x

ρ− +=

( ) ( ) ( ) ( )n kq x g x x d x xρ−= +

Example• Construct a systematic (7,4) cyclic code

using a generator polynomial.SolutionAs we know g(x) = x3 + x2 + 1

Consider a data vector d = 1010

d(x) = x3 + x

so xn – kd(x) = x6 +x4

• x6 + x5 + x3

x5 + x4 + x3

x5 + x4 + x2

x3 + x2

x3 + x2 + 11

3 2

3 2 6 41

1x x

x x x x+ +

+ + +( )q x←

( )xρ←

Hence

and

C = 1010001

3

3 3

6 4

( ) ( ) ( )( ) 1

1

c x x d x xx x xx x

ρ= +

= + +

= + +

• There is one shortcut to make code table for cyclic codes by making generator matrix G.

• Find out code words only for four combinations of inputs 1000, 0100, 0010, 0001, these are 1000110, 0100011, 0010111, 0001101.

• Now recognize these four code words are the four rows of G.

• Once we make generator matrix , then code table can be create using equation

c = d.G

1000110010001100101110001101

G

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

data Code1111 11111111110 11100101101 11010001100 11001011011 10111001010 10100011001 10010111000 10001100111 01110010110 01101000101 01011100100 01000110011 00110100010 00101110001 00011010000 0000000

Generator Polynomial and Generator Matrix of Cyclic Codes

• Once the generator matrix G = [ I p] by determining the parity submatrix P

1st row of P:

2nd row of P:

Kth row of P:

1

Re( )

nxmg x

−

2

Re( )

nxmg x

−

Re( )

n kxmg x

−

• For example

• 1st row of P:

• 2nd row of P:

• 3rd row of P:

• 4th row of P:

3( ) 1g x x x= + +

62Re 1

( )xm x

g x= +

52Re 1

( )xm x x

g x= + +

42Re

( )xm x x

g x= +

3

Re 1( )xm x

g x= +

1000101010011100101100001011

G

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

111010001110101101001

H⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

Decoding• Every valid code polynomial c(x) is a

multiple of g(x). In other words c(x) is divisible by g(x). When an error occurs during the transmission the received polynomial r(x) will not be a multiple of g(x). If the number of errors in r is correctable. Thus

1( ) ( )( )( ) ( )

r x s xm xg x g x

= +

Where the syndrome polynomial s(x) has a degree n – k – 1 or less.

If e(x) is the error polynomial thenr(x) = c(x) + e(x)

Remembering that c(x) is a multiple of g(x)

( )( ) Re( )

r xs x mg x

=

( )( ) Re( )

( ) ( )Re( )

( )Re( )

r xs x mg x

c x e xmg x

e xmg x

=

+=

=

Example

• Construct the decoding table for the single error correcting (7,4) code. Determine the data vectors transmitted for the following received vectors r: (a) 1101101 (b) 0101000 (c) 0001100

• Solution:The first step is to construct the decoding table.

• Because n – k – 1 , the syndrome polynomial of the second order, and there are seven possible nonzero syndroms.

• There are seven possible correctable single-error patterns because n = 7, we can use to find error

s = e. HT

• s = 001,010,011,100,101,110,111

• For example s = [1 1 0] then e = [1 0 0 0 0 0 0]

101110011100100111001

H⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

error (e) syndrome (s)1000000 1100100000 0110010000 1110001000 1010000100 1000000010 0100000001 001

• r = [1 1 0 1 1 0 1]s = r. HT

= [1 1 0 1 1 0 1]

= [ 1 0 1]e = [ 0 0 0 1 0 0 0] for s = [1 0 1]

d = 1100

101110011100100111001

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1101101 00010001100101

c r e= ⊕ = ⊕=