CYCLIC CODES

Dr. Musbah Shaat EELE 6338 08/10/2012

Outline

EELE 6338 2

Definition

• An (n,k) linear code C is cyclic if every cyclic

shift of a codeword in C is also a codeword in

C.

If c c c …. c c is a codeword, thenIf c0

c1

c2

…. cn-2

cn-1

is a codeword, then

cn-1

c0

c1

…. cn-3

cn-2

cn-2

cn-1

c0

…. cn-4

cn-3

: : : : :

c1

c2

c3

…. cn-1

c0

are also codewords.

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Example

• The (7,4) Hamming code discussed before is cyclic:

1010001 1110010 0000000 1111111

1101000 01110011101000 0111001

0110100 1011100

0011010 0101110

0001101 0010111

1000110 1001011

0100011 1100101

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Code Polynomial

• Let c = c0 c1 c2 …. cn-1.

The code polynomial of c:c(X) = c0 + c1X+ c2 X

2 + …. + cn-1 X

n-1

where the power of X corresponds to the bit

position, and the coefficients are 0’s and 1’s.position, and the coefficients are 0’s and 1’s.

• Example:

1010001 ; 1+X2+X6

0101110 ; X+X3+X4+X5

Each codeword is represented by a polynomial of

degree less than or equal n-1.

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deg[ ]c(X) ≤ −n 1

Implementing the Shift

Let c = c0 c1 c2 …. cn-1

and c(i) = cn-i cn-i+1 … c

n-1 c0 …. cn-i-1 (i shifts to the right)

Then, c(X) = c0 + c1X+ c2 X2 + …. + c

n-1 Xn-1 , and

c (i)(X) = c + c X + …. + c Xi-1 + c Xi +…. +c Xn-1c (i)(X) = cn-i + c

n-i+1 X + …. + cn-1 X

i-1 + c0Xi +…. +cn-i-1 X

n-1

What is the relation between c(X) and c (i)(X)?

Apparently, shifting a bit one place to the right is equivalent

to multiplying the term by X. For i shifts:

Xic(X)= c0Xi +c1X i+1 + ….+ cn-i-1 X

n-1 + cn-i X

n ….+ cn-1 X

n+i-1

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Implementing the Shift (�)

Rearranging,

Xic(X) = cn-i X

n +…+ cn-1 X

n+i-1 + c0Xi + c1X i+1 + …+ cn-i-1 X

n-1

Add to the polynomial the zero-valued sequence:

(cn-i + c

n-i ) + (cn-i+1 + c

n-i+1 )X + …. + (cn-1 + c

n-1 )Xi-1(c

n-i + cn-i ) + (c

n-i+1 + cn-i+1 )X + …. + (c

n-1 + cn-1 )X

i-1

Xic(X) = cn-i (X

n +1) + cn-i+1 X (Xn +1) +…. + c

n-1 Xi-1 (Xn +1) +

cn-i + c

n-i+1 X +…. + cn-1 X

i-1 +

c0Xi +c1X i+1 + …. + cn-i-1 X

n-1

That is: Xic(X) = q(X)(Xn +1) + c(i)(X)

Where q(X)= cn-i + c

n-i+1 X +…. + cn-1 X

i-1

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Implementing the Shift (�)

c(i)(X) is the remainder from dividing Xic(X) by (Xn +1).

c(i)(X) = Rem[Xic(X)/ (Xn +1)] = Xic(X) mod (Xn +1).

Example: c = 0101110. c(3) = 1100101 Example: c = 0101110. c(3) = 1100101

Mathematically: c(X) = X + X3 + X4 + X5.

X3c(X) = X4 + X6 + X7 + X8

Rem[X3c(X)/ (X7 +1)] = 1 + X + X4 + X6

c(3) = 1100101

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The Code Polynomial of Minimum Degree

• TH 5.1: The nonzero code polynomial of minimum degree

in a cyclic code C is unique.(If not, the sum of the two polynomials will be a code polynomial of

degree less than the minimum. Contradiction)

• TH 5.2: Let g(X) = g + g X +….+ g Xr-1 +Xr be the nonzero • TH 5.2: Let g(X) = g0 + g1X +….+ gr-1Xr-1 +Xr be the nonzero

code polynomial of minimum degree in an (n,k) cyclic

code. Then the constant term g0 must be equal to 1.(If not, then one cyclic shift to the left (n-1 to right) will produce a

code polynomial of degree less than the minimum. Contradiction)

• For the (7,4) code given in the Table 5.1, the nonzero code

polynomial of minimum degree is g(X) = 1 + X + X3

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EELE 6338 10

Generator Polynomial

Let g(X), the minimum-degree polynomial in C, have degree r

• Since the code is cyclic: Xg(X), X2g(X),…., Xn-r-1g(X) are code

polynomials in C. (Note that deg[Xn-r-1g(X)] = n-1).

• Since the code is linear: Since the code is linear:

(a0 + a1X + …. + an-r-1 X

n-r-1)g(X) is also a code

polynomial, where ai= 0 or 1.

• TH 5.3: A binary polynomial of degree ≤ n-1 is a code

polynomial if and only if it is a multiple of g(X).(if a code polynomial c(X) is not a multiple of g(X), then Rem[c(X)/g(X)]

will be a code polynomial of

degree less than the minimum. Contradiction)

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Generator Polynomial (�)

• All code polynomials are generated from the multiplication c(X) = a(X)g(X).

deg[c(X)] n-1, deg[g(X)] = r, ==> deg[a(X)] ≤ n-r-1

There is a total of 2n-r polynomials of degree ≤ n-r-1

Therefore the number of codewords = 2n-r = 2k

≤

Therefore the number of codewords = 2n-r = 2k

Therefore, r = deg[g(X)] = n-k

• Since deg[a(X)] ≤ k-1, the polynomial a(X) may be taken to be the information polynomial u(X) (a polynomial whose coefficients are the information bits). Encoding is performed by the multiplication c(X) = u(X)g(X).

• g(X), generator polynomial, completely defines the code.

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• TH 5.4:

In an (n,k) cyclic code, there exists one and

only one code polynomial of degree n-k,

1+ g1X + g

2X2 + …+ g

n-k-1Xn-k-1 +Xn-k1+ g

1X + g

2X2 + …+ g

n-k-1Xn-k-1 +Xn-k

Every code polynomial is a multiple of g(X),

and every binary polynomial of degree n-1 or

less that is a multiple of g(X) is a code

polynomial.

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(7,4) Code Generated by 1+X+X3

Infor. Code Code polynomials

0000 0000000 0 = 0 . g(X)

1000 1101000 1 + X + X3 = 1 . g(X)

0100 0110100 X + X2 + X4 = X . g(X)0100 0110100 X + X + X = X . g(X)

1100 1011100 1 + X2 + X3 + X4 = (1 + X) . g(X)

0010 0011010 X2 + X3 + X5 = X2 . g(X)

1010 1110010 1 + X+ X2 + X5 = (1 + X2) . g(X)

0110 0101110 X+ X3 + X4 + X5 = (X+ X2) . g(X)

1110 1000110 1 + X4 + X5 = (1 + X + X2) . g(X)

0001 0001101 X3 + X4 + X6 = X3 . g(X)

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(7,4) Code Generated by 1+X+X3 (�)

Infor. Code Code polynomials

1001 1100101 1 + X + X4 + X6 = (1 + X3) . g(X)

0101 0111001 X+ X2 + X3 + X6 = (X+ X3) . g(X)

1101 1010001 1 + X2 + X6 = (1 + X + X3) . g(X)

0011 0010111 X2 + X4 + X5 + X6 = (X2 + X3). g(X)0011 0010111 X2 + X4 + X5 + X6 = (X2 + X3). g(X)

1011 1111111 1 + X + X2 + X3 + X4 + X5 + X6

= (1 + X2 + X3) . g(X)

0111 0100011 X + X5 + X6 = (X + X2 + X3). g(X)

1111 1001011 1 + X3 + X5 + X6

= (1 + X + X2 + X3) . g(X)

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Properties of the Generator

Polynomial • TH 5.5: The generator polynomial g(X) of an (n,k) cyclic

code is a factor of Xn+1.

• TH 5.6: If g(X) is a polynomial of degree n-k and is a factor

of Xn+1, then it generates an (n,k) cyclic code. of X +1, then it generates an (n,k) cyclic code.

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Constructing g(X)

(1) To construct a cyclic code of length n, find the factors of

the polynomial Xn+1.

(2) The factor (or product of factors) of degree n-k serves as

the generator polynomial of an (n,k) cyclic code. Clearly, a

cyclic code of length n does not exist for every k.cyclic code of length n does not exist for every k.

(3) The code generated this way is guaranteed to be cyclic.

But we know nothing yet about its minimum distance. The

generated code may be good or bad.

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Cyclic Codes of Length 7

Example (5.1-mod): What cyclic codes of length 7 can be constructed?

Rembember r=n-k � k=n-r.

X7+1 = (1 + X)(1 + X + X3)(1 + X2 + X3)X +1 = (1 + X)(1 + X + X )(1 + X + X )

g(X) Code g(X) Code

(1 + X) (7,6) (1 + X)(1 + X + X3) (7,3)

(1 + X + X3) (7,4) (1 + X) (1 + X2 + X3) (7,3)

(1 + X2 + X3) (7,4) (1 + X + X3)(1 + X2 + X3) (7,1)

What if u=(1010) � u= 1 + X2 � v(x)= 1 + X + X2 + X5

• � 1110010

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Encoding of Systematic Cyclic Codes

Systematic: b0 b1 b2 …. bn-k-1 u0 u1 u2 …. u

k-1

b(X) = b0 + b1X+….+bn-k-1Xn-k-1, u(X) = u0+u1X+ ….+u

k-1Xk-1

then c(X) = b(X) + Xn-k u(X)

a(X)g(X) = b(X) + Xn-k u(X)

Xn-k u(X)= a(X)g(X) + b(X)Xn-k u(X)= a(X)g(X) + b(X)

Or b(X) = Rem[Xn-k u(X)/g(X)]

Encoding Procedure:

1. Multiply u(X) by Xn-k

2. Divide Xn-k u(X) by g(X), obtaining the remainder b(X).

3. Add b(X) to Xn-k u(X), obtaining c(X) in systematic form.

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Encoding of Systematic Cyclic Codes (�)

Example

Consider the (7,4) cyclic code generated by g(X) = 1 + X + X3.

Find the systematic codeword for the message 1001.

u(X) = 1 + X3u(X) = 1 + X3

X3u(X) = X3 + X6

b(X) = Rem[X3u(x)/g(X)] = X + X2

Therefore, c = 0111001

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Generator Matrix of Cyclic Codes

• The generator matrix of a cyclic code can be

formed as follows:

1st row: Coefficients of g(X) followed by k-1

zeroszeros

Other rows are the k-1 cyclic shifts of first row

• The matrix can be arranged in systematic form

by row operations.

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Generator Matrix of Cyclic Codes (�)

EELE 6338 22

Generator Matrix in Systematic form

EELE 6338 23

Example 5.4

EELE 6338 24

Encoder Circuit

g2

g1

++ + +

Gate

gr-1

• Switch Down & Gate ON: k message bits are shifted into the channel. The parity bits are formed in the register.

• Switch Up & Gate OFF: Contents of register are shifted into the channel.

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(7,4) Encoder Based on 1 + X + X3

+ +

Gate

Input 1 1 0 1

Register : 000 110 101 100 100initial 1st shift 2nd shift 3rd shift 4th shift

Codeword: 1 0 0 1 0 1 1

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Decoding of Cyclic Codes

• STEPS:

(1) Syndrome computation

(2) Associating the syndrome to the error pattern(2) Associating the syndrome to the error pattern

(3) Error correction

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Syndrome Computation

• Received word: r(X) = r0 + r1X +….+ rn-1Xn-1

• If r(X) is a correct codeword, it is divisible by g(X).

Otherwise: r(X) = q(X)g(X) + s(X).

• deg[s(X)] ≤ n-k-1.

• s(X) is called the syndrome polynomial.• s(X) is called the syndrome polynomial.

• s(X) = Rem[r(X)/g(X)] = Rem[ {a(X)g(X) + e(X)}/g(X)] =

Rem[e(X)/g(X)]

• The syndrome polynomial depends on the error pattern

only.

• s(X) is obtained by shifting r(X) into a divider-by-g(X) circuit.

The register contents are the syndrome bits.

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Syndrome Circuit

EELE 6338 29

Circuit for Syndrome Computation

++

Gate

r = 0010110

Shift Input Register contents

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Shift Input Register contents

0 0 0 (initial state)

1 0 0 0 0

2 1 1 0 0

3 1 1 1 0

4 0 0 1 1

5 1 0 1 1

6 0 1 1 1

7 0 1 0 1 (syndrome s)

More on Syndrome Computation

• TH5.8 Let s(X) be the syndrome of a received polynomial r(X) = r0 +

r1X +….+ rn-1Xn-1 . Then the remainder resulting from dividing Xs(X)

by g(X) is the syndrome of r(1)(X), which is a cyclic shift of r(X).

• Proof: r(X) = r0 + r1X +….+ rn-1Xn-1

r(1)(X) = rn-1 + r0X +….+ r

n-2Xn-1 = rn-1 + Xr(X) + r

n-1Xnr(1)(X) = rn-1 + r0X +….+ r

n-2Xn-1 = rn-1 + Xr(X) + r

n-1Xn

= rn-1(Xn+1) + Xr(X)

c(X)g(X) + y(X) = rn-1 g(X)h(X)+ X{a(X)g(x) + s(X)}

where y(X) is the syndrome of r(1)(X) .

Xs(X) = {c(X) + Xa(X) + rn-1 h(X)}g(X) + y(X)

Therefore, Syndrome of r(1)(X)= Rem[Xs(X)/g(X)]. Q.E.D.

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More on Syndrome Computation (�)

• Note: for simplicity of notation, let Rem[Xs(X)/g(X)] be denoted by s(1)(X). s(1)(X) is NOT a cyclic shift of s(X), but the syndrome of r(1)(X) which is a cyclic shift of r(X).

• Example:

r(X) = X2 + X4 + X5; g(X) = 1 + X + X3

s(X) = Rem[r(X)/g(X)] = 1 + X2

r(1)(X) = X3 + X5 + X6

s(1)(X) = Rem[r(1)(X)/g(X)] = 1 (polynomial)

Also, s(1)(X) = Rem[Xs(X)/g(X)] = 1.

• It follows from TH 5.8 that, in general, the remainder s(i)(X) resulting from dividing Xis(X) by g(X) is the syndrome of r(i)(X).

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More on Syndrome Computation

(�)

Shift Input Register contents

++

Gate

Gate

r = 0010110

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Shift Input Register contents

0 0 0 (initial state)

1 0 0 0 0

2 1 1 0 0

3 1 1 1 0

4 0 0 1 1

5 1 0 1 1

6 0 1 1 1

7 0 1 0 1 (syndrome s)

8 (input gate off) - 1 0 0 (syndrome s(1)

)

9 - 0 1 0 (syndrome s(2)

)

General Decoding Circuit

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Meggit Decoder

1. Shift r(X) into the buffer B and the syndrome register R simultaneously. Once r(X) is completely shifted in B, R will contain s(X), the syndrome of r(X).

2. Based on the contents of R, the detection circuit yields the output e (0 or 1). the output e (0 or 1).

3. During the next clock cycle:

(a) Add e to the rightmost bit of B while shifting the contents of B. (The rightmost bit of B may be read out). Call the modified content of B: r1

(1)(X).

35

Meggit Decoder (�)

(b) Add e to the left of R while shifting the contents of R.

The modified content of R is s1(1)(X), the syndrome of r1

(1)(X)

[will be shown soon].

4. Repeat steps 2-3 n times.

36

More on Syndrome ComputationLet r(X) = r0 + r1X +….+ r

n-1Xn-1 has the syndrome s(X). Then

r(1)(X) = rn-1 + r0 X + ….+ r

n-2Xn-1 has the syndrome:

s(1)(X) = Rem[r(1)(X)/g(X)].

Define r1 (X) = r(X) + Xn-1 = r0 + r1X +….+ (rn-1+1)Xn-1

The syndrome of r1 (X), call it s1 (X):The syndrome of r1 (X), call it s1 (X):

s1 (X)= Rem[{r(X)+ Xn-1}/g(X)] = s(X) + Rem[Xn-1/g(X)]

r1(1)(X), which is one cyclic shift of r1 (X), has the syndrome

s1(1)(X) = Rem[X s1 (X)/g(X)] = Rem[Xs(X)/g(X)+ Xn/g(X)]

= s(1)(X) + 1 (since Xn +1 = g(X)h(X))

Therefore, adding “1” to the left of R removes the effect of the erroneous digit on the syndrome

37

Worked Example

Consider the (7,4) Hamming code generated by 1+X+X3.

Error pattern Syndrome polynomial. Syndrome vector

X6

1 + X2

1 0 1

X5

1 + X + X2

1 1 1

X4

X + X2

0 1 1

Let c = 1 0 0 1 0 1 1 and r = 1 0 1 1 0 1 1

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X4

X + X2

0 1 1

X3

1 + X 1 1 0

X2

X2

0 0 1

X1

X 0 1 0

X0

1 1 0 0

Cyclic Decoding of the (7,4) Code

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