III Cyclic Codes Generator Matrix of Cyclic Codes

III. Cyclic Codes

Generator Matrix of Cyclic Codes Any Codeword is a Linear Combination of: g(X), Xg(X), …, Xk-1 g(X) Example: (7, 4) Cyclic Code with g(X)=1+X+X 3 Could be converted to systematic form with the help of row operations © Tallal Elshabrawy 2

Parity Check Matrix of Cyclic Codes (Xn+1) is divisible by g(X), i. e. , let (Xn+1)=g(X)h(X) where h(X)=h 0+h 1 X+h 2 X 2+…+hk. Xk Therefore v(X)h(x) = a(X)g(X)h(X) = a(X)(Xn+1) The degree of a(X)≤k-1, therefore, The powers Xk, Xk+1, …, Xn-1 do not appear in a(X)(Xn+1) v(X)h(X)=(v 0+v 1 X+…vn-1 Xn-1) (h 0+h 1 X+…hk. Xk), therefore vkh 0 +vk-1 h 1+ … + v 0 hk =0 vk+1 h 0+vkh 1 + … + v 1 hk =0. vn-1 h 0+vn-2 h 1+ … + vn-k-1 hk=0 © Tallal Elshabrawy 3

Parity Check Matrix of Cyclic Codes Example: (7, 4) Cyclic Code with g(X)=1+X+X 3 h(X)=(X 7+1 )/g(X) h(X)=1+X+X 2+X 4 © Tallal Elshabrawy 4

Cyclic Codes in Systematic form Consider an (n, k) cyclic code g(X) is the generator polynomial u(X) is a message polynomial of degree<k Xn-ku(X)=u 0 Xn-k+u 1 Xn-k+1+… +uk-1 Xn-1 Divide Xn-ku(X) by g(X), Therefore Xn-ku(X)=a(X)g(X)+b(X) The degree of b(X) must be n-k-1 or less, i. e. , b(X)=b 0+b 1 X+…+bn-k-1 Xn-k-1 Rearranging: b(X)+Xn-ku(X)=a(X)g(X) b(X)+Xn-ku(X) is a multiple of g(X) and therefore is a code polynomial in the cyclic code b(X)+Xn-ku(X) corresponds to the codeword: (b 0, b 1, …, bn-k-1, u 0, u 1, …, uk-1) The code is systematic © Tallal Elshabrawy 5

Example Consider a (7, 4) cyclic code with g(X)=1+X+X 3 Given u=(1 0 0 1). Find the corresponding codeword if l The code is nonsystematic v(X)=u(X)g(X)=(1+X 3)(1+X+X 3)=1+X+X 4+X 6 v=(1 1 0 0 1) l The code is systematic X 3 u(X)=X 3+X 6. Dividing by g(X). The remainder b(X)=X 2+X v(X)=b(X)+X 3 u(X)=X+X 2+X 3+X 6 V=(0 1 1 1 0 0 1) © Tallal Elshabrawy 6

Encoding of Cyclic Codes in Systematic Form Three Steps: 1. Multiply the message polynomial u(X) by Xn-k 2. Dividing Xn-ku(X) by g(X) to obtain the remainder b(X) 3. Forming the codeword b(X)+Xn-ku(X) Encoding Circuit is a Division Circuit Gate g 2 g 1 b 0 + b 1 + gn-k-1 b 2 . . + bn-k-1 Xn-ku(X) + Codeword Information Bits Parity Check Bits © Tallal Elshabrawy 7

Encoding Circuit Example: Implementation Encoding Circuit of (7, 4) Cyclic Code with g(X)=1+X+X 3 Gate b 0 + Assume u=(1 0 1 1) Input Register Contents b 1 b 2 Xn-ku(X) + Codeword Information Bits Parity Check Bits 0 0 0 (Initial State) 1 1 1 0 (First Shift) 1 1 0 1 (Second Shift) 0 1 0 0 (Third Shift) 1 1 0 0 (Fourth Shift) © Tallal Elshabrawy Codeword: (1 0 0 1 1) 8

Encoding Circuit Example: Verification u=(1 0 1 1) u(X) = 1+X 2+X 3 1. X 3 u(X) = X 3+X 5+X 6 2. X 3 u(X)/g(X) = (1+X+X 2+X 3) +1/g(X) b(X) = 1 3. v(X) = 1+X 3+X 5+X 6 v = (1 0 0 1 1) © Tallal Elshabrawy 9

Syndrome Polynomial and Error Detection Let Received Polynomial represent the received vector r(X)=a(X)g(X)+s(X) is called the syndrome polynomial Error Detection All codewords MUST BE divisible by g(X). i. e. , s(X) is identical to zero if and only if r(X) is a code polynomial If s(X)≠ 0 An Error Pattern is Detected © Tallal Elshabrawy 10

Syndrome Circuit is a Division Circuit Gate g 2 g 1 gn-k-1 r(X) + Received Vector © Tallal Elshabrawy s 0 + s 1 + s 2 . . + sn-k-1 11

Encoding Circuit Example: Implementation Syndrome Circuit of (7, 4) Cyclic Code with g(X)=1+X+X 3 Gate + s 0 + s 1 s 2 Assume r=(0 0 1 1 0) Input Register Contents 0 0 0 (Initial State) 0 000 1 110 0 011 1 011 0 101 © Tallal Elshabrawy Syndrome: (1 0 1) 12