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Cyclic codes
By: Nihal Gupta(Ass. Professor)
Cyclic codes
• Definition
• Cyclic codes are subpart of linear block code
• It follows two properties:
1. Linear Property
If we have two code words Ci, Cj
Cp= Ci+Cj
where Cp must be a code word
2. Cyclic Shifting
Code word = (C1,C2,C3,…….Cn,)
after shifting left or right by any number of bits
resultant should be a code word.
Example: code C={0000, 0110, 1001, 1111} is it cyclic codes?
So we will follow two properties
1. Linearity Property
0110 0110 1001
+1001 +1111 +1111
1111 1001 0110
(Here + represents modulo 2 sum)
all code-words are available so we can observe that it follows linearity property.
2. Cyclic Shifting
Let me take a code word 0 1 1 0
And do the left shifting
It clearly shows that shifted code does not exist so it isn’t a cyclic code.
0 0 1 1
Question: {0000, 0101, 1010, 1111} is it a cyclic code?
Cyclic codes divided into two categories:
Systematic Cyclic Codes (message & parity bits are fixed)
Non- Systematic Cyclic Codes (message & parity bits are random)
Non-systematic Cyclic Codes
Cyclic Code word can be described in polynomial form
𝑐 𝑥 = 𝑐1(𝑥 𝑛−1)+𝑐2 𝑥 𝑛−2 + 𝑐3 𝑥 𝑛−3 + … … … . . +𝑐 𝑛
And the code polynomial 𝑐 𝑖
(𝑥) for the code vector 𝑐 𝑖
is
𝑐 𝑖
(𝑥) = 𝑐𝑖+1(𝑥 𝑛−1
)+𝑐𝑖+2 𝑥 𝑛−2
+ 𝑐𝑖+3 𝑥 𝑛−3
+ ⋯ + 𝑐 𝑛 𝑥 𝑖
+ 𝑐1 𝑥 𝑖−1
+ ⋯ … + 𝑐𝑖
If we divide 𝑥. 𝑐 𝑖
(𝑥) from 𝑥 𝑛
+ 1 the remainder will give us the next code word 𝑐 𝑖
(𝑥)
From last slide we can conclude that g(x) will exits under xn +1
A cyclic code polynomial c(x) can be generated by the data polynomial d(x) of
degree k-1 and a generator polynomial g(x) of degree n-k as
c(x) = d(x).g(x)
Where g(x) is an (n-k)th order factor of (xn + 1)
for a data vector (d0,d1,d2,…..,dn) , the data polynomial is
𝑑 𝑥 = 𝑑1 𝑥 𝑘−1 + 𝑑2 𝑥 𝑘−2+𝑑3 𝑥 𝑘−3 + ⋯ … . +𝑑 𝑘
So c(x) = d(x).g(x)
= 𝑑1 𝑥 𝑘−1
𝑔 𝑥 + 𝑑2 𝑥 𝑘−2
𝑔(𝑥)+𝑑3 𝑥 𝑘−3
𝑔(𝑥) + ⋯ … . +𝑑 𝑘 𝑔(𝑥)
Example: Find a generator polynomial g(x) for a (7,4) cyclic code, and find code vectors
for the following data vectors 1010, 1111, 0001 and 1000.
Hence,
c = 1110010
Similarly, code words for other data words can be found
From the table we can easily identify the structure of code words. The first k digits are
not necessarily the data digits. Hence this is not a systematic code.
Data (d) Code word (c)
1010 1110010
1111 1001011
0001 0001101
1000 1101000
Cyclic code non systematic

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Cyclic code non systematic

  • 1. Cyclic codes By: Nihal Gupta(Ass. Professor)
  • 2. Cyclic codes • Definition • Cyclic codes are subpart of linear block code • It follows two properties: 1. Linear Property If we have two code words Ci, Cj Cp= Ci+Cj where Cp must be a code word 2. Cyclic Shifting Code word = (C1,C2,C3,…….Cn,) after shifting left or right by any number of bits resultant should be a code word.
  • 3. Example: code C={0000, 0110, 1001, 1111} is it cyclic codes? So we will follow two properties 1. Linearity Property 0110 0110 1001 +1001 +1111 +1111 1111 1001 0110 (Here + represents modulo 2 sum) all code-words are available so we can observe that it follows linearity property. 2. Cyclic Shifting Let me take a code word 0 1 1 0 And do the left shifting It clearly shows that shifted code does not exist so it isn’t a cyclic code. 0 0 1 1
  • 4. Question: {0000, 0101, 1010, 1111} is it a cyclic code? Cyclic codes divided into two categories: Systematic Cyclic Codes (message & parity bits are fixed) Non- Systematic Cyclic Codes (message & parity bits are random) Non-systematic Cyclic Codes Cyclic Code word can be described in polynomial form 𝑐 𝑥 = 𝑐1(𝑥 𝑛−1)+𝑐2 𝑥 𝑛−2 + 𝑐3 𝑥 𝑛−3 + … … … . . +𝑐 𝑛 And the code polynomial 𝑐 𝑖 (𝑥) for the code vector 𝑐 𝑖 is 𝑐 𝑖 (𝑥) = 𝑐𝑖+1(𝑥 𝑛−1 )+𝑐𝑖+2 𝑥 𝑛−2 + 𝑐𝑖+3 𝑥 𝑛−3 + ⋯ + 𝑐 𝑛 𝑥 𝑖 + 𝑐1 𝑥 𝑖−1 + ⋯ … + 𝑐𝑖 If we divide 𝑥. 𝑐 𝑖 (𝑥) from 𝑥 𝑛 + 1 the remainder will give us the next code word 𝑐 𝑖 (𝑥)
  • 5. From last slide we can conclude that g(x) will exits under xn +1 A cyclic code polynomial c(x) can be generated by the data polynomial d(x) of degree k-1 and a generator polynomial g(x) of degree n-k as c(x) = d(x).g(x) Where g(x) is an (n-k)th order factor of (xn + 1) for a data vector (d0,d1,d2,…..,dn) , the data polynomial is 𝑑 𝑥 = 𝑑1 𝑥 𝑘−1 + 𝑑2 𝑥 𝑘−2+𝑑3 𝑥 𝑘−3 + ⋯ … . +𝑑 𝑘 So c(x) = d(x).g(x) = 𝑑1 𝑥 𝑘−1 𝑔 𝑥 + 𝑑2 𝑥 𝑘−2 𝑔(𝑥)+𝑑3 𝑥 𝑘−3 𝑔(𝑥) + ⋯ … . +𝑑 𝑘 𝑔(𝑥)
  • 6. Example: Find a generator polynomial g(x) for a (7,4) cyclic code, and find code vectors for the following data vectors 1010, 1111, 0001 and 1000.
  • 7. Hence, c = 1110010 Similarly, code words for other data words can be found From the table we can easily identify the structure of code words. The first k digits are not necessarily the data digits. Hence this is not a systematic code. Data (d) Code word (c) 1010 1110010 1111 1001011 0001 0001101 1000 1101000