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By: Nihal Gupta(Ass. Professor)
For the Hamming Code(n,k)
In a systematic code, the first k digits are message bits and
last (n-k) digits are parity check bits.
So for systematic cyclic code word c(x) corresponding to
data polynomial d(x) is given by
c(x)= xn-k d(x)+ P(x)
Were P(x) is:
P(x) = Rem{xn-k d(x)/g(x)}
Example: Construct a systematic cyclic code (7,4) cyclic code
using a generator polynomial g(x)= x3+x2+1 and consider a data
word 1010.
n= 7 and k=4
So n-k=3
If data d= 1010
Then d(x) = x3+x2
So P(x)= Rem{xn-k d(x)/g(x)}
P(x)=Rem{x3(x3+x2)/ x3+x2+1}
P(x)
Hence
c(x)= x3 d(x)+ P(x)
c(x)=x3 * (x3 + x) + 1
c(x) = x6 + x4 + 1
and code word is
C = 1010001
We can construct the code table using the procedure but
this a time taking process so from earlier knowledge,
we will try to make generator matrix g[]
because c[] = d[] * g[]
and g = [ Ik : Pn-k]
so the first thing we will calculate code words for data
vectors 1000, 0100, 0010 & 0001 all are rows of Identity
matrix
so the code words are
Now recognize these four code words are the four rows of G
once we have generator matrix, then code table can be
created using
c = d * g
data Code word
1000 1000110
0100 0100011
0010 0010111
0001 0001101
and there is also another method for forming generator
matrix
as g = [ Ik : Pn-k]
by determining the parity submatrix P
Cyclic code systematic
data code word
0000 00000000
0001 0001101
0010 0010111
0011 0011010
0100 0100011
0101 0101110
0110 0110100
0111 0111001
1000 1000110
1001 1001011
1010 1010001
1011 1011100
1100 1100101
1101 1101000
1110 1110010
1111 111111
Cyclic code systematic

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Cyclic code systematic

  • 1. By: Nihal Gupta(Ass. Professor)
  • 2. For the Hamming Code(n,k) In a systematic code, the first k digits are message bits and last (n-k) digits are parity check bits. So for systematic cyclic code word c(x) corresponding to data polynomial d(x) is given by c(x)= xn-k d(x)+ P(x) Were P(x) is: P(x) = Rem{xn-k d(x)/g(x)}
  • 3. Example: Construct a systematic cyclic code (7,4) cyclic code using a generator polynomial g(x)= x3+x2+1 and consider a data word 1010. n= 7 and k=4 So n-k=3 If data d= 1010 Then d(x) = x3+x2 So P(x)= Rem{xn-k d(x)/g(x)} P(x)=Rem{x3(x3+x2)/ x3+x2+1}
  • 5. Hence c(x)= x3 d(x)+ P(x) c(x)=x3 * (x3 + x) + 1 c(x) = x6 + x4 + 1 and code word is C = 1010001 We can construct the code table using the procedure but this a time taking process so from earlier knowledge, we will try to make generator matrix g[] because c[] = d[] * g[] and g = [ Ik : Pn-k] so the first thing we will calculate code words for data vectors 1000, 0100, 0010 & 0001 all are rows of Identity matrix
  • 6. so the code words are Now recognize these four code words are the four rows of G once we have generator matrix, then code table can be created using c = d * g data Code word 1000 1000110 0100 0100011 0010 0010111 0001 0001101
  • 7. and there is also another method for forming generator matrix as g = [ Ik : Pn-k] by determining the parity submatrix P
  • 9. data code word 0000 00000000 0001 0001101 0010 0010111 0011 0011010 0100 0100011 0101 0101110 0110 0110100 0111 0111001 1000 1000110 1001 1001011 1010 1010001 1011 1011100 1100 1100101 1101 1101000 1110 1110010 1111 111111