Cyclic codes error detection

[Arhel]

Hi guys!
Could you please help me with next question:
How to plot probability of error detection as a function of the Hamming eight?(hamming weight = 1,2,3,...,31), like barchart. For cyclic code

Many thanks!

 

[andre_luis]

I done in MATLAB a routine to determine to the Hamming(7,4,1) code, the amount of syndromes equal to Zero.
You could perform some few changes to fit to your code.

% Program to generate all the input error patterns from 0000000 to 1111111
% and then calculate corresponding syndromes

d =[ 0 0 0 0 0 0 0 ] ;
h =[ [ 1 0 0 1 0 1 1 ] ; [ 0 1 0 1 1 1 0 ] ; [ 0 0 1 0 1 1 1 ] ] ;
ht =h’ ;
for x = 0 : 127
% pause
c=1 ;
for i = 7 : -1 : 1
a=d( i );
d( i ) = xor ( a , c) ;
c = a & c ;
end ;
d
s = d * ht
end ;

The following input patterns generated null syndrome :

0 0 0 0 0 0 0
0 0 0 1 1 0 1
0 0 1 0 1 1 1
0 0 1 1 0 1 0
0 1 0 0 0 1 1
0 1 0 1 1 1 0
0 1 1 0 1 0 0
0 1 1 1 0 0 1
1 0 0 0 1 1 0
1 0 0 1 0 1 1
1 0 1 0 0 0 1
1 0 1 1 1 0 0
1 1 0 0 1 0 1
1 1 0 1 0 0 0
1 1 1 0 0 1 0
1 1 1 1 1 1 1

Due pattern 0 0 0 0 0 0 0 could not be accounted on the calculus, we should eliminate from analysys.
So, the probability of detection in the case of Hamming(7,4,1) code was : 1- (16-1)/(128) = ~89%

+++

 

[Arhel]

Thanks!!!
So , the probability of the detection is equal to 1 - (syndrome - 1)/(input sequence) ?

 

[andre_luis]

Thanks!!!
So , the probability of the detection is equal to 1 - (syndrome - 1)/(input sequence) ?

For inference, I would say :

1 - ( ( (# of syndrome equal to zero) - 1)/(# of input sequence) )

+++

 

[andre_luis]

It sounds like if it is matching the following rule to the Hamming(7,4,1) code:

DETECTION_PROBABILITY = 1 - ( ((4)^2-1) / ((7)^2) )

Chek if it is true.


+++

 

[Arhel]

Mny thanks!
Also, what will happend if there are several different Hamming weights?