Ask abt parity check matrix of linear code

[honhungoc]

When i solve problems, i face this expression but i dont know what it imply

Anyone please help me

a (6,3) linear code is constructed as follows

bit 2 is a parity check on bits 5 and 4
bit 1 is a parity check on bits 4 and 3
bit 0 is a parity check on bits 5 and 3

Thank u very much

 

[bulx]

let transmitted word = [t5, t4, t3, t2, t1, t0]
let noise = [n5, n4, n3, n2, n1, n0]
rcvd word = [t5 Xn5, t4 X n4, t3 X n3, t2 X n2, t1 X n1, t0 X n0] ( X = XOR)

we find syndromes at reciever:
S1 = r2 X r5 X r4
S2 = r1 X r4 X r3
S3 = r0 X r5 X r3

take S1 for example
S1 = (t2 X n2) X (t5 X n5) X (t4 X n4)

but from encoding scheme
t2 X t5 X t4 = 0

so

S1 = n2 X n5 X n4

Now assume only one noise bit can be zero (capacity of code, more might not be corrected)

If n5 is 1, then S1 and S3 will be one, S2 will be zero
If n4 is 1, then S1 and S2 will be one, S3 will be zero
If n3 is 1, then S2 and S3 will be one, S1 will be zero
If n2 is 1, then S1 will be one, S2 and S3 will be zero
If n1 is 1, then S2 will be one, S1 and S2 will be zero
If n0 is 1, then S3 will be one, S1 and S2 will be zero

So any 1 bit error can be corrected

You can also see if any two noise bits are one, at least one syndrome will be on, so error can be detected, but not corrected, beyond 2 only some error patterns can be detected. hope it is clear

The trick is that each redundant bit looks at two bits, and every bit is looked at at twice. Thats how the redundant bits are formed.
-b