E.g. Consider the parity bit P1 and we have to find the position of message bits which we’ll cover with this parity bit.
Firstly write the binary equivalents of positions of message bit
Bit1 bit2 bit3 bit4 bit5 bit6 bit7
Parity parity parity
P1 P2 M1 P3 M2 M3 M4
001 010 011 100 101 110 111
Now let’s see in the binary equivalent of position of parity bit P1 that at which position we have 1and we see 1 is at LSB so we select the message bits which have positions with 1 at LSB which are M1, M2 and M4 So P1 bit would check the parity for M1, M2 and M4
E.g. Consider the parity bit P2 and we have to find the position of message bits which we’ll cover with this parity bit
We have 1 at second position from left so we choose message bits which have 1 at 2nd position n their position’s binary equivalent. Hence we get message bits M1 M3 and M4. So P2 checks parity for message bits of M1 M3 and M4
Similarly we have 1 at 3rd position of P3 message bits with 1 at 3rd position are M2M3 M4
So we now have
P1 Checks bit number 1,3,5,7
P2 Checks bit number 2,3,6,7
P3 Checks bit number 4,5,6,7
These parity bits can be either even or odd parity bits but all parity bits must be same i.e. all odd or all even
Your article helped me a lot, is there any more related content? Thanks!
Can you be more specific about the content of your article? After reading it, I still have some doubts. Hope you can help me.