Problem in CRC

Pensu

Pensu

@pensu-8tNeGU • Oct 27, 2024
Hello ppl,
I have a problem regarding CRC codes. I understood that we have a k bit dataword and n bit codeword and n=k+r where r is the number of zeros augmented to dataword to make it a codeword. By looking at some examples i got that r, i.e. number of zeros, is minus one of divisor. If divisor consist of 4 digits then we augment 3 zeros to dataword. But i am not able to get the logic behind it. Can anyone explain it to me?
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  • eternalthinker

    eternalthinker

    @eternalthinker-Kvsi5Y • Jan 25, 2011

    I don't remember much about CRC. But the number of zeroes being one less than divisor has to do with the fact that the remainder of the division will have one bit less.
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  • Pensu

    Pensu

    @pensu-8tNeGU • Jan 25, 2011

    Hmmm......and one more reason might be that the left most bit will always be cancelled, so we will be left with one bit less. Anyways thanks for the reply....😀
    0