Problem in CRC
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?
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?
Replies
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eternalthinkerI 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. -
PensuHmmm......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....😀
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