Understanding two different ways of implementing CRC generation with LFSR

Question

There are two ways of implementing CRC generation with linear feedback shift registers (LFSR), as shown in this figure

. The coefficients of generator polynomial in this picture are 100111, and the red "+" circles are exclusive-or operators. The initialization register values are 00000 for both.

For example, if the input data bit stream is 10010011, both A and B will give CRC checksum of 1010. The difference is A finishes with 8 shifts, while B with 8+5=13 shifts because of the 5 zeros appended to the input data. I can understand B very easily since it closely mimics the modulo-2 division. However, I can not understand mathematically how A can give the same result with 5 less shifts. I heard people were talking A took advantage of the pre-appending zeros, but I didn't get it. Can anyone explain it to me? Thanks!

Answer 1

Here is my quick understanding.

Let M(x) be the input message of order m (i.e. has m+1 bits) and G(x) be the CRC polynomial of order n. CRC result for such a message is given by

C(x) = (M(x) * xⁿ) % G(x)

This is what the circuit B is implementing. The additional 5 cycles it takes is because of the xⁿ operation.

Instead of following this approach, circuit A tries to do something smarter. Its trying to solve the question

If C(x) is the CRC of M(x), what would be the CRC for message {M(x), D}

where D is the new bit. So its trying to solve one bit at a time instead of entire message as in case of circuit b. Hence circuit A will take just 8 cycles for a 8 bit message.

Now since you already understand why circuit B looks the way it does, lets look at circuit A. The math, specifically for your case, for the effect of adding bit D to message M(x) on CRC is as below

Let current CRC C(x) be {c4, c3, c2, c1, c0} and new bit that is shifted in be D
NewCRC = {M(x), D}*x⁵) % G(x) = (({M(x), 0} * x⁵) % G(x)) XOR ((D * x⁵) % G(x))
which is ({c3, c2, c1, c0, 0} XOR {0, 0, c4, c4, c4}) XOR ({0, 0, D, D, D})
which is {c3, c2, c1^c4^D, c0^c4^D, c4^D}

i.e. the LFSR circuit A.