help related to CRC -8 checksum calculation method

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tanmoykar

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help related to CRC -8

Can any one help me regarding CRC-8? Now i came accross a CRC-8 implementation : According to this
generator polynomial x^8+x^2+x+1 with procedure
1)All 7 bits over which CRC will be applied, are complemented and become the coefficients of a polynomial, M(x), of degree 6,
2) the remainder, R(x), is calculated from [(M(x) * x^8 + x^7)/G(x)],
3) R(x) is complemented to become the CRC.

Can any one please explain me why X^7 is added? As any other CRC defination includes only 1st part (M(x)*x^8)/G(x). This document provides the reference of ANSI x3.66-1979.but i didn't find this document. Can any one please provide me this document?
 

my take:
ignore the inversion, initial bit stream could have been anything. The multiply by x^8 is to "make space" (7 bits) for the CRC bits, so that the whole poly becomes a multiple of the CRC poly after the CRC bits are appended. That would have been the case if there was no x^7. Since the CRC poly has degree 8, the x^7 will remain after division, thus it amounts to inverting the msb of the 'correct' CRC word. Not sure why all these manipulations are reqd.
- B
 

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