RE: [usb] crc algorithm

[Bo <bo_ye@yahoo.com>]

Hi Kwanie,
     Thanks for the diagram. 
     I agree on the second part. Xor the X0 with
the MSB will yield the correct when the last bit
is shifted in. But need to preset all the register
to 1's. (correct me if I am wrong) 
     The first part, looks like using two-string
feed back instead. It will result in
Y = 1 + X^3 + X^5 instead of Y = 1 + X^2 + X^5 in
USB protocol for token packet checksum.

regards
bo
 

--- "kw@nie" <kwanie@pacific.net.sg> wrote:
> I think I know why the binary representations, in
> the USB specification, are
> different from the polynomials. Attached is a brief
> explanation based on the
> 5-bit CRC but the principle applies for both CRC
> generators.
> 
> Cheers,
> kwanie
> 
> 
> 
> 
>  -----Original Message-----
> From: 	owner-usb@opencores.org
> [mailto:owner-usb@opencores.org]  On Behalf
> Of kebloo@mitra.net.id
> Sent:	21 February 2002 08:34
> To:	usb@opencores.org
> Subject:	[usb] crc algorithm
> 
> Hello,
> 
> Sorry for this beginner question, but I wish some
> one could
> answer my curiosity...
> 
> USB 1.1 specification defines generator polynomial
> of 5-bit crc
> as G(x) = x^5+x^2+1 and binary pattern that
> represent it as
> 00101B and 16-bit crc as G(x) = x^16+x^15+x^2+1 and
> binary pattern that represent it as
> 1000000000000101B.
> My question is why the binary pattern stated like
> that and
> not 100101B for 5-bit crc and 11000000000000101B ?
> Thanks....
> 
> 
> 
> Best Regard,
> 
> 
> RE Tambunan
> --
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> 

> ATTACHMENT part 2 image/jpeg name=crc5.JPG

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