Re: Two questions about USB CRC

On 8月10日, 下午4时11分, riderofgiraffes <mathforum..org...@xxxxxxxxxxxxxx>
wrote:
"Without the remainder bit-wise inverted,
trailing zeros at the end of a packet could
not be detected as data transmission errors."

As you say, two streams with trailing zeros
at the end could have the same encoding results.
For example, two streams, one is 10000000, and
one is 1000000000000000000. The shift register
is preloaded with all 1s.
The result of the former is 10000.
The result of the latter is 11100.
I've got the different results.
Could you please share your idea about it with me?

When the transmitter computes the CRC it is appended
to the data stream. If you use an uninverted remainder
then the result of the check on reception should be 0.
You don't compute the CRC of only the payload and then
compare, you simply compute the CRC of the entire data
stream, including the CRC. That way you can compare
against a constant, which is hardware more efficient.

If on transmission there are trailing 0s (after the
appended CRC) these won't be detected by the decoding.

If you invert the remainder before appending then the
final answer will no longer be 0, but should be the
same on all transmissions. You can precalculate this
as described in the paper, or you can simply run a few
blocks of data through the system and see that the
result is always the same.

Try it. Try several examples. Notice how with the
reminder uninverted, computing the CRC of the entire
stream is zero. Then append the inverted CRC and see
that the result is always the same.

Thank you for your answer. I'm very appreciated for your detailed
explanations.
Let's go back to the first topic.
As you say, the shift register is preloaded with 0s
and then encode 00000000 00000000 000000001, the same
answer is got as encoding 00000001.

But even if the shift register is preloaded with 1s
and then encode 00000000 00000000 00000000 00000000 000000001,
the same answer is got as encoding 000000001, because
the period of the CRC5 shift register (x^5+x^2+1) is 32.

Could you please share your idea about it with me?
.

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