R
RCIngham
Guest
[snip]
Unless 'length' is limited, your worst case has header "0000001111111111
(with an extra bit stuffed) followed by 16 * 1023 = 16368 zeros, which wil
have 2728 ones stuffed into them. Total line packet length is 1911
symbols. If the clocks are within 1/19114 of each other, the same number o
symbols will be received as sent, ASSUMING no jitter. You can't assum
that, but if there is 'not much' jitter then perhaps 1/100k will be goo
enough for relative drift to not need to be corrected for.
So, for version 1, use the 'sync' to establish the start of frame and th
sampling point, simulate the 'Rx fast' and 'Rx slow' cases in parallel, an
see whether it works.
BTW, this is off-topic for C.A.F., as it is a system design problem no
related to the implementation method.
---------------------------------------
Posted through http://www.FPGARelated.com
A frame is defined as follows:
- sync :'111'
- header: dtype (4) - n.u.(2) - length (10)
- data : (16) * length
in principle between frames there can be any number of zeros (with bit
stuffing). An 'all zero' pattern in this sense might be of any number of
bits.
[snip]
Unless 'length' is limited, your worst case has header "0000001111111111
(with an extra bit stuffed) followed by 16 * 1023 = 16368 zeros, which wil
have 2728 ones stuffed into them. Total line packet length is 1911
symbols. If the clocks are within 1/19114 of each other, the same number o
symbols will be received as sent, ASSUMING no jitter. You can't assum
that, but if there is 'not much' jitter then perhaps 1/100k will be goo
enough for relative drift to not need to be corrected for.
So, for version 1, use the 'sync' to establish the start of frame and th
sampling point, simulate the 'Rx fast' and 'Rx slow' cases in parallel, an
see whether it works.
BTW, this is off-topic for C.A.F., as it is a system design problem no
related to the implementation method.
---------------------------------------
Posted through http://www.FPGARelated.com