serial protocol specs and verification

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Unless 'length' is limited, your worst case has header "0000001111111111" (with an extra bit stuffed) followed by 16 * 1023 = 16368 zeros, which will have 2728 ones stuffed into them. Total line packet length is 19113 symbols. If the clocks are within 1/19114 of each other, the same number of symbols will be received as sent, ASSUMING no jitter. You can't assume that, but if there is 'not much' jitter then perhaps 1/100k will be good 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 the sampling point, simulate the 'Rx fast' and 'Rx slow' cases in parallel, and see whether it works.

BTW, this is off-topic for C.A.F., as it is a system design problem not related to the implementation method.

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Reply to
RCIngham
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or do like many USB PHYs, assume the 2x clock in reasonably 50/50 and use a DDR input flop to sample at 4x

-Lasse

Reply to
langwadt

Yes, that would be interesting to design actually, the logic gets two bits at the same time rather than one bit, I guess it makes the machine a bit more complicated in that you have to deal with four states and four possible input combinations. Still, not a big deal, just a bit of work on paper to understand the logic needed.

--

Rick
Reply to
rickman

On 31/07/2013 13:44, rickman wrote: []

A data stream which is *exactly* flowing with a frequency f can be

*exactly* sampled with a clock frequency f, it happens continuously in your synchronous logic. What happened to Nyquist theorem?

If you have a protocol with data and clock, does it mean that you will recognize only half of the bits because your clock rate is just equal to your data rate? I'm confused...

IMO calling a signal 'asynchronous' does not make any difference. Mr. Nyquist referred to reconstructing an analog signal with a discrete sampling (no quantization error involved). How does that applies to digital transmission?

It does not work not because of Nyquist limit, but because the recovery of a phase shift cannot be done with just two clocks per bit.

[]

This is what I meant indeed. I believe I confused DPLL with ADPLL...

Reply to
alb

may not technically be Nyquist limit, but like so many things in nature the same relations are repeated

and if you take NRZ you'll notice that the highest "frequency" (0101010101..) is only half of the data rate

-Lasse

Reply to
langwadt

Why you excluded the sync symbol?

If the clocks are within 1/19114 of each other, the same number of

5*10e-5 is a very large difference. We are using 0.5 ppm oscillators. The amount of symbols received has to take into account phase shift otherwise bits will be lost or oversampled.

Still not sure what you are trying to say.

by saying 'in parallel' you mean a data stream with some bits slower and some faster? I think the main problem lies on the slight difference in clock frequencies which lead to increasing phase shift to the point where a bit is lost or oversampled.

IMO is an implementation issue, no specs will tell me how many times I need to sample the data stream. The system design does not have a problem IMO, it simply specify the protocol between two modules. But I will be more than happy if you could point me out to some more appropriate group.

Reply to
alb

Since you can resynchronize your sampling clock on each transition received, you only need to "hold lock" for the maximum time between transitions, which is 7 bit times. This would mean that if you have a nominal 4x clock, some sample points will be only 3 clocks apart (if you are slow) or some will be 5 clocks apart (if you are fast), while most will be 4 clock apart. This is the reason for the 1 bit stuffing.

Reply to
Richard Damon

"0000001111111111"

will

of

good

the

and

The bit-stuffing in long sequences of zeroes is almost certainly there to facilitate a conventional clock recovery method, which I am proposing not using PROVIDED THAT the clocks at each end are within a sufficiently tight tolerance. Detect the ones in the as-sent stream first, then decide which are due to bit-stuffing, and remove them.

Deciding how tight a tolerance is 'sufficiently tight' is probably non-trivial, so I won't be doing it for free.

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Reply to
RCIngham

On 01/08/2013 11:56, RCIngham wrote: []

What is the gain of not using 'conventional clock recovery'?

Reply to
alb

I think the point is that if the sequences are short enough that the available timing tolerance is adequate, then you just don't need to recover timing from the bit stream.

I've been looking at this, then working on other issues and have lost my train of thought on this. I believe that a PLL (or DPLL) is not needed as long as the input can be sampled fast enough and the reference frequency is matched closely enough. But it is still important to correct for "phase" as the OP puts it (IIRC) so that you can tell where the bits are and not sample on transitions, just like a conventional UART does it. We frequent enough transitions, the phase can be detected and aligned while the exact frequency does not need to be recovered.

--

Rick
Reply to
rickman

Yes, you are right about the rates. I was not thinking of this correctly. The Nyquist theorem looks at *frequency* content which is not the same as bit rate.

I am not familiar with ADPLL. What is that?

--

Rick
Reply to
rickman

Since a 4x clock allows for a 25% data period correction, and we will get an opportunity to do so every 7 data periods, we can tolerate about a 25/7 ~ 3% error in clock frequency. (To get a more exact value we will need to know details like jitter and sampling apertures, but this gives us a good ball-park figure). Higher sampling rates can about double this, the key is we need to be able to know which direction the error is in, so we need to be less than a 50% of a data period error including the variation within a sample clock.

To try to gather the data without resynchronizing VASTLY decreases your tolerance for clock errors as you need to stay within a clock cycle over the entire message.

The protocol, with its 3 one preamble, does seem like there may have been some effort to enable the use of a PLL to generate the data sampling clock, which may have been the original method. This does have the advantage the the data clock out of the sampler is more regular (not having the sudden jumps from the resyncronizing), and getting a set a burst of 1s helps the PLL to get a bit more centered on the data. My experience though is that with FPGAs (as would be on topic for this group), this sort of PLL synchronism is not normally used, but oversampling clocks with phase correction is fairly standard.

Reply to
Richard Damon

On 02/08/2013 06:19, rickman wrote: []

It is an All Digital PLL:

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all the elements of a PLL are implemented in the digital domain.

Reply to
alb

Hi Lasse,

On 01/08/2013 00:03, snipped-for-privacy@fonz.dk wrote: []

A signal traveling on a physical channel (be it on a cable, a PCB route, an FPGA interconnection...) will have sharp transitions at the beginning of its journey and sloppier ones at the end due to losses, but if you take a comparator and discriminate a '1' or '0', then you do not 'need' higher frequencies than half the data rate itself (or symbol rate to be precise). If you take a sinusoidal waveform and put a threshold at 0, then you have two symbols per cycle.

Why sampling at the data rate is not sufficient then? Because there are several other factors. First of all encoding and decoding are processes which do introduce 'noise' as well as 'limitations'. Having a comparator to discriminate 0/1 does introduce noise in the time of transaction, therefore distorting the phase of the signal. The medium itself might be source of other jitter since it is sensitive to the environment (temperature, pressure, humidity, ...).

TRANSMITTER MEDIUM RECEIVER

+-------------------+ +-------------------+ | +---| |---+ | | '10100101' -> |ENC| -\/\/\/\/->|DEC| -> '10101110' | | +---| physical |---+ | +-------------------+ signal +-------------------+ ^ ^ | | +-----+ +-----+ | clk | | clk | +-----+ +-----+

You do not care about reconstructing a physical signal (like in ADC sampling), you *do* care about reconstructing a data stream. Another source of troubles are the two clock generators on the TX and RX. They cannot be assumed to be perfectly matching and any difference will lead to a phase drift which eventually will spoil your data sampling.

that is why a clock frequency = to data rate is sufficient to 'sample' the information.

the NRZ is a line code, i.e. a translation of your data stream with appropriate physical signal (light, current, sound, ...) for the chosen physical medium (fiber, cable, air, ...) and has nothing to do with a toggling bit.

Reply to
alb

Hi Richard,

On 02/08/2013 06:22, Richard Damon wrote: []

According to your math it looks like a 2x clock allows for a 50% data period correction and therefore a 50/7 ~6% error in clock frequency, which seems to me quite counter intuitive... Am I missing something?

[]

This is indeed what I'm looking for, oversampling (4x or 8x) and phase correct.

Reply to
alb

which

number

assume

parallel,

not

you

to

not

tight

which

Some form of clock recovery is essential for continuous ('synchronous') data streams. It is not required for 'sufficiently short' asynchronous data bursts, the classic example of which is RS-232. What I am suggesting is that the OP determines - using simulation - whether these frames are too long given the relative clock tolerances for a system design without clock recovery.

As I previously noted, this is first a 'system design' problem. Only after that has been completed does it become an 'FPGA design' problem.

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Reply to
RCIngham

I guess I wasn't aware that a digital PLL wasn't *all* digital. That is what I have been referring to as digital.

--

Rick
Reply to
rickman

I don't think the frame length is the key parameter, rather it is the 6 zero, one insertion that guarantees a transition every 7 bits.

--

Rick
Reply to
rickman

On 02/08/2013 16:16, rickman wrote: []

you might find this article interesting:

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Reply to
alb

The details are that for a Nx sampling clocks, every time you see a clock, you can possibly shift N/2-1 high speed clock cycles every adjustment. For example, with a 16x clock, you can correct for the edge being between -7 and +7 sampling clocks from the expected point. If it is 8 clocks off, you don't know if is should be +8 or -8, so you are in trouble. If N is odd, you can possibly handle (N-1)/2 cycles. (Note that this assumes negligible jitter.) So our final allowable shift in data clocks is (N/2-1)/N which can also be written as 1/2-1/N, which leads to my 6% for N large (50% correction) and 3% for N=4. For N-2 this gives us 0%.

Reply to
Richard Damon

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