[802.3ae_Serial] MIME-Version: 1.0
Mike,
Your ability to draw in ASCII has always been very impressive. This is
one of your best yet, and clearly illustrates your points. Thanks.
The latest XAUI jitter values have changed a bit from your example.
Jitter ouput at the end of the channel can now be 0.47 UI DJ and 0.65 UI
TJ. 0.10 UI of those is due to "other", which I am willing to assume to
be mean-centered; for tolerance testing, that 0.1 UI is done with SJ,
which is definitely mean-centered. So, to relate to your response below,
the relevant values should be 0.37 UI DJ and 0.65 UI TJ.
DJ/2+TJ/2=0.51
I am not going to push the DJ/2 factor too hard - it's a limit from
which we may want to back off a little for this discussion. However, in
any case, we probably still get to a value that is not far from 0.50,
and that concerns me. I always thought CDRs needed approx 0.3 UI pk-pk
to cover those items.
So, 2 questions to CDR folks (including you, Mike) are:
1. How much room is needed for Rx sampling uncertainty/noise and offet?
2. Offset from what - is the normal sampling point typically biased 0.5
UI from the means (or medians) of the crossings?
To the latter, Mike's points about finite sensitivity and rise/fall
times make sense, and I can see where they will create an offset from
the bias point. I agree with Mike that this should help, but it also
seems somewhat orthogonal to the jitter/pdf questions.
Thanks, Tom
Tom Lindsay
Stratos Lightwave
425/672-8005
-----Original Message-----
From: Mike Jenkins [mailto:jenkins@lsil.com]
Sent: Thursday, June 14, 2001 12:47 PM
To: 802.3ae Serial PMD
Subject: [Fwd: Re: XAUI update from Interim meeting]
Pat,
Thanks much for your detailed response. Below, I will amplify on it
to make my point in better detail. For what it's worth, I was part
of the Fibre Channel group (MJS) on whose work a lot of the Ethernet
jitter specs are founded. Also, I work for a serdes manufacturer,
so I'm quite motivated to have specs that work.
Regards,
Mike Jenkins
pat_thaler@agilent.com wrote:
>
> Mike,
>
> I find it difficult to understand how a receiver can work with a
floating
> mask with the center of the transition allowed to occur over a 0.6 UI
> window. Consider the following pathological transmitter case:
>
> 999 of 1000 transitions out of the transmitter have jitter that falls
> withing a .1 UI window
> 1 out of 1000 transitions falls .5 UI after the .1 UI window.
That's the reason for existence of the deterministic jitter
(DJ) component of total jitter (TJ). DJ is the pathological
part and it is limited to 0.36 UI. The remainder of the
jitter distribution is comprised of random jitter (Gaussian
with zero mean).
>
> This transmitter's output would meet a 0.6 UI floating window spec.
Below, I have recast your example with the bimodal distribution
limited to 0.36 UI, convolving it with random jitter to get to
0.60 UI total Jitter. I sketched in the distribution under
the far end template. (Hopefully, you're viewing this with
fixed width font.) The sketch is roughly to scale.
>
> A receiver PLL locking to that signal is going to locate the bit cell
edge
> somewhere within the .1 UI window. The receiver will sample the level
half a
> UI from where it believes the bit cell center is. When the transition
occurs
> .5 UI after the window it will occur after the sampling point and an
error
> will occur.
_____ ______
\ / \
\ / \
\ :<--- TJ/2 --->:<--- TJ/2 --->: / \
\: : :/ \
\ : / \
/ : \ /
/: : :\ /
/ : 0 UI : \ /
/ : : : \ /
______/ : :< DJ/2 >:< DJ/2 >: : \______/
: * : :
: * : :
: *** : :
: *** : :
: ***** * :
: ******* *** :
************ *********
**************** **************
^
|
~mean
The mean of this jitter will be just to the right of the left
peak (and very close for your 999-to-1 example. Now, if the
receiver lined up on this mean, the most trailing edge would
be at DJ/2 + TJ/2 = 0.18 + 0.3 = 0.48 UI. Close, but no error.
Below, I will go into why it's not even quite this extreme.
HOWEVER, if the spec requires the waveform to be lined up
such that the mean is at 0 UI, that pushes the right peak of
the distribution inside the template (hexagon), forcing it to
fail the test. This is why I'm railing against this change.
>
> Now this case may be a bit extreme, but I have seen signals in a
copper
> system where most of the transitions were relatively tightly grouped
and
> there were a few outlying transitions to one side so it isn't all that
> unusual.
I agree. What is more, the typical low-pass mechanisms in
copper (and my polemic is solely addressed to copper specs)
cause the tails to be skewed something like this example,
with the mean somewhat left of center between the extremes.
>
> The receive PLL will locate the average bit cell edge. Jitter affects
the
> receiver to the extent that edges are skewed from the average bit cell
edge.
> The receiver's ability to withstand jitter doesn't depend on the
center of
> the distribution of edges. It depends on the distance of edges from
the
> mean.
A receiver might use mean or median or some other statistics.
I'm not prepared to discuss such details. However, any real
receiver has a finite sensitivity and is presented with finite
input risetime. This causes the effective transition time to
be somewhat later than (i.e., to the right of) the point marked
by histograms on twenty-something GHz 'scopes.
For the above reasons:
* Adjusting the waveform in time relative to the template is
very appropriate, and
* Forcing the mean to be positioned at the 0 UI point of the
template will break that spec.
I hope you agree, and will join me in trying to remove this
recent change to P802.3ae. I'd appreciate your comments.