[802.3ae_Serial] RE: peak to peak jitter

["Lindsay, Tom" <tlindsay@stratoslightwave.com>]

Petar - 

Here are my thoughts.

From a pk-pk view, 99% should fall within and 1% would be outside of the
300mUI pk-pk boundaries. The distribution should not be wildly skewed,
and therefore approx half of the outliers should be on each side.

The intentional jitter sources we are adding (sine jitter and sine
interference) should be symmetrical about center (mean), and if
achieved, a peak veiw from center may also be considered. For this peak
view, 99% should fall within and 1% would outside the 150 mUI peak
boundary on each side. I probably wouldn't actually use a peak method,
except hoping it clarifies the pk-pk method.

Your last interpretation (1%-99% inside) leave 2% total outside, which
is different from my interpretation. Sounds like clarification may be
required.

For vertical eye closure, I have been thinking a little differently. The
vertical opening would be 99.9% down from the nominal logic 1 value
minus 99.9% up from the nominal logic0 value. Vertical closure would be
OMA minus opening, where OMA is nominal logic1 minus nominal logic0.

I understand the implied challenges in here on symmetry and linearity.

Indeed we need to guard against tails, as you say!

Tom


-----Original Message-----
From: Petar Pepeljugoski [mailto:petarp@us.ibm.com]
Sent: Monday, April 15, 2002 12:14 PM
To: stds-802-3-hssg-serialpmd@ieee.org; piers_dawe@agilent.com; Lindsay,
Tom; dkabal@picolight.com; peter.ohlen@optillion.com; John Ewen;
krahn@lucent.com; mdudek@cieloinc.com; greg_lecheminant@agilent.com;
stretch_camnitz@agilent.com; chris_miller@agilent.com;
buchheit@lightlogic.com; rsavara@networkelements.com
Subject: peak to peak jitter


Hi all,

I would like to ask for clarification on the jitter calibration for the
stressed signal. Does the 1% apply to each side of the histogram, or it
is
divided in 0.5% on each side? What about 0.75% on one side or 0.25% on
the
other? (I guess that we also need to guard against tails)

My interpretation is that there is no more than 1% of the total
histogram
area on each side (i.e. you are looking at the cumulative values, so one
side will be 1%, the other will be 99%).

Any thoughts?

Thanks,

Peter

Petar Pepeljugoski
IBM Research
P.O.Box 218 (mail)
1101 Kitchawan Road, Rte. 134 (shipping)
Yorktown Heights, NY 10598

e-mail: petarp@us.ibm.com
phone: (914)-945-3761
fax:        (914)-945-4134