I should know this, but it's easy to ask.
Assume the rising edge of a pulse that is delayed from a trigger, and that has 1 ns p-p jitter relative to trigger, with a uniform probability distribution. We'd get that if we used a 1 GHz unsynchronized clock to generate the delay, which we're now doing in an FPGA.
What is the RMS jitter? I recall there being something like a square root of 12 in there somewhere.
Should be the same as the RMS value of a triangle wave, no? That would be peak value * 1/(3^0.5) or in your case, about 0.57 ns.
-- Rick
(peak from center) / sqrt(3), or (p-p) / sqrt(12)
Or about 0.29 ns, more or less, with an 0.5ns fixed delay.
you get the mean square error by integrating x^2 from x=-1/2 to x=1/2, which is (1/2)^3/3 - (-1/2)^3/3 = (wait for it) 1/12.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
This is one geared to troll that Really Smart Linux dood that knows everyth ing about everything and then some then goes on and tells everyone to go to hell after bragging about its infinite intelligence in some area totally n ot at issue et cetera et cetera, right?
It's actually X*SQR(n/12) to get your answer one way and X/SQR(n/12) the other direction.
289 ps RMS. Thanks.
A bit over three inches, at any rate.
-- www.wescottdesign.com
I get 1/sqrt(12) ns RMS deviation from ideal, rms jitter between events may be worse
-- \_(?)_
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