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Understanding Jee Measurement

FPMKh

Hi,
My Virtuoso and Spectre Version: ICADVM20.1-64b.NYISR30.2

Different vendors and groups have different definitions of jitter; some are contradictory. I want to stick to the definition used in Virtuoso and try to understand it better.
In this link: https://support.cadence.com/apex/ArticleAttachmentPortal?id=a1O3w000009bh2QEAQ&pageName=ArticleContent different jitter measurements are mentioned. For the Jee, it says Jee or absolute jitter. When I use Jee (using PSS+Pnoise, sampled noise) versus using transient noise (and then std of abs_jitter) I actually get the same result. So the definition seems to be consistent within Virtuoso. However, I still don't know how the measurement is done. According to the link above, Jee looks at the time difference at time tn (nth rising edge) and compares it with ideal nT (Jee(n) = tn – nT, n=1,2,…, infinity). What is the final Jee number that is being reported? The standard deviation of what distribution? If it's the standard deviation of the nth component in Jee(n), what "n" is chosen when it reports the number in the pnoise simulation?


Thank you

Parents

  • Hi,

    this is my understanding...

    In the example you gave:

    absJ_plot = abs_jitter(clip(vtime('tran "/Out1") 400.0p 8.0004u ) "rising" 1.65 ?yUnit "s" ?Tnom 400.0p )

    Jee_rms = stddev(absJ_plot)

    you have a signal with 400ps period measured during 8us, which means that you have 20k rising edges in total. This is your sample space. You are interested in edge-to-edge or absolute jitter, so every "noisy" edge is compared against its "clean" counterpart. Therefore, you get 20k samples of jitter in seconds in the span from -100fs to 100fs (just a random example). Your mean will be around 0 and those 100fs extremes will be around 3sigma. Jee is basically 1sigma.

    You could actually plot that absJ_plot in ViVA. Then (still in ViVA) go Measurements -> Histogram -> Put 10 bins and add Std Dev Lines -> There is your distribution!

    Hope this helps!

    BR

    /Janko

  • in reply to JankoK

    Thank you, Janko, for your reply.
    When you say there are 20K rising edges and hence 20K samples (from which we can get the distribution), I understand it. But my question is how jitter is calculated for each of these samples. For example, for the 10000th sample, what is the reported jitter? Is it the actual timing where this 10000 rising edge happens minus where it was supposed to happen (i.e. tn-nT = tn-10000X400pS)? If that's the case, then the higher the sample gets, the more jitter will be expected as it gets accumulated. If the jitter calculation differs from what I mentioned here, then I'd like to know what it is.

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  • in reply to JankoK

    Thank you, Janko, for your reply.
    When you say there are 20K rising edges and hence 20K samples (from which we can get the distribution), I understand it. But my question is how jitter is calculated for each of these samples. For example, for the 10000th sample, what is the reported jitter? Is it the actual timing where this 10000 rising edge happens minus where it was supposed to happen (i.e. tn-nT = tn-10000X400pS)? If that's the case, then the higher the sample gets, the more jitter will be expected as it gets accumulated. If the jitter calculation differs from what I mentioned here, then I'd like to know what it is.

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