[email protected] said: > There is always an implied clock recovery loop that the the jitter is > measured against. The loop may itself affect the jitter measurement either by > cleaning up jitter or contributing to it.
Interesting. I hadn't thought about it that way. Suppose I measure the edge to edge times and make a histogram. Can I get jitter out of that? Where is the clock recovery loop? -- These are my opinions. I hate spam. _______________________________________________ time-nuts mailing list -- [email protected] To unsubscribe, go to http://lists.febo.com/mailman/listinfo/time-nuts_lists.febo.com and follow the instructions there.
