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is it true that absolute jitter is less than period jitter?

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liusupeng

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according to some lecture notes:
absolute jitter < 2*period jitter
period jitter < 2*adjacent period jitter.
Is it true?
It is easy to understand period jitter<2*adjacent period jitter. But why absolute jitter<2*period jitter?

---------- Post added at 02:20 ---------- Previous post was at 01:47 ----------

while some other reference suggest absolute jitter increase with measurement time
 

Hi,

just to make clear if I understand the jitter definition correct
Jitter - Wikipedia, the free encyclopedia
Absolute jitter is the absolute difference in the position of a clock's edge from where it would ideally be.
Period jitter (aka cycle jitter) is the difference between any one clock period and the ideal clock period.

I would say
If you have a defined absolute Jitter of +- something
your equation absolute jitter < 2*period jitter is correct

but if you have a defined period jitter of +- something
you can not know how your absolute jitter is.
Because in this case a small period jitter can integrate to a large absolute jitter

while some other reference suggest absolute jitter increase with measurement time

I think it means for gaussian random jitter the peak-peak value increases for longer measurement times.

regards
 

Hi,

just to make clear if I understand the jitter definition correct
Jitter - Wikipedia, the free encyclopedia


I would say
If you have a defined absolute Jitter of +- something
your equation absolute jitter < 2*period jitter is correct

but if you have a defined period jitter of +- something
you can not know how your absolute jitter is.
Because in this case a small period jitter can integrate to a large absolute jitter




regards

Good point!

but for free running oscillator, the absolute jitter should be unbounded and the period jitter should be a finite value. Is this correct?
 

Good point!

but for free running oscillator, the absolute jitter should be unbounded and the period jitter should be a finite value. Is this correct?

I think you are right

regards
 

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