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From: gretzteam on 31 May 2010 11:40 Hi, Say I have a simple x-bit NCO running at 'fin' (100MHz). M and N are programmable. The output frequency of the NCO is given by: fout = fin*(N/M) Assuming a perfect input clock, how can I specify the jitter of 'fout'? I understand there are different measure of jitter but can't really figure out how they work with this system. With 'nice' M and N, the NCO is performing an exact power of two divider and the jitter should be 0 (again assuming the input clock is perfect). I understand that for arbitrary M and N, 'fout' will have the following shape: '6 6 6 6 7 6 6 6 6 7 6 6 6 6 7 etc...' (Here 6 and 7 being the number of input cycles). How is jitter specified for such a clock? Thanks!
From: Tim Wescott on 31 May 2010 13:06 On 05/31/2010 08:40 AM, gretzteam wrote: > Hi, > Say I have a simple x-bit NCO running at 'fin' (100MHz). M and N are > programmable. The output frequency of the NCO is given by: > fout = fin*(N/M) > > Assuming a perfect input clock, how can I specify the jitter of 'fout'? I > understand there are different measure of jitter but can't really figure > out how they work with this system. > With 'nice' M and N, the NCO is performing an exact power of two divider > and the jitter should be 0 (again assuming the input clock is perfect). > > I understand that for arbitrary M and N, 'fout' will have the following > shape: > > '6 6 6 6 7 6 6 6 6 7 6 6 6 6 7 etc...' (Here 6 and 7 being the number of > input cycles). This makes sense if the NCO is some sort of a fractional N divider -- is that what you're dealing with? Certainly that's not the behavior of a phase accumulating NCO. > How is jitter specified for such a clock? Compare the actual with the ideal, and measure the magnitude of the error. Part of your confusion probably arises because you need to specify whether your jitter is timing jitter, phase jitter, or frequency jitter. USUALLY questions like this aren't difficult because the actual math is difficult: USUALLY questions like this are difficult because the specification is fuzzy, leading the practitioner to being confused about WHAT math to use. I think this is your problem, so I am going to gloss over a whole bunch of concerns, pull a specification out of my hat, and show how easy the math is -- but you still have to figure out what _your_ specification should be. So: Assuming that you mean timing jitter, and taking your sequence ('shape') of the output signal, your ideal sequence would be 6.2 6.2 6.2 ... So your ideal minus your actual would go: 0.2 0.4 0.6 0.8 0 0.2 0.4 ... If you take the average of this it'd go: -0.2 0.0 0.2 0.4 -0.4 -0.2 0.0 ... From this you can see immediately that your peak to peak jitter is 0.8 (of whatever units you're using), and you can fairly easily calculate your RMS jitter. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: Eric Jacobsen on 31 May 2010 14:11 On 5/31/2010 8:40 AM, gretzteam wrote: > Hi, > Say I have a simple x-bit NCO running at 'fin' (100MHz). M and N are > programmable. The output frequency of the NCO is given by: > fout = fin*(N/M) > > Assuming a perfect input clock, how can I specify the jitter of 'fout'? I > understand there are different measure of jitter but can't really figure > out how they work with this system. > With 'nice' M and N, the NCO is performing an exact power of two divider > and the jitter should be 0 (again assuming the input clock is perfect). > > I understand that for arbitrary M and N, 'fout' will have the following > shape: > > '6 6 6 6 7 6 6 6 6 7 6 6 6 6 7 etc...' (Here 6 and 7 being the number of > input cycles). > > How is jitter specified for such a clock? > > Thanks! I've not actually heard of M and N being associated directly with an NCO before. Are you using an NCO or some sort of hybrid synthesizer? An NCO by itself, by definitions that I'm used to, will contain just the oscillator with a Phase Increment Register. I don't know why you'd need a divide-by-N when you can just reprogram the PIR for a lower frequency. Can you clarify? -- Eric Jacobsen Minister of Algorithms Abineau Communications http://www.abineau.com
From: Jerry Avins on 31 May 2010 14:21 On 5/31/2010 2:11 PM, Eric Jacobsen wrote: > On 5/31/2010 8:40 AM, gretzteam wrote: >> Hi, >> Say I have a simple x-bit NCO running at 'fin' (100MHz). M and N are >> programmable. The output frequency of the NCO is given by: >> fout = fin*(N/M) >> >> Assuming a perfect input clock, how can I specify the jitter of 'fout'? I >> understand there are different measure of jitter but can't really figure >> out how they work with this system. >> With 'nice' M and N, the NCO is performing an exact power of two divider >> and the jitter should be 0 (again assuming the input clock is perfect). >> >> I understand that for arbitrary M and N, 'fout' will have the following >> shape: >> >> '6 6 6 6 7 6 6 6 6 7 6 6 6 6 7 etc...' (Here 6 and 7 being the number of >> input cycles). >> >> How is jitter specified for such a clock? >> >> Thanks! > > I've not actually heard of M and N being associated directly with an NCO > before. Are you using an NCO or some sort of hybrid synthesizer? > > An NCO by itself, by definitions that I'm used to, will contain just the > oscillator with a Phase Increment Register. I don't know why you'd need > a divide-by-N when you can just reprogram the PIR for a lower frequency. Working the NCO at a frequency that allows a final division by a large power of two is a dandy way to reduce jitter. Not very tunable, though. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
From: Steve Pope on 31 May 2010 14:30
gretzteam <gretzteam (a)n_o_s_p_a_m.yahoo.com> wrote:>Hi, >Say I have a simple x-bit NCO running at 'fin' (100MHz). M and N are >programmable. The output frequency of the NCO is given by: >fout = fin*(N/M) > >Assuming a perfect input clock, how can I specify the jitter of 'fout'? I >understand there are different measure of jitter but can't really figure >out how they work with this system. RMS jitter is as follows: Form a reference clock signal that is jitter free and of the same exact frequency and same average phase as the NCO. The RMS value of the time diffeence of the NCO clock edges relative to the reference clock, divided by the period, is the RMS jitter as a dimensionless ratio. (If you don't divide it by the period, it's the RMS jitter in time units.) For a given NCO you might be able to compute this analytically from parameters such as your N and M, if it is important to do so. Steve |