[time-nuts] 20logN was Re: phase noise questions (long)
christophe.huygens at cs.kuleuven.ac.be
Thu Jan 24 04:55:17 EST 2008
I like everybodies so answers so far, as it helps to form a more thorough
understanding... I ll revist the math over the weekend. Thanks for yr
Another unrelated topic, I seem to get emails from time-nuts only
sporadically, and with sender time-nuts-bounces. However I
don t have any bounce status... Does anybody else see this
phenomenon. It is a bit irritating to have to go to the archive
page to read up.
Christophe Huygens wrote:
> Hi John, Steve, et al,
> While I am not a phase noise buff at all, in talking to many on this
> I feel that this is not well understood. When I ask where the 6db/Hz for
> doubling or 20logN in general comes from, I very often get an
> answer and I have seen strange notes on this mailing list on this
> subject as
> For me to understand what happens in a simplified way 2 things are key:
> 1. Phase noise is subject to FM theory - you can think of the carrier
> being FM modulated with a very low modulation index, with
> modulation frequency the offset from the carrier. This is easy
> enough to accept for most. The noise phasor sits on top of the carrier.
> This give amplitude noise, that can be limited away, and well...
> phase noise. The actual modulation index in our case is always
> very small I guess, except when you looking real close to the
> carrier, but then still - if the oscillator is good, the deviation will
> still be small hence low modulation index theory still applies..
> 2. What happens with an FM signal when applied to an ideal doubler -
> this is a bit of a trickier. Say I have a narrowband (low modulation
> index) signal of 200Hz, modulated by 20Hz.
> a. The spectrum is:
> sideband 1 (180) - carrier (200) - sideband 2 (220).
> AFTER the doubler the spectrum is:
> sideband 1 (380) - carrier (400) - sideband 2 (420).
> I have a hard time to find an intuitive explanation for this,
> but it only takes 20 lines of octave/matlab code to verify...
> I am getting too old (or I m too young) to get into the Bessel
> functions myself.
> So no need to multiply the offset also by N as sometimes seen.
> b. The amplitude of the sidebands does grow with respect to
> the carrier (all this for small modulation indexes) by about
> 6 db. Also easy to show in a a simulation.
> The duality of multiplication in the time domain and convolution
> in the frequency domain also explains this I think, like it
> can explain a.
> Maybe somebody on the list can step in and give a clear and
> concise explanation for the above.
> John Miles wrote:
>> Doubling your clock frequency adds 6 dBc/Hz to whatever the noise
>> level was
>> at the input, at all offsets within the doubler's bandwidth. Only if
>> input noise level is near or below the multiplier's own residual
>> noise floor
>> will the increase be worse than 6 dBc/Hz.
>> That will not happen when ordinary crystal oscillators and conventional
>> Schottky-diode multipliers are used together; high-performance active
>> multipliers are needed only when working with exceptionally clean
>> At input noise levels higher than -155 to -160 dBc/Hz, ordinary diode
>> multipliers will not usually contribute any additional noise.
>> -- john, KE5FX
>>> I followed with some interest a discussion about a NIST doubler circuit
>>> using matched FET's and I was wondering if you could get similar
>>> using an analog multiplier chip from Analog Devices. It would seem that
>>> they take some care about device matching and have parts that work up
>>> to pretty high frequencies. Of course there would need to be some
>>> employed. Oh, and I think those parts do pretty well with temperature.
>>> Also, when using a doubler that is rated in dBc how do you apply that
>>> number to get an expectation from a given starting dBc oscillator. So
>>> if my 10 MHz clock is -125dBc and I use the NIST circuit, what would
>>> I see at 20 MHz in dBc?
>>> thanks in advance,
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