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calculating phase noise of pll / sampling loop - RF Cafe Forums

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Post subject: calculating phase noise of pll / sampling loop Posted: Fri Nov 02, 2007 3:52 am


Joined: Fri Nov 02, 2007 3:45 am
Posts: 8
Location: germany
consider I want to phase lock a 2-3 ghz VCO
to a 20mhz reference ( -120dBc/Hz) and i am thinking about how to get the best phase noise performance-

if i do it with a single loop pll i can roughly calculate phase noise to be
20log(N) (with n= 150) for 3ghz output frequency at -76dBc/Hz inside the loop bandwith
with a perfect phase detector

found some publications using mixing techniques including comb generators and pretuning and i wondered how they are
superior to the single loop solution with that division ratio.

if i use a sampling phase detector with wide bandwith as a clean up loop
in a configuration of : comb generator driven at 20mhz and mix 3ghz vco output with the 149th harmonic ( 2980mhz)
what phase noise can i expect ? - how strong is the influence of the
phase noise of the 149 harmonic of 20mhz onto overall system noise ??
i calculate phase noise of 2980mhz mixing signal to be
-120dBc/hz -(43.5dB)=76.5dBc/Hz

am i right to say inside the loop bandwith i can expect phase noise of my loop to be
average of reference phase noise and harmonic phase noise , calculating
sqrt( -120dbc*76.5dBc)= 95.8dBc/hz
thus would result in an improvement of nearly 20dB comparing to the single loop solution ?

does someone has got experience with phase locked loops using harmonic locking ? how do i calculate the phase noise i can expect
if i use mixing / harmonic locking techniques ?

mny thanks for comments



Tony Kurlovich
Post subject: Posted: Thu Nov 08, 2007 5:10 pm


Joined: Thu Oct 19, 2006 6:02 pm
Posts: 7
The answer to your last question is no. For this application, you can't average dbs. If you could, a large PN could be reduced by adding a lesser PN.

If the phase noise components to be added are not correlated, it would be the sum of the noise powers. If the components are correlated (this case), it is the sum of the noise voltages. This will yield the same answer as 20log(150). Which is only for small deviation PM.

If your output is 20MHz*150, 0.1 Degree at 20MHz is worth 15 Degrees at 3000Mhz. You don't get one without the other.

Posted  11/12/2012

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