[time-nuts] New topics (was Re: He is a Time-Nut Troublemaker....)

[Magnus Danielson]

Didier skrev:
>> -----Original Message-----
>> On Behalf Of Magnus Danielson
>>
>> Didier skrev:
>>> John,
>>>
>>> When you add two (statistically independent) 5 MHz signals 
>>> and get a 10MHz signal, the 10 MHz signal's *relative* noise 
>>> and drift will be the average of the *relative* noise and 
>>> drift of the two 5 MHz signals.
>> Not to ruin your analogy here, but what I was discussing on 
>> interlocking was intended as means to phase-lock the two 
>> oscillators (say 5 MHz but I was thinking 10 MHz) and you 
>> could then just add their sines, not mix them up. Thus, you 
>> do not get the sum frequency, you get the average frequency. 
>> Of course you could go for the frequency multiplication 
>> variant if you want.
>>
> 
> Specifically, John was suggesting adding the two 5MHz signals, instead of
> locking them, that's why I added "statistically independent".

They are statistically independent regardless, except for within (or in 
the vincinity of) the PLL bandwidth.

>>> So as when you average n signals, the noise and drift are 
>>> reduced by sq.rt of n, in this case, 1.4, or about 2dB 
>>> (if I am correct), a modest improvement.
>> Square root of 2 is about 1,414 or about 3,01 dB.
> 
> I am always confused when considering noise, is it 10*log(p1/p0) or
> 20*log(p1/p0)?

It is very simple. bel is a power ratio in base 10 logarithm, thus 
log10(P/Pref) where P is the power (in watts) and Pref is the reference 
Power by which we normalize. A decibel is the same thing but scaled to a 
tenth of the scale, thus 10*log10(P/Pref). For voltage we recall that 
P=UI and Ohm's law gives us U=RI giving I=U/R and thus P=U^/R. Inserting 
this gives 10*log10(U^2/R/Pref). Since log(x^2)=2log(x) we can shift the 
2 out and form 20*log10(U/sqrt(R*Pref)) which we can simplify to 
20*log10(U/Uref) by letting Uref = sqrt(R*Pref).

So 10*log... is for powers and 20*log is for voltages.

Double power gives 10*log10(2) = 10*0.301 = 3,01 dB.
Double voltage gives 20*log10(2) = 20*0.301 = 6,02 dB.

>>> Combining more than 2 signals that way (to get more than 2dB 
>>> improvement) gets complicated in a hurry.
>> Actually no. Not really. You can build pairs and then 
>> interconnect them together the same way to form a quad, and 
>> so you go on. The neat thing is that the combined oscillators 
>> behave as a new oscillator. This is a very traditional way of 
>> combining sources to reduce noise. It is certainly not new.
>>
> 
> First of all, that only works for a number of oscillators that is a power of
> two. Then I suppose that as you increase the number of pairs, it will become
> harder for individual oscillators to lock themselves to the output (unless
> you add circuitry), so you move away from an interlocked system and closer
> to a purely added system. Adding the outputs from a bunch of independent
> oscillators does not give you one clean output, it gives you a narrow band
> of noise. I must be missing something?

If you look at the grouping that I described you will see that for 
combining N oscillators it takes N-1 diffrential locking loops, meaning 
N-1 phase comparators and N-1 loopfilters etc. I was also giving the 
specific hint that this was a simplified description in that I assumed 
non-weighed result, where as you could fairly easily create a weighed 
form. Consider want to lock one oscillator two a pair. Then weigth 2/3 
of the signal to the pair input and 4/3 of the signal to the single 
oscillator, thus slightly less to the pair and slightly more to the 
single oscillator.

The actual trick here is that the signal and the noise will add 
differently together. Recall, that for uncorrelated signals we add 
powers, so Ptot = P1 + P2 or from a voltage perspective Utot^2 = U1^2 + 
U2^2, assuing U1 = U2 gives Utot = U1 * sqrt(2) or Ptot = P1 * 2 thus 
giving +3,01 dB. For correlated signals we add amplitude (in voltage) 
such that Utot = U1 + U2 which for U1=U2 gives Utot = U1 * 2 thus giving 
+6,02 dB. Now, for two frequency-locked 10 MHz sines which is also 
phase-locked to be nominally at 0 degrees from each other we have two 
correlated signals and the phase-alignment ensures that their amplitudes 
adds correctly to the maximum value, thus giving us a +6,02 dB gain. The 
noise of the two oscillators is uncorrelated to each other, and thus 
giving us a +3,01 dB gain, assuming more or less equalent oscillators.
The net gain in S/N ratio is 6,02-3,01 = 3,01 dB.

You can combine the two sines through a mixer rather than through 
additive combination, but that was not what was originally described.

This technique have been used for ages for low-noise amplifiers etc. I 
have only adapted it to the fields of oscillators. It is a crude 
approximation which gives some improvement (if done correctly) but does 
not really comes THAT cheaply.

Cheers,
Magnus