FREQUENCY MULTIPLIER PHASE NOISE 20*LOG(N) QUESTION

Because phase is just the integral of frequency, so if you double the frequency you must be doubling the phase. Any frequency noise will be doubled in the nonlinearity, and phase noise is directly linked to frequency noise. Anything that doesn't get filtered out by the following bandpass filter will show up on the output.

That depends on the doubler, although if you're not giving it a pure tone (take FM as a pure tone, eh?) it isn't really doubling any more. If it's a single diode and the amplitudes are right you'll get 10MHz,

100MHz, 110MHz, 200MHz, 210MHz, 220MHz. Then you'll get 20MHz, 30MHz, 90MHz, 120MHz, etc. You should expect to see frequency components at f_out = n * f_1 + m * f_2 for n and m ranging over all positive and negative integers.

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Doubling the phase would be adding 3 dB. This is more like quadrupling the phase noise.

Does anyone have a simple math derivation of how they get 20log(N)?

So it's really just a mixer in this case. But it's never really a 100% pure tone anyhow.

S.

Doubling the phase doubles the _amplitude_ of the noise, which quadruples the power, which is a 6dB increase.

Amplitude goes with N, power goes with N^2, 10 log(N^2) = 20 log(N).

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Agreed..

but another way to look at it is...

a doubler doubles the carrier frequency and also doubles the FM deviation...

Mark

And another way is to note that X picoseconds of jitter at frequency F multiplies to the same X picoseconds at F*N, but the relative damage is N times worse.

(Within the active bandwidth of the multiplier, of course.)

John

You can, of course, build a frequency multiplier that has less jitter/phase noise at its output than at its input.

John

Are we assuming that the phase noise drops

-6dB/octave? Is this a good assumption?

Can you mathematically show how the amplitude of the noise doubles when you double the phase?

S.

Nicely done!

Marc

Doubling the phase doubles the _amplitude_ of the noise, which quadruples the power, which is a 6dB increase. Amplitude goes with N, power goes with N^2, 10 log(N^2) = 20 log(N).

And another way is to note that X picoseconds of jitter at frequency F multiplies to the same X picoseconds at F*N, but the relative damage is N times worse.

but another way to look at it is...

a doubler doubles the carrier frequency and also doubles the FM deviation...

No, I don't assume that phase noise drops 6dB per octave, because it's a bad assumption. Phase noise is predictable, and there will often be a region where it drops 6dB/octave of remove from the carrier, but it doesn't do that everywhere. Search on "oscillator" and "phase noise" and you'll find some info, I'm sure.

For small phi, cos(wt + phi(t)) approximately equals cos(wt) + phi(t)sin(wt).

So the amplitude of the noise (phi(t)sin(wt)) follows the amplitude of the phase (phi(t)), modulated by a carrier that's 90 degrees out from the main carrier (sin(wt)).

I will let you figure out the doubling part.

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Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

PLL?

Mark

Or just a high-Q resonator.

John

?????

By basic trig identity:

cos(wt + phi(t))= cos(wt)*cos(phi(t))-sin(wt)*sin(phi(t))

For small phi:

cos(wt + phi(t))=cos(wt)*1-sin(wt)*0=cos(wt)

No, please, you first...

But I don't expect a DSP/control systems expert to know that much about phase noise.

Anyone else?

S.