Multiplier Phase Noise - RF Cafe Forums

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Peter Raynald
Post subject: Multiplier Phase Noise
Unread postPosted: Thu Apr 07, 2005 10:37 am
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I have been looking for an explanation in textbook about the justification why is a multiplier giving 20logN phase noise deterioration.

I have always beleived that this was caused by the S(f) = k(1/f)^2 nature of the noise assumed in this statement, and that would mean that the noise deterioration is due to a speading effect of the multiplication.

Now I have been proven wrong in my belief by direct measurement on a PLL oscilator output that gets multiplied on which the flat part of the response (not 1/f) and the loop bandwidth transition region is identical before and after multiplication that is occuring outside of the loop (no spreading), with exeption that the level is 20logN higher than before the mulitplication. That means no spreading.

Does that make sense?

I was looking for a source where this phenomenon is put to equation, without success. Anybody have some source to suggest?

Thank you very much


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mcp
Post subject:
Unread postPosted: Thu Apr 07, 2005 4:18 pm
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Isn't this due to the period decreasing and the jitter remaining the same?

If you divide an oscillator signal by N, the phase noise improves by 20logN because the jitter remains the same while the period increases ( the error in the "zero crossings" are a smaller fraction of the period).


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Guest
Post subject:
Unread postPosted: Thu Apr 07, 2005 5:20 pm

If you take the inverse of the cosine of that you should get your answer. Easy!


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Peter Raynald
Post subject:
Unread postPosted: Fri Apr 08, 2005 11:43 am
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But how to you relate this to the 20logN formula?

Jitter is frequency deviation taken in time domain, RMS, and refered to the period of the signal. Why doesn't jitter double like is frequency deviation increase on a modulated signal that gets multiplied?

Phase noise is the spectal distribution of the random variable that represent this phase noise, it an inversed power serie for a free running oscillator, but a more complex response for a closed loop oscillator.

Simple multiplication rule shows that cos^2(x) = 1/2(1+cos(2x)

If X = wt + m(x) then multiplied argument is 2wt + 2m(x)

That means that the frequency deviation gets multiplied also so the phase noise level is raised by spreading not by level shift.

If you look at in in terms of FM modulation even there you find that doubling the modulation doesn't make the bessel function terms give an increase of 6dB for a doubling.

Now 20logN represents the increase of the phase noise in reference to the power level of the carrier. How do you get from that intuitive jitter explanation to the 20logN term in frequency domain?

Any clue?


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Another Guest
Post subject: Noise multiplication
Unread postPosted: Fri Apr 08, 2005 11:55 am

In my experience, there are two kinds of jitter:
1. Systematic/correlated - includes leakage at the phase detector frequency
2. Noise-like/uncorrelated - random but generally weighted in the frequency domain.

It can be hard to separate these out sometimes - especially in the time domain, where frequency weighting effects can look like systematic effects.

It seems obvious that these behave differently under multiplication conditions.

What do you think?


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Guest
Post subject:
Unread postPosted: Sat Apr 09, 2005 10:06 am

I cant offer any numbers to subtantiate my claim, but when you double a signal you are actually cutting the pulse repetition in half. Therefore, the amount of signal and noise is doubled in the same time slot. Dividing will have the opposite effect. I actually read an article that proofed it with the numbers and I remembered that it made sense at the time, but I can't remember where I put the article. You got me thinking about it so I probably will be obsessed with it and wont stop hunting for the article until I find it. Will post if I ever find the article.





Posted  11/12/2012