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Peter Raynald Post subject: Multiplier Phase Noise Unread
postPosted: Thu Apr 07, 2005 10:37 am Offline Captain
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Joined: Tue Sep 07, 2004 3:09 pm Posts: 11
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 Offline Lieutenant
Joined: Thu Apr 07, 2005 4:03 pm Posts: 2 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 Offline Captain
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Joined: Tue Sep 07, 2004 3:09 pm Posts: 11
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. Noiselike/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



