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Below are all of the forum threads, including all
the responses to the original posts.
Peter Raynald
Post subject: Multiplier Phase Noise
Unread postPosted:
Thu Apr 07, 2005 10:37 am
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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
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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
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Joined: Tue Sep 07, 2004 3:09 pm
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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. 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