Multiplier Phase Noise - RF Cafe Forums
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Post subject: Multiplier Phase Noise
Unread postPosted: Thu Apr 07, 2005 10:37 am
Joined: Tue Sep 07, 2004 3:09 pm
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
Unread postPosted: Thu Apr 07,
2005 4:18 pm
Joined: Thu Apr 07, 2005 4:03 pm
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).
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!
Unread postPosted: Fri Apr 08,
2005 11:43 am
Joined: Tue Sep 07, 2004 3:09 pm
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
Post subject: Noise multiplication
Unread postPosted: Fri Apr 08, 2005 11:55 am
In my experience, there are two kinds of jitter:
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?
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.