Doubling your clock frequency adds 6 dBc/Hz to whatever the noise level was
at the input, at all offsets within the doubler's bandwidth.  Only if the
input noise level is near or below the multiplier's own residual noise floor
will the increase be worse than 6 dBc/Hz.

That will not happen when ordinary crystal oscillators and conventional
Schottky-diode multipliers are used together; high-performance active
multipliers are needed only when working with exceptionally clean inputs.
At input noise levels higher than -155 to -160 dBc/Hz, ordinary diode
multipliers will not usually contribute any additional noise.

-- john, KE5FX

Steve
steve the knife wrote:

No, there is no local RF feedback in such multipliers, consequently they
are very noisy.
Use a conventional diode, FET or BJT low phase noise doubler.
Dont use the transistor doubler circuits in the ARRL handbook they are
noisy.
Smooth switching action combined with local RF feedback to suppress
flicker noise is necessary for low phase noise.
N.B. a finite source impedance can provide sufficient RF feedback.
The input signal level should be high to maximise the conversion efficiency.
Using small input signals and exploiting the approximate square law
characteristics of devices like JFETs will degrade the output phase
noise significantly over that achievable with a large signal input
switching doubler. The output current waveform (before filtering) should
resemble a rectified sinewave for a NIST style doubler.

Device matching has little effect on phase noise, however it does
improve odd harmonic and fundamental suppression in the output.
If your input frequency is 10MHz you dont need particularly high ft
transistors for efficient low phase noise frequency doublers.

An ideal doubler degrades this by 6dB.
i.e to -119dBc/Hz at the (unspecified) offset from the carrier.
Real frequency doublers degrade this further by a few dB.

Bruce

Hi Steve!,
I know this is off topic, but I would be facinated to know why you are
called "Steve the knife" [Please don`t tell me if it is anything
illegal]
Yours in extremely low noise,................Don.

----- Original Message -----
From: "steve the knife" poolshark@disinfo.net
To: time-nuts@febo.com
Sent: Wednesday, January 23, 2008 1:22 PM
Subject: [time-nuts] phase noise questions

Just to let you all know that Steve the knife did reply to me off group - it
is a facinating story.
Very, Very low noise to you all,........Don C.

----- Original Message -----
From: "Don Collie" donmer@woosh.co.nz
To: "Discussion of precise time and frequency measurement"
time-nuts@febo.com
Sent: Wednesday, January 23, 2008 5:04 PM
Subject: Re: [time-nuts] phase noise questions

Hi John, Steve, et al,

While I am not a phase noise buff at all, in talking to many on this subject
I feel that this is not well understood. When I ask where the 6db/Hz for
doubling or 20logN in general comes from, I very often get an unsatisfying
answer and I have seen strange notes on this mailing list on this
subject as
well...

For me to understand what happens in a simplified way 2 things are key:

1. Phase noise is subject to FM theory - you can think of the carrier
   being FM modulated with a very low modulation index, with
   modulation frequency the offset from the carrier. This is easy
   enough to accept for most. The noise phasor sits on top of the carrier.
   This give amplitude noise, that can be limited away, and well...
   phase noise. The actual modulation index in our case is always
   very small I guess, except when you looking real close to the
   carrier, but then still - if the oscillator is good, the deviation will
   still be small hence low modulation index theory still applies..

2. What happens with an FM signal when applied to an ideal doubler -
   this is a bit of a trickier. Say I have a narrowband (low modulation
   index) signal of 200Hz, modulated by 20Hz.

a. The spectrum is:
sideband 1 (180) - carrier (200) - sideband 2 (220).
AFTER the doubler the spectrum is:
sideband 1 (380) - carrier (400) - sideband 2 (420).

I have a hard time to find an intuitive explanation for this,
but it only takes 20 lines of octave/matlab code to verify...
I am getting too old (or I m too young) to get into the Bessel
functions myself.

So no need to multiply the offset also by N as sometimes seen.

b. The amplitude of the sidebands does grow with respect to
the carrier (all this for small modulation indexes) by about
6 db.  Also easy to show in a a simulation.

The duality of multiplication in the time domain and convolution
in the frequency domain also explains this I think, like it
can explain a.

Maybe somebody on the list can step in and give a clear and
concise explanation for the above.

Christophe

John Miles wrote:

Christophe Huygens wrote:

Christophe

1. There's no need to resort to using Matlab, simple trigonometric
   identities will suffice.

For a multiplier type frequency doubler with small modulation index FM:
fc(t) = (1- 0.5betabeta)coswct - 0.5beta[cos ((wm-wc)t) - cos
((wc+wm)t)]

fc(t)fc(t) = (1- 0.5betabeta)coswct(1- 0.5betabeta)coswct - (1-
0.5betabeta)coswct0.5beta[cos ((wm-wc)t) - cos ((wc+wm)t)]  +
0.5beta[cos ((wm-wc)t) - cos ((wc+wm)t)] 0.5beta[cos ((wm - wc)t) -
cos ((wc + wm)t)]

fc(t)fc(t) = (1- 0.5betabeta)(1- 0.5betabeta)[0.5(1 + cos(2wct)]

• (1- 0.5betabeta)0.5beta*[0.5*[cos((wm-2wc)t) + cos((2wc+wm)t)]]
• 0.25betabeta*[0.5*(1 + cos(2*(wm-wc)t)) + 0.5*(1 + cos(2*(wm +
  wc)t)) + (cos ((2wc)t) + cos ((2wm)t)]

fc(t) = (0.5 - 0.25beta^2  + 0.125beta^4) +
(0.5 - 0.25beta^2 + 0.125beta^4) cos(2wct) +
(0.25beta - 0.125beta^3 )* cos((2wc-wm)t) +
(0.25beta - 0.125beta^3 )* cos((2wc+wm)t) +
(0.125beta^2)cos(2(wm-wc)t) +
(0.125beta^2)cos(2(wm + wc)t)

When beta <<1, the AC components can be approximated by

0.5cos((2wc)t) +
0.25betacos((2wc - wm)t) +
0.25beta*cos((2wc + wm)t)

Thus the ratio of the sideband components to the carrier has increased
by 6 dB at the output of the frequency doubler.

It is perhaps even simpler to use the complex exponential representation
of the signal (this is especially true for multiplication by N) as
exponentiation is an easier operation (by hand at least) than complex
trigonometric identities.

fc(t) = Re[exp(jwct)(1+ 0.5beta*{exp(jwmt) - exp(-jwmt)})]

fc(t)^N = Re[exp(jNwct)(1+0.5beta*{exp(jwmt) - exp(-jwmt)})^N)

Re[exp(jNwct)(1 + 0.5betaN{exp(jwmt) - exp(-jwmt)}]
or
Re[exp(jNwct) + 0.5betaNexp(jNwct + jwmt) + 0.5betaNexp(jNwct -jwmt)]
or
cos (Nwct) + 0.5mN cos((Nwc+wm)t) + 0.5mN*cos((Nwc +wm)t)

2. Bessel functions are only necessary to calculate the amplitudes of
   the various components, their functional form is unaltered.

3. The even harmonic FM sidelobes are in effect amplitude modulation
   components of the signal.

4. The phase sensitive detector used in phase noise measurements is
   insensitive to AM.

Bruce

Hi,

I did the same thing using the Euler relation for sin(w1+sin(w2t)) and straight
multiplication and collecting terms gave 1-0.5cos(2w1+2*sin(w2t)). I wanted
to actually contribute but was too slow :(

-steve

Quoting Bruce Griffiths bruce.griffiths@xtra.co.nz:

I'm not really a pool shark but I can beat YOU!

steve the knife wrote:

Steve

Please post your derivation it may be easier to follow for some of us
than mine.

If anyone (including me) was wondering where the factor of 2 difference
between the analytic signal method and the trigonometric derivation
arose see:

http://en.wikipedia.org/wiki/Analytic_signal

Bruce

Christophe -

While all of the mathematics to prove the 20logN behavior can be found in
most elementary texts, the concept of what is really happening may not be so
obvious from the math, even though the math of course supports the 20logN
behavior. You, yourself said it in your assumption in 2a. When a carrier is
phase modulated by a single frequency you do get the +/- sidebands about the
carrier as stated. The sidebands have two items contributing to the
sidebands location and their amplitude. Their location is due to the
frequency of modulation, and their amplitude to the deviation of that
frequency about the carrier. When doubled, the carrier is doubled but the
modulation frequency remains the same, yet the deviation is also doubled,
hence the 6 dB increase relative to the carrier of said sideband amplitude,
or the level it was before. This was a very common way to get increased
deviation in WWII military radios. They started out at a low frequency and
FM modulated it one way or another, however with a resultant low deviation.
After numerous multiplier stages to get to the final frequency, the
deviation was sufficient to recover adequate audio in the receivers
discriminator, yet the audio frequency was the same as initially injected.
Regards - Mike

Mike B. Feher, N4FS
89 Arnold Blvd.
Howell, NJ, 07731
732-886-5960

-----Original Message-----
From: time-nuts-bounces@febo.com [mailto:time-nuts-bounces@febo.com] On
Behalf Of Christophe Huygens
Sent: Wednesday, January 23, 2008 3:13 PM
To: Discussion of precise time and frequency measurement
Subject: [time-nuts] 20logN was Re: phase noise questions (long)

Hi John, Steve, et al,

While I am not a phase noise buff at all, in talking to many on this subject
I feel that this is not well understood. When I ask where the 6db/Hz for
doubling or 20logN in general comes from, I very often get an unsatisfying
answer and I have seen strange notes on this mailing list on this
subject as
well...

For me to understand what happens in a simplified way 2 things are key:

1. Phase noise is subject to FM theory - you can think of the carrier
   being FM modulated with a very low modulation index, with
   modulation frequency the offset from the carrier. This is easy
   enough to accept for most. The noise phasor sits on top of the carrier.
   This give amplitude noise, that can be limited away, and well...
   phase noise. The actual modulation index in our case is always
   very small I guess, except when you looking real close to the
   carrier, but then still - if the oscillator is good, the deviation will
   still be small hence low modulation index theory still applies..

2. What happens with an FM signal when applied to an ideal doubler -
   this is a bit of a trickier. Say I have a narrowband (low modulation
   index) signal of 200Hz, modulated by 20Hz.

a. The spectrum is:
sideband 1 (180) - carrier (200) - sideband 2 (220).
AFTER the doubler the spectrum is:
sideband 1 (380) - carrier (400) - sideband 2 (420).

I have a hard time to find an intuitive explanation for this,
but it only takes 20 lines of octave/matlab code to verify...
I am getting too old (or I m too young) to get into the Bessel
functions myself.

So no need to multiply the offset also by N as sometimes seen.

b. The amplitude of the sidebands does grow with respect to
the carrier (all this for small modulation indexes) by about
6 db.  Also easy to show in a a simulation.

The duality of multiplication in the time domain and convolution
in the frequency domain also explains this I think, like it
can explain a.

Maybe somebody on the list can step in and give a clear and
concise explanation for the above.

Christophe

John Miles wrote:

was

floor