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FREQUENCY MULTIPLIER PHASE NOISE 20*LOG(N) QUESTION

Hi,

In 10 years of RF experience, i've never had to use frequency
multipliers until now. (Dividers for PLLs, most certainly, yes!)

So we all know that the phase noise at each offset
(dBc/Hz @ offset in Hz) should increase by 20*log(N),
or 6 dB for a frequency doubler.

But i'd like an explanation of how they got this?

Also, if you put two tones into a frequency doubler,
like 100 MHz and 110 MHz, will you get out 200 MHz
and 220 MHz, plus harmonics? Or will you get 200MHz
and 210 MHz, plus intermod products from the non-
linearity of the diode that they often use in these?

Thanks for the help!

Slick
 
T

Tim Wescott

Jan 1, 1970
0
Hi,

In 10 years of RF experience, i've never had to use frequency
multipliers until now. (Dividers for PLLs, most certainly, yes!)

So we all know that the phase noise at each offset
(dBc/Hz @ offset in Hz) should increase by 20*log(N),
or 6 dB for a frequency doubler.

But i'd like an explanation of how they got this?

Because phase is just the integral of frequency, so if you double the
frequency you must be doubling the phase. Any frequency noise will be
doubled in the nonlinearity, and phase noise is directly linked to
frequency noise. Anything that doesn't get filtered out by the
following bandpass filter will show up on the output.
Also, if you put two tones into a frequency doubler,
like 100 MHz and 110 MHz, will you get out 200 MHz
and 220 MHz, plus harmonics? Or will you get 200MHz
and 210 MHz, plus intermod products from the non-
linearity of the diode that they often use in these?
That depends on the doubler, although if you're not giving it a pure
tone (take FM as a pure tone, eh?) it isn't really doubling any more.
If it's a single diode and the amplitudes are right you'll get 10MHz,
100MHz, 110MHz, 200MHz, 210MHz, 220MHz. Then you'll get 20MHz, 30MHz,
90MHz, 120MHz, etc. You should expect to see frequency components at
f_out = n * f_1 + m * f_2 for n and m ranging over all positive and
negative integers.
 
Tim said:
Because phase is just the integral of frequency, so if you double the
frequency you must be doubling the phase. Any frequency noise will be
doubled in the nonlinearity, and phase noise is directly linked to
frequency noise. Anything that doesn't get filtered out by the
following bandpass filter will show up on the output.

Doubling the phase would be adding 3 dB. This
is more like quadrupling the phase noise.

Does anyone have a simple math derivation of
how they get 20log(N)?


That depends on the doubler, although if you're not giving it a pure
tone (take FM as a pure tone, eh?) it isn't really doubling any more.
If it's a single diode and the amplitudes are right you'll get 10MHz,
100MHz, 110MHz, 200MHz, 210MHz, 220MHz. Then you'll get 20MHz, 30MHz,
90MHz, 120MHz, etc. You should expect to see frequency components at
f_out = n * f_1 + m * f_2 for n and m ranging over all positive and
negative integers.

So it's really just a mixer in this case. But it's never
really a 100% pure tone anyhow.


S.
 
T

Tim Wescott

Jan 1, 1970
0
Doubling the phase would be adding 3 dB. This
is more like quadrupling the phase noise.

Does anyone have a simple math derivation of
how they get 20log(N)?
Doubling the phase doubles the _amplitude_ of the noise, which
quadruples the power, which is a 6dB increase.

Amplitude goes with N, power goes with N^2, 10 log(N^2) = 20 log(N).
 
M

Mark

Jan 1, 1970
0
Doubling the phase doubles the _amplitude_ of the noise, which
quadruples the power, which is a 6dB increase.

Amplitude goes with N, power goes with N^2, 10 log(N^2) = 20 log(N).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Agreed..

but another way to look at it is...

a doubler doubles the carrier frequency and also doubles the FM
deviation...

Mark
 
J

John Larkin

Jan 1, 1970
0
Agreed..

but another way to look at it is...

a doubler doubles the carrier frequency and also doubles the FM
deviation...

Mark

And another way is to note that X picoseconds of jitter at frequency F
multiplies to the same X picoseconds at F*N, but the relative damage
is N times worse.

(Within the active bandwidth of the multiplier, of course.)

John
 
M

Marc Popek

Jan 1, 1970
0
Nicely done!

Marc

Doubling the phase doubles the _amplitude_ of the noise, which
quadruples the power, which is a 6dB increase.
Amplitude goes with N, power goes with N^2, 10 log(N^2) = 20 log(N).

And another way is to note that X picoseconds of jitter at frequency F
multiplies to the same X picoseconds at F*N, but the relative damage
is N times worse.

but another way to look at it is...

a doubler doubles the carrier frequency and also doubles the FM
deviation...
 
J

John Larkin

Jan 1, 1970
0
You can, of course, build a frequency multiplier that has less
jitter/phase noise at its output than at its input.

John
 
Tim said:
Doubling the phase doubles the _amplitude_ of the noise, which
quadruples the power, which is a 6dB increase.

Amplitude goes with N, power goes with N^2, 10 log(N^2) = 20 log(N).

Are we assuming that the phase noise drops
-6dB/octave? Is this a good assumption?

Can you mathematically show how the amplitude
of the noise doubles when you double the phase?


S.
 
T

Tim Wescott

Jan 1, 1970
0
Are we assuming that the phase noise drops
-6dB/octave? Is this a good assumption?

No, I don't assume that phase noise drops 6dB per octave, because it's a
bad assumption. Phase noise is predictable, and there will often be a
region where it drops 6dB/octave of remove from the carrier, but it
doesn't do that everywhere. Search on "oscillator" and "phase noise"
and you'll find some info, I'm sure.
Can you mathematically show how the amplitude
of the noise doubles when you double the phase?

For small phi, cos(wt + phi(t)) approximately equals
cos(wt) + phi(t)sin(wt).

So the amplitude of the noise (phi(t)sin(wt)) follows the amplitude of
the phase (phi(t)), modulated by a carrier that's 90 degrees out from
the main carrier (sin(wt)).

I will let you figure out the doubling part.
 
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