# [time-nuts] 20logN was Re: phase noise questions (long)

Bruce Griffiths bruce.griffiths at xtra.co.nz
Wed Jan 23 17:13:42 EST 2008

```Christophe Huygens wrote:
> 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
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- 0.5*beta*beta)*coswct - 0.5*beta[cos ((wm-wc)t) - cos
((wc+wm)t)]

fc(t)*fc(t) = (1- 0.5*beta*beta)*coswct*(1- 0.5*beta*beta)*coswct - (1-
0.5*beta*beta)*coswct*0.5*beta[cos ((wm-wc)t) - cos ((wc+wm)t)]  +
0.5*beta[cos ((wm-wc)t) - cos ((wc+wm)t)] *0.5*beta[cos ((wm - wc)t) -
cos ((wc + wm)t)]

fc(t)*fc(t) = (1- 0.5*beta*beta)*(1- 0.5*beta*beta)*[0.5*(1 + cos(2wct)]
+ (1- 0.5*beta*beta)*0.5*beta*[0.5*[cos((wm-2wc)t) + cos((2wc+wm)t)]]
+ 0.25*beta*beta*[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 - 0.25*beta^2  + 0.125*beta^4) +
(0.5 - 0.25*beta^2 + 0.125*beta^4) cos(2wct) +
(0.25*beta - 0.125*beta^3 )* cos((2wc-wm)t) +
(0.25*beta - 0.125*beta^3 )* cos((2wc+wm)t) +
(0.125*beta^2)*cos(2*(wm-wc)t) +
(0.125*beta^2)*cos(2*(wm + wc)t)

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

0.5*cos((2wc)t) +
0.25*beta*cos((2wc - wm)t) +
0.25*beta*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+ 0.5*beta*{exp(jwmt) - exp(-jwmt)})]

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

```