[time-nuts] 20logN was Re: phase noise questions (long)
Bruce Griffiths
bruce.griffiths at xtra.co.nz
Wed Jan 23 18:13:30 EST 2008
Perhaps the results can be summarised by:
(1+x)2 = 1+2x + x2
For |x| << 1 the term in x2 is negligible.
For FM the term in x determines the amplitude of the (2fc-fm), (2fc+fm)
components
whereas the x2 term determines the amplitude of the (2fc-2fm) and
(2fc+2fm) components.
The 1 term determines the amplitude of the dc and 2fc components.
When the carrier amplitude is much larger than the sideband components
the amplitudes of the 2(fc - fm) and 2(fc+fm) components are much
smaller than the amplitudes of the (2fc-fm) and the (2fc+fm) components
in the doubler output spectrum.
Similarly for frequency multiplication by N the amplitudes of the (Nfc -
fm) and (Nfc + fm) components are much larger than all other components
other than the Nfc component.
In this case (1+x)^N = 1 + Nx + (N(N-1)/2)x2 + ....
The real lesson is that speculation is no substitute for analysis, its
just too easy to arrive at an incorrect conclusion.
Bruce
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