I need a mathematical description of what happens to the spectrum of a signal when that signal is processed to divide its frequency.

As an example, construct an input signal as the sum of two sine waves of different frequencies (as commonly used in a "two tone" test). Processing this signal with a buffer that changes state at the rising zero crossings of the input signal produces a squared wave whose frequency can be divided by a flip-flop: enter image description here

The spectrum of the the input signal comprises tones at 102kHz and 105kHz: enter image description here "Squaring" the input preserves the two equal tones, albeit with added side tones (spaced at the difference between the two original tones) that result from removal of the amplitude envelope and added harmonics from the squaring: enter image description here

Passing the squared signal through a flip-flop produces a second square wave which in which the two-tone nature is no longer present: enter image description here

The single dominant tone is at 1/2 the frequency of the higher of the original two tones, while the spacing between the peaks continues to be the spacing between the two original tones.

Passing the squared result through an LC low-pass filter produces a sine wave with cycles that are half as frequent as the input: enter image description here

with the same general spectral distribution as the squared output: enter image description here

Can someone, please, explain what happened to the other tone?

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    \$\begingroup\$ Possible duplicate of Effect of frequency division on a signal's spectrum? \$\endgroup\$ – Chris Stratton Mar 10 '19 at 18:06
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    \$\begingroup\$ You already asked this yesterday. This should be an edit to illustrate your existing question, not a new posting which discards all the relevant history there. As suggested yesterday, if you want to see the effect of division try one tone at a time. What you are doing is getting confused by the non-linearity of your flip-flop. \$\endgroup\$ – Chris Stratton Mar 10 '19 at 18:07
  • \$\begingroup\$ If you want to understand Fourier spectrum, you can draw arbitrary waves with mouse or choose sin square etc. . But check boxes Mag/Phase and Log views falstad.com/fourier but if you want to design a SSB modulator then you need to search "SSB or vestigial filter" like those used for TV signal Tx's \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 10 '19 at 18:53

I would observe that your two tones can alternately be seen as a DSB modulated carrier at the average frequency, 100% modulated by half the difference.

$$\cos (\theta) + \cos (\phi) = 2 \cos ((\theta+\phi)/2) \cos ((\theta-\phi)/2)$$

But your clipper effectively converts the phase modulation component to pulse width modulation (this is much more obvious if you re run the maths with the carriers much further apart, try a 50kHz spacing) in a horribly non linear way, kind of like jitter on a clock, I think there be bessel functions hiding in there somewhere.

The divide by two however is an edge triggered event but the place the edges really move a lot is right around the modulation zero as the difference component swaps polarity and you will usually get an extra edge at this point, giving you the single tone (The divider does not care about width, only either rising or falling edges) with phase discontinuities at the difference frequency as the $$\cos((\theta-\phi)/2)$$ term changes sign.

Possibly a bit handwavy, but it has been a long time since school.

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If you input 103k and 108k, then slicing the signal (comparator) produces the average f and sidebands from the input Δf then f/2 +/- Δf/2 after the 1/2 divider and all the odd harmonics of each input f/2 as it just divides the spectrum in half.

read more about VSB or SSB filters here and there


Consider a more intuitive case using 95kHz and 105kHz sines of equal amplitude.

If only one signal at 100kHz is input into a slicer and /2 FF you get 50kHz.

Superposition of 2 sines gives f1+f2 when sliced into a square wave each has odd harmonics to declining amplitude ( neglect Vdd/2 for now)

enter image description here

Thus 1/2 divide FF yields frequencies of :
Δf= 1/2( nf1+-mf2 ) for n,m = 1 to x harmonics of each carrier.

Peak spectrum output of 1/2 FF is 50kHz
with 1st sideband @ 45kHz , slightly bigger than 3rd peak @ 55 kHz
and so forth every +/- 5khz since the initial Δf=10kHz is now 5kHz after f/2
and then intermodulation peaks rising from 5k, 10k 15k, 20k then declining ..30k etc

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