IIRC (I could be wrong), the transfer function coefficients contribute non-equally to HD and IMD, so I believe you can manipulate the coefficients to get what you want. For example, 2nd harmonic doesn't contribute to IMD as much as 3rd harmonic does. So system B could be all 3rd harmonic, which would have low THD but high IMD, and system C could be all 2nd harmonic which could high THD with low IMD.

Anyway, that's the idea I had in mind.

Not the math I have seen. I thought I had presented someplace but couldn't find it with a quick search. I have it in an old notebook; not hard, just have to grunge through it. All third harmonic would lead to third-order IMD with no second-order terms, and of course second-order HD would create only second-order IMD terms. There is some debate in which is worse; IMD3 creates tones close to the two fundamental tones so is usually worse than IMD2 tones that appear near DC and at twice the fundamentals, but if the two fundamental tones are very close they may be masked by our hearing.

BTW, isn't an observed fact that we do see in measurements, equipment or other devices having varying levels of HD and IMD? That is, the ratio of HD to IMD is not constant. Different non-linear transfer functions have different ratios of HD and IMD.

Hmmm... Generally not in my experience (which certainly does not cover everything). The times they did not track, it was related to bandwidth rolling off higher-order higher-frequency terms. I suppose it depends upon what sort of transfer function you can create (assuming it matches reality -- easy to create one that is not realizable by a physical circuit).

<Pause>

I actually have a copy of my old notes, from when I derived the terms way back then (and most certainly it was done many decades before I did it!) Since you asked for it, and bearing in mind I may have made copy mistakes from my old hand-written notes:

Let Vout = a + b*Vin + c*Vin^2 + d*Vin^3 where Vin = A*cos(w1*t) + B*cos(w2*t) -- stupid of me to use lower- and upper-case A/B but oh well...

Note w1 and w2 are the two input tones, in rad/s, so w1 = 2*pi*f1 and w2 + 2*pi*f2 where f1 and f2 are in Hz.

Plug in and grind through it using a few (three) trig identities:

cos^2(x) = 1/2(cos(2x) + 1); cosx*cosy = 1/2*[cos(x+y) + cos(x-y)]; cos^3(x) = 1/4*(cos(3x) + 3cosx)

At the end of a page or two of algebra and trig identity swapping you (or I, anyway) get for output Vout = four terms:

- a = DC term
- b*Vin = b[Acos(w1*t) + Bcos(w2*t)] <linear term>
- c*Vin^2 =
- 1/2*c*(A^2 + B^2) <dc>
- + 1/2*c*[A^2*cos(2*w1*t) + B^2*cos(2*w2*t)] <2HD>
- + c*A*B*[cos((w1+w2)*t) + cos((w1-w2)*t)] <2IMD>

- d*Vin^3 =
- 3/2*d*[(1/2*A^3 + A*B^2)cos(w1*t) + (A^2*B + 1/2*B^3)cos(w2*t)] <fundamental terms, typically phase-shifted>
- + 3/4*d*{A*B^2*[cos((w1+2*w2)*t) + cos((w1-2*w2)*t)] + A^2*B*[cos((2*w1+w2)*t) + cos((2*w1-w2)*t)]} <3IMD>
- + 1/4*d*[A^3*cos(3*w1*t) + B^3*cos(3*w2*t)] <3HD>

Now look at the distortion terms:

- 2nd-order distortion gives 2HD at 2*f1 and 2*f2; 2IMD at f1+f2 and f1-f2; and a DC term
- 3rd-order distortion gives additional fundamental energy at f1 and f2; 3HD at 3*f1 and 3*f2; and, 3IMD at f1+/-2f2 and 2f1+/-f2

Looking at the amplitudes, 2IMD will be 6.021 dB higher than 2HD, and 3IMD will be 9.542 dB higher than the 3HD, tracking with input and distortion terms. You can design circuits to suppress even harmonics, for example, and even-order IMD products will fall correspondingly, but according to the formulae. Of course, if distortion does not behave classically, anything goes.

I may have a made a mistake, but it matches classical amplifier design theory and practice (measurements), which was my goal at the time since I was designing high-speed amplifiers and such. I had measured the relationships empirically and my mentor wisely challenged me to prove it in theory. There were deviations, but they fairly were minor and ultimately explainable (by math a lot worse than this and I have no desire to dredge

*that *up!).

My eyes are glazing, HTH - Don