Unfortunately I cannot edit this post any more but in addition I have one more question:
When I have vout = c1*vin + c3*vin^3, IIP3 is defined as the voltage level at which 3rd level distortion equals first order: viip3^2 = 4/3 * c1/c3
However, most of the time, IIP3 is given in dBm. I can think now about three possible conversions:
1.) Voltage square is already power, so P = 10*log10(viiip^2) -> dBW; P = 10*log10(viiip^2 * 1000) -> dBm
2.) Supposing input voltage is amplitude of a sinusoid and assuming R=1 (as often), P = 10*log10(viiip^2/2) -> dBW; P = 10*log10(viiip^2/2 * 1000) -> dBm
3.) But why R=1? Maybe R=50? P = 10*log10(viiip^2/50) -> dBW; P = 10*log10(viiip^2/50 * 1000) -> dBm
Which one is the correct conversion? Keep in mind that I am not asking from an abstract point of view (NOT circuits/physical!), i.e., if I would just do some simulations in MATLAB where I need to generate some numbers. They can be anywhere between eps and 1/eps. So if I assume all my numbers have unit Volts and if I write vout = c1*vin + c3*vin^3 I mean Voltage-in/Voltage-out amplifier, which from the 3 options is the correct way to transform e.g. IIP3=30dBm ?
Last but not least, nearly all RF papers quote IIP3 as quality of "linearity FOM". I still cannot wrap my head around how this could be useful without specifying the input signal level at the same time. For example,
https://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=1377345 says it has an IIP3=-dBm but it does not say anything about the input signal level and it not even specifies intended application (from which an input signal level could be deducted?). How can this be useful? Is -9dBm good or bad? Comparing with the plot above it looks terrible but they still call it "highly linear". WHY?