Suppose I have an unknown system and I would like to know its linearity.

I am confused by the proper measurement of the IIP3. The gain is defined as "total output power" divided by "total input power".

By sweeping a single input tone and obtaining a the power of the single output tone, I know that the gain is about 24.

But to create an IIP3 plot, one uses TWO tones and measures the power of ONE output tone and ONE IM3 (e.g., https://www.maximintegrated.com/en/app-notes/index.mvp/id/5429).

Option 1: Use one output bin and subtract 6dB from the total input power (since it consists of two tones)

enter image description here

  • The gain is correct
  • The IIP3 is worse than I would expect
  • The input power is the DUT is actually wrong ... in reality it is twice as powerful (since I have two tones but only consider one).

Option 2: Add the power of both output signal bins and both output IM3 bins (in linear domain) - use total input power:

enter image description here

  • The gain is off by 3dB
  • The IIP3 comes closer to the region I would hope for
  • I do not like that because the amplifier is frequency dependent and in general, the two tones (as well as the two IM products) can be (slightly) different.

Option 3: Use one output bin and one IM3 output bin and total input power

enter image description here

  • The gain is off by 6dB (This is expected because I am considering only half the power at the output than I have at the input)
  • The IIP3 is closer where I could hope it to be

Option 4: Use total input power. For output power, use sum (in linear units) from both signal bins ... for the IM power, only use one bin

enter image description here

  • The gain is again off by 3dB
  • The IIP3 is where I would hope :-)
  • It does not sound right to take the sum of the output fundamentals but only consider one output IM3.

In any case, none of these 4 options looks "correct" to me.


1 Answer 1


You need to understand what the VSG is calling its Pout, the amplifier's Pin, when there are two tones present. It might be the total (thermal) power (+3dB(single)), it might be the peak power (+6dB(single)), or it might be the power of one individual tone (0dB(single)). You should be able to find out in its operating manual. Or, you could find out by measuring it with your ADC without the amplifier in circuit.

Presumably the Pout obtained from spectral analysis of the ADC on the output is measuring the power of any individual tone, regardless of how many there are.

I am a mistrustful curmudgeon, and will routinely check what a source or analyser is doing when making/analysing a complicated signal by direct measurement, when there is the slightest possibility of uncertainty. It's also nice to 'warm up' by making a measurement for which I know what the answer should be, just to check I understand everything before trying to measure a device I know nothing about. I continue to be surprised at how often the warm up measurement throws up an unexpected problem in the gear, the cables, or my concept of what their readings mean.

  • \$\begingroup\$ It's less about the devices: I know (and measured/confirmed) that the VSG Pout is the total RMS output. My question is: The input signal to my DUT consists of 2 tones with a total power of "Pin". At the output I am only measuring one of these tones - "P1". Do I manually need to subtract 6dB from "Pin" in order to to make it consistent (compare one input tone to one output tone)? (What confuses me here that this would be wrong: The input power to my DUT is NOT one tone but it is TWO tones) \$\endgroup\$
    – divB
    Commented May 10, 2018 at 18:15
  • \$\begingroup\$ If the VSG outptu is total rms output, then you need to add the power of the two tones you've received on your spectrum analyser to make the total rms amplifier output. If the two tomes are equal, then that's 3dB above either of the single tones. You have to measure both ports according to the same definition, or you get the wrong answer. \$\endgroup\$
    – Neil_UK
    Commented May 10, 2018 at 19:56
  • \$\begingroup\$ But this still does not give me the correct result. I think you misunderstood my question. I completely revised it along with better descriptions and my IIP3 plots. \$\endgroup\$
    – divB
    Commented May 10, 2018 at 20:42
  • \$\begingroup\$ I'm not going to flog through your confusing plots, and your incorrect logic where you subtract 6dB to reduce indicated power, or add power as linear units. Do this. Determine the output power of each of your VSG output tones, whether it's making one, two or N tones, by measuring each tone with your ADC/specANA through a direct cable, ignore your misunderstanding of what it claims. Do the same to the output of the amplifier. Take ratio. Relax in certainty. \$\endgroup\$
    – Neil_UK
    Commented May 11, 2018 at 5:19
  • \$\begingroup\$ To be clear here, if I was mentoring a junior engineer through this sort of problem, once I'd told him to go off and calibrate his VSG like I did 24 hours ago, if the problem remained, I'd lean over his shoulder to watch exactly what he was doing. Unfortunately I can't do that to do, so I have zero faith in the correspondance of any of your reported figures with what the instruments say. Which is why I'm not going to trawl through the figures, second guessing where you might have errored. Note that 3dB is twice power, and 6dB is twice voltage. \$\endgroup\$
    – Neil_UK
    Commented May 11, 2018 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.