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The cubic difference frequency (or 3rd degree intermodulation product) is given by:

$$D_3 = - 10\log\left(\frac{P_2\left(2 f_{1,2} - f_{2,1}\right)}{P_2\left(f_1,f_2\right)}\right)$$

where $P_2$ is the output power. As far as I understand, the 3rd degree IM products are at frequencies $2f_1 - f_2$ and $2f_2-f_1$. So the formula for the IM product of $2f_1 - f_2$ should be:

$$D_3 = - 10\log\left(\frac{P_2\left(2 f_{1} - f_{2}\right)}{P_2\left(f_1\right)}\right)$$

When the power is given in an logarithmic scale, this would simply be the difference between the two power values.

  • But what does this actually tell me?
  • Is this the power of my 3rd degree intermodulation product? (if it is, $D_3$ should be equal for both frequencies)
  • And is this formula only valid for an input signal of two different frequencies?

According to a paper D3 can also be expressed in terms of gain compression c and change in AM-PM characteristic kp due to only one carrier: $$D_3 = -10\log(\frac{c_1}{2}^2 + k_{p1}^2)$$

The gain compression is given by $c = 1 - \frac{\Delta P_2 / P_2}{\Delta P_1 / P1}$ , where $\Delta P_1$ is a constant change of input power and $\Delta P_2$ is a not constant change of output power. I am not sure however, if $P_1$ or $P_2$ refers to the input/output power before the change or after.

  • I cannot see $P_1$ in your formula, and the index notation does not seem consistent ($P2$ or $P_2$?). Could you also provide a reference for the notions and formulae you are using? – Laurent Duval Dec 29 '16 at 10:56
  • you're right, $P_1$ is not being used in this formula. As a reference, I am using a german paper: Bretting, J., and P. Treytl. "Größen zur Charakterisierung der Nichtlinearitäten von Mikrowellen-Sendeverstärkern und deren Zusammenhänge." NTZ-Kurier, March (1972). I can send you a copy, if you want, but since it's in german I don't know how useful it will be. I haven't found an english paper about this topic, but if you know any, I appreciate any help ! –  Dec 29 '16 at 14:30
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    There is too little context to understand the question (at least for me). I do not understand the notation. $P_2(f_1,f_2)$ looks like a function of two variables, but that doesn't square with the notation in the numerator. What is $f_{1,2}$, etc.? The input and output of what? If you like provide a link to the corresponding paper, or - even better - add sufficient context to the question. – Matt L. Dec 29 '16 at 17:16
  • I don't have a link. But I could send you a copy. –  Dec 29 '16 at 19:04
  • @Luk Share the paper, or at least a crop of the page, if you can. German is not a problem as long as there are some maths in it – Laurent Duval Dec 29 '16 at 20:32
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    You keep on modifying your question, and you unaccept an accepted answer just because you actually ask a new question. You should have asked a new question instead of changing your old question, because it's not possible to answer an ever changing question. – Matt L. Jan 05 '17 at 10:03
  • ok, sorry. I thought it was easier to see what my question is this way. I will ask a new question then. –  Jan 05 '17 at 10:38

1 Answers1

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After skimming through the paper, I can see more clearly now. The measure $D_3$ quantifies the relative strength of the 3rd order intermodulation product. If two sinusoidal signals with frequencies $f_1$ and $f_2$ are input to a non-linear device (such as a microwave amplifier, as referred to in the paper), there will be intermodulation products at the output. Third-order intermodulation products occur at frequencies $2f_1+f_2$, $2f_1-f_2$, $2f_2-f_1$, and $2f_2+f_1$. Problematic are the ones at frequencies $2f_1-f_2$ and $2f_2-f_1$ because they can end up in the frequency band of interest.

The quantity $D_3$ is defined as the ratio of the power of one of these undesired frequency components and the power of one of the desired frequencies (in dB) at the output of the nonlinear device. There are four possible definitions:

$$D_{3,1}=-10\log\left(\frac{P_2(2f_1-f_2)}{P_2(f_1)}\right)\\ D_{3,2}=-10\log\left(\frac{P_2(2f_1-f_2)}{P_2(f_2)}\right)\\ D_{3,3}=-10\log\left(\frac{P_2(2f_2-f_1)}{P_2(f_1)}\right)\\ D_{3,4}=-10\log\left(\frac{P_2(2f_2-f_1)}{P_2(f_2)}\right)$$

where $P_2(f_x)$ is the output power at frequency $f_x$. These four definitions do not necessarily result in the same value for $D_3$. It's just important to specify which of them is used, and to consistently stick to that definition.

To answer your questions: no, $D_3$ is not the power of the 3rd order intermodulation product, but it is the ratio (in dB) of that power to the power at one of the signal frequencies. This quantity is defined based on the assumption that the input signal is the sum of two sinusoidal signals, but it is used to quantify the strength of 3rd order nonlinearities at the output of a nonlinear system regardless of the input signal.

Concerning the power compression factor, you can view the changes $\Delta P_1$ and $\Delta P_2$ as infinitesimal changes, and so the question if $P_1$ and $P_2$ are the values before or after the change becomes irrelevant. At a given frequency, an ideal linear amplifier would satisfy

$$P_2=g\cdot P_1$$

where $g$ is the gain factor at the given frequency. So you have

$$g=\frac{P_2}{P_1}=\frac{dP_2}{dP_1}\Longrightarrow\frac{dP_2}{dP_1}\frac{P_1}{P_2}=1$$

That's why the power compression factor

$$c=1-\frac{dP_2}{dP_1}\frac{P_1}{P_2}$$

is a measure for the nonlinearity of power amplification.

Matt L.
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  • allright, thank you so much! Assuming I am not given the power in the numerator such as $P_2(2f_2-f_1)$ , according to formula (14) I should still be able to calculare $D_3$, right? I would have to calculate the compression c and AM-PM conversion $k_p$ according to (1a) and (2) for one carrier e.g. $f_1$ first. –  Dec 30 '16 at 17:31
  • @Luk: That's what the authors claim; I haven't checked the formula, but note that it is an approximation, obtained by a Taylor series expansion. – Matt L. Dec 30 '16 at 17:42
  • I edited my question as I don't understand (1) completely either. It'd be awesome if you could have a look at that as well! –  Dec 30 '16 at 17:43
  • @Luk: Have a look at my edited answer. – Matt L. Dec 30 '16 at 18:50
  • cool, that makes sense! Say I know the corresponding output power of two different input powers: $P_1(1) = -7 dBm$, $P_1(2) = -6 dBm$ and $P_2(1) = 51.74 dBm$, $P_2(2) = 51.787 dBm$. Is it possible to calculate c from these known values? I mean the change $dP_1$ and $dP_2$ are now not infinitesimal. When the compression is determined they should be infinitesimal however. When I calculate the compression from the given data, $c = 1 - \frac{0.047 * (-7)}{1*51.787} = 1.0064$ –  Dec 30 '16 at 19:28
  • sorry, what I meant was: with the above mentioned values, c can either be $c = 1 - \frac{(P_2(2)-P_2(1))P_1(1)}{(P_1(2)-P_1(1))P_2(1)} = 1.0064$ or it can be $c = 1 - \frac{(P_2(2)-P_2(1))P_1(2)}{(P_1(2)-P_1(1))P_2(2)} = 1.0054$. But that's just provided it was possible at all to calculate the compression from such values. –  Dec 31 '16 at 11:24
  • @Luk: First of all, you need to use linear power values, not values in dB. Second, if you approximate infinitesimal $dP$ by a finite number $\Delta P$, it doesn't really matter if you use $P$ or $P+\Delta P$, or some value in between as a reference value, as long as you're clear about it, so others can reproduce what you've done. – Matt L. Dec 31 '16 at 15:44
  • I am given the input and output power in dBm as well as the according amplitudes. And I am also given the change in phase of my input signal in degree. Since you say I need to use linear power values, I assume the amplitudes will do. And kp should be equal to the change in phase, I assume. Is that right so far? Because then I finally understand. –  Dec 31 '16 at 15:56
  • @Luk: Yes, the amplitudes should do. As for $k_p$, see my answer to your other question, but as far as I understand your comment, it looks like you've understood it correctly. – Matt L. Dec 31 '16 at 16:21
  • May I suggest you to add a link or the first illustration from http://electronicdesign.com/communications/understanding-intermodulation-distortion-measurements that seems quite clear – Laurent Duval Jan 01 '17 at 16:46
  • @LaurentDuval: Sure, feel free to add anything that you consider useful information. – Matt L. Jan 01 '17 at 16:52
  • @LaurentDuval: thx for the link! I consider a two-tone input signal. First the input power is -26dBm, then -16dBm and then -6dBm. According to the link, the power of my 3rd-order IM-products should increase by 3dB per 1-dB increase of input power. In my simulation the power of the 3rd-order IM-products increase by 26.528 dB in the first step and by 14.43 dB in the second step. Is this a hint that my simulation results are wrong? –  Jan 04 '17 at 15:43
  • I believe that @MattL. is better able to answer this – Laurent Duval Jan 05 '17 at 20:40