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I'm trying to theoretically analyze the resulting field strength, R, from two devices with given field strengths (say r0 and r1), both represented in dBuV/m.

My first intuition was to use

R = 20*log( 10^(r0/20) + 10^(r1/20) )

But now, I'm starting to think that maybe I should use

R = 10*log( 10^(r0/20)^2 + 10^(r1/20)^2 )

, since if I were to first convert the dBuV/m to V/m I would have to sum them using (v0^2+v1^2) right?

To sum the question.

How to sum two field strengths given in dBuV/m?

EDIT (after answer by @Andy aka ): Is below then the correct computation for resulting field strenght in dBuV/m? r, R in [dBuV/m] v, V in [V/m].

Given (1), (2) and (3)

(1) \begin{equation} v = 10^{( (r - 120) / 20)} \end{equation} (2) \begin{equation} r = 20*log(V) + 120 \end{equation} (3) \begin{equation} V = \sqrt[]{v_0^2 + v_1^2} \end{equation} Gives (4).

(4) \begin{equation} R = 20*log(V) + 120 = 20*log(\sqrt[]{v_0^2 + v_1^2}) + 120 = \\ 10*log((10^{( (r_0 - 120) / 20)})^2 + (10^{( (r_1 - 120) / 20)})^2) + 120 = \\ 10*log((10^{(r_0 - 120) / 10)} + 10^{( (r_1 - 120) / 10)}) + 120 = \\ 10*log(10^{r_0 / 10} + 10^{ r_1 / 10}) \end{equation}

niCk cAMel
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2 Answers2

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You do have to convert to real voltages and then sum the fields as per \$\sqrt{A^2+B^2}\$ where A is one field and B is the other field. If the two sources are frequency locked then you will have to take note of their respective phases and add them vectorially but for different frequencies the Pythagoras formula above holds good.

Andy aka
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  • Thank you for the awesomely quick reply. I've edited my question. – niCk cAMel Apr 20 '16 at 09:30
  • Why have you edited your answer - is there another question resulting from my answer that I'm missing? – Andy aka Apr 20 '16 at 10:27
  • Well... my edit clarifies my question of "How to add/sum..." to "How to mathematically add/sum...". Your answer gave a very good hint, so now I just want to ask wether the derivation is correct. I.e. whether the final equation (4) is correct for use of summing two field strengths – niCk cAMel Apr 20 '16 at 10:38
  • Try suggesting two field strengths numerically and see if I come up with the same result as you. – Andy aka Apr 20 '16 at 10:59
  • LOL, ok. How about... 44.3 dBuV/m + 37.9 dBuV/m ? I get 45.2 dBuV/m – niCk cAMel Apr 20 '16 at 11:04
  • Exactly the number I get!!! – Andy aka Apr 20 '16 at 11:10
  • Excellent! Answer accepted! ... Brings me to a spin-off to your answer (which you may choose to reply to if it makes any sense). I guess having two sources that are frequency locked AND are in phase would also bring the computation down to Pythagoras formula, right? I.e. Pythagoras formula gives the theoretical maximum sum of two field strengths. In other cases, e.g. adding them vectorically, could only result in same or less value? – niCk cAMel Apr 20 '16 at 11:20
  • Two signals that are same frequency and in phase are just additive i.e. 1Vp-p and 1Vp-p = 2Vp-p. However, if they are antiphase you get zero. – Andy aka Apr 20 '16 at 11:34
  • Ok, in order to wrap this up. What number do you get using the same numerical examples used above, assuming same frequency and phase from both fields? I get 47.7 dBuV/m. – niCk cAMel Apr 20 '16 at 11:47
  • It's a bit trivial but yes I get 47.7. – Andy aka Apr 20 '16 at 12:56
  • Why trivial? If you have a radiation limit of say 50dBuV/m in this case at a certain frequency, and you want to check wether you at least theoretically pass the test or not... isn't this a good approach, i.e. to assume phase-matched? – niCk cAMel Apr 20 '16 at 12:59
  • It's a trivial math the addition of two numbers rather than all the square root stuff. – Andy aka Apr 20 '16 at 13:38
  • If you mean that it is mathematically 'simpler' to perform the summation without the 'square toor stuff' I might agree. Otherwise I'd say that 2dB difference in the methods used is not at all trivial. Also, it could be 100 numbers, and not just two, and if I'm looking for the theoretical maximum, I'd say that the choice of method is not trivial. With 'theoretical' meaning that I couldn't say for sure anything about the phases and have to make assumptions. I'm guessing that you are suggesting to use the 'square root stuff' since this reflects reality better? – niCk cAMel Apr 20 '16 at 13:49
  • Use the square root stuff when the two frequencies are unconnected BUT any possibility that they have the same frequency, the combined signal could be 3 dB higher than the sq root math implies. But, conversely the two signals can be subtractive thus cancelling to a flat line. Or anywhere in between. – Andy aka Apr 20 '16 at 13:58
  • Steady on now chap LOL!! – Andy aka Apr 20 '16 at 14:24
  • Yes you are right. Changed back. Your answer AND discussion gave more in return. I just figured @Fionn answer was more direct/layman-styled for somebody else looking for an answer. – niCk cAMel Jun 27 '16 at 15:11
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Depending the origin of these field, one solution could be applied or another.

If both fields are in phase, you should add them in therms of voltage. (1st option) If both fields are noise, it means that they are not in phase, so you should add them in terms of power. (2nd option)

Fionn
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