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If you have two antennas \$T_1\$ and \$T_2\$ each broadcasting 10GHz Gaussian White Noise at the same receiver, each with an EIRP of say 20 dBW with a random phase offset between them, then is there an equation that governs the average received power at the receiver?

This question makes it clear that the sum of two independent white noise sources is still white noise, but it says nothing about the amplitude/received power.

My assumption is the result will be on average the same power as a single transmitter, and if this is the case, is there an alternative type of noise that can be additive despite a random phase offset?

Freddie R
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  • I would think, docking a ship into port, that you'd see more energy from *two* lighthouses as opposed to one. – rdtsc Feb 03 '22 at 13:21
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    You receive twice the power (assuming the same distance to each TX). So you can expect sqrt(2) the received amplitude. As it's statistical, you may have to observe for a while to measure accurately... –  Feb 03 '22 at 13:35
  • As @user_1818839 wrote, you'll get double the power when receiving two **independent** white noise sources. Note that for independent sources there is no concept of "a random phase offset". That phrase suggests you have two copies of the **same** white noise source with some phase offset between them. If you want to know the statistics of that I suggest you ask on the [Signal Processing](https://dsp.stackexchange.com/) stack. – Graham Nye Feb 09 '22 at 20:43

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is there an equation that governs the average received power at the receiver?

Two unrelated signals (voltage or current sources for instance) add up as the sum of squares: -

$$\text{Total RMS} = \sqrt{A_{RMS}^2 + B_{RMS}^2}$$

My assumption is the result will be on average the same power as a single transmitter

No, that isn't the case.

Andy aka
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  • What about the phase difference, frequency difference ? – Wireless Learning Feb 04 '22 at 10:14
  • @LucasVivi the two sources are gaussian white noise hence they have equal spectral densities. In that respect they have the same RMS voltage for each 1 Hz part of the spectrum hence, you can say there are no frequency differences but, they are both noises so, to ask what the frequency difference is is irrelevant. Ditto phase difference. Phase as a measure of the difference between two signals only has relevance when talking about sinewaves (and not noise). – Andy aka Feb 04 '22 at 10:24
  • @LucasVivi did you downvote - was there something I didn't explain sufficiently? – Andy aka Feb 04 '22 at 11:43
  • Noise could be seen as multiple sinewaves superimposed no? (Fourier series) Then if you consider a 1 Hz part of the spectrum you have then two sinewaves from two different emitter which may or may not be in phase? – Wireless Learning Feb 04 '22 at 14:11
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    @LucasVivi no, gaussian white noise cannot be regarded this way and, as I and others have said (and there are many online references that say the same thing), noises add as \$\sqrt{A^2+B^2}\$. [Example](https://www.tempoautomation.com/wp-content/uploads/2018/07/adding_noise_sources-300x188.jpg). [Read also these answers](https://electronics.stackexchange.com/questions/456954/why-are-different-rms-noise-sources-added-in-this-way). If you downvoted, can you please explain why? – Andy aka Feb 04 '22 at 14:21
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    No references... – Wireless Learning Feb 07 '22 at 08:37
  • Put it this way, if I hadn't made an answer, I wouldn't have received a downvote from you but, because I did answer and gave you partial information (i.e. you still gained) you see it fit to then downvote me. Do you think that is fair or reasonable @LucasVivi - I think it is totally unfair and unreasonable and you should think about your logic here. – Andy aka Feb 07 '22 at 09:11
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    @Andyaka As the phrase you've quoted mentions power it would be helpful if you clarified that the equation you've given relates to voltage or current, but not power. The powers of independent noise sources just add together, using simple addition, of course. – Graham Nye Feb 09 '22 at 20:48
  • @GrahamNye my formula equates RMS quantities. RMS quantities are not powers. – Andy aka Feb 09 '22 at 22:04
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    @Andyaka I'm aware of that. It would be an improvement if you clarified that point in your answer for the benefit of those who aren't. It is normal to define the terms used in an equation. Your equation is sandwiched between two mentions of power so it's easy to misinterpret it as also dealing with power. – Graham Nye Feb 09 '22 at 22:19
  • Just for you I've made an amendment although I have to say that it is a form of dumbing down because there is no such thing as RMS power @Graham – Andy aka Feb 09 '22 at 23:18