Can two or more RF noise signals (or noise sources) be attenuated below the thermal noise floor, and still add together to be higher than TNF?

If a noise signal experiences path loss or attenuation great enough to reduce the original noise signal below the thermal noise floor, will the signal still add to the noise of other noise sources, and if there are enough of these noise sources adding together, can they cumulatively add together to raise the total noise received above the thermal noise floor at the Rx site?

Example question to illustrate further:

There are 100 signal boosters acting as white RF noise sources (uncorrelated, non-coherent, etc.) and all are pointing to the same RF receive site, but they are all at different locations. The noise power level from each of them is measured at exactly -40dBm @ 10kHz RBW from their Tx output.

Each noise source transmitted then experiences a "total path loss" attenuation of exactly 100dB on the way to the Rx site (for simplicity, we'll say this total path loss includes all FSPL, antenna gains, feedline and connector losses, etc.).

Assume the noise at the Rx site is perfect, just the Thermal Noise Floor @ 10kHz RBW = -134dBm (or -174dBm @ 1Hz)

log100 x 10 = 20dB (This should be the total increase/gain in noise signal power for the 100 equal noise signals adding together)

In the above example, what is the final RF noise received at the site from all the signal boosters?

A) -120dBm @10kHz RBW (-140dBm + 20dB -100dB = -120dBm) (This answer implies that each noise signal adds to the value of the other noise signals, EVEN if they fall below the Thermal Noise Floor)

B) -134dBm @10kHz RBW, the Thermal Noise Floor (This answer implies that once each noise signal drops below the thermal noise floor, it can no longer be added to the noise from the other noise sources)

C) Some other value (Please explain!)

This answer is important to that owner of the receive site, if the correct answer is A, the noise floor at the Rx site has been raised well above the TNF (I refer to this as "noise creep"), just a little bit by each one of those signal boosters, but when taken altogether, this can result in a significant desense to the receiver.

Source article to consider : https://combausa.com/en/tech-briefs/whats-all-the-fuss-about-uplink-noise

I assume the article information is correct, however, the noise signal levels mentioned are all above TNF. So, its still not quite clear if this would still apply to noise signals that are "theoretically" below the noise floor.

Thanks for reading my lengthy question, and to anyone who can help answer this! Please, if possible, site a reliable source material or website if you can.

  • \$\begingroup\$ Look up the radiometer equation! It describes how long you need to integrate to be able to detect a certain noise power (below your receiver thermal noise). For example, when taking the temperature of Mars or Jupiter, you don't expect to see a big, visible change in the received noise power. But integrated over minutes or hours, and carefully calibrated, you can measure the tiny contribution from the planet. \$\endgroup\$
    – tomnexus
    Commented Aug 31, 2023 at 0:20

1 Answer 1


power doesn't disappear just because it's under the kTB noise floor, the contribution is still there. Any noise sources (noise defined liberally here) will add to the kTB thermal noise floor. If you have enough of them, you'll increase the noise floor. The 2.4 GHz and 5.8 GHz ISM bands apparently have this problem now.

  • \$\begingroup\$ Thanks! I believe your concise answer is pretty much exactly what I was looking for. PLEASE, Do you have any reference to cite for this??? It was very difficult to find even a similar question on threads like this, no less anything from a text book or official reference source of any kind. \$\endgroup\$
    – M P
    Commented Aug 30, 2023 at 15:33
  • \$\begingroup\$ law of conversation of energy? Noise is energy. The kTB noise floor is a thermal noise contribution, but it's not magic, it's just energy that's always present, but its only one contribution to a receiver noise floor. In full duplex transceivers, the transmitter portion will output broadband noise because of the PA, which will leak into the receiver(Remember, they're connected to a single antenna) and add to the kTB thermal noise floor. That effect is what sets duplexer rejection specifications. \$\endgroup\$
    – rfdave
    Commented Sep 1, 2023 at 1:54
  • \$\begingroup\$ Answer B is incorrect, BTW. The noise floor with 1 additional booster signal is the sum of -134 dBm + -140dBm. -> 39.811e-18W + 10e-18W = 49.811e-18W = -133.03 dBm so even with 1 booster you degrade the receiver noise floor by ~ 1 dB. \$\endgroup\$
    – rfdave
    Commented Sep 1, 2023 at 2:00
  • \$\begingroup\$ Yes, B was assumed incorrect. According to your response, it is answer A that is correct. Answer B was only suggested as a correct answer if the false assumption that the noise was "gone" if it fell below the thermal noise floor was actually true. Since it is not, then we have answer A: -140dBm x100 = -120dBm. Technically, we could also add the -134 from the TNF to the -120dBm here, which would give us about -119.8305, but I was just looking for a round answer and constructed the problem so as to get the answer about the concept, not the exact number to 4 decimal places, useless in the field. \$\endgroup\$
    – M P
    Commented Sep 18, 2023 at 23:40
  • \$\begingroup\$ 1st law of thermodynamics makes perfect sense. I was just looking for something like a specific textbook or reference example spelling out something closer to this scenario. \$\endgroup\$
    – M P
    Commented Sep 18, 2023 at 23:41

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