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I have recently taken part of noise floor measurements for emission tests from two separate EMC labs, both well known and international companies. The thing that shocked me is the dramatic difference in the noise floor. We are talking up to 25-30 dBm difference at many frequencies in range of 30MHz-13GHz, with one setup being far superior to the other.

However, given that both noise floors are well below the standard limit lines – how beneficial is a low noise floor in terms of passing below an emission limit? Are the emissions caused by a test object additive relative to the noise floor or are they absolute and the lower noise floor will make little or no difference, at least in terms of compliance?

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  • \$\begingroup\$ The quality of the spectrum analyzer matters a lot. But they need not have a lower noise floor than what the standards tested at demands (with margins). Normal commercial EMC you can probably verify with any old spec. A normal limit is typically some -54dBm iirc, or for spurious emissions of a radio DUT, just some -37dBm. But if going for military EMC, you will need far better equipment. So it depends on the application and requirements. \$\endgroup\$
    – Lundin
    Commented Nov 9, 2023 at 11:01

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However, given that both noise floors are well below the standard limit lines – how beneficial is a low noise floor in terms of passing below an emission limit?

The easiest thing to do is a calculation.

Let's say the noise floor is 20 dB below the limit at a certain spot frequency of interest. Let's arbitrarily say that we have a signal at the limit (corresponding to 1 mV RMS). The noise floor will be at 0.1 mV RMS and, if you vectorially add the two voltages \$(\sqrt{A^2+B^2})\$ you get 1.005 mV or an increase in decibels of 0.043 dB.

If the noise floor were 6 dB lower than the limit, the net effect is a 0.97 dB error.

Are the emissions caused by a test object additive relative to the noise floor or are they absolute and the lower noise floor will make little or no difference, at least in terms of compliance?

You have the numbers so you can do the calculations based on the above.

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  • \$\begingroup\$ I think I don't understand what the vector sum of the two voltages gives. The more I think of it the more I lean toward that the noise floor will not make any difference for compliance as implied in Lundins comment. If you have a emission of 0.5 mV RMS it will always correspond to 0.5 mV RMS independent of what the noise floor was from the start. And this is turn would give same result in both chambes as 0 dB reference shoudl be the same. Would you agree with this? \$\endgroup\$ Commented Nov 9, 2023 at 15:22
  • \$\begingroup\$ No, of course I don't agree. \$\endgroup\$
    – Andy aka
    Commented Nov 9, 2023 at 17:03
  • \$\begingroup\$ Then I think there is some concept I simply do not understand. I could accept that an emitted signal would add vectorially with the noise floor (in contrast with my previous comment), but why add the limit and noise floor vectorially as in your answer? Could you please elaborate on the operations you propose and the net error effect that you mention? \$\endgroup\$ Commented Nov 10, 2023 at 9:38
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    \$\begingroup\$ I chose to imagine that the emitted signal was at the limit. Maybe I rushed it a tad. I'll make that clearer @NoobPointerException \$\endgroup\$
    – Andy aka
    Commented Nov 10, 2023 at 9:42
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    \$\begingroup\$ Now I finally understand the concept, thank you. \$\endgroup\$ Commented Nov 10, 2023 at 13:44

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