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If a device claims a noise figure of 0.3 dB NF such as the Qorvo QPL9547, can you calculate the dBm value of a signal that is at the noise floor (ie, a barely detectable signal)?

Otherwise stated: I know the noise figure, can I use that to estimate the minimum detectable signal?

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    \$\begingroup\$ No, it can only tell you the ratio of SNR Out/In. Sensitivity depends on your matched filter = signal BW and impedance and connection losses \$\endgroup\$ Dec 24, 2021 at 5:08
  • \$\begingroup\$ You can if you know the characteristic impedance and bandwidth. \$\endgroup\$
    – user16324
    Dec 24, 2021 at 16:24

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can you calculate the dBm value of a signal that is at the noise floor

You can calculate the power of the noise in any specified bandwidth in dBm.

The noise figure of the device will give you the input referred noise, when terminated in the system impedance. The noise of a correctly terminated system is about -174 dBm/Hz in that system impedance. If a device claimed a NF of 0.3 dB, then you'd have an input referred noise of -173.7 dBm/Hz.

Note that the datasheet has de-embedded the loss of the PCB input trace when specifying that, so you'll never actually achieve that figure. All losses from the antenna to the device will increase the antenna-referred noise figure.

(ie, a barely detectable signal)?

Whether a signal whose power is equal to the noise in the signal's transmitted bandwidth is 'detectable' is a function of the signal design and the definition of 'detectable'. For instance, a GPS signal is designed to have a much smaller information bandwidth than its transmitted bandwidth, and use that apparent inefficiency to make it detectable way below the noise in its transmitted bandwidth. A CW signal, or a broadband modulated signal carrying analogue, or forward error corrected digital, would all have different 'detection' criteria.

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  • \$\begingroup\$ Is rtHz sqrt(Bandwdith-in-Hz)? If so, then understand correctly the noise level for a 15kHz bandwidth FM signal would be -173.7/sqrt(15000) = -1.4 dBm? This doesn't seem right. (I realize that ideal conditions are unattainable, I'm just trying to understand the math.) \$\endgroup\$
    – KJ7LNW
    Dec 28, 2021 at 2:46
  • \$\begingroup\$ After a bit of searching I found this to be about what I am looking for for a 15 kHz bandwidth: −174+0.3+10×log(15×10^3)=−131.9 from en.wikipedia.org/wiki/Minimum_detectable_signal . I'm guessing your explanation has the same information, how does your sqrt term come into play here? \$\endgroup\$
    – KJ7LNW
    Dec 28, 2021 at 2:53
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    \$\begingroup\$ @KJ7LNW Sorry for confusing you, your -131.9 figure is correct. I was using the sqrt(Hz) lazily and wrongly. dBm is always power, and power is per Hz. However frequently when people talk about detectable signal, they are using volts, which is of course sqrt(power) and so needs sqrt(Hz) to normalise. I've got so used to correcting people, I over-corrected, wrongly. I've fixed my answer. Good for you for not believing the obviously wrong -1.4dBm, and having the patience to find the wiki article. My noise_floor != MDS is still valid, and depends on the design of the signal, and SNR for detection \$\endgroup\$
    – Neil_UK
    Dec 28, 2021 at 9:23

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