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I'm just a bit confused. Consider an example where an antenna receives two signals - one with a power of -25dBm and one with a power of -47dBm and we are asked to find the total power received by the antenna

Now, if I just add -25 with -47 dBm, I get -72dBm which is a lower total power. How come? Am I not allowed to just add these two? Intuitively, we should get a higher total power surely? Maybe something closer to 1 or 2dBm maybe but not -72dBm?

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    \$\begingroup\$ dBm is a logarithmic scale so addition and multiplication work a bit differently than you'd think. Adding logarithms is the same as multiplying normal numbers. 10dBm + 10dBm does not equal 20dBm, it equals 13dBm. You see, 10dBm = 10^(10dBm/10) = 10mW, 20dBm = 100mW and 20mW = 10log(20mW) dBm = 13dBm. To add signal powers you need to get out of logarithms, convert to mW, add the powers then convert back to dBm (hint: the answer will be really close to -25dBm because -25dBm is ~160 times bigger than -47dBm and a big number + a small number ~= the same big number) \$\endgroup\$
    – Sam
    Sep 28, 2018 at 23:53
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    \$\begingroup\$ Why did you post it as a comment, Sam? It looks like a good answer. \$\endgroup\$
    – Transistor
    Sep 28, 2018 at 23:56

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As the difference in levels between the two signals is 'large' (more than 20dB, more than 100x in power), you can pretty much ignore the smaller signal.

The total power of two signals of -25dBm and -47dBm will be 'about' -25dBm.

If you want to convert both to power, add, and convert back to dB to get the exact figure, as Transistor has done, then you get -24.975dBm. The difference of 0.025dB is insignificant.

You'd have to make very careful measurements in a wired system to believe measurements to that precision. In an antenna system, you can be many dBs adrift without even trying due to direction of arrival, impedance mismatch, path loss, proximity effects etc.

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I haven't done this for a long time so check my answer.

Remember that the decibel, dB, is a logarithmic scale. Adding logs is the same as multiplication of the "un-logged" version of the numbers so that isn't going to work.

Instead we need to get the anti-log of each, sum them and get the log of the result.

$$ \begin{align} P & = 10 log_{10}( 10^{\frac {-25}{10} } + 10^{\frac {-47}{10} } ) \\ & = 10 log_{10}( 0.003162 + 0.00001995 ) \\ & = 10 log_{10}(0.003182) \\ & = -24.975 \ \text {dBm} \end{align} $$

I'm expecting it to make a little bit more of a difference than that. Can anyone see an error?


The benefits of using dB are explained well in the Wikipedia decibel article:

Level values in decibels can be added instead of multiplying the underlying power values, which means that the overall gain of a multi-component system, such as a series of amplifier stages, can be calculated by summing the gains in decibels of the individual components, rather than multiply the amplification factors; that is, log(A × B × C) = log(A) + log(B) + log(C). Practically, this means that, armed only with the knowledge that 1 dB is approximately 26% power gain, 3 dB is approximately 2× power gain, and 10 dB is 10× power gain, it is possible to determine the power ratio of a system from the gain in dB with only simple addition and multiplication.

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    \$\begingroup\$ I'm expecting it to make a little bit more of a difference than that. Can anyone see an error? Simply, no. You said it yourself, it's a logarithmic scale. You are adding something that is 100x smaller. Not going to matter much. \$\endgroup\$
    – mike65535
    Sep 29, 2018 at 0:34
  • \$\begingroup\$ maybe correct it by the RMS of the two not the sum \$\endgroup\$ Sep 29, 2018 at 1:17
  • \$\begingroup\$ Go ahead, Tony, and write an answer. I have very seldom had to work in dB / dBm. I'm doing this as a mental challenge. \$\endgroup\$
    – Transistor
    Sep 29, 2018 at 1:20
  • \$\begingroup\$ I think just square root of (sum of watt^2) then convert back to dB \$\endgroup\$ Sep 29, 2018 at 1:25

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