I'm reading about a signal to noise ratio (SNR), I'm quite new to this kind of stuff. $$SNR=10\log_{10}\left(\frac{P_{SIGNAL}}{P_{NOISE}}\right)=20\log_{10}\left(\frac{A_{SIGNAL}}{A_{NOISE}}\right)$$ I don't know why I read two times that the SNR is set to be: $$SNR=20\log_{10}\left(\frac{A_{SIGNAL}}{A_{NOISE}}\right)-3dB$$

Why to put as a baseline -3dB .

I read online that is a recurring number for this I ask the question but I don't know where it comes from

  • 1
    \$\begingroup\$ Can you share a link to exactly where you read this? \$\endgroup\$
    – The Photon
    Oct 24, 2019 at 21:05

1 Answer 1


-3dB is the point of 50% power. The point when the tipping point changes from a majority of the power to a minority of the power gets through.

The native form of the equation is: $$ SNR=10\log_{10}\left(\frac{P_{SIGNAL}}{P_{NOISE}}\right) $$

Power is proportional the square amplitude (think \$P= I^{2}R\$ or \$P = \frac{V^{2}}{R}\$). Which is why if you use dB to calculate amplitude, you get:

$$ SNR=20\log_{10}\left(\frac{A_{SIGNAL}}{A_{NOISE}}\right) $$

Changing the coefficient from "10" to "20" applies an extra "2" which works it's way into the exponent through the log function which effectively squares the amplitude to end up with the dB which is supposed to represent power.

So the thing to understand is that dB ALWAYS talks about power. ALWAYS. There is no such thing like dB for amplitude. It's always dB for power. There's no such thing as "it increased in amplitude by 3dB". It's always "it increased in power by 3dB", but you can use that to calculate the equivalent increase in amplitude.

So now, to answer your question:

$$ SNR=20\log_{10}\left(\frac{A_{SIGNAL}}{A_{NOISE}}\right)-3dB$$

is wrong as far as I can tell. Subtracting 3dB just makes the SNR half the power which makes no sense. If you provided more context maybe we can be more clear. I am also really curious where you saw this if you saw it more than once (especially if you saw it in more than one paper).

EDIT: I think I know what's going on. It's still wrong though.

I think whoever wrote that was thought that the 10log() equation was dB for power, the 20log() equation was dB for amplitude (something that doesn't exist), but knew that SNR in dB was always in terms of power.

And then they tried to come up with a quick shortcut between the dB equations for power and amplitude. Their minus 3dB was supposed to be a shortcut way to undo the 20log() and turn it back into a 10log() without replacing the amplitude ratios with power ratios inside the log function.

However, they messed up because the multiplication by two which turns the 10log() into a 20log() actually represents a power of two since the log() function places it in the exponent. It appears in the equation as a multiplication of two but isn't actually a multiplication of two on a linear scale.

Meanwhile, subtracting 3dB equates to a division of two on a linear scale, which was supposed to undo the previous multiplication by two...except as we established, the previous multiplication by two wasn't actually a multiplication of two on a linear scale so it doesn't actually undo the squaring that took place.

Of course, none of that was necessary since the 10log() and 20log() equations both represent the same thing: dB which is a measurement of power, not amplitude.

  • 2
    \$\begingroup\$ I'm not sure it's quite right to say "there's no such thing as dB for amplitude". It makes sense to say something increased in amplitude by 3 dB, that just means it increased by a factor of √2. (equivalently, the power increased by 1.5 dB) \$\endgroup\$
    – Hearth
    Oct 24, 2019 at 21:29
  • \$\begingroup\$ @Hearth I understand what you mean, but that's a nuance that should really be avoided by beginners who have not yet grasped things. And even with that wording, the dB is still directly referring to power, not amplitude. \$\endgroup\$
    – DKNguyen
    Oct 24, 2019 at 21:33
  • \$\begingroup\$ @Hearth It's kind of like saying "that resistor is dissipating 10A worth of power" Even though your words refer to current, the content of your words are actually referring to the amount of power" \$\endgroup\$
    – DKNguyen
    Oct 24, 2019 at 22:11
  • \$\begingroup\$ @DKNguyen I love hearing your mind work. You hold precision meanings for your terms, not sloppy ones, and that's important. The idea when applied to voltage is sloppy and assumes a particular repeating wave type (sinusoidal, for example) in order to come up with things like \$\frac{\sqrt{2}}{2}\$. It's better to work outward while standing on solid, clear concepts. Then "things just work right." I appreciate seeing your clarity here. FYI. +1 \$\endgroup\$
    – jonk
    Oct 24, 2019 at 22:16
  • \$\begingroup\$ @jonk Interesting. I never even thought about what dB being used for amplitude really means. I always thought of it as just "amplitude" and nothing more. It had never even occurred to me it specifically refers to a sinusoid and would get messy really fast for any other waveform (except for maybe DC). \$\endgroup\$
    – DKNguyen
    Oct 24, 2019 at 22:26

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