Let $S = X + N$ be the sum of two audio signals $X$ and $N$ which are both stationnary (let's think X is a constant volume 440 Hz sinusoid and N is constant volume noise).

If the sum S has a -20 dB volume and N has a volume of -30 dB, what is the volume of X? (could be RMS or peak volume, it doesn't matter here).

The answer is

20*log10(10^(-20/20)-10^(-30/20)) ~ -23.3

i.e. X has a peak volume of -23.3 dB.

(of course this is not true if X and N have phase-cancellation, but except this case, it works).

enter image description here

Question: what is the name of this computation? i.e. find the volume of X given the volume of N and X + N?

I'm using an application of this for STFT denoising (with a noise template which is N):

  • if a FFT bin has -20 dB amplitude (signal S), and the noise has amplitude -30 dB amplitude for the same bin (signal N), what would be the amplitude of the denoised signal X ? Answer: -23.3 dB, thus this FFT bin should be lowered by 3.3 dB.

It works quite well for my noise reduction application (again here X is nearly a constant sinusoid and N constant noise), but I haven't found a name for this simple technique. What would be the name?


1 Answer 1


Estimating the power of a sum of signals depends on coherence or incoherence assumptions. Details are given for instance on incoherent signal summing or coherent signal summing. Those can provide an estimate of the noise level. They can be called power sums or voltage sums. Other details in Voltage and power add differently. I'll be digging for other names.

The expectation of the energy of a sum $v$ of variables $v_1$ and $v_2$ is:

$$ E((v_1+v_2)^2) = E(v_1^2)+E(v_2^2)+2E(v_1 v_2)\,.$$

Three specific cases. If $v_1=v_2$, you get $+10\log 4 = +6 $ dB. If sources are uncorrelated, $E(v_1 v_2)=0$, so you get $+10\log 2 = +3 $ dB. If sources are oppositive, they cancel each other, and the total energy becomes $-\infty$ dB.

Then, subtracting the noise level to the observed signal is an instance of (scalar) hard thresholding, called spectral subtraction in the Fourier domain. As eluded to in your Spectrogram with square or non-square magnitude of STFT: power vs. magnitude question, scalar spectral subtraction on the observed STFT ${S}(\tau,\omega)$ can be performed with a power law, as:

$$|\overline{X}(\tau,\omega)|^\alpha=\max\left(|{S}(\tau,\omega)|^\alpha - \lambda|\overline{N}(\tau,\omega)|^\alpha,0\right)\,.$$

The recovery of the noiseless phase is a full different world.

  • $\begingroup$ Thanks for your answer. Just a little something, are you sure about this source sengpielaudio.com/calculator-leveladding.htm? Adding of two incoherent (noncoherent) sound pressure levels or voltage levels: 10*log10(10^(A/10)+10^(B/10)). Adding of two values of the same level results an increase of the total level of 3 dB. This isn't true for voltage, even if incoherent, right? When I add two constant-volume incoherent signals peaking at -30 dB in an audio software, I get -24 dB, and not -27 dB. $\endgroup$
    – g6kxjv1ozn
    Sep 3, 2018 at 23:16
  • $\begingroup$ I cannot speak for the case you mention. I have added some computations and three cases $\endgroup$ Sep 4, 2018 at 9:55

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