If I put next to my cooler another similar cooler that produces similar white noise, will the overall noise level increase?

I want to point out that I am speaking about adding another independent, non-correlated white noise source, not just another pair of speakers that produce sound of the same phase and curve. It is evident for me that adding another speaker will increase amplitude twofold compared to one speaker, but this is not evident for coolers which produce non-correlated sound.

  • $\begingroup$ Yes it will. There is no reason for the noise of the two coolers to cancel. $\endgroup$ – Greg P Apr 22 '11 at 23:55
  • $\begingroup$ If it will increase, then by what magnitude? Twice? $\sqrt{2}$? $\endgroup$ – Anixx Apr 22 '11 at 23:58
  • 2
    $\begingroup$ you might want to edit that into the question. This is actually an interesting thing to ask about, but you kind of have to prompt people to go deeper than just "yes" ;-) $\endgroup$ – David Z Apr 23 '11 at 0:35
  • $\begingroup$ Generating white noise is some task! Are You shure Your "cooler" (what ever that might be) procuces white noise? $\endgroup$ – Georg Apr 23 '11 at 10:11
  • 1
    $\begingroup$ @GregP Thats an over-simplistic response. The waves will likely cancel each other several thousand times a second as they'll occasionally be in opposite phase. However the opposite is also true. The real question is what is the overall effect of the second sound source being added, as perceived by the ear. My guess is it won't double it, but neither will it cancel out or in any way quieten it overall. $\endgroup$ – NickG Jun 17 '13 at 15:13

I assume you're talking about "coolers" as in CPU-coolers or other cooling systems in computers. The noise-spectrum from such a fan is not white strictly speaking, you can see it measured on this page. The spectrum is fairly level if you look at small parts though and ignore the tonal components.

Anyway, by your definition, two uncorrelated white noise sources add without interference. The "peaks and troughs" (sound pressures) don't add up coherently (in ideal sources of course :), but they don't cancel each other perfectly either. Therefore the sound power is doubled. By definition the RMS-amplitude (proportional to the sound pressure) is then multiplied by $\sqrt 2$. The sound intensity is proportional to the sound power and hence is doubled as well.

I actually simulated this in Matlab, because the terminology with sound power, sound energy, sound levels, sound pressures etc is daunting, misleading and confusing.. I'm specifically avoiding talking about decibels.

>> w1 = wgn(8192, 1, 10, 'real');  % generates real white noise, 8192 samples
>> w2 = wgn(8192, 1, 10, 'real');
>> mean(([w1 w2]).^2)              % show mean powers of both noise signals

ans =

   10.0410   10.0832

>> mean(([(w1+w1) (w1+w2)]).^2)    % add one of them to itself, add both

ans =

   40.1640   20.6295

So yes you do get a doubling of power when adding uncorrelated white-noise sources.. and when perfectly adding a single source, you quadruple the power. By definition the RMS amplitude is the root of the above.

  • $\begingroup$ Yes, but I doubt that in real life if to add two identical sources, the resulting pressure amplitude will double. Mathematically you double it and you get what you get but in real life I think the result will be different (in the both cases). $\endgroup$ – Anixx Apr 25 '11 at 9:05
  • $\begingroup$ yes, that's why I wrote "perfectly" adding. In practice this is difficult if you try to do it with two speakers. In any case, with uncorrelated sources, you should get 2x the power for the white-noise parts. The tonal parts of the fan noise will maybe add with correlation and in some cases you'll get "beat" frequencies (slowly varying modulations of vibrations in the chassis for example). But also note that the human ear is not linear so the perceived sound loudness is quite different from the power. Two fans won't sound like a doubling in loudness I guess (this is a different question). $\endgroup$ – BjornW Apr 25 '11 at 12:55

The white noise of both coolers will add up and appear to be one white noise with stronger intensity than the white noise of each individual cooler.

The reason is that the white noise produced by each cooler can be seen as a sound wave that has a large amount of "random walk" behaviour to it. The two similar white noise sources will both have a "random walk" behaviour but as they are just similar but otherwise not related to each other, they can both (at some points in time) cancel each other or (at some other points in time) amplify each other in some direction. When they work against each other and when they "cooperate" is, well, random. So the result when combining them is essentially... white noise. However, as they sometimes "cooperate" and result in wave spikes in the same direction, you end up having bigger spikes and alterations back and forth in the combined white noise than in the individual white noises, so you end up with a more intense white noise.

(IIRC, the formal definition of white noise is such that if you add two white noise waves with the same frequency together, you end up with a result which mathematically is white noise)

  • $\begingroup$ What is the "frequency" of white noise? $\endgroup$ – Greg P Apr 23 '11 at 2:03
  • 1
    $\begingroup$ @Greg ideal white noise it has a flat spectrum, so it contains equal contributions from all frequencies. $\endgroup$ – David Z Apr 23 '11 at 4:32
  • $\begingroup$ But what will be the intensity of the resulting white noise? $\endgroup$ – Anixx Apr 23 '11 at 8:24
  • $\begingroup$ If you think of the amplitude of the white noise at any given time as a Gaussian (I'm assuming you can do this), then you are talking about adding two independent Gaussians. Then their variances (the mean of the square of the amplitude) are additive. So the variance of the sum is the sum of the variances (only because they are independent!). And I think that the 'intensity' corresponds to the mean square amplitude (or variance). If all of this is correct, the intensity will be twice that of each 'cooler.' $\endgroup$ – Greg P Apr 23 '11 at 16:36
  • $\begingroup$ If the means of squares of amplitudes are additive, then the resulting amplitude will be: $\sqrt{a^2+a^2}=\sqrt{2}a$ $\endgroup$ – Anixx Apr 23 '11 at 17:07

Yes. The noise from the two different sources should be uncorrelated, so they add in power. Adding a second identical noise source increases the power by a factor of 2. Since the power is proportional to the square of the amplitude, that means the amplitude increases by a factor of $\sqrt{2}$.

In terms of decibels, this means that the power increases by 3 dB, and the amplitude increases by 1.5 dB. (Power is more commonly measured in dB than amplitude, but there's no reason you can't use dB for both power and amplitude.)

  • $\begingroup$ No. Adding a second source increases the amplitude by 2 only if they both have the same wave curve and phase. Since the both coolers are not connected to each other and produce independent white noise, I would not expect twofold increase in amplitude (some peaks cansel each other and some increase each other). I wonder why people vote for this answer. $\endgroup$ – Anixx Apr 23 '11 at 8:22
  • $\begingroup$ @Anixx: he did write a factor of sqrt(2) on the amplitude, not 2. $\endgroup$ – BjornW Apr 24 '11 at 11:47
  • $\begingroup$ ""should be uncorrelated, so they add in power"" Even "correlated" noise will ad in power, because inerference will be destructive as well as additive. To have destructive inerference totally needs more than correlation. This works for frequencies much shorter in wavelength than distance of the sources only. $\endgroup$ – Georg Apr 24 '11 at 12:13
  • $\begingroup$ "amplitude increases by 1.5 dB" No, they will both increase by 3 dB. Power decibels are 10*log10() and amplitude decibels are 20*log10() so it works out the same en.wikipedia.org/wiki/Field,_power,_and_root-power_quantities $\endgroup$ – endolith Jan 25 '17 at 17:12

What you very probably have in mind is noise cancelling. If you have two sounds which are exactly the same, but one of them is inversed (phase and anti-phase) and you play them together, they cancel each other out. The problem is that having two exactly same sounds in real life is almost impossible, without using any microphones and/or digital processing that is.

Also, to have two sounds with aforementioned qualities cancelled out, you'd need to listen to them together perfectly, that means they'd need to point to your ears in a way so that you listen to the phase without any offset to any of them. This is, again, in real life, impossible.

So your idea is based on something solid, but it can't work in your case, because (to sum it up):

  • the two sounds of the coolers are different
  • the two sounds of the coolers are not in an inversed phase to each other
  • the two sounds of the coolers aren't pointing perfectly to your ears :)

Therefore, your two coolers cannot cancel each other out, nor can they together produce the same level of volume. The volume of course increases.

sengpielaudio.com says:

20 dB gain modification should give the ratio of 4 (four times) for sensed volume and loudness
20 dB gain modification gives the ratio of 10 for measured voltage and sound pressure and
20 dB gain modification gives the ratio of 100 for calculated sound power and acoustic intensity

Doubling of the volume (loudness) should be felt by a level difference of 10 dB
Doubling the sound pressure (voltage) corresponds to a measured level change of 6 dB
Doubling of acoustic power (sound intensity) corresponds to a level change of 3 dB

3 dB = twice the power (Power respectively intensity - mostly calculated)
6 dB = twice the amplitude (Voltage respectively sound pressure - mostly measured)
10 dB = twice the perceived volume or twice as loud (Loudness nearly sensed - psychoacoustics)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.