# Why do two coherent sounds add up $+6db$?

I read that two incoherent sounds add up $$+3db$$, and two coherent $$+6db$$. Tell me how does this not break law of conservation of energy.

$$+3 db =$$ double power

$$+6 db =$$ quadruple power

If we have one sound source that creates 1 Watt sound power,and add another 1 Watt sound source,we have total of 2 Watts of sound power,it does not matter if the sounds emitted are coherent,thats absolute maximum limit before breaking law of conservation of energy.

Please note,two speakers using pistonic motion of fixed size surface ( all normal speakers ) to create sound do add up +6db if they are closer together than 1/2 of wavelenght,that is becose the effective radiation surface doubles,so efficieny doubles,there is effect called mutual coupling.

Two times efficiency + two times power does indeed result in +6db,but as far as I am aware,this doesnt work when sound sources are more than 1 wavelenght away and nowhere did it mentioned that coherent sounds add up +6db only when sound sources are close together,it seemed to be written as if its universal law that exist no matter the position of sound source.

• Its not very clear where you are headed to, but interference seems to be the case. Here energy is simply redistributed and not destroyed/created. – King Tut Feb 27 '18 at 12:36

## 4 Answers

You have several good points, especially this one:

[3 dB incoherent addition] seems to be written as a universal law that exists no matter the position of the sound source.

It's definitely not a universal law. Only a 'rule-of-thumb'. But it's a rule of thumb that has a solid basis when we consider how most sound measurements are made.

Here's an intuitive picture of why the rule of thumb exists.

If we have two speakers producing the exact same signal over time we'd have a standing, invariant pattern of constructive and deconstructive interference: If the two signals are not in phase then the interference pattern shifts and moves with time. Convince yourself of this before moving on. If it helps, imagine the diagram above but on one side the heavy circles are at the positions of the light circles. How would the diagram look different?

Now the amplitude part.

I made a quick animation that shows two 4 Hz waves of exactly the same amplitude slowly going in-and-out of phase with each other. The green waveform is what you'd sample in one second's time if you had a microphone at that point in the sound field and the red and blue signals had their current phase relationship: The time-averaged sound pressure level for the green waveform (i.e., the equivalent sound pressure level, Leq) for one full cycle of phasing turns out to be

$$L_{eq} = L_{in \ phase} - 3.0 \ dB$$

Or, in other words, due to phasing interference, the time-averaged measurement for two incoherent (out-of-phase) sources is three decibels less than that expected for coherent (perfectly in-phase) sources.

How well this rule-of-thumb works will depend on how your speakers are placed, but for the diagram above, it will work quite well for the central area between the two speakers.

• I want to thank you for writting such nice answer,I really appriciate that you didnt ignore my question.I have few things to say,when I written about the "universal law"I was thinking +6db,not +3db......... I cant accept your answer becose it does not answer my question,I was asking how it is possible that two coherent sounds combine to +6db when +6db requires quadruple energy and two coherent sounds combine to double energy at best if they are perfectly in phase.It seems to me to break the law of conservation of energy,unless the "+6db" is meant speakers close together due to mutual coupling – wav scientist Feb 28 '18 at 21:07
• Why thank you! I think I can see why it appears as though I haven't answered your question. Here's how my logic works. I start by rejecting the premise that two identical speakers dissipate quadruple the energy, and instead start with them dissipating twice the energy, as expected. We know that for one speaker doubling the energy will increase SPL by 6 dB. But the smaller observed increase for two speakers is because of the 3 dB 'discount' due to interference. 6 dB - 3 dB = 3 dB. In other words Energy_Doubling - Interference_Effects = 3 dB. – D. Betchkal Feb 28 '18 at 21:31
• With all due respect,I am 99.9999% sure that doubling sound pressure requires not double,but quadruple kinetic energy.One speaker playing 6db louder will consume 4x as much energy if its efficiency curve does not change depending on its output level.Sound pressure is momentum of air molecules,to double momentum requires double velocity,double velocity is quadruple kinetic energy. – wav scientist Feb 28 '18 at 21:43
• You say: "One speaker playing 6db louder will consume 4x as much energy", and I agree. In your question you're concerned about two sounds "adding up", though, and that's talking about Sound Pressure Level which is a property of the sound field, not the speaker. Check out the "+6 dB" row of this table. Both of our assertions are on here. Your 4x scaling is the "Power Gain Factor". My 2x scaling is the "Voltage Gain Factor", which is the correct scaling factor for SPL addition. That's why the above answer is correct. – D. Betchkal Feb 28 '18 at 22:19
• It is true that +6db is double "Voltage" or sound pressure,that does not change the fact that double sound pressure requires quadruple energy and since two sounds when coherent can add up to two times the energy,thats means its impossible gain +6db. – wav scientist Feb 28 '18 at 22:30

If you are listening to two coherent sources, then if they are in phase at your location they will be +6 dB louder than either one alone. If they are out of phase at your location, then they cancel out and you hear nothing at all: -∞ dB. Different locations will have different sound intensities, so how loud things are at your location doesn't tell you how loud things are a few wavelengths away.

It is important to specify the type of decibels. For example a speaker producing 100 Watts of power might be said to be producing 20 dBW or 50 dBm (referenced to 1 W and 1 mW respectively).

In audio you sometimes use dBspl, referenced to a sound pressure level of 20 microPascals. dBspl can be converted to W/m2. Multiply by the area of your eardrum to get the amount of power that you actually 'hear'.

Example: A speaker produces 10 Watts of audio power (10 dBW = 40 dBm). It spreads out and hits the (perfectly-absorbing, spherical) wall of your listening chamber where the sound level is e.g. 100 dBspl, and the amount of power absorbed by the wall is (multiplying the area of the walls in m2 by the power in W/m2 that you calculate from the dBspl) the same 10 Watts.

Now have 2 coherent speakers each producing 10 Watts (20 Watts total: 13 dBW). At the wall there will be places that have up to 106 dBspl of sound, and other places that are dead quiet. If you add up all the power absorbed by all the square meters of wall, the loud regions absorb more power and the quiet regions absorb less, but the total will turn out to be 20 Watts.

• If 100db is 1 Watt of sound power and 106db is 4 Watt of sound power,how is it possible than two 1 Watt sound sources combine to 4 Watt of sound power except the case when they are close together and mutual coupling doubles efficiency. – wav scientist Feb 28 '18 at 21:11

Firstly is important to define Decibel (Bel x 10). Decibel expresses the ratio between two homogeneous magnitudes. Power, voltage, SPL.... whatever... And express it in a logarithmic scale. Having P1=130w and P2=37w, the ratio expressed in dB is dB=10log130/37; dB=5,45dB. P1 is 5,45dB higher than P2 For P1=2w and P2=1w (double) dB=10log2/1; dB=3,01dB. Double power is 3dB more. This formula is used for Power Magnitudes For Tension Magnitudes (voltages, SPL...) we must to use dB=20logV1/V2. Why?

Having a Power Gain Gp,

Gp=10logP2/P1. knowing P=V²/R (P=power, V=voltage, R=resistance)

Gp=10log ((V₂)²/(V₁)²)/(R/R); of course R/R=1 Gp=10log(V₂/V₁)² Gp=20logV₂/V₁

If V₂= 2 volts and V₁= 1 volt

dB=20log2/1=20log2=6,02

Double voltage, SPL current.... always 6dB

About acoustics, always two identical signals summing in perfect conditions (never happens) will increase on 6dB. For this both of signals must have same phase, and this is called coherent. Incoherent is when nobody knows the relative phase on each signal, so can sum energy from 6dB (same phase) or the opposite effect, can produce a total cancellation (when phases have 180 degree difference)... All the possibilities between can happen. So why Incoherence is 3dB? Is an average adopted for some circumstances like in a stadium a lot of people Clapp and is impossible control the phase. We assume one clap maybe is 90dB two claps 93dB I hope this clarifies this question a little bit

• Hi German Rodriguez, welcome to PSE and thank you for your answer! Please note that this site supports mathjax, which is an engine that allows you to typeset equations (with a syntax similar to latex). See here for a tutorial: math.meta.stackexchange.com/q/5020/289977 If you use it to fancify your answer, it will look much better. Cheers! – AccidentalFourierTransform May 15 '18 at 3:38
• Hi AccidentalFourierTransform... Thanks a lot for your help to improve my experience here. Cheers!! – German Rodriguez May 15 '18 at 16:20

I'd like to try and add to this. I think there is a misconception here surrounding the physical properties of intensity and pressure.

In your first question, you discuss adding Watts together, which results in a 3 dB increase to the overall sound power. Power is the amount of work done over time (expressed as the Watt) and intensity is that Power over some unit area (W/m^2).

In your next portion you discuss the synchronous movement of piston-driven speakers that produce a sound pressure wave. Pressure is a Force over some unit area. With sound, we usually measure with uPa (micropascals) or Pa (Pascals).

Intensity is proportional to pressure squared (I proportional to P^2)

So when you are comparing between these, you will note that a doubling of intensity (when converted into a dB measurement) derives a dBintensity = 10 * log10 (I1/I0) equation - where I1 is an incident intensity and I0 is a reference intensity. If you double the intensity of some reference (I0), your ratio is 2:1, dBintensity = 10 * log10 (2) = ~3 dB. When you convert this to a pressure value - you must remember that I is proportional to P^2, which alters the equation: dBpressure = 10 * log10 (P1^2/P0^2) - where P1 is some incident pressure, compared to some reference pressure. If we work this example a little more, you can see that the equation simplifies to:

dBpressure = 10 * log10 (P1/P0)^2

dBpressure = 10 * 2 * log10 (P1/P0)

dBpressure = 20 * log10 (P1/P0) (and with a pressure ratio of 2, as in your example)

dBpressure = 20 * log10(2) = +6 dBpressure

I hope this is helpful.

You can read more here on why intensity is proportional to pressure squared: Why is sound intensity proportional to the square of sound pressure not to sound pressure alone?