Adding dB

Goal

Understand how to add dB, the correlated and uncorrelated sound equations and where they comes from.

Correlated and Uncorrelated Sound

The distinction of if a signal is correlated or not correlated is very important. It will determine which equation we should use.

Correlated Sound

If multiple sound sources are given the same material and are both excited by the same event then they will both produce identical waves that will be correlated and hence shall add. (I am greatly simplifying this, ignoring the room, time delays and other complex variables). This is because constructive interference is taking place, and this means that the amplitudes are correlated.

Examples of correlated sound is a set of speakers all playing the same signal, or a reflection off a wall that combines with the direct sound. They are correlated because it is the same signal.

Uncorrelated Sound

If multiple signals are not identical and they are not triggered by the same exciter then they are uncorrelated.

For example, say I have 2 singers. I give a cue and they both start singing the same note. Since the singers have separate vocal cords which are excited by separate sources (meaning the thing that causes the vocal cords to vibrate) therefore they are uncorrelated. Another example would be multiple machines in a room all making noise.

Adding Signals

Say we have a signal at 60 dB SPL and another at 60 dB SPL. If we simply add them together we may think we would get 60 dB SPL + 60 dB SPL = 120 dB SPL. This is wrong! Remember the dB uses a log! Logs take us from a number line that moves in a "plus or minus" kind of way to one that moves in a "multiply or divide" kind of way! So we cannot just add them because that would be as if we are on a linear number line. Each increase of 10 dB is a 10 fold increase! So to go to 120 dB SPL would imply we have increased by 1 MILLION times! This is not at all what is happening.

I heard it put like this, if you add a 3 digit number and another 3 digit number do you expect a 6 digit number as the result? No! 100 + 100 is not 200000 its 200! Adding a digit is like multiplying by 10. You have to move in a "plus or minus" kind of way not a multiplicative one! dB however move in a multiplicative way. Adding them linearly makes no sense just like adding our 3 digit numbers. Every increase of 10 for a dB is an increase of 10 fold! So adding 2 dB will work very differently than you expect! In fact often 2 dB combined leads to an increase of 3 or 6 at the max!

Adding Correlated Sound

We need to take another look at intensity in order to understand adding dB's. Recall that sound intensity is sound power divided by area. So intensity is power over an area. Remember we record the pressure and the pressure squared is what varies with intensity, therefore it also varies with the power. This is where the 20 log rule came from.

Say we had a pressure such as

2 \ Pa

. In order for

2 \ Pa

to be related to intensity we must square it, hence we get 4 in order for it to be related to intensity. So an increase by a factor of

2

. Instead of squaring directly we opted to take the square out and move it to the front of the equation in the form of

2

, making the

10

become a

20

. We could have left the

10

in the front and squared the pressure directly!

Each of these works for pressure:

dB_{SPL} =10\log\left(\frac{x^{2}}{x_{ref}^{2}}\right) =20\log\left(\frac{x}{x_{ref}}\right)

Now consider two instances of pressure, say

2 \ Pa

and

2 \ Pa

. First we sum them. It is critical we sum them before squaring them. Summing them first reflect the doubling of pressure that is happening. If we squared each pressure and then added then it would be akin to summing the power of the waves, something we don't want. We want the pressure. This gives

4 \ Pa

. (Assuming they are perfectly in phase.)

Plugging this in you will see an increase in

6 \ dB_{SPL}

! This is because when pressure is doubled power scales by a factor of four! See how before with a single pressure we only had

2

and it increased to

4

, but now it increased from

2

16

! (We start at 2 because this is the reference pressure, it doubles to 4 and then the square takes it to 16.) This is due to the inverse square law. Remember the power is spread out over an area and so a lot more power is required to increase the pressure at every point in the entire area. Since the pressure doubled it lead to a much more dramatic increase in power to maintain that through the area.

So in this case when they are in phase we can simply use the

20 \log

rule and thus the

6 \ dB

rule. However things are often not in phase. In fact just by changing from where you a measuring you could go from in phase to out of phase! If the waves had a phase shift of say 180° this can dramatically change the power depending on the wave form. Phase shifts for some waves, such as asymmetrical waves are different than those from from pure sines, and complex waves will almost certainly not cancel from a phase shift even if they are identical. In some cases the volume may even go down! So keep in mind that we are calculating the specific condition of the pressure getting doubled in these cases.

In simple cases for sine waves you could write out the sine wave with the phase shift, find their instantaneous amplitude at a point and use that to deduce a figure for the pressure but we are not going to follow such a process. I just want you to understand that when dealing with early reflections off of walls combining with direct sound from the speaker that these are correlated cases and they will change based on your measuring position because changing your position will change the phase relationship between them, so while a

6 \ dB_{SPL}

increase is what we will calculate this is the "high" end and typically we care about potential maxes more than potential leasts.

Derivation for summing correlated sound

Consider how we add pressures for correlated sound. We assume the sounds are in phase and pressure will therefore double:

total\ pressure\ =p_{T} =p_{1} +p_{2} +...+p_{n}

Now with the new pressure we plug into our dB equation for SPL:

dB_{SPL} =20\log\left(\frac{p_{T}}{p_{ref}}\right)

Another form is:

dB_{SPL} =20\log\left(\frac{p_{1} +p_{2} +...+p_{n}}{p_{ref}}\right)

We typically have measurements in terms of dB. So we would like a form that uses dB as the input rather than pressure. To do this we will work around the dB until we have it as a definition of pressure. Then we can replace the pressure in our definition with the dB version.

First we divide the 20 to the other side. Remember our goal is to get one side with just the measured pressure.

\frac{dB_{SPL}}{20} =\log\left(\frac{p}{p_{ref}}\right)

Next we must get rid of the log. One way to do so is to use an exponent similar to how multiplication undoes division, exponents undo logs. If we do the same thing to both sides equally then we have not changed the balance of the equation and so it will still be true, therefore we raise both sides to the power of the base of the log which in this case is 10.

10^{\frac{dB_{SPL}}{20}} =10^{\log\left(\frac{p}{p_{ref}}\right)}

The log and the 10 cancel out and we have:

10^{\frac{dB_{SPL}}{20}} =\frac{p}{p_{ref}}

Now we multiply by the reference to both sides and get:

p_{ref} \times 10^{\frac{dB_{SPL}}{20}} =p

Now we have a pressure definition. Consider our original definition for total dB for correlated sound:

dB_{SPL} =20\log\left(\frac{p_{1} +p_{2} +...+p_{n}}{p_{ref}}\right)

Also note that the reference for many pressures added can be factored out:

total\ pressure\ =p_{T} =p_{ref} \times 10^{\frac{dB_{1SPL}}{20}} +p_{ref} \times 10^{\frac{dB_{2SPL}}{20}} +...+p_{ref} \times 10^{\frac{dB_{nSPL}}{20}})

total\ pressure\ =p_{T} =p_{ref}\left( 10^{\frac{dB_{1SPL}}{20}} +10^{\frac{dB_{2SPL}}{20}} +...+10^{\frac{dB_{nSPL}}{20}}\right)

Subbing in we obtain:

dB_{SPL} =20\log\left(\frac{p_{ref}\left( 10^{\frac{dB_{1SPL}}{20}} +10^{\frac{dB_{2SPL}}{20}} +...+10^{\frac{dB_{nSPL}}{20}}\right)}{p_{ref}}\right)

dB_{SPL} =20\log\left( 10^{\frac{dB_{1SPL}}{20}} +10^{\frac{dB_{2SPL}}{20}} +...+10^{\frac{dB_{nSPL}}{20}}\right)

There it is! We can check it. If we plug in two of the same dB SPL value we expect it to go up by 6 dB SPL, lets choose 50dB SPL.

dB_{SPL} =20\log\left( 10^{\frac{50\ dB_{1SPL}}{20}} +10^{\frac{50dB_{2SPL}}{20}}\right) \approx \boxed{56\ dB_{SPL}}

Adding Uncorrelated Sound

Sound waves are often not perfect copies. If 2 people are talking in a room, even if they are saying the same exact same thing or even singing the same note their sound waves will be totally different looking, hence they are uncorrelated. So the pressure at every point will vary. In some parts of the wave it will get softer and in others it will get louder depending on the phase interference at that particular part of the wave. Just viewing it as "pressure" is not a good solution because the wave will effectively "average out". This makes the 20 log equation unfit for summing situations like this.

When adding pressures that are uncorrelated the phase cancelation causes the pressure to "average out". The summation follows the increase in power so we will need to go from intensity to power. The reason for this change can be more fully described using statistics but is outside the scope of this course. Essentially we can treat uncorrelated noise as a random variable. We could actually determine how correlated the signals are and bunch of other stuff with stats but in practice we just need to understand that when we add pressures we are going to take the square root of the summed pressures in order to go back to a power relationship.

So for summing powers we have the following:

total\ pressure\ =p_{T} =\sqrt{p_{1}^{2} +p_{2}^{2} +...+p_{n}^{2}}

This formula no longer cares about phases since the waves have significantly more complex phase relationships and it no longer has as dramatic an effect as it did before. In fact because the phase is now more random it means that on average the signal will increase. In correlated waves this was not so leading to some phase relationships potentially resulting in a decrease.

Each pressure is squared before hand so that it is related to power. Then we sum the waves. To get back to regular old pressure we then take the square root.

Derivation of the Uncorrelated Sound Equation

Start with a single source just like before:

dB_{SPL} =20\log\left(\frac{p}{p_{ref}}\right)

Next by substitution

dB_{SPL} =20\log\left(\frac{\sqrt{p_{1}^{2} +p_{2}^{2} +...+p_{n}^{2}}}{p_{ref}}\right)

Rewrite the square root in its exponent form:

dB_{SPL} =20\log\left(\frac{\left( p_{1}^{2} +p_{2}^{2} +...+p_{n}^{2}\right)^{\frac{1}{2}}}{p_{ref}}\right)

Square both the top and bottom and raise the whole ratio to 1/2, this way we can bring the 1/2 down.

dB_{SPL} =20\log\left(\frac{\left( p_{1}^{2} +p_{2}^{2} +...+p_{n}^{2}\right)}{p_{ref}^{2}}\right)^{\frac{1}{2}}

Use the log property to bring the 1/2 down giving us a 10 in the front which matches up with our expectation that we are relating powers.

dB_{SPL} =10\log\left(\frac{p_{1}^{2} +p_{2}^{2} +...+p_{n}^{2}}{p_{ref}^{2}}\right)

This is the formula for summing multiple uncorrelated pressures. However we again desire an equation that will take dB as input and give us the total so again we solve for pressure in the form of dB. First lets replace the pressure summation with a total pressure variable:

dB_{SPL} =10\log\left(\frac{p_{T}}{p_{ref}^{2}}\right)

Divide the 10 out to the other side:

\frac{dB_{SPL}}{10} =\log\left(\frac{p_{T}}{p_{ref}^{2}}\right)

Raise both side to the 10:

10^{\frac{dB_{SPL}}{10}} =10^{\log\left(\frac{p_{T}}{p_{ref}^{2}}\right)}

The log and the exponent cancel out:

10^{\frac{dB_{SPL}}{10}} =\frac{p_{T}}{p_{ref}^{2}}

Multiply out the reference to both sides:

p_{ref}^{2} \times 10^{\frac{dB_{SPL}}{10}} =p_{T}

Now we have an expression for pressure. Using this we can substitute into

dB_{SPL} =10\log\left(\frac{p_{1}^{2} +p_{2}^{2} +...+p_{n}^{2}}{p_{ref}^{2}}\right)

Substituting each pressure we get:

dB_{SPL} =10\log\left(\frac{p_{ref}^{2} \times 10^{\frac{dB_{1SPL}}{10}} +p_{ref}^{2} \times 10^{\frac{dB_{2SPL}}{10}} +...+p_{ref}^{2} \times 10^{\frac{dB_{nSPL}}{10}}}{p^{2} ref}\right)

Factor out the reference:

dB_{SPL} =10\log\left(\frac{p_{ref}^{2}\left( 10^{\frac{dB_{1SPL}}{10}} +10^{\frac{dB_{2SPL}}{10}} +...+10^{\frac{dB_{nSPL}}{10}}\right)}{p^{2} ref}\right)

Now simplify:

dB_{SPL} =10\log\left( 10^{\frac{dB_{1SPL}}{10}} +10^{\frac{dB_{2SPL}}{10}} +...+10^{\frac{dB_{nSPL}}{10}}\right)

This is the equation for summing multiple dB for uncorrelated sound! Here we use the 3dB rule since a 10 log is used.

Gaining an intuition of adding dB

Adding dB can be very confusing because dB are multiplicative. There are a few principles that can help you.

If 2 dB's are the same then use the respective doubling rule (3 or 6).
If 2 dB's are similar in level it will remain around the doubling value.
If 2 dB's are far apart the level will be much closer to whatever the higher level is.

Examples

Example

A horn section has a trumpet, sax, and trombone. The trumpet is playing at 76 dB SPL, the sax at 83 dB SPL, and the trombone at 85 dB SPL. What level is the section together?

Show Answer

First we note the sources are uncorrelated. Plugging the measurements into the equation we obtain:

dB_{SPL} =10\log\left( 10^{\frac{76dB_{SPL}}{10}} +10^{\frac{83dB_{SPL}}{10}} +10^{\frac{85dB_{SPL}}{10}}\right) \approx \boxed{87.45\ dB_{SPL}}

Example

You are in a room with a trumpet player who is playing at a solid 73 dB SPL. A second trumpet player enters the room and also begins to play at 73 dB SPL. How much did the level increase by?

Show Answer

Here we can just use the doubling rule. In this case the signals are uncorrelated hence the 3dB rule is used. Thus it would be 76 dB SPL, so it increased by about 3 dB SPL.

We could also use the equation:

dB_{SPL} =10\log\left( 10^{\frac{73dB_{SPL}}{10}} +10^{\frac{73dB_{SPL}}{10}}\right) \approx \boxed{76.01\ dB_{SPL}}

\therefore 3\ dB_{SPL} \ increase

Example

4 singers are in a recording session. Their respective levels are 76, 67, 79, and 80 dB SPL. What number do you expect? How loud is the group?

Show Answer

dB_{SPL} =10\log\left( 10^{\frac{76dB_{SPL}}{10}} +10^{\frac{67dB_{SPL}}{10}} +10^{\frac{79dB_{SPL}}{10}} 10^{\frac{80dB_{SPL}}{10}}\right) \approx \boxed{83.51\ dB_{SPL}}

Isn't it amazing how close to 83 it is! It is as if the only level the really mattered was 79! 76 contributed barely anything! The 67 is an entire order of magnitude softer! 10 times softer! So it contributed practically nothing.

Example

You are in a room with two speakers both playing the same sine wave at 80 dB SPL. What is the dB SPL in the room?

Show Answer

Here we can just use the doubling rule. In this case the signals are correlated hence the 6dB rule is used. Thus it would be about 86 dB SPL.

Example

8 people are all talking. Each person has a volume of 60 dB SPL. What level is the room?

Show Answer

You could plug in all 8 60 dB SPL measurements and get the answer, but the faster way is to just use the doubling rule. The first person gives 60 dB SPL, the second adds 3 dB so 63 dB SPL, the next 2 people add another 3 dB so 66 dB SPL, finally the last 4 people add another 3 dB SPL therefore the final level is 69 dB SPL.

Example

A stereo speaker system is playing a square wave with each speaker playing at a level of 50 dB SPL. What level is the system? Should we expect this level at all points in the room? Explain. What kind of variation do we expect? Would sound power follow the same logic? Explain.

Show Answer

The signals are correlated therefore we expect a level increase of 6 dB SPL, so 56 dB SPL. For a square wave a phase shift could completely cancel the wave out so we expect a variation in room of silence all the way up to 56 dB SPL! It depends on the phase relationship at the place of measurement.

Sound power would not follow the same logic, sound power measures the power through an area. All phase cancelation would be contained the measurement around the sources causing the sound power to not vary with measurement position because it would measure an entire surface around the source, not just a single point.

Example

Two flutes are playing. One has a volume of 50 dB SPL and the other 70 dB SPL. What level is the room?

Show Answer

Again we could use the equation but it is overkill. The 50 dB SPL would contribute next to nothing, so the answer will be approximately 70 dB SPL.

Example

Imagine you’re in a super noisy crowd. I’m talking like a football stadium that is half full. The side you are on is packed together and screaming at full blast! Now imagine the other side of the entire stadium gets filled. We are talking thousands more people and they all come in yelling every bit as loud as your half. What is the level change?

Show Answer

Despite the tremendous increase in the number of people it is still only doubled so it would only increase in level by about 3 dB SPL.

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