dB calculations

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Sean Thomas

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Nov 8, 2013
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Having a debate on a dB Technology forum about adding 2 subs together and what the ouput would be.

Sub is 18", self powered at 1500w, and 134dB output.

Some one said that doubling of power = +3dB AND adding another speaker = +3dB. So adding another sub 1 + 1 = +6dB.

I said that none of the specs matter except for the dB specs and mathmatics. In other words, it's irrellevent how many speakers and amps you have.
Math says 134dB + 134dB = +3dB = 137dB.

This of course does not take in phase and freq. issues that would make that different in the real world.

Is my math correct? Please elaborate.
 
Re: dB calculations

Sean Thomas said:
Having a debate on a dB Technology forum about adding 2 subs together and what the ouput would be.

Sub is 18", self powered at 1500w, and 134dB output.

Some one said that doubling of power = +3dB AND adding another speaker = +3dB. So adding another sub 1 + 1 = +6dB.

I said that none of the specs matter except for the dB specs and mathmatics. In other words, it's irrellevent how many speakers and amps you have.
Math says 134dB + 134dB = +3dB = 137dB.

This of course does not take in phase and freq. issues that would make that different in the real world.

Is my math correct? Please elaborate.
Click to expand...

It's a little different, I think, with powered speakers. You have to treat the amp & transducer as one unit. Ivan?

In a conventional system, if you add another loudspeaker wired in parallel each speaker "sees" the same drive voltage but divides the current between them. That doubling of piston area should give you +3dB. If you were to add the additional speaker and provide separate power to it, what do you think will happen, and why?

Homework.
 
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Re: dB calculations

Tim McCulloch said:
It's a little different, I think, with powered speakers. You have to treat the amp & transducer as one unit. Ivan?

In a conventional system, if you add another loudspeaker wired in parallel each speaker "sees" the same drive voltage but divides the current between them. That doubling of piston area should give you +3dB. If you were to add the additional speaker and provide separate power to it, what do you think will happen, and why?

Homework.
Click to expand...
No-it works the same way. You are simply doubling the power and the speaker and you get 6dB-assuming the speakers are close enough to each other within the freq range of interest.

It doesn't matter if the amp is inside the speaker or outside.

The reason for 6dB gets a bit "interesting". Some claim it is due to basically "double the voltage applied". WIth the same voltage applied to 2 speakers this "could" seem as a "logical" explanation.

The real answer is a little different.

Yes twice the power will result in a 3dB increase in level. The other 3dB basically comes from a change in the "radiation resistance" of the air-caused by the additional driver.

Since the way sound "works" is by compression and rarefraction. With more drivers acting-this "resistance" changes.

Yes it is small change-but it is not the amount of the change that makes a difference (just like wattage). Let's say it is a 1% change. That is not very much-UNTIL you consider what happens if you go from 1% to 2%. You have DOUBLED the amount.

Just like wattage. A change of 1 watt may not seem like much-unless you start with 1 watt. Starting with 100 watts-the additional 1 watt makes no difference.

It is for the same reason that it more and more subs to keep getting the same amount (dB) of change.
 
Re: dB calculations

It would seem that the number of drivers or the number of power amps and the location of the power amp is all irrelevent. This is math. Do we really need to know all of this to calculate (x)dB + (x)dB? I don't think so.

So what is 134dB + 134dB?

This link shows what I have always understood as the principles for adding multiple sound sources: https://www.osha.gov/dts/osta/otm/noise/health_effects/decibels.html
 
Re: dB calculations

Sean Thomas said:
It would seem that the number of drivers or the number of power amps and the location of the power amp is all irrelevent. This is math. Do we really need to know all of this to calculate (x)dB + (x)dB? I don't think so.

So what is 134dB + 134dB?

This link shows what I have always understood as the principles for adding multiple sound sources: https://www.osha.gov/dts/osta/otm/noise/health_effects/decibels.html
Click to expand...
It is NOT irrelevant.

There are OTHER factors that come into play-namely PHASE.

Here is a an example. Take 2 subs that are 134dB each. Now flip the polarity on ONE of them.

We don't need any math for people to realize that it will be QUIETER than either one alone-NOT louder.

It is NOT a simple matter of a couple of numbers.


Any complicated question is easily answered by a simple to understand WRONG answer.


In sound sources it, depends on the spacing of the devices relative to the listening position and the freq involved.

Different freq will give different answers if everything else stays the same.

I still stand by my previous answer that it will be 6dB louder-due to the decreases radiation resistance caused by the multiple drivers acting on the same "body" of air.

I do agree that the location of the amp (inside or outside the cabinet) is of little consequence-except for possible cable losses over small long runs.
 
Re: dB calculations

Sean Thomas said:
It would seem that the number of drivers or the number of power amps and the location of the power amp is all irrelevent. This is math. Do we really need to know all of this to calculate (x)dB + (x)dB? I don't think so.

So what is 134dB + 134dB?

This link shows what I have always understood as the principles for adding multiple sound sources: https://www.osha.gov/dts/osta/otm/noise/health_effects/decibels.html
Click to expand...

Except, as Ivan said, empirical results show n correlated sources of measured output Lspl = 20 log10 ( p / 0.000020 )

add as Ltot = Lspl + 20(log10(n))

or Ltot = 20log10( 10^(Lspl1/20) + 10^(Lspl2/20) + .. + 10^(Lspln/20) )

where Lspln is a measurement of an individual correlated source.
 
Re: dB calculations

I was looking for the basic math equation. I realize there are many other factors such as phase and frequency that will change the real world results. But for a mathmatical starting point, you are saying that the OSHA link is wrong?

134dB + 134dB is not +3 but +6? Again, as a basic math formula/starting point, disreguarding other factors such as phase, freq. etc.
 
Re: dB calculations

Sean Thomas said:
I was looking for the basic math equation. I realize there are many other factors such as phase and frequency that will change the real world results. But for a mathmatical starting point, you are saying that the OSHA link is wrong?

134dB + 134dB is not +3 but +6? Again, as a basic math formula/starting point, disreguarding other factors such as phase, freq. etc.
Click to expand...

Here's what is missing - going back to my original example, when you double the piston area you get +3dB increase in output. When you remove the current divider (by providing an independent power source) you get an additional +3dB of headroom that may or may not make a difference due to factors like power compression.
 
Re: dB calculations

Tim, I understand your example, it's the same as the post on the dB Technologies forum. So you are saying that the two 134dB subs will put out 140dB (+6dB) and that the OSHA web site is wrong?

Mark, I have been reading about log10 vs log20 calculations. Very interesting. So the next question is when do you use which calculation? The OSHA site is obviously using log10 for it's calcs, why 10 and not 20?
 
Re: dB calculations

Sean Thomas said:
Tim, I understand your example, it's the same as the post on the dB Technologies forum. So you are saying that the two 134dB subs will put out 140dB (+6dB) and that the OSHA web site is wrong?

Mark, I have been reading about log10 vs log20 calculations. Very interesting. So the next question is when do you use which calculation? The OSHA site is obviously using log10 for it's calcs, why 10 and not 20?
Click to expand...

I'm saying that OSHA is right, and so is Ivan because they're talking about different things. OSHA is showing the addition of uncorrelated noise, where changing the WAY the noise is created has no impact on how much noise there is... it's simple addition of single factors.

In combining subs, you have 2 factors - additional radiator area and doubling of the available current (if you don't run the subs in parallel). Now the kicker - the additional current doesn't really make a difference until you run out the initial current. That's why I referred to it as headroom.

The 10 and 20 base logs are used because power and voltage do not have the same logarithmic curve.
 
Re: dB calculations

Is this a good summary:

100dB + 100dB = +3dB or 103dB IF the sound is uncorrelated. OSHA uses this for random items making noise, aka log10.

100dB + 100dB = +6dB or 106dB IF the sound is correlated (the same sources as in music to each speaker), aka log20.

Of course, real world measurements could vary due to phase, frequency, etc.
 
Re: dB calculations

http://www.who.int/occupational_health/publications/noise1.pdf has a pretty good explanation of the whole thing. As does 'Handbook for Sound Engineers' by Glen Ballou. Remember that we are normally measuring dBSPL (Sound Pressure Level). Many noise measurements are done using sound intensity probes (to determine sound power, as in the osha document) and the sources are uncorrelated.
 
Re: dB calculations

Still thinking.... as a guitar player, I have a 50w head plugged into a 1x12 cab and have 100dB of sound, what do I get when I plug into a 2x12, and then a 4x12?

This can't be right: 1 speaker = 100dB / 2 speakers = 106dB / 4 speakers = 112dB

This would assume a correllated source AND 100dB for one speaker AND 100dB for the second speaker, using the consensus from above it would be 106dB.

Since that doesn't seem logical, I now have to assume that the equation is wrong. 100dB measured from the first speaker, then changing to a 2x12 does NOT add another 100dB of correllated sound.

What is the equation in the case of 2x12 and 4x12?
 
Re: dB calculations

Ian, I don't exaclty understand "10dB is for power. 20dB is for voltage and distance."

Is this the same thing as this: http://www.sengpielaudio.com/calculator-coherentsources.htm

It seems to clarify that you use the log10 or +3dB equation for non-similar sounds AND log20 or +6dB for same sounds
as in the sub or musical examples.
 
Re: dB calculations

The relevant equation from the previously attached PDF:

for two sources is p_t^2 = p1^2 + p2^2 + 2*p1*p2*cos(ph1-ph2)

add more sources and the polynomial gets bigger, plug in all a=b=c=1 into:
(a+b+c)^2 = a^2 + 2 a b + b^2 + 2 a c + 2 b c + c^2

only in the ideal case of course, assuming everything sums in phase, which it never will in any bandwidth.
 
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Re: dB calculations

Sean Thomas said:
Still thinking.... as a guitar player, I have a 50w head plugged into a 1x12 cab and have 100dB of sound, what do I get when I plug into a 2x12, and then a 4x12?

This can't be right: 1 speaker = 100dB / 2 speakers = 106dB / 4 speakers = 112dB

This would assume a correllated source AND 100dB for one speaker AND 100dB for the second speaker, using the consensus from above it would be 106dB.

Since that doesn't seem logical, I now have to assume that the equation is wrong. 100dB measured from the first speaker, then changing to a 2x12 does NOT add another 100dB of correllated sound.

What is the equation in the case of 2x12 and 4x12?
Click to expand...

Let's just assume all 12in drivers move the same way, at the same time, and are close enough to couple properly (i.e. a 4x12 cab). You use the same amount of power, and they are all passively wired. You now have four times the radiator area at the same power. Do the math - forgive me, I'm tired and am still learning this myself.
 
Re: dB calculations

Am I to understand that 10 watts into a 10" speaker will produce less dbSPL than the same 10 watts into a 100" speaker?

Aside from physical differences that result in forces such as friction and larger mass, 10W should produce the same amount of SPL in any speaker (efficiency not withstanding).

I think there's a whole lotta confusion regarding units of measure. dB SPL is not o let the electrical power being consumed by the driver, but also the mechanical resistance of the cone, cabinet design, and environmental factors such as other drivers nearby.

I think a pure theoretical world, double the power is an additional 3 dB, regardless of the units. In the real world, though, that may result in more than +3 dB SPL.

Yeah?
 
Re: dB calculations

Mitch Miller said:
Aside from physical differences that result in forces such as friction and larger mass, 10W should produce the same amount of SPL in any speaker (efficiency not withstanding).

I think there's a whole lotta confusion regarding units of measure. dB SPL is not o let the electrical power being consumed by the driver, but also the mechanical resistance of the cone, cabinet design, and environmental factors such as other drivers nearby.

I think a pure theoretical world, double the power is an additional 3 dB, regardless of the units. In the real world, though, that may result in more than +3 dB SPL.

Yeah?
Click to expand...

Yes, dBSPL is about the RMS pressure at a single spot in space, and it's units make it confusing for complex waveforms. It just happens to be very easy to measure.

Like you said, there is a lot of system that needs to be modeled to go from, voltage input to the amplifier, and then talk about the sound pressure at a single spot in the room. People have spent their entire careers working on one tiny piece of that puzzle.
 
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Re: dB calculations

Max Warasila said:
Let's just assume all 12in drivers move the same way, at the same time, and are close enough to couple properly (i.e. a 4x12 cab). You use the same amount of power, and they are all passively wired. You now have four times the radiator area at the same power. Do the math - forgive me, I'm tired and am still learning this myself.
Click to expand...
But 4x12" speakers in the guitar range ARE NOT close enough to couple properly.

Consider that 2 speakers need to be within 1/4 wavelength in order to couple well.

If 1,000Hz is1.13' long, then 1/4 wavelength would be a tad over 3". Higher freq are shorter.

The math works well IF the drivers are occupying the same physical space-but since they cannot do that-there are the issues.

And that is the "crux" of most of the problem PA cabinets out there.
 
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