getting the feel for dB calculations

A primer on room ratios and other matters related to acoustic analysis

getting the feel for dB calculations

Postby Eric.Desart » Fri Apr 25, 2008 7:13 pm

The dB scale & dBs: getting some feel

A summary:
dBs are nothing but a factor expressed on a logarithmic scale.
They don't mean a thing, being just a factor, in absolute sense without referring them to something (pressure, power, velocity, intensity, acceleration, an emission or immission level, whatever)
In the audio world they are often or mostly interpreted as sound pressure levels as measured by audio or measurement tools or levels to be read on an SLM (Sound Level Meter).

One could make an analogy (poetic license) with the use of the notion kilo which represents the factor 1000.
If one orders a kilo apples at the groceries, there is very little chance the shopkeeper will pack anything else than 1 kg (kilogram), and won't consider neither 1 km (kilometer) nor a kHz (kilohertz) worth of apples.
In popular speech a kilo refers to a weight, however kilo is a general prefix combined with numerous metric units.

When just calculating with dB's in dB's you don't need to wonder about this factor 10 or 20 times the log which just refers to the conversion of arithmetic units to dBs.
The difference between this 10 and 20 is the squared relationship between pressure and power/intensity.
You don't need to bother about all this when you're converting general factors in dBs.

Arithmetic operators versus logarithmic operators:
Note that this factor 10 is exclusively related to the d in dB
The scales in dBs are in fact buildup and calculated in Bs (Bells or bels) but expressed in dBs (decibels = 1/10 bel).
Therefore in acoustic formulas you find these repeatedly divisions and multiplications by 10 (toggling between the logarithmic calculation scale and the way Bells or bels are expressed as a convention in dBs for pragmatic reasons).

Logarithmic operators lower 1 level compared to Arithmetic operators.
  1. An arithmetic power becomes a logarithmic multiplication:
    10*log(6^2) = (10*log(6))*2 = 10*2*log(6) = 20*log(6)
  2. An arithmetic root becomes a logarithmic division:
    10*log(sqrt(6)) = (10*log(6))/2 = 10/2*log(6) = 5*log(6)
  3. An arithmetic multiplication becomes a logarithmic addition:
    10*log(6*2) = 10*log(6) +10*log(2)
    When you calculate energetically, ignoring phase relationship (hence uncorrelated signals):
    If you have 14 sources of 74.5 dB or if you increase a level with a factor 14 then:
    Lp = 74.5 dB*14 = 74.5 + 10*log(14) = 74.5 + 11.5 dB = 86 dB
  4. An arithmetic division becomes a logarithmic subtraction:            
    10*log(6/2) = 10*log(6) - 10*log(2)
    If you want to divide the number of radiating sources by 5 or decreasing your level with a factor 5 then:
    Lp = 74.5 dB/5  = 74.5 - 10*log(5) = 74.5 - 7.0 dB = 67.5 dB
    This is in fact equal to following multiplication:
    74.5 dB at 20% = 74.5 * 0.2 = 74.5 + 10*log(0.2) = 74.5 + -7.0dB = 74.5 - 7 dB = 67.5 dB
  5. An arithmetic addition becomes a logarithmic "??? Hé ..., there is no lower level operator available ...":  
    You can't directly add dB values.
    If you somewhere see dB values being added, that operation shows an arithmetic multiplication.

    In order to add dB values you must  first convert them to the arithmetic equivalent of the bel (not dB) value*, then adding, then converting the resulting sum back to the equivalent logarithmic dB value.
    * remember: the logarithmic scale calculates in bels not decibels.
    Let's add 74.5 dB + 83.2 dB:
    The antilog (arithmetic equivalent) of a dB value of e.g. 74.5 dB is 10^(74.5/10) where the (74.5/10) is the conversion from dB to bels. The 1st 10 is the base of the logarithmic scale.

      This can be compared with ln(...) and exp(...). ln refers to a logarithmic scale with base e = 2.7182818284590452353603.......
      The antilog then equals exp(...) = e^(...) = 2.7182818284590452353603^(...)
    Lp = 74.5 dB + 83.2 dB = 10*log(10^(74.5/10) + 10^(83.2/10)) = 10*log(28,183,829 + 208,929,613) = 10*log(237,113,442) = 83.7 dB
  6. An arithmetic subtraction becomes a logarithmic "??? Hé ..., again no lower level operator available ...":
    You can't directly subtract dB values.
    If you somewhere see dB values being subtracted, that operation shows an arithmetic division with a factor > 1 or > 100% OR a multiplication with a factor >0 and <1.

    E.g. an amplifier set to -30 dB means that the amplifier is set to 10^(-30/10) = 0.001 = 0.1% of its nominal output.
    This minus dB scale is nothing else than a logarithmic, in dB expressed, percentage versus the nominal maximum of 100%.
    The solution for a subtraction is the comparable with the above addition. A roundabout via the arithmetic equivalents (antilogs) of the bel (not dB) values is required.
    The only difference is that the addition + operator is substituted by the subtraction - operator.
    Let's remove 1 sound source of 79.3 dB from a room with 5 sources totaling 84.8 dB (number of sources doesn't matter, the total level at the measurement point is defining)
    Again this is an energetic calculation, not taking phase influence into account (uncorrelated sound)
    84.8 dB - 79.3 dB = 10*log(10^(84.8/10)-10^(79.3/10)) = 10*log(301,995,172-85,113,804) = 10*log(216,881,368) = 83.4 dB
      Note: if the 2nd term is equal or larger than the 1st term resulting in a 0 or negative (olive colored) arithmetic value the calculation WILL return an error for the same simple fact that logarithms just can't work with, or express 0 or negative values in the arithmetic sense of the concepts.
      You just can't express in dB that there is no sound and you certainly can't express negative sound in dB.
Every dB value one ever sees represents a POSITIVE arithmetic value.
  • A negative dB value represents an arithmetic factor >0 AND <1.
  • 0 dB represents the arithmetic factor 1.
  • A positive dB value represents an arithmetic factor larger than 1.
When calculating with dBs as factors (main operators) one doesn't need a workaround via  pressure, intensity or whatever the scale can refer to.
Logarithmic calculations (hence and therefore dBs as well) are just a mathematical way of calculating in itself.
As long as one calculates with uncorrelated sources the logarithmic math can stand on its own.
  • Calculating 5% expressed in dB is nothing more than 10*log(0.05) = -13.0103 dB
  • Calculating the factor 1/5% = 20 is nothing more than 10*log(20) = 13.0103 dB
  • Calculating the factor 95% in dB is nothing more than 10*log(0.95) = -0.22276395 dB
  • Calculating the factor 1/95% is nothing more than 10*log(1/0.95) = 0.22276395 dB
  • When log. adding -13.0103 dB and -0.22276395 dB one gets back 0 dB equaling the original 100% signal.
Once one needs to take phase relationships and superposition into account things become different, because then the physics and math related to interference, superposition and so on comes into the picture, forcing you to start from the basics.
It's logical that if you add or remove sources altering the interference pattern and you want to calculate levels on certain spots you have to take these possible reflections, reinforcements and cancelations, etc. into account.

It's the lack of feeling for logarithmic math that causes often/sometimes wrong advices on the net.
This logarithmic math is used to cover for the non-linearity of our hearing and the enormous dynamic range involved between low and high levels.
People are that used to arithmetic reasoning that conclusions are drawn which are wrong or only correct in an insignificant manner and therefore not or hardly to the point.

Have fun
Eric Desart

If you want more in-depth details and math about dBs and sound propagation, as well as the relationship of dB's versus their respective reference units (in APPENDIX) look in this document:
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