A decibel (dB) represents the ratio of two variables on a logarithmic scale, and has no base unit (e.g. metres). Using a logarithmic scale is a much better approximation of human hearing than the linear variables. As well, the gigantic ratio of barely perceptible sound pressure (the auditory threshold) to the loudest tolerable sound pressure (pain threshold) of 1: 3,000,000 is compressed into a much more manageable scale of 0 to 130 dB. The general calculation is as follows: log (value/reference value). We use the logarithm to base 10, which is generally given as ‘log’ on calculator keypads. The result is the Bel, one-tenth of which is one deci-bel, i.e. a decibel. These are power ratios. For sound pressures, voltages and currents, the factor is 20.
Power ratio in dB: 10 x log_{10} (power/reference power) or 10 x log_{10} (P/P_{0})
Sound pressure, voltage or current ratios in dB: 20 x log_{10} (value/reference value)
In the case of sound pressure ratios, the auditory threshold is used, having a value of 20 μPa. Because there is a defined reference value, in this case ‘SPL’ is appended to the unit ‘dB’. Nowadays, however, it has become common to omit the ‘SPL’ when discussing sound pressure levels. Other references:
Reference value | 1 μV | 1 mV | 0,775 V | 1 V | 20 μPa |
Decibel | dB μV | dB mV | dBu | dBV | dB SPL |
The following table shows a few relationships governing the calculation of physical values and decibel values, and the conversion between these types of values:
Physical | Multiplication | Division | < 1 | 1 | > 1 | Negative |
⇩ | ⇩ | ⇩ | ⇩ | ⇩ | ⇩ | |
Decibels | Addition | Subtraction | Negative | 0 | Positive | Not possible |
Example 1: An amplifier amplifies an input signal of 1 mV (millivolts) to an output signal of 1,000 mV. The gain is thus 1000-fold (1000: 1), or 20 x log (1,000 / 1) = +60 dB.
Example 2: An attenuator attenuates a voltage to one-tenth. The ratio between output and input is 0.1/1 = 0.1. Expressed in dB: 20 x log (0.1 / 1) = -20 dB.
Example 3: The attenuator (example 2) is connected to the output of the amplifi er (example 1). The gain is thus: 1,000 x 0.1 = 100. Stated in dB: 60 dB + (-20 dB) = 60 dB – 20 dB = 40 dB.
If the sound pressure level is stated in dB, this information can be used in calculations. For instance, a loudspeaker datasheet provides us with information for the characteristic sound pressure level (1 W/1 m): 95 dB. This means that at 1 watt of power, the loudspeaker generates a sound pressure level of 95 dB at a distance of 1 meter. The following table indicates by how many decibels the sound pressure level of the loudspeakers increases at a given power.
Power (W) | 1 | 2 | 5 | 6 | 10 | 15 | 20 | 30 | 50 | 100 |
Increase in the | 0 | 3 | 7 | 8 | 10 | 12 | 13 | 15 | 17 | 20 |
The table shows that at 6 watts, you need to add 8 dB to the 95 dB. Consequently, at 6 watts of power we obtain 103 dB SPL at a distance of 1 metre. There is also a mathematical formula for this calculation that yields the same result.
p_{1} = p_{n} + 10 x log (P)
p_{1}: Sound pressure level (dB) p_{n}: Characteristic sound pressure level (dB) P: supplied power (W)
Each doubling of power gives us an additional 3 dB of SPL.
If you would like to calculate the sound pressure level produced by the loudspeaker not at a distance of 1 meter, but at e.g. 6 meters, there is a table/formula for this purpose as well.
Distance (m) | 1 | 2 | 3 | 4 | 5 | 10 | 20 | 50 | 100 |
Decrease (dB SPL) | 0 | 6 | 9,5 | 12 | 14 | 20 | 26 | 34 | 40 |
Based on the same example, we will have to subtract an amount, corresponding to the distance, from the calculated figure of 103 dB. The reduction resulting from a distance of 5 metres from the loudspeakers is 14 dB – which corresponds to a sound pressure level of 89 dB. The formula for the calculation is as follows: p = p_{1} - 20 x log (d)
p: Sound pressure level at a defined distance (dB characteristic sound pressure)
d: Distance (m)
p_{1}: Sound pressure level at a distance of 1 m
With each doubling in distance, the sound pressure level drops by 6 dB SPL.
The formulas for sound pressure at a defined power and at a defi ned distance are combined. The sound pressure level at a given power and distance is calculated as follows: p = p_{n} + 10 x log (P) - 20 x log (d)
p: Sound pressure level (dB SPL) p_{n}: Characteristic sound pressure level of the loudspeaker (dB)
d: Distance from the loudspeaker (m) (m) P: supplied power (W)
Example: We want to install a loudspeaker in a room. The greatest distance to the audience is 8 m. The loudspeaker has a characteristic sound pressure level of 90 dB 1 W/1 m and an input power of 30 watts. How high is the sound pressure level at the maximum distance?
Sound pressure level
= 90 dB + 10 x log (30) - 20 x log (8)
= 90 dB + 15 dB - 18 dB
= 87 dB
If you use the values from the two tables provided above (the distance is composed of 4 m x 2m = 8 m, physical multiplication turns into addition of the decibel values) this yields:
Sound pressure level
= 90 dB + 15 dB (at 30 watts) - 12 dB (at 4 m) - 6 dB (at 2 m)
= 87 dB
A perceived doubling in volume requires around 10 times the amplifier power.
Distance and minimum sound pressure level between TOA’s standard ceiling loudspeakers at different degrees of speech intelligibility and 6 W of power:
Ceiling height (m) | 3 | 3,5 | 4 | 4,5 | 5 | 5,5 | 6 | |
Best intelligibility | Distance between loudspeakers (m) | 2,3 | 3,1 | 3,8 | 4,6 | 5,4 | 6,1 | 6,9 |
min. Sound pressure level (dB) | 92 | 90 | 88 | 86 | 85 | 84 | 83 | |
Good intelligibility | Distance between loudspeakers (m) | 3,6 | 4,8 | 6 | 7,2 | 8,3 | 9,5 | 10,7 |
min. Sound pressure level (dB) | 90 | 88 | 86 | 84 | 83 | 82 | 81 | |
Background music | Distance between loudspeakers (m) | 8,2 | 11 | 13,7 | 16,5 | 19,2 | 22 | 24,7 |
min. Sound pressure level (dB) | 85 | 82 | 81 | 79 | 78 | 76 | 75 |