A ruler is a linear scale: it has marks on it corresponding to equal quantities of distance. One way of expressing this is to say that the ratio of successive intervals is equal to one. A logarithmic scale is different in that the ratio of successive intervals is not equal to one. Each interval on a logarithmic scale is some common factor larger than the previous interval. A typical ratio is 10, so that the marks on the scale read: 1, 10, 100, 1000, 10000, etc. Such a scale is useful if you are plotting a graph of values which have a very large range.

Since many aspects of perception are related to proportional change, logarithmic scales are very common in psychophysics. A graph of many perceptual scales against the logarithm of the stimulus size is a straight line over some range (this is known as Weber's law). A scale of perceptual pitch against log (Hz) is a good example.

Loudness is a perceptual or subjective quality of a sound; intensity is a physical or objective property. Although changes in intensity can cause changes in loudness, they are clearly two different scales. In particular, sounds which are below the threshold of audibility have a non-zero intensity, but zero loudness.

Intensity is measured in Wm^-2, but we usually prefer to use the Sound Pressure Level scale (dBSPL). Loudness can be measured in units called phons, where 10 phons is the perceived loudness associated with a pure tone of 1000Hz at 10dB above the threshold of audibility.

The number 20 has two causes: one that gives us a multiplier of 10, and one that gives us a multiplier of 2.

The factor of 10 is easy - we are working in decibels not bels. One bel (named after Alexander Graham Bell, by the way) is rather a large unit, roughly equal to a tripling in amplitude. So we multiply by 10 and work in tenths of a bel to give us more sensitivity.

The factor of 2 is there because we have ignored the fact that the correct definition of decibels is as a logarithmic ratio of powers not amplitudes. The correct definition would look like:

In sound, we prefer to work in Pascals: units of pressure rather than units of power. Fortunately, it is fairly easy to show that the power in a sound signal is proportional to the square of the pressure (imagine the air pressing against a membrane with some force F and moving it through some distance d; if the membrane is elastic, then d is proportional to the pressure; and the energy transferred is then just the force on the membrane F times d. But since both F and d are proportional to pressure, then the energy is proportional to the pressure squared). Thus the power ratio is numerically equal to the square of the pressure ratio, and we can write:

or, with the magic of logarithms:

The decibel scale is a logarithmic ratio scale: we start with a ratio of pressures, then take the logarithm and finally multiply by 20. If the the ratio is a number greater than one, then the logarithm must produce a value greater than 0 (e.g. log(10)=1). If the ratio is a number less than one, then the logarithm must produce a value less than 0 (e.g. the log(0.1)=-1). If the ratio is equal to one (i.e. that the two pressures are equal) then the logarithm returns 0 (since log(1)=0). Thus the zero point on the decibel scale is simply the point at which the measured amplitude is equal to the reference amplitude.

Note that it doesn't make sense to do the calculation when there is no measurable pressure change, since the logarithm of zero is not defined (minus infinity).

Answer A: Well all this is saying is that 10^1.3 is equal to 20. Is this really so strange? After all 10^1.0 = 10, and 10^2.0 = 100, so 10^1.3 has got to be a number between 10 and 100 hasn't it?

Answer B: We can try to work out the logarithm of 20 approximately by the following method. We are seeking a value x in the expression:

Let's first divide out a factor of 10:

And now raise to the tenth power:

Since 2¹⁰ = 1024, we can now make the following approximation (~= means approximately):

Or, in other words:

Or

We need to invert the formula for decibels:

OK, first divide by 20:

Raise each side to a power of 10:

And, finally multiply by 20µPa:

For example, 60dBSPL is