dB: What is a decibel?

Decibels: dB, dB(A), dBA, dB(C), dBV, dBm and dBi? What are they all? How are they related to loudness, to phons and to sones? This page describes and compares them all and gives sound file examples. A related page allows you to measure your hearing response and to compare with standard hearing curves. This is a background page to the multimedia chapters Sound and Quantifying Sound.

graph of dB scale


Definition and examples

The decibel ( dB) is used to measure sound level, but it is also widely used in electronics, signals and communication. The dB is a logarithmic way of dscribing a ratio. The ratio may be power, sound pressure, voltage or intensity or several other things. Later on we relate dB to the phon and the sone (related to loudness). But first, to get a taste for logarithmic expressions, let's look at some numbers. (If you have forgotten, go to What is a logarithm?)

For instance, suppose we have two loudspeakers, the first playing a sound with power P1, and another playing a louder version of the same sound with power P2, but everything else (how far away, frequency) kept the same.

The difference in decibels between the two is defined to be

10 log (P2/P1) dB        where the log is to base 10.

If the second produces twice as much power than the first, the difference in dB is

10 log (P2/P1) = 10 log 2 = 3 dB.

as is shown on the graph, which plots 10  log (P2/P1) against P2/P1. To continue the example, if the second had 10 times the power of the first, the difference in dB would be

10 log (P2/P1) = 10 log 10 = 10 dB.

If the second had a million times the power of the first, the difference in dB would be

10 log (P2/P1) = 10 log 1,000,000 = 60 dB.

This example shows one feature of decibel scales that is useful in discussing sound: they can describe very big ratios using numbers of modest size. But note that the decibel describes a ratio: so far we have not said what power either of the speakers radiates, only the ratio of powers. (Note also the factor 10 in the definition, which puts the 'deci' in decibel).

Sound pressure, sound level and dB. Sound is usually measured with microphones and they respond (approximately) proportionally to the sound pressure, p. Now the power in a sound wave, all else equal, goes as the square of the pressure. (Similarly, electrical power in a resistor goes as the square of the voltage.) The log of the square of x is just 2 log x, so this introduces a factor of 2 when we convert to decibels for pressures. The difference in sound pressure level between two sounds with p1 and p2 is therefore:

20 log (p2/p1) dB   =  10 log (p22/p12) dB   = 10 log (P2/P1) dB       where again the log is to base 10.

What happens when you halve the sound power? The log of 2 is 0.3, so the log of 1/2 is -0.3. So, if you halve the power, you reduce the power and the sound level by 3 dB. Halve it again (down to 1/4 of the original power) and you reduce the level by another 3 dB. If you keep on halving the power, you have these ratios.

examples of dB, pressure ratios and

What happens if I add two identical sounds? Do I double the intensity (increase of 3 dB)? Or do I double the pressure (increase of 6 dB)? This is a frequently asked question, and it is a little subtle, so it is here on our FAQ.

Sound files to show the size of a decibel

We saw above that halving the power reduces the sound pressure by root 2 and the sound level by 3 dB. That is exactly what we have done in the first graphic and sound file below.

The first sample of sound is white noise (a mix of all audible frequencies, just as white light is a mix of all visible frequencies). The second sample is the same noise, with the voltage reduced by a factor of the square root of 2. The reciprocal of the square root of 2 is approximately 0.7, so -3 dB corresponds to reducing the voltage or the pressure to 70% of its original value. The green line shows the voltage as a function of time. The red line shows a continuous exponential decay with time. Note that the voltage falls by 50% for every second sample.

Note, too, that a doubling of the power does not make a huge difference to the loudness. We'll discuss this further below, but it's a useful thing to remember when choosing sound reproduction equipment.

Sound files and flash animation by John Tann and George Hatsidimitris.

How big is a decibel? In the next series, successive samples are reduced by just one decibel.

One decibel is close to the Just Noticeable Difference (JND) for sound level. As you listen to these files, you will notice that the last is quieter than the first, but it is rather less clear to the ear that the second of any pair is quieter than its predecessor. 10*log10(1.26) = 1, so to increase the sound level by 1 dB, the power must be increased by 26%, or the voltage by 12%.

What if the difference is less than a decibel? Sound levels are rarely given with decimal places. The reason is that sound levels that differ by less than 1 dB are hard to distinguish, as the next example shows.

You may notice that the last is quieter than the first, but it is difficult to notice the difference between successive pairs. 10*log10(1.07) = 0.3, so to increase the sound level by 0.3 dB, the power must be increased by 7%, or the voltage by 3.5%.

Standard reference levels ("absolute" sound level)

Logarithmic response, psychophysical measures, sones and phons

The filters used for dBA and dB(C)

Loudness, phons and sones

Recording level and decibels

Intensity, radiation and dB

sketch of spherically symmetric radiation
    How does sound level (or radio signal level, etc) depend on distance from the source?

    A source tht emits radiation equally in all directions is called isotropic. Consider an isolated source of sound, far from any reflecting surfaces – perhaps a bird singing high in the air. Imagine a sphere with radius r, centred on the source. The source outputs a total power P, continuously. This sound power spreads out and is passing through the surface of the sphere. If the source is isotropic, the intensity I is the same everywhere on this surface, by definition. The intensity I is defined as the power per unit area. The surface area of the sphere is 4πr2, so the power (in our example, the sound power) passing through each square metre of surface is, by definition:

      I = P/4 πr2.
    So we see that, for an isotropic source, intensity is inversely proportional to the square of the distance away from the source:
      I2/I1 = r12/r22.
    But intensity is proportional to the square of the sound pressure, so we could equally write:
      p2/p1 = r1/r2.
    So, if we double the distance, we reduce the sound pressure by a factor of 2 and the intensity by a factor of 4: in other words, we reduce the sound level by 6 dB. If we increase r by a factor of 10, we decrease the level by 20 dB, etc.

    Be warned, however, that many sources are not isotropic, especially if the wavelength is smaller than, or of a size comparable with the source. Further, reflections are often quite important, especially if the ground is nearby, or if you are indoors.

Pressure, intensity and specific impedance

dBi and radiation that varies with direction

Example problems

Occupational health and safety

Some FAQs


Related pages

What is a logarithm? A brief introduction.

    First let's look at exponents. If we write 102 or 103 , we mean
      102 = 10*10 = 100   and    103 = 10*10*10 = 1000.
    So the exponent (2 or 3 in our example) tells us how many times to multiply the base (10 in our example) by itself. For this page, we only need logarithms to base 10, so that's all we'll discuss. In these examples, 2 is the log of 100, and 3 is the log of 1000. In a multiplication calculation like those above, 101 would mean that there is only one 10 in the product, so 1 is the log of 10, or in other words
      101 = 10.
    We can also have negative logarithms. When we write 10-2 we mean 0.01, which is 1/100, so
      10-n = 1/10n
    Let's go one step more complicated. Let's work out the value of (102)3. This is easy enough to do, one step at a time:
      (102)3 = (100)3 = 100*100*100 = 1,000,000 = 106.
    By writing it out, you should convince yourself that, for any whole numbers n and m,
      (10n)m = 10nm.
    But what if n is not a whole number? Since the rules we have used so far don't tell us what this would mean, we can define it to mean what we like, but we should choose our definition so that it is consistent. The definition of the logarithm of a number a (to base 10) is this:
      10log a = a.
    In other words, the log of the number a is the power to which you must raise 10 to get the number a. For an example of a number whose log is not a whole number, let's consider the square root of 10, which is 3.1623..., in other words 3.16232 = 10. Using our definition above, we can write this as
      3.16232 = (10log 3.1623)2 = 10 = 101.
    However, using our rule that (10n)m = 10nm, we see that in this case log 3.1623*2 = 1, so the log of 3.1623... is 1/2. The square root of 10 is 100.5. Now there are a couple of questions: how do we calculate logs? and Can we be sure that all real numbers greater than zero have real logs? We leave these to mathematicians (who, by the way, would be happy to give you a more rigorous treatment of exponents that this superficial account).

    A few other important examples are worth noting. 100 would have the property that, no matter how many times you multiplied it by itself, it would never get as large as 10. Further, no matter how many times you divided it into 1, you would never get as small as 1/10. Using our (10n)m = 10nm rule, you will see that 100 = 1 satisfies this, so the log of one is zero. The log of 2 is used often in acoustics, and it is 0.3010 (see graph at right). Hence, a factor of 2 in power corresponds to 3.01 dB, which we should normally write as 3 dB because, as you can discover for yourself in hearing response, decimal points of decibels are usually too small to notice.

    Go back to top of page.

    Joe Wolfe / J.Wolfe@unsw.edu.au.

    title

    log10x vs x.