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? And how loud is loud? 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.
The decibel (dB) is a logarithmic unit used to
measure sound level. It is also widely used in electronics,
signals and communication. The dB is a logarithmic way of describing 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 to the sone, which measures 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 P_{1}, and another playing a louder
version of the same sound with power P_{2}, but everything
else (how far away, frequency) kept the same.
Using the decibel unit, the difference in sound level, between the two is defined to
be
10 log (P_{2}/P_{1}) 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 (P_{2}/P_{1}) =
10 log 2 = 3 dB (to a good approximation).
This is shown on the graph, which
plots 10 log (P_{2}/P_{1}) against
P_{2}/P_{1}. To continue the example, if the second
had 10 times the power of the first, the difference in dB would be
10 log (P_{2}/P_{1}) =
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 (P_{2}/P_{1}) =
10 log 1,000,000 = 60 dB.
This example shows a 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: level difference in bels (named for Alexander Graham Bell) is just log (P_{2}/P_{1}).)
Sound pressure,
sound level and dB. Sound is usually measured with microphones and they
respond 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
x^{2} is just 2 log x, so this introduces a factor of 2
when we convert pressure ratios to decibels. The difference in sound pressure
level between two sounds with p_{1} and p_{2} is therefore:
20 log (p_{2}/p_{1}) dB
= 10 log (p_{2}^{2}/p_{1}^{2}) dB
= 10 log (P_{2}/P_{1}) dB
(throughout, the log is to base 10).
What happens when you halve the sound power? The log of 2 is 0.3010, so the log
of 1/2 is -0.3, to a good approximation. 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.
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 frequently asked question is a little subtle, so it is discussed 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 √2 and
the sound level by 3 dB. That is what we have done in the first
graphic and sound file below.
The first sample of sound is white noise (a mix of a broad range of
audible frequencies, analogous to white light, which is a mix of all visible
frequencies). The second sample is the same noise, with the voltage
reduced by a factor of √2. Now 1/√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 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 of the same order as 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*log_{10}(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.
(This makes the dB a convenient size unit.)
0.3 dB steps. You may notice that the last is quieter than the first, but
it is difficult to notice the difference between successive pairs.
10*log_{10}(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)
We said above that the decibel is
a ratio. So, when it is used to give the sound level for a single sound rather
than a ratio, a reference level must be chosen. For sound pressure level, the
reference level (for air) is usually chosen as p_{ref} = 20 micropascals (20 μPa),
or 0.02 mPa. This is very low: it is 2 ten billionths of an atmosphere.
Nevertheless, this is about the limit of sensitivity of the human ear, in its
sensitive range of frequency. (Usually this sensitivity is only found in
rather young people or in people who have not been exposed to loud music or
other loud noises. Personal music systems with in-ear speakers
are capable of very high sound levels in the ear, and are believed by some to
be responsible for much of the hearing loss
in young adults in some countries.)
So if you read of a sound pressure level of 86 dB, it means that
20 log (p_{2}/p_{ref}) = 86 dB
where
p_{ref} is the sound pressure of the reference level, and p_{2}
that of the sound in question. Divide both sides by 20:
log (p_{2}/p_{ref}) = 4.3
p_{2}/p_{ref} = 10^{4.3}
4 is the log
of 10 thousand, 0.3 is the log of 2, so this sound has a sound pressure 20
thousand times greater than that of the reference level
(p_{2}/p_{ref} = 20,000) or an intensity 400 million times the reference intensity. 86 dB is a loud sound but not dangerous—provided that exposure is brief.
What does 0 dB mean? This level occurs when the measured intensity
is equal to the reference level. i.e., it is the sound level corresponding to
0.02 mPa. In this case we have
Remember
that decibels measure a ratio. 0 dB occurs when you take the log of a
ratio of 1 (log 1 = 0). So 0 dB does not mean no sound, it
means a sound level where the sound pressure is equal to that of the reference
level. This is a small pressure, but not zero. It is also possible to have
negative sound levels: - 20 dB would mean a sound with pressure 10
times smaller than the reference pressure, i.e. 2 μPa.
Not all sound pressures are equally loud. This
is because the human ear does not respond equally to all frequencies: we are
much more sensitive to sounds in the frequency range about 1 kHz to
7 kHz (1000 to 7000 vibrations per second) than to very low or high
frequency sounds. For this reason, sound meters are usually fitted with a
filter whose response to frequency is a bit like that of the human ear. (More
about these filters below.) If the "A weighting filter" is used, the sound
pressure level is given in units of dB(A) or dBA. Sound pressure
level on the dBA scale is easy to measure and is therefore widely used. One reason why it is different from loudness is because the filter does not
respond in the same way as the ear. To understand the loudness of a
sound, the first thing you need to do consult some curves representing the frequency response of
the human ear, given below. (Alternatively, you can measure your own hearing response.) Another reason is that human hearing is not logarithmic.
Logarithmic measures
Why do we use decibels? The ear is capable of hearing a very large
range of sounds: the ratio of the sound pressure that causes permanent damage
from short exposure to the limit that (undamaged) ears can hear is more than a
million. To deal with such a range, logarithmic units are useful: the log of a
million is 6, so this ratio represents a difference of 120 dB.
Hearing is not inherently logarithmic in response. (Logarithmic measures are also useful when a sound (briefly) increases or decreases exponentially over time. This happens in many applications involving proportional gain or proportional loss.)
The filters used for dBA and dB(C)
The most widely used sound level filter is the A scale, which roughly
corresponds roughly to the inverse of the 40 dB (at 1 kHz) equal-loudness
curve. Using this filter, the sound level meter is thus less sensitive to very
high and very low frequencies. Measurements made on this scale are expressed
as dBA. The C scale varies little over several octaves and is thus
suitable for subjective measurements only for moderate to high sound levels.
Measurements made on this scale are expressed as dB(C). There is also a (rarely
used) B weighting scale, intermediate between A and C. The figure below shows
the response of the A filter (left) and C filter, with gains in dB given with
respect to 1 kHz. (For an introduction to filters, see RC filters, integrators
and differentiators.)
On the music acoustics and
speech acoustics sites, we
plot the sound spectra in dB. The reason for this common practice is that the
range of measured sound pressures is large.
dB(G) measurements use a narrow band filter that gives high weighting to frequencies between 1 and 20 Hz, and low weighting to others. It thus gives large values for sounds and infrasounds that cannot readily be heard. ISO 7196:1995
Loudness, phons and sones, hearing response curves
The phon is a unit that is related to dB by the psychophysically
measured frequency response of the ear. At 1 kHz, readings in
phons and dB are, by definition, the same. For all other frequencies, the phon
scale is determined by the results of experiments in which volunteers were
asked to adjust the loudness of a signal at a given frequency until they
judged its loudness to equal that of a 1 kHz signal. To convert from dB
to phons, you need a graph of such results. Such a graph depends on sound
level: it becomes flatter at high sound levels.
This graph, courtesy of Lindosland, shows the 2003 data from the International
Standards Organisation for curves of equal loudness determined experimentally.
Plots of equal loudness as a function of frequency are
often generically called Fletcher-Munson curves after the original work by
Fletcher, H. and Munson, W.A. (1933) J.Acoust.Soc.Am. 6:59. You can make your own curves using our hearing
response site.
The sone is derived from psychophysical measurements which involved
volunteers adjusting sounds until they judge them to be twice as loud. This
allows one to relate perceived loudness to phons. One sone is defined to be
equal to 40 phons. Experimentally it was found that, above 40 phons, a 10 dB
increase in sound level corresponds approximately to a perceived doubling of
loudness. So that approximation is used in the definition of the sone:
1 sone = 40 phon, 2 sone =
50 phon, 4 sone = 60 phon, etc.
This relation implies that loudness and intensity are related by a power law: loudness in sones is proportional to (intensity)^{log 2} = (intensity)^{0.3}.
Wouldn't it be great to be able to convert from dB (which can be measured
by an instrument) to sones (which approximate loudness as perceived by
people)? This is sometimes done using tables that you can find in acoustics
handbooks. However, if you don't mind a rather crude approximation, you can
say that the A weighting curve approximates the human frequency response at
low to moderate sound levels, so dB(A) is very roughly the same as phons, over a limited range of low levels.
Then one can use the logarithmic relation between sones and phons described above.
Recording level and decibels
Meters measuring recording or output level on audio electronic gear
(mixing consoles etc) are almost always recording the AC rms voltage (see
links to find out about AC and rms). For a given
resistor R, the power P is V^{2}/R, so
difference in voltage level =
20 log (V_{2}/V_{1}) dB
= 10 log (V_{2}^{2}/V_{1}^{2})
dB = 10 log (P_{2}/P_{1}) dB,
or
absolute voltage level = 20 log (V/V_{ref})
where V_{ref} is a reference voltage. So what is the
reference voltage?
The obvious level to choose is one volt rms, and in this case the level is
written as dBV. This is rational, and also convenient with analog-digital cards whose maximum range is often about one volt rms. So one
has to remember to the keep the level in negative dBV (less than one volt) to
avoid clipping the peaks of the signal, but not too negative (so your signal
is still much bigger than the background noise).
Sometimes you will see dBm. This used to mean decibels of electrical
power, with respect to one milliwatt, and sometimes it still does. However,
it's complicated for historical reasons. In the mid twentieth century, many
audio lines had a nominal impedance of 600 Ω. If the impedance is purely
resisitive, and if you set V^{2}/600 Ω = 1 mW,
then you get V = 0.775 volts. So, providing you were using a
600 Ω load, 1 mW of power was 0 dBm, which was 0.775 V, so you calibrated your level meters thus. The problem arose because, once a
level meter that measures voltage is calibrated like this, it will read
0 dBm at 0.775 V even if it is not connected to 600 Ω So,
perhaps illogically, dBm will sometimes mean dB with respect to 0.775 V.
(When I was a boy, calculators were expensive so I used dad's
old slide rule, which had the factor 0.775 marked on the cursor window to
facilitate such calculations.)
How to convert dBV or dBm into dB of sound level? There is no simple
way. It depends on how you convert the electrical power into sound power. Even
if your electrical signal is connected directly to a loudspeaker, the
conversion will depend on the efficiency and impedance of your loudspeaker.
And of course there may be a power amplifier, and various acoustic
complications between where you measure the dBV on the mixing desk and where
your ears are in the sound field.
Intensity, radiation and dB
How does sound level (or radio signal level, etc) depend on distance
from the source?
A source that 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 4πr^{2}, so the power (in our
example, the sound power) passing through each square metre of surface
is, by definition:
I = P/4πr^{2}.
So we see that, for an isotropic
source, intensity is inversely proportional to the square of the
distance away from the source:
I_{2}/I_{1} =
r_{1}^{2}/r_{2}^{2}.
But intensity
is proportional to the square of the sound pressure, so we could equally
write:
p_{2}/p_{1} = r_{1}/r_{2}.
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
For acoustic waves, the specific acoustic impedance z is defined as the
ratio of the acoustic pressure p to the average particle velocity u, due to the sound ave,
z = p/u . In Acoustic impedance, intensity and power, we show how to
relate RMS acoustic pressure p and intensity I:
I = p^{2}/z
For air, the specific
acoustic impedance z is
420 kg.s^{−1}.m^{−2} =
420 Pa.s.m^{−1}. For (fresh) water, the specific acoustic
impedance for water is 1.48 MPa.s.m^{−1}. So a sound wave in
water with the same pressure has a much lower intensity than one in air.
dBi and radiation that varies with direction
Radiation that varies in direction is called anisotropic. For many
cases in communication, isotropic radiation is wasteful: why emit a
substantial fraction of power upwards if the receiver is, like you, relatively
close to ground level. For sound of short wavelength (including most of the
important range for speech), a megaphone can help make your voice more
anisotropic. For radio, a wide range of designs allows antennae to be highly
anisotropic for both transmission and reception.
So, when you interested in emission in (or reception from) a particular
direction, you want the ratio of intensity measured in that direction, at a
given distance, to be higher than that measured at the same distance from an
isotropic radiator (or received by an isotropic receiver). This ratio is
called the gain; express the ratio in dB and you have the gain in
dBi for that radiator. This unit is mainly used for antennae, either
transmitting and receiving, but it is sometimes used for sound sources and
directional microphones.
Example problems
A few people have written asking for examples in using dB in calculations.
So...
All else equal, how much louder is loudspeaker driven (in its linear
range) by a 100 W amplifier than by a 10 W amplifier?
The powers differ by a factor of ten, which, as we saw above, is
10 dB. All else equal here means that the frequency responses are equal
and that the same input signal is used, etc. So the frequency dependence
should be the same. 10 dB corresponds to 10 phons. To get a
perceived doubling of loudness, you need an increase of 10 phons. So the
speaker driven by the 100 W amplifier is twice as loud as when driven by
the 10 W, assuming you stay in the linear range and don't distort or
destroy the speaker. (The 100 W amplifier produces twice as many sones as
does the 10 W.)
I am standing at a distance R from a small source of sound (size much less
than R), at ground level out in the open where reflections may be neglected.
The sound level is L. If I now move to a distance nR (n a number, and nR still
much greater than the size of the source), what will be the new sound level?
First, note that the neglect of reflections is very important. This
calculation will not work inside a room, where reflections from the wall
(collectively producing reverberation) make the calculation quite difficult.
Out in the open, the sound intensity is proportional to 1/r^{2}, where
r is the distance from the source. (The constant of proportionality depends on
how well the ground reflects, and doesn't concern us here, because it will
roughly cancel in the calculation, provided r is reasonably large.) So, if we increase r from R to nR, we decrease the
intensity from I to I/n ^{2}.
The difference in decibels between the two signals of intensity I
_{2} and I _{1} is defined above to be
For example, if n is 2 (ie if we go twice as far away), the intensity is
reduced by a factor of four and sound level falls from L to
(L − 6dB).
If, in ideal quiet conditions, a young person can hear a 1 kHz tone
at 0 dB emitted by a loudspeaker (perhaps a softspeaker?), by how much
must the power of the loudspeaker be increased to raise the sound to
110 dB (a dangerously loud but survivable level)?
The difference in decibels between the two signals of power P_{2}
and P_{1} is defined above to be
ΔL = 10 log (P_{2}/P_{1}) dB so,
raising 10 to the power of these two equal quantities: 10^{L/10}
=
P_{2}/P_{1} so: P_{2}/P_{1}
= 10^{110/10} = 10^{11} = one hundred thousand
million.
which is a demonstration that the human ear has a remarkably
large dynamic range, perhaps greater than that of the eye.
An amplifier has an input of 10 mV and and output of 2 V. What is its
voltage gain in dB?
Voltage, like pressure, appears squared in expressions for power or
intensity. (The power dissipated in a resistor R is V^{2}/R.) So, by
convention, we define:
gain = 20 log (V_{out}/V_{in})
= 20 log
(2V/10mV) = 46 dB
(In the acoustic cases given above, we saw that the pressure
ratio, expressed in dB, was the same as the power ratio: that was the reason
for the factor 20 when defining dB for pressure. It is worth noting that, in
the voltage gain example, the power gain of the ampifier is unlikely to equal
the voltage gain, which is defined by the convention used here. The power is proportional to the square of the voltage in a
given resistor. However, the input and output impedances of amplifiers are
often quite different. For instance, a buffer amplifier or emitter follower
has a voltage gain of about 1, but a large current gain.)
What is the difference, in dB, between the irradiance (light intensity) on earth (8.3 light minutes from the sun) and on Uranus (160 light minutes)?
Like sound, isotropic light intensity decreases as r^{−2}, so the intensity ratio is (160/8.3)^{2} = 20 log (160/8.3) = 26 dB.
Occupational health and safety
Different countries and provinces obviously have different laws concerning
noise exposure at work, which are enforced with differing enthusiasm. Many
such regulations have a limit for exposure to continuous noise of 85 dB(A), for
an 8 hour shift. For each 3 dB increase, the allowed exposure is halved. So,
if you work in a nightclub where amplified music produces 100 dB(A) near your
ears, the allowed exposure is 15 minutes. There is a limit for impulse noise
like firearms or tools that use explosive shots. (e.g. 140 dB peak should not
be exceeded at any time during the day.) There are many documents providing
advice on how to reduce noise exposure at the source (ie turn the music level
down), between the source and the ear (ie move away from the loudspeakers at a
concert) and at the ear (ie wear ear plugs or industrial hearing protectors).
Noise
management and protection of hearing at work is the code of practice in
the state of New South Wales, Australia (the author's address).
Some FAQs
How loud is an aircraft? A train? A person singing? A dog barking? A power
tool? The answers to this question vary considerably. It depends strongly upon
how far away you are, whether you are indoors or not, whether there is
reverberation, how strong the particular source is and what its spectrum is.
To give values, without being very specific about the conditions, would be
somewhat misleading. Because the rest of this page is intended to be reliable,
as far as it goes, I'd rather not give values here.
How does one "add decibels"?, meaning What sound level do you get when you
add level a to level b? If the sources are coherent (which usually means that
they ultimately come from the same source), then there may be complicated
interference effects. In most cases, where the sources are independent, one
can add the intensities and then convert to decibels. However, if you
are given the sound levels in dB(A), it is not so easy to go back to intensity,
and one must know something about the spectrum of the sound. If you know the
distribution of the sound in different frequency bands, you can use the applet
on this link.
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, 10^{1} would mean that there is
only one 10 in the product, so 1 is the log of 10, or in other words
10^{1} = 10.
We can also have negative logarithms. When
we write 10^{-2} we mean 0.01, which is
1/100, so
10^{-n} = 1/10^{n}
Let's go one step more complicated. Let's work out the value of
(10^{2})^{3}. This is easy enough to do, one step at a
time: