The measurement of sound amplitude is complicated by the

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					                   THE LIFE AND TIMES OF THE DECIBEL
The measurement of sound amplitude is complicated by the fact that sound may be considered either in
terms of its pressure or its intensity. Intensity is the amount of power acting on an area (i.e., against a
surface), and, in terms of sound, is measured as watts/cm2 (where a watt is the unit of power). Sound
pressure on the other hand is the amount of force acting on an area, and, again in terms of sound, is
measured in Pascals where a Pascal represents 1 newton/m2 (where a newton is the unit of force).

Since watts/cm2 and Pascals can both to measure sound amplitude, it follows that intensity (I) and pressure
(p) must somehow be related. In fact, I = p 2. Remember this relationship - we‟ll be referring to it again

Let‟s consider two pure tones (frequency needn‟t be considered - the explanation that follows is valid for all
sound frequencies). The first pure tone is hereby designated as the „quiet‟ (Q) sound - it‟s the quietest
sound that the average human ear can hear. The second pure tone is henceforward to be known as the
„loud‟ (L) sound - it‟s the loudest sound that the average human ear can tolerate without pain. The table
below shows the measurements for these two sounds in terms of both intensity and pressure 2.

                                                                        Q          L
                                       Intensity (in watts/cm2)       10-16       10-2
                                       Pressure (in Pascals)         2  10-5   2  102

                                     Table 1. Intensity and pressure value for Q and L
In common speech, sound pressure for low amplitude sounds is usually referred to in terms of millionths of
a Pascal or micro-Pascals (P). Thus, the 2  10-5 P above is more commonly referred to as 20 micro-
Pascals (= 20 P). Sounds with even greater pressures are measured in milli-Pascals or even directly in
Pascals. For example, 2  102 P is the same as 200 P.

The next stage to the birth of the decibel is to consider the ratio of the amplitude of the L sound to that of
the Q sound (i.e., L  Q. We need to do this for both intensity and for pressure.

                    L 10 2 watts / cm2  10 2
                      16               16  10 2  10 16  10 216  1014
                    Q 10 watts / cm2 10
                                                        Equation 13
Thus the L sound is 1014 times as intense as the Q sound (in long form, that‟s 100,000,000,000,000 or 100
trillion times as intense).

                          L 2  102 Pascals 102
                                 5
                                            5  102  105  10 25  107
                          Q 2  10 Pascals 10
                                                         Equation 2
This indicates that the L sound has 10 (10,000,000 or 10 million) times as much pressure than the Q sound.

   For you mathematical purists out there, intensity does not, strictly speaking, equal sound pressure
squared, but is proportional to pressure squared. For the sake of simplicity, I‟m ignoring that nicety.
  For anyone unfamiliar or uncomfortable with the exponential notation used in the following table, there
are some notes at the end explaining it as well as arithmetic operations involving exponentials.
  Note: the watts/cm2 divides out of the equation since it is in both the numerator and denominator of the
equation. Anything divided by itself equals 1 and thus cancels out of the equation.

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Why this exercise in arithmetic? Well, first remember that dividing one quantity by another is
mathematically equivalent to comparing the one to the other. These sound amplitude ratios therefore
compare the L sound to the Q sound in terms of their intensity and their pressure respectively. These ratios
demonstrate the wide range of sound amplitudes that the ear can process effectively - this is known as the
„dynamic range‟ of hearing. However, the numbers involved in the measurement of hearing and with the
dynamic range of hearing are very small or very large for the most part - not at all comfortable numbers to
understand or have a feel for. Who of us really comprehends what 10 -16 signifies, or 100 trillion or 20
millionths? Certainly not the engineers and physicists. Accordingly, they exercised their minds to
establishing a way of expressing these very large and small numbers in a way that is more convenient and
easier to understand. Convenience they succeeded with. You can be the judge of ease.

The method they settled on did, nonetheless, depends on comparing one sound to another.

            The first sound is the one they wish to measure. I‟ll call this the test sound.
            The second, the comparison sound is known as the reference sound.

In other words, their method of expressing sound measurement does not give an absolute measurement of
pressure or intensity. Rather it give a relative measurement, a measurement that tells how much more (or
less) sound intensity or sound pressure the test sound has relative to the reference sound. This is a very
effective way of dealing with sound amplitude (it really is) … (really and truly) … (believe me) …
(please), but which has been a curse to generations of students (me included) ever since. Let‟s see how it‟s

First of all, we already know how to compare one sound intensity (or pressure) to another sound intensity
(or pressure) - divide the intensity (or pressure) of the sound of interest (the test sound) by the reference
sound‟s intensity (or pressure). That‟s what we did in equations 1 and 2 above where L was the test sound
and Q was the reference sound.

In acoustics and audiology, the intensity of the Q sound is very commonly (but by no means always) used
as the reference in measuring the intensity of a test sound. We shall use it here. In fact, we already did in
equation 1 where we calculated the ratio L/Q = 1014. This is one of those uncomfortable numbers - a „1‟
with a string of fourteen zeros after it. But can we find a more comfortable number with 10 14 - hopefully
one for which we have more of a natural „feel‟. Of course we can - what about the 14! But can one just pull
the exponent away from the base number 10 like that. Of course we can - that‟s exactly what the
mathematical function known as a logarithm does. In other words:

                                                 log 1014 = 14
                                                    Equation 3
Some dyslexic boffin decided to name the result of this operation on sound intensity - he named it the Bel
after Alexander Graham Bell. Thus, the L sound is 14 Bels more intense than the Q sound. Notice the
last sentence carefully - its wording is critical. Often it gets shortened to “the L sound‟s intensity is 14
Bels”. However, that abbreviated form is meaningless unless the listener already knows what the reference
sound is that L is being compared to.

An important point number one to remember: THE COMPARISON OF THE TEST SOUND’S
THE Bel IS USED. This will also be true of the decibel when we reach it. Remember also that while the
Q sound frequently is used as the reference sound, it isn‟t always. The reference intensity can be and often
is the intensity of some other sound - the choice of the Q sound as reference is completely arbitrary.

Let‟s look at the general equation for the Bel, and at one other example:

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                                                Intensity in Bels  log
                                                        Equation 4
where It and Ir refer to the intensities of the test and reference sounds respectively.

As an example, the average intensity of conversational level speech at about three feet from the speaker is
about 10-10 watts/cm2. Applying the above equation, and using the Q sound intensity again as reference:

                                                                        10 10 watts / cm 2
                               Intensity of conversation  log
                                                                        10 16 watts / cm 2
                                                                 log 10 10  1016   
                                                                 log10  6

                                                                 6 Bels
                                                        Equation 5
Thus, conversation speech has, on average, an intensity of 6 Bels with reference to the Q sound, the quietest
sound the average ear can detect.

Let‟s reconsider the dynamic intensity range of the ear, this time as measured in Bels. It is, of course, 14
Bels as calculated in equation 1. Having put us through all this misery, one would think the physicists
would be happy. No way. They now decided that, whereas 10 14 was too large a number with which to
meaningfully describe the ear‟s dynamic range, 14 on the other hand was too small! So, totally arbitrarily,
they decided to multiply Bels by 10.

The result of this, and our ultimate destination, is the decibel, or dB. Consequently, the dynamic intensity
range of the ear is 140 dB relative to Q, or 10  log (IL  IQ) = 10  log (1014) = 10  14 = 140 dB.
Similarly, conversational speech is, on average, 60 dB relative to Q.

The general equation for dB is:

                                               Intensity in dB = 10  log
                                                        Equation 6

Do you remember that equation from earlier: I = p 2. I warned you it would come back to haunt this
discussion. In calculating dB for pressure (I‟m going to leave Bels for pressure as and exercise for you), we
must take this equation into account and use one additional feature of logarithms. If we let n stand for any
number we choose, then:

                                                       
                                                   log n 2  2  log n

                                                        Equation 7
Recall to the general equation for intensity in dB (equation 6). We can rewrite equation 6 substituting p2
wherever there‟s and I as follows:

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                                                                      p2 
                                           Pressure in dB  10  log  2 

                                                                     pr 
                                                                        p 
                                                              10  log  t 
                                                                        pr 
                                                                            p 
                                                              2  10  log  t 
                                                                            pr 
                                                                        p 
                                                              20  log  t 
                                                                        pr 
                                                         Equation 8
Equation 8 is the general equation for pressure in dB. Let‟s put some numbers in. Earlier, in equation 2, we
calculated that the dynamic pressure range of the ear (L  Q) is 107. If we let 2  102 Pascals (from Table
1) = pt, and if we let 2  10-5 (also from Table 1) = pr, then apply the general equation, we get:

                                                       
                                               20  log 107  20  7  140 dB

                                                         Equation 9
Finally everything has come together! Through this tortuous route, we have learned that the dynamic range
of the ear is 140 dB whether measured as intensity or as pressure. Without doing the calculation, it should
come as not surprise that the average pressure level of conversation is 60 dB re 20 P („re 20 P‟ simply
indicates the reference pressure used in the measurement).

You‟ll be pleased to hear that, for our purposes and for most of the time, sound pressure rather than sound
intensity. Hence, for us, equation 8 is more relevant than equation 6.4

Let‟s review and summarize the most crucial things so far discussed about the decibel.

1. It was introduced to transform the very small and very large values that arise from the measurement of
   sound into more manageable numbers.
2. The decibel is NOT a unit of sound pressure or sound intensity. A sound of interest is always compared
   mathematically to an arbitrary reference sound by dividing its pressure (or intensity) by the pressure (or
   intensity) of the reference sound.
3. A frequently used reference sound is the Q sound, the lowest pressure (or intensity) sound that the
   average human ear can hear. Remember, however, that other reference pressures are also used for
   specific purposes.
4. First, in order for the decibel to have any meaning, it is essential that the reference pressure (or intensity)
   be explicitly stated. Beware of any articles, whether scholarly or popular, that fail to mention the
   reference pressure when reporting decibel measurements.

With regard to point 4, there are a number of conventions used to declare the reference pressure (or
intensity). For example, you will very often encounter „dB SPL‟. This stands for „decibels Sound Pressure
Level‟. This is a shorthand way of saying that the reference sound is the Q sound and hence the reference
pressure is 20 P. (For sound intensity, the equivalent is „dB IL‟ or „dB Intensity Level‟ where 10-16
watts/cm2 is the reference intensity.)

  If you decide to take the instrumentation module next year, be warned - equation 6 will again surface if
only momentarily.

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Another very common shorthand method of declaring the reference pressure is „dBA‟ or „dBC‟.5 The „A‟
and „C‟ refer to settings on a sound level meter used for measuring sound. Since sound level meters are
normally calibrated to use the Q sound as reference, dBA and dBC can be considered shorthand for dBA
SPL and dBC SPL respectively. Of the two, we are most likely to encounter and/or use the dBA setting.

„dB HL‟ and possible „dB SL‟ are other declaration types you will encounter in studying audiology. We
won‟t get into these here (they‟ll rear their heads in audiology) other than to issue the warning that for HL
and SL, the Q sound is not the reference. (HL stands for „Hearing Level‟ and SL for Sensitivity Level.)

Otherwise, when the Q sound is not the reference, and when dB HL and dB SL are not being used, the
reference should be written out. This is usually done in the format I used earlier, viz., dB re n P (or dB re
n watts/cm2 for intensity).

There are now one or two burning questions you are asking yourselves. Aren‟t there? So let‟s conclude by
briefly answering them. First, what if the pressure of the test sound is less than that of the reference sound?.
For example, consider a sound of 2 P (most of us wouldn‟t be able to hear it, but it is a sound nonetheless).
First, let‟s compare it to the Q sound which has more pressure:

                                         p t 2  10 6 P
                                                   5
                                                           10 6  105  10 1
                                         p r 2  10 P
                                                      Equation 10
Converting this ratio to dB:

                                      20  log (10-1) = 20  (-1) = -20 dB SPL
                                                      Equation 11
So if a sound has less pressure (or intensity) than the reference sound, its measurement in dB will be a
negative number. Or to do the obvious and turn the statement around, negative decibels signify a sound
pressure (or intensity) that is less than that of the reference sound.

Next question: what does 0 dB mean? Well, if positive numbers mean that pt > pr, and if negative numbers
mean that pt < pr, then intuitively, one might conclude that 0 dB means that p t = pr. Why? If pt = pr, then
their ratio must = 1 and log (1) = 0 and 20  0 = 0 dB irrespective of the reference pressure.

Question no. 3 arises from the fact that so far, all the sounds discussed have had pressure ratios that were
obvious powers of 10. Very neat, but also very unrealistic in practical terms. What, for example, if a sound
has twice the pressure of the reference. As a concrete example, take a sound that is 40 P. The ratio with
the Q sound by definition must be 2, but let‟s work it out anyway:

                           p t 4  10 5 P
                                     5
                                             2  10 5  105  2  100  2  1  2
                           p r 2  10 P
                                                      Equation 12
and the pressure in decibels:

                                                     20  log (2)

  Yes, there is also a „dBB‟ and even a „dBD‟, but you are less likely to encounter these so I‟ll ignore them

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                                                 Equation 13
Uh oh - previously, we‟ve just been able to casually pull off the exponent from 10 (recall that log (10 n) = n).
Here there ain‟t no exponent of 10 to pull off - or is there? Is there a number, call it n, such that 10n = 2?
Of course there is - it just happens to be 0.301029995664… . In other words, 100.301029995664… = 2. So let‟s
redo equation 13 as:

                  20  log ( 100.301029995664…)  20  log (100.3) = 20  .3 = 6 dB SPL
                                                 Equation 14
I‟ll leave the arithmetic for you, dear readers, but can you see that if you double the sound pressure of any
sound, this will be the same as increasing it by 6 dB. For example, if a sound has a pressure of 37 dB SPL,
and if you double its pressure, it will then have a pressure of 37 + 6 = 43 dB SPL. One the other hand, if we
were to halve the pressure, this equates to reducing its pressure by 6 dB, i.e., by subtracting 6 dB.

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Mindful that some readers may be completely new to exponentiation, let‟s start at the beginning. Exponents
are a shorthand method of indications repeated multiplication of a number by itself. So 10  10 is
multiplying 10 two times, and is signified by 102 (aka „10 squared‟ and „10 to the 2nd power‟).

                                                10  10 = 102 = 100
                                                     Equation 15

                        10  10  10  10 = 104 = 10,000 (or 10 to the 4th power)
                                                     Equation 16

                  2  2  2  2  2  2  2  2  2 = 29 = 512 (or 2 to the 9th power)
                                                     Equation 17
Since the discussion of decibels mainly uses powers of 10, I‟ll stick to these. Exponents really come into
their element with very large (or, as we shall see, with very small numbers).                       Thus,
100,000,000,000,000,000,000 is more easily representated as 1020 (10 to the 20th power).

A brief table of exponentiated 10‟s follows:

           100 = 1
           101 = 10
           102 = 100
           103 = 1,000
           104 = 10,000
           105 = 100,000
           106 = 1,000,000

How would 7,000 be represented exponentially? We could use 10 3.84509804. More usually, however, it is
written in the form 7  103 (i.e., 7  1000). Similarly, 93,000,000 (the distance of the earth from the sun in
miles) would be 93  106 (93  1,000,000). It could also be written as 930  105 (930  100,000) or 9.3 
107 (9.3  10,000,000) - all three forms equal 93,000,000.

Multiplication of exponential numbers
The multiplication of exponentiated numbers is quite simple. Take the example 1,000  1,000,000 =
1,000,000,000 or 109. Let‟s look at it:

                                               103  106 = 103+6 = 109
                                                     Equation 18

 As a side-track, do you know what a „googol‟ is? Seriously - look it up in a dictionary! It‟s 10100. Written
out, a googol is a 1 followed by 100 zeros. Even larger is the „googolplex‟ or 10 googol. Written out, it would
be 1 with a googol of zeros following. I‟ve seen a googolplex written out - it takes several hundred pages of
small print.

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In other words, to multiply 10n by 10m, you simply add the exponents m + n. The general rule for the
multiplication of exponential number is:

                                                      10n  10m = 10n+m
                                                           Equation 19
A slightly more complicated example would be to multiply 4,700  300. This is done as follows:

                                                                         
                       4700  300  4.7  10 3  3  10 2  4.7  3  10 3  10 2       
                                          14.1  10 32   14.1  105
                                          1,410,000
                                                           Equation 20
Easy isn‟t it? If in doubt, try working it out longhand - that should convince you. By the way, the brackets
used above aren‟t really necessary - they can be omitted.

Division of exponential numbers
Division is no less simple. For example, divide 10,000,000 by 1,000. The quotient is obviously 10,000 or
104. Let‟s look at it exponentially:

                                    10 7
                                          10 7  10 3  10  7  3  10 4  10,000
                                                           Equation 21
So, in division, take the denominator (i.e., the divisor), change the sign of its exponent, then multiply it
times the numerator (dividend) using the addition-of-exponents rule from above. In general form, the
division rule is:

                                        10 n
                                              10 n  10  m  10 n   m  10 n  m
                                                           Equation 22
Try the following division using exponents yourself:

                                                                 2,400

One further example is interesting and important - what is 100  200? The answer is obviously 1, but let‟s
look at the exponential division:

                                               10 2
                                                     10 2  10 2  10 2  2  10 0
                                                           Equation 23
So, 10 = 1. Similarly, any number raised to the zero power = 1. Thus, 5 0 = 4,7650 = 1,000,0000 = n0 = 1.

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                      THE LIFE AND TIMES OF THE DECIBEL
Negative exponents
In the discussion of the decibel, there are many examples of negative exponents, e.g., 10 -16 watts/cm2. What
does this number mean? A little thought about the division rule should indicate the truth. Let‟s work
backwards - take 10-2. This can be rewritten as follows:

                                 10 2  1  10 2 since any number multiplied by 1 equals itself
                                                                              Equation 24
By applying the division rule in reverse:

                                                                                           10 0    1
                                                            1  10 2  10 0  10 2         2
                                                                                                      0.01
                                                                                           10     100
                                                                              Equation 25
Note the change of the sign of 10 when it moves to the denominator. It can be seen, then, that a number
with a negative exponent is merely a fraction less than one. Another brief table may be of help:

           10-1 = 0.1
           10-2 = 0.01
           10-3 = 0.001
           10-4 = 0.0001
           10-5 = 0.00001

Frequency and exponential numbers
Exponentiation also typically finds a role in graphing frequency. The frequency range that the human ear
can hear is approximately from 20Hz to 20,000Hz. This is called the frequency bandwith of hearing. As
we saw with the ear‟s dynamic range, the frequency bandwidth is also very broad. You will frequently
encounter graphs that plot sound frequency against sound amplitude (e.g., audiologists use the „audiogram‟
which is a graph of a person‟s hearing sensitivity). Commonly such charts will use a logarithmic axis for
frequency. Below is a typical chart (since it‟s the chart itself that‟ of interest for the moment, there‟s no
data plotted on it):

                   Relative amplitude (dB SPL)

                                                       10            100        1,000          10,000    100,000

                                                                           Frequency (Hz)

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Note the labels on the horizontal axis showing frequency - the numbers progress exponentially as 10, 100,
1,000, 10,000, 100,000 or 101, 102, 103, 104, 105. This use of a logarithmic scale for the axis allows a very
wide bandwidth to be displayed on a relatively small graph. (Imagine how wide this page would have to be
if the axis labels increased by a contant 100Hz - there would have to be 1,000 ticks on the x axis).

One other point about the graph - why do you think the vertical axis is labelled „Relative amplitude‟? Hint:
if you don‟t know, think about the meaning of the decibel again.

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         Force: a mass of approximately 100g can produce a force of 1 Newton (N).

         Pressure: when a force acts (pushes) against a surface or area (A), the result is pressure.
          Mathematically, this is represented as:

          For a force of 1N acting against a surface of 1m2 will produce a pressure of 1 Pascal (Pa) or:

                                   1N                         N
                                         1Pa or, more simply, 2  Pa
                                   1m 2                       m

         Work: to move a mass of 10.2kg a distance of 1cm requires 1 joule (J) of work, or:

                                               1J  10.2kgcm
         Power: work (w) done in a given time (t) is power (P), or:

          1J applied for 1 second produces 1 watt (W) of power, or:

                                               1J                        J
                                    1W            or, more simply, W 
                                               sec                      sec
         Intensity: If power is applied to a surface, the result is intensity (I), or:

          1W applied to a surface of area 1cm2 produces an intensity of 1W/cm2.

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