Introduction to Logarithmic Scales

The purpose of this series of web pages is to introduce you to the concept of

logarithmic scales, and in particular to the decibel scales commonly used in

acoustics to measure loudness. There are four web documents (including this one)

that cover the necessary topics. The second document is a mathematical review of

logarithms. Even if you are confident in your mathematical skills with logs, it is

worth trying some of the worked examples just to verify you know how to use your

calculator! After going through all of the documents you should take the

associated online quiz. It is your responsibility to build your understanding to the

appropriate level to attempt the quiz. I am available via e-mail, after class, and in

office hours to answer your questions. Start reviewing the material early so that

you will have sufficient time to get your questions answered. Good Luck.

Logarithmic versus Linear Scales

The purpose of the following paragraphs is to explain the difference between a

linear and a logarithmic scale and to indicate conceptually why we use logarithmic

scales as the measurement method in stimuli/sensation situations such as

quantifying loudness.

First, we must clearly understand the idea of a linear scale. Our experience of the

everyday world makes us familiar with linear scales. A simple example is a ruler

used to quantify distance. Each centimeter step on the ruler is equally spaced, that

is, the distance between 0 cm and 1 cm is the same as the distance between 11 cm

and 12 cm. This concept may seem obvious and trivial; however, contrast this

with the following example of the logarithmic nature of the the frequency

separation of the octaves on a piano keyboard.

Like the centimeter spacings on a ruler, the keys on a piano are equally spaced

apart; thus, playing two notes an octave apart requires the same spread of the hand

at the bass end of the piano as at the treble end. Now consider the frequency

difference between the lowest and highest octaves on a piano. We know that to

raise a note an octave we must double its frequency. Note that this is a

multiplicative process rather than an additive process. What difference does this

make? Consider the frequencies of the successive octaves of the lowest note on a

piano. On my piano at home the lowest note is A. This A is four octaves below

the A above middle C. Some quick math should convince you that if A above

middle C has a frequency of 440 Hz then the lowest A has a frequency of 27.5 Hz

(divide 440 by 2 four times). The octaves of the notes A thus have frequencies

given by the sequence 27.5, 55, 110, 220, 440, 880, 1760, 3520 Hz. Note that at

the bass end the frequency difference between the notes of an octave is 55-

27.5=27.5 Hz. In contrast at the highest octave the frequency difference is 3520-

1760=1760 Hz. Even though we would perceive each interval as an octave the

frequency difference between the notes in the two cases is vastly different because

of the multiplicative relation between the notes in an octave. We will see that not

only are the octaves related by a multiplicative factor but also the individual

semitones that make up a scale, but that's the subject of later lectures.

The key point to understand from the previous example is that the difference

between a logarithmic and a linear scale rests on whether the natural steps increase

in an additive fashion (linear scale) or in a multiplicative fashion (logarithmic

scale). As a quick self-test, identify which of the following sequences are linear

and which are logarithmic. The answers are given at the bottom of the page.

Self-Test 0

1, 2.5, 4, 5.5, 7, 8.5, 10...

3, 9, 27, 81, 243...

10, 100, 1000, 10000 ...

50, 550, 1050, 1550, 2050...

Logarithmic Nature of Loudness

You may well be wondering why we are spending so much time developing an

understanding of logarithmic scales. Well, it so happens that our senses respond in

a fashion that is logarithmic (at least to a pretty good approximation). We are

concerned here primarily with the sense of hearing; however, a very similar

relation holds for vision. Sensation is the reaction of our ear/brain to an incoming

stimulus (e.g. a sound). Now it is possible to precisely measure stimuli because (in

the case of sound) the size of the stimulus is related to the pressure amplitude of

the sound wave. A microphone can translate the pressure variations into an

electrical signal that can be digitized and measured precisely. In contrast, the

sensation produced by a sound is a subjective quantity. Sensation can only be

determined by asking a human test subject questions such as "Which of these

sounds is louder?". These experiments are referred to as being in the realm of

psychophysics, i.e. the combination of psychological testing methods with physics.

One of the early workers in the field who explored the psychophysical relation

between sensation and stimulus was named Fechner. He documented the

logarithmic nature of the senses in a book The Elements of Psychophysics

published in 1860. The relation between stimulus and sensation--often called

Fechner's law--states that "As stimuli are increased by multiplication, sensation

increases by addition." For example, if a series of sounds are played with

quantitatively measured stimuli that increase logarithmically in the ratio 1, 2, 4, 8,

16 our ear/brain combination would measure a loudness increase that went up in

equal steps. In other words the last sound would be perceived as roughly 5 times

as loud as the first sound rather than 16 times as loud as would be implied from the

quantitative measurement. I should stress here that the multiplicative factor

between steps is not necessarily 2, I just chose the number for this example. The

important point is that our perception of loudness is logarithmic not linear.

You may wonder why we would evolve with a logarithmic response to loudness.

One reason is related to the fact that our ability to hear spans an enormous range of

pressure amplitudes—a very large dynamic range. A logarithmic response helps to

compress this range so that our response to variations in weak sounds is similar to

the response to variations in loud sounds. We will explore this compression in

some of the following modules. First, we will review the mathematics of

logarithms in the next section.

Answers to the Self-Test 0

Linear, addition of 1.5 with each step.

Logarithmic, multiplication by 3 with each step.

Logarithmic, multiplication by 10 with each step.

Linear, addition of 500 with each step.

Scientific Notation, Powers of Ten, and Logarithms

This dapper dude is John Napier the "inventor" of logarithms. If you're

curious you can learn more about him from the web. He was a Scot, like my

distant ancestors (my nostrils flare when I hear bagpipe music!). The following

section explains the concept of expressing numbers in scientific notation. If you

understand scientific notation then it's not too big a stretch to comprehend how

logarithms work. There are a number of self-test problems on this page. Be sure

to attempt these problems before looking at the answers! The answers are listed at

the bottom of the page. If these problems give you no trouble, then the quiz should

be straightforward. On the other hand if you can't figure out how to do the

problems it's time to find help.

Scientific Notation

To more compactly express and manipulate very large or very small numbers

it is convenient to use the concept of scientific notation. We did some of this in the

first classes of the semester. Using this notation a number such as 95,000,000 can

be expressed as 9.5x107 or the number 0.00006 can be written 6x10

-5. Writing

numbers in this fashion is based on the use of powers of ten to replace a large

number of zeros either before or after the decimal point. The powers of ten just

tells you how many times the number was multiplied or divided by 10. In the first

example above 95 million is written as 9.5 multiplied by 10 seven successive

times. Each multiplication by 10 moves the decimal point along one place so it

should be easy to convince yourself that the two representations are equivalent.

The case of 6x10-5

implies that the number 6 is divided by 10, five successive

times. The number to which 10 is raised to is called the exponent (the little uppy

number to the right of the 10). Thus, the exponent in the first example is 7 and in

the second example -5. Remember a negative exponent means divide by 10. This

concept is pretty straightforward; however, in practice folks tend to get confused

over the number of zeros. If you keep in mind the concept of the exponent as the

number of times you multiply or divide the original number by 10 you should not

go wrong. Try these self-test examples.

Self Test 1

Convert the following numbers into scientific notation:

5670000

0.0000546

35

Self Test 2

Convert the following numbers from scientific to regular notation:

3.14 x 107

9.765x10-3

6x1024

(This example makes you appreciate the usefulness of scientific

notation!)

One last item to note is that while 101 means multiply by 10, and 10

-1 means

divide by 10, 100 means multiply by 1! Actually any number raised to the power 0

equals 1.

Multiplying Numbers in Scientific Notation

How is scientific notation used in operations such as division and

multiplication? Let's examine this simple example: 100 x 6000. The answer by

straightforward multiplication is 600,000. Now consider the same problem stated

in scientific notation: 1 x 102 x 6 x 10

3 = 6 x 10

5. Note that although we multiply

the numbers in the normal way (1x6=6), in order to get the appropriate power of 10

we add the exponents (2+3=5). For the process of division we perform in a similar

manner only using subtraction instead of addition. Again a simple example: 4 x

104 / 2 x 10

2 = 2 x 10

2 . Again we divide the regular numbers 4/2 = 2. For the

exponents we subtract the denominator's exponent from that of the numerator 4-

2=2. Of course for the division case you must get the order correct.

Again try these simple examples to make sure that you understand and believe

what I've just described.

Self Test 3

103 x 10

7

2 x 102 x 5 x 10

-2

106 / 10

3

102 x 10

2 / 10

5

Exponents and Logarithms

We have seen that when two numbers are multiplied we just add the

corresponding exponents. You should also be getting a vague sense of how this

relation is connected to the logarithmic nature of hearing described in the

introductory module in that a multiplicative relation becomes an additive one.

Now we want to introduce the logarithm. The concept of a logarithm is to merely

replace a number by the exponent to which 10 would have to be raised to get that

number. For example, consider the number 100. To what power would 10 have to

be raised to get 100? I hope you cried out 2, since 102 = 100. Thus, the logarithm

of 100 is 2. In mathematical terms we would write this as log(100) = 2. Quick

now! What is the logarithm of 1000? 10? 0.1? If you answered 3, 1, -1 award

yourself a major prize.

Now we come to the big leap in understanding. What is the logarithm of a

number like 57? This number cannot be expressed in a nice easy format of 10 to

some integer exponent. Nevertheless there is still some number that when used as

the exponent will give 57. In the old days to find this number required the use of

tables of logarithms. Nowadays we use our trusty calculators. Get your calculator

and try these numbers to see for yourself. Not only will this reinforce your

understanding—it will also make sure you know where all the appropriate buttons

are on your calculator! Type in 57 and then find the "log" button on your

calculator and press it. You should be returned the value 1.75587... This number is

the logarithm of 57. If you think about it this value makes sense because we know

that 57 lies between 10 and 100 and the logs of these two numbers are 1 and 2

respectively.

Finally, if we have the log of a number how do we recover the number itself?

Let’s continue with the example of 57. We now know that 1.75587 is the log of

57. To get from the log to the original number we must use the log value as the

exponent of 10. Some calculators have a 10x button. In this case enter 1.75587

and hit the 10x button. Voila! You should get 57 (or something pretty close

depending on how many significant digits you entered). Unfortunately not all

calculators have the 10x button. Some require that you enter 1.75587 and then hit

INV and then the LOG buttons. Make sure you figure out how to use your

calculator to take the log of a number and to get from the log value back to the

number. You can ask me, I've dealt with every manner of calculator over the

semesters and I love a challenge. Try the following self-test to verify your

calculator savvy.

Self Test 4

Find the logs of 50, 500, 5000, 50000. Do you see a pattern?

Find the number whose log is 3.30059548389 (i.e. find the antilog of this value).

Find the log of 81. Divide this log value by 2 and then find the antilog of this

value. How are the initial and final numbers related?

Try finding the logs of 1, 0, and -1.

The Rules of Logarithms

Now that you understand what a logarithm means I want to show you a few

simple mathematical manipulations that can be done using logarithms. These are

often called the Rules of Logarithms; however, they should not be mysterious to

you given what we have covered in the previous sections. The first rule is typically

expressed as

log(XY) = log X + log Y

In words this rule states that if I take two numbers X and Y and multiply them

together and then take the logarithm I obtain the same result as if I had added the

logarithms of the two numbers separately. Consider the simple example of 102

times 103. Multiplied together these come to 10

5 and the log of 10

5 is 5 (that is

what the left hand side of the equation tells us to do). Now the log of 102 is 2 and

the right hand side of the equation instructs us to add this to the log of 103 which is

3 to get 5. The rule works! Of course this relation is nothing more than putting

together the two ideas (1) that the log of a number is the exponent of 10 required to

equal that number and (2) when we multiply powers of 10 we add the exponents.

The second rule of logarithms is a straightforward extension of the first--it

states that

log (Xn) = n log X

that is, if the number X is raised to the power n the result is the same as n times log

of X. To understand this rule imagine that n=2. In that case

log (X2) = log (XX) = log (X) + log (X) = 2 log X

where I merely wrote out X2 as XX in the first step and then used the first rule

above. The extension to higher n should be obvious—try writing it out for n=3.

Finally there is a rule for division as well as for multiplication. This rule

states that

log(X/Y) = log X - log Y.

Using these rules one can do some nifty manipulation to reduce the need for

calculation. I know that in these days of calculators this may not seem a big deal

but it's still important in understanding the manipulation of logarithms. For

example, in the previous sections self-test I asked you to find the logs of a series

starting 50, 500 etc. You should have seen the pattern that resulted and now using

Rule 1 you can understand why this pattern came about. The log of 50 is 1.69897.

If I want the log of 500 and I don't have access to a calculator I can use rule 1 by

recognizing that 500 = 50 x 10. Thus,

log(500) = log(50x10) = log(50) + log(10) = log(50) + 1 = 2.69897

making use of the fact that we know that the log of 10 is 1. The following

self-test--the last in this module--will allow you to practice the use of some of the

rules of logarithms. There are hints located a few steps below the questions. No

calculators for this section!!!

Self Test 5

Given that log(2)=0.301 and log(4)=0.602 find log(8).

Using the information in the previous question find log(16).

Find the log of the square root of 2.

Find the log of 400.

Hints to the last self-test

Use Rule 1 with X=2 and Y=4.

Remember 42 =16.

Remember square root of 2 squared is equal to 2. Look to Rule 2!

Look at the example at the end of the section.

Answers to self-tests

Self-test 1

5.67 x 106

5.46 x 10-5

3.5 x 10

Self-test 2

31400000

0.009765

600000000000000000000000

Self-test 3

1010

10 x 100 = 10 (Remember 100 = 1)

103

10-1

Self-test 4

1.69897, 2.69897, 3.69897, 4.69897

1998

1.90848, 0.954243, 9. The last number 9 is the square root of 81.

0, ERROR, ERROR. Zero and negative numbers don't have valid logarithms!

Self-test 5

log(8) = log(4x2) = log(4) + log(2) = 0.301 + 0.602 = 0.903

log(16) = log (42) =2 log(4) = 1.204

log(root(2)2) = 2 log (root(2)) = log 2. Thus log (root(2)) = log(2) /2 =0.1505.

log(400) = log(4x100) = log(4) + log (100) = log(4) + 2 =2.602.

Pressure Amplitude: Quantitative Measurement of Sound

As we will discuss in this section decibels are a means of creating a logarithmic

scale relative to some reference. Decibel scales are by no means confined to

acoustics, there are decibel scales defined for use in electronics and optics. Even in

acoustics there are a variety of different definitions of the decibel scale depending

upon the quantity being used as a reference. Thus, to begin our discussion I want

to cover the question of what parameters are measured in determining sound levels.

The most physically understandable quantity used in determining the size of a

sound signal is the Pressure Amplitude. Pressure amplitude is a measure of the

size of the variation in air pressure caused by a sound wave. In pure silence there

is a constant pressure--atmospheric pressure. Atmospheric pressure is measured in

Newtons/meter2 and is approximately 10

5 N/m

2. The atmospheric pressure varies a

small amount over the space of hours or days--this is what they mean by high and

low pressure centers on the weather map--however, we can consider it to be

essentially constant on the time scale of sound waves.

A sound wave creates a variation in the normally constant air pressure that

oscillates above and below the normal atmospheric level of 105 N/m

2. The average

size of the pressure variation away from the constant background level is the

Pressure Amplitude of the sound wave. I am hiding a bit of complication in the

word average; the pressure goes above (positive) and below (negative) the constant

atmospheric background so a straight average would tend to give zero. The real

average is a root mean square average which you might be familiar with if you

have done some electronics. For a pure tone--a pure sine wave--the averaging

process leads to a pressure amplitude that is reduced from the maximum amplitude

by a factor of 0.707 (equivalent to dividing by the square root of 2). The relation

between amplitude of the sine wave (measured from the background level to one

extreme) to the pressure amplitude (average pressure excursion from the

background) is shown in the figure below.

It is fairly simple to understand how a calibrated measurement of the pressure

amplitude can be made using a microphone to convert the pressure variations into an

electrical signal. By applying known pressure variations to the microphone the

electrical signal can be calibrated to directly measure the air pressure variations. With

suitable processing this pressure variation can be converted into the pressure

amplitude. This function is performed by Sound Pressure Level (SPL) meters.

Our next concern is to address the question: how large is the pressure amplitude for

typical sounds? To answer we begin by considering the pressure amplitude for the

weakest sound that is audible to the average person. This value is referred to as the

threshold of hearing (for obvious reasons). The pressure amplitude for the threshold

of hearing is 2 x 10-5 N/m2. This is a standard value defined for a pure sine wave at a

frequency of 1000 Hz. As we will see later in this class, our ability to hear is a strong

function of frequency so the value is quite different at other frequencies. Of course

the ability to hear also varies from person to person so this is an average for the

human population.

It is instructive to compare the threshold of hearing value to the background air

pressure. Remember the pressure amplitude is the average size of the pressure

variation on the constant atmospheric background. By taking the ratio of the

threshold of hearing pressure amplitude to atmospheric pressure we see that the

weakest audible sound waves only create a variation in the background pressure of

0.00000003%. Pretty small. Now how about very loud sounds. The threshold of

pain corresponds roughly to a pressure amplitude of 30 N/m2 (a million times larger

than the threshold of hearing). This value translates to pressure variations of 0.03%

about the atmospheric background value. Even loud sounds do not vary the

background air pressure by even 1%!

Sound Pressure Level: A Decibel Scale defined in Terms of

Pressure Amplitude

Now that we understand pressure amplitude as one measure of the size of a sound

signal lets look at the decibel scale defined for this quantity. The decibel scale for

pressure amplitude is called Sound Pressure Level, typically abbreviated SPL, and

designated in our text and in this course by the symbol Lp. The decibel SPL value for

a sound with pressure amplitude P is given by the relation

Note that there is one quantity in the equation that I have not defined yet namely

P0. Remember that I said at the beginning of this page that decibels are a relative

measurement scale--in this case the sound pressure amplitude being measured P is

determined relative to the value P0. P0 is an agreed upon value of 2 x 10-5 N/m2. You

might recognize this value as the pressure amplitude for the threshold of hearing at

1000 Hz mentioned a couple of paragraphs above. This choice of reference values is

of course intentional.

Lets look at two examples of how to move from a pressure amplitude to a SPL decibel

reading and from a decibel value back to the pressure amplitude P.

Example 1

Find the decibel reading corresponding to a pressure amplitude P = 0.2 N/m2.

Using the above definition for Lp we begin with the term in brackets on the right hand

side of the equation by dividing P by P0.

P/P0 = 0.2/2x10-5 = 10000

Next we take the log of this value

log(10000) = 4

and multiply by 20 to get Lp

Lp=20 x 4 = 80 dB.

Note that we write dB standing for decibels after the final answer. Decibels are not

actually a measurement unit because the decibel value depends on the agreed upon

reference. Note that in the first step of the calculation we divide one pressure

amplitude value by another eliminating all units!

Example 2

Find the pressure amplitude corresponding to a decibel reading of 35 dB.

In this case we know the left hand side of the equation ands as always we know the

value of P0=2 x 10-5, thus we can write the equation

To get P we first divide both sides by 20 to get

Now the step that often stumps folks whose algebra is rusty--we want to remove the

log from the left hand side. From our review of logarithms we saw that the log of a

number is the exponent that 10 would have to be raised to to get that number. Thus,

10 raised to the power 1.75 is P/2x10-5! Think this step through to make sure it's

clear. The result is that

For the last step we race to our calculators, put in 1.75, hit the 10x button, and then

multiply by 2x10-5 to get the final answer 1.12x10-3 N/m2. Note that our answer has

the units of pressure N/m2.

Self Test #6

Try the following questions that test your ability to work with the SPL decibel

equation. These questions also illustrate some important properties of decibel scales

so make sure you work and understand the problems.

1. Find the pressure amplitude that corresponds to the threshold of pain decibel

reading of 120 dB.

2. Calculate the decibel reading for a pressure amplitude of P = 7 x 10-3 N/m2.

3. Calculate the decibel reading for a sound at the threshold of hearing, i.e. P = 2 x

10-5 N/m2.

4. Calculate the decibel reading for a sound with P = 1.5 x 10-5 N/m2. This sound

pressure amplitude is below the threshold of hearing.

5. Calculate the pressure amplitudes for SPL readings of 26 dB and 20 dB. How

are the two pressure amplitudes related?

6. Calculate the pressure amplitudes for SPL readings of 56 dB and 50 dB. Again

how are the two pressure amplitudes related? Notice a pattern?

Intensity and its Relation to Pressure Amplitude

Although pressure amplitude is one measure of the presence of a sound wave it is not

the only one, nor is it always the most convenient parameter to use. A second

important quantity in characterizing sound waves is called Intensity. Intensity is a

measure of the sound energy that passes through a given area each second. Now

energy per second is measured in Watts (a 60 Watt light bulb gives out 60 Joules of

energy each second) thus intensity has units of Watts/m2. In class we should already

have talked about (or we will cover soon) the fact that waves carry energy and how,

from a point source, the energy is spread over an increasingly larger area as the wave

spreads out. From this discussion we concluded that in the free field limit the

intensity of a wave drops off as the inverse square of the distance from the source.

The point that I want to make here is that the intensity is related to the pressure

amplitude. In particular the energy in a wave is proportional to the square of the

pressure amplitude. In other words if I double the amplitude of the wave I quadruple

the energy carried by that wave. You will see how this relation becomes important in

the coming comparison of the decibel scale defined with intensity to the SPL scale we

have already examined.

Sound Intensity Level

The decibel scale defined in terms of the sound wave intensity is given by the relation

where I is the intensity of the sound wave and I0 is the reference value. The form of

the equation looks similar to that for SPL that we defined above except that the

multiplier in front of the log is 10 instead of 20. The reference value I0 is the

threshold of hearing intensity at 1000 Hz and is 10-12 W/m2. The intensity decibel

scale is called the Sound Intensity Level abbreviated SIL and, in common with our

text, the corresponding symbol is LI.

Because the SPL and SIL scales are both referenced to the threshold of hearing value

the decibel reading for the two scales can usually be assumed to be identical in most

simple situations. In practice there are some subtleties because the pressure amplitude

at a point is the result of waves coming from all directions whereas the intensity

implies that we have to define a surface and a direction through which the energy

flows. We will not concern ourselves with these points but it's good to be aware that

they exist.

We can understand the origin of the factor of 2 difference in the multipliers for the

definitions of SPL and SIL using the square relation between pressure amplitude and

intensity and our rules for logarithms. Lets begin with the SPL definition but lets

write 20 = 10 x 2.

In the second step we took the 2 inside the logarithm and used it to raise the pressure

amplitude ratio to the power 2. Look at Rule of Logarithm number 2. Finally,

because the intensity is related to the square of the pressure amplitude we replace the

pressure amplitude ratio squared with the intensity ratio.

We will see in the next unit why it is sometimes easier to work with intensity rather

than pressure amplitude; however, the mathematics of manipulating the two equations

is very similar because of the parallel forms of the equations. Test yourself with these

quick self test questions.

Self-Test #7

1. What is the intensity of a 90 dB sound?

2. What is the intensity of a -3 dB sound?

3. What is the intensity of a 0 dB sound?

4. What is the SIL of a sound with intensity of 5 x 10-6 W/m2?

5. What is the SIL of a sound with 0.5 W/m2?

Answers to the Self-Test #6

1. 20 N/m2.

2. 50.9 dB.

3. 0 dB.

4. -2.5 dB

5. 4 x 10-4 N/m2 and 2 x 10-4 N/m2. One is double the other.

6. 0.0126 N/m2 and 0.0063 N/m2. Again a factor of 2 between the SPL values

that are 6 dB apart.

Answers to Self-Test #7

1. 1x10-3 W/m2.

2. 0.5 x 10-12 W/m2.

3. 10-12 W/m2.

4. 67 dB.

5. 117 dB.

Combining Decibel Quantities

We have defined the SPL and SIL decibel scales, now I want to examine the

mathematics of combining decibel readings. In particular we will see that it is not

possible to merely add decibel quantities and get anything sensible. For example,

if I have two uncorrelated (we'll define this) sources each creating sounds of 30 dB

then when played together they emit 33 dB! What kind of crazy math is that you

say? Well it all goes back to the fact that we are dealing with a multiplicative

rather than additive scale when we use logarithmic quantities such as decibels. The

central notion of this section is that to combine two or more decibel readings

requires converting each value back to the corresponding intensity or pressure

amplitude, adding these values appropriately, and finally converting the sum back

into a decibel quantity. The tricky part comes from the fact that there is a different

procedure for combining intensity and pressure amplitude values. Read on.

Combing SIL Readings

We will start with the procedure for combining intensity values because it is the

most straightforward. To combine two SIL readings requires that I follow the

following simple prescription: (1) Convert the two SIL values into the

corresponding intensities I1 and I2, (2) Add the two intensities I1 + I2 = Itotal, (3)

convert Itotal back into a decibel reading.

This procedure assumes that the two sources of sound that give rise to the SIL

values are uncorrelated. We will define correlated and uncorrelated sources a bit

more carefully below, however, for now you can take uncorrelated to mean that the

sources create sound independently of each other e.g. two radios tuned to different

stations. Let's look at an example and then you can try a few self test questions.

Example 1

Two electric guitars create SIL values of 45 dB and 50 dB respectively. What dB

reading would you expect when both are played simultaneously.

Following the above procedure:

(1) 45 dB corresponds to an intensity I1 = 3.16 x 10-8

W/m2, 50 dB corresponds to

an intensity of I2 = 1 x 10-7

W/m2. If you can't figure out how I got these values

you need to review the previous web page!

(2) Add the two intensity values to get the total intensity Itotal = I1 + I2 = (3.16 +

10) x 10-8

= 13.16 x 10-8

W/m2.

(3) Convert the total intensity back into a dB reading. You should be good at this

by now. The answer is 51.2 dB.

Combining the SIL values by adding the corresponding intensities is fairly

simple. Try these self test exercises.

Self Test 8

1. Find the combined dB reading for two uncorrelated sources each with with

SIL of 100 dB.

2. Find the combined dB reading for two uncorrelated sources with SIL values

of 0 dB and 3 dB.

3. Find the combined dB reading for two uncorrelated sources with SIL values

of 0 dB and -3 dB.

4. Find the combined dB reading for 10 sources each with SIL of 40 dB.

Combining SPL Readings

The process of combining SPL values is a little more complex. I will outline the

process in step-wise fashion as I did for the SIL section. As you will see the step

of combining pressure amplitudes is not as simple as the addition we used for the

intensities. I will discuss why there is this difference and on the way we will learn

more about the concept of correlated and uncorrelated sources.

First the procedure for combining SPL values: (1) Convert the two SPL values into

the corresponding pressure amplitudes P1 and P2, (2) Combine the two pressure

amplitudes according to P12 + P2

2 = P

2total, (3) convert Ptotal back into a decibel

reading.

The only tricky part is step (2) in which we add the squares of the pressure

amplitude values to obtain the square of the total pressure amplitude. Don't forget

to take the square root of the total pressure amplitude obtained in step (2) before

using it in step (3). You should be asking yourself why do we need this procedure

to combine the pressure amplitudes? One mathematical way of looking at the

process is to reason that because the intensity is proportional to the square of the

pressure amplitude and because we add intensities we must therefore add the

squares of the pressure amplitudes. Okay, that's one line of reasoning but it does

not give an understanding of the physical reason underlying this combination

process. I want to look at the subject a little more closely.

To understand the physical reason for the averaging process we must examine the

process of combining wave amplitudes. In the examples below I assume that we

are combining two equal amplitude pure sine waves of the same frequency. Of

course the waveforms associated with real sounds consist of many frequencies and

amplitudes; however, as we will understand later in the semester, these complex

waveforms can always be decomposed into a sum of individual pure sines.

Let’s begin by assuming that we have correlated sound sources. By correlated we

mean that each source creates exactly the same wave amplitude at the same time at

a given point. How would we make a correlated source in practice? Well if we

took a one electrical signal, split it, and fed it to two identical speakers AND we

stood equidistant from each speaker we would have correlated sounds. By

splitting one signal we ensure that each speaker is creating high and low pressure

wave amplitudes at the same time. The reason that we stand equidistant from the

speakers is to ensure that the sound waves that reach us have traveled the same

distance and thus arrive in phase. A pictorial representation of such in-phase

waves is given in the figure below.

The blue curve has been offset by a small amount because otherwise the two

curves would overlap precisely. Unless you are racing ahead of the pack, we

should by now have talked in class about the Superposition of Waves which states

that when two waves occupy the same point in space the total wave amplitude at

that point is just the sum of the amplitudes of the individual waves. Thus, for

correlated waves the pressure amplitudes add, i.e. P1 + P2 = Ptotal! This result goes

against what we covered previously for combining pressure amplitudes using the

squared values; however, we were also careful to say above that our combination

rules for decibel quantities only worked for uncorrelated signals and the direct

addition is only valid in this particular case of correlated signals.

So let's consider the case of uncorrelated signals. There is now no guarantee that

two sound waves will arrive at the same point in phase. Sometimes they will be,

but they are just as likely to be in complete anti-phase (180 degrees out of phase)

as shown in this figure,

or any value in between as in this figure.

As time goes by the net result will be some average value between the extreme

cases of completely in-phase and out-of-phase.

Does the sum of the squares method of adding the pressure amplitudes make

sense? The complete mathematical treatment is a little too involved for our

purposes but let’s look at whether the result is within the correct range. Imagine

we are combining two sound waves each with pressure amplitude 1. If they are in

phase then at the peak the maximum amplitude is 1 + 1 = 2. Similarly if they are

in complete antiphase as in the second graph above the sum of the amplitudes is 1 -

1 = 0. Our average value should lie somewhere in the range of 0 to 2. Well the

square and add method give P2

total = 12 + 1

2 = 2. Or in other words Ptotal is 1.414

(the square root of 2) which does lie between 0 and 2 as we expected.

I know this last material can be a bit bewildering the first time through. However,

I want to give you an accurate sense of the complications that occur when one is

combining quantities associated with waves. To test your understanding of the SPL

combination formula try the following self-test.

Self-Test 9

1. Two uncorrelated sources have SPLs of 30 dB and 32 dB. What is the

combined SPL?

2. Two uncorrelated sources have SPLs of 100 dB each. What is the combined

SPL of both played together?

3. Now consider the same problem but with 2 correlated sources of 100

dB. What change do you make in the calculation?

4. What is the SPL of 1 million uncorrelated sources each with SPL of 0 dB?

Here endth the section on Logarithms and decibel scales.

Answers to Self-Test 8

1. 103 dB

2. 4.77 dB

3. 1.76 dB.

4. 50 dB.

Answers to Self-Test 9

1. 34.12 dB.

2. 103 dB.

3. 106 dB.

4. 60 dB.