INDEX

College of Santa Fe Auditory Theory

Lecture 003- Sound II

INSTRUCTOR: CHARLES FEILDING

  1. Sound intensity, power and pressure level
  2. Logarithms
  3. Sound Intensity Level
  4. Sound Power Level
  5. Sound Pressure Level
  6. The Decibel Scale
  7. The level when correlated sounds add
  8. Phase alignment
  9. Overlapping Note Ons
  10. The level when two uncorrelated sounds add
  11. Adding Decibels
  12. The inverse square Law
  13. The effect of boundaries
  14. Brain Bullets

Real sounds vary over an range of pressure amplitudes which is greater than a million. In the range of human hearing sound pressure can vary from from less than 20 micro Pascals (20/1,000,000) to greater than 20 pascals. These two pressure correspond roughly to the threshold of hearing to the threshold of pain. Because of this and the way we perceive sound it is convenient to express sound pressure levels on a logarithmic scale.


Logarithms

The logarithm is perhaps the single, most useful arithmetic concept in all the sciences; and an understanding of them is essential to an understanding of many scientific ideas. Logarithms may be defined and introduced in several different ways. But for our purposes, let's adopt a simple approach. This approach originally arose out of a desire to simplify multiplication and division to the level of addition and subtraction. Of course, in this era of the cheap hand calculator, this is not necessary anymore but it still serves as a useful way to introduce logarithms. The question is, therefore:

Q: Is there any operation in mathematics which produces a multiplication by the performance of an addition?

A: What is 23 x 24.

The answer is 2 7 which is obtained by adding the powers 3 and 4. This is correct, of course, since 23 x 24 is just seven 2s multiplied together. Note that this addition trick does not work for the case of 3 3 x 24. The base numbers must be the same, as in the first case, where we used 2.

In general, this addition trick can be written as pa x pb = pa+b. This expression will do our job of multiplying any two numbers, say 1.3 and 6.9, if we can only express 1.3 as pa and 6.9 as pb.

What number will we use for the base p? Any number will do, but traditionally, only two are in common use:

Ten (10) and the transcendental number e (= 2.71828...), giving logarithms to the base 10 or common logarithms (log), and logarithms to the base e or natural logarithms (ln).

Let's first talk about logarithms to the base 10 or common logs. We thus choose to let our number 1.3 be equal to 10a.

1.3 = 10a

`a' is called "the logarithm of 1.3". How large is `a'? Well, it's not 0 since 100 = 1 and it's less than 1 since 101 = 10. Therefore, we see that all numbers between 1 and 10 have logarithms between 0 and 1. If you look at the table below you'll see a summary of this.

Number range Logarithm Range
1 - 10 or 100 - 101 0 -1
10 - 100 or 101 - 102 1 - 2
100 -1000 or 102 - 103 2 - 3
etc. etc.

As you see, we have the number range listed on the left and the logarithm range listed on the right. For numbers between 1 and 10, that is between 100 and 101, the logarithm lies in the range 0 to 1. For numbers between 10 and 100, that is between 101 and 102, the logarithm lies in the range 1 to 2, and so on. Now in the bad old days before calculators, you would have to learn to use a set of logarithm tables to find the logarithm of our number, 1.3, that we asked for earlier. But nowadays, you can get it at the press of a button on your calculator.

Most calculators are very straightforward in obtaining the logarithm. They either have two logarithm keys or a dual function key. In any case, the labels will be `log' and `ln' which is often pronounced `lon'.

Log is the key for logs to the base 10 and ln is for natural logs. We want logs to the base 10 in our example so we use `log'. Enter 1.3 on your calculator, and then press the log key.

Do you have 0.113943? You should have. This number then is `a' back in our previous expression and therefore the logarithm of 1.3. Pause now and determine `b' in that expression, the logarithm of 6.9.

You should have 0.838849 for the log of 6.9. If not, review what we have done and try again. Now we are going to do something silly in view of the fact that you have a calculator. We're going to use the two logarithms you have evaluated to find the product of 1.3 and 6.9. Of course, you can do it quickly with your calculator, but this will show that logarithms do what they are supposed to do. According to our original idea, the sum of the two logarithms was supposed to be the logarithm of the answer.

Now add the two logarithms. The sum is 0.952792. This is the logarithm of the answer. If we only knew what number had 0.952792 as its logarithm, we would know the value of 1.3 x 6.9.

The problem of finding a number when you know its logarithm is called finding the "antilogarithm" or sometimes "exponentiation". Again, lets look at your calculator. Here is where calculators differ a lot and I hope I mention one that is something like yours.

You should look for a key on your calculator that says something like 10x or 10y. If so, then pressing that key will take the antilog of the number in the display. Alternatively, your calculator may have an "inverse" key. If so, then pressing inverse and then log will take the antilog of the number in the display. Enter 0.952792 into your calculator and find the antilog.

Did you get 8.97? If not, try again. Of course, you might have got something like 8.96999 but of course that really is 8.97. Now, multiply 1.3 x 6.9 on your calculator and you'll see that 8.97 is indeed the correct answer.

The whole operation could be done with natural logarithms as well as shown below.

1.3 x 6.9 = ?  
ln 1.3 = 0.262364 i.e. 1.3 = e0.262364
ln 6.9 = 1.931521 i.e. 6.9 = e1.931521
total = 2.193885 i.e. 1.3 x 6.9 = e2.193885
antiln 2.193885 = 8.97  

 

If the sum of logarithms gives the product of two numbers, then the difference gives the quotient.

In the table below, I've taken the difference between the ln of 1.3 and the ln of 6.9. Check it on your calculator.

1.3/6.9 = ?
ln 1.3 = 0.262364
ln 6.9 = 1.931521
ln 1.3/6.9 = -1.669157
antiln (-1.669157) = 0.1884

Don't be afraid of the negative sign. It simply means that the answer is less than 1. Enter -1.669157 on your calculator, then find its antiln. Note we are working with natural logs in this example.

If you didn't get 0.1884, try again. Of course, this is just 1.3 divided by 6.9. In the table below, I have done the whole problem over again using common logs. Pause here and check it.

log 1.3 = 0.113943 log 6.9 = 0.838849 log(1.3/6.9) = -0.724906 antilog (-0.724906) = 0.1884

The logarithmic and exponential functions are very important since many physical and biological processes can be described by them.

Logarithms

Input decimal number >0=
Logarithm=

Logarithms and the nonlinearity of our senses.

In Venice in 1609 Galileo first started to look at the night sky with a telescope of his own design. It soon became obvious to him that some method was needed for categorizing the brigtness of stars. He established and arbitrary scale from 1 to 10 where 10 was the brightest and 1 was the faintest and he began making a catalog of all the stars that were visible to him. This system was refined over the years but it was strictly subjective and led to a great deal of haggling over which stars belonged in which level of brightness.

It was not until the invention of the Charge Coupled Device (CCD) in the 20th century that objective measurement of brightness became possible. When CCD were applied to the task of evaluating the brightness of stars they go the following result

 

Galileo CCD value Difference Antilog of CCD value
10 1 0.0458 10
9 0.9542 0.0512 9
8 0.9031 0.0580 8
7 0.8451 0.0669 7
6 0.7782 0.0792 6
5 0.6990 0.0969 5
4 0.6021 0.1249 4
3 0.4771 0.1761 3
2 0.3010 0.3010 2
1 0.000   1

This demonstrates that although Galileo thought he was seeing in a a linear manner he was actually perceiving the brightness of stars in a logarithmic manner. Sound works the same way and what we think is linear in our hearing is in fact logarithmic.

 


Energy

The energy of a sound is a measure of the total sound present and is measured in Joules.

Power

If we measure the rate of energy released over time we can get the power of the sound: Joules/seconds = Watts.

Sound Intensity

Sound expands spherically from the point of origin and our ears only occupy one segment of that sphere. If we measure the sound passing through an area of the sphere we come up with a value called Sound Intensity: the power density of a sound wave propagating in a particular direction..

Fig 1.7 Sound Intensity

1.2 Sound intensity, power and Pressure level

The energy of a sound wave is a measure of the amount of sound present. However, in general we are more interested in the rate of energy transfer, instead of the total energy transferred. Therefore we are interested in the amount of energy transferred per unit of time, that is the number of joules per second (watts) that propagate. Sound is also a three-dimensional quantity and so a sound wave will occupy space. Because of this it is helpful to characterise the rate of energy transfer with respect to area, that is, in terms watts per unit area. This gives a quantity known as the sound intensity which is a measure of the power density of a sound wave propagating in a particular direction, as shown in Figure 1.7.

1.2.1 Sound Intensity level

Sound intensity is defined as the sound power per unit area. The usual context is the measurement of sound intensity in the air at a listener's location. The basic units are watts m-2 or watts cm-2 . Many sound intensity measurements are made relative to a standard threshold of hearing intensity:

SIL is measured relative to a standard intensity level of 1 picowatt per square meter

SIL = 10 log10(Iactual / Ireference )

(1.10)

where

The factor of 10 arises because this makes the result a number in which an integer change is approximately equal to the smallest change that can be perceived by the human ear. A factor of 10 change in the the power density ratio is called the bel. The integer unit that results from Equation 1.10 is therefore called the decibel (dB). It represents a 10 change in the power density ratio, that ratio is about 1.26.

Sound Intensity Level (SIL)

SIL = 10 log10(Iactual / Ireference )

Emitter Diameter= cm
Radiated Power= milliWatt
Sound Intensity = Wm-2
I reference = 10-12 Watts m-2 (1 pico watt per square meter)
Sound Intensity Level = dB

1.2.2 Sound Power Level

The sound power level is a measure of the total power radiated in all directions by the source of a sound and is expressed as SWL. Once again it is expressed as a ratio of the actual power level to a reference power level of 1 picowatt. If you input 1 watt into the following calculator you will see that 1 watt is a very loud sound if you were to receive all of it. However because sound expands sperically you only receive a small part of it.

SWL = 10 log10(Wactual/ Wreference )

(1.11)

Sound Power Level (SWL)

SWL = 10 log10(Wactual/ Wreference )

Actual Power Watts
W reference = 10-12 Watts (1 pico watt)
Sound Power Level = dB

1.2.3 Sound Pressure level

Sound pressure is measured in Pascals. 1 pascal = 1 newton per square meter (1 Nm2 )

The sound intensity is one way of measuring and describing the amplitude of a sound wave at a particular point. However, although it is useful theoretically, and can be measured, it is not the usual quantity used when describing the amplitude of a sound. Other measures could be either the amplitude of the pressure, or the associated velocity component of the sound wave. Because human ears are sensitive to pressure, which will be described in Chapter 2, and because it is easier to measure, pressure is used as a measure of the amplitude of the sound wave. This gives a quantity which is known as the sound pressure, which is the root mean square (rms) pressure of a sound wave at a particular point.

Root Mean Square

Circuit currents and voltages in AC circuits are generally stated as root-mean-square or rms values rather than by quoting the maximum values. The root-mean-square for a current is defined by:

Irms = (I2)average

That is, you take the square of the current and average it, then take the square root.

The sound pressure for real sound sources can vary from less than 20 microPascals (20 μPa or 20 x l0-6 Pa) to greater than 20 Pascals (20 Pa). Note that 1 Pa equals a pressure of 1 N m-2. These two pressures broadly correspond to the threshold of hearing (20 μPa) and the threshold of pain (20 Pa) for a human being, at a frequency of 1 kHz, respectively. Thus real sounds can vary over a range of pressure amplitudes which is greater than a million. Because of this, and because of the way we perceive sound, the sound pressure level is also usually expressed on a logarithmic scale. This scale is based on the ratio of the actual sound pressure to the notional threshold of hearing at 1 kHz of 20 μPa. Thus the sound pressure level (SPL) is defined as:

SPL = 20 log10 (Pactual/Preference)

(1.12)

where

 

Sound Pressure Level (SPL)

SPL = 20 log10(Pactual/ Preference)

Actual Pressure Pascal
I reference = 10-6 Pascal /m2 (1 micro Pascal)
Sound Pressure Level = dB

The multiplier of 20 has a twofold purpose. The first is to make the result a number in which an integer change is approximately equal to the smallest change that can be perceived by the human

About pressure

I asked my brother in England (who is a physicist) to give me some real world examples of these values. He wrote to me as follows:

100,000 Newton/sq metre (or Pascal) is about 1 atmosphere, or about 2088 lb per sqare foot, or 14.5 pounds per square inch (a low pressure tire).

A punch is a fist travelling at a certain speed, having a given momentum. If it lands with the width of the knuckle, the pressure will bruise but not break the skin. If the pressure is concentrated on a knife point, it will penetrate deep. If it lands on a pointed finger (martial arts style) the effects are in between these extremes.

The whole Newtonian concept is about momentum (lb-ft/sec). Everything has it's intrinsic momentum at any point. It takes the action of a force to change this. The momentum of one pound at one foot per second is half that of one pound at two feet per second. It is also half that of two pounds at one foot per second. It is a quarter of two pounds at two feet per second. The pressure at landing depends on the area of contact.

An exercise ball of 1 kilo (2.2 lb) at one metre per second (3.25 ft/sec) is volleyball stuff. To stop it very quickly needs a lot of work. A hard punch is 4 sq ins (0.0026 sq metre) of knuckle absorbing 1 kilometre/sec over a half a second of contact: 1 Newton divided by the half second divided by the area is 769 pascal. Ouch. To stop it slowly takes less: catch it while bending arms and transferring the ball's momentum to yourself, stepping backwards a pace to slowly remove the momentum - 48 sq ins (0.03 sq m) of hand stopping the ball over two seconds is 1/(2x0.03)=17 pascal. No pain.

If you understand this, I'll explain it again until you don't . .

End Quote

The decibel scale

The decibel (dB) is used to measure ofsound level, but it is also widely used in electronics, signals and communication. The dB is a logarithmic unit used to describe 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 (other units related to loudness). But first, to get a taste for logarithmic units, let's look at some numbers:

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 etc) kept the same.

The difference in decibels between the two is defined to be

10 log10 (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 log10 (P2/P1) = 10 log10 (2) = 3 dB.

If the second had 10 times the power of the first, the difference in dB would be

10 log10 (P2/P1)= 10 log10(10) = 10 dB.

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

10 log10 (P2/P1) = 10 log10(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.

It might be a good idea to get some kind of intuitive feel for how the decibel scale sounds.

The Decibel Scale (Audio Demos)

1.3 Adding sounds together

There are two situations in which sounds can be added

1.3.1 The level when correlated sounds add

When the sounds are correlated the sources simply add

Ptotal correlated (t) = P1(t) +P1(t) +....Pn(t)

(1.13)

Phase alignment.

However exactly what happens depends upon the phase of the correlated sounds. Phase alignment is a very important concept in sound and we should take some time to define it precisely.

If we were to attach a pen to a wheel and then rotate the wheel in front of a sheet of paper which moved in such a way as to keep the pen speed constant relative to the paper the result would be a sine wave. Therefore the phase angle of a wave is defined in degrees representing the angle at which the hypothetical wheel would be at that point of drawing the wave.

If two correlated waves start at different times their phase angles will be different

How correlated waves add is enormously affected by the phase angle.

What does this mean to your life? A lot. In the 1970's Stevie wonder was recording a song called "Living for the city" which began with a very funky flat 9 figure played on a Fender Rhodes Satellite model which had two speakers for setting up on either side of the stage. This gave a huge pan when the player used the phaser and the engineer had mic'd each speaker separately. As long as they listened in stereo everything sounded fine but as soon as the engineer threw it into mono to check the Radio Mix the piano vanished completely. Soloing left and soloing right showed audio was present and the VU meters were jumping on both channels. However as soon as the two channels were combined in the air the piano was gone.

Can anyone tell me what happened?

The Fender Rhodes was using phase alignment to create a super wide stereo between the speakers. When summed to mono the correlated sources cancelled each other to silence.

NOTE Use Signal Suite to demonstrate out of phase and between phase effects.

With the advent of MIDI tone generators and sequencers musician have be more alert than ever for phase problems. Wave ROM is expensive and manufacturers will rarely put in two of an instrument when one will suffice. This means that your dark piano is probably the bright piano with a filter. Any unison part played between those two is now correlated data probably with a phase offset caused by the fact that Tone Generators are notoriously slow.

Here is an example.

In the following sound file I play a short passage of Alberti Bass on my bright piano. Hoping to fatten it up I then layer the same part with the dark piano. Notice how phase cancellation between the identical waves not only fails to fatten it up it significantly reduces the bass on the second pass.

Alberti Bass (Audio Demo)

Memorize this nasty little sound folks, it is the sound of phase cancellation and it can be all over your MIDI studio if you are not careful.

Overlapping Note Ons.

The quickest way to get phase cancelling happening in a MIDI composition is to allow simultaneous MIDI note ons to occur on the same note. Aha! I hear you cry, that's OK because it is impossible to play the same note twice in MIDI without the system sending a note off in between. This is true but there are many sequencer programs that will quantize NOTE ON and leave NOTE OFF alone.

Where they cross you will have phase cancelling.

This can also happen when you use cut and paste or merge to combine MIDI parts to a single instrument.

1.3.2 The level when uncorrelated sounds add.

If sound waves are uncorrelated they do not add algebraically. Instead we must add the powers of the individual waves together. The power of a waveform is proportional to the square of the pressure levels so we must square the pressure amplitudes before adding them together. If we want the result as pressure we must take the square root of the result.

Ptotal uncorrelated = ( P12 + P22+....Pn2)

(1.14)

1.3.3 Adding Decibels

Decibels are a logarithmic scale and this means adding dB is not the same as adding the sources amplitudes. This is because adding logarithms is the equivalent of MULTIPLYING the quantities that they represent. dB have to be converted back to ratios before they are added together.

SPL = 20 log10((P12 + P22+....Pn2)/Preference)

(1.15)

Adding Decibels

SPL = 20 log10((P12 + P22+....Pn2)/Preference)

Sound 1 Level: dB
Sound 2 Level: dB
Summed SPL: dB

Adding two uncorrelated sounds of equal volume is easy..you get a 3 db increase.

Adding equal uncorrelated sounds

Ptotal = P1+10* log10N
Sound level of one source: dB
Number of Sources:
Total level: dB

 

1.4 The inverse square law 28

Up to now we have only studied sound travelling in one direction.

In fact sound expands radially from a point source. Something like this

To be precise it is not perfectly spherical at first..the sphere is oblate in the first part of the expansion rounding out about two feet from the source. This gives rise to a considerable bass boost close to the sound source which is known as the Proximity Effect. Those of you who have worked in bands will doubtless recall the singer apparently trying to swallow the microphone as they intuitively try to take advantage of the proximity effect.

As the sound spreads out from a point source it weakens, not only because of absorption but because it's energy must be spread more thinly to cover the expanding wave front.

Intensity = Powersource÷ 4π r2

Every time the distance from a sound source is doubled the intensity of the sound reduces by a factor of four, there is an inverse square relationship between the sound intensity and the distance from the sound source

The area of a sphere = 4 x π x r2

Sound intensity is defined as the power per unit area. Therefore the sound intensity as a function of distance from a source is expressed as:

I = Wsource /Asphere = Wsource/4πr2

(1.16)

where

Equation 1.16 shows that the sound intensity for a sound wave that spreads out in all directions from a source reduces as the square of the distance. Furthermore this reduction in intensity is purely a function of geometry and is not due to any physical absorption process. In practice there are additional sources of absorption in air, for example impurities and water molecules, or smog and humidity. These extra sources of absorption have more effect at high frequencies and, as a result sound not only gets quieter, but also gets duller, as one moves away from a source, due to the extra attenuation these cause at high frequencies. The amount of excess attenuation is dependent on the level of impurities and humidity and is therefore variable.


Example 1.13 An omnidirectional loudspeaker radiates one hundred milliwatts (100 m W). What is the sound intensity level (SIL) at a distance of 1 m, 2 m and 4 m from the loudspeaker? How does this compare with the sound power level (SWU at the loudspeaker?

The sound power level can be calculated from Equation 1.11 and is given by:

SWL = 10 log10 ( Wactual / Wreference) = 10 log10 (100mW/(1 x 10-12W)

=10 log10(1 x 1011) = 110dB

The sound intensity at a given distance can be calculated using Equations 1.10 and 1.16 as:

SIL = 10 log10 (Iactual /Ireference) = 10 log10 ( (Wsource/4πr2)/Ireference)

this can be simplified to give:

SIL = 10 log10 ( Wsource/Wreference ) - 10 log10(4π) - 10 log10(r2)

which can be simplified further to:

SIL = 10 log10 ( Wsource/Wreference ) - 20 log10(r) - 11 dB

(1.17)

This equation can then be used to calculate the intensity level at the three distances as:

SIL1ml = 10 log10 (100 mW/10-12W) - 20 log10(1) - 11 dB= 110 dB - 0 dB - 11 dB = 99 dB

 

SIL2ml = 10 log10 (100 mW/10-12W) - 20 log10(2) - 11 dB= 110 dB - 6 dB - 11 dB = 93 dB

 

SIL4ml = 10 log10 (100 mW/10-12W) - 20 log10(4) - 11 dB= 110 dB - 12 dB - 11 dB = 87 dB


From these results we can see that the sound at 1 m from a source is 11 dB less than the sound power level at the source. Note that the sound intensity level at the source is, in theory, infinite because the area for a point source is zero. In practice, all real sources have a finite area so the intensity at the source is always finite. We can also see that the sound intensity level reduces by 6 dB every time we double the distance; this is a direct consequence of the inverse square law and is a convenient rule of thumb. The reduction in intensity of a source with respect to the logarithm of distance is plotted in Figure 1.12 and shows the 6 dB per doubling of distance relationship as straight line except when one is very close to the source. In this situation the fact that the source is finite in extent renders Equation 1.16 invalid. As an approximate rule the nearfield region occurs within the radius described by the physical size of the source. In this region the sound field can vary wildly depending on the local variation of the vibration amplitudes of the source.

Equation 1.16 describes the reduction in sound intensity for a source which radiates in all directions. However, this is only possible when the sound source is well away from any surfaces that might reflect the propagating wave. Sound radiation in this type of propagating environment is often called the free field radiation, because there are no boundaries to restrict wave propagation.

Intensity as a function of Distance

SIL = 10 log10(Wsource /Wattsreference )-20 log10(r) - 11db
Sound source intensity: Watts
Distance from source meters
Distant SIL: dB

By experimenting with the above calculator we can observe that sound diminishes 6dB Everytime we double the distance from the source. This is a useful rule of thumb:

Fig 1.12 Sound intensity as a function of distance from the source

The Decibel Scale (Audio Demo)

1.4.1 The effect of boundaries

The effect of boundaries is to concentrate the sound power of the source into a smaller range of angles. This can be calculated using the following equation where Q represents directivity of the sound relative to a sphere. The presence of 1 2 and 3 boundaries reduce the sphere to hemisphere, half hemisphere and quarter hemisphere which correspond to a Q of 2 4 and 8 respectively.

SIL = 10* log10(Wactual/Wreference) +10 * log10(Q) -10 * log10(4*π ) - 20* log10 (r)

The effect of boundaries

SIL = 10* log10(Wactual/Wreference) +10 * log10(Q) -10 * log10(4*π ) - 20* log10 (r)
Power of source: Watts
Distance from Source: meters
Q:
About Q: When boundaries = 1 Q=2, when boundaries = 2 Q = 4, when boundaries = 3 Q = 8,
Sound Intensity at distance: dB

But how does a boundary affect Equation 1.16? Clearly many acoustic contexts involve the presence of boundaries near acoustic sources, or even all the way round them in the case of rooms and some of these effects will be considered in Chapter 7. However, in many cases a sound source is placed on a boundary, such as a floor. In these situations the sound is radiating into a restricted space, as shown in Figure 1.13. However, despite the restriction of the radiating space, the surface area of the sound wave still increases in proportion to the square of the distance, as shown in Figure 1.13. The effect of the boundaries is to merely concentrate the sound power of the source into a smaller range of angles. This concentration can be expressed as an extra multiplication factor in Equation 1.16. Therefore the equation can be rewritten as:

Idirective source= (QWsource)/4πr2

(1.18)

where

Equation 1.18 can be applied to any source of sound which directs its sound energy into a restricted set of angles which are less than a sphere. Obviously the presence of boundaries is one means of restriction, but other techniques can also achieve the same effect. For example the horn structure of brass instruments results in the same effect. However, it is important to remember that the sound intensity of a source reduces in proportion to the square of the distance, irrespective of the directivity.

The effect of having the source on a boundary can also be calculated; as an example let us examine the effect of boundaries on the sound intensity from a loudspeaker.


Example 1.14 A loudspeaker radiates 100 mW. Calculate the sound intensity level (SIL) at a distance of 2 m from the loudspeaker when it is mounted on 1, 2 and 3 boundaries.

The sound intensity at a given distance can be calculated using Equations 1.10 and 1.18 as:

SIL = 10 log10 (Iactual/Ireference) = 10 log10 ((QWsource)/4πr2)/Wreference )

which can be simplified to give:

SIL = 10 log10 ( Wsource /Wreference ) + 10 log10(Q) - 10 log10(4π) - 20 log10(r) ref

This is similar to Equation 1.18 except for the addition of the term for the directivity, Q. The presence of 1, 2 and 3 boundaries reduces the sphere to a hemisphere, half hemisphere and quarter hemisphere, which corresponds to a Q of 2,4 and 8, respectively. As the only difference between the result with the boundaries is the term in Q, the sound intensity level at 2m can be calculated as:

SIL1 boundary = SIL2m + 10 log10(Q) = 93 dB + 10 log10(2) = 93 dB + 3 dB = 96 dB

SIL2 boundary = SIL2m + 10 log10(Q) = 93 dB + 10 log10(4) = 93 dB + 6 dB = 99 dB

SIL3 boundary = SIL2m + 10 log10(Q) = 93 dB + 10 log10(4) = 93 dB + 9 dB = 102 dB

From these calculations we can see that each boundary increases the sound intensity at a point by 3 dB, due to the increased directivity. Note that one cannot use the above equations on more than three boundaries because then the sound can no longer expand without bumping into something.

 


Reading Assignment

Before next class please read Section

1.5.n Sound Interactions

Pages 33 to 51 of Acoustics and Psychoacoustics. We may have a brief quiz on this section at the beginning of the next class.

You Need to Know