Chap 3 Outdoor Sound Propagation

8/18/2019 Chap 3 Outdoor Sound Propagation

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Learning Outcome

CLO 3: Apply physical and mathematical

model of sound wave to analyse outdoorsound propagation.Mathematical description of the propagation of

sound. Physical and mathematical description of the

propagation of sound outdoors and indoors; transmission

through and around barriers; effects of cut-outs, weak links

and flanking transmission paths.

EM412 Acoustic & Noise Control


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The Geometric Spreading of Sound

• The sound intensity is sound power per unit area.

• If we assume an ideal sound source (radiates equally inall direction) and free field condition (no reflecting or

absorption surface nearby), then the surface area is a

sphere of radius, S = 4πr 2.

• Then

S W I /

24/ r W I (2)

(1)



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Inverse Square Law

• A law that relates sound intensity, I and distance,r from the source.

• The equation (2) tells us that I is inverselyproportional to the r from the source.

• Thus, if r is doubles than I reduce to one quarterof its value.

21r

I (3)



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• This relationship can be used to find the I 2 at any

distance r 2 provided that the intensity I 1 at oneparticular distance r 1 is known

2

2

1

1

2

r

r

I

I (4)



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Sound pressure with distance from the source

• Since

2

1

1

2

r

r

p

p

•Thus, this relationship can be used to find the sound

pressure, p 2 at any distance r 2 provided that the sound

pressure p 1 at one particular distance r 1 is known

2 p I 2

2 1r

p 2

1r

I

(5)



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Example 1

Given that the sound pressure at a distance of 10 m

from the source is 0.578 Pa, what is the sound

pressure at a distance of 25 m?

Using equation (5):

p2 /0.578 = 10/25 from which p2 = 0.23 Pa



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• In decibel terms, if the distance from the source

is double (or half) then it correspond to an

decrease ( or increase) of 6 dB.

• Hence, in inverse square law corresponds to ‘6

dB per doubling of distance’ rule.

• From Example 1, we can deduce the SPL atfurther distance as follow:

– 89.2 dB at 10 m

– 83.2 dB at 20 m

– 77.2 dB at 40 m

– 71.2 dB at 80 m

– 65.2 dB at 160 m



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Expressing the inverse square law in dB

• The decibel version of I = W/4πr

2

is:

Where Lp

is the sound pressure level at a

distance of r meters from the sound source level,

Lw

• The decibel version of (p2/p1) = (r 1/r 2)

11log20 r L L W p

1

212 log20 r

r L L



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Example 2



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Example 3

• If the sound pressure level at a distance of 10 m from a

source sound is 89 dB, what will be the sound pressure

level at a distance of 160 m from the source?



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……

……


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• For the simple theory of a point source in a freefield this is 6 dB per doubling the distance.

• In practice the attenuation rate can be greaterthan this (7 or 8 dB per double) due to groundattenuation OR

• May be less than 6 dB per double distance due

to reflection down a street for example.• In such a case, the inverse square law formulamay be adapted;

• Where N = 20 is the theoretical inverse lawcase.

1221 log r r N L L



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Fig: Illustrating change in sound intensity and sound level

with distance arising from geometrical spreading, i.e.

inverse square law. Note linear distance scale (left) for

intensity, I proportional to (1/r 2), and log distance (right) for

sound pressure level, L p proportional to (1/r).EM412 Acoustic & Noise Control


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Example 4

In short



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Cont…



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Sound Sources

• Spherical wave radiator – A monopole source would radiate spherical wave

– Dodecahedron sources are available involving 12

identical loudspeakers arranged in a spherical array

which give better approximation to an ideal monopolesource.

Fig: Spherical wavesEM412 Acoustic & Noise Control


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• Spherical and Plane waves



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• Dipole source

– Consists of 2 identical simple sourceseparated by a short distance and operating

half a cycle out of phase with each other, to

create compression and rarefaction.



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Fig: Sound radiation from a piston source (left) and dipole source (right)



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Near and Far Fields of Sound Sources

• Near field of the source – Close to the source of sound where the sound from

all points can combine to produce complicated pattern

of sound.

– Sound pressure may vary with position and distancefrom the source in a complicated way and hard to

predict.

• Far field of the source

– Sound field settle down into a succession of smoothwaveform.

– Intensity falls off in accordance with inverse square

law and ‘6 dB /doubling of distance.



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Fig: Near and Far Field Region



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Important

• SPL may only be used to predict sound pressurelevel at far field positions, but not in near field.

• Only SPL measured in far filed may be used to

predict levels at other far filed distances.

• The simple prediction methods based on inverse

square law may be extended to real sources, but

only for far field positions.

• To take into account the possible directivity ofthe sound source when making prediction.



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Directivity of Sound Sources

• The directivity pattern of an ideal source iscircular (spherical radiation, equal in alldirection)

• For non-ideal sound source, the pattern isnon-circular directivity

• Directivity of the radiation of a source can

be quantified in 2 ways: – The Directivity Factor – The Directivity Index



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1) The Directivity Factor

• The ratio of the sound intensity in a given direction

measured at a certain distance from the source dividedby the average sound intensity at the same distance

• Iavge and p2avge are the average values which would be

obtained from imaginary omni-directional source emittingthe same sound power.

• If Q = 2, twice I in a certain direction as compare toaverage,

if Q = 0.5, half I in a certain direction as compare toaverage

2

avgeavge p p

I I Q



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2) The Directivity Index

• Express in decibels (dB)

• It is difference between the SPL, Lθ in a certaindirection (defined by the angle θ) measured at a

certain distance from the source and the averageSPL, Lavge which would be produced at the samedistance by the omni-directional source of the samesound power.

• Lavge is from logarithmic average over levels over alldirections

avge L L D



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• High Directivity Index and Factor



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Example 5



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and Q = 10D/10avge L L D

Using



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Similarly for all the other directions:



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Sound near Reflecting Planes

• Reflecting planes act as the reflector to thesound by reflecting all the sound reach it.

• Increase the sound intensity in certain direction

and produce directional source of sound with

high Q and D

Table: Directivity associated with source position



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The Prediction of Sound Level from Real

Sound Sources

• To adapt simple point source model of prediction

based on inverse square law to real sound

sources by including the directivity index, D of

the source in the equation

Dr L L w p 11log20



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• Ex: The sound radiating from freely flowing traffic

moving on a road produces a good approximation to

cylindrical spreading• SPL and I are given:

Where W and LWL are sound power per unit length and

sound power level per unit length

8log10 r L L WL p

1221 log10 r r L L

r W I

2/



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Fig: Line Source with Cylindrical Spreading



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Plane Source

• Ex: Large open aperture in the wall of a factory.

• Close to the sound, there is no beam spreading.

• At great distances, the wavefronts become

spherical and thus SPL falls at 6 dB per doubling

of distance.• In between, there is a region where the

reduction may approximate to 3 dB per doubling

of distance of line source



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Plane Source• Rathe Method

– a fuller treatment of finite planar source of height a and length b(b > a) identifies 3 zones with a rate of fall 0, 3 and 6 dB per

doubling of distance respectively.

Fig: The radiation of sound from a

plane area with variation of Lp

with

distance



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Example 6



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Outdoor Sound Propagation

• Effects which modify propagation in anoutdoor situation = Excess attenuation.

• Excess attenuation is the excess over

beam spreading or geometricalattenuation.

• All the effects are summed into a single

excess attenuation term to be added to theequation.



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• It involves with absorption either by the body of air

through which the sound passes or as a result of

interaction of the sound waves with the terrain over whichthe sound travels, with obstacles such as topographical or

man-made barriers, those associated with meteorological

effects (temperature and wind velocity gradients) and

those associated with vegetation.

Where Aexcess is tot effect of atmospheric and environmentalpropagation effects.

excessW p A Dr L L 11log20



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Absorption of Sound in Air, Aa

• Under normal circumstances, absorptionof sound by air is not considered very

important.

• Air absorption only becomes significantwhen considering propagation outdoors

over long distances (more than a few

hundred meters) and when highfrequency sound is involved.



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Refraction of Sound in Atmosphere

• Refraction is the change of direction of a

sound wave which occurs when sound

moves from one medium to another medium

with a different sound velocity.• For instance, change is sound speed with

height above the ground caused by changes

in air temperature and wind speed orcombination of both.



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Temperature Gradient• When the sound wave travels from a layer of warm air in

the atmosphere to a layer of cooler air, a sound raybends towards the normal.



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• The change in temperature is gradual but

continuous with the result that sound rays

are gradually bent into a curved path.



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• During the night time: air near the ground

cooler and the propagation path bendstowards the ground.



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• During the daytime: air temperature fall

with increasing height above the surface of

the earth.



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Wind Effect

• Sound level is greater upwind than

downwind from the source.

• It is a consequences of the gradient of the

wind speed above the ground plane which

lead to the bending of the sound wave.

• In the direction of the wind the sound wave

from the source are bent downwards.

• In the opposite direction the sound wave

are bent upwards, towards the sky.



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• Wind Effect



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Anomalous Sound Propagation in the Atmosphere

• Sound may be trapped in atmospheric refraction channels

and travel for very long distances with very little attenuationbefore returning to earth, while the sound may not be heard

at much shorter intermediate distances from the sources.



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• On 1 June 1666, during the military engagement

of English and Dutch armies in English Channel

the sounds of the guns were heard in London,but not on South Downs or Dover.



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Barrier

• To reduce sound transmission between

the source and receiver.

• The attenuation/insertion loss from barrier

is the reduction in noise level at the

receiver as a result of the barrier.

• It is limited by diffraction of sound over the

top and around the side of the barrier,

which is determined by the size of barrier.



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• One of the simplest method in predicting

barrier attenuation for a long thin barrier isbased on path difference (Maekawa)

• The path difference, δ is the additional

sound path for the rays travelling over asimple barrier

– δ = ( a + b ) - c

• The wavelength, λ of the incident – N = 2δ / λ.



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• The path difference, δ (a + b –c)



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2 versions to predict barrier attenuation, Abarrier in

receiver;

a)Illuminated zone there is direct line of sight of

the sound source from the receiver position

- Attenuation = 10log(3 – 20N) dB

b)If the receiver in the shadow zone of barrier (no

line of sight)

– Attenuation = 10 log(3 + 20 N) dB



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• In practice, the attenuation achieved with a

barrier will be less due to a combination of all

the effect mentioned before.

• Max attenuation obtained with a practical barrier

is about 15 to 20 dBA for very high barrier orbarrier very close to either the source of

receiver.



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Example



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Solution



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Solution

Note how the predicted barrier attenuation varies considerably

across the frequency spectrum, increasing in frequency from

7.6 dB at 63 Hz to 25.4 dB at 8 kHz.EM412 Acoustic & Noise Control


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Solution



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Other Sound Propagation Effects



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Prediction of Outdoor Sound Levels


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Prediction of Outdoor Sound Levels

and Noise Mapping

• The total outdoor sound level is determined by

considering each of the factor considered before

(air, barrier, ground etc)

.....11log20 barrier ground air W p A A A Dr L L



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The Doppler Effect

• The apparent change in frequency,(f) due

to the relative motion between the source

and the observer.

• Ex: An emergency vehicle sounding asiren approaches, passes and recedes

from the observer.

• Arise from the change of wavelength (λ) orsound speed.



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• Scenario



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Observer moving towards a source• The λ that reach the observer is unchanged but due to

the motion of observer the frequency with which theobserver receives successive peaks is increased.

• The apparent frequency, f

– a)

– b)

f cvc f '

cv f f 1'


E l


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ExampleThe security alarm on a parked car goes off and produces

a frequency of 952 Hz. The speed of sound is 343 m/s. As

you drive toward this parked car, pass it, and drive away,

you observe the frequency to change by 97 Hz. At what

speed are you driving?

Apparent frequency, f = (952-97)=855 Hz

Use this formula;

855= 952 [(343-V)/(343)

v = 34 9485 m/s

vcc f f '

Chap 3 Outdoor Sound Propagation

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