Basics of Noise

8/11/2019 Basics of Noise

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Hearing by numbers

Dr Max Graham

Faculty of Advanced TechnologyUniversity of Glamorgan


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First principles: Generation of a sound wave from a vibrating source

A simple plane progressive wave


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Relationship between velocity (v), frequency (f), wavelength () and period (T)

V = f m/s

T = 1/f s


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Some velocities of sound

MEDIUM VELOCITY (m/s)

AIR 340

WATER 1500

STEEL 5200

RUBBER (HARD)

RUBBER (SOFT)

1400

50

SAND 95200


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Frequency range of human hearing

Roughly between 20 Hz and 20 kHz

Below 20 Hz is known as infrasound

Above 20 kHz is known as ultrasound Perception of maximum loudness is

around 4 kHz


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Equal loudness contours (humans)


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Frequency ranges of animals

Species Frequency range (Hz)

Dog 5045 000

Cat 4565 000

Horse 5034 000Sheep 10030 000

Mouse 100091 000

Bat 2000110 000

Whale 1000123 000

Goldfish 203000

Tuna 501000

Bullfrog 1003000


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Noise measurement:

Single and third octave band centre frequencies


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Noise spectrum based on single octave band frequency analysis


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Wavelengths of frequency limits of

human hearing

V = f m/s, so = V/f m = 340/f m

At 20 Hz, =340/20 = 17 m

At 20 kHz, =340/20 000 = 17 mm And beyond..

At 20 MHz, =340/20 000 000 = 0.017 mm

(less than the thickness of a human hair)


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A wavelength of 100 mm

Frequency = V/= 340/0.1 = 3400 H z


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Measurement of sound levels?

Decibels (dB)

But what are they?


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Range of sound powers

Acoustic power

(watts)

Sound power

level (dB)

Typical source

100 000 000 200 Rocket

10 000 160 Boeing 707full power

100 140 75 piece orchestra

1 120 Chain saw

0.001 90 Motor car

0.000001 60 Normal voice

0.000000001 30 Whisper

0.000000000001 0 Threshold ofhearing


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Range of sound pressures

Threshold of hearing: 2 x 10-5N/m2or Pa

Threshold of pain: 200 Pa

Atmospheric pressure = 101325 N/m2


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A linear scale?

Adopting a linear scale for say sound

power over the range from the threshold

of hearing (10-12W) to the threshold of

pain (100 W)

A scale of 1014increments

If each increment was represented by an

atom (typical diameter 10-10m), the scale

would stretch for 10 km!


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A logarithmic scale

By definition:

if logab = c,

then ac= b

Number Common

logarithm

1 0

10 1

100 2

1000 3

. .

. .

1000 000 6 etc


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The deci-Bel (dB) scale

The Bel scale was too limited. It is opened

up by a factor of 10 to provide the decibel

scale:

LW= 10 log (W/Wo) dB re Wo = 10-12W

We now have a more realistic range ofbetween 0 dB (threshold of hearing) and

140 dB (threshold of pain)


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Alternative decibel scales

Power (W) of a noise source is expressed as the

sound power level (dB)

Intensity (W/m2) at some distance is expressedas the sound intensity level (dB)

Pressure (N/m2

) at some distance is expressedas the sound pressure level (dB)


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Definition of decibel (dB)

parameters

Sound power level (LW)

LW= 10 log (W/Wo) dB re Wo = 10-12W

Sound intensity level (LI) LI= 10 log (I/Io) dB re Io = 10

-12W/m2

Sound pressure level (Lp)

Lp = 20 log (P/Po) dB re 2 x 10-5N/m2(Pa) or 20 Pa


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Sound pressure level (Lp)

Where does the 20 come from?

P = Iz N/m2 or P2= Iz

Remember log xn= n logx

Z is the characteristic impedance of the medium through whichthe sound is travelling

Z = density of the medium (kg/m3) x velocity of sound throughit (m/s)

Zair = 1.2 x 340 = 408 kg/m2

s (Rayls), roughly 400 Rayls Consider reference values:

Po = Ioz = 10-12x 400 = 4 x 10-10= 2 x 10-5N/m2


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What are the limits?

How quiet can we go?

How loud can we go?


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Decibel manipulation 1

Combination of multiple sources

(equal noise levels)

1 noise source = x dB

2 noise sources = x + 3 dB

L = 10 log(2I/Io)10 log(I/Io) dB

= 10 log 2 = 3 dB 10 noise sources = x + 10 dB

L = 10 log(10I/Io)10 log(I/Io) dB

= 10 log 10 = 10 dB

N noise sources = x + 10 logN dB

L = 10logN dB


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Addition of decibels

A general expression:

L = 10 log(10L1/10+ 10 L2/10+ . 10 Ln/10) dB


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Subtraction of decibels

Subtract a noise level L2 from a noise

level L1

New level, L = 10 log(10L!/1010 L2/10) dB

Example: 80 dB + 85 dB = 10 log (108.0 + 108.5) dB = 86.2 dB

86.2 85 dB = 10 log (108.62108.5) dB = 80 dB


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Decibel manipulation 3(Averaging decibels of equal time periods)

If noise levels are sounded over equal time

periods the average noise level over the total

period they have been sounding for is given by:

LAVG= 10 log((10L1/10+ 10L2/10+ 10Ln/10)/n) dB

Example, the average of 80 dB and 100 dB is given by:

LAVG = 10 log((108.0

+ 1010

)/2) dB = 97 dB

NOT (80 + 100) / 2 = 90 dB (the arithmetic average)


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(Time weighted averages)

The average noise level (having the same

acoustic energy) for several noise levels

sounding over different time periods can

be calculated and expressed over anytime period (T)

Leq,T = 10 log ((t1 x 10L1/10+ t2 x 10L2/10+.+ tn x 10Ln/10)/T) dB

This can be applied to a single source or any number of sources


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Time weighted averages

An example

A noise level of 90 dB sounds continuously over

a period of one hour.

The noise level this is equivalent to over a period of 15 minutes isgiven by:

L = 10 log ((t x 10L/10)/T dB

= 10 log (60 x 109/15) dB = 96 dB

Similarly, the noise level this is equivalent to over a period of 8 hours

is given by:

L = 10 log(1 x 109/8) dB = 81 dB


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Dont forget your basics

90 dB for 1 hour = 96 dB for 15 minutes

Why?

4 x energy for a quarter of the time

90 dB for 1 hour = 81 dB for 8 hours

1/8 of the energy for 8 x the time

We are equating the energy in all cases.


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Similar principles apply to heat and

light


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The analogy between sound and heat (also applies to light)


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Distribution of energy from airborne sound striking a partition


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Introducing the transmission

coefficient ()

The transmission coefficient () is the

fraction of the acoustic energy which is

transmitted through a partition

TL and are related:

TL = 10 log(1/) dB


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Measures of sound insulation

Transmission loss Transmission

coefficient ()

Zero dB (air gaps) 1

20 dB (single glazing) 0.01

50 dB (cavity brickwork

with ties)

0.00001


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Implications for composite

construction

TL = 10 log (ATOT/(A11+ A22++Ann)) dB

The presence of a relatively small area of

construction with a high transmission coefficientwill seriously reduce the overall sound insulation


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An example

The incorporation of air gaps to the extent

of 1% of the area of a concrete block wall

which should, if properly built, have an

insulation of 50 dB, will reduce this to 20dBthe equivalent of cheaply fitted single

glazing!


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Sound transmission characteristics of a partition:

The mass law


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Resonances

Prediction of resonant frequencies:

F = 0.45Vh((Nx/Lx)2+ (Ny/Ly)2) Hz

F = resonant frequency (Hz)

V = velocity of sound (m/s)

H = thickness of element (m)

Nx, Ny are integers1,2,3 etc Lx = width (m)

Ly = height (m)


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An example

Calculate the lowest resonant frequency

for 6 mm single glazing which is 2 m high

and 1m wide:

F = 0.45x5300x0.006x((1/1)2+ (1/2)2) Hz

= 18 Hz

Second and third resonant frequencies

are: 72 Hz and 161 Hz


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Critical frequency(the lowest frequency at which coincidence occurs)

Critical frequency = VA2/1.8hVL Hz

VA= velocity of sound in air (340 m/s) H = thickness of element (m)

VL = longitudinal velocity of sound through

the element (m/s)


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An example

Calculate the critical frequency for a 100

mm brick wall. The longitudinal velocity of

sound through brickwork is 2350 m/s

Critical frequency = 3402/ (1.8 x 0.1 x 2350) Hz

= 273 Hz


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Thank you

Documents

Basics of Noise