EE1.el3 (EEE1023): Electronics III

Acoustics lecture 18

Room acoustics

Dr Philip Jackson

www.ee.surrey.ac.uk/Teaching/Courses/ee1.el3

Room acoustics

Objectives:

• relate sound energy density to rate of absorption

• compare direct and reflected sound levels

• measure and calculate reverberation time

• explain differences between free and diffuse sound fields

• define an expression for a room’s critical distance

Topics:

• Direct and reflected sound echoes

• Sound energy stored in a room

• A room’s characteristic measurements

• Worked examples

R.1

Preparation for room acoustics

• What is the absorption coefficient α?

– find a definition

– give two examples of materials

with different values of α

• How can we characterise the amount

of reflection or absorption in a room?

– find out what is the meaning

of reverberation time

– how can it be measured?

R.2

Reflections in a room

Direct sound

• original waveform from source

Early reflections

• reinforce direct sound• inform on room size & shape

Reverberation (late reflections)

• strengthens room impression

acousticalh(m)filter,source

soundsound

pressure

received

R.3

Echoes

All surfaces both absorb and reflect sound. The absorptioncoefficient α gives the ratio of absorbed to incident sound:

Wab = αWin Wre = (1− α)Win

Echoes need delay >50 ms to be heard distinctly

Parallel reflective surfaces produce flutter echoes:

• large separation → fast repetitions

• medium separation → buzzing

• small separation → ringing

Combination of multiple echoes produces reverberation:

• sound reflected by all surfaces

• from all directions with equal probability

• impulse response exhibits an exponential decay

R.4

Diffuse sound stored in a room

Absorption of sound in a room

Rooms provide means of storing and absorbing sound.

By analysing the effect of sound

in a small volume, δV , on a

small patch of wall, δS, it can

be shown that the intensity of

incident sound depends on the

energy density, E, in the room:

rδV

δS

Iin =Ec4

where E =p2

rms

ρ0c2(1)

With surface area S and average absorption α, the rateof energy absorption by the room determines its absorbedsound power:

Wab =Sα Ec

4(2)

R.5

Calculating sound absorption

Frequency (Hz)Material 125 250 500 1000 2000 4000

concrete .01 .01 .02 .02 .02 .02glass .35 .25 .18 .12 .07 .04wood .30 .25 .20 .17 .15 .10carpet .02 .06 .14 .35 .60 .65fibreglass .08 .25 .45 .75 .75 .65acoustic tile .40 .50 .60 .75 .70 .60audience .40 .55 .80 .95 .90 .85

Table 1. Typical sound absorption coefficients α

For a room 4 m (L) × 3 m (W) × 2.5 m (H) with woodenwalls, acoustic tiled ceiling and carpet on the floor, we cancalculate the total absorption at 1 kHz:

A = (2× (4 + 3)× 2.5× .17) + (4× 3× .75) + (4× 3× .35)

= 5.95 + 9.00 + 4.20 = 19.15 sabin

R.6

Growth of sound in a room

From eq. 2, we obtain a differential equation to describe

the growth of sound energy within a closed volume V :

VdEdt

= Wsrc −Sα Ec

4(3)

⇒4V

Sαc

dEdt

+ E =4Wsrc

Sαc

If the sound source is started at t = 0, solution yields

E(t) =4Wsrc

Sαc

(1− e−

Sαc4V t

)(4)

As t→∞, the energy density reaches equilibrium:

E∞ =4Wsrc

Sαc(5)

R.7

Absorption rate at equilibrium

Energy in a reverberant room acts like a first-order system:

t( )

S α c

srcW4=

ON OFF t

At equilibrium, the rate of energy absorption matches the

input power:

Wsrc = W∞ab =Sα E∞ c

4(6)

R.8

Rate of reverberation decay

If the source is switched off, eq. 3 gives an expression for

the rate of decay of the reverberation:

dEdt

= −Sα Ec

4V(7)

The initial conditions then provide the full equation:

E(t) = E0 exp(−Sαc

4Vt

)(8)

Also, from eq. 1, we relate energy density to sound level:

SPL(t) = 10 log10

(p2

rms

p2ref

)= 10 log10

(E(t)ρ0c

2

p2ref

)

= 10(log10 E(t) + log10 ρ0c

2 − log10 p2ref

)= 10 log10 E(t) + C

= 10(

log10 E0 −Sαc

4Vt log10 e

)+ C (9)

R.9

Reverberation time

The reverberation time, T60, is the time for the SPL in aroom to drop by 60 dB, i.e., when t = T60 in eq. 9:

−10Sαc4V T60 log10 e = −60

T60 =KV

Sα(10)

where K = 24/(c log10 e) ≈ 0.16 sm−1 gives Sabine’s Eq.

Note: T60 can depend on frequency but is otherwise inde-pendent of original sound source R.10

Measurement of reverberation time

Reverberation time is commonly measured with pink noise

from a loudspeaker:

With limited dynamic range, T30 is measured to avoid the

noise floor and doubled to give T60. As reverberation time

varies with frequency, band-limited noise can be used.

R.11

Soundfields in an acoustic environment

Free field, no reflections, sound comes direct from source

Near field (close), more direct sound than reflections

Diffuse field (reverberant field), more reflected sound fromall directions than direct sound

Critical distance (a.k.a. room radius), dc, equal levels ofdirect and reflected sound

R.12

Level of reverberant sound

As in eq. 3, we have the ODE for reverberant energy:

VdErev

dt= (1− α)Wsrc −

Sα Erevc

4

⇒4V

Sαc

dErev

dt+ Erev =

4 (1− α)

SαcWsrc

⇒ Erev(t) =4 (1− α)Wsrc

Sαc

(1− e−

Sαc4V t

)

⇒ E∞rev =4 (1− α)Wsrc

SαcFrom eq. 1, we can obtain the reverberant SPL:

SPLrev = 10 log10

(E∞rev ρ0c

2

p2ref

)

= 10 log10

(4 (1− α)Wsrc ρ0c

Sα p2ref

)

= 10 log10

(4Wsrc

RAIref

)R.13

Critical distance calculation

The critical distance dc is reached when SPLrev = SPLdir.

For direct sound from a point source in free field, we have

SPLdir = SILdir = 10 log10

(Idir

Iref

)= 10 log10

(Wsrc

4πd2c Iref

)

which matches reverberant sound level at critical distance

SPLrev = 10 log10

(4Wsrc

RAIref

)where RA = Sα/(1− α) is the room absorption constant.

Source power Wsrc and reference intensity Iref cancel out,

to yield the final result:

4

RA=

1

4πd2c

⇒ dc =1

4

√RAπ

(11)

R.14

Worked example: factory noise

A machine with a sound power of Wsrc = 1 W is situated ina room that has dimensions 7 m (L) × 5 m (W) × 3 m (H).

1. Treating it as a point source, what is the intensity at adistance of 2 m?

2. The machine is left running continuously and reflectionscontribute to build up a reverberant field. If the absorp-tion coefficient is α = 0.1, what will be the asymptoticenergy density?

3. What is the corresponding pressure level of the rever-berant sound?

4. What changes would occur if the ceiling was treatedwith a material having α = 0.6?

R.15

Room acoustics

• Reflection and absorption of sound in a room

• ODE of sound stored in a room

• Measurement of reverberation time

• Calculation of critical distance

Reference

L. E. Kinsler, A. R. Frey, A. B. Coppens and J. V. Sanders,

“Fundamentals of Acoustics”, 4th ed., Wiley, 2000.

Chapter 12, [shelf 534 KIN]

R.16

Preparation for musical acoustics

• For one instrument from each class, identify what cre-

ates a sound source, and a resonator that enhances it?

– voice

– woodwind

– brass

– percussion

– string

R.17

Appendix: Sound stored in a room

Absorption of sound in a room

Sound from a compact source

impinges on a patch δS and in a

reverberant field:

δW = δS cos θWsrc

4πr2

δE = δS cos θE δV4πr2

(12)

where sound energy density Eis related to the RMS pressure

of the diffuse field:

E =p2

rms

ρ0c2, (13)

assuming the constituents act

like plane waves (eq. 7a on N.6).

δS

r

δV

δSδS cosθ

θr

R.18

Derivation of the reverberant energy density

Considering a hemispherical shell of thickness δr and radius

r centred on δS, we first obtain the energy in a ring:

δV = 2πr sin θ δr rδθ

The total energy in time interval δt = δr/c is:

E =∫dE =

∫δS cos θ

E4πr2

dV

= δSE∫ π/2

0cos θ

2πr sin θ δr r

4πr2dθ

=δS E δr

2

∫ π/2

0cos θ sin θ dθ

=δS E δr

4

∫ π/2

0sin 2θ dθ

=δS E δr

4

[−

1

2cos 2θ

]π/2

0=

δS E δr4

(14)

δS

δV

θr

δSδθ

δr

rθ

R.19

Surface intensity from a diffuse field

From the total sound energy on a small patch of the wall

E =δS E δr

4the rate of incident energy per unit area gives the intensity

Iin =E

δt δS=Ec4

(15)

The result is 1/4 of that for a normally-incident plane wave:

I⊥ = Ec

With total sound absorption of surfaces A =∫α δS = Sα

with area S and average absorption α, the rate of energyabsorption defines the absorbed sound power:

Wab =Sα Ec

4(16)

R.20