Chapter Ten
Sound in enclosures
Abstract
Stationary and standing waves, normal modes and frequencies, and wave patterns in rectangular structures are treated first. The effects of damping materials on the strength of particular modes are demonstrated. Sound-pressure decay curves and the shape of a resonant curve come next. Illustrated is a normal frequency diagram for a rectangular space followed by two-part network for small enclosures. The second part is on sound in large enclosures. Here diffuse sound field, mean free path, rate of sound decay, reverberation time, direct sound, reverberant sound, strength of sound field, and power requirements needed for amplification of speech and music are presented.
Keywords
Acoustic design; Air absorption; Effective area; Impedance; Resonance curve; Reverberant sound; Stationary waves
Part XXX: Sound fields in small regularly shaped enclosures
10.1. Introduction
The study of sound in enclosures involves not only a search into how sounds are reflected backward and forward in an enclosure but also investigations into how to measure sound under such conditions and the effect various materials have in absorbing and controlling this sound. Also of great importance in applying one's engineering knowledge of the behavior of sound in such enclosed spaces is an understanding of the personal preferences of listeners, whether listening in the room where the music is produced or listening at a remote point to a microphone pickup. Psychological criteria for acoustic design have occupied the attention of many investigators and should always be borne in mind. This chapter is confined to physical acoustics.
Two extremes to the study of sound in enclosures can be analyzed and understood easily. At one extreme, we have small enclosures of simple shape, such as rectangular boxes, cylindrical tubes, or spherical shells. In these cases, the interior sound field is describable in precise mathematical terms, although the analysis becomes complicated if the walls of the enclosures are covered in whole or in part with acoustical absorbing materials.
At the other extreme, we have very large irregularly shaped enclosures where no precise description can be made of the sound field but where a statistically reliable statement can be made of the average conditions in the room. This is analogous to a study that a physician might make of a particular man to determine the number of years he will live, as opposed to a study of the entire population on a statistical basis to determine how long a man, on the average, will live. As might be expected, the statistical study leads to simpler formulas than the detailed study of a particular case.
10.2. Stationary and standing waves
One type of small regularly shaped enclosure, the rigidly closed tube, has been discussed already in Part IV. This case provides an excellent example of the acoustical situation that exists in large enclosures.
First, we noted that along the x axis of the tube, the sound field could be described as the combination of an outward-traveling wave and a backward-traveling wave. Actually, the outward-traveling wave is the sum of the original free-field wave that started out from the source plus the outward-going waves that are making their second, third, fourth, and so on round trips. Similarly, the backward-traveling wave is a combination of the first reflected wave and of waves that are making the return leg of their second, third, fourth, and so on round trips. These outward- and backward-traveling waves add in magnitude to produce what is called a stationary wave if the intensity along the tube is zero. If there is some, but not complete, absorption at the terminating end of the tube so that power flows along the tube away from the source (intensity not equal to zero), it is called a standing wave. In the case of complete absorption, we have a traveling or progressive wave.
10.3. Normal modes and normal frequencies
We saw from Eq. (2.70) that whenever the driving frequency is such that sin kl → 0, the pressure in the tube reaches a very large value. That is to say, the pressure is very large whenever
kl=nπ
(10.1)
Then, because
k=2πfc=2πλ,
(10.2)
we have
fn=nc2l
(10.3)
or
lλn=n2
(10.4)
where
n=1,2,3,4,…∞
Eq. (10.3) tells us that the pressure is very large whenever the length of the tube equals some integral multiple of a half wavelength (λ/2).
The condition where the frequency equals nc/2l so that a very large sound pressure builds up in the tube is called a resonance condition or a normal mode of vibration of the air space in the tube. The frequency f
n
of a normal mode of vibration is called a normal frequency. There are an infinite number of normal modes of vibration for a tube because n can take on all integral values between 0 and ∞. We may look on the tube, or in fact on any enclosure, as a large number of acoustic resonators, each with its own normal frequency.
In the closed-tube discussion of Part IV, we made no mention of the effect on the results of the cross-sectional shape or size of the tube. It was assumed that the transverse dimensions were less than about 0.1 wavelength so that no transverse resonances would occur in the frequency region of interest.
If the transverse dimensions are greater than one-half wavelength, we have a small room which, if rectangular, can be described by the dimensions shown in Fig. 10.1. Waves can travel in the room backward and forward between any two opposing walls. They can travel also around the room involving the walls at various angles of incidence. If these angles are chosen properly, the waves will return on themselves and set up stationary or standing waves. Each standing wave is a normal mode of vibration for the enclosure.
In Sec. 7.18, we solve such a rectangular enclosure mathematically and describe exactly the distribution of sound as determined by the strength of a piston source in one of the walls. In this section, however, we shall describe the simplest cases to gain insight into the problem.
The number of modes of vibration in a rectangular enclosure is much greater than that for the rigidly closed tube whose diameter is small compared with a wavelength. In fact, the normal frequencies of such an enclosure are given by the equation
fn=ωn2π=c2√(nxlx)2+(nyly)2+(nzlz)2
(10.5)
where
As an example, let us assume that the z dimension, l
z
, is less than 0.1 of all wavelengths being considered. This corresponds to n
z
being zero at all times. Hence,
fnx,ny,0=c2√(nxlx)2+(nyly)2
(10.6)
Let l
x
=
4
m and l
y
=
3
m. Find the normal frequencies of the n
x
=
1, n
y
=
1, and the n
x
=
3, n
y
=
2 normal modes of vibration. We have
f1,1,0=344.82√116+19=71.8Hz
and
f3,2,0=344.82√916+43=237Hz
The sound-pressure distribution in a rectangular box for each normal mode of vibration with a normal frequency ω
n
is proportional to the product of three cosines:
pnx,ny,nzαcosπnxxlxcosπnyylycosπnzzlzejωnt
(10.7)
where the origin of coordinates is at the corner of the box. It is assumed in writing Eq. (10.7) that the walls have very low absorption. If the absorption is high, the sound pressure cannot be represented by a simple product of cosines.
If we inspect Eq. (10.7) in detail, we see that n
x
, n
y
, and n
z
indicate the number of planes of zero pressure occurring along the x, y, and z coordinates, respectively. Such a distribution of sound pressure levels (SPLs) can be represented by forward- and backward-traveling waves in the room. This situation is analogous to that for the closed tube (one-dimensional case). Examples of pressure distributions for three modes of vibration in a rectangular room are shown in Fig. 10.2. The lines indicate planes of constant pressure extending from floor to ceiling along the z dimension. Note that n
x
and n
y
indicate the number of planes of zero pressure occurring along the x and y coordinates, respectively.
The angles θ
x
, θ
y
, and θ
z
at which the forward- and backward-traveling waves are incident on and reflect from the walls are given by the relations
θx=arctan√(ny/ly)2+(nz/lz)2nx/lx=arccosnxc2lxfn
(10.8)
θy=arctan√(nx/lx)2+(nz/lz)2ny/ly=arccosnyc2lyfn
(10.9)
θz=similarly
(10.10)
For the examples where n
x
=
1, n
y
=
1 and n
x
=
3, n
y
=
2, the traveling waves reflect from the x
=
0 and x
=
l
x
walls at
(θx)1,1,0=arctanlxly=arctan43=53.1°
(θx)3,2,0=arctan2lx3ly=arctan89=41.6°
The angles of reflection at the y
=
0 and y
=
l
y
walls are
(θy)1,1,0=arctanlylx=arctan34=36.9°
(θy)3,2,0=arctan3ly2lx=arctan98=48.4°
The wave fronts travel as shown in (a) and (b) of Fig. 10.3. It is seen that there are two forward-traveling waves (1 and 3) and two backward-traveling waves (2 and 4). In the three-dimensional case, there will be four forward- and four backward-traveling waves.
When the acoustical absorbing materials are placed on some or all surfaces in an enclosure, energy will be absorbed from the sound field at these surfaces and the
sound-pressure distribution will be changed from that for the hard wall case. For example, if an absorbing material were put on one of the l
x
l
z
walls, the sound pressure at that wall would be lower than at the other l
x
l
z
wall and the traveling wave would undergo a phase shift as it is reflected from the absorbing surface.
All normal modes of vibration cannot be excited to their fullest extent by a sound source placed at other than a maximum pressure point in the room. In Fig. 10.2, for example, the source of sound can excite only a normal mode to its fullest extent if it is at a 1.0 contour. Obviously, because the peak value of sound pressure occurs on a 1.0 contour, the microphone also must be located on a 1.0 contour to measure the maximum pressure.
If the source is at a corner of a rectangular room, it will be possible for it to excite every mode of vibration to its fullest extent, provided it radiates sound energy at every normal frequency. Similarly, if a microphone is at the corner of the room, it will measure the peak sound pressure for every normal mode of vibration, provided the mode is excited.
If either the source or the microphone is at the center of a rectangular room, only one-eighth of the normal modes of vibration will be excited or detected because at the center of the room, seven-eighths of the modes have contours of zero pressure. In Fig. 10.2, as an illustration, two out of the three normal modes portrayed have contours of zero pressure at the center of the room. In fact, only those modes of vibration having even numbers simultaneously for n
x
, n
y
, and n
z
will not have zero sound pressure at the center.
Examples of the transmission of sound from a point source to an observation point in a model sound chamber are shown in Figs. 10.4 and 10.5. The curves were obtained using the following equation for the pressure at the observation point (x, y, z):
˜p(x,y,z)=−4ρ0c˜U0lxly∑∞m=0∑∞n=0kcos(mπx0/lx)cos(nπy0/ly)cos(mπx/lx)cos(nπy/ly)kmn(1+δm0)(1+δ0n)×kmnZskρ0ccoskmnz+jsinkmnzcoskmnlz+jkmnZskρ0csinkmnlz
(10.11)
where
kmn=√k2−(mπlx)2−(nπly)2
(10.12)
and k
=
ω/c
=
2π/λ, which is derived in the same way as we derive the 2-port network for a bass reflex enclosure in Part XXIV, except that the rectangular pistons are replaced by a point source of volume velocity
˜U0
at a point (x
0, y
0, l
z
) described by the Dirac delta function
at a point (x
0, y
0, l
z
) described by the Dirac delta function
δ(x−x0)δ(y−y0).
The absorbing material at z
=
0 has a specific impedance Z
s
, which is related to the flow resistance R
f
of the material by
Zs=Rfd3+P0jωd
(10.13)
where d is the thickness for the material, which is subtracted from l
z
. The eightfold increase in the number of modes of vibration that were excited with the source at the corner over that with the source at the center is apparent. It is apparent also that the addition of sound-absorbing material decreases the height of resonance peaks and smoothes the transmission curve, particularly at the higher frequencies, where the sound-absorbing material is most effective.
(x,y,z)/(ρ
0
c
˜U
0)), where
˜p
(x,y,z) is calculated from Eq. (10.11). This result has also been verified experimentally [1].
10.4. Steady-state and transient sound pressures
Sound pressure at normal modes
When a source of sound is turned on in a small enclosure, such as that of Fig. 10.1, it will excite one or more of the stationary-wave possibilities, i.e., normal modes of vibration in
the room. Let us assume that the source is constant in strength and is of a single frequency and that its frequency coincides with one of the normal frequencies of the enclosure. The sound pressure for that normal mode of vibration will build up until the magnitude of its rms value (averaged in time and also in space by moving the microphone backward and forward over a wavelength) equals [1]
|pn|=Kkn
(10.14)
where
- K is a source constant determined principally by the strength and location of the source and by the volume of the room.
- k n is damping constant determined principally by the amount of absorption in the room and by the volume of the room. The more absorbing material that is introduced into the room, the greater k n becomes, and the smaller the value of the average pressure. The value of k n is inversely proportional to the value of Q n .
Blocked-tube impedance and equivalent circuit
To illustrate what happens when the driving frequency does not necessarily coincide with the normal frequency, we shall simplify the problem by considering only those modes of vibration which occur in one direction only. Hence, we may model the room as a one-dimensional tube. Furthermore, although absorption mainly occurs at boundary surfaces, we may simplify the problem even further by assuming that it occurs everywhere. Also, we assume the acoustic resistance to have the same value at all frequencies, although this is unlikely in practice. However, if the variation of resistance with frequency is known, the resistance value at each normal frequency may be used to improve accuracy.
ZT=−jZscotkl
(10.15)
which is expanded using Eq. (A2.43) from Appendix II:
ZT=−jZs∑∞n=0(2−δ0n)kl(kl)2−n2π2
(10.16)
where from Eqs. (2.115), (2.119), and (2.120), the complex wave number
k
and characteristic impedance Z
s
are given by
k=ω√1P0(ρ0+Rfjω)
(10.17)
Zs=√P0(ρ0+Rfjω)
(10.18)
where P
0 is the static pressure, ρ
0 is the density of air, and R
f
is the flow resistance per unit length of the filling material. Hence, the impedance of the tube may be written as
ZT=1C0s+∑∞n=1Zn
(10.19)
where each impedance term is represented by a parallel resonance circuit in which
Zn=1Cn·s+RnLns2+RnLns+1LnCn
(10.20)
where s
=
ω, and the specific compliance C
n
, mass L
n
, and resistance R
n
element values are given by
C0=lP0,Cn=l2P0,Ln=2ρ0ln2π2,Rn=2Rfln2π2
(10.21)
or
Zn=1Cn·s+ωnQns2+ωnQns+ω2n
(10.22)
where the angular normal frequency ω
n
and Q
n
values are given by
ωn=1√LnCn=nπc1/√γ,Qn=ωnρ0Rf=nπρ0c√γRfl
(10.23)
The equivalent circuit for a blocked tube using this impedance expansion is shown in Fig. 10.6a.
Alternatively, we may use the expansion of Eq. (A2.42) from Appendix II for the admittance:
ZT=−jZstankl=(∑∞n=0Yn)−1
(10.24)
Yn=1Lnss2+RnLns+1LnCn
(10.25)
and
Cn=2l(n+12)2π2P0,Ln=ρ0l2,Rn=Rfl2
(10.26)
so that
ωn=1√LnCn=(n+12)πcl√γ,Qn=ωnρ0Rf=(n+12)πρ0c√γRfl
(10.27)
The equivalent circuit for a blocked tube using this admittance expansion is shown in Fig. 10.6b. In general, we use the impedance expansion to calculate the time response of the pressure as a function of an input velocity and the admittance expansion to calculate the time response of the velocity as a function of an input pressure.
Open-tube impedance and equivalent circuit
Although we shall only consider the decay of sound in a blocked tube, the equivalent circuit of an open tube is derived here just for completeness as it is frequently encountered in the field of acoustics.
ZT=jZstankl
(10.28)
which is expanded using Eq. (A2.42) from Appendix II:
ZT=∑∞n=0Zn
(10.29)
where
Zn=1Cn·s+RnLns2+RnLns+1LnCn
(10.30)
and
Cn=l2P0,Ln=2ρ0l(n+12)2π2,Rn=2Rfl(n+12)2π2
(10.31)
or
Zn=1Cn·s+ωnQns2+ωnQns+ω2n
(10.32)
where
ωn=1√LnCn=(n+12)πc1/√γ,Qn=ωnρ0Rf=(n+12)πρ0c√γRfl
(10.33)
The equivalent circuit for an open tube using this impedance expansion is shown in Fig. 10.7a.
Alternatively, we may use the expansion of Eq. (A2.43) from Appendix II for the admittance:
ZT=jZscotkl=(1L0s+R0L0+∑∞n=1Yn)−1
(10.34)
where
Yn=1Lnss2+RnLns+1LnCn
(10.35)
Cn=2ln2π2P0,L0=ρ0l,Ln=ρ0l2,R0=Rfl,Rn=Rfl2
(10.36)
so that
ωn=1√LnCn=nπcl√γ,Qn=ωnρ0Rf=nπρ0c√γRfl
(10.37)
The equivalent circuit for an open tube using this admittance expansion is shown in Fig. 10.7b.
Resonance curve
When the driving frequency does not coincide with the normal frequency, the pressure for that particular mode of vibration builds up according to a standard resonance curve as shown in Fig. 10.8. The maximum value of the resonance curve is given by
Zn|ω=ωn=(Qn+j)QnRn≈Q2nRn,Qn≥3
(10.38)
The width of the resonance curve at the half-power (3
dB down) points is equal to Ref. [2].
f″−f≈fnQn
(10.39)
When driven by an excitation velocity u
0, the magnitude of the sound pressure p
n
for a single mode as a function of frequency is given by
|pn|=u0|Zn|=u0Cn√Q2nω2+ω2nQ2n(ω2n−ω2)2+ω2nω2
(10.40)
where ω is the angular driving frequency and ω
n
is the angular normal frequency given approximately by Eq. (10.5).
Obviously, if the driving frequency lies between two normal frequencies or if k
n
is large so that the resonance curve is broad, more than one normal mode of vibration will be excited significantly, each to the extent shown by Eq. (10.40). Because the phase above a normal frequency is opposite to that below it, there will be a cancellation at some frequency between a pair of adjacent normal frequencies, leading to a minimum impedance value. These minimum impedance frequencies correspond to the resonance frequencies ω
n
in the admittance expansion.
Transient response
When the source of sound is turned off, each normal mode of vibration behaves like an electrical parallel resonance circuit in which energy has been stored initially. The pressure for each normal mode of vibration will decay exponentially at its own normal frequency as shown in Fig. 10.9. To simulate the decay of sound, let us apply an impulse to our tube model, rather like a hand clap in a room. We simply take the expression for the impedance of each mode given by Eq. (10.22) and apply the inverse Laplace transform given by Table 6.2 in Sec. 6.17 to obtain
pn(t)=u0Zn(t)=u0Cne−ωntcosθnsin(θn+ωntsinθn)sinθn
(10.41)
where cos θ
n
=
1/(2Q
n
). If only one mode of vibration is excited, the decay is as shown in Fig. 10.9a. Stated differently, on a log p
n
scale versus time, the magnitude of the rms SPL decays linearly with time.
If two or more modes of vibration are decaying simultaneously, beats will occur because each has its own normal frequency (Fig. 10.9b). However, as we superimpose an ever greater number of modes, the waveform becomes a series of impulses (Fig. 10.9c), as we would expect, due to the original impulse being reflected at each end of the tube and thus making multiple round journeys along it. In a real room, as opposed to a simple one-dimensional tube, early reflections would behave in a similar manner, being distinct and thus specular in nature. However, later reflections resulting from random reflections of multiple surfaces tend to cluster together and are termed diffuse.
In this illustration, each mode has the same decay constant (ω
n
/2Q
n
=
R
f
/2ρ
0) because the specific flow resistance per unit length R
f
has been assumed to be independent of frequency. However, it is very possible that each will have its own decay constant, dependent on the position of the absorbing materials in the room.
In actual measurements of sound in rooms, it is quite common to use fast Fourier transforms (FFTs) to create waterfall plots of the sound pressure decay against both time T and frequency f, which in this case is obtained as follows:
pn(f,T)=T+δt∫T−δt(0.54−0.46cos(t−T+δtδtπ))pn(t)ej2πftdt
(10.42)
where the integration is performed over a sliding interval or “window” of width 2δt centered on the time of interest T. The term in parenthesis is the Hamming window function, which minimizes any unwanted frequency components that may otherwise appear in the spectrum because of the finite integral limits. In this way, we can plot the variation of the frequency spectrum with time and thus see how the individual normal modes of vibration decay relative to each other, as shown in Fig. 10.10.
In summary, we see that when a sound source of a given frequency is placed in an enclosure, it will excite one or more of the infinity of resonance conditions called normal modes of vibration. Each of those normal modes of vibration has a different distribution of sound pressures in the enclosure, its own normal frequency, and its own damping constant. The damping constant determines the maximum height and the width of the steady-state sound-pressure resonance curve.
In addition, when the source of sound is turned off, the sound pressure associated with each mode of vibration decays exponentially with its own normal frequency and at a rate determined by its damping constant. The room is thus an assemblage of resonators that act independently of each other when the sound source is turned off. The larger the room and the higher the frequency, the nearer together will be the normal frequencies and the larger will be the number of modes of vibration excited by a single-frequency source or by a source with a narrow band of frequencies.
10.5. Examples of rectangular enclosures
Example 10.1. Determine the normal frequencies and directional cosines for the lowest six normal modes of vibration in a room with dimensions 5
×
4
×
3
m.
Solution. From Eq. (10.5), we see that
f1,0,0=348.82×15=34.9Hzf0,1,0=348.82×14=40.4Hzf1,1,0=348.82√125+116=55.8Hzf2,0,0=348.82×25=69.8Hzf2,1,0=348.82√425+116=82.3Hzf0,0,1=348.82×13=58.1Hz
(1,0,0)mode:θx=0;θy=90°;θz=90°(0,1,0)mode:θx=90°;θy=0°;θz=90°(1,1,0)mode:θx=arccos348.82×5×55.8=51.3°θy=arccos348.82×4×55.8=38.6°θz=90°(2,0,0)mode:θx=0;θy=90°;θz=90°(2,1,0)mode:θx=arccos2×348.82×5×82.3=32.1°θy=arccos348.82×4×82.3=58.0°θz=90°(0,0,1)mode:θx=90°;θy=90°;θz=0°
Example 10.2. A rectangular room with dimensions l
x
=
3
m, l
y
=
4
m, and l
z
=
5
m is excited by a sound source located in one corner of the room. The SPL developed is measured at another corner of the room. The sound source produces a continuous band of frequencies between 450 and 550
Hz, with a uniform spectrum level, and a total acoustic power output of 1
W. When the sound source is turned off, a linear decay curve (log p vs. t) is obtained, which has a slope of 30
dB/s. (a) Determine graphically the number of normal modes of vibration excited by the source; (b) determine the approximate angle of incidence of the traveling-wave field involving the walls at x
=
0 and x
=
l
x
in each of the principal groupings of normal frequencies shown in the graphical construction.
Solution. (a) A graphical solution to Eq. (10.5) is given in Fig. 10.11. The frequency of any given normal mode of vibration is the distance from the origin of coordinates to one of the black spheres shown. That frequency will be made up of three components given
by cn
x
/2l
x
, cn
y
/2l
y
, and cn
z
/2l
z
. Notice that along the vertical coordinate, the normal frequencies occur in increments of 348.8/6 along the right-hand axis in increments of 348.8/8 and along the remaining axis in increments of 348.8/10. On the layer labeled N
0, there are 53 normal frequencies. The total number of normal frequencies between 450 and 550
Hz for this room is 507. The average frequency is 500
Hz.
Solution. (b) The θ
x
angles of incidence can be divided into 10 principal groups as shown in Fig. 10.11. The angles are as follows:
θx(0,ny,nz)≈90°θx(1,ny,nz)≈cos−1(3456·1500)≈83°θx(2,ny,nz)≈cos−1(2.3456.500)≈77°θx(3,ny,nz)≈cos−1(0.345)≈70°θx(4,ny,nz)≈cos−1(0.46)≈63°θx(5,ny,nz)≈cos−1(0.575)≈55°θx(6,ny,nz)≈cos−1(0.69)≈46°θx(7,ny,nz)≈cos−1(0.805)≈36°θx(8,ny,nz)≈cos−1(0.92)≈23°θx(9,ny,nz)≈cos−1(0.995)≈6°
Part XXXI: Sound in large enclosures
10.6. Basic matters
When a sound source, having components that extend over a band of frequencies, radiates sound into a large irregular enclosure, a microphone that is moved about will experience fluctuations in sound pressure. The maxima and minima of these fluctuations will lie much closer together in such an enclosure than in a small or regular enclosure because there are a large number of room resonances in all bands except for the very lowest frequency bands. Thus, in these enclosures, the mean square sound pressure can be determined by moving the microphone back and forth over a short distance. The sound field is largely a superposition of plane waves traveling in all directions with equal probability. This condition is called a diffuse sound field. To avoid the influence of the direct sound, this condition is experienced at a reasonable distance from the source.
The number of reflections from surfaces in such a room per second is equal to c/d, where d is the mean free path of the wave and c is the speed of sound. By actual measurements in rooms of varying shapes and sizes, it has been found that mean free path is equal to
d=4VSm
(10.43)
where V is the volume of the room in m3 and S is the total area of the surfaces of the room in m2. If, after establishing a steady-state sound field, the source of sound is turned off, the sound energy stored in the enclosure will decrease with each reflection (See Fig. 10.12) according to
D(n)=D′(1−α)n
(10.44)
where D′ is the steady-state energy density before the source was turned off, n is the number of reflections that have occurred, and α is the sound absorption coefficient, which is taken to be averaged for all angles of incidence. By replacing n with ct/d = (cS/4V)t, the decay formula is
D(t)=D′(1−α)(cS/4V)t=D′e−(cS/4V)(−ln(1−α))t
(10.45)
where ln is the logarithm to the base e. In a reverberant sound field, the energy density is proportional to the mean square sound pressure. Hence,
p2av(t)=p2av(0)e−(cS/4V)(−ln(1−α))t
(10.46)
Because 10 log10 of the exponential function equals
10(cS/4V)(log10(1−α))t,
−10cS4Vlog10(1−α)dB/s
(10.47)
10.7. The reverberation equations
The reverberation time of the enclosure is defined as the time required for the SPL to fall 60
dB. Thus, the well-known Eyring equation [3], which gives the reverberation time T for an energy drop of 60
dB (often denoted by T
60), is obtained from Eq. (10.47), with S = S
tot
=
∑S
i
, where S
i
’s are areas of particular surfaces in the room, such as audience area and ceiling area:
T=24V−cStotlog10(1−αey)s
(10.48)
where V is the volume of room in m3, S
tot
is the area of all surfaces in the room, and α
ey
is the average sound absorption coefficient for the surface S
i
as shown in Fig. 10.12. The Eyring equation is usually presented with either the natural logarithm or log10 in the denominator and with c taken as 343.5
m/s at 20°C so that
T=0.161VStot(−ln(1−αey))=0.161VStot(−2.30log10(1−αey))s
(10.49)
Note that if the surfaces are perfectly absorbing, i.e., α
ey
=
1.0, the reverberation time T goes to zero.
The Sabine equation [4] was derived by Wallace Sabine from measurements he made in a number of rooms at Harvard University:
T=0.161VStotαtots(metricunits)
(10.50)
T=0.049VStotαtots(Englishunits)
(10.51)
Note that, in the Sabine equation, T only goes to zero if α
tot
approaches infinity. Even today, most published data on acoustical materials and the absorption of audiences and the like have been obtained using the Sabine equation, partly because the formula is simpler to use and partly because for α
ey
less than 0.26, α
tot
is decreasingly less than 0.3.
It is possible to derive the absorption coefficients in one equation from the absorption coefficients in the other equation [5]. In the Sabine equation, let
αtot=∑αs,iSiStot
(10.52)
where α
s,i
is the Sabine absorption coefficient for a particular area S
i
and S
tot
=
ΣS
i
.
In the Eyring equation, let
αey=∑αe,iSiStot
(10.53)
where α
e,i
is the Eyring absorption coefficient for a particular area S
i
.
Then, we find
αeyαtot=∑αe,iSi∑αs,iSi
(10.54)
Hence,
αe,i=(αey/αtot)αs,i.
(10.55)
10.8. Air absorption
As a sound wave travels from one reflection to another in a room, some energy is lost in the air itself. Such absorption in all but very large rooms is appreciable only at frequencies above 1000
Hz. When the reverberation equations are corrected to account for air absorption, they read as follows.
Eyring equation, metric units:
T=0.161VStot(−ln(1−αey))+4mV=0.161VStot(−2.30log10(1−αey))+4mV.
(10.56)
Sabine equation, metric units:
T=0.161TStotαtot+4mV,
(10.57)
where m is the energy attenuation constant in units of reciprocal length. Measured values of 4m under some typical atmosphere conditions are shown in Table 10.1.
10.9. Total steady sound pressure level
We are now in a position to incorporate the direct sound field from a source into the energy equations and calculate the total steady-state SPL.
Table 10.1
Direct steady-state sound pressure
The space average sound pressure in a room (determined by moving a microphone back and forth over at least one wavelength) at a distance r from a small directional source radiating W watts is
p2(r)=ρ0cW4πr2QN2/m4,
(10.58)
where Q is the directivity index (not in decibels) (see Sec. 4.16).
Reverberant steady-state sound pressure
The sound power absorbed by the first reflection is Wα; hence, the power remaining for the reverberant field is W
r
=
W(1
−
α). Let t′ be the length of time it takes for the sound to travel one mean free path length:
t′=4VcSs.
(10.59)
Let the steady-state value of the reverberant energy density be D
r
′. Then, the total energy per second removed from the room is
D′rVαt′=Wr,
(10.60)
p2r=4ρ0cWSα(1−α)N2/m4.
(10.61)
Total steady-state sound pressure
p2(r)=Wρ0c[Q4πr2+4(1−α)Sα]N2/m4.
(10.62)
The restrictions on this equation are that α not be too large and the mean free path is about 4V/S. The absorption coefficient α is the Eyring coefficient.
10.10. Optimum reverberation time
The following formula [6–11] gives the average optimum reverberation time T for a given auditorium volume V based on subjective results:
log10V=5.72+log10T−2.43√T,
(10.63)
10.11. Sound strength G
It is now customary in auditorium acoustics to express Eq. (10.61) in terms of sound strength G [12]. Sound strength G, in decibels, is the ratio of the sound energy that comes from a nondirectional source (Q
=
1) measured at a distance r in the auditorium to the same sound energy from the same source but measured in an anechoic chamber at r
=
10
m. Thus, the reference sound pressure is
p2ref=Wρ0c4π∗100
(10.64)
G=10log10(100r2+1600π(1−αtot)Stotαtot)dB.
(10.65)
The reason α
tot
is used here instead of α
ey
because it has been found that if T is measured in an actual hall and if Sα is determined from the Sabine formula (T
=
0.161V/Stotα
tot
) and if this value for Sα is used in the G equation to calculate G, the calculated G equals the actual measured values of G in the hall very closely (when using the reverberation method of calibrating the standard dodecahedral source) (see Fig. 10.14). If the Eyring equation is used, this means that the [−2.30 log(1
–
α
ey
)] must be used and not just α
ey
to calculate G. If α
ey
is used, the calculated G will be about 2.5
dB higher than the measured G.
The second term in Eq. (10.65) would seem to indicate that the reverberant sound field is uniform in an auditorium, but SPLs measure larger in the front part of an auditorium than toward the rear (see Section 10.12). This term actually indicates the average of the SPLs determined from measurements at a large number of positions in the auditorium (with r large enough that the first term does not appreciably influence the second).
10.12. Early and reverberant sound in concert halls
It can be shown that the second term of Eq. (10.65) may be divided into two parts, one for early sound (that arriving within 80
ms of the direct sound) and the other for late (reverberant) sound (after 80
ms), both varying with distance r [13]. These equations are
Eearly=31200TVe−0.04r/T(1−e−1.11/T),
(10.66)
Ereverberant=31200TVe−0.04r/T(e−1.11/T).
(10.67)
As an example, these equations, with V
=
20,000
m3 and T
=
2
s, are plotted in Fig. 10.15. Zero on the ordinate is set for the direct sound with r
=
10
m. For r between 10 and 40
m, the top curve predicts the difference in G(total) to be 3.8
dB. Measurements made in nine shoebox-shaped halls, with average V
=
16,500
m3 and T
=
2.5
s, found that for r between 10 and 40
m, G(total) drops about 2
dB, while in 11 surround halls, with average V
=
23,000
m3 and T
=
2.2
s, it drops by about 5
dB. The quantity of 3.8
dB above for V
=
20,000
m3 and T
=
2
s is correctly between these two numbers. In addition, measurements show that the levels drop off faster if the reverberation times are less than about 1.5 s—the drop-off rate significantly increasing (nearer the drop in direct sound level) as RT’s become less than 0.7
s.
10.13. Distance for equality of direct and reverberant sound fields
We will define the distance r
rev
at which the reverberant field takes over as the distance at which the direct and reverberant fields are equal. Hence,
rrev=14√QStotαtotπ(1−αtot).
(10.68)
From Barron M, Lee LJ. Energy relations in concert auditoriums. J Acoust Soc Am 1988;84(2):618–628.
The total absorbent area S
tot
and absorption coefficient α
tot
are both related to the volume of the auditorium. On average [8]
Stot=2.2V2/3.
(10.69)
Let us also assume the reverberation time is the optimum value given by Eq. (10.63) and plotted in Fig. 10.13, and that we have a point source with Q
=
1. From Eqs. (10.48) and (10.50),
αtot=0.161VStotT,αey=1−e−0.161V/(StotT)
(10.70)
We note that r
rev
will be greater for sources which are more directional than a point source (Q
>
1). It is common to use directional loudspeakers such as horns or column arrays in more reverberant spaces where satisfactory speech intelligibility is needed. The reference distance is generally taken as 10
m, which is valid for even the larger concert halls.
10.14. Sound levels for speech and music
When designing a sound system for a specific auditorium, we need to know how much sound pressure is required to produce realistic volumes for music or speech or both. The second column of Table 10.2 shows the maximum peak SPL at 10
m from various sources. However, conversational speech at such a distance is too quiet so the third column gives an SPL value adjusted for a distance of 1
m, which is more natural. The orchestra is adjusted for a distance of 3
m, which represents a good seat a few rows back from the stage. For speech, the crest factor (the difference between the maximum peak SPL and average rms SPL) is about 13
dB. For music, it is about 20
dB.
Knowing the required pressure from the third column of Table 10.2, S
tot
from Eq. (10.69), and α
ey
from Eq. (10.70), we can evaluate the acoustic power required [17] from Eq. (10.62) as follows:
W=4×10(SPL/10)−10ρ0c(Q4πr2ref+4(1−αey)Stotαey)−1,
(10.71)
where r
ref
=
10
m and the SPL value is taken from the third column of Table 10.2. The maximum peak acoustic power is plotted against auditorium volumes in Fig. 10.17. Of
course, the required amplifier output power will depend on the choice of loudspeaker. For example, a living room with a volume of 60
m3 will require a stereo amplifier with a power rating of 6
W per channel to reproduce a large orchestra if the loudspeakers have an efficiency of 1%. If loudspeakers with an efficiency of 10% can be employed, the power rating of the amplifier can be reduced to 0.6
W per channel.
Table 10.2
| Sound source | Maximum peak SPL (dB) at 10 m from source | Maximum peak SPL (dB) adjusted for 1 m (conversational speech) and 3 m (others) |
|---|---|---|
| Conversational speech [15] | 56 | 76 |
| Declamatory speech [15] | 67.5 | 78 |
| Large orchestra [16] | 92 | 102.5 |
Problem 10.1. An auditorium is 12.5 m long by 10 m wide by 8 m high. We assume all surfaces to be perfectly reflective except for those in the table below for which the Sabine absorption coefficients α
s
are given.
| Surface | Area (m2) | αs @ 500 Hz |
|---|---|---|
| Seating (half occupied) | 100 | 0.7 |
| Acoustic ceiling tiles | 125 | 0.6 |
| Drapes | 32 | 0.5 |
Calculate the total volume V of the auditorium, the total absorptive area S
tot
, the average Sabine absorption coefficient α
tot
, the reverberation time T, sound strength G, the distance r
rev
from a source to equal direct and reverberant fields, and the average Eyring absorption coefficient α
ey
. Is the reverberation time optimum for the size of auditorium? Also, calculate the amplifier power needed in a sound system to reproduce the sound of a large orchestra assuming the loudspeakers have an efficiency of 0.5%.
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