Reverberation Time

Much attention has been paid to the acoustics of concert halls and other enclosed spaces for lectures and the performing arts. Many factors, both physical and psycho-logical, influence the judgment of the acoustic quality of rooms and many descriptors have been introduced and used in an effort to quantify various aspects of this concept.

Systematic work in room acoustics began almost 100 years ago with the pioneering studies by Sabine, then a physics professor at Harvard. For further comments on Sabine, see Section 1.2.2.

We start by deriving Sabine’s formula for the reverberation time in a room. The sound field is assumed to be diffuse which means that at a point in the room sound arrives from all direction with equal probability and intensity. Thus, if the point of observation is surrounded by a spherical control surface, the contribution to the acoustic energy density within the sphere from every solid angle element d on the sphere will be the same and we express it as (I0/c)d, where I0is an intensity and c the sound speed (see Chapter 3). The fact that the field is diffuse means that these contributions are all uncorrelated so that their energies add. Then, with a total solid angle of the spherical control surface of 4π, the total energy density becomes

E= 4πI0/c. (6.1)

Again, using the intensity I0, we can express the acoustic power incident on a wall element of unit area in terms of I0as follows. Place a spherical control surface of unit radius with the center at the wall element and consider a wave that strikes the wall at an angle of incidence φ. The solid angle between φ and φ+ dφ is a ring of radius sin φ on the sphere with an area 2π sin φ dφ (see Fig. 4.4). Thus, the total intensity that strikes unit area of the wall at an angle of incidence φ will be I02π sin φ cos φ dφ since the power intercepted by a unit area of the wall will be I0cos φ.

Integrating over the entire control surface, we obtain the total power per unit area, the diffuse field intensity Id,

I_d=

_π/2

0

I₀2π sin φ cos φ= πI0= Ec/4, (6.2)

175

where we have used Eq. 6.1; E is the acoustic energy density introduced above. Thus, the average intensity on the wall in the diffuse field is one quarter of the intensity of a wave at normal incidence with the same energy density E.

We define the diffuse field absorption coefficient αdsuch that Idα_dis the absorbed power per unit area. Thus, with Id = Ec/4 and with the physical area of an absorber on a wall being A, the absorbed power will be (Ec/4)Aαd and the quantity Aαd is the absorption area. The coefficient αdis the same as the angle averaged absorption coefficient given in Chapter 4, Eq. 4.53.

These considerations imply an infinitely extended surface so that edge effects (diffraction) of the absorber can be ignored. In reality, with a finite absorptive wall el-ement, this is no longer true, and the effective absorptive area will be larger than Aαd. It is generally quite difficult to calculate, even for an absorber of simple shape. The actual absorption area will be denoted Asand a corresponding absorption coefficient αs is defined by As == αsA. The absorbed power is then As(E/4c).

In this context of room acoustics, we shall call Asthe absorption area or the Sabine (area) and the corresponding absorption coefficient αs = As/Awill be called the Sabine absorption coefficient. It is this coefficient that is measured by the rever-beration method, to be described below. It can exceed unity, particularly at low frequencies when diffraction effects play a significant role.

With the volume of the room denoted V , the total acoustic energy in the room is EV and the rate of decrease of it must equal the absorbed energy, i.e.,

Decay of energy density in a room V dE/dt= −AsEc/4) i.e.,

E(t )= E(0) e^{−c(As/4V ) t}

. (6.3)

In other words, the energy density and hence the mean square sound pressure in the room decays exponentially, the decay constant being proportional to the total absorption area (cross section) As = Aαs and inversely proportional to the volume.

If the absorption coefficient varies over the area, Ashas to be replaced by the average value.

The reverberation time is defined as the time in which the average sound pressure level in the room decreases 60 dB. It follows from Eq. 6.3 that

10 log[E(0)/E(t)] = cA_s

4V t10 log(e). (6.4)

Thus, with the left side put equal to 60 and the reverberation time denoted Tr, we get

Reverberation time

T_r = 240/[10 log(e)] (V /cAs)≈ 55 V /cAs (6.5) [V : Room volume. As: Absorption area (Sabine) (product of absorption coefficient and absorber area. c: Sound speed.]

Introducing the numerical value for the sound speed c≈ 342 m/s, the numerical expression for the reverberation time becomes Tr ≈ 0.16 V /As, where V and Asare expressed as m³and m², respectively.

For example, for a cubical room with a side length L, the internal surface area is 6L², and the reverberation time becomes Tr = 0.027 L/αs. Thus, with L = 10 m and a Sabine absorption coefficient of 10 percent, the reverberation time becomes 2.7 seconds.

A reverberation room in an acoustical testing facility typically may have a reverber-ation time of about 10 seconds at low frequencies (200 Hz, say). Among the quantities that have been proposed and used for the description of the acoustics of a room, the reverberation time is still regarded as a primary parameter. For example, there is good correlation between the intelligibility of speech in a lecture room and the rever-beration time so that an optimum value can be established. Such a correlation can be determined by the fraction of randomly selected spoken words from the podium that can be understood by a listener in the audience. The optimum reverberation time depends on the room size, but typically is about one second.

To determine an optimum reverberation time for music is a more subjective matter and depends on the character of the music; typically this optimum is between 1 and 3 seconds.

The expression for the reverberation time in Eq. 6.5 accounts only for the ab-sorption at the walls (by acoustic treatment) and formally goes to infinity when the absorption coefficient of the absorptive material on the walls goes to zero. In reality, even a rigid, impervious wall yields some absorption because of visco-thermal losses, and sound transmission through the wall is equivalent to absorption. From the dis-cussion in Section 4.2.4, the diffuse field average absorption coefficient of a rigid wall can be shown to be

α≈ 1.7 × 10⁻⁴

f , (6.6)

where f is the frequency in Hz.

There is also sound absorption throughout the volume of the room due to visco-thermal and molecular relaxation effects. With the spatial decay of the intensity expressed as I0 ∝ exp(−βx), the corresponding temporal rate of decrease of the energy density in a diffuse sound field in a room, following Eq. 6.1, will be βcE. This means that the right-hand side of Eq. 6.3 has to be replaced by−Ec(βV + αA/4), where α is the visco-thermal absorption coefficient and A the total wall area.

From Eq. 6.3 it follows that the decay of the acoustic energy density due to visco-thermal losses at the walls and losses throughout the volume of a room is given by

E(t )= E(0)e−(β+αA/4V )ct. (6.7) For sufficiently large rooms and high frequencies, the absorption in the volume of the room dominates.

In regard to numerical results, we already have an expression for the frequency dependence of the absorption coefficient in Eq. 6.6. For β, we refer to the discussion in the chapter on atmospheric acoustics, in particular Fig. 10.2, where, at a temper-ature of 20^◦C, the wave attenuation is plotted in dB/km as a function frequency. For relative humidities less than 50 percent, the vibrational relaxation effects (of Oxy-gen) dominate, and at 50 percent, the attenuation can be written approximately as (f/1000)²dB/km. The corresponding expression for β is then

β≈ 2.29 (f/1000)²10⁻⁴ m⁻¹, (6.8)

where we have used 10 log(e)≈ 4.36.

The condition that the two absorption effects contribute equally to the decay is β= αA/4V , and with the numerical values given in Eqs. 6.6 and 9.16, this corresponds to f/1000 = 3.27(A/V )^2/3, where the length unit is 1 meter. For a cubical room with a side length L, this means f/1000≈ 10.9(1/L)^2/3. Thus, with L= 10 m, the volume absorption will exceed surface absorption at frequencies above 1000 Hz.