Last time, we discussed the unit area (1 m²) of the surrounding wall of a diffuse sound field with energy density E. 2) Acoustic energy incident on it per second I We have explained up to the point where we can obtain it as shown in equation (1).
.................................(1)
E:拡散音場のエネルギ密度
c:音速
Using the energy density in this diffuse sound field, we derive the theoretical formula for reverberation time from the equilibrium equation between the energy incident on the entire surrounding walls of the room and the energy absorbed.
The formula for reverberation time used in standards such as the reverberation chamber method and sound absorption coefficient (for example, JIS A 1409:1998) is Sabine 's formula (2) below.
.................................(2)
T: Reverberation time (seconds)
K = 0.161
V: Room volume (m 3)
A: Equivalent sound absorption area of the room (m²)
However, even experts find it difficult to explain how this simple equation is derived, for example, why the constant K is 0.161. This time, we will explain the process by which equation (2) is derived. First, from equation (1), the energy incident on the entire perimeter wall is obtained by multiplying equation (1) by the total surface area S of the perimeter wall of the chamber,
If the average sound absorption coefficient of the surrounding wall is α, then the energy absorbed by the entire surrounding wall is, further multiplied by α in the above equation:
This is the result. Here, if we denote the output of the sound source (energy supplied to the sound field per unit time) as W, and the change in acoustic energy in the room as V(dE/dt), then this change can be expressed by the following differential equation.
.................................(3)
Let's consider this by dividing it into the growth process, steady state, and decay process of the sound field.
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Figure 1 Time waveforms of sound source and receiver in a room.
Figure 1 shows the time waveforms of a sound source and receiver in a room. The sound source always emits a constant output, and at the receiver, the direct sound arrives first, followed by the reflected sound, and the energy gradually increases. However, since the sound source continues to emit a constant output throughout this process, the energy at the receiver increases further and eventually reaches a steady state. This is called sound growth. After reaching a steady state, if the output of the sound source is stopped, no new energy is supplied, and the energy at the receiver undergoes a decay process. This is reverberation.
(1) Growth process
As an initial condition, if we set E = 0 at t = 0, we can obtain the growth equation. That is, the condition is that the acoustic energy density in the room was 0 until the sound source began emitting sound at t = 0.
Solving the differential equation in (3),
.................................(4)
exp is the exponential function
For example, equation (4) shows the process by which the energy density in a room grows from zero over time.
(2) Steady state
In equation (4), if we set E → E0 (where E0 is the energy density in the steady state) as t → ∞, we obtain equation (5) as the steady-state equation.
.................................(5)
(3) Damping process
Stopping the sound source in a steady state, the differential equation t = 0 in, W = 0, E = E 0 If we set this and solve, the damping equation is (6) It can be found using the formula shown.
.................................(6)
This shows the process by which the energy density decreases exponentially over time from a steady state value E 0 (exponential decay). Equation (6) is the fundamental equation for considering reverberation.
(6) The decay formula in the equation represents the level of decay per unit time. D (Damping) year, (7) It can be expressed as an equation.
.................................(7)
Therefore, the reverberation time T is the time it takes for the sound to decay by 60 dB.
.................................(8)
.................................(9)
This yields Sabine's reverberation formula. Also,
.................................(10)
At a standard temperature of 20°C, K ≈ 0.161
(Excerpt from the email newsletter issued on July 22, 2010)
