**7. Sound Waveguide and Cavity Modes**

The subsections within this section are:

- analysis (immediately below)
- reflecting plane wave model of waveguide modes
- room resonant frequencies
- resonant build-up (room gain)
- mode spatial patterns
- resonances in irregularly shaped rooms
- limitations of modal theory.

**Analysis**

The propagation of sound waves in
tubes of rectangular or circular cross-section is a classic physics problem,
and provides the basis for analyzing the tones produced by organ pipes and for
calculating room resonant frequencies. We assume the harmonic case with time
variation e^{jωt } suppressed. The wave equation for pressure in
this case is:

where k is the free-space propagation constant. For a tube extending an infinite distance in the ±x direction with a cross section as illustrated below,

the pressure has a general solution

where n and m are a pair of integers, including zero, which define the nm-th waveguide "mode." The pressure reaches a maximum at the walls of the tube. The propagation constant β is given by

This solution assumes that resistive losses are negligible.

The wavelength in the waveguide is
λ_{g}=2π/β
which is always greater than the free-space wavelength λ_{0}=2π/k,
except for the 00-th mode, where λ_{g}=λ_{0}.
As frequency decreases, the guide wavelength increases until it becomes
infinite, at a cutoff frequency, which is given by

where c is the speed of sound. Below cutoff the propagation constant β becomes imaginary, and the mode decays rapidly instead of propagating without loss. The 00-th mode has a cutoff frequency of zero.

**Reflecting
Plane Wave Model of Waveguide Modes**

Waveguide modes can be thought of as consisting of sets of plane waves reflecting from the walls. The figure below illustrates the ray-paths of two plane waves.

The superposition of the reflecting plane wave in the top part of the figure and the wave in the bottom part creates a sound wave which is mathematically identical to the nm-th waveguide mode with n=0, and m related to the angle θ by:

Other modes are created by other combinations of plane waves. This is a mathematically rigorous solution of the wave equation, even when the wavelength is larger than b. I have read statements to the effect that an optics-like reflection of a sound wave in a room only occurs for wavelengths much smaller than the dimensions of the room. It is true that an optical-type reflection in general depends on the size of the wavelength, but a rectangular room is an exception. This is an important fact used in the development of image theory in the section following this one.

The main utility of modal analysis
for the purpose of room acoustics is to calculate the resonant frequencies of
a room. A waveguide can be thought of as a tunnel that extends an infinite distance
in the ±x direction. We now introduce two more walls, at x=0 and x=d, to create
a rectangular room. The waveguide modes will reflect from the two new walls.
At discrete frequencies these reflections will add in-phase to create resonances,
and form cavity modes. The required condition is that the room is an integral
multiple of half of the guide wavelength λ_{g}, leading to resonant
frequency solutions for the qnm-th cavity mode

** **

** **

** **

** ****Resonant
Build-up (Room Gain)**

Suppose a subwoofer is radiating
a constant-frequency tone, and the sound wave is bouncing back and forth from
the far wall, a distance d away. If I am sitting in my chair against the far
wall, some sound pressure reaches me directly from the speaker without any reflections;
write this part s_{o}e^{-jβd}. Other sound reaches me after
N round-trip reflections; it has the form:

where a reflection loss α is
now included. Energy is proportional to the square of equation (53), so the
energy loss per round trip is (1-α)^{2}. When all of the sounds
are added coherently, the total sound I hear is:

The summation is (in principle) taken over an infinite number of reflections. For a distance d=5.75 meters, there will be 30 round-trip reflections in one second. If α =0.2, the pressure of the 30-th term will be more than 58 dB below the 1st term. Summing over 30 terms or an infinite number of terms will provide a virtually identical result.

At resonance, e^{-2jβd}=1;
at anti-resonance e^{-2jβd}= -1. Therefore the sound level I hear
will vary between 1/α and 1/(2-α ) as the frequency varies. For α
=0.2 the total at a resonance peak is a factor of 5 in pressure, or 14 dB, above
the direct signal from the subwoofer. This is the resonant build-up, or "room
gain," for this mode series. At anti-resonance the total is 5 dB lower than
the direct signal. The responses for the q00-th cavity mode for a room 5.75
meters long is shown below. The four peaks represent modes with q=0 through
q=3. The figure clearly illustrates why room gain is potentially a major acoustic
problem; the gain occurs in a very spiky pattern, which causes unwanted perturbations
in the frequency response.

Fortunately there are a large number of modes, and each has a different pattern of resonances. The classic attempt to fix this problem is to space out the resonances of the room modes to obtain the smoothest possible overall response. This is discussed in the section on room acoustics.

The section following this one presents the technique of image analysis. From the image analysis viewpoint the room gain computed above arises from a series of image speakers spread along one line perpendicular to the two walls separated by the distance d. There are many other images as well, which correspond to the modes where m and n are not equal to zero, and the total room gain is given by the sum of all of the image contributions.

At the mode resonance frequency, given by equation (52), equation (49) can be used to solve for β . Then substituting into equation (48) yields the mode spatial pressure pattern,

Modes which have zero values for
two of the three indices are called "axial modes," modes which have a zero value
for one index are called "tangential modes," and modes with non-zero values
for all indices are called "oblique" modes. The velocity component **u**
of the mode is in phase quadrature with the pressure, and the three vector components
are related to the gradient of equation (55); See
equation (34). Therefore at a point where there is a pressure null, the
velocity is maximum, and vice versa. So it turns out that at a pressure null
you do actually hear some sound; See Section 5.

**Resonances
in Irregularly Shaped Rooms**

It might be thought (and quite a few people mistakenly think) that designing a room with walls at odd angles would eliminate resonances. The resonances of such rooms have been numerically calculated using finite-element techniques [Handbook for Sound Engineers, Sect. 3.9]. The solutions are two-dimensional, corresponding to walls which are perpendicular to the floor and ceiling, but with oddly-shaped floor plans. Although the modal pressure patterns were irregular, it was found that the resonant frequencies were almost the same as a rectangular room with the same floor area.

Curving the walls also does not work; a spherical cavity, for example, has resonances that can be found by solving the wave equation in spherical coordinates; the modes have characteristics very similar to rectangular modes. See Resonances in a Spherical Cavity.

Simply knowing the resonant frequencies of a room really does not provide a lot of information, because the response in one part of a room is typically very different than another part. To compute the actual sound response in a room using the modal approach, it would be necessary to know the amplitude and phase of each mode, and then coherently combine the mode responses. If a subwoofer is in a corner of the room, it will excite all modes, but not with equal amplitudes. An alternate and easier way to reach a more accurate answer for the room response is image theory, which is the subject of the next section.

In addition, rooms are normally filled with furniture and other mathematically inconvenient stuff, which can have a major influence on the response.