**5. Coupling Power out of a Plane
Wave**

The primary motivation for this section is to shed light on how a human eardrum receives sound. The complete real-world problem is formidable, so we take the time-honored approach of analyzing simple models that provide insight into the basic mechanisms involved.

The first model is a piston inside a tube, which
is a crude model of the eardrum inside the ear canal. The second model is a
small open-ended tube inside a large tube. This is an even cruder model of the
way the ear canal would receive sound from unbounded space, since there is not
even a rudimentary representation of the head. However the model still provides
some interesting results. For example, in the section on acoustics
of small rooms, it is tacitly assumed that at a pressure null no sound can
be heard. But for a standing wave a __pressure__ null p=0 coincides with
a __velocity__ **u**=maximum. The ear responds to both, so you
will in fact hear something at a pressure null - but how much? The second model
addresses this question.

**Analysis of a Power Coupling to a Piston in a
Tube**

A piston in a tube, harmonically vibrating with peak
velocity v_{p}, delivers energy to a gas at a rate given by the product
of the piston velocity and the pressure on the piston face. If the pressure
is 180^{0} out of phase with the velocity, then the product is negative,
and the piston receives energy from the gas.

To quantify the reception of sound power, we start with a piston location at x=0, a piston velocity of zero, and a sound wave in the tube generated by a source at x=∞ . When this wave hits the piston, it will be perfectly reflected, and will create a standing wave in the tube, which, for x ≥ 0 we write as

The piston geometry is shown below.

The boundary condition at the piston face at x=0
is u_{x}=v_{p}=0, which is satisfied by equation (39). Assuming
that the piston excursion is much smaller than the wavelength, we can take x=0
as the location of the piston face even when it is moving. (An analysis which
includes the effect of piston excursion can be found
here). We also assume that the static pressure is the same on both sides
of the piston face, so the net force is due only to the variable sound pressure.
In this case the constant b_{o}
velocity term discussed in the previous section, cancels out in equation
(39), and will not contribute to the RMS power. Thus it can be omitted from
the wave generated by the moving piston. The generated wave can be superimposed
on the incoming wave, creating a total wave for x ≥ 0

Then the product of the piston velocity and pressure at the piston face is

__Maximum Power Transfer from the Wave to the
Piston__

Positive W means the piston is providing energy to
the gas, and negative W means it is receiving energy. The minimum of the RMS
value of equation (41) occurs when v_{p}=-u_{o}, when the RMS
value is:

So under these conditions the piston wave is exactly canceling the reflected wave, and all of the energy flux of the incoming wave is being absorbed by the piston.

**Piston Conditions
for Maximum Power Coupling**

The general equation of motion for the piston illustrated above is:

where p is the pressure at the piston face, M_{p} is the piston mass
per unit area, γ is the frictional force coefficient, and κ is the spring
constant, both also normalized with respect to area. The conditions under which
the piston velocity will match the value for power absorption, v_{p}=-u_{o
}are:

The condition relating κ and M_{p} means that the resonant frequency
of the piston equals the frequency of the incoming wave, and the condition on
γ is the condition for a "matched load," as it is called in electrical engineering.
(I wonder how close the human eardrum comes to this optimum condition).

A little more than a year after writing the above sentence I found out. Voss
and Allen have measured the eardrum reflection coefficient of 10 people,
and at 5 kHz the eardrum typically absorbs 75% of the incident sound energy.
The Q of the piston is ω_{o}M _{p}/γ where ω_{o}
is the resonant frequency. The lower the Q value, the broader the frequency
range over which power is absorbed efficiently. For Q=2.4 the coupling loss
of a piston with a resonant frequency of 4 kHz is shown
here [42 kb]. A comparison of this graph to a graph of sensitivity of human
hearing, the lowest curve of the Fletcher-Munson
contours [102 kb], reveals quite a bit of similarity.

**Analysis of Power Coupling to an Open-ended Tube**

The second model we consider is shown
here [1.6kb]. The tube cross-sections are circular, and the wall thickness
of the small tube is assumed to be negligible. The small tube extends from z=-∞
to 0, and the large tube spans z=±∞. As long as the small tube radius satisfies
ρ_{1}≤ ρ_{2}/10, the response of the small tube
is very nearly the same as if it were in unbounded space.

If a planar wave of sound pressure exp(jkz) is incident from plus infinity, the same planar wave will exist in the small tube, and in the coaxial section, for z ≤ 0. So there is a uniform planar wave everywhere. The non-trivial case is for a wave in the opposite direction exp(-jkz) within the coaxial section, incident from minus infinity.

Let the wave amplitude induced in the small tube
by pressure, and velocity, respectively, be denoted a_{p} and a_{v}.
From the plus-infinity-incidence case we have a_{p}+a_{v}=1.
The response in the minus-infinity-coaxial-incidence case is a_{p}-a_{v}.
Thus by solving for the latter case we can determine both a_{p} and
a_{v}.

**Problem Solution**

J_{0} and Y_{0 }are Bessel functions of the first and second
kind. (The zero subscript is the only order that can exist in the present case
where the pressure is axially symmetric; in the completely general case solutions
of order m, with a exp(±;jmθ ) axial variation term, must also be
included).

First consider the small tube for z ≤ 0, or the large tube for z ≥
0. Only the J_{0} function can be present, since the Y_{0} function
is infinite at ρ=0. The radial velocity component must be zero at the wall,
so β_{n} is specified such that ρ_{1}β_{n},
or ρ_{2}β_{n}, is the nth root of the first derivative
of J_{0}, for the small or large tube respectively. Thus the potential
function is a modal series with discrete values of β_{n}, rather
than a continuous spectrum of β values which can exist in unbounded space,
as described in the section
on wave spectra.

For the coaxial section, the ratio a_{n}/b_{n} and β_{n}
are jointly specified such that the first derivative of the combined J_{0}
and Y_{0 }functions are zero at __both__ ρ_{1}β_{n}
and ρ_{2}β_{n}. Matlab programs were developed to find
all of the required ratios and roots.

Three series are used; one for the small tube for z ≤ 0, one for the large tube for z≥ 0 and one for the coax section for z ≤ 0. The additional boundary conditions at z=0 are that the potential function and its derivative with respect to z are both continuous - which is the same as saying that pressure and the z-component of velocity are continuous. So each boundary point creates two equations. A solution for the coefficients can be obtained via a matrix equation. It was determined that 500 terms each in the coax and large tube series, and 50 in the small tube series, provides an accurate solution; 1000 boundary points were used (thus 2000 equations), and a pseudo-inverse solution was obtained for the 1050 unknowns. A typical solution is illustrated here for the real part [60kb], and here for the imaginary part [58kb]. The wavelength is 100 times the radius, and a close-up of the interesting region is shown. The real part has a maximum of .99 at z=.05, and the magnitude of the imaginary part has a minimum value of .08 slightly closer to the tube.

**Relative Responses to Pressure and Velocity**

The response to pressure, compared to the response
to velocity, is the ratio a_{p}/a_{v}. Ratios were determined
for various wavelengths, and are plotted
over the curve of the ratio of pressure to velocity computed for an elementary
spherical wave, as discussed in the next section. It is seen that the two curves
are similar, as expected. For wavelengths up to 10 times the tube radius the
response to velocity is comparable to the response to pressure; for wavelengths
more than 100 times the tube radius the pressure response dominates.

**Response to a Standing Wave Pressure Null**

Assume that the wavelength is large compared to the
radius of the small tube, and consider the following process: (1) with the small
tube absent, block the large tube at z=∞ such that a standing wave is created,
with a perfect pressure null at z=0; (2) unblock the large tube, insert the
small tube, and note the pressure in the small tube; (3) re-block the large
tube. Result: the pressure in the small tube drops significantly, but it does
__not__ go to zero.

However, if the null in step 1 is about ρ_{1}/2
in front of the small tube, then the pressure in the small tube does drop very
close to zero. In other words, perfect pressure nulls do produce near-perfect
silence, but at a slightly different location than one might expect.

This is the end of the primary part of the analysis. For completeness, the next section very briefly presents the equations for a spherical wave solution of the wave equations.