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 vp, 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 1800 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 ux=vp=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 bo 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 vp=-uo, 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, Mp 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, vp=-uo are:
The condition relating κ and Mp 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 ωoM p/γ where ωo 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 ρ1≤ ρ2/10, 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 ap and av. From the plus-infinity-incidence case we have ap+av=1. The response in the minus-infinity-coaxial-incidence case is ap-av. Thus by solving for the latter case we can determine both ap and av.
Problem SolutionIn addition to the planar wave, an infinite series of cylindrical waveguide modes can propagate in each tube section (although most waves are highly attenuated). The velocity and pressure can be derived from the potential function described in the previous section, which for our problem is of the form
J0 and Y0 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 J0 function can be present, since the Y0 function is infinite at ρ=0. The radial velocity component must be zero at the wall, so βn is specified such that ρ1βn, or ρ2βn, is the nth root of the first derivative of J0, for the small or large tube respectively. Thus the potential function is a modal series with discrete values of βn, 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 an/bn and βn are jointly specified such that the first derivative of the combined J0 and Y0 functions are zero at both ρ1βn and ρ2βn. 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 ap/av. 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 ρ1/2 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.
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