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Wave Impedance in a Cone

From Eq.(C.167) we have that the traveling-wave solution of the wave equation in spherical coordinates can be expressed as

$\displaystyle p(t,x) \eqsp \frac{f\left(t \mp \frac{x}{c}\right)}{x} $

where the minus sign is associated with an expanding spherical wave, and the plus sign corresponds to a converging wave. The spatial derivative of this function is

$\displaystyle p' \eqsp \frac{f'}{x}-\frac{f}{x^2} \eqsp \mp \frac{{\dot f}}{c x}-\frac{f}{x^2} \eqsp \mp \frac{{\dot p}}{c}-\frac{p}{x} \protect$ (C.170)

i.e., it can be expressed in terms of its own time derivative. This is a general property of any traveling wave.

Figure C.48: Cone parameters.
\includegraphics[width=\twidth]{eps/fconeparams}

Referring to Fig.C.48, the area function $ A(x)$ can be written for any cone in terms of the distance from its apex as

$\displaystyle A(x) \eqsp \alpha x^2 $

for some $ \alpha$ . (We chose $ x=0$ at the tip of the cone in arriving at the basic traveling-wave solution to the wave equation.) This formula holds for both the planar cross-section indicated in Fig.C.48 and the spherical section having the same perimeter (i.e., spherical surface area is proportional to radius squared). The logarithmic derivative of $ A(x)=\alpha x^2$ is then

   ln$\displaystyle ' A \eqsp \frac{ A' }{A} \eqsp \frac{2}{x}, $

which is independent of the conical taper angle $ \theta = \tan\left(\sqrt{\alpha/\pi}\right)$ . That is, one conical section of a spherical wave is like any other, as it must be due to spherical symmetry.

Substituting the logarithmic derivative of $ A(x)$ and $ p'$ from Eq.(C.171) into the momentum-conservation equation Eq. (C.169) yields

$\displaystyle \mp \frac{{\dot p}}{c}-\frac{p}{x} + p\frac{2}{x} \eqsp - \rho {\dot u} $

or

$\displaystyle \mp \frac{{\dot p}}{c} + \frac{p}{x} \eqsp - \rho {\dot u}, $

where the minus sign is for an expanding spherical wave, and the plus sign is for a converging spherical wave. Taking the Laplace transform of the above expression gives

$\displaystyle \mp \frac{s P(s)}{c} + \frac{P(s)}{x} \eqsp - \rho s U(s) $

in the case of zero initial conditions $ p(0,x)=u(0,x)=0$ for all $ x$ .

We can now solve for the wave impedance in each direction, where the wave impedance may be defined (§7.1) as the Laplace transform of the traveling pressure divided by the Laplace transform of the corresponding traveling velocity wave:

$\displaystyle R^{+}(s)$ $\displaystyle \isdef$ $\displaystyle \frac{P^+(s)}{U^{+}(s)}$  
$\displaystyle R^{-}(s)$ $\displaystyle \isdef$ $\displaystyle \frac{P^-(s)}{U^{-}(s)}$  

We introduce the shorthand

$\displaystyle P^\pm(s)\eqsp \pm R^\pm (s)U^\pm(s) $

where all of the upper signs or all of the lower signs are taken together, corresponding to expanding or converging spherical waves, respectively. In this notation, we may solve for the conical wave impedance as

$\displaystyle R^\pm (s)\isdefs \frac{P^\pm(s)}{\pm U^\pm(s)} \eqsp \frac{\rho c}{1 \mp \frac{c}{sx}}. $

Along the frequency axis $ s=j\omega$ we get

$\displaystyle R^\pm (j\omega) \eqsp \frac{\rho c}{1 \pm j\frac{c}{\omega x}} \eqsp \frac{\rho c}{1 \pm j\frac{1}{kx}} $

where $ \omega = 2\pi f$ is radian temporal frequency, and $ k \isdef \omega/c$ is the radian spatial frequency, or wavenumber.

Note that for a cylindrical tube, the wave impedance in both directions is $ R^\pm (s)= \rho c$ , and there is no frequency dependence. A wavelength or more away from the conical tip, i.e., for $ x\gg 1/k = \lambda/2\pi$ , where $ \lambda$ is the spatial wavelength, the wave impedance again approaches that of a cylindrical bore. However, in conical musical instruments, the fundamental wavelength is typically twice the bore length, so the complex nature of the wave impedance is important throughout the bore and approaches being purely imaginary near the mouthpiece. This is especially relevant to conical-bore double-reeds, such as the bassoon.

Writing the wave impedance as

$\displaystyle R^\pm (s)\isdefs \frac{P^\pm(s)}{\pm U^\pm(s)} \eqsp \frac{1}{\frac{1}{\rho c} \mp \frac{1}{s\rho x}}, $

the reader familiar with circuit theory (§7.2.3) will recognize this as the parallel combination of the wave impedance $ \rho c$ of a cylindrical bore and the lumped impedance of a mass formally equal to $ \mp\rho x$ . This equivalent circuit is explored in [37]. As $ x$ goes to 0 at the tip, the mass goes to zero and ``shorts out'' the traveling wave. Note that in the case of a converging cone, the mass is negative. This will give rise to an unstable one-pole filter in the waveguide model, and various solutions have been proposed for this problem [532,50].

Up to now, we have been defining wave impedance as pressure divided by particle velocity. In acoustic tubes, volume velocity is what is conserved at a junction between two different acoustic tube types. Therefore, in acoustic tubes, we define the wave impedance as the ratio of pressure to volume velocity

$\displaystyle R_A^\pm (s)\isdefs \frac{P^\pm(s)}{\pm U^\pm(s)A(x)} \eqsp \frac{R^\pm (s)}{A(x)} \eqsp \frac{\rho c}{A(x)} \frac{1}{1 \pm \frac{c}{sx}} $

or, defining $ t_x \isdef x/c$ as the time to propagate over the distance $ x$ ,

$\displaystyle \fbox{$\displaystyle R_A^\pm (s)\eqsp \frac{\rho c}{A(x)} \frac{s}{s \pm \frac{1}{t_x}}.$} \protect$ (C.171)

This is the wave impedance we use to compute the generalized reflection and transmission coefficients at a change in cross-sectional area and/or taper angle in a conical acoustic tube. Note that it has a zero at $ s=0$ and a pole at $ s=\pm 1/t_x$ .

In this case, the equivalent mass is $ \pm\rho x/A(x) = \pm\rho/(\alpha x)$ . It would perhaps be more satisfying if the equivalent mass in the conical wave impedance were instead $ \pm\rho x A(x)$ which is the mass of air contained in a cylinder of radius $ A(x)$ projected back to the tip of the cone. However, the ``acoustic mass'' cannot be physically equivalent to mechanical mass. To see this, consider that the impedance of a mechanical mass is $ ms$ which is in physical units of mass per unit time, and by definition of mechanical impedance this equals force over velocity. The impedance in an acoustic tube, on the other hand, must be in units of pressure (force/area) divided by volume velocity (velocity $ \times$ area) and this reduces to

$\displaystyle \frac{\mbox{force/area}}{\mbox{velocity}\cdot\mbox{area}} \eqsp \frac{\mbox{mass}}{\mbox{time}} \cdot \frac{1}{\mbox{area}^2} \eqsp \frac{\mbox{mass-volume-density}}{\mbox{time}\cdot\mbox{distance}} $

which is what we found.

The real part of the wave impedance corresponds to transportation of wave energy, the imaginary part is a so-called ``reactance'' and does not correspond to power transfer. Instead, it corresponds to a ``standing wave'' which is created by equal and opposite power flow, or an ``evanescent wave'' (§C.8.2), which is a non-propagating, exponentially decaying, limiting form of a traveling wave in which the ``propagation constant'' is purely imaginary due to being at a frequency above or below a ``cut off'' frequency for the waveguide [297,122]. Driving an ideal mass at the end of a waveguide results in total reflection of all incident wave energy along with a quarter-cycle phase shift. Another interpretation is that the traveling wave becomes a standing wave at the tip of the cone. This is one way to see how the resonances of a cone can be the same as those of a cylinder the same length which is open on both ends. (One might first expect the cone to behave like a cylinder which is open on one end and closed on the other.) Because the impedance approaches a purely imaginary zero at the tip, it looks like a mass (with impedance $ \pm\rho x s$ ). The ``piston of air'' at the open end similarly looks like a mass [287].

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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4
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Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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