Cylinder with Conical Cap

Consider a cylindrical acoustic tube adjoined to a converging conical cap, as depicted in Figure C.48a. We may consider the cylinder to be either open or closed on the left side, but everywhere else it is closed. Since such a physical system is obviously passive, an interesting test of acoustic theory is to check whether theory predicts passivity in this case.

Figure C.48: Cylindrical acoustic tube with a conical cap. a) Physical outline. b) Continuous-time waveguide formulation.

It is well known that a growing exponential appears at the junction of two conical waveguides when the waves in one conical taper angle reflect from a section with a smaller (or more negative) taper angle [7,300,8,160,9]. The most natural way to model a growing exponential in discrete time is to use an unstable one-pole filter [506]. Since unstable filters do not normally correspond to passive systems, we might at first expect passivity to not be predicted. However, it turns out that all unstable poles are ultimately canceled, and the model is stable after all, as we will see. Unfortunately, as is well known in the field of automatic control, it is not practical to attempt to cancel an unstable pole in a real system, even when it is digital. This is because round-off errors will grow exponentially in the unstable feedback loop and eventually dominate the output.

The need for an unstable filter to model reflection and transmission at a converging conical junction has precluded the use of a straightforward recursive filter model [406]. Using special ``truncated infinite impulse response'' (TIIR) digital filters [540], an unstable recursive filter model can in fact be used in practice [528]. All that is then required is that the infinite-precision system be passive, and this is what we will show in the special case of Fig.C.48.

Scattering Filters at the Cylinder-Cone Junction

As derived in §C.18.4, the wave impedance (for volume velocity) at frequency $ \omega $ rad/sec in a converging cone is given by

$\displaystyle Z_\xi(j\omega) = \frac{\rho c}{S(\xi)} \frac{j\omega}{j\omega-c/\xi}$ (C.152)

where $ \xi$ is the distance to the apex of the cone, $ S(\xi)$ is the cross-sectional area, and $ \rho c$ is the wave impedance in open air. A cylindrical tube is the special case $ \xi=\infty$, giving $ Z_\infty(j\omega) = \rho c/S$, independent of position in the tube. Under normal assumptions such as pressure continuity and flow conservation at the cylinder-cone junction (see, e.g., [300]), the junction reflection transfer function (reflectance) seen from the cylinder looking into the cone is derived to be

$\displaystyle R(s) = -\frac{c/\xi}{c/\xi - 2s}$ (C.153)

(where $ s$ is the Laplace transform variable which generalizes $ s=j\omega$) while the junction transmission transfer function (transmittance) to the right is given by

$\displaystyle T(s) = 1 + R(s) = -\frac{2s}{c/\xi - 2s}$ (C.154)

The reflectance and transmittance from the right of the junction are the same when there is no wavefront area discontinuity at the junction [300]. Both $ R(s)$ and $ T(s)$ are first-order transfer functions: They each have a single real pole at $ s=c/(2\xi)$. Since this pole is in the right-half plane, it corresponds to an unstable one-pole filter.

Reflectance of the Conical Cap

Let $ {t_{\xi}}\isdef \xi/c$ denote the time to propagate across the length of the cone in one direction. As is well known [22], the reflectance at the tip of an infinite cone is $ -1$ for pressure waves. I.e., it reflects like an open-ended cylinder. We ignore any absorption losses propagating in the cone, so that the transfer function from the entrance of the cone to the tip is $ e^{-s{t_{\xi}}}$. Similarly, the transfer function from the conical tip back to the entrance is also $ e^{-s{t_{\xi}}}$. The complete reflection transfer function from the entrance to the tip and back is then

$\displaystyle R_{t_{\xi}}(s) = - e^{-2s{t_{\xi}}}$ (C.155)

Note that this is the reflectance a distance $ \xi=c{t_{\xi}}$ from a conical tip inside the cone.

We now want to interface the conical cap reflectance $ R_{t_{\xi}}(s)$ to the cylinder. Since this entails a change in taper angle, there will be reflection and transmission filtering at the cylinder-cone junction given by Eq.$ \,$(C.154) and Eq.$ \,$(C.155).

From inside the cylinder, immediately next to the cylinder-cone junction shown in Fig.C.48, the reflectance of the conical cap is readily derived from Fig.C.48b and Equations (C.154) and (C.155) to be

$\displaystyle R_J(s)$ $\displaystyle =$ $\displaystyle \frac{R(s) + 2 R(s) R_{t_{\xi}}(s) + R_{t_{\xi}}(s)}{1 - R(s)R_{t...
= \frac{1 + (1+2s{t_{\xi}})R_{t_{\xi}}(s)}{2s{t_{\xi}}-1-R_{t_{\xi}}(s)}$  
  $\displaystyle =$ $\displaystyle \frac{1 - (1+2s{t_{\xi}})e^{-2s{t_{\xi}}}}{2s{t_{\xi}}-1+e^{-2s{t_{\xi}}}}
\isdef \frac{N(s)}{D(s)}$ (C.156)


$\displaystyle N(s) \isdef 1 - e^{-2s{t_{\xi}}} - 2s{t_{\xi}}e^{-2s{t_{\xi}}}$ (C.157)

is the numerator of the conical cap reflectance, and

$\displaystyle D(s) \isdef 2 s{t_{\xi}}- 1 + e^{-2s{t_{\xi}}}$ (C.158)

is the denominator. Note that for very large $ {t_{\xi}}$, the conical cap reflectance approaches $ R_J = -e^{-2s{t_{\xi}}}$ which coincides with the impedance of a length $ \xi=c{t_{\xi}}$ open-end cylinder, as expected.

Stability Proof

A transfer function $ R_J(s) = N(s)/D(s)$ is stable if there are no poles in the right-half $ s$ plane. That is, for each zero $ s_i$ of $ D(s)$, we must have re$ \left\{s_i\right\}\leq 0$. If this can be shown, along with $ \left\vert R_J(j\omega)\right\vert\leq 1$, then the reflectance $ R_J$ is shown to be passive. We must also study the system zeros (roots of $ N(s)$) in order to determine if there are any pole-zero cancellations (common factors in $ D(s)$ and $ N(s)$).

Since re$ \left\{s{t_{\xi}}\right\}\geq 0$ if and only if re$ \left\{s\right\}\geq 0$, for $ {t_{\xi}}>0$, we may set $ {t_{\xi}}=1$ without loss of generality. Thus, we need only study the roots of

N(s) &=& 1 - e^{-2s} - 2s e^{-2s} \\
D(s) &=& 2s - 1 + e^{-2s}

If this system is stable, we have stability also for all $ {t_{\xi}}>0$. Since $ e^{-2s}$ is not a rational function of $ s$, the reflectance $ R_J(s)$ may have infinitely many poles and zeros.

Let's first consider the roots of the denominator

$\displaystyle D(s) = 2s - 1 + e^{-2s}.$ (C.159)

At any solution $ s$ of $ D(s)=0$, we must have

$\displaystyle s = \frac{1-e^{-2s}}{2}$ (C.160)

To obtain separate equations for the real and imaginary parts, set $ s=\sigma+j\omega$, where $ \sigma$ and $ \omega $ are real, and take the real and imaginary parts of Eq.$ \,$(C.161) to get

\mbox{re}\left\{D(s)\right\} &=& (2\sigma - 1) + e^{-2\sigma}\...
...}\left\{D(s)\right\} &=& 2\omega - e^{-2\sigma}\sin(2\omega) = 0

Both of these equations must hold at any pole of the reflectance. For stability, we further require $ \sigma\leq 0$. Defining $ \tau = 2\sigma$ and $ \nu=2\omega$, we obtain the somewhat simpler conditions

$\displaystyle e^\tau ( 1 - \tau)$ $\displaystyle =$ $\displaystyle \cos(\nu)$ (C.161)
$\displaystyle e^\tau$ $\displaystyle =$ $\displaystyle \frac{\sin(\nu)}{\nu}$ (C.162)

For any poles of $ R_J(s)$ on the $ j\omega $ axis, we have $ \tau=0$, and Eq.$ \,$(C.163) reduces to

$\displaystyle \nu = \sin(\nu)$ (C.163)

It is well known that the ``sinc function'' $ \sin(\nu)/\nu$ is less than $ 1$ in magnitude at all $ \nu$ except $ \nu=0$. Therefore, Eq.$ \,$(C.164) can hold only at $ \omega=\nu=0$.

We have so far proved that any poles on the $ j\omega $ axis must be at $ \omega=0$.

The same argument can be extended to the entire right-half plane as follows. Going back to the more general case of Eq.$ \,$(C.163), we have

$\displaystyle \frac{\sin(\nu)}{\nu} = e^\tau$ (C.164)

Since $ \left\vert\sin(\nu)/\nu\right\vert\leq 1$ for all real $ \nu$, and since $ \left\vert e^\tau\right\vert>1$ for $ \tau>0$, this equation clearly has no solutions in the right-half plane. Therefore, the only possible poles in the right-half plane, including the $ j\omega $ axis, are at $ s=0$.

In the left-half plane, there are many potential poles. Equation (C.162) has infinitely many solutions for each $ \tau<0$ since the elementary inequality $ 1-\tau \leq e^{-\tau}$ implies $ e^\tau ( 1
- \tau) < e^\tau e^{-\tau} = 1$. Also, Eq.$ \,$(C.163) has an increasing number of solutions as $ \tau $ grows more and more negative; in the limit of $ \tau=-\infty$, the number of solutions is infinite and given by the roots of $ \sin(\nu)$ ( $ \nu = k\pi$ for any integer $ k$). However, note that at $ \tau=-\infty$, the solutions of Eq.$ \,$(C.162) converge to the roots of $ \cos(\nu)$ ( $ \nu = (2k+1)(\pi/2)$ for any integer $ k$). The only issue is that the solutions of Eq.$ \,$(C.162) and Eq.$ \,$(C.163) must occur together.

Figure: Locus of solutions to Eq.$ \,$(C.162) (dashed) (and Eq.$ \,$(C.163) (solid). The dotted lines are lines of construction for an analytic proof that there are infinitely many roots in the left-half $ s$-plane.

Figure C.49 plots the locus of real-part zeros (solutions to Eq.$ \,$(C.162)) and imaginary-part zeros (Eq.$ \,$(C.163)) in a portion the left-half plane. The roots at $ s=0$ can be seen at the middle-right. Also, the asymptotic interlacing of these loci can be seen along the left edge of the plot. It is clear that the two loci must intersect at infinitely many points in the left-half plane near the intersections indicated in the graph. As $ \left\vert\nu\right\vert$ becomes large, the intersections evidently converge to the peaks of the imaginary-part root locus (a log-sinc function rotated 90 degrees). At all frequencies $ \nu$, the roots occur near the log-sinc peaks, getting closer to the peaks as $ \left\vert\nu\right\vert$ increases. The log-sinc peaks thus provide a reasonable estimate number and distribution in the left-half $ s$-plane. An outline of an analytic proof is as follows:

  • Rotate the loci in Fig.C.49 counterclockwise by 90 degrees.

  • Prove that the two root loci are continuous, single-valued functions of $ \nu$ (as the figure suggests).

  • Prove that for $ \left\vert\nu\right\vert>3$, there are infinitely many extrema of the log-sinc function (imaginary-part root-locus) which have negative curvature and which lie below $ \tau<-1$ (as the figure suggests). The $ \tau=-1$ and $ \nu=\pm 3$ lines are shown in the figure as dotted lines.

  • Prove that the other root locus (for the real part) has infinitely many similar extrema which occur for $ \tau>-1$ (again as the figure suggests).

  • Prove that the two root-loci interlace at $ \tau=-\infty$ (already done above).

  • Then topologically, the continuous functions must cross at infinitely many points in order to achieve interlacing at $ \tau=-\infty$.

The peaks of the log-sinc function not only indicate approximately where the left-half-plane roots occur

Reflectance Magnitude

We have shown that the conical cap reflectance has no poles in the strict right-half plane. For passivity, we also need to show that its magnitude is bounded by unity for all $ s$ on the $ j\omega $ axis.

We have

$\displaystyle R_J(j\omega) = \frac{1 - e^{-2j\omega} - 2j\omega e^{-2j\omega}}{...
...mega} - 1 + 2j\omega}
= e^{-2j\omega} \frac{\overline{D(j\omega)}}{D(j\omega)}

so that $ \left\vert R_J(j\omega)\right\vert = 1$, which is exactly lossless, as expected.

Poles at $ s=0$

We know from the above that the denominator of the cone reflectance has at least one root at $ s=0$. In this subsection we investigate this ``dc behavior'' of the cone more thoroughly.

A hasty analysis based on the reflection and transmission filters in Equations (C.154) and (C.155) might conclude that the reflectance of the conical cap converges to $ -1$ at dc, since $ R(0)=-1$ and $ T(0)=0$. However, this would be incorrect. Instead, it is necessary to take the limit as $ \omega\to0$ of the complete conical cap reflectance $ R_J(s)$:

$\displaystyle R_J(s) = \frac{1 - e^{-2s} - 2s e^{-2s}}{2s - 1 + e^{-2s}}$ (C.165)

We already discovered a root at $ s=0$ in the denominator in the context of the preceding stability proof. However, note that the numerator also goes to zero at $ s=0$. This indicates a pole-zero cancellation at dc. To find the reflectance at dc, we may use L'Hospital's rule to obtain

$\displaystyle R_J(0) = \lim_{s\to0} \frac{N^\prime(s)}{D^\prime(s)} = \lim_{s\to 0}\frac{4s e^{-2s}}{2-2e^{-2s}}$ (C.166)

and once again the limit is an indeterminate $ 0/0$ form. We therefore apply L'Hospital's rule again to obtain

$\displaystyle R_J(0) = \lim_{s\to0} \frac{N^{\prime\prime}(s)}{D^{\prime\prime}(s)} = \lim_{s\to 0}\frac{(4-8s) e^{-2s}}{4e^{-2s}} = +1$ (C.167)

Thus, two poles and zeros cancel at dc, and the dc reflectance is $ +1$, not $ -1$ as an analysis based only on the scattering filters would indicate. From a physical point of view, it makes more sense that the cone should ``look like'' a simple rigid termination of the cylinder at dc, since its length becomes small compared with the wavelength in the limit.

Another method of showing this result is to form a Taylor series expansion of the numerator and denominator:

$\displaystyle N(s)$ $\displaystyle =$ $\displaystyle 2 {s^2} - {{8 {s^3}}\over 3} + 2 {s^4} + \cdots$ (C.168)
$\displaystyle D(s)$ $\displaystyle =$ $\displaystyle 2 {s^2} - {{4 {s^3}}\over 3} + {{2 {s^4}}\over 3} + \cdots$ (C.169)

Both series begin with the term $ 2s^2$ which means both the numerator and denominator have two roots at $ s=0$. Hence, again the conclusion is two pole-zero cancellations at dc. The series for the conical cap reflectance can be shown to be

$\displaystyle R_J(s) = 1 - {{2 s}\over 3} + {{2 {s^2}}\over 9} - {{4 {s^3}}\over {135}} - {{2 {s^4}}\over {405}} + \cdots$ (C.170)

which approaches $ +1$ as $ s\to0$.

An alternative analysis of this issue is given by Benade in [37].

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