# 11.3.3 Impedance of a conical tube

The first step in modelling a soprano saxophone, or any other conical reed instrument, is to calculate the input impedance for a truncated cone. This will play the same role as the impedance of a cylindrical tube, which we calculated in section 11.1.1 and used when building an idealised model of the clarinet. Figure 1 shows the geometry: measured from the point of the complete cone, the two ends of the truncated cone are at positions $x_1$ and $x_2$, so the length of the tube is $L=x_2-x_1$. Although the diagram is drawn with a rapidly flaring cone, in reality the instruments we are interested in have much more gentle flares: the angle at the apex of the cone is about $4^\circ$ for a soprano saxophone, about $1.5^\circ$ for an oboe and as little as $0.8^\circ$ for a bassoon. These angles are all small, so in our analysis we will make no distinction between horizontal distance $x$ and radial distance $r$.

We already know from section 4.1.2 that the pressure inside the tube, for sinusoidal excitation at frequency $\omega$, can be written as a sum of right-travelling and left-travelling spherical waves:

$$p=A_R \dfrac{e^{i \omega t -ikr}}{r} + A_L \dfrac{e^{i \omega t +ikr}}{r} \tag{1}$$

where $A_R$ and $A_L$ are (complex) coefficients and the wavenumber $k=\omega/c$ as usual, where $c$ is the speed of sound. The particle velocity $v$ then must satisfy equation (2) from section 4.1.3, so that

$$\rho_0\dfrac{\partial v}{\partial t}=-\dfrac{\partial p}{\partial r} \tag{2}$$

where $\rho_0$ is the density of air. Substituting,

$$\rho_0 i \omega v = A_R \dfrac{e^{i \omega t -ikr}}{r^2} + A_L \dfrac{e^{i \omega t +ikr}}{r^2}$$

$$+ ik A_R\dfrac{e^{i \omega t -ikr}}{r}- ik A_L\dfrac{e^{i \omega t +ikr}}{r} \tag{3}$$

so that

$$v=\dfrac{1}{i \omega \rho_0}\left[ \dfrac{p}{r}+\dfrac{ik}{r}\left(A_R e^{i \omega t -ikr} – A_L e^{i \omega t +ikr} \right) \right] . \tag{4}$$

The tube is open at the right-hand end, so $p=0$ at $r=x_2$, giving

$$A_R e^{-ikx_2} + A_L e^{ikx_2} = 0 \tag{5}$$

so that

$$A_L=-A_R e^{-2ikx_2} . \tag{6}$$

Now at $r=x_1$ we have

$$p=\dfrac{e^{i \omega t}}{x_1} A_R \left[ e^{-ikx_1} – e^{ik(x_1-2x_2)} \right] \tag{7}$$

and

$$v= \dfrac{e^{i \omega t}}{ik\rho_0 c x_1^2} A_R \left[ e^{-ikx_1} – e^{ik(x_1-2x_2)} \right]$$

$$+ \dfrac{e^{i \omega t}}{i\rho_0 c x_1} A_R \left[ e^{-ikx_1}+ e^{ik(x_1-2x_2)} \right] . \tag{8}$$

So the input impedance $Z$, and its inverse the admittance $Y$, satisfies

$$\dfrac{1}{Z}=Y=\dfrac{S_1v}{p}=\dfrac{S_1}{\rho_0 c} \dfrac{e^{-ikx_1} \left(1+\dfrac{1}{ikx_1}\right) + e^{ik(x_1-2x_2)} \left(1-\dfrac{1}{ikx_1} \right)}{e^{-ikx_1} – e^{ik(x_1-2x_2)}}$$

$$=\dfrac{1}{Z_t}\left[ \dfrac{1+e^{-2ikL}}{1-e^{-2ikL}}+ \dfrac{1}{ikx_1}\right]=-\dfrac{i}{Z_t}\left(\cot kL + \dfrac{1}{kx_1} \right) \tag{9}$$

where $S_1$ is the cross-sectional area at position $x_1$ and $Z_t=\rho_0 c/S_1$ is the characteristic impedance of a cylindrical tube with that area.

This gives the input impedance expression we want, but we should also note a further approximation. For low frequencies such that $kx_1 \ll 1$, then $kx_1 \approx \tan kx_1$ so that

$$Y \approx -\dfrac{i}{Z_t}\left(\cot kL + \cot kx_1 \right) . \tag{10}$$

This is the “cylindrical saxophone” approximation: it is precisely the expression for the input admittance at distance $x_1$ from the end of a cylindrical pipe of length $x_1+L$ which is open at both ends. Using a standard trigonometric formula for $\cot A + \cot B$ we deduce that the zeros of $Y$, which are the peaks of $Z$, occur where $\sin k(x_1+L)=0$, so that they are exactly harmonically spaced. The same expression (10) (apart from needing a different constant in place of $Z_t$) gives the impedance at position $x_1$ on a string of length $x_1+L$, which is the basis of the analogy with a bowed string.

If we want to couple our truncated cone to a mouthpiece cavity with volume $V$, we simply need to replace $Y$ from equation (9) by

$$Y_m=Y+\dfrac{i \omega V}{\rho_0 c^2} \tag{11}$$

because the pressure has to be the same at the tube mouth and in the cavity, while the total volume flow rate is the sum of the two contributions, so the two admittances must be added. The expression for the admittance of the enclosed air volume follows from equations (2) and (3) of section 4.2.1. The required value of $V$ to match the volume of the missing portion of the cone is

$$V=x_1 S_1/3 . \tag{12}$$

However, equation (11) is only valid at low frequency, because it relies on the assumption $kx_1 \ll 1$. At very high frequencies the influence of the mouthpiece should tend to zero, so for the purposes of computation equation (11) is replaced by an ad hoc version including a low-pass filter:

$$Y_m=Y+\left( \dfrac{\omega_c^2}{\omega^2+\omega_c^2} \right) \dfrac{i \omega V}{\rho_0 c^2} \tag{13}$$

with a cutoff frequency $\omega_c$ chosen suitably: for the saxophone model discussed here this cutoff was placed at 1 kHz.

For the purposes of simulation, we need to deduce a reflection function from the input impedance. The procedure was described in section 11.3.2. It involves using an inverse FFT, and for that purpose the damping needs to be sufficient to avoid numerical difficulties. As was done in the clarinet case, the baseline level of damping was determined by textbook expressions for visco-thermal damping, but this was enhanced in an ad hoc way at higher frequencies. The resulting input impedances are shown in Fig. 2, for models with and without the mouthpiece compensation. The difference between the two looks very small in this plot, but plots shown in section 11.3 demonstrate that it is big enough to have a significant effect on the results of simulations.

The assumed level of damping was set in such a way as to resemble measured results. Figure 3 shows a measured input impedance of a soprano saxophone fingered for its lowest note, with all tone-holes closed, again taken from Joe Wolfe’s web site. The general level and the extent of peak-to-valley excursions at low frequencies give quite a good match to Fig. 2. But the real instrument has far higher damping above about 2.5 kHz, largely as a result of the radiation characteristics of the bell, not included in the simplified model here.

For comparison, Fig. 4 shows the “cylindrical saxophone” impedance from equation (10), as a green curve superimposed on the two impedance curves from Fig. 2.

From the impedance functions shown in Fig. 2, it is straightforward to calculate the corresonding impulse responses and the reflection functions, as described in section 11.3.2. The results are shown in Figs. 5 and 6, for the models with and without the mouthpiece, using the same colour code as in Fig. 2. The reflection functions are more spread out in time than the corresponding result for a cylindrical tube (shown in Fig. 4 of section 11.3.2).

Finally, while discussing the ingredients of the saxophone simulations this is a convenient place to summarise the parameter values chosen for the reed model. The width $w$ was set at 12 mm, the reed stiffness $K_r$ was chosen to be significantly softer than the clarinet reed at 5 kPa/mm, the reed resonance frequency was set at 3 kHz, and the reed Q-factor at 2.5.