EXERCISES
4.3 Uniform Tube Model
4.3.2 Effect of Energy Loss
4.3 Uniform Tube Model 127
where, because the corresponding response is real, the poles occur in complex conjugate pairs.
There are then an infinite number of poles that occur on the j axis and can be written as
sn = ±j
(2k +1)π c 2l
, k = 0,±1,±2, . . .
corresponding to the resonant frequencies of the uniform tube. The poles fall on thej axis, and thus the frequency response at resonant frequencies is infinite in amplitude, because there is no energy loss in the simple configuration. In particular, it was assumed that friction of air particles does not occur along the vocal tract walls, that heat is not conducted away through the vocal tract walls, and that the walls are rigid and so energy was not lost due to their movement.
In addition, because we assumed boundaries consisting of an ideal volume velocity source and zero pressure at the lips, no energy loss was entailed at the input or output to the uniform tube.
In practice, on the other hand, such losses do occur and will not allow a frequency response of infinite amplitude at resonance. Consider by analogy a lumped electrical circuit with a resistor, capacitor, and inductor in series. Without the resistor to introduce thermal loss into the network, the amplitude of the frequency response at resonance equals infinity. We investigate these forms of energy loss in the following two sections. Energy loss may also occur due to rotational flow, e.g., vortices and other nonlinear effects that were discarded in the derivation of the wave equa- tion, such as from Equation (4.3). These more complex forms of energy loss are discussed in Chapter 11.
Portnoff then assumed that small, differential pieces of the surface of the wall, denoted
by d
, are independent, i.e., “locally reacting,” and that the mechanics of each piece are modeled by a massmw, spring constantkw, and damping constantbw per unit surface area illustrated in Figure 4.9. Because the change in cross-section due to pressure change is very small relative to the average cross-section, Portnoff expressed the cross-section along the dimension x as A(x, t ) = Ao(x, t )+A(x, t ), whereA(x, t )is a linear perturbation about the averageAo(x, t ). Using the lumped parameter model in Figure 4.9, the second-order differential equation for the perturbation term is given by
mwd2A
dt2 +bwdA
dt +kwA = p(x, t ) (4.21) where the mass, damping, and stiffness per unit surface area of the wall are assumed to be constant, i.e., not a function of the variablexalong the tube, and where we have used the relation A(x, t ) = So(x, t )r,So(x, t )being the average vocal tract perimeter at equilibrium and r being the displacement of the wall perpendicular to the wall. (The reader should perform the appropriate replacement of variables in the model equation of Figure 4.9.) The pressure and perturbation relations in Equation (4.21) are coupled to the relations in the modified wave equation pair, Equation (4.20), and thus the three equations are solved simultaneously, under the two boundary conditions of the known volume velocity sourceu(0, t )and the output pressure p(l, t ) = 0. Discarding second-order terms in the expansion for Au = Ao+uA, the three coupled equations in variablesp,u, andAfor the uniform tube reduce approximately to [26]
−∂p
∂x = ρ Ao
∂u
∂t
−∂u
∂x = Ao ρc2
∂p
∂t + ∂A
∂t p = mwd2A
dt2 +bwdA
dt +kwA (4.22)
where we assume that the average cross-section Ao is constant, i.e., the vocal tract is time- invariant being held in a particular configuration.
Under the steady-state assumption that sound propagation has occurred long enough so that transient responses have died out, and given that the above three coupled equations are linear and time-invariant, an inputug(t ) = u(0, t ) = U ()ej t results in solutions of the formp(x, t )= P (x, )ej t,u(x, t ) = U (x, )ej t, andA(x, t )= A(x, )eˆ j t along the tube. Therefore, the time-dependence of the three coupled equations is eliminated;
the resulting three coupled equations are then functions of the variablexfor each frequency:
−∂P (x, )
∂x = ρ
AoU (x, )
−∂U (x, )
∂x = Ao
ρc2P (x, )+ A(x, )ˆ
P (x, ) = −2mwA(x, )ˆ +j bwA(x, )ˆ +kwA(x, ).ˆ (4.23)
4.3 Uniform Tube Model 129
So Ao
ΔA
ΔA = Sor
r x
bwdΣ mwdΣ
pdΣ
Damper Spring
Mass kwdΣ p = mwr + b: w r + k. wr
r
Equilibrium (r = 0)
Figure 4.9 Mechanical model of differential surface element d of vibrating wall. Adjacent wall elements are assumed uncoupled.
SOURCE: M.R. Portnoff,A Quasi-One-Dimensional Digital Simulation for the Time-Varying Vocal Tract[26]. ©1973, M.R. Portnoff and the Massachusetts Institute of Technology. Used by permission.
Portnoff has solved these coupled equations using standard numerical simulation tech- niques12invoking finite-difference approximations to the spatial partial differential operator ∂x∂ . Suppose that the length of the tube isl. Then we seek the acoustic pressure and volume velocity atN equally spaced points over the interval [0, l]. In particular, ∂x∂ was approximated by a central difference with averaging; that is, for a partial differential with respect toxof a function f (x), first compute the central difference over a spatial differencex= N−l1:
g[n] = 1
2x(f[(n −1)x]−f[(n+ 1)x]) and then perform a first backward average:
∂
∂xf (t ) ≈ 1
2(g[nx]+g[(n−1)x]).
It can be shown that this transformation from the continuous- to the discrete-space variable is equivalent to a bilinear transformation.13 This partial differential approximation leads to
12We can also obtain in this case an approximate closed-form, frequency-domain solution [26],[28], but we choose here to describe the numerical simulation because this solution approach is the basis for later-encountered nonlinear coupled equations.
13The bilinear transformation for going from continuous to discrete time is given by thes-plane toz-plane mapping ofs= t2[zz−+11] and has the property that the imaginary axis in thes-plane maps to the unit circle in thez-plane [22]. The frequencyin continuous time is related to the discrete-time frequency by the tangential mapping t 2 =tan(ω2). Likewise, the bilinear mapping in going from continuous space to discrete space is given byK=x2 [kk−1+1] and has the same tangential mapping as the time variable.
2N−2 equations in 2N unknown pressure and velocity variables. The two additional equa- tions necessary for a unique solution are supplied by the boundary conditions at the lips and glottis.
The resulting frequency responseVa()= U (l,)Ug() of the numerical simulation is shown in Figure 4.10a (pressure and velocity computed at 96 samples in space along thexvariable) for a uniform tube with yielding walls and no other losses, terminated in a zero pressure boundary condition [26]. The tube is 17.5 cm in length and 5 cm2 in cross section. The yielding wall parameters are given by Flanagan [8] to bemw = 0.4 gm/cm2,bw = 6500 dyne-sec/cm3, and kw = 0. The result is markedly different from the lossless case. Because of energy loss due to the wall vibration, the poles of the transfer function are no longer on thej axis and so the bandwidth is nonzero. In addition, the resonant frequencies have slightly increased. Finally, these effects are most pronounced at low frequencies because the inertial mass of the wall results in less wall motion at high frequencies.
Viscosity and Thermal Loss —The effects of both viscous and thermal loss can be represented by modification of Equations (4.20) and (4.21) with the introduction of a resistive term, rep- resenting the energy loss due to friction of air particles along the wall, and a conductive term, representing heat loss through the vibrating walls, respectively [8],[26]. The resulting coupled equations are solved numerically for the steady-state condition, again using a central difference approximation to the spatial partial derivative [26]. With viscous and thermal loss only, effects are less noticeable than with wall vibration loss, being more pronounced at high frequencies where more friction and heat are generated. The addition of viscosity and thermal conduction to the presence of vibrating walls yields slight decreases in resonant frequencies and some broad- ening of the bandwidths. Figure 4.10b gives the result of Portnoff’s numerical simulation with all three losses for the uniform tube with zero pressure termination [26].