Lossless Case

EXERCISES

4.3 Uniform Tube Model

4.3.1 Lossless Case

circuits. On the left end of the tube, the moving piston provides an ideal particle velocity source v(x, t ), i.e., the piston moves independently of pressure. This is analogous to an ideal cur- rent source in electrical circuits. We shortly further describe and refine these acoustic/electric analogies.

In the case of one-dimensional (planar) sound propagation in confined areas, it is con- venient to use the velocity of a volume of air rather than particle velocity. We denotevolume velocitybyu(x, t )and define it as the rate of flow of air particles perpendicularly through a specified area. Therefore, the relation between volume velocity and particle velocity for our uniform tube is given byu(x, t ) = Av(x, t ). We can then rewrite the various forms of the wave equation in terms of volume velocity using the relationv(x, t )=u(x, t )/A. In partic- ular, the second-order differential equation, Equation pair (4.5), (4.6), remains intact except for a replacement ofv(x, t )byu(x, t ), while the Equation pair (4.1), (4.4) becomes

−∂p

∂x = ρ A

∂u

∂t

−∂u

∂x = A ρc²

∂p

∂t. (4.7)

We can show that two solutions of the following form (Exercise 4.1) satisfy Equa- tion (4.7):

u(x, t ) = u⁺(t − x/c)−u⁻(t +x/c) p(x, t ) = ρc

A^[u⁺(t −x/c) +u⁻(t +x/c)] (4.8) which represent velocity and pressure, each of which is comprised of a forward- and backward- traveling wave (Figure 4.5), denoted by the superscript + ^and −, respectively. To see the forward-backward traveling wave property of Equation (4.8), consider a velocity wave at

u₊(t₀ – x/c) u₊(t₁ – x/c)

c (t₁ – t₀)

x

Velocity

Figure 4.5 Forward-traveling velocity waves. The quantityc(t₁−t₀)is the distance moved in the timet1−t0, wheret1> t0and wherecis the speed of sound. Backward-traveling waves are similar, but moving toward the left.

4.3 Uniform Tube Model 121

timet₀and timet₁. To show that the wave travels a distancec(t₁−t₀), consider the forward going wave

u⁺(t₁ −x/c) = u⁺

t₀ + (t₁ −t₀)c

c − x/c

= u⁺

t₀ + (t₁ −t₀)c −x c

= u⁺

t₀ − x −x₀ c

wherex₀ =(t₁−t₀)c. Thus, the wave at timet₁ is the wave at timet₀ shifted in space byx₀ to the right. A similar argument can be made for the backward traveling wave. Therefore, the forward and backward waves propagate without change in shape.

It is instructive to observe that the wave equation for plane-wave sound propagation is similar to that for plane-wave propagation along an electrical transmission line, where voltage and current are analogous to pressure and volume velocity [2],[28]. The voltagee(x, t )and currenti(x, t )for a lossless transmission line are related by the coupled equations

−∂e

∂x = L∂i

∂t

−∂i

∂x = C∂e

∂t ^(4.9)

whereLis inductance andCis capacitance per unit length of the transmission line. Figure 4.6 illustrates the acoustic and electrical analogies where ^ρ_A, defined as theacoustic inductance, is analogous to the electrical inductance L, and _ρc^A2, defined as theacoustic capacitance, is analogous to the electrical capacitanceC. These quantities reflect the “inertia” (mass or density) and the “springiness” (elasticity) of the air medium, respectively. In the context of electrical transmission lines, steady-state solutions are often written in the frequency domain using a com- plex sinewave representation. For example, a sinusoidal current wave of frequency, traveling forward at speedcin the transmission line medium, is written asi(x, t )=I (x, )e^{j (t−x/c)}. Motivated by this electrical/acoustic analogy, we will use similar representations in describing pressure and volume velocity in acoustic propagation.

We have seen above the general form of the solution for a uniform-tube configuration.

In order to determine a specific solution, we observe that at the boundaryx = 0 the volume velocity is determined by the sinusoidally vibrating piston, i.e.,u(0, t ) = U ()e^{j t}, where by the above convention we use the complex form of the real driving velocity Re[U ()e^{j t}].

This is analogous to a sinusoidal current source for a transmission line. We write the source as a function of frequency because we are ultimately interested in the frequency response of the uniform tube. Because this volume velocity represents the glottal airflow velocity, we also write this boundary condition asu(₀, t )=u_g(t )=U_g()e^{j t}. The second boundary condition is that the pressure is zero at the tube termination (the lips), i.e.,p(l, t )=0, which is analogous to placing a short circuit at the end of a transmission line. We will see later in this section that neither of these boundary conditions is accurate in practice with a real oral tract, i.e., the

p– Pressure u– Volume Velocity ρ/A– Acoustic Inductance A/(ρc²)– Acoustic Capacitance

v– Voltage i– Current

L– Electric Inductance C– Electric Capacitance

Acoustic Electric

∂p

∂x ρ A

∂u

∂t

=

∂u

∂x

∂v

∂x

∂i

∂x

∂i

∂t

∂v

∂t A

ρc²

∂p

∂t

=

= L

= C –

– –

–

Figure 4.6 Acoustic tube/electric transmission line analogies. Induc- tance and capacitance represent quantities in per-unit length.

SOURCE: L.R. Rabiner and R.W. Schafer,Digital Processing of Speech Signals[28]. ©1978, Pearson Education, Inc. Used by permission.

glottal airflow velocity source is not ideal, and there is energy loss due to radiation from the lips, resulting in a nonzero pressure drop at the lips.

We have already seen the general solution to the wave equation in Equation (4.8). Because our velocity driving function is sinusoidal,⁹we assume the specific velocity and pressure wave solutions to be of the form

u(x, t ) = k⁺e^{j (t−x/c)} −k⁻e^{j (t+x/c)} p(x, t ) = ρc

A^[k⁺e^{j (t−x/c)} +k⁻e^{j (t+x/c)}] (4.10) which are sinewaves traveling forward and backward in time and wherek⁺ andk⁻represent the amplitudes of the forward- and backward-going waves, respectively.

9Observe that the vibrating piston can take on a more general form which, from the Fourier transform pair for continuous-time signals, can be written as a sum (integral) of complex exponentials of the formU ()e^{j t}, i.e.,u(0, t )=_2π¹ _∞

−∞U ()e^{j t}d. When the uniform tube configuration is linear, the solution can be written as a sum of the individual solutions to the differential componentsU ()e^{j t}d. With the boundary condition at the tube termination of zero pressurep(l, t )=0, the uniform tube can be shown to be linear (Exercise 4.2).

4.3 Uniform Tube Model 123

The next objective, given our two boundary conditions, is to solve for the unknown constantsk⁺andk⁻. The boundary condition at the source gives

u(0, t ) = k⁺e^{j t} −k⁻e^{j t} = U_g()e^{j t} resulting in

k⁺ −k⁻ = U_g(). (4.11)

Likewise, the boundary condition at the open end gives p(l, t ) = ρc

A^[k⁺e^{j (t−l/c)} + k⁻e^{j (t+l/c)}] = 0 yielding

k⁺e^{−j l/c}+ k⁻e^{j l/c} = ⁰. (4.12)

Solving simultaneously for the two expressions ink⁺ _andk⁻, from Equation (4.11),k⁻ =

−U_g()+k⁺, which we substitute into Equation (4.12) to obtain k⁺e^{−j l/c} −^[U_g()−k⁺]e^{j l/c} = ⁰, which is rewritten as

k⁺ = U_g()e^{j l/c}

e^{−j l/c} +e^{j l/c}. (4.13)

Likewise, from Equation (4.11),k⁺=U_g()+k⁻, which we substitute into Equation (4.12) to obtain

[U_g()+k⁻_]e^{−j l/c} +k⁻e^{j l/c} = 0, which is rewritten as

k⁻ = −U_g()e^{−j l/c}

e^{j l/c} +e^{−j l/c}. (4.14)

Substituting Equations (4.13) and (4.14) into the Equation pair (4.10), we obtain u(x, t ) = ^cos[(l −x)/c]

cos[l/c] U_g()e^{j t} p(x, t ) = jρc

A

sin[(l − x)/c]

cos[l/c] U_g()e^{j t} (4.15)

which expresses the relationship between the sinusoidal volume velocity source and the pressure and volume velocity at any point in the tube. The particular solution is referred to asstanding wavesbecause the forward- and backward-traveling waves have added to make the wave shape stationary in time; the differencet − ^x_c or the sumt + ^x_c no longer appears as an argument in the complex exponential function. Since the volume velocity and pressure solutions are in complex form, we can think of the cosine and sine multiplier functions as theenvelopeof the volume velocity and pressure functions along the tube, given respectively by ^cos[_cos[^(l_l/c^−x)/c_] ^] and

Volume velocity Pressure 1

0.5 0 –0.5

0 0.1 0.2 0.3 0.4 0.5

Position Along Tube (Normalized)

0.6 0.7 0.8 0.9 1

Amplitude

Figure 4.7 Envelopes of pressure and volume velocity standing waves resulting from sound propagation in a uniform tube with zero pressure termination.

ρc A

sin[(l−x)/c]

cos[l/c] . We see that the two envelope functions are 90^◦out of phase (Figure 4.7), and so velocity and pressure are “orthogonal” in space. The two functions are also orthogonal in time because of the imaginaryjmultiplier of the pressure function. We can see this relation by writing the real counterpart of Equation (4.15):

Re[u(x, t )] = ^cos[(l −x)/c]

cos[l/c] U_g()cos(t ) Re[p(x, t )_] = ρc

A

cos[(l − x)/c+π/2]

cos[l/c_] U_g()_cos(t + π/₂), _(4.16) which reveals that at any time instant the velocity and pressure variations are 90^◦out of phase. An interesting consequence of this phase relation is that the acoustic kinetic and potential energies along the tube are also “orthogonal” (Exercise 4.3).

Consider now the specific relationship between the volume velocity at the open end of the tube (the lips) and the volume velocity at the source (the glottis). From the volume velocity component of Equation (4.15), we have atx=l

u(l, t ) = ^cos[(l −l)/c]

cos[l/c] U_g()e^{j t}

= ¹

cos[l/c]U_g()e^{j t}. (4.17)

LetU (l, )= _cos[l/c¹ _]U_g(), i.e., the complex amplitude at the open tube end for complex inputU_g()e^{j t}. Then we can write

U (l, )

U_g() = ¹ cos(l/c)

= V_a() (4.18)

where V_a() denotes the frequency response relating input to output volume velocities of the analog uniform-tube model of our simple speech production configuration. The roots of

4.3 Uniform Tube Model 125

the denominator ofV_a(), i.e., the uniformly-spaced frequencies at which the peak (infinite) amplitude occurs, are the resonances of the analog system. These roots are given by solutions to ^l_c =k^π

2, wherekis an integer or = kπ

2 c

l, k = 1,₃,₅. . .

EXAMPLE4.2 Consider the frequency response in Equation (4.18) and suppose that a uniform tube has lengthl=35 cm. With the speed of soundc=350 m/s, the roots of our particular uniform tube are

f =

2π = k 82000

= ²⁵⁰,750,1250, . . .

The frequency response is shown in Figure 4.8. Observe that the resonances have zero bandwidth since there is no energy loss in the system (the reader should argue this using the definitions of bandwidth in Chapter 2). As we decrease the lengthlof the tube, the resonant frequencies increase. We have all had the experience of filling a tall bottle under a faucet. The bottle (tube) is open at the top, and we can think of the water hitting the bottom as providing a volume velocity source. As the water runs, the open volume decreases, reducing the tube length and increasing the aurally perceived resonances.

An example is shown in Figure 4.8 in which the effective bottle length decreases froml =35 cm tol=17.5 cm. Resonances can also be thought of as “natural frequencies” which are excited by a disturbance. Why should a slug of air have natural frequencies? As we have seen, air has mass and springiness, and, as with lumped mechanical or electrical systems having inductance and capacitance, such a “mass-spring” system is more responsive to certain frequencies. An alternate analogy is the transmission line that we reviewed earlier as having inductance and capacitance per unit length; the acoustic “line” also has inductance and capacitance per unit length. 䉱

250 750 1250 1750 Ω

Frequency (Hz) Increasing

Resonances

Volume Velocity

Figure 4.8 Illustration of time-varying resonances as the effective length of a bottle decreases as it is filled with running water (Example 4.2). We aurally perceive the increasing resonances as the tube length decreases.

An important quantity relating pressure and volume velocity is their ratio along the tube. We call this quantity theacoustic impedance, analogous to electric impedance given by voltage divided by current, and from Equation (4.15) we can express it as

Z_A() = p(x, t ) u(x, t )

= jρc

A ^tan[(l −x)/c].

When the tube is very short with length denoted byx, using a Taylor series expansion of the tangent function, we obtain the approximation (looking into the tube fromx=⁰⁾

Z_A() ≈ jρx

A

where ^ρx_A can be thought of as an acoustic mass, and the short plug of air acts as a lumped acoustic inductance¹⁰where_A^ρ is inductance per unit length. As with electrical and mechanical impedance, a large value ofZ_A()implies that a large pressure is required to obtain a specified volume velocity, i.e., the medium “impedes” the flow of particles when acted upon. Impedance will become particularly important at glottal and lip boundaries because the impedance gives the pressure and velocity relations, coupled with the wave equation, that must be maintained in crossing these boundaries.

Returning to our solution Equation (4.18), we can change the frequency response to a transfer function by replacingby ^s_j to obtain¹¹

V_a(s) = ¹ cos(_{j c}^sl)

= ¹

(e^sl/c +e^−sl/c)/2

= ²e^−sl/c

1 +e^−2sl/c. _(4.19)

From functional analysis, if a complex functionf (x)of a complex variablex meets certain restrictions, then, using a Taylor expansion, we can writef (x)in terms of an infinite product of its zero factors, i.e., in terms of factors of the form(x−x_k), wherex_kis a zero. Flanagan has used this relation to show that Equation (4.19) can be written as [8]

V_a(s) = _∞ ¹

k=1(s− s_k)(s −s^∗_k)

10When the short tube is closed on the right, rather than open, then its acoustic impedance is of the form ZA≈ −j_Ax^ρc2 and acts rather as a lumped acoustic capacitance (Exercise 4.8).

11The Laplace transform of the continuous-time impulse response gives its transfer function. Recall that in the Laplace transform the complex variables=σ+j . Whensis evaluated on the imaginary axisj , i.e.,σ=0, thens=j . Therefore,=_j^s|σ=0.

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

s_n = ±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.