It was assumed that the pressureat the lips is zero and the volume velocity source is ideal no energy loss at the input and output.

For radiation impedance: Morse and Ingard (1986),

Beranek (1954), Flanagan (1972). Lip radiation is

modelled by a piston in an infinite wall. The

acoustic impedance is a resistance Rr and an

inductor Lr in parallel.

For the infinite wall Rr = 1289 / 92 and Lr =8a / 3c

Boundary Effects

rr

rr

rr

R LjR

LRj

LjRlU

lPZ

11

1

,

,

The impedance equation can be converted to a differential equation via Laplace transform.

Portnoff (1973) numerically

simulated the above equation

coupled to the wave equation.

Broader bandwidths and lowering of the resonances

are observed. Higher frequencies are affected most.

(This can be seen by considering Zr 0 when 0 (small)

p(l,t) = 0 but for large , Lr >>Rr Zr = Rr)

Boundary Effects

rr

rrR sLR

LsR

sU

sPsZ

Glottal Source and Impedance

Flanagan-Isızaka (1978) proposed a linear model for the impedance.

This is a differential equation in the time-domain. The solution of this equation together with the wave equation yields broadening of the bandwidths at low frequencies.

This can be seen from Zg = Rg + j Lg becomes Zg Rg for small ; (purely resistive).

Overall Frequency Response

introduces a highpass filter effect

ggg LjRZ

g

g Z

PUU

,0,0

aR

gg

VZ

U

lU

lU

lP

U

lPH

,

,

,,

Summarizing the results

1) Resonances are due to vocal tract. Resonant frequenciesshift as a result various losses.

2) Bandwidths of lower resonances are controlledby wall vibration and glottal impedance loss.

3) Bandwidths of higher resonances are controlled by radiation, viscous and thermal losses.

(Energy loss is assumed to be only at the lips.)

For the kth tube

(*)

Boundary conditions are

Model of Concatenated Tubes

c

xtu

c

xtu

A

ctxp

c

xtu

c

xtutxu

kkk

k

kkk

,

,

tptlp

tutlu

kkk

kkk

,0,

,0,

1

1

The boundary conditions and (*) yield :

Let k= lk /c

Using in (2)

Model of Concatenated Tubes

)2(

)1(

111

11

tutuc

ltu

c

ltu

A

A

tutuc

ltu

c

ltu

kkk

kk

kk

k

kkk

kk

k

)3()1( 11 tutututu kkkkkk

)4(2

11

1

1

11 tu

AA

AAtu

AA

Atu k

kk

kkkk

kk

kk

(4) – (3)

Let

Both equations contain a component due to reflection and one component due to transmission.

Model of Concatenated Tubes

)5(2

111

1 tuAA

Atu

AA

AAtu k

kk

kkk

kk

kkkk

kk

kkk AA

AAr

1

1 turturtu

turturtu

kkkkkkk

kkkkkk

1

11

1

1

Lips: Suppose radiation impedance is real. At the Nth tube Since

There is no backward going wave from free space.

Boundary Relations

tluZtlp NNrNN ,,

c

xtu

c

xtu

A

ctxp

c

xtu

c

xtutxu

,,

NNNNrNNNNN

tutuZtutuA

c

NNrN

NNrN

tuZA

ctuZ

A

c

NNLNN

rN

rN

NN turtu

ZA

c

ZA

c

tu

Boundary Relations

NNLNNNNNN turtututlu 1,Therefore

Outgoing wave from the lips: Let the outer space be represented by an infinite length tube of cross section AN+1 , and in particular, if

(Also, if Zr () = 0 rL=1, AN+1 no radiation from the lips.

NN

NNL

Nr AA

AAr

A

cZ

1

1

1

Glottis: Suppose the glottal impedance is purely real, Zg=Rg

In particular, if glottal impedance is modeled by a tube of cross section A0 and A0 is chosen such that

By chosig Ak s properly it is possible to approximate formant bandwidths.

Boundary Relations

1

111

111

11

11

2

1

1

,0,0

Ac

Z

Ac

Z

rturtur

tu

tutuA

c

Ztututu

Z

tututu

g

g

gggg

gg

gg

01

01

0 AA

AAr

A

cZ gg

Consider a vocal-tract model consisting of N lossless concatenated tubes with total length l.

length of a tube = x = l/N

propagation time in a tube = = x/c

Let va(t) be the impulse samples of the impulse response with 2 intervals.

At the end of the tube va(t) can be written as:

Since the samples up to time N will be zero.

The Laplace transform and the frequency response are

The frequency response is periodic with 2/2; Va(+ 2/2) = Va()

10 2

kka kNtbNtbtv

Discrete Time Realization

0

2

0

2

1

20

k

kjka

k

ksk

sN

k

kNsk

sNa

ebV

ebeebebsV

If the impulse response of the system is not band limited then aliasing will occur.

The discrete-time frequency response can be obtained by = /2

Discrete Time Realization

00

2

k

kka

k

ska zbzVebsV

k

The effect of aliasing can be reduced by decreasing the individual tube lengths and hence the sampling period.

It can be shown that the transfer function of the discrete-time model has the form

Discrete Time Realization

zDzD

NkzDzrzDzD

zD

zazDzD

zAzV

N

kk

kkk

kN

kk

N

,2,1;

1

1;

111

0

1

2

When Ak > 0 all poles are inside the unit circle.

Ex: Let l=17.5 cm, c=350 m/sec. Find N to cover a bandwidth of 5000 Hz. (Assume that vocal tract impulse response and excitation are bandlimited to 5000 Hz.)

Soln: /2 is the cutoff bandwidth.

5000 Hz 10000 rad/sec

/2 = 10000 = 1/20000

N = l/c = 10

Since N is the order of the system, there are 5 complex conjugate poles

There can be 5 resonances in the given bandwidth.

Discrete Time Realization

Comparison of numerical simulation of differential equations and concatenated tube model (N=10)

Discrete Time Realization

Reflection coefficientsTube cross sections

SimulationConcatenated tubes