Digital Speech ProcessingDigital Speech Processing——Lectures 5Lectures 5--66

Sound Propagation in Sound Propagation in ggthe Vocal Tractthe Vocal Tract

1

BasicsBasics• can use basic physics to formulate air flow equations for vocal tract• need to make simplifying assumptions about vocal tract shape

and energy losses to solve air flow equationsgy q• some complicating factors:

– time variation of the vocal tract shape (we will look mainly at fixed shapes)

– losses in flow at vocal tract walls (we will first assume no loss then a– losses in flow at vocal tract walls (we will first assume no loss, then a simple model of loss)

– softness of vocal tract walls (leads to sound absorption issues)– radiation of sound at lips (need to model how radiation occurs)

l li ( li t th t b d l it l d t lti t b– nasal coupling (complicates the tube models as it leads to multi-tube solutions)

– excitation of sound in the vocal tract (need to worry about vocal source coupling to vocal tract as well as source-system interactions)

Bottom Line: simplify as much as possible and see h t l b t th h i f d

2

what we can learn about the mechanics of sound propagation in the human vocal tract

Sound in the Vocal TractSound in the Vocal Tract

• Issues in creating a detailed physical model– time varying acoustic systemtime varying acoustic system– losses due to heat conduction and friction in the

walls.– radiation of sound at the lips and nostrils– softness of the walls

l li3

– nasal coupling– excitation of sound in the vocal tract

Schematic Vocal TractSchematic Vocal Tract

PDEs mustbe solved in this region

• simplified vocal tract area => non-uniform tube with time varying cross section

• plane wave propagation along the axis of the tube (this assumption

4

plane wave propagation along the axis of the tube (this assumption valid for frequencies below about 4000 Hz)

• no losses at walls

Sound Wave PropagationSound Wave Propagation• using the laws of conservation of mass, momentum and energy, it can be

shown that sound wave propagation in a lossless tube satisfies the equations:

( / )

1 ( )

p u Ax tu pA A

ρ∂ ∂− =∂ ∂∂ ∂ ∂

21 ( )

where( ) d i h b i i d i

u pA Ax c t tρ∂ ∂ ∂

− = +∂ ∂ ∂

•( , ) sound pressure in the tube at position and time ( , ) volume velocity flow at position and

p p x t x tu u x t x= == = time

the density of air in the tubet

ρ =the velocity of sound

( , ) the 'area function' of the tube, i e the cross sectional area normal to the axis of the tube

cA A x t

ρ== =

5

i.e., the cross-sectional area normal to the axis of the tube, as a function of the distance along the tube and as a function of time

Solutions to Wave EquationSolutions to Wave EquationSolutions to Wave EquationSolutions to Wave Equation• no closed form solutions exist for the

propagation equationspropagation equations– need boundary conditions, namely u(0,t) (the

volume velocity flow at the glottis), and p(l,t), (the sound pressure at the lips) to solve the equationssound pressure at the lips) to solve the equations numerically (by a process of iteration)

– need complete specification of A(x,t), the vocal tract area function; for simplification purposes we willtract area function; for simplification purposes we will assume that there is no time variability in A(x,t) => the term related to the partial time derivative of Abecomes 0

– even with these simplifying assumptions, numerical solutions are very hard to compute

6Consider simple cases and extrapolate results

to more complicated cases

Uniform Lossless TubeUniform Lossless Tube• Assume uniform lossless tube => A(x,t)=A

(shape consistent with /UH/ vowel)

u(0,t)p(l,t)

piston

ρ∂ ∂− =∂ ∂∂ ∂

p ux A tu A p

∂ ∂− =

∂ ∂∂ ∂

v iLx ti v

72ρ

∂ ∂− =∂ ∂

u A px c t

∂ ∂− =

∂ ∂i vCx t

AcousticAcoustic--Electrical AnalogsElectrical AnalogsAcousticAcoustic Electrical AnalogsElectrical AnalogsAcoustic ElectricalAcoustic Electrical

pressurevolume velocity

pu==

voltagecurrent

vi==

2

y/ acoustic inductance/( ) acoustic capacitanceA

A cρ

ρ

=

=

currentinductancecapacitance

iLC==

uniform acoustic tube

0

lossless transmission line terminated in a short circuit, ( ) at one end excitedv l t 0

0( , ) , at one end, excited

by a current source ( , ) ( )at the other end

G

v l ti t i t

==

8

at the other end

Traveling Wave SolutionTraveling Wave SolutionTraveling Wave SolutionTraveling Wave Solution assume traveling wave solution

( ) ( / ) ( / )+

•

⎡ ⎤t t t ( / )u t x c+ ( / )u t x c+ ( , ) ( / ) ( / )

( , ) ( / ) ( / )ρ

+ −

+ −

⎡ ⎤= − − +⎣ ⎦

⎡ ⎤= − + +⎣ ⎦

u x t u t x c u t x c

cp x t u t x c u t x cA

0( / )u t x c+ − 1( / )u t x c+ −

-- ( / )wave travelling forward-- ( / )wave travelling backward

+

−

−

+

Au t x cu t x c

boundary conditio• ns at the glottis and at the lips gives:(0, ) ( )( , ) 0

Ω= Ω

=

j tGu t U e

p t( , ) 0since the differential equations are linear with constant

coefficients, the solutions must be of the form•

p t

9

( / ) ( / )(

+ + Ω −

−

− = j t x cu t x c K eu ( / )/ ) − Ω ++ = j t x ct x c K e

Traveling Wave SolutionTraveling Wave SolutionTraveling Wave SolutionTraveling Wave Solution solve for and + −• K K

( / ) ( / )

(0, ) ( )

( ) 0 ρ

Ω + Ω − Ω

+ Ω − − Ω +

= Ω = −

⎡ ⎤= = +⎣ ⎦

j t j t j tG

j t c j t c

u t U e K e K ecp t K e K e

2 /

2 / 2 /

( , ) 0

( )( ) ;1 1

Ω+ −

Ω Ω

⎡ ⎤= = +⎣ ⎦

Ω= Ω = −

j cG

G j c j c

p t K e K eA

UeK U K2 / 2 /

(2 ) / /

( )1 1

solve for ( , ) and ( , )

Ω Ω

Ω − Ω

+ +•

+

G j c j c

j x c j x c

e eu x t p x t

e e⎡ ⎤( )

( , ) ( )1

Ω += Ω

+

j jj t

Ge eu x t U e 2 /

(2 ) / /ρ

Ω

Ω − Ω

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤

j c

j x c j x c

e

c e e10

( )

2 /( , ) ( )1

ρ ΩΩ

⎡ ⎤−= Ω ⎢ ⎥+⎣ ⎦

j jj t

G j c

c e ep x t U eA e

Traveling Wave SolutionTraveling Wave SolutionTraveling Wave SolutionTraveling Wave Solution

look at solution for ( )• u t

2

21

/

/

look at solution for ( , )

( , ) ( ) ( , )Ω

Ω ΩΩ

•

⎡ ⎤= Ω = Ω⎢ ⎥+⎣ ⎦

j cj t j t

G j c

u t

eu t U e U ee1

1 giving for the transfer function of volume velocity

( )

+⎣ ⎦•

Ω

je

U 1( , ) ( )( ) cos( / )Ω

Ω = =Ω Ωa

G

UVU c

11

Overall Transfer FunctionOverall Transfer Function• consider the volume velocity at the lips (x=l) as a

function of the source (at the glottis)

1 ( , ) ( , )

( )

Ω

Ω

= Ω

Ω

j t

j t

u t U e

U e

1

( )cos( / )

( , ) ( )

Ω= ΩΩ

Ω= Ω =

j tGU e

cU V

formants of uniform tube

2 ; 35 000 17 5cm/sec; = cmf cπΩ = = ( )( ) cos( / )

ΩΩ Ωa

G

VU c

Frequency response of

2 ; 35,000 17.52

2 2 3 5

cm/sec; cm

h

f cf

c cf f

ππ

π π π π π

ΩΩ

=

⎛ ⎞Frequency response of uniform tube in terms of

volume velocities

2 2 3 5cos 0 , , ,...2 2 2

(2 1), 0,1, 2,...4

when

i.e., when n

f fc c

cf n n

π π π π π⎛ ⎞= =⎜ ⎟⎝ ⎠

= ⋅ + =

120 1 2500 1500 2500 ...Hz; Hz; Hz,f f f= = =

Hardwalled Tube and BuzzerHardwalled Tube and Buzzer

13

Spectrum SlicesSpectrum Slices

The formants are not at theare not at the frequencies 500, 1500, 2500 H2500, … Hz. What are some possiblesources of error in thiserror in this experiment?

14

Summary of Solution of Sound Propagation Summary of Solution of Sound Propagation Equations in the Vocal TractEquations in the Vocal Tractqq

( )/Step 1--Basic sound wave propagation equations

ρ∂∂

− =∂ ∂

u Apx t

( / )

( / )

( / )( / )

Step 6--Simplified forward and backward waves+ + Ω −

− − Ω +

− =

+ =

j t x c

j t x c

u t x c K eu t x c K e

( )2

1

(0 ) ( )Step 2--Boundary conditions

sound source at glottis

ρ

∂ ∂∂∂ ∂

− = +∂ ∂ ∂

⇒

x tpAu A

x c t t

u t u t

(2 ) / /

2 /

( / )( , ), ( , )

( , ) ( )1

Step 7--Determine and , and solve for + −

Ω − ΩΩ

Ω

+ =

⎡ ⎤+= Ω ⎢ ⎥+⎣ ⎦

j x c j x cj t

G j c

u t x c K eK K u x t p x t

e eu x t U ee

(0, ) ( )( , ) 0

sound source at glottisno pressure at lips

Step 3--Simplifying assumption

= ⇒

= ⇒Gu t u t

p t fixed area

ρ⇒

∂ ∂− =∂ ∂

Ap u

A t

( , )p x t(2 ) / /

2 /( )1

( , )( , ) ( , )

Step 8--Solve for

ρ Ω − ΩΩ

Ω

Ω

⎡ ⎤−= Ω ⎢ ⎥+⎣ ⎦

= Ω

j x c j x cj t

G j c

j t

c e eU eA e

u tu t U e

2

Step 4--Assume travelling waves form of solutionρ

∂ ∂∂ ∂

− =∂ ∂

x A tu A px c t

( , ) ( , )

( , ) 1( )( ) cos( / )

Step 9--Solve for transfer function of volume velocity

Step 10--Determine tube resonan

ΩΩ = =

Ω ΩaG

UVU c

ces( , ) ( / ) ( / )

( , ) ( / ) ( / )

Step 5--Simplified boundary conditions

ρ

+ −

+ −

= − − +

⎡ ⎤= − + +⎣ ⎦

u x t u t x c u t x ccp x t u t x c u t x c

A

Step 10 Determine tube resonan17.5 35,000(2 1) 500,1500,2500,...

4

ces cm; cm/sec

Hz

Step 11--Interpretation of traveling wave solution

= = ⇒+

= =n

cn cf

15

(0, ) ( )= ΩGu t U e( , ) 0

Ω

=

j t

p t/

2 /

( , ) 2 1( )( ) 1 cosh( / )

Step 11 Interpretation of traveling wave solution−

−= = =+

s c

a s cG

U s eV sU s e s c

Traveling Wave SolutionTraveling Wave SolutionTraveling Wave SolutionTraveling Wave Solution

using -transform notation ( - ) we getσ• = Ωs s j

2

2 11

/

/

using transform notation ( ) we get( , )( )

( ) h( / )

σ−

−

• = Ω

= = =s c

a s c

s s jU s eV sU 21 /( ) cosh( / )+a s c

GU s e s c

16

Frequency Domain RepresentationFrequency Domain Representationq y pq y p

we can alternatively express ( , ) and ( , ) assin( ( ) / )

•Ω

p x t u x tx c

0sin( ( ) / ) ( , ) ( )

cos( / )cos( ( ) / )( ) ( )

Ω

Ω

Ω −= Ω

ΩΩ −

Ω

j tg

j t

x cp x t jZ U ec

x cu x t U e( ( ) ) ( , ) ( )cos( / )

where

Ω= ΩΩ

•

j tgu x t U e

c

0 characteristρ= =

cZA

ic acoustic impedance of tube

17

Alternative Wave Equation Solution (avoids solution Alternative Wave Equation Solution (avoids solution f f d d b k d t lli )f f d d b k d t lli )

express ( , ) and ( , ) as complex transfer functions of the form:) Ω

•j t

p x t u x tp(x t) P(x Ω e

for forward and backward travelling waves)for forward and backward travelling waves)

) ( , ) ( , ) inserting these representations

Ω

Ω

=

= Ω•

j t

j t

p(x,t) P(x,Ω eu x t U x e

back into the wave equation gives:

2

, / acoustic impedance per unit length

, /( ) acoustic admittance per unit length

ρ

ρ

− = = Ω =

− = = Ω =

dP ZU Z j Adx

dU YP Y j A c

Solution 1

, /( ) acoustic admittance per unit length

can show that solutions of wa

ρΩ

•

YP Y j A cdx

ve equation have form: ( , ) γ γ−Ω = +x xP x Ae Be

0

( , )

/- by using boundary conditions ( )

γ γ

γ

−Ω = +

= = ΩΩ =

x xU x Ce De

ZY j cP 0( ) ( ) can solve forΩ = ΩU U A B C D

18

0 by using boundary conditions, ( , ) , Ω =P 0( , ) ( ), can solve for , , ,Ω = ΩgU U A B C D

Effects of Losses in VTEffects of Losses in VT• several types of losses to be considered

– viscous friction at the walls of the tube– viscous friction at the walls of the tube– heat conduction through the walls of the tube– vibration of the tube walls

• loss will change the frequency response of the tubeloss will change the frequency response of the tube• consider first wall vibrations

– assume walls are elastic => cross-sectional area of the tube will change with pressure in the tubechange with pressure in the tube

– assume walls are ‘locally’ reacting => A(x,t) ~ p(x,t)– assume pressure variations are very small

0 ( , ) ( , ) ( , )A x t A x t A x tδ= +

19

Effects of LossEffects of Loss there is a differential equation relationship between area

perturbation ( , ) and the pressure variation, ( , ) of A x t p x tδ•

2

2

the form:( ) ( ) ( ) ( , ) whereW W W

d A d Am b k A p x tdt dtδ δ δ+ + =

( ) mass/W

dt dtm x = unit length of the vocal tract wall

( ) damping/unit length of the vocal tract wall( ) ff / f

Wb x =

( ) stiffness/unit length of the vocal tract wall neglecting second order terms in / and , the ba

Wk xu A pA

=

• sic wave equations become

0

1

q( / )

( ) ( )

u Apx t

pA Au A

ρ

δ

∂∂− =∂ ∂

∂ ∂∂ ∂

20

0 02

1 ( ) ( ) pA Au Ax c t t t

δρ

∂ ∂∂ ∂− = + +∂ ∂ ∂ ∂

Losses in Frequency DomainLosses in Frequency Domain

0

consider a time-invariant constant area tube excited by a complex volume velocity source

( ) ( ) ( ) j tu t u t U e Ω

•

= = Ω0 ( ) ( , ) ( ) since the loss differential equation is linear and time-invariant,

the form for

G Gu t u t U e= = Ω

• and is:

j t

p(x,t) u(x,t)Ω ( , ) ( , )

( , ) ( , ) ( , ) ( , ) substituting into the wave equations yields the following:

j t

j t j t

p x t P x eu x t U x e A x t A x eδ δ

Ω

Ω Ω

= Ω

= Ω = Ω•

( , ) , ( , )P Z x U Z x jx

ρ∂− = Ω Ω = Ω∂ 0

0

( )( )

A xA xU∂

Solution 2—same as Solution 1 with

new term, YW, 0

2

1

( ) ( , ) ( , ) , ( , )

( , ) ( )

W

W

A xU Y x P Y x P Y x jx c

Y x k x

ρ∂

− = Ω + Ω Ω = Ω∂

Ω =

and with Z and Yterms being

functions of x

21

( )( ) ( )

WW W

k xj m x b xj

Ω + +Ω

Effects of Loss on FREffects of Loss on FR

• using estimates for mW, bW, and kW from measurementsand kW from measurements on body tissue, and with boundary condition at lips of (l t) 0 tp(l,t)=0, we get:

( , ) ( )( )Ω

Ω =Ωa

UV jU ( )Ωa

GUcan similarly account for effects of viscous friction and thermal

• complex poles with non-zero bandwidths

li h l hi h f i f

conduction at the walls

• increases bandwidth of complex poles

22

• slightly higher frequencies for resonances

• most effect at lower frequencies

p p

• decreases resonance frequency (slightly)

Friction and Thermal Conduction LossesFriction and Thermal Conduction LossesVi f i ti b t d f i th f• Viscous friction can be accounted for in the frequency domain by including a real, frequency dependent term in the expression for the acoustic impedance, Z, of the form:

20 0

( )( , ) / 2[ ( )] ( )

( ) is the circumference of the tube in cm

S xZ x jA x A x

S x

ρλρμΩ = + Ω

• Heat conduction accounted for by adding a real

3 ) is the coefficient of friction (0.000186) is the density of air in the tube (0.00114 gm/cm

μ

ρ

• Heat conduction accounted for by adding a real frequency dependent term to the acoustic admittance, of the form:

( )( )( 1) A xS x η λΩ 02 2

( )( )( 1)( , )2

is the specific heat at constant pressure (0.24)i th ti f ifi h t t t t

p

p

A xS xY x jc c c

c

η λρ ρ ρ

− ΩΩ = + Ω

23

is the ratio of specifiec heat at constant pressure to that at constant volume (1.4) is the coef

η

λ ficient of heat conduction (0.000055)

Friction and Thermal Conduction LossesFriction and Thermal Conduction Losses

Main effect of friction andMain effect of friction and thermal conduction losses is that the formant bandwidths increase

i f i ti d th l• since friction and thermal losses increase with Ω1/2, the higher frequency resonances experience a pgreater broadening than the lower resonances• the effects of friction and thermal loss are smallthermal loss are small compared to the effects of wall vibration for frequencies below 3-4 kHz

24

Effects of Radiation at LipsEffects of Radiation at LipsEffects of Radiation at LipsEffects of Radiation at Lips• we have assumed p(l,t)=0 at the lips (the acoustical analog of a

short circuit) => no pressure changes at the lips no matter how ) p g pmuch the volume velocity changes at the lips

• in reality, vocal tract tube terminates with open lips, and sometimes open nostrils (for nasal consonants)

• this leads to two models for sound radiation at the lips

headmodel valid when

lip opening is small compared to size of spherelips

head

to size of spherelips

25

Radiation at LipsRadiation at LipsRadiation at LipsRadiation at Lips• using the infinite plane baffle model for radiation at the lips, can

replace the boundary condition for a complex sinusoid input with the following:

( , ) ( ) ( , ) where

( ) -- 'radiation impedance' or 'radiation load' at lips

Ω = Ω Ω

ΩΩ =

+ Ω

L

r rL

r r

P Z Uj L RZ

R j L this 'radiation load' is the equivalent of a parallel connection of a radiation

resista•

nce, , and a radiation inductance, . Suitable values for thesecomponents are:

r rR L

2

128 89 3

components are:

, , where is the radius of the opening and is the

velocity of soundπ π

= =r raR L a cc

26

velocity of sound

Behavior of Radiation LoadBehavior of Radiation LoadBehavior of Radiation LoadBehavior of Radiation Loadradiationradiation

losses most significant at

hi hhigher frequencies

( ) ΩΩ =

+ Ωr r

Lr r

j L RZR j L+ Ωr rR j L

0 at low frequencies, ( ) (short circuit termination) old solutionLZ• Ω ≈ ⇒

27

q , ( ) ( ) at mid-range frequencies, ( ) (inductive load) at higher frequencies, ( ) (resistive load)

L

L r r r

L r r r

Z j L R LZ R L R

• Ω ≈ Ω ⇒ Ω

• Ω ≈ ⇒ Ω

Overall Transfer FunctionOverall Transfer Function

• for the case of a uniform, time-invariant tube with yielding walls, friction and thermal losses, and radiation loss of an i fi it l b ffl linfinite plane baffle, can solve the wave equations for the transfer function:

( , ) ( )( )

i i t t l tti f f

ΩΩ =

ΩaG

UV jU

assuming input at glottis of form: (0, ) ( ) Ω

•

= Ω j tGU t U e

• higher bandwidths, lower resonance frequencies

28

higher bandwidths, lower resonance frequencies

• first resonance is primarily determined by wall loss

• higher resonance bandwidths are primarily determined by radiation losses

Vocal Tract Transfer FunctionVocal Tract Transfer Function• look at transfer function of pressure at the lips and

volume velocity at the glottis, which is of the form:

( , ) ( , ) ( , )( )( ) ( , ) ( )Ω Ω Ω

Ω = = ⋅Ω Ω Ωa

G G

P P UHU U U

( ) ( )= Ω ⋅ ΩL aZ V

Notice:

• zero at Ω=0

• high frequency h i (

29

emphasis (compare with previous chart)

Vocal Tract Transfer Functions Vocal Tract Transfer Functions f V lf V lfor Vowelsfor Vowels

• using the frequency domain equations,using the frequency domain equations, can compute the frequency response functions for a set of area functions of the vocal tract for various vowel sounds, using all the loss mechanisms, assuming:– A(x), 0≤x≤l (glottis-to-lips) measured and

knownstead state so nds (dA/dt 0)– steady state sounds (dA/dt=0)

– measure U(l,Ω)/UG(Ω) for the vowels /AA/ /EH/ /IY/ /UW/

30

/AA/,/EH/,/IY/,/UW/

Area Function from XArea Function from X--Ray PhotographsRay Photographs

Gunnar Fant, Acoustic Theory of Speech P d i

31

Production,Mouton, 1970

Area Functions and FR for Vowels /AA/ d /EH//AA/ and /EH/

32

Area Functions and FR for Vowels Area Functions and FR for Vowels /IY/ d /UW//IY/ d /UW//IY/ and /UW//IY/ and /UW/

33

VT Transfer FunctionsVT Transfer FunctionsVT Transfer FunctionsVT Transfer Functions• the vocal tract tube can be characterized by

a set of resonances (formants) that depend on the vocal tract area function-with shifts d t l d di tidue to losses and radiation

• the bandwidths of the two lowest (F1 d F2) d d i ilresonances (F1 and F2) depend primarily on

the vocal tract wall lossesthe band idths of the highest resonances• the bandwidths of the highest resonances (F3, F4, ...) depend primarily on viscous friction thermal losses and radiation losses

34

friction, thermal losses, and radiation losses

Nasal Coupling EffectsNasal Coupling EffectsNasal Coupling EffectsNasal Coupling Effects• at the branching point

sound pressure the same as at input of• sound pressure the same as at input of each tube

• volume velocity is the sum of the volume velocities at inputs to nasal and poral cavities

• can solve flow equations numerically

• results show resonances dependent on di ti

pshape and length of the 3 tubes

• closed oral cavity can trap energy at certain frequencies, preventing those frequencies from appearing in the nasal output => anti

radiation impedance

open circuitfrom appearing in the nasal output => anti-resonances or zeros of the transfer function

• nasal resonances have broader bandwidths than non-nasal voiced sounds => due to

35

greater viscous friction and thermal loss due to large surface area of the nasal cavity

Sound Excitation in VTSound Excitation in VTSound Excitation in VTSound Excitation in VT1. air flow from lungs is modulated by vocal cord g y

vibration, resulting in a quasi-periodic pulse-like source

2 i fl f l b b l i2. air flow from lungs becomes turbulent as air passes through a constriction in the vocal tract, resulting in a noise-like sourceresulting in a noise like source

3. air flow builds up pressure behind a point of total closure in the vocal tract => the rapid release of this pressure, by removing the constriction, causes a transient excitation (pop-like sound)

36

like sound)

Vocal Cord SimulationVocal Cord Simulation

J. L. Flanagan and K. Ishi aka did the firstIshizaka, did the first detailed simulationsof vocal cord oscillators. Subsequent researchersSubsequent researchers have refined the model for singing voice.

37

Voiced Excitation in VTVoiced Excitation in VT

• lung pressure is increased, causing air to flow out of the lungs and through the opening between the vocal cords (the glottis)

• according to Bernoulli’s law, if the tension in the vocal cords is properly adjusted, the reduced pressure in the constriction allows the cords to come together, thereby constricting air flow (see dotted lines above)

• because of closure of the vocal cords, pressure increases behind the vocal cords and eventually reaches a level sufficient to force the vocal cords to open and allows air to flow through the glottis again

38sustained Bernoulli oscillations => rate of opening and closing is controlled by air pressure in the lungs, tension and stiffness of the vocal cords, and area of the glottal opening; the vocal

tract area at the glottis also effects the rate

Glottal Excitation ModelGlottal Excitation ModelGlottal Excitation ModelGlottal Excitation Model• vocal tract acts as a load on the vocal cord oscillator

• time varying glottal resistance and inductance-both functions of 1/AG(t) => when AG(t)=0 (total closure), impedance is infinite and volume velocity is zero

build-up of glottal pulses

39

Rosenberg Glottal Pulse and Rosenberg Glottal Pulse and SSSpectrumSpectrum

1 10 5 1 02

π= − ≤ ≤≤ ≤

[ ] . [ cos( / )][ ( ) /( )]

g n n N n NN N N N N1 2 1 1 220

π= − ≤ ≤ +=

cos[ ( ) /( )]otherwise

n N N N n N N

40

Note the high frequency fall off due to the lowpass pulse shape

Other Excitation SourcesOther Excitation SourcesOther Excitation SourcesOther Excitation Sources• voiceless excitation occurs at a

constriction of the vocal tract when volume velocity exceeds a critical value (called the (Reynolds number) => this can be modeled using a randomly time varying source at the point y g pof constriction

• a combination of voiced and voiceless excitation is used forvoiceless excitation is used for voiced fricatives

• a total closure of the tract is used for stop consonants

41

used for stop consonants

SourceSource--System ModelSystem ModelSourceSource System ModelSystem Model

Pitch, Formants, Vocal Tract AreaVoiced/Unvoiced,

Amplitude

Formants, Vocal Tract Area Functions, Articulatory

Paramameters

42

Summary of Losses, Radiation Summary of Losses, Radiation d B d C di i Effd B d C di i Effand Boundary Condition Effectsand Boundary Condition Effects

• considered losses due to friction at walls, heat conduction through walls, vibration of walls

• losses introduce new terms into sound propagation equations• effects of losses are increased bandwidth of complex poles (from 0• effects of losses are increased bandwidth of complex poles (from 0

to a finite quantity) and changes in the regular spacing of the resonance (formant) frequencies of the tract

• radiation at lips adds a parallel resistance and inductance t d i t i ifi t t hi h f icomponent and is most significant at higher frequencies

• nasal coupling adds components to solution which include anti-resonances (frequency response zeros)

• sound excitation models lead to simplified model with a distinctsound excitation models lead to simplified model with a distinct glottal pulse (for voiced speech) with strong high frequency drop-off in level

• the overall vocal tract is well modeled as a variable excitation generator exciting a time varying linear system

43

generator exciting a time-varying linear system

Lossless Tube ModelsLossless Tube ModelsLossless Tube ModelsLossless Tube Models• approximate A(x) by a

i f l lseries of lossless, constant cross sectional area, acoustic tubes of the form shown at thethe form shown at the right

• as the number of tubes becomes larger (smallerbecomes larger (smaller approximation error for the vocal tract area function), the ),approximation error for modeling the vocal tract goes to zero

44How do we use the lossless tube model to solve for various

vocal tract transfer functions

Concatenated Tube ModelsConcatenated Tube ModelsConcatenated Tube ModelsConcatenated Tube Models

45

Wave Propagation in Lossless TubesWave Propagation in Lossless TubesWave Propagation in Lossless TubesWave Propagation in Lossless Tubes

• since each individual tube is lossless, can solvesince each individual tube is lossless, can solve the basic wave equation for each individual tube, giving:

for tube:ρ

•

⎡ ⎤

thkc 0

0

( , ) ( / ) ( / ) ,

( , ) ( / ) ( / ),

ρ + −

+ −

⎡ ⎤= − + + ≤ ≤⎣ ⎦

= − − + ≤ ≤

k k k kk

k k k k

cp x t u t x c u t x c xA

u x t u t x c u t x c x

0

th where is the distance measured from the left-hand end of the tube and (+•

≤ ≤ k k

x k( x ) u

th

) and ( ) are positive-going and

ti i t li i th t b

−ku

k

46

th negative-going traveling waves in the tubek

Wave Propagation in Lossless TubesWave Propagation in Lossless Tubes

boundary conditions at edges of adjacent tubes state that both pressure and volume velocity must be continuous in both•

1

p y

time and space consider junction between and ( ) tubesth stk k• +

47

Lossless Tube JunctionLossless Tube JunctionLossless Tube JunctionLossless Tube Junction1

0 at the junction between and ( ) tubes, we get:

( ) ( )• +th stk k

t t1

1

00

( , ) ( , ) ( , ) ( , ) substituting from the previous set of equations, we get:

+

+

=

=

•

k k k

k k k

p t p tu t u t

1 ( ) ( )τ τ+ −+ ⎡ ⎤− + + =⎣ ⎦k

k k k k kk

A u t u t uA 1 1( ) ( )

( ) ( ) ( ) ( )τ τ

+ −+ +

+ − + −

+

− − + = −

kt u t

u t u t u t u t1 1 ( ) ( ) ( ) ( ) where / is the time for a wave to travel the length

of the tube

τ ττ

+ ++ =

• =k k k k k k

k kth

u t u t u t u tc

k at the junction between tubes, part of the positive g• oing wave

is propagated to the right while part is reflected back to the left similarly part of the negative going wave is propagated to the•

48

y p g g g p p g left while part is reflected back to the right

Lossless Tube ModelLossless Tube Model1 1 solve for ( ) and ( ) in terms of ( ) and ( )

to see how forward and reverse travelling waves propagatek k k k k ku t u t u t u tτ τ+ − − ++ +• + −

1 11 1

1 1

2

to see how forward and reverse travelling waves propagate

( ) ( ) ( )k k kk k k k

k k k k

A A Au t u t u tA A A A

τ+ + −+ ++ +

+ +

⎡ ⎤ ⎡ ⎤−= − +⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤11

1 1

2 ( ) ( ) ( )

the quantity

k k kk k k k k

k k k k

A A Au t u t u tA A A A

τ τ− + −++

+ +

⎡ ⎤ ⎡ ⎤−+ = − − +⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

•

11

1

q y

amount of ( ) that is reflected at the junctionk kk k

k k

A Ar u tA A

−++

+

⎡ ⎤−= =⎢ ⎥+⎣ ⎦

th is called the 'reflection cokr•

1 1

1 1

1

efficient' for the junction, with , and rewriting above equations as

( ) ( ) ( ) ( )

th

k

k k k k k k

kr

u t r u t r u tτ+ + −+ +

− ≤ ≤

= + − +

49

1 1

11

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) k k k k k k

k k k k k k ku t r u t r u tτ τ+ +

− + −++ = − − + −

Lossless Tube ModelLossless Tube ModelLossless Tube ModelLossless Tube Model

1 11( ) ( ) ( ) ( ) τ+ + −+ += + − +k k k k k ku t r u t r u t1 1

11

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) τ τ+ +

− + −++ = − − + −

k k k k k k

k k k k k k ku t r u t r u t

50

Boundary ConditionsBoundary ConditionsBoundary ConditionsBoundary Conditions• for a 5-tube model there are 5 sets of forwardfor a 5 tube model there are 5 sets of forward

and backward delays, 4 junctions (each characterized by a reflection coefficient), and a set of boundary conditions at the lips and glottis

• assume N sections (indexed 1 to N) starting at the glottis

• want to relate pressure (pN(lN,t)) and volume l i ( (l )) f Nth b hvelocity (uN(lN,t)) at output of Nth tube to the

radiated pressure and volume velocity (at the lips)

51

lips)

Boundary ConditionsBoundary ConditionsBoundary ConditionsBoundary Conditions

earlier we saw that at the lips we have• earlier we saw that at the lips we have ( , ) ( , )

giving the time domain relation

•Ω = ⋅ Ω

•N N L N NP Z U

giving the time domain relation

( ) ( ) ( ) ( )ρ τ τ τ τ+ − + −

•

⎡ ⎤ ⎡ ⎤− + + = − − +⎣ ⎦ ⎣ ⎦N N N N L N N N NN

c u t u t Z u t u tA

solving for ( ) we get

(

τ−

−

• +N

N N

N

u t

u t ) ( )τ τ++ = − −N L N Nr u t(N

1 1

) ( ) where the reflection coefficient at the lips is

/ρ

•

⎡ ⎤−≤ ≤⎢ ⎥

N L N N

N Lc A Zr52

1 1 /

ρρ

− ≤ = ≤⎢ ⎥+⎣ ⎦N L

LN L

rc A Z

Termination at LipsTermination at LipsTermination at LipsTermination at Lips( ) ( )N N L N Nu t r u tτ τ− ++ = − −

output volume velocity at the lips is( ) ( ) ( )τ τ+ −

•

+u t u t u t

1

( , ) ( ) ( )

( ) ( )

τ τ

τ+

= − − +

= + −N N N N N N

L N N

u t u t u t

r u t

53

Adding the Excitation SourceAdding the Excitation Source• assume that the excitation source and the vocal tract

are linearly separable, then at the first tube we have the boundary relation:

1 1

1 11 1

0 0 ( , ) ( ) ( , ) /

( ) ( ) ( ) ( ) ( ) ρ + −+ −

Ω = Ω − Ω

⎡ ⎤+− = − ⎢ ⎥

⎣ ⎦

G G

G

U U P Z

u t u tcu t u t u tA Z1 1

1

1

1 solving for ( ) we get:

( )( ) ( ) ( )

+

+

⎢ ⎥⎣ ⎦

•

+

GG

G

A Z

u trt t t1 1

12

( ) ( ) ( ) ( )

where the reflection coefficient, , is

+ −+= +

•

GG G

G

ru t u t r u t

r

⎡ ⎤c

1 −

− ≤ =G

G

Zr 1 1

ρ

ρ

⎡ ⎤⎢ ⎥⎢ ⎥ ≤⎢ ⎥+⎢ ⎥⎣ ⎦

G

cAcZ

A

54

1⎢ ⎥⎣ ⎦

G A

Lossless Two Tube ModelLossless Two Tube Model

2 2 volume velocity at lips is ( ) ( , ) transfer function from glottis to lips is• =

•Lu t u t

( )

( ) ( )( )

)( )( ) τ τΩ +

ΩΩ =

ΩL

aG

j

UVU

( 1 2

1 2 1 2

12 2 2

1 1

0 5 1 1 11

( )

( )

)( )( ) τ τ

τ τ τ τ

− Ω +

− Ω − Ω − Ω +

+ + +=

+ + +

•

jG L

j j jG L L G

. ( r r r er r e r r e r r e

1 2 note total delay of ( ) is total propagation delay from τ τ+

55

1 2y ( ) p p g y glottis to lips

TwoTwo--Tube Model for Vowel /AA/Tube Model for Vowel /AA/TwoTwo Tube Model for Vowel /AA/Tube Model for Vowel /AA/

56

TwoTwo--Tube Model for Vowel /IY/ Tube Model for Vowel /IY/ (L Li )(L Li )(Losses at Lips)(Losses at Lips)

57

TwoTwo--Tube Model for Vowel /IY/ Tube Model for Vowel /IY/ (L Gl i )(L Gl i )(Losses at Glottis)(Losses at Glottis)

58

Two Tube Model ResonancesTwo Tube Model Resonances

59

Summary of VT ModelSummary of VT ModelSummary of VT ModelSummary of VT Modelu(0 t) u(l t)

Vocal Tract zLzGuG(t)

u(0,t) u(l,t)

60

Lossless VT SectionsLossless VT SectionsLossless VT SectionsLossless VT Sections1+rk Lips: 1+rL

1-rk

-rk rk

k A l k+1 A l

-rL rk

f1

1

; reflection coefficientk kk

k k

A ArA A

+

+

−=

+; reflection coefficient

LN

L

c zAr

ρ−

=

k, Ak, lk k+1, Ak+1, lk+1 N, AN, lN Infinite tube, zL

;L

LN

c zAρ

+(1+rG)/2Glottis:

rG

I fi it t b

1 ; reflection coefficientG

G

G

c zAr c z

ρ

ρ

− +=

+

61

1, A1, l1Infinite tube, zG1

GzA

+

Relationship to Digital FiltersRelationship to Digital FiltersRelationship to Digital FiltersRelationship to Digital Filters• observation that lossless tube model appears similar to digital filter

implementations => consider a system of N lossless tubes, each ofimplementations consider a system of N lossless tubes, each of length Δx=l/N where l is the overall length of the vocal tract

62 all delays equal to , the time to traverse the length of one tubex/cτ• = Δ

Relation to Digital FiltersRelation to Digital FiltersRelation to Digital FiltersRelation to Digital Filters consider response to a unit impulse, ( ) ( )Gu t tδ• =

impulse propagates down the series of tubes, being partially reflected and partially propagated at the junctions; can show that the impulse respo

•

nse (the volume velocity at the lips) is of the form:p

01

2

( y p )

( ) ( ) ( )a kk

v t t N t N kα δ τ α δ τ τ∞

=

= − + − −∑

initial propagation with delay N

x Tau

successive impulses with one or more

reflections

2 2( )

overall system function is:

s N k sN s k∞ ∞

+

•

∑ ∑63

2 2

0 0

( ) ( ) s N k sN s ka k k

k kV s e e eτ τ τα α− + − −

= =

= =∑ ∑

Relation to Digital FiltersRelation to Digital FiltersRelation to Digital FiltersRelation to Digital Filters2 ( ) sN s k

a kV s e eτ τα∞

− −= ∑0

( )

ˆ ˆ ( ) ( ) ( )

a kk

sNa a ae V s v t v t Nτ τ=

−= ↔ = +

∑

64

Relation to Digital FiltersRelation to Digital FiltersRelation to Digital FiltersRelation to Digital Filtersˆ frequency response ( ) is• ΩaV

2

0

ˆ ( )

can easily show that

τα∞

− Ω

=

Ω =

•

∑ j ka k

kV e

22

2

can easily show that

ˆ ˆ ( ) ( ) periodic in frequency

if input at glottis is bandlimited to frequencies below / (

πτ

π τ

Ω + = Ω =>

•

a aV V

)2 if input at glottis is bandlimited to frequencies below / (π τ•2

) we can sample the input at the glottis with period and filter the sampled signal with a digital filter whose IR is

τ=T

00 0

2

ˆ [ ]

for , the delay of

α

τ

= ≥

= <• =

nv n nn

T 2corresponds to a shift of / samplesτN N

65

2 for , the delay ofτT 2 corresponds to a shift of / samples giving the digital filter implementation of part b.

τN N

Signal Flow Graph ModelSignal Flow Graph ModelSignal Flow Graph ModelSignal Flow Graph Model• ˆˆz-transform of ( ) is simply ( ) with replaced by , givingsTv n V s e z

α∞

−

=

=

•

∑0

z transform of ( ) is simply ( ) with replaced by , giving

ˆ ( )

signal flow graph for the discrete system can be obtained from the flow

a

kk

k

v n V s e z

V z z

g g p y graph of the analog system by re

τ τ• =

placing each node variable in the analog system by the corresponding sequence of samples each second delay is replaced by a 1/2 sample delay since / 2T

−••

1/ 2 section propagation delay is bette

zr digital signal flow graph model is of the form of a ladder with the

delay elements only in the upper and lower paths; in the upper paths signals propagate to the right and in the lower paths the =>y propagate to the left can move delay elements in the lower paths to the upper paths; this only changes the overall delay from input to output, but has no practical

66

significance

Signal Flow GraphsSignal Flow Graphs

Analog model

D-T equivalent for bandlimited inputs

67D-T equivalent without half-sample delays

Transfer Function of Lossless Tube ModelTransfer Function of Lossless Tube Model

want to determine( )( )( )

•

= LU zV zU z

( )1/ 21/ 2

1/ 21 1 1 1

( ) at junctions we have the relations

( ) ( ) 1 ( ) ( ) ( ) ( )1 1

+ + − − + + −+ + + +

•

= + + ⇒ = −+ +

G

kk k k k k k k k

U z

r zzU z U z z r U z r U z U z U zr r

( )1 1/ 21

1 1

( ) ( ) ( )(1 ) ( )−

− + − − − −+

+ +

−= − + − ⇒ =

k k

kk k k k k k

r r

r zU z U z z r U z r z U z1/ 2 1/ 2

1 1( ) ( )1 1

at lips use same formulation with fictitious ( 1) tube that is infinitely

−+ −+ ++

+ +

• +

k kk k

zU z U zr r

N st

1

at lips use same formulation with fictitious ( 1) tube that is infinitelylong (no negative going wave)=> ( 1) tube terminated in its characteristicimpedance

• ++

N

N stN st

U ( ) ( )+ = Lz U z)( zU k

+ )(1 zU k++

1+NU

1

1

( ) ( )

( ) 0

,ρ

−+

+

=

= =

L

N

N N LL

z U z

U zcA r r

Z)( zU k

−)(1 zU k

−+

68

L

Transfer Function of Lossless Tube Transfer Function of Lossless Tube M d lM d lModelModel

• at the glottis22 + −= −

+ +

•

1 122( ) ( ) ( )

1 1putting it all together gives

GG

G G

rU z U z U zr r

=

⎡ ⎤ ⎡ ⎤= = −⎢ ⎥ ⎢ ⎥+ + ⎣ ⎦⎣ ⎦

∏1

putting it all together gives1( ) 21 2 ,

( ) ( ) 1 1 0

NG G

kkL G G

U z r QU z V z r r

−⎡ ⎤⎢ ⎥+ +⎢ ⎥= =⎢ ⎥

1/ 2 1/ 21 1

11 1 ˆ

k

k kk k

rr r

Q z z Q− −⎢ ⎥−

⎢ ⎥+ +⎣ ⎦

1 1

1 1

1 2

k kk

k k

r z zr r

⎡ ⎤ ⎡ ⎤12 Nr

69

=+

/ 21 2( ) 1

NzV z r =

⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦

∏1

12 ˆ,1 0

NG

kkG G

r Qr

Transfer Function of Lossless Tube Transfer Function of Lossless Tube M d lM d lModelModel

• = consider a 2-section tube ( 2)N

( )( )( )( )( )( )

− − −+ + +=

+ + +

1 1 21 2 1 2

1 2

1

2(1 )1 ( ) 1 1 1

G G

G

r r z r r z r r z zV z r r r

( )( )( )( )

−

− −

+ + +=

+ + +

11 2

1 21 2 1 2

0.5 1 1 1( )

1i l

G

G G

r r r zV z

r r r r z r r z

( ) ( ) −

•

⎡ ⎤+ +⎢ ⎥

⎣ ⎦∏ / 2

in general

0.5 1 1N

NG kr r z

[ ]

=⎣ ⎦=

−⎡⎢

1

1

( )( )

1( ) 1

kV zD z

rD

−⎤ ⎡ ⎤ ⎡ ⎤⎥ ⎢ ⎥

1 1Nr

70

[ ] − −

⎡= − ⎢−⎣

11 1

1

( ) 1, GD z rr z z − −

⎤ ⎡ ⎤ ⎡ ⎤⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦⎦ ⎣ ⎦

1 1...0

N

Nr z z

Transfer Function of Lossless Tube Transfer Function of Lossless Tube M d lM d lModelModel

1( )N

kkD z zα −= −∑

1

0

11

special case of ( )( )

k

G Gr ZD z

=

• = = ∞

=1

1 1 1 2( ) ( ) ( ), , ,...,( ) ( )

Examples:

kk k k k

N

D z D z r z D z k ND z D z

− −− −= + =

=

•1 1 1 1

1 1 0 1 0 1

1 1 2 2 12 1 1 2 2 1 2 1

1 1 1

1

Examples:( ) ( ) ( ) ( )

( ) ( ) (

D z r z D z r z D z r z

D z r z r r z r z D z r z D z

− − − −

− − − − −

•

= + = + = +

= + + + = + )2 1 1 2 2 1 2 1( ) ( ) (1 2 1 1 2

1 2 1 1 1 2 21 1 110

)

( ) choose as a reasonable number of tubes for model

r z r z r z r z r r z r zN

− − − − −= + + + = + + +

• =

71

1

1

1

0 714 28 2

(infinite tube at lips)

. cmN N

N N

r A

r A+

+

= ⇒ = ∞

= ⇒ =

Transfer Function of Lossless Tube Transfer Function of Lossless Tube M d lM d lModelModel

72

Summary of Lossless Tube ModelsSummary of Lossless Tube Modelsyy1. The vocal tract area function, , is now a function of , ( )A x A x

2. Solve the wave equation for the tube:

( ) ( / ) ( / ) 0ρ + −⎡ ⎤ ≤ ≤

thkcp t t c t c( , ) ( / ) ( / ) , 0

( , ) ( / ) ( / ) , 0

ρ +

+ −

⎡ ⎤= − + + ≤ ≤⎣ ⎦

⎡ ⎤= − − + ≤ ≤⎣ ⎦

k k k k

k k k k

p x t u t x c u t x c xA

u x t u t x c u t x c x

73

⎣ ⎦

Summary of Lossless Tube ModelsSummary of Lossless Tube ModelsSummary of Lossless Tube ModelsSummary of Lossless Tube Models3. add boundary conditions at the edges of adjacent tubes - both pressure and volume velocity must be continuous in both time and space at boundaries

1 ( ) (0 )p t p t

74

1 1. ( , ) (0, )2

+=k k kp t p t

1. ( , ) (0, )+=k k ku t u t

Summary of Lossless Tube ModelsSummary of Lossless Tube Models4. at each junction: - part of the positive going wave is propagated to the right while part is reflected back to the leftp - part of the negative going wave is propagated to the left while part is reflected back to the right

A A1

1

1 1

5. reflection coefficient for the junction

( ) (1 ) ( ) ( )

thk kk

k k

k k k k k k

A Ar kA A

u t r u t r u tτ

+

+

+ + −+ +

−= =

+

= + − +

( ) ( ) (1 )k k k k k k ku t r u t r uτ τ− +++ = − − + − 1( )t−

75

Summary of Lossless Tube ModelsSummary of Lossless Tube Models6 for an tube model there are ( 1) junctions with reflectionN N6. for an -tube model there are ( 1) junctions with reflection coefficients - there are boundary conditions at the lips and glottis

−N N

7. need to relate ( , ) and ( , ) to pressure anN N N Np l t u l t d volume velocity at the lips via:

/ρ −N Lc A z/ = /

where is lip impedance, i.e., ( ) ( )

ρρ +

Ω Ω

N LL

N L

L

c A zrc A z

zP U ( , ) ( , )Ω = ΩN N L N NP z U

76

Summary of Lossless Tube ModelsSummary of Lossless Tube Models

8. boundary condition at the glottis:/ A1

1

/ /

GG

G

z c Arz c A

ρρ

−=

+ 1

where is glottal impedanceG

Gzρ

77

Summary of Lossless Tube ModelsSummary of Lossless Tube Models9. for the special case of a 2-tube model we get:

( )jΩ 1 2

1 2 1 2

( )1

2 2 2( )1 1

0.5(1 )(1 )(1 )( ) ( )( ) 1

jG LL

a j j jG G L L G

r r r eUVU r r e r r e r r e

τ τ

τ τ τ τ

− Ω +

− Ω − Ω − Ω +

+ + +ΩΩ = =

Ω + + +

78

Digital Models for SpeechDigital Models for Speechg pg p• sound is generated in 3 ways—each yielding a distinctive output• vocal tract imposes resonance (and anti-resonance) structure so as to

produce different speech soundsproduce different speech sounds=> this leads to a “terminal analog” speech model—correct at the terminal

points, but without mimicing the physics of speech production

1 N

kk

GV zzα −

=− ∑

︵ ︶

79

11

need to model time varying,

kk

G

z

u n

α=

•

∑︵ ︶

Digital Model FeaturesDigital Model Features poles of ( ) correspond to vocal tract resonances missing zeros of ( ) for nasals

V zV z

•• modify ( ) increase to large number to model zeros roots of ( ) are at:

V zN

V z

−−

•

, 2 (k k k ks s j Fσ π∗ = − ±2

)

,

cos 2 ) sin 2 )

k k

k k

T j F Tk k

T Tk k

s plane

z z e e

e ( F T je ( F T

σ π

σ σπ π

− ±∗

− −

−

=

= ±

2| | kTz e F Tσ θ π− 2| | ,kk k kz e F Tθ π= =

80

Other Synthesis ImplementationsOther Synthesis ImplementationsOther Synthesis ImplementationsOther Synthesis Implementations

• direct form differencedirect form difference equation

• cascade of second order systemsy

=∏( ) ( )M

kV z V z

ππ

=

− −

− +=

− +

12

1 2 2

1 2 | | cos(2 ) | |( )1 2 | | cos(2 ) | |

k

k k kk

k k k

z F T zV zz F T z z z

81

Radiation at LipsRadiation at LipsRadiation at LipsRadiation at Lips

( ) ( ) ( ) -- high pass filteringP z R z U z=1

0 1

( ) ( ) ( ) -- high pass filtering

( ) ( ) -- crude differentiatorL LP z R z U z

R z R z−

=

= −

82

Excitation ModelExcitation ModelExcitation ModelExcitation Model

83

Glottal Pulse ModelGlottal Pulse ModelGlottal Pulse ModelGlottal Pulse Model

[ ]1 1

1 2 1 1 2

0 5 1 0

2

[ ] . cos( / ) , cos( ( ) / ),

π

π

= − ≤ ≤

= − ≤ ≤ +

g n n N n N

n N N N n N N0 otherwise=

-lowpass filtering effect

84

General Synthesis ModelGeneral Synthesis ModelGeneral Synthesis ModelGeneral Synthesis Model

[ ]Lp n[ ]u n

11( )R z zα −= −

[ ]Lu n

85

Components of Speech ModelComponents of Speech Model

86

Summary of LectureSummary of LectureSummary of LectureSummary of Lecture• Derived sound propagation equations for vocal p p g q

tract– first considered uniform lossless tube– added simple models of lossadded simple models of loss– added model for radiation at lips– added source model at glottis

dd d l d l f l t t– added nasal model for nasal tract– broadened the model to N-tube approximation—

lossless casel k d 2 b d l f i l l– looked at 2-tube models for simple vowels

– examined range of digital speech production/synthesis models

87