# Uniform Tube Modeling of Speech Processing – VII

ok so ah we have derived that the all pole

model for vocal track so vz is nothing but a i can assign a g by dz ok so where dz is

nothing but a one plus k sorry summation of k equal to one to n alpha k z to the power

minus k ok now if you see that d ah [vocalized noise] that second the for the two tube model

the equation vz is nothing but a zero point five one plus rg one plus r one so for two

tube model vz is n equal to two n equal to two is two tube model zero point five one

plus rg one plus r one one plus r two into z to the power minus one divided by i can

say one plus r one r two r one r two r one rg into z to the power minus one plus r two

rg z to the power minus two so dz for two tube model is nothing but a one plus r one

r two plus r one rg z to the power minus one plus r two rg z to the power minus two now

if i say rg is equal to one if i say rg is equal to one what all impedance is infinite

zg is equal to infinite then rg equal to one so rg is nothing but a you know that rg is

r rt minus rg divided by rt plus rg ah or whatever rg minus rt or rt minus rg rg minus

rt so to that that is there in constant when

when we model in that ah boundary condition when you model in boundary condition we said

rg boundary condition modeling where here so here i can say rg is nothing but a zg minus

zd divided by zg plus so rg is nothing but a here i write capital rg lets consider small

rg rg is zg minus zd divided by zg plus zd zg is infinite then rg is equal to one ok

now if i consider rg is equal to one then dz is nothing but a dz is nothing but a one

plus r one r two plus r one plus z to the power minus one plus r two z to the power

minus two ok so if it is nth order i if i can write that if it is instead of two tube

it is n tube or i can say it is nothing but a one plus r one one plus r one r two r one

z to the power minus one plus r two z to the power minus two ok so if i say that ok i can

want to write down the recursive equation so d one z if it is d two z so d one z if

i write d one z is nothing but a one plus r one z to the power minus one if rg is equal

to one d two z is one plus r one z to the power minus one plus r one r two z to the

power minus one plus r to z to the power minus two that i have derived here z to the power

minus two now if i say this is nothing but a d one z

one plus r one z to the power minus one is nothing but a d one z ok so d one z plus r

two z to the power minus two into d one z to the power minus one r one r two z to the

power minus one plus r two z to the power minus two this term can be written like this

way r two z to the power minus two into d one z to the power minus one because d one

z to the power minus one will be one plus r one z to the minus one instead of z so z

so r one z into this one so r one r two z to the power minus one and r one z ok so this

can be also written by d zero z there is no junction no nothing is there is one so it

is nothing but a d zero z into plus r one i can say z to the power minus one d zero

z to the power minus one ok similarly i can write down this way because this is nothing

but a d one we put put d one value and this will come same value will here ok similarly

i can write dnz or dkz dnz i can write nth nth tube n number of tube this is nothing

but a d n minus one z plus r two z to the power minus n d n minus one z to the power

minus one or i can say dk z is nothing but a d k minus one z plus r two z to the power

minus one dk minus one z to the power minus one which is same as if i write the equation

dz is equal to one plus k equal to one to n alpha k z to the power minus k same way

i can write so this can be written as like this way ok so if i write this one if n is

eq number of section is equal to ten and lets rn is is equal to one load but rn is equal

to rl so lets load is rl is equal to one ok so no registive load for low frequency rn

is equal to rl which is equal to one ok now i can say i can find out the what is the

value of if i know that rn is equal to one then from that side last side this a last

last rn is equal to one so i can find out rn minus one rn minus two rn minus three if

it is ten tu ten tube model so rn means r ten is equal to one i can find out r nine

r eight r seven how because using n is equal number of tube then model rn is equal to one

rn is equal to lets this one n plus one is equal to eight eighteen centimeter so what

is saying that that if i know the reflection coefficient i can implement the tube digital

implement the tube or other hand if i know the signal which is coming out from the tube

from there i can estimate the area function of different junction because if i know r

r is rn is nothing but a what is rk rk is nothing but a a k plus one minus ak divided

by a k plus one plus ak so ak a are the area func cross sectional area function of the

tube different tube so if it is a ten number of tube consist of the vocal track is model

within a ten number of tube last rn rn is equal to one last one is equal to one then

i can find out ak so rn i know rn so this is rk so so i know rn rn is equal

to a n plus one minus an divided by a n plus one plus a n so if i know the cross sectional

area then i can know the derive the reflection coefficient and i can model the tube in digital

domain on the other hand if i know the pc signal and if i able to find out the value

of reflection coefficient alpha k in somehow i can able to find out the alpha k value then

i can find out the cross sectional orea area of the different tube i can give you one example

from ravinand say for that that is written there is a different cross sectional area

for a vowel i think the vowel which vowel i dont know that there is a vowel so this

vowel area functions is written either i know the area function i can generate the vowel

if i know the vowel i can derive the area function ok so how do we digitally implement

this tube i want to digitally implement this tube so i know i have to implement dz is nothing

but a one plus k equal to one to n alpha k z to the power minus k this equation i have

to implement so if i able to implement how to implement very simple so there is a un

lets n unn impulse then i have a output which is uln ok so this will be simple delay by

z to the power minus one alpha one and has to be added with here then z to the power

minus two alpha two added in here dot dot dot z to the power minus n alpha n added here

this is one ok so i can implement in digital systems i can

implement it digital filter the line nothing but a digital filter the equation looks like

a nothing but a digital filter so i can easily implement it using that things that vocal

vocal track i can implement using digital domain ok so now there is some concept poles

of vocal track so you that poles all pole models are vocal track tube so i can say the

vz is model using g by dz where dz is equal to one plus k equal to one to n alpha k z

to the power minus k ok so dz has an n if it is order is n so n number of pole so i

can say vz is a all pole model which has n number of pole so if i have a ten junction

tube ten tube ten ah ten ten section tube then i can say ten poles will be there ten

poles will be there so now if dz is real so the the dz have either the all re real real

pole will be the real functions so there is a real pole will be occur or there

all pole have a complex conjugate in nature so if dz has a n number of pole then i can

say that is a n by two complex conjugate pole will be there n by two pair of complex conjugate

pole so i can say vz is nothing but a g by k equal to one to n by two one minus alpha

k z to the power minus one into one minus alpha k star z to the power minus one so it

is nothing but a n by two complex conjugate pole product ok so what is the if you see

that then if i say the this is my unit circle ok so complex conjugate pole so this is the

real axis this is the imaginary axis if a pole occur in here with a angle of theta there

will be a another pole which is conjugate this pole minus theta and here so if there

is a pole in here there is a another pole will be in here if the pole in here there

is another pole will be here so every pole has an complex conjugate pole

ok if it is complex number then alpha k i can say it is nothing but a rk e to the power

j theta so a complex number has a two a pole let this is the pole so these as a theta and

this as a amplitude r so alpha k can be represented using rk and e to the power j theta this rk

is not reflection coefficient ok so there is a amplitude of the pole and e to the power

j theta so if i say alpha k complex conjugate part it will be nothing but a rk amplitude

will be same e to the power minus j theta value of alpha complex conjugate pole ok now

interpreted the value of value of alpha so if i do that if i put the value of alpha in

here then the dz will becomes or you can say vz will become g by one minus alpha k z to

the power minus one into one minus alpha k star z to the power minus one lets this right

down this product i am not i am product of k equal to one to n by two so if i do this

one instead of alpha k i put the value of rk e to the power of j theta instead of alpha

star k i can put the ar rk e to the power minus j theta if i put that value it will

come g by one minus two rk cos theta k z to the power minus one plus rk square z to the

power minus two for this complex conjugate product ok so i can now if you see the vz vkz in term

of this one this is vkz vz if i want vz right vz then there will be a product of pi equal

to k k equal to one to n so vk i can say this two only this is v if i write vk this one

then i dont write this product term ok now if it is that then what is contribution of

rk and bk lets rk is equal to e to the power minus bk and bk is equal to so bk is equal

to minus ln rk ok if i put that value then vkz will become one by one minus two rk so

two e to the power minus bk cos theta k z to the power minus one plus e to the power

minus two bk ask r square k so z to the power minus two now importance of this bk is nothing

but a produce a band width bk produce a band width and theta k give the formant position

when the value of rk approach to unit i can get the formant frequency resonance frequency

so if it is rk approach to unit cir close to unit circle then i can get the resonant

frequency and if the bk right bk value is non zero then i get a band width pole there

will be a pole there will be a formant which has a band width ok so bk provide me the band

width theta k provide me the formant position so this information will be used when we develop

the model using linear prediction model for speech production system ok so if i know the

theta k if i know the theta k and value of rk then i can model the system theta k give

me the formant frequency position and if rk tends to zero tends to unit circle amp amplitude

of that pole is tends to your close to unit circle that give you the formant frequency

ok so if there is a n number of pole so n by two complex conjugate pole will be there

so each pair of complex conjugate pole give me a formant frequency so if i have a five

formant frequency if i give you a spectrogram or let the i told you the spectrum of the

speech signal is like this one two three four lets f one f two f three f four then if i

aks how many complex conjugate pole will be there four pair how many pole will be there

complex pole will be eight so if n tube model is there n number of pole is there in a transfer

function then i can say n by two complex conjugate pole will be there so literally i can get

n by two formant frequency in a spectrogram ok so if i give you the formant frequency

and formant band width if i able to find out for a speech event i can able to derive the

transfer function for that speech event the vocal track transfer function now if i say

that that event is given so speech event lets say speech event the steady state vowel r

if i take the steady state vowel and analyze the spectro [vocalized noise] frequency analysis

and find pout the formant frequency and formant band width then i can able to find out the

transfer function of that vocal track ok so complex conjugate num so dont confuse to the

number of complex conjugate pole and pole number of normal pole ok so if there is a

n tube model n tube model is there then i can get n by two complex conjugate

pole so there will be a n by two format so if i able to lets there is a n equal to ten

tube i take ten ten tube to model this the ten this whole vocal track system is divided

by ten parts and model it then i can get ten by two complex conjugate pole so i can get

five formant frequency ok on the other hand lets i derive that mathematics that is a example

i have given in the ah i in the slides lets the length of the vocal track l is equal to

seventeen centimeter and the velocity of sound is three forty meter per second find the number

of section required to generate five kilo hertz band width voiced signal so i have to

generate five kilo hertz band width voiced signal what should be the sampling frequency

fs should be ten kilo hertz ok or not fs should be ten kilo hertz then what is my tau tau

is nothing but a x by c so what is x x is a length of each section ok i have to find

out number of section required so lets i require a n number of section so

if the length of the tube is l then the x is equal to l by n ok so then tau is equal

to l by n into c ok now if it is l is equal to seventeen point five centimeter and c is

equal to three five centimeter per second then i can find out it is nothing but the

seventeen point five divided by n into three five triple zero which is nothing but a one

seven five divided by n into three fifty lets right three fifty into ten to the power three

so it is two so it is nothing but a one by two n into ten to the power three ok now what

is tau so what is t t is equal to one by fs so that t is nothing but a t is equal to two

tau so tau is equal to t by two tau is equal to t by two what is t t is nothing but a one

by fs so it is nothing but a two by fs so i can write one by two fs is equal to one

by two n ten to the power three two two cancel so it is n is equal to fs by ten to the power

three so fs is equal to ten kilo hertz which is ten kilo hertz nothing but a ten sections

ok or i can say n is nothing but a fs by n by two i can say n by two is nothing but a

band width if it is a fs is equal to two b five kilo hertz ok so ten ten section is required

to model that tube five kilo hertz band width i require ten sections so similar kind of

mathematics you can expect that ok if i want to generate four kilo hertz band width signal

and fs is equal to lets eight kilo hertz sub frequency is eight kilo hertz then find out

the number of section is required minimum number of section is required ok so i can

find out the number of section is required so this way i can find out and this way you

can model that signal so since the vocal track can be model as a

all pole model i am not going details of that things this is the formant trajectory so i

can see that whole vz can be so this is the vk is the product of this if it is like vz

then product will be here so i can say this digital implement is an again same each of

the chunk represent the formant frequency this is the first formant one formant two

formant three so that will come so each stage represent one formant frequency each stage

represent one formant frequency so that we n by two number of stage will be there so

if it is n by two number of stage then n by two formant frequency will be there so i can

say that loss less tube model lossless vocal tube model can be done by a linear system

which is nothing but a vz is a linear systems vz is equal to one plus k is equal to one

to n n alpha k z to the power minus k lt so will discuss the about that lta system so

this i can say in summary this is the vocal track vz radiation is rz and glottal pulse

is gz so if i say who what is the total transfer function is nothing but a hz whole speech

production transfer function gz vz into rz whole hz the thus if i say thus from the speech

i want to find out hz which is the product of glottal pulse transfer function glottal

generate glottal transfer function plus vocal track transfer function into vocal track transfer

function into vocal track transformation into lip radiation transfer function ok so if there

is a voice in impulse train will be there that will be modified by glottal transfer

function and that modified signal fed to the vocal track and after lip radiation i get

the speech signal if it is unvoiced speech then i can say it is connected to the random

voice and it will only modify by the vocal track and lip radiation and i get the speech

signal ok so random noise passes through vocal track

and included lip radiation produce the speech for unvoice speech if it is voice speech impulse

ten will you modify by the glottal pulse generator and that will pass through the vocal track

and lip radiation produce the speech so if you see three transfer function gz can be

approximate by the ah second order ah ah this two pole two pole function for radiation there

will be a one single pole is required so two plus one three pole if i do that then i can

get hz ok so will discuss about that how many pole ah ah what are the linear prediction

analysis that things during the linear prediction analysis function what are the how do you

get what are of the linear prediction that time we use this information ok so this is

the whole vocal track tube modeling so in summary i can say the vocal track this human

vocal track can be model using a digital signal processing or [vocalized noise] all can be

implemented using a digital linear filter based on that requirement so we have shown that if i consider this vocal

track is a fun or you can is a simulated using a number of junction or number of section

lossless tube section lets n number of lossless tube section that then it can be implemented

using a linear time linear system which is vz and which can be implemented in digital

domain so thats why this is called uniform tube modeling the throughout the section if

i say n number of section throughout the each section the cross sectional area of the vocal

chord i said uniform so i can say throughout the whole vocal track can be single tube whole

vocal track can be two tube or whole vocal track can be n number of tube but if i say

n number of tube each tube cross sectional area is constant thats why uniform and loss

less also because if it is loss lossy then all kinds of complex it will come up so if

it is loss less uniform tube model then it can be implemented using digital system that

is vz once i know that then i know yes if it is implementable by a vz then can i think

that output speech which has collected using a microphone cannot be can it be analyzed

using linear signal processing yes so that from there the concept of linear prediction

analysis come so this system can be linearly model so from the signal if i want to find

out the transfer function that sys the thats why i want to predict the vocal track constriction

so if i know the signal yes it is possible to imple find out the area cross sectional

area of different section if i say it is n tube model n different section cross sectional

area is possible to find out from the behavior of output speech or if i know the area function

i can implement it digitally it is possible to implement it if i excited by impulse response

i can able to prodsssuce the speech ok so this is the tube modeling so this is called

uniform tube model or loss less tube modeling of speech production system ok thank you