# Uniform Tube Modeling of Speech Processing – II

ok so we have derived that two wave equation

this two wave equation we have derived now here one things is there v is the particle

velocity ok now if i say volume velocity so if if you say volume velocity if the velocity

is v of lets tube the cross sectional area is a then what is the volume velocity a into

v the whole total volume v is the length of the tube and a is the cross sectional area

so v is the a into v is the volume so if i say volume velocity is u then u is equal to

v by a or sorry u equal to v into a or i can say v is equal to u by a

now if i [pit/put] put this thing on put this thing on this two equation where i can say

ah that ah del square or del p by del x is equal to minus rho del v by del t so this

will become v is equal to u by a so del p by del x is equal to minus rho p u by a rho

by a del u by del t similarly del p by del t is equal to minus rho c square del v by

del x this will become del p by del t is equal to minus rho c square ah rho c square minus

rho c square or i can say del u sorry this is del u by del t not del p del u by del t

is or rho c square by a del v by del x or i can say del u by del x is equal to minus

a by rho c square del p by del t ok now if you see this equation and this equation

these two equation if you see this is analogous to electrical transmission line equation so

electrical transmission line voltage v is a function of x and t and current i is also

a function of x and t transition line so in that case in electrical transmission line

minus del v by del x is equal to l into del i by del t similarly minus del i by del x

is equal to c into del v by del t ok now this is analogous to so these two equation analogous

to this is the electrical transmission line equation lossless electrical transmission

line equation then i can say if this this equation is analogous to this equation then

i can say that p is analogous to v so pressure in acoustic domain is analogous

to voltage in electrical domain similarly u is analogous to i volume velocity in acoustic

domain is equivalent to electrical current in electrical domain and if i say l is equal

to rho by a so this rho by a is called acoustical inductance acoustical inductance similarly

i can say that a by rho c square is analogous to c so it is called acoustical capacitance

a by rho c square is acoustical capacitance and rho by a is the acoustical inductance

p is analogous to voltage current is analogous to volume velocity ok

so if you see this slide i have written this slide same things i have written in this slide

ok now so this i know so i can say that ok if i know the equation i can analogous electric

transition line i can also draw ok this same as electrical transmission line lossless transmission

line now so what i want to know what i want to do is that i want to find out the transfer

function of a uniform tube which is i can say this side is the vocal cords which acts

as a piston and so x is equal to zero where the force is applied in the tube and x is

equal to l where the tube radiate the sound energy in the air

ok now if i see that that this one this the this tube whole tube is nothing but a this

kind of switches of pressure or you can say the pressure is applied in here the acoustical

pressure is applied in here so i can say the particle velocity in this point i can say

or volume velocity in this point is u zero t and volume velocity at this point is u x

t pressure at this point is nothing but p zero t pressure at here is p x t and cross

section of the tube is a the area cross sectional area of the tube is a ok i will consider throughout

the tube a remain constant ok so now a remain constant now try to find out the traveling

wave solution in the tube ok what is travelling wave solution now if i see that minus del

u by del x is equal to a by rho c square del p by del t or or i can write down these two

equations where i said this is the this is the pressure wave equation or not this not

this one where it is this will be here so this is the pressure wave equation is the

pressure wave equation and this is the velocity wave equa[tion] ok now i have to find out

the solution of this second order differential equation and again solution of this second

order differential equation will give me the velocity so if i draw the solution of the

second order differential equation you know that second order differential is total solution

is nothing but a so i can say that u x t is equal to u plus t minus x by c minus u plus

t plus x by c what is that t minus x by c so i can say this is called forward wave this

is called backward wave so in the tube there is a forward wave and there is a backward

wave in the forward wave i can say the if the time

axis this is the time axis in the forward wave then t is minus x by c if it is the backward

wave so negative direction t t plus x by c c is the velocity of the sound x is the position

now similarly what is p x t p x t is nothing but a rho c by a u plus t minus x by c plus

u minus t plus x by c ok so i this i can say is nothing but a z z t which is called impedance

of the transmission line or the impedance of the acoustic line is nothing but a u plus

t minus x by c plus u minus t plus x by c ok so what z d is equal to rho c by a so u

plus is the forward wave u minus is the backward traveling wave ok now if it is that then there

is there is a two boundary condition in the tube if you see what is the boundary condition

at the glottis at the here the glottis point where the vocal cords are vibrating here u

zero t is nothing but a glottal excitation which is nothing but a lets glottal excitation

is u g omega and equal to it into e to the power j omega t

where this omega is the continuous frequency so all frequency the frequency of u so i can

say i excited the tube with unit impulse or i guess every frequency and i have to find

out the every frequency response at the output ok thats why i said omega e j omega t and

this excitation is if it is impulse then that or you can say the excitation contain all

the frequencies equal amplitude and that amplitude is modified as for the glottal response glottal

frequency response thats these u g omega ok so i can say the the glottal frequency response

which is u g omega has to be passed through the tube and the tube output of the tube i

have to find out and then output by input is the tube transfer function

so u zero t u zero t is nothing but a u g omega e to the power j omega t and at the

lips lets there is a radiation load is ah for the low frequency radiation load is completely

not there so what will happen at the lips it it again it is atmospheric pressure so

i can say p x p l t at x equal to l is equal to zero so if i put p l t so what is u zero

t so u zero t is u g omega e to the power j omega t is equal to u plus so x is equal

to zero so i can say u plus t minus u minus sorry this will be minus u minus t x equal

to zero so x equal to zero will be not there so i can say u u plus t u minus t

and if it is p l t then p l t is equal to zero so zero is p l t is equal to zero so

i can say z t into u plus t minus l y c plus u minus t plus l y c ok so there is a differential

equation so since the differential equation are linear with constant coefficient the solution

must be in the form of k plus k minus represent the amplitude and the forward and backward

wave so i can say if it is linear wave equation forward wave u plus t or you can say that

u plus t minus x by c can be represented in the form of amplitude lets this is k plus

e to the power j omega t minus x by c differential equation second order differential equation

and u minus i can say t plus x by c is nothing but a k minus e to the power j omega t plus

x by c exponential solution ok so this is the exponential solution now if i put that

value in two equation equation number one in here and equation number two in here

so in one i can say u g omega e to the power j omega t is nothing but a k plus e to the

power j omega t because x is equal to zero minus k minus e to the power j omega t x equal

to zero ok and similarly zero is equal to z t into u plus k plus e to the power j omega

t minus l y c plus k minus e to the power j omega t plus l y c ok from here i can say

e to the power j omega t e to the j omega t cancel so i can say k plus minus k minus

is equal to u g omega and here i can say z t is equal to k to this is equal to this so

i can this is the equation number one and this is equation number two again i can solve

for k plus and k minus im not deducing the solution because you can do it easily

so you can just deduce the solution and find out the k plus and k minus so k plus and k

minus will be in the form of k plus will be equal to u g omega e to the power two j omega

l by c divided by one plus e to the power two j omega l y c this is the k plus and k

minus will be minus u g omega divided by one plus e to the power two j omega l by c ok

once i know the k plus and k minus then i can write down the u x t u x t is nothing

but a k plus e to the power j omega t minus x by c minus e to the k minus e to the power

j omega t plus x by c now i put the value of k k plus is nothing but a u g omega this

value in here and k minus i can put the value in here then if i put the value then i can

say u g u g omega e to the power two j omega l by c divided by one plus e to the power

two j omega l y c ok into e to the power j omega t minus x by c minus u g so it is minus

minus plus u g omega u g omega divided by one plus e to the power two j omega l by c

into e to the power j omega t plus x by c ok

now i can say that u g omega is common so u g omega e to the power j omega t will be

same ok e to the power j omega t then i can say it is nothing but a e to the power j omega

two l two l minus x divided by c ok minus plus e to sorry plus e to the power j omega

x by c divided by one plus e to the power two j omega l by c ok now if it is that then

i can solve for that this is the u g u g u omega so u g omega if i simplify it then u

j omega sorry this is a u x t so ultimately if i again simplify it u x t will be in the

form of cos omega l minus x by c divided by cos omega l by c cos omega l by c into u g

omega e to the power j omega t here if i say that e to the power j theta

look that way if i do it l minus l minus ah x if i do that then it will cos omega l minus

x by c into cos omega l if i simplify it similarly i will get the p x t p x t will be j z t sin

omega l minus x divided by c divided by cos omega l by c into u g omega e to the power

j omega t ok where z t is equal to rho c by a ok now i get this to u and p now what is

if you see p is analogous to voltage u is analogous to current now if i say what is

the ah you can say the impedance z is nothing but a p x t divided by u x t if i do that

p x t divided by u x t so it will come j z t into tan omega l minus

x by c tan omega l minus x by c ok this divided by this so that this divided by this so cos

by sin sin by cos tan this will come so this is nothing but a acoustics impedance inside

the tube at any point of x so if x is equal to something i can find out the acoustic impedance

ok so now if the del x if the x is variation of x is very small then i can say that z is

equal to nothing but a z z is nothing but a j omega a rho by c rho by a into del x omega

so simplification of that for a small x del x is very small then i can do that ok now

this is a z now if i interpreted this thing the part this

is particle velocity velocity this is pressure wave this is volume velocity and this is pressure

wave now if i interpreted that things what is the envelope here envelop is z t pressure

wave z t sin omega l minus x by c this is part is the envelop part this part will be

same variation e to the j omega t that is that is like a envelop part and this is the

envelop part in here now if this portion is constant if the l is constant this portion

is constant so one is sin another is cos so if i say how the pressure amplitude pressure

wave amplitude and volume velocity amplitude is very inside the tube then i can say only

x axis here x only i am drawing the envelop portion

so one is sin another is cos so if this is cos is the volume velocity see the particle

velocity is maximum then pressure wave is minimum zero so particle velocity will vary

like this then pressure wave for a volume velocity sorry u is the volume velocity pressure

wave will be look like this one is cos another is sin ok now if i say real part of this part

the real part of this so i can say if i write u x t is equal to cos of omega l minus x by

ah by c divided by cos omega l by c into u g omega e to the power j omega t so e to the

power j omega t is nothing but a cos omega t plus j sin omega t ok

so cos part is the real part sin part is the imaginary part so i can say the real part

of u x t is nothing but a cos omega l minus x divided by c divided by cos omega l by c

into u g omega cos omega t similarly i can get the real part of the pressure wave ok

or not so real part of this u u g omega t is this one only ok now i come to the transfer

function of the tube so in the tube i can say this is x equal to zero this is x equal

to l so if i say input is u zero t and output is u x t volume velocity so i can say the

v a transfer function v a omega is nothing but a u x sorry u l t u l t divided by u zero

t x is l and x is zero here ok so what is that

now if i say that u is equal to cos omega l minus x by c so at u at l x is equal to

l cos omega l minus at x equal to l x equal to l divided by c divided by cos omega l by

c into u g omega e to the power j omega t which is nothing but u l t ok so u l t it

is l minus l is zero so it is nothing but a one by cos omega l by c u g omega e to the

power j omega t now what is u zero t u zero t is nothing but a u g omega it is what g

omega t which is bottle impulse at the glottis point here so if i say u l t by u ah u zero

t so b a omega is nothing but a u l t one by cos omega l by c into u g omega e to the

power j omega t divided by u g omega e to the power j omega t so this this this will

cancel so it is nothing but a one by cos omega l by c

so transfer function of the uniform tube where this side is closed and mouth is open x equal

to zero and x equal to l is one by cos omega l by c ok now if it is one by [co/cos] cos

omega l by c then what is the pole position what do you mean by pole position pole you

know that pole so any transfer function h omega is nothing but consistence of p omega

divided by q omega so there is some pole and some solution of p will give me that zero

solution of q will give me the pole so if you see solution of q when q omega q omega

is equal to zero then what is the amplitude value value is infinite that means at the

pole the resonance of the system is occur so i get the highest energy maximum energy

resonance will be occur at pole every pole so if if i have a five pole i get five resonance

frequency because so if the fifth order q is the order is the fifth that means i can

get the fifth solution five solution five solution give me the five resonant frequency

how many solution i can get here cos theta is equal to zero when when omega l by c is

equal to two n plus one into pi by two cos pi by two cos pi by two zero again it will

be zero and wave pi ok so cos pi by two is equal to zero so it cos pi by two is equal

to this way again it will be zero but it is three pi by two when it is five pi by two

that way if it is happened then it will be zero ok

so one by the b a is nothing but a one by cos omega l by c ok so what i described that

any transfer function can be written any any any system transfer function h omega can be

written as p omega divided by q omega p omega divided by q omega now if you know that solution

of p omega this polynomial will give me the what zero and solution of this omega will

give me the pole thats why p h mega is called the pole zero function so zero and pole now

what is the physical significance of pole suppose i said ok there is a pole what do

mean my pole physically what will happen now if you see what do you mean by solution of

q omega at solution point q omega will be zero if i put it zero then h omega value will

be infinite at solution point of p omega p omega will be zero the solution point of p

omega then h omega will be zero so i can see that if each solution point of

q omega that means in every pole the system response will be infinite that means system

will resonant and every solution for p omega which is nothing but a zero the system response

will be zero thats why it is called pole zero so every pole corresponding to a resonance

in speech those resonance are called formant ok in speech those resonance are called formant

so i know if i want to know the formant response or frequency response of b omega b a omega

then i have to find out where its resonance is occur when the resonance will be occur

when cos omega l by c is equal to zero so if i want to know at which frequency this

cos omega l by c is equal to zero ok at which frequency so what is omega is continuous frequency

is nothing but a two pi f so i put so omega omega is equal to two pi f when cos theta

will be zero cos theta will be zero if theta is equal to pi by two ok theta is equal to

pi by two cos theta is equal to zero so i know whats omega l by c should be pi by two

ok now when pi by two ok lets i put that two pi f l by c is equal to pi by two so i know

at theta is equal to pi by two cos theta is equal to zero so omega l by c is equal to

pi by two pi if pi by two then cos omega l by c is equal to zero so resonance will be

occur so first resonance because cos theta can be

zero not only pi by two but it cos theta is equal to zero if theta is equal to pi by two

not only pi by two if theta is equal to three pi by two then also it is zero if it is five

pi by two then also it is zero so lets minimum frequency first thats why we call first resonance

frequency is theta is equal to pi by two so pi pi cancel it is nothing but a two into

f l by c is equal to half or i can say f is equal to c by four l ok now what is c c is

the velocity of sound what is l l is the length of the tube average human vocal vocal cord

length is equal to l is equal to call seventeen point five centimeter so l is equal to seventeen

point five centimeters what is c velocity of sound velocity of sound is three five centimeter

per second so i can say f is equal to three five triple

zero divided by four into seventeen point five which is equal to five hundred hertz

ok if omega l by c is equal to three pi by two then f will becomes one point five kilo

hertz so it is f one first formant second formant if it is third formant then omega

l by c is equal to five pi by two then it will become two point five kilo hertz if you

see distance between the first formants and second formant is one kilo hertz second formant

and third formant is one kilo hertz it is fixed because tube length is fixed ok this

is similar with your earlier physics theory that if this side is closed then the first

harmonics will occur at lambda by four so l is equal to lambda by four so f is equal

to c by lambda which becomes c by four l since the velocity and l is the length of the tube

which is same in here also then it will be odd multiples of so it is nothing f is equal

to nothing but a two n plus one c by full length lambda ok so here also same so f one

f two f three i can get now can i draw the frequency response of the uniform tube like

there is no loss no nothing any loss so i can say if this is my frequency axis this

is five hundred hertz this is one point five kilo hertz this is two point five kilo hertz

so at five hundred hertz power is maxima so on it will going so it is i can put an arrow

in here so it is infinite power ok infinite impulse with the impulse power at

five hundred hertz and bandwidth is equal to zero this is a real pole so bandwidth is

equal to zero bandwidth is equal to zero with zero bandwidth infinite power resonance will

occur at five hundred hertz one point five kilo hertz two point five kilo hertz if the

length of the vocal vocal tube is seventeen point five centimeter and velocity of the

sound is three five triple zero centimeter per second ok

so that i know this is the frequency response for the uniform tube model with without considering

any kinds of loss or any walls is rigid wall not variable wall means walls is rigid so

next lecture i will discuss how this response will be affected if i consider one loss at

a time if i consider one loss how it will be modified second loss how it will be modified

then ultimately we can get what should be the frequency response or what should be the

response of this vocal tract tube if i get it and what will be the transfer function

ok thank you