# Uniform Tube Modeling of Speech Processing – III

ok so now we have derived that ah transfer

function of lossless tube without considering any loss infinite power at five hundred hertz

one point five kilo hertz two point five kilo hertz now if i consider that the tube the

wall which i i supposed to be rigid if this wall is not rigid this wall can flexible this

wall can modify if pressure is high if there is a pressure wall can change so what is happening

instead of cross sectional area a is fixed now a is also function of x and t a is also

function of x and t so anytime the cross sectional area a x t is nothing but a average area or

you can say the a x t plus delta change of a x t due to the pressure at that position

at that time delta a so if you see the slides if you see the slides

ah the slides is there so inside it is a zero t i can say it is a zero t fixed and then

it is a delta a x t is expansion so now if i instead of u a and rho a the wave equation

i have modified because the wave equation instead of fixed a a is modified then if i

put this one a zero x t into del a x t for a zero x t then the modified equation will

be like this so i am not going details modified equation mathematical things then it will

be take lot of time so i this is not not objective to derive the details details can be available

in the papers also mathematical details so there is a complex equation will come out

if i see the complex so i can say this tube is made of a muscles i can say muscle is nothing

but a spring mass action so it is nothing but a spring mass mechanical

oscillator if i consider the second order equation of the spring mass mechanical oscillator

then this will be like this then again if i one two three if i consider and then the

wave equation instead of to wave equation with this three equation will derive and from

that three equation i get an analytical solution of what will happen in the frequency response

which is v a omega u l omega divided by u g omega now they said if the length of the

tube is seventeen point five centimeter and five centimeter square is the cross sectional

area and m omega is zero point five gram per centimeter square b omega is six five dyne

that means m omega is nothing but a mass per unit length b omega damping per unit length

and stiffness per unit length of the vocal tract wall ok

if i consider that value then it is said it was found in the frequency response earlier

it was infinite energy five hundred hertz one point five kilo hertz two point five kilo

hertz like that it is said that complex pole with nonzero bandwidth so here it is zero

bandwidth here it is zero bandwidth bandwidth is zero in tube lossless condition now if

i introduce the loss what will happen the instead of zero bandwidth some bandwidth will

be generated so the poles are complex with nonzero bandwidth slightly higher frequency

the frequency will be formant instead of five hundreds hertz it maybe five hundred and five

hertz slightly formant equation is shifted towards the higher frequency side most affected

in the lower band lower frequency will be the most so lower

frequency bandwidth will be increase more compared to the higher frequency so most affected

is the lower band ok now if i consider so i said that earlier i am not considered the

friction loss viscous loss and thermal loss now if i consider the all kinds of friction

loss thermal conduction on the wall viscosity all kind of losses then i found it increase

the bandwidth of the complex pole and decrease the resonance frequency slightly so ultimately

i can say that yes if i consider the losses and not rigid to wall so instead of a infinite

zero bandwidth infinite impulse at every resonant frequency it becomes a finite bandwidth with

a slightly shifting higher direction of the formant

so formant is shifted slightly higher direction so instead of this figure ultimately i will

get some bandwidth kind of formant but if you see more or less it will be on an around

of five hundred hertz slightly shifted so i can easily say normally if it is tube is

totally open first formant is five hundred hertz second formant is one point five kilo

hertz third formant is two point five kilo hertz ah third formant is two point five kilo

hertz forth formant is three point five kilo hertz fifth formant is four point five kilo

hertz ok so now if you see if a signal is band limited with four kilo hertz then i can

say i only can get up to fourth formant i cannot get the fifth formant because fifth

one four point five kilohertz so i will not get it ok so if i cut the signal in here i

will get only three formant if i cut the signal in there i can get the three or four formant

so depends on the sampling frequency format will come i will come this will discuss later

on also ok so i am not detail discuss about the this kind of losses now second difficulty

is the effect of radiation at lips so what i said once the air is radiated from the mouth

so i can say at the opening of the tube the acoustic wave is radiated so the radiation

this radiation why it is radiated in the atmospheric air

so what will be the radiation losses what kind of effect i will get due to this radiation

ok so now if you see the assume that p l t is equal to zero at lips this is the we have

assumed the acoustical analogue this short circuit the output couple this thing if i

say transmission line i said the output is totally short circuit p is equal to zero t

means the voltage is equal to zero so output is short circuit here i said there is a no

no load no acoustic load so it is loaded zero output is short circuit it is ideal condition

but it may not be the ideal condition so what will be the effect of air load on the lip

or you can say the radiation effect on the acoustics wave transmission

so then i if i put the microphone then i can say what kind of signal i am expecting from

the mouth ok now if i consider that how do you do that so i can say lets this is the

tube and this is the mouth here it is radiated so whole load is the atmospheric load is there

in the lip so lets the load is nothing but a inductive and resistive load l r and r r

acoustics mass inductive load and resistive load ok now it is there then i can say here

it is p l omega is nothing but a p l omega is nothing but a lets this constitute this

load constituted z r so z r into u l omega so i have not writing small u keep it capital

u l omega means length omega means instead of time it is frequency it is length and frequency

capital p frequency response p l omega is nothing but a z r into u l omega or i can

say that p l omega p l t is nothing but a small p l t is nothing but a z r into u l

t ok now what is z r z r is nothing but these two are in parallel so j omega you can say

the j omega l r r divided by j omega omega l r plus r r because inductance j omega l

r parallel with r so j omega l r r r divided by j omega l r plus r r ok now if you see

if omega is very omega tends to zero at very low frequency z r is equal to zero

that means the acrostics load in the radium radiation load in here is not there it is

totally short circuit which is the idle condition we said p l t is equal to zero so the frequency

response whatever we have get due to the considering the loss and that things that will be remain

constant remain same that means that implies or that means that low frequency components

are not affected by the radiation load because at low frequency radiation load doesnt effect

the frequency response of the acoustical tube so low frequency are less affected now if

i say j omega or omega l r is much much greater than r r at high frequency much much greater

than r r then i can say it is nothing it is nothing

but a r r j omega l j omega l cancel so it is nothing but a r r so load is totally resistive

so radiation load is resistive means there is a loss radiation loss so i can say high

frequency are affected due to the radiation loss ok so i can say the frequency response

which i will get after the considering the losses with the bandwidth if it is passed

through the radiation loss then low frequency will not affected much more but high frequency

will be lost ok so i will explain it in here if you see this is the figure with the no

loss condition at five hundred hertz infinite energy zero bandwidth

if i consider the wall vibration then i say the bandwidth is increases slightly amplitude

increases in high frequency and band to bandwidth is introduced and bandwidth in the low frequency

are much more then if i say the viscous loss friction law loss consider then again the

bandwidth is increase the bandwidth and slightly decrease the formant frequency it slightly

increase the formant freque[ncy] so increase decrease if i cancel out i can say that ok

formant frequency is and around what about five hundred hertz i have getting here but

bandwidth is here here is zero bandwidth now if i consider the radiation loss then if you

see high frequency are resistive radiation so this circuit this will be suppressed ok

so radiation loss due to the radiation loss high frequency components amplitude will be

decrease ok now if you see that also here also then in mathematics h omega transfer

function of the tube is nothing but a p l omega divided by u g omega which is p l omega

u l omega so it is nothing but a z r omega will be multiply with the v omega z r omega

is one or no effect if omega is very low if z r omega is sorry if the omega is low radiation

effect is not negligible if it is z r omega is omega is very high then there is a lot

of attenuation in the high frequency so frequency response of the z r omega is nothing but like

this this stepper high if the frequency increases loss is increases ok so thats why if i see

any speech signal if you see the speech signal it will look like that high frequency or amplitude

are less it is due to the radiation loss high frequency amplitude are very less ok

so this is the frequency response of the uniform tube if i consider whole tube vocal tract

in a single tube then i say this is the frequency response this is a transfer function of the

single tube model where the frequency response is that formant frequency are five hundred

hertz one point five kilo hertz two point five kilo hertz but the response is high frequency

are attenuated due to the radiation loss and it is not zero bandwidth because of the if

i consider the losses inside the tube ok now so vocal tract can be characterized by

a set of resonance that depends on the vocal tract area function with shifts due to the

loss and radiation the bandwidth of the two lowest resonance depend primarily on the vocal

tract wall losses f one and f two and the bandwidth of the highest resonance depends

the highest frequency resonance depend primarily on the viscous friction loss thermal loss

and radiation loss because if you see i said the bandwidth is most affected frequency or

low frequency in case of wall variation in the wall in case of viscous law high frequency

bandwidth are increase ah introduced now next one is nasal coupling effect ok this is oral

one cavity i have done this is the one cavity i have done now if you see you experience

that this is a cavity and it produces sound now if i put a hole in here sound will change

if you see the flute there is a lot of hole in the top of the flute and by pressuring

the closing the ah hole i can change the sound if you see that flute or saxophone ah just

closing the hole means i am closing the hole in the tube so if i put a hole here you know

the frequency response is change the two frequency response will be change ok now in case of

our case in vocal tract also there is a nasal coupling in here so once the velam is open

so nasal coupling is coupled with the oral tract give structure is changed frequency

response will be change so sound pressure the same as at input for each tube the volume

velocity is the sum of the volume velocity at the input to nasal and oral cavity so closed

oral cavity can trap energy at certain frequency preventing those frequency from appearing

in the nasal output so if you think about what i producing a nasal

consonant or nasal sound if my oral cavity is totally closed then the air is coming through

the nasal now if you oral cavity is not totally closed then the nasal and oral both cavity

are so if i consider the nasal case sound is coming in this path then sound pressure

will be trapped by these oral cavity so that trap will produce some anti resonance frequency

so nasal resonance have border bandwidth loss is much more so bandwidth so in case of the

formant band width will be border in case of nasal sound and the anti resonance will

be there so i can say deep if i say this this point

will be much more larger point anti resonance will be introduced so bandwidth will be larger

also so compared to the oral sound nasal sound bandwidth is larger ok now lets i stop here

because next class i will start about the sound another topic which is called how the

excitation will excited the vocal tract then will derive the two boundary condition which

is at the lip at the glottis after derived considering the two boundary condition will

derive the total transfer function in digital domain and then try to implement is uniform

tube model then you go for the multi tube modeling of the human vocal tract ok

thank you