Uniform Tube Modeling of Speech Processing – III

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

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