Uniform Tube Modeling of Speech Processing – II

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

Leave a Reply

Your email address will not be published. Required fields are marked *