Uniform Tube Modeling of Speech Processing – VI

# Uniform Tube Modeling of Speech Processing – VI

ok so last class we dis derived that junction
effect if i consider the vocal track is nothing but a junction of several tube and of the
cross sectional area across the tube is different what should be the junction effect we have
discussed now if i we said that if the each tube has a fixed length del x then this tau
the delay line tau can be represented by z to the power minus one because z to the power
minus if the t one is equal to tau ok so this is the single junction k th junction signal
flow diagram now if i consider the same if the tube has n number of section so suppose
i have a tube length is l and i cut this tube in n number of section so the length of the
each section x is equal length l by n so tau is nothing but a tau is nothing but a x by
c so it is nothing but a tau is nothing but
a x is nothing but a l by nc so if i I have a tube of seventeen point five second five
let say five centimeter tube long and if i ah divided the tube n n equal to ten junction
then the tau is nothing but a seventeen point five centimeter divided by n into c c is the
speed of the sound that much of delay is required ok now i come if my tube is n number of tube
is there so there will be a n minus one are the junction one to n minus one number of
junction so i can draw the equation like this way lets this is ugn then the first junction
second junction dot dot dot in a n minus one junction and this is the and if you see the
last one is the boundary condition at list and this one is the boundary condition at
glottis so it a rg it is half of one plus rg it is minus rl one plus rl i can draw the
n two equivalent signal diagram signal flow diagram and is delay is represented by z to
the power minus one ok now if this is my tube lets ugn is nothing
but a delta signal delta n ok delta n delta signal or delta n if it is delta signal then output if this
is the my signal flow diagram then output i should get bat at the output is nothing
but a b zero delta t minus n tau plus k equal to one to infinity bk delta t minus n tau
minus two k two tau why so if you see if i have a tube with n number of junction of length
lets n ah the sec n section of junction so each section create a delay tau so n section
create a delay n tau but that is the first first signal first delta thats why b zero
into delta n but after all a signal the after the first round the the delay will be there
so what is the delay two tau is the round tip delay so this is the first arrival of
the signal second will be arrival you know due to the back propagation so there will
be a two tau delay ok so earliest arrival is n tau which is n number
of tube is section delayed by tau next signal will come by two tau delay so i can next signal
will be integer multiple of two tau delay so thats why i said two k sorry k two tau
k into two tau k is the integer and two tau is the delay ok now if i consider to avoiding
the alysing of digitization process if i consider my sampling period t is equal to two tau so
my sampling period t is equal to two tau what i am saying suppose i want to create a band
limited signal or lets p signal whose maximum frequency is five hundred hertz the five five
kilo hertz or five kilo hertz lets five kilo hertz is the maximum band width what is the
sampling frequency ten kilo hertz ten kilo hertz is the sampling frequency so band limited
is five kilo hertz nyquist criteria is ten kilo hertz ok to if i if i want to reach the
nyquist criteria in this two tau delay so two tau must be equal to the sample period
ok so two tau is equal to sampling period if two tau is equal to sample period then
how do i derive this vz if we if i consider this is a z domain vz
which is nothing but a ulz divided by ug z ok and if z to the power minus one now z to
the power minus one is single sample delay now my sample delay is t is equal to two tau
so i can say instead of z to the power minus it will minus one to z to the power minus
half so all the z value will be replaced by z to the power minus half z to the power minus
half z to the power minus half z to the power minus half so i get this signal flow diagram
once i get this signal flow diagram can i derive the vz transfer function of the tube
which is nothing but a output which is uz ulz divided by ugz z domain ok so ugz i know
it is nothing but a impulse or gotal or or gotal response now what is ulz so i could
derive this transfer function so how do i derive it using a signal flow
diagram i can derive it so lets there is a procedural the for derive it so lets use some
procedure of its a some simplest this may be the simplest form procedure for deriving
he transfer function of vz which is nothing but the ulz divided by ugz how do we derived
it if i consider each one of the junction is acts as a lattice if you see the symmetrical
this is r one this is r two this is r n minus one but all are symmetrical ok now lets i
draw in here because for our understanding if i draw it here it will be very clear to
you now lets this is nothing but a my ugn ok coming to the first junction lets first
junction is here which is nothing but a i will first draw the junction and then i write
down the signal flows this is this way this is this way this is this way and this is ah
this way so this i write in the red pen so this is nothing but a one minus r one this
is nothing but a one plus r one this is r one this is minus r one ok now i have a ug so i have a ug is here ugn
is coming to here and then there is a terminal at glottis which is nothing but a so there
will be z to the power ok ah so lets there will be z to the power minus one delay z to
the power minus half here will be delay z to the power minus half and this will be rg
and this will be one plus rg divided by two ok and then the same junction i can continue
there will be a delay z to the power minus half there will be a delay z to the power
minus half and again there will be a junction so junction lets signal flow diagram this
is this this is this this is this again this will be one minus r two this will be one plus
r two this is r two this is minus r two ok then i can say there will be again delay again
delay z to the power minus half z to the power minus half ok again there will be a junction
lets this is nth junction n minus one junction there will be a junction this is the lets
n minus one junction so there will be a lot of junction in here so this if it is n minus
one junction this is one plus r n minus one this is one minus rn minus one this is rn
minus one this is rn minus one ok and at the end lip there will be boundary condition so
there will be a z to the power minus one then put the boundary condition which is like this
z to the power minus half z to the power minus half and this is nothing but a minus rl this
is one plus rl i get vln ok this is output so i have to find out you if it is z domain
the uz ulz divided by ugz that i have to find out so how do we find out lets calculate lets
calculate stage by stage so if i say any u so you know that uk plus one plus z is nothing
but a one plus rk that this oh only we derived one plus rk z to the power minus half ukz
uk plus z plus one z plus rk into uk plus one minus z that we have derived last day
uk plus one z the volume velocity which is injected in the k plus one tube is nothing
but this one ok from here i can find out what is uk plus z is nothing but in term of k plus
one tube so it is nothing but a so z to the power minus half if i say it will come z to
the power plus half divided by one plus rk into uk plus one plus z minus this will be
come this side so minus rk z to the power half divided by one plus rk uk plus one minus
z ok so this is you can say uk plus z uk plus at kth tube uk plus z is nothing but this
one ok similarly i can find out what is uk minus z ok so what is uk minus z already we
have done uk minus z is nothing but a minus rk into uk minus rk into uk plus z into z
to the power minus half plus one minus rk uk plus one minus z this we have already done
already derived ok so from here i can say uk minus z is nothing but a minus rk z to
the power minus half divided by one plus rk uk plus one plus z z plus z to the power minus
half one plus rk uk plus one minus z ok i can do that k plus one so i this i I represent
uk plus z uk minus z in term of uk plus one z uk ba ba uk pla uk plus one plus z uk plus
one minus z ok now once i do that then if you see here this is my uk plus equation and
this is my uk minus equation once i do that now if you see that if i say matrix this this
this all us uk one so u one u one plus u one minus the u one minus z u one plus u one minus
z you can see that blue and yellow so those are the equation now if i say lets uk is nothing
but a consist of two plus uk plus z and uk minus z whole uk is nothing but a forward
wave and backward wave and from here i can write uk is nothing but a rk into uk plus
one if i write that if you see this is the two
equation uk minus and uk plus uk so if it is uk plus one is nothing but a uk plus one
plus z and uk plus one minus z ok this has to be multiply by this matrix equation what
is the matrix z to the power half divided by one plus rk minus rk into z to the power
minus half a my z to the power half divided by one plus rk and another one is minus minus
rk z to the power minus half divided by one plus rk and z to the power minus half divided
by one plus rk so those coefficients i write in matrix form if i write this matrix from
ok then i can say i simplify this matrix z to the power minus half one plus rk i can
write down here so it is one minus rk divided by this one a one minus rk ah one minus rk
minus rk z to the power minus one z to the power minus one now if i see these matrix
rk cap and this whole matrix qk is equal to rk so i can write rk is nothing but a z to
the power minus half so qk is equal to rk i can write ok so rk is nothing but this one
ok now if it is that uk equal to uk equal to rk pla into uk plus one so if i say k varies from one to n thats zero
to one to n one to n if i k varies varies with the one to n then what will happen then
u one i can say is nothing but a r one r two r three dot dot dot rn into u n minus one
sorry u u n plus one i can write k varies from one to n lets if it is k varies from
one to n u one is nothing but a r one r two r three dot dot dot r n into un plus one ok
now if you see that this diagram there is a nth n minus one symmetry junction so this
is one two and this is n minus one here the junction is not symmetry in in rl cases only
problem is that there is a no backward wave in here lets i put a backward wave with zero
so there will be a i can put a backward wave here which is nothing but a zero so it is
one minus rm this is one plus rm one plus rl lets rl is equal to rn so it is one plus
rn and this is minus rn if this is plus rn this is minus rn i can write i can write so
see this here in the slides if you see i added this block so now there is a n number of block
instead of n minus one block i can get n number of block with input here backward wave is
zero so i can write u n plus one un plus one is equal to nothing but a ulz and zero because
here ulz is only zero is added with l ulz so ulz plus and minus wave so i can write
it is nothing but a one zero ulz ok and u one equal to r one r two r three dot dot dot
dot rn u n minus one so it is nothing but a r one r two r three dot dot dot rn into
one zero ulz ok now u one I get now i have to find out what is the relation
between ugz and u one z so i have to consider the boundary condition at the glottis so what
is ug relation between the ugz and u one z ugz so you one plus z is equal to zero point
five one plus rg one plus rg ugz plus rg ug u one minus z i can write so i can say ugz
is equal to two by one plus rg so zero point half means or you can write zero point five
by one plus rg two this is two two by one plus rg because this is half only ok two by
one plus rg u one plus z minus two rg by one plus rg u one minus z so i can say it is nothing
but a two coefficient two by one plus rg one and another is two rg divided by one plus
rg into u one z ok so ugz is nothing but a two by one plus rg it is constant it is nothing
but a one this is minus i think this is minus minus so it is minus rg into u one z so i can write u one is equal to r one r two
r three r four rn into one zero ulz so i can write ugz is equal to r one r two r three
sorry ugz equal to two by one plus rg one minus rg so instead of u one z i can write
r one r two r three dot dot dot rn one zero ulz now i get ugz divided by ulz is equal
to two by one plus rg one minus rg r one r two r three dot dot dot dot rn into one zero
i can say that so which is nothing but a one by vz one by the transfer function so vz i
want to find out vz which vz is equal to ulz divided by ugz so it is nothing but a one
by vz so one by vz i can write in term of two by one plus rg one minus rg r one r two
r three rn one zero ok now if i write that then if i put the value of r one and r two
the equation will becomes two by one plus rg one minus rg z to the power n by two so
if you see the rk is z to the power half so i can say if it is n so it is z to the power n by two k equal to
one to n one by one plus rk r one z to the power minus half for r two z to the power
minus z to the power half for r three z to the if it is rn product of z to the power
half z to the power half z to the power half so n number of so it is nothing but a z to
the power n by two k equal to one to n one plus one plus rk k equal to one to n cap rk
cap so rk is like the matrix is this and one zero so now if i say my n is equal to two
then i can find out the impulse or transfer function of the tube model so if i simplify
this thing for two tube if i consider that my length of tube whole vocal chord is vocal
track is simulated using two tube so if it is two tube so n is equal to two if i put
n equal to two then one by vz will become two by one plus rg will be same this will
be one minus rg will be there so there is no effect on here now is z to the power n
by two n equal to two so two by two ok then product of k equal to one to n k equal to
one to n means if n equal to two so k instead of k equal to one to n i can write one by
one plus r one into one by one plus r two ok then i can write k equal to two in this
equation so k equal to two k equal to one one minus r one one so i can write one minus r one another one
is minus r one z to the power minus one z to the power minus one into one minus upto
minus upto z to the power minus one z to the power minus one then one zero if i evaluate
it it will come so z to the power half z so z two z divided by one plus rg into one minus
rg this will be one by one plus r one into one plus r two then i can matrix multiply
this in this one so this one is one plus r one r two z to the power minus one one minus
r one z to the power minus one minus r two z to the power minus two if i evaluate this
matrix will be one zero i can say first evaluate this i two then come to the here this side
ok so if i write down the equation then whole equation one by vz is nothing but a i can
multiply this thing one plus r r one one plus r two and this thing i can multiply together
and i can find out so vz one by vz and then vz will becomes if i derive these things and
then the vz will become vz will be zero point five one plus rg into one plus r one into
one plus r two z to the power minus one divided by 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 this will be the two two tube
vocal track model if you see this transfer function i have a zero at origin z to the
power minus one but i have a two pole this is second order equation so i have a second two second order you can
say two pole or it is a second order equation so i have a single zero but two tube model
i have a zero at center one zero and two pole so at origin one zero and two pole so if i
have a n tube model i can say n by two zero at origin and nth this this equation become
n model so i can say then we n number of pole this equation become nth order so i can say
this whole vz can be same this is all zero are in center and this is all pole model so
i can write vz is nothing but a g by dz where dz is nothing but a pole model so if it is
if it is nth pole nth order so it is nothing but a one plus k equal to one to n alpha k
z to the power minus k so kth pole n nth pole will be the n number of pole will be there
so i can write down the equation one one plus k equal to one to n alpha k z to the power
minus k and n by two zero at origin so instead of writing this ah ah the all zero model z
to the power minus one minus have been here z to the power minus half in here i can write
this dz in a signal flow diagram as a all pole model and z equal to this vz z to the
power minus so all pole model then it will be z to the
power minus n by two so i can write down the signal flow diagram in all pole model so i
can say mathematically it is proved that a vocal track transfer function can be simplify
as a all pole digital filter all pole digital filter i can simplify it so if i able to simplify
all pole digital filter and if i know alpha k if i know alpha k which is nothing but in
term of r one r two and the r one r two rg i have to know so what is r one r two all
are reflection coefficient if i know all reflection coefficient i can simulate this dz so it is
impossible to implement this dz using digital filter ok so now i can say dz dz is nothing
but a there is a error in the slides please correct it dz is nothing but a one plus k
equal to one to n alpha k z to the power minus k so lets rg is equal to one that means the
the lets there is no gotal impedance rg rg is equal to one ok
thank you