hello friends welcome to my lecture on convergence
of z transform ah before discussing the convergence
of z transform let us look at some more ah
methods where we can find the z transform
ah we can find the inverse z transform of
uz ah the partial diffraction method we have
discussed earlier also here we have one more
ah case where we can discuss the we can find
the inverse z transform of uz so in this method
as i have said in my previous lecture when
we want to use the partial different ah diffraction
method ah we consider uz by z and break uz
by z into partial fractions and then multiply
the resulting expansion by z and then we invert
that so suppose we have the example of uz
equal to two z square plus three z over z
plus two z minus four then and we are to find
the inverse z transform of this function of
z so then what we will do is let us say let
uz equal to two z square plus three z divided
by z plus two z minus four so then we consider
ah uz by z let us write uz by z his will be
two z plus three divided by z plus two z minus
four ah we will break this two z plus three
over z plus two z minus four into partial
fractions
so let us write this as a over z plus two
and b over z minus four then the values of
ah a and b can be found easily so then a will
be equal to ah two z plus three divided by
z minus four evaluated at z equal to minus
two and this will be giving you minus four
plus three so minus one divided by minus two
minus four so that is minus six so we will
get one by six and b will be equal to ah similarly
two z plus three divided by ah z plus two
and we will have to evaluate this at z equal
to four so when we put z equal to four here
four into two eight plus three is eleven divided
by four plus two is six so b is six so thus
uz by z is equal to one by six z plus two
and then eleven by six into one over z minus
four so we will then multiply ah this equation
by z so multiplying by z we get uz as z over
one over six into z over z plus two and then
eleven over six oh z over z minus four now
let us recall that we know we know that z
inverse of z over z minus a this is equal
to a to the power n provided mod of z is greater
than mod of a so so z inverse of z over z
plus two will be equal to minus two to the
power n provided mod of z is greater than
mod of minus two which is two and similarly
z inverse of z over z minus four will be equal
to four to the power n provided mod of z is
greater than four and hence z inverse of uz
will be equal to one by six z inverse of z
over z plus two which is minus two to the
power n and plus eleven by six ah z over z
minus four inverse in z inverse of z over
z minus four which is four to the power n
and the common region is mod of the in case
of z inverse z over z plus two the region
of convergence was mod of z greater than two
and here it is mod of z greater than four
so we will have this is valid when mod of
z is greater than four
so here un is equal to one by six ah two to
the power minus two to the power n and plus
eleven by six four to the power n where mod
of z is greater than four so this is how we
can find the inverse z transform in this case
ah let us now there is another method which
is known as inversion integral method ah here
we will need um analysis of complex analysis
knowledge of complex analysis that is the
contour integration so i think i hope that
you are familiar with this contour integration
ah so if you know contour integration you
can use this method also to find the inverse
z transform so the suppose ah we have sequence
un ah the inverse z transform we we know the
z transform of a sequence un it is given as
uz then the inverse z transform f uz is given
by un equal to one over two pi i integral
over c ah ez uz into z to the power n minus
one dz where c is the contour which encloses
the similarities of the function ah uz so
and we calc
so to evaluate un we calculate the sum of
residuals of the analytical function uz into
z to the power n minus one at the poles of
uz which lie inside the contour c according
to the given region of convergence so by the
given region of convergence we decide how
many poles of uz lie inside the ah contour
c and then calculate the residues at those
ah poles of uz and take their sum to get the
ah sequence un so let us look at this ah method
suppose we want to determine the inverse z
transform of ah three z over z minus one z
minus two so z inverse of three z over z minus
one z minus two here let us say u un uz is
equal to three z over z minus one z minus
two so ah let us find the poles of uz ah the
similarities of uz are the simple poles 
at z equal to ah one end z equal to two now
we are not given the region of convergence
so we shall assume ah that the region of convergence
is such that it encloses all the similarities
ok so z equal to one z equal to two are enclosed
inside the contour c ok so lets see the contour
c enclose both the poles z equal to one z
equal to two then we have to find the residue
at z equal to one so residue at z equal to
one of the function ah uz into z to the power
n minus one is equal to limit z tends to one
z minus one into uz into z to the power n
minus one
we know that when he function fz in complex
analysis has a simple pole at the point z
equal to a and we want to determine the residue
there then we multiply fz by z minus a and
take the limit as z tends to a so here we
multiply uz into z to the power n minus one
by the factor z minus one and find the limit
at z tends to one so this is equal to limit
z tends to one z minus one and then uz is
three z into z to the power n minus one divided
by z minus one z minus two
so this z minus one we can cancel and when
z tends to one it is ah three into one to
the power n divided by ah three minus two
so one this is three into one to the power
m and similarly limit z tends to two similarly
the residue at z equal to two of the function
uz into z to the power ah n minus one is equal
to ah limit z tends to two z minus two into
ah uz into z to the power n minus one which
will be equal to limit z tends to two z minus
two into three z divided by z minus one z
minus two into z to the power n minus one
so this will cancel with this and when z tends
to two what we get is ah ah three into two
to the power n ok so thus we get thus ah un
by the formula un equal to sum of residues
at the poles z equal to one and z equal to
two so three into one to the power n plus
ah three into two to the power here we are
getting ah one minus two minus one we are
getting so this will be minus so this is minus
here minus three into one to the power n plus
three into two to the power n so we can write
it three times two to the power n minus one
so this is the ah this how we can find this
sequence un ah by using the invergent integral
method ah ok so let us discuss one more example
z inverse of ah two z over ah z minus one
z square plus one ok let us say let us say
uz be equal to ah two z over z minus one z
square plus one so then we write uz by z so
then uz by z will be equal to two over z minus
one z square plus one which we can write as
a over z minus one b over z minus i and c
over z plus i because z square plus one can
be broken into ah linear factors z minus i
z plus i now so then a will be equal to ah
put z equal to one and two over z square plus
one so in two over z square plus one when
we put z equal to one we will get two over
two we get one then we we can find so two
over z minus one and here z minus two we shall
leave z plus i we will take
so this evaluated at z equal to i this will
give you two over i minus one into two i so
we shall get these two will cancel with his
two and we shall get ah i square which is
minus one so minus one minus i so minus one
upon one plus i and similarly c will be equal
to ah two over z minus one into z minus i
at z equal to minus i so this will be equal
to two over minus i minus one and then minus
two i so ah two will cancel with two when
we multiply by minus i here so minus i into
minus i is i square which is minus one so
minus one and then we get plus i so we get
minus one upon one minus i and hence uz will
be equal to ah z we multiply in this equation
and a is one so z over z minus one we have
and then we have v is equal to minus one over
one plus i into z over z minus i and then
c is minus one over one minus i so minus one
over one minus i into z over z plus i and
this implies u one the inverse z transform
of uz will be z over z minus one means ah
one to the power n because inverse z transform
of z over z minus i is ah a to the power n
ah and then minus one over one plus i inverse
z transform here will be i to the power n
so here mod of z is greater than one here
mod of z is greater than mod of i mod of i
is also one so region of convergence is same
mod of z greater than one and here we have
minus one over one minus i into minus i to
the power n and the region of convergence
is mod of z greater than ah mod of minus i
which is again mod of z greater than one so
we get mod of z greater than one so we have
one minus ah i to the power n over one plus
i minus minus i to the power n over one minus
i where mod z is greater than one so this
is the solution of example two now let us
go to a two sided z transform so far we consider
the z transform of which was one sided z transform
where one sided z transform means ah we consider
the values of n to be non negative ah n equal
to zero one two three and so on and we considered
uz equal to sigma n equal to zero to infinity
un z to the power minus n but n can take negative
values also so when n takes values from minus
infinity to plus infinity the z transform
will be defined as sigma n equal to minus
infinity to un z to the power minus n and
this transforms then called as two sided z
transform ah in the case of one sided z transform
where we had taken n greater than or equal
to zero the region of convergence as we have
seen is always of the form mod of z greater
than mod of a which is the region outside
the circle ah mod of z equal to mod of a the
circle where the center is at the origin and
a dash is mod of a so the region of convergence
will look like this 
if this is the circle with mod of with ds
mod of a in the z plane then ah the region
outside the circle mod of z equal to mod of
a is this shaded region so the region of convergence
is always the circle is mod of z equal to
mod of a
so in the case of the ah values of n greater
than or equal to zero the region of convergence
is always outside the circle it is of the
form mod z greater than mod of a but when
ah n takes the negative values that is from
minus infinity to zero the region of convergence
we shall see is of the form mod z less than
ah mod of b so it will be of this form suppose
this is your mod z equal to mod of b circle
with center at origin and ready as mod of
b then it is interior of this circle the region
of convergence ah lies to the interior of
the circle ah so in the case of two sided
z transform ah for the negative values of
the ah negative powers of z we get the ah
region of convergence which is outside the
circle for the power of z which are positive
we the region of convergence will look like
ah it is inside the circles so it so for the
z two sided z transform the region of convergence
will be of the form ah mod of a less than
mod of z less than mod of b that is a the
region of convergence lying between two concentric
circles so such a region of convergence is
known as annular region and the inner circle
than inner circle is mod of z equal to a mod
of z equal to mod of a so it bounce the comes
in negative powers of z and the outer circle
which is mod of z equal to mod of b it bounds
the terms in the positive powers of z so for
example let us consider this two less than
mod of z less than three so it is an anywhere
region ah the region concentric to two circles
mod of z equal to mod of two mod of z equal
to two and then mod of z equal to three
so this region is the given region of convergence
so here two is less than mod of z less than
three let us find the inverse z transform
of the function ah uz equal to here so uz
is equal to two times z square minus five
z plus six point five divided by z minus two
z minus three whole square so you can put
it as a over z minus two let us break it into
partial fractions b over z minus three and
then c over z minus three whole square so
we can determine the values of a b c very
easily ah a is equal to two times z square
minus five z plus six point five divided by
z minus three whole square evaluated at z
equal to two so let us put z equal to two
here so then two times we have four minus
ten plus six point five divided by two minus
three whole square so this is one and here
we have ten point five minus ten so we get
point five point five into two is equal to
one and similarly we can find directly the
value of c c will be two times z square minus
five z plus six point five divided by z minus
two and let us put z equal to three in this
so that we will get two times ah nine minus
fifteen and then plus six point five divided
by three minus two is one
so we get here ah fifteen point five minus
fifteen is point five point five into two
is equal to one now to determine b we have
to write the identity so two times let us
take the lcm two times z square minus five
z plus six point five this is equal to i am
taking lcm this is equal to a times z minus
three whole square plus b times z minus two
z minus three plus c times ah z minus two
now let us equate the terms containing z square
on both sides so here the term containing
z square is a and then here b so a plus b
the coefficient of z square on the right side
is a plus b and the left side it is two so
a plus b equal to two a we have got already
equal to one so a equal to one gives us b
also equal to one so a b c here are all equal
to one ok so thus we have so thus uz is equal
to z over ah a say sorry one over z minus
two and then one over z minus three and then
one over z minus three square ok now we are
given the region of convergence as two or
less than mod of z less than three so here
we can see that mod of z greater than two
implies ah two over mod of z two over mod
of z is less than one so what we will do here
two ah expand one over z minus two by using
the binomial theorem his i can write as one
over z times one over one minus two over z
and mod of two over z is less than one so
we can expand it in the binomial by binomial
theorem and here ah this is mod of z is less
than three so mod of z divided by three is
less than one so what i will do is i will
take minus one over three out and write it
as one minus z upon three
so that mod of z less than three gives us
the infinite series expansion of one over
minus z by three so here again here we shall
write one over minus three whole square and
then i will write one over one minus z by
three whole square so now let us expand by
binomial theorem so if we we do that i will
write it as one over z since mod of two by
z is less than one i will write it sigma n
equal to zero to infinity two by z raise to
the power m and then minus one by three again
this is my mod of z by three is less than
one
so sigma n equal to zero to infinity z by
three raise to the power n and here plus one
by nine summation n equal to zero to infinity
ah i think i should write like first i would
write it in expanded form so i should write
it as this is one minus z by three raise to
the power minus two so i will get here one
plus two times z by three plus two into three
by two factorial and then i will have z y
three m z by three whole square and then two
three four divided by three factorial minus
z by three whole cube and so on ah ok here
sorry minus minus minus and minus becomes
plus so i will get this here so ok so this
will be equal to ah so this will be one plus
two times z by three and then two into three
by two factorial ah z by three whole square
and then two into three into four divided
by three factorial z by three to the power
three and so on ok now let us see at this
ah this further let us expand this so one
by z this is what when n is equal to zero
first term is one then two by z we have then
we have four by z two square by z square ah
two cube by z cube and so on and so these
terms give us negative powers of z by these
two series are in the positive powers of z
so minus one by three and then here we have
ah one plus z by three plus z square by three
square z cube by three cube and so on this
the series in the second term in the third
term we get one by nine i can multiply one
by nine ah are ok one by nine one by nine
into ah plus two by nine into z by three and
then we have two into three divided by nine
into ah two factorial z by three whole square
and so on
now let us determine the sequence un from
here so for n greater than or equal to zero
we will get the value of un from this so because
sequence un we have to find un for the values
of n ah greater than or equal to zero so when
n is ah z to the power minus n when n is equal
to one and on wards so n greater than or equal
to one gives you because when n is equal to
one we get one here when n is equal to two
here we get two to the power two minus one
so we get two here the coefficient of z to
the power minus two is two and the coefficient
of z to the power minus three is two square
so ah un is two to the power minus n and we
get here minus n plus two into three to the
power n minus two so this is for n less than
or equal to zero let us take for n equal to
zero what do we get when n is equal to zero
we have here minus two over three square so
this is minus two over three square minus
two over three square means minus two over
nine and here what do we get minus one over
three minus one over three plus one over nine
so this is nine minus three plus one so minus
two over three square so here also we get
minus two over three square and the coefficient
of then z ah z here is minus one by three
square and here it is ah two by three cube
so coefficient of z is two by twenty seven
minus one by nine which is we can say it is
ah six minus one six minus three so we get
three by twenty seven or one by nine and what
we get here i think minus one by nine and
here we get two by twenty seven two by twenty
seven minus one by nine so we get here twenty
seven this will be two here two here and here
we will get minus three so we will get minus
one by twenty seven minus one by twenty seven
and here what we get so when n is equal to
minus one it will be ah minus minus one plus
two and three to the power minus three so
this will be minus one over three cube which
is minus one by twenty seven so the terms
which contain positive powers of z there un
is minus n plus two three to the power n minus
two and when the negative powers of z give
you un equal to two to the power n minus one
where n is bigger than or equal to one
so in the case of this annular region ah which
we were get the two sided z transform ah and
the ah sequence un then is two sided ah for
n greater than or equal to one we get two
to the power n minus one and for n less than
or equal to zero we get minus n plus two three
to the power n minus two ah we shall be applying
the knowledge of z transform to the difference
equations as we know that the laplace transforms
are very useful in solving linear differential
equations we shall show in our next two lectures
the z transforms are also very useful to solve
liner difference equations when we have a
discreet system the performance of the discreet
system is expressed by a difference equation
and z transforms are very useful in the analysis
and representation of discreet time systems
so the solution of the difference equation
is required to determine the ah for frequency
response of such a discreet system so in our
next two lecture we shall be discussing how
we can solve a ah difference equation by applying
the z transforms ah with that i would like
to conclude my lecture
thank you very much for your attention
