hello friends welcome to my lecture on convergence of z transform before discussing the convergence
of z transform let us look at some more methods where we can find the z transform
we can find the inverse z transform of uz the partial diffraction method we have
discussed earlier also here we have one more 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 diffraction method 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 uz by z let us write uz by z his will be
two z plus three divided by z plus two z minus four 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
a and b can be found easily so then a will be equal to 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 similarly two z plus three divided by 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 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 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 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 let us now there is another method which
is known as inversion integral method 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 so if you know contour integration you
can use this method also to find the inverse z transform so the suppose we have sequence
un 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 ez uz into z to the power n minus one dz where c is the contour which encloses
the similarities of the function 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 contour c and then calculate the residues at those
poles of uz and take their sum to get the sequence un so let us look at this method
suppose we want to determine the inverse z transform of 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 let us find the poles of uz the
similarities of uz are the simple poles at z equal to one end z equal to two now
we are not given the region of convergence so we shall assume 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 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 three into one to
the power n divided by 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 n minus one is equal to limit z tends to two z minus two into
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 three into two
to the power n ok so thus we get thus 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
three into two to the power here we are getting 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 this how we can find this sequence un by using the invergent integral
method ok so let us discuss one more example z inverse of two z over z minus one
z square plus one ok let us say let us say uz be equal to 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 linear factors z minus i
z plus i now so then a will be equal to 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 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 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 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 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 one to the power n because inverse z transform
of z over z minus i is a to the power n 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 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 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 we consider
the values of n to be non negative 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 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 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 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 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
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 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 lies to the interior of
the circle so in the case of two sided z transform for the negative values of
the negative powers of z we get the region of convergence which is outside the
circle for the power of z which are positive we the region of convergence will look like
it is inside the circles so it so for the z two sided z transform the region of convergence
will be of the form 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 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 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 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 nine minus fifteen and then plus six point five divided
by three minus two is one
so we get here 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 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 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 two over mod of z two over mod of z is less than one so what we will do here
two 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 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 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 ok here
sorry minus minus minus and minus becomes plus so i will get this here so ok so this
will be equal to so this will be one plus two times z by three and then two into three
by two factorial 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
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
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
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 are ok one by nine one by nine into plus two by nine into z by three and
then we have two into three divided by nine into 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 greater than or equal to zero so when n is 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 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 z here is minus one by three
square and here it is two by three cube so coefficient of z is two by twenty seven
minus one by nine which is we can say it is 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 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 which we were get the two sided z transform and
the sequence un then is two sided 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 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 for frequency response of such a discreet system so in our
next two lecture we shall be discussing how we can solve a difference equation by applying
the z transforms with that i would like to conclude my lecture
thank you very much for your attention