hello friends welcome to my second lecture on properties of z transforms so first
we will discuss what happens when we know the z transform of a sequence un say
uz and then the z transform of n into un ok n into un that is un is multiplied by n hen
we see that z transform of n un is minus z into d over dz of uz so this result is also
going to be very useful when we solve difference equations so by the definition of z transform
z and un z and un is sigma n equal to zero to infinity n un into z to the power minus
n and this we can write as minus z times sigma n equal to zero to infinity un into minus
z n minus n times z to the power minus n minus one so applying the definition of z transform
we will we are able to write z transform of n un in this form
now but when we see that this can be written as minus z times sigma n equal to zero to
infinity un d over dz of z to the power minus n when you differentiate z to the power minus
n with respect to z what you get is minus n times z to the power minus n minus one so
we express in this form in this form but then this is equal to minus z d over dz of
sigma n equal to zero to infinity even z to the power minus n and therefore it is equal
to minus z times d over dz of uz now if you apply mathematical induction on p we can see
that z transform of n to the power p un is equal to minus z to the power p dp over dzp
uz where p is the positive integer now say suppose we have un equal to n cos
n theta then n we want to find the z transform of u one then we can use the known result
for z transform of cos n theta and make use of the theorem which we have just proved
that is the z transform of n un is minus z times d over dz of uz so let us see we
let us recall that z transform of cos n theta is equal to we we know that z transform
of cos n theta this we have proved in the previous lecture that it is z into z
minus cos theta divided by z square minus two z cos theta plus one so now let us apply
the theorem multiplication by n so then z transform of n cos n theta will be equal
to minus z times d over dz of let us call it uz so d over dz of uz
so this is minus z times d over dz of z square minus z cos theta divided by z square
minus two z cos theta plus one ok so this will be equal to minus z times the derivative
of the numerator is two z minus cos theta multiplied by z square minus two z cos
theta plus one minus two z minus two cos theta into z square minus z cos theta divided
by z square minus two z cos theta plus one whole square ok now when we simplify the
numerator we will have this as minus z times minus z times z square sorry
so this will be if you simplify the numerator you will get z cube cos theta minus two
z square plus z cos theta divided by z square minus two z cos theta plus one whole square
so on simplifying we get the z transform of un z cube cos theta minus two z square
plus z cos theta over z square minus two z plus cos z cos theta plus one whole square
now suppose we have un equal to n plus one square we want to find the z transform
of un equal to n plus one square then here we can use we know that z transform of n square
z transform of n square is z square plus z divided by z minus one raise to the power
three ok so and we know we know this shifting theorem if z of un is equal to uz then
z transform of un plus one z transform of un plus one is z times uz z times times
uz minus u naught so here we will get z transform of
un plus one so here un you can take as n square if you take un as n square then z transform
of n square is z square plus uz is z square plus z over z minus one whole cube and then
z transform of n plus one whole square so z transform of n plus one whole square will
be z times uz uz is z square plus z my over z minus one whole cube so this is minus
u naught now here we need the value of u naught ok
so for that i need to know what is u naught and this u naught we shall require we shall
we shall do the initial value theorem there we shall know that if we know z transform
of a sequence un that is uz then by taking the limit of uz z tends to infinity we get
u naught so we know it it we we shall show that
u naught is equal to limit z tends to infinity uz and here uz is z transform of n square
so z square plus z over z minus one whole cube so u naught is equal to limit z tends
to infinity z square plus z divided by z minus one whole cube and this is a polynomial in
z of degree two here we have a polynomial in z of degree three so the limit is call
obviously zero ok so u naught is equal to zero and hence z transform of n plus one whole
square is equal to z cube plus z square divided by z minus one whole cube
now you may wonder how do you determine how we prove that limit of uz as z tends to
infinity is u naught so this is very simple we know that since uz
is equal to z transform of un is sigma n is equal to zero to infinity un z to the power
minus n which is u naught plus u one by z plus u two by z square and so on so when you
take the limit of this z tends to infinity this gives you u naught so this implies so
uz equal to naught plus u one by z plus u two by z square and so on this implies limit
as z tends to infinity uz is equal to u naught this is called initial value theorem which
we will prove in which we will prove in the next lecture so we can use this that initial
value theorem to arrive at the z transform of n plus one whole square this is z cube
plus z square over z minus one whole cube so this is this result uses z transform of
that the value of u naught which we get by initial value theorem but if we do not want
to use the initial value theorem then we can write by by linearity property so we can
do this result by applying in by initial value by linearity property z transform of
plus one whole square is equal to z transform of n square plus two times z transform of
n plus z transform of one and z transform of n square is z square plus z divided by
z minus one whole cube plus two times z transform of n which is z over z minus one whole
square and this is z transform of one is z over z minus one
so simplifying this we shall have z cube plus z square divided by z minus one so by using
the initial value property but by using the linearity property we can obtain the z
transform of n plus one whole square and if you do like this that is you use the shifting
property then you need to know the value of u naught which we will get by using the
definition of uz as sigma n equal to zero to infinity un z to the power minus
taking the limit as z tends to infinity to get the value of u naught
now let us go to unit impulse sequence it is given by we define a unit impulse sequence
by delta n equal to one when n is zero n zero for all n different from zero so if you find
z transform of delta n let us apply the definition the definition sigma n equal to zero to
infinity delta n z to the power minus n when is equal to zero we are given delta n to be
equal to one otherwise it is zero so all other terms are zero except the term for n equal
to zero where delta n is one so we get one into z to the power zero which is one so zero
transform of delta n is equal to one so whenever we need the z the inverse z transform of
one we will write there delta n then we need the z transform unit step function
so unity step function we defined as un equal to zero when n is less than zero one for all
n greater than or equal to z that is for all non negative integers we define un to be equal
to one and for all negative integers we define it as zero so z transform of the unity
step function un is then again by definition sigma n equal to zero to infinity un zero
to the power minus n which is sigma n equal to zero to infinity z to the power minus n
because un is equal to one for all n bigger than or equal to zero now this is z to the
power so this is one plus z plus z square and so on unit step function this is sigma
n equal to zero to infinity z to the power minus n
so this is one plus one by z plus one by z square one by z cube and so on so this series
converges when mod of z is greater than one and the sum of the series is one over
one minus one by z so this is true when this is greater than this is less than one or this
is z over z minus one when mod of z is greater than one so there is enough convergence of
this series is mod of z greater than one the z transform of unity step function
is z upon z minus one now let us consider a an example of a rational function in z and
see how we determine the inverse z transform so here we know uz so let uz be equal to
four z square minus two z divided by z cube minus five z square plus eight z minus four
ok so we will now let us recall that z
transform of n is z over z minus one whole square z transform of n square is z square
plus z divided by z minus one whole cube ok now these tendered technique to determine
the inverse z transform is that when uz is a rational function of z we write first uz
by z so uz by z is equal to four z minus two divided by z cube minus five z square
minus eight z minus four it will be clear in when we do the solution why we are taking
uz bt z so let us break uz by z into partial fractions we see that when you put z equal
to one here this denominator becomes zero so z minus one is the factor of this so we
have to find factors of this polynomial z cube minus five z square plus eight z minus
four so let us find the factors of this z minus one is the factor and it is a polynomial
so what we do is we write it as z minus one z minus one we multiply by z square we get
z cube minus z square then minus four z minus four z square minus four z square then
plus four z plus four so we get z minus one into z square minus
four z plus four which is z minus one into z minus two whole square so we get here four
z minus two divided by z minus one into z minus two whole square and the factors then
then the partial fractions we can write as a over z minus one plus b over z minus two
plus c over z minus two whole square corresponding to z minus one and z minus two whole square
now the values of a and c can be found directly so a is equal to we remove z minus one
from here so four z minus two over z minus two whole square at z equal to one so this
will be four minus two that is two when minus two is minus one whole square there is one
so we get the value yes two and then c can also be obtained directly c is equal to z
minus two whole square we remove from here so four z minus two divided by z minus one
and when we put z equal to two so this will be four into two eight minus
two six so c is equal to six we have to determine now the value of b for that we write
the lcm so lcm is z minus one z minus two whole square and so then four z minus two
is equal to a times z minus two whole square plus b times z minus one z minus two and c
times z minus one now on the right side we get terms in z square but n the left side
there is no term in z square so let us collecting the terms in z square here the coefficient
of z square is a here here the coefficient of z square is b so a plus b is equal to zero
so we get a plus b equal to zero and a we have already found so this implies b is equal
to minus two so thus we get the value of uz by z so hence uz by z is equal to
a a is two two over z minus one and then b over z minus two so minus two over z minus
two and then c is six so six over z minus two whole square
ok now k now we multiply by z so we get r
uz equal to two z upon z minus one then minus two z upon z minus two and then six z upon
z minus two whole square now let us see why we see we bring we consider uz by z
because then the partial fraction will be some constant divided by z minus one and constant
divided by z minus two square and what we need when we find the inverse z transform
of uz we need here the forms the standard forms z over z minus one whole square or z
minus z over because we know that z transform of a to the power n is equal to z over z minus
a provided mod of z greater than mod of a so here we need z over z minus there if if
we do not consider uz over there and instead consider uz equal to uz four z square minus
two z upon this and then we bracket into partial fractions we shall not get the a standard
forms where z should come in the numerator so we consider uz by z now this is two z over
z minus one so we can take the inverse transform now and let us see when we take the inverse
z transform so z inverse of uz is equal to z inverse two times z inverse of z over z
minus one then minus two times z inverse of z over z minus two and then plus six times
z inverse of z over z minus two whole square z transform of a to the power n is z over
z minus a so when mod of z is greater than one this transform will be put a equal to
one so two times one to the power n for this and then minus two times z transform of
n is z over z minus one whole square we need the z transform z inverse z transform
of z over z minus two so if we take z transform of n is z so this implies so then z transform
of n into a to the power n if we want this then we multiply then we divide using the
dumping rule we will get z over a divided by z over a minus one whole square and
so alright k ok so this will be ok his is az divided by z minus a whole square ok
now here here we can get it directly so minus two times z over z minus taking mod of z greater
than two ok mod of z get two to the power n so here mod of z greater than we want ere
we want mod of z greater than two and then plus six times and then z inverse of ok
so z inverse of az divided by z minus a whole square this is equal to n into a to the power
n so we can here we have a in place of a we have two here so this will be n into a to
the power n divided by a so six into n into two to the power divided by two we
can multiply and divide by two here six by two and then we can bring two here like this
ok so this will be so this is this first inverse z transform is valid when mod of z
is greater than one second transform is valid when mod of z is greater than two so the common
region we have to take so mod of z greater than one and mod of z greater than two so
mod of z greater than one if we have mod of z greater than than two we have mod of z greater
than two then both will be valid so we have two minus two to the power n plus
one plus three times n into two to the power n so this is valid for mod of z greater than
the two we have taken the common region of convergence mod of z greater than one and
mod of z greater than two so in so this is how we find the inverse z transform un ok
now we take up example two where suppose un is equal to one over n plus one then z
transform of one over n plus one we have to find so we know that z transform of one by
n if we find the z transform of one by n z transform of one by n then z transform of
one by n is given by sigma n equal to zero to infinity one by n z to the power minus
n and z transform of one by n plus one for this we can use the shifting property
if you take un equal to i by n then un plus one can be considered as one by n plus one
so this is z times uz minus u naught u naught is the term in z n minus one which
does not contain your your z does which does not contain n your z term so here we
will consider n to begin with i think i should have taken z directly one by n plus
one let me write it again z transform of one by n plus one is sigma n equal to zero to
infinity one by n plus one z to the power minus n and if you expand this what we will
get n equal to zero so one minus n equal to one by two over z then n equal to two one
by three here also plus one by three z to the power two one by four z to the power three
and so on now what we will do is we can write it
as see let us recall that as log ln one plus z let us recall the extension of ln z plus
one this is z minus z square by two plus z cube by three and so on minus z cube z four
by four and so on ok so ln one minus one by z let us so this is mod of z less than one
and ln one minus one by z if we write this then what we have minus one by z minus
one by two z square minus one by three z q minus one by four z four and so on
so this so this i can write as minus z times if you identify with this minus one upon z
minus one by two z square minus one by three z q we can write like this so here mod
z less than one here mod of one by z will be less than one this means mod of z must
be greater than one so mod of z must be greater than one so this is minus z ln one minus one
by z provided mod of z is greater than one and this is raised to the power ln z minus
ok so this is z times ln one minus one by z raised to the power minus one or z times
ln z over z minus one so this is the z transform of one over n plus one provided mod of z is
greater than one and then we can find the z transform of one over in to n plus one z
transform by using linearity property so this is equal to z transform of one by
n minus z transform of one by n plus one if we know the z transform of un we can
also find the z transform of one by n then we will need that n equal to zero will term
we will cannot take z equal to z transform of we know that if z transform of un is equal
to uz then z transform of un minus k z to the power k into uz z to the power minus k
into u z so z transform of we know z transform of one over n plus one so z transform of
one over n z transform of one over n plus one we know this is z of ln z over z minus
one so from here if we write z transform of one by n using this result we have z to the
power minus one into uz that is a z in to ln z over z minus one so we have ln
z over z minus one so this will be ln z over z minus one by using
the shifting theorem so this is ln one minus z times ln z over z minus one provided
mod z is greater than one so by using the shifting theorem so this is ln one minus
z times ln z over z minus one so y z mod z is greater than one so by using shifting theorem
we can find this now let us consider this problem example three so example three
we have to find inverse z transform of one over z minus five whole to the power three
inverse z transform of one over z minus five to the power three let us write one over
z minus five to the power three as one over z cube times one minus five by z to the power
three so then one by z q one minus five by z to the power three minus three
now we are given that mod of z is greater than five so mod of five by z is less than
one so we can apply the one angle theorem and write it as one over z q and one plus
three times five by z then three in to four divided by one in to two two factorial five
by z whole square then three into four into five divided by one into two in to three five
by z whole cube and so on three four five six divided by one two three four five by
z whole cube and so on ok i can write it as one by two into one by z cube then two into
five by z to the power zero then two into three five by z to the power one then three
into four five by a to the power two here we shall have four into five five by z to
the power three and so on five in to six five into six five by z to the power
here five by z to the power zero five by z to the power one five by z to the power two
five by z to the power three there five by z to the power four so five by z to the power
four and so on so this is this can be written as one by two zq sigma we have two two
into three so we have m plus two five by z to the power
let us say m here m plus three m here m m equal to if i write it for say from zero to
infinity then what do we get here m equal to five m plus two m plus three we get here
n is equal to m plus three so m plus three means m plus two m plus one let me see m is
equal to zero if i put here i get here two into one m equal one i put here two into three
m equal to two i put here so three into four yeah so m plus i can write here m plus one
and here m plus two m plus two and then five by z to the power m now i think it is ok m
equal to zero put two into five by a to the power zero then two in to three five by a
to the power one then three in to four five by a to the power two and so on ok so we can
write like this and then this will be equal to one by two summation m equal to zero till
infinity ok so m plus one in to m plus two into five to the power m z to the power minus
m minus three z to the power three we have in the denominator
so z to the power m minus minus m minus three we get now let us put n equal to m plus three
let us put n equal to m plus three so then this will be one by two times summation n
is equal to three to infinity and m is equal to n minus three so we will get n minus two
here and here we will get n minus three plus two so n minus one and then we get five to
the power n minus three five to the power n minus three into z to the power z to the
power minus n where we are putting n equal to m plus three
so from here we can see that now this is nothing but this summation n equal to three to infinity
n minus one n minus two divided by two five to the power n minus three into z to the power
minus n so when n is greater than or equal to of this now let us compare it with the
z uz is equal to sigma n is equal to zero to infinity u n z to the power minus n
this is z transform of un sequence so if we compare with this then we can say that here
un is equal to one by two n minus one and n minus two five to the power n minus three
when n is greater than or equal to three and this is equal to zero when n is less than
three so this is the inverse z transform of z minus five over q with that i would like
to conclude my lecture
thank you for your attention