hello
friends this is my first lecture on z transforms so in this lecture we shall discuss
the z transform and inverse z transform of some elementary functions the z transforms
are useful in solving difference equations which arise in in the when in the
discrete systems the development of the communication branch of engineering is actually based on
the theory of z transforms the z transforms has the same role in discrete systems as
the laplace transform in continuous systems many properties of the z transforms are similar
to the properties of the laplace transforms z transform operates on a sequence of u n
of discrete discrete integer valued arguments n equal to zero plus minus one plus minus
two and so on and like the laplace transform which operates on continuous functions z transform
is we can say is the discrete analog of the laplace transform for every operational rule
and application of laplace transform there is a there is an operational rule and application
of z transforms a discrete system can be expressed as a difference equation and then its solution
can be obtained by using z transforms so let us see how we define the z transforms
the z transform of the sequence u n defined for discrete values n equal to zero one two
three and so on and e one equal to zero for n less than zero is denoted by z u n and is
defined as z u n equal to sigma n equal to zero to infinity u n z to the power minus
n which is a function of z so we write it as u z provided the infinite series convergence
so whenever this infinite series converges it will give us a function of z so it is u
z the inverse z transform is written as z inverse oh of u z z inverse of u z equal to
u n so whenever we know the z transform of a sequence its inverse z transform will
give us back the sequence even like in the case of laplace transform we have we define
the inverse laplace transform here again the inverse z transform is defined in a similar
manner so z inverse u z equal to u n now let us discuss z transforms of some
a standard sequences so let us begin with u n equal a to the power n so when u n is
a to the power n then where a is a complex number then z of a to the power n is sigma
n equal to zero to infinity so u n here is a to the n so n equal to zero to in between
u n z to the power minus n so we have a at to the power n and z to the power minus n
this capital z this is small z now this is summation n equal to zero infinity a by
z raise to the power n now we know that here z is a complex number a is also a a is
a complex constant so a by z is a complex number so the series of complex numbers sigma
n equal to zero to infinity a by z to the power n converges provided mod of z a by z
is less than one and we have this is equal to one over one minus a by z provided mod
of a by z is less than one so this gives you z over z minus a provided mod of a by z is
less than one or mod of z is greater than a so the condition for the convergence
of the series here is that mod of z must be greater than mod of a and now let us
look look at the formula which is known as the recurrence formula if you take u n
to be to the power p let us say u n equal to n to the power p where p is a positive
integer then we have a z of n to the power p this formula is very useful in determining
the z transforms of other sequences so minus z d over d z of z and to the power
p minus one ok so here what do we let us prove this formula ok
so to prove this formula we have z of n to the power p z of n to the power p by definition
is sigma n equal to zero to infinity n to the power p z to the power minus n ok so we
have this we have given here the proof this can be written as z times summation a
equal to zero to infinity n to the power p minus one into n into z to the power minus
n plus one now since by definition z of n to the power p minus one is equal to summation
n equal to zero to infinity n to the power p minus one z to the power minus n we have
d over d z of d over d z of z n to the power p minus one equal to summation n equal to
zero to infinity n to the power p minus one into minus n z to the power minus n plus one
ok now let us combine this equation one and two so when you combine one and two what you
find is sigma n equal to zero to infinity n to the power p minus one into n into z to
the power minus n plus one that is z of n to the power p so we get z of n to the power
p ok equal to minus z into d over d z of n to the power p minus one so combining one
and two we have the [re/required] required recurrence relation from here so z of n
to the power p minus z d over d z z of n to the power p minus one now this result is
going to be very useful in finding the z transforms of some other sequences
so let us first look at the special cases of the result z of a to the power
n equal to z over z minus a here if you take a equal to one we will have
see if you take let us say we take let a b equal to one then u n is equal to one
for all n and z of u n is equal to z of one will be equal to z over z minus one from this
result so this is valid one mod of z is greater than one so z of one z of one can be found
z of a to the power n equal to z over z minus a by taking a equal to one so z of let us
see how we find z of n equal to z over z minus one whole square so let us use this result
so let us take p equal to one in the recurrence result in the recurrence formula in the
recurrence formula so then z of n to the power p that we are taking p equal to ones z of
n equal to minus z d over d z of z to the power n to the power zero which is one and
z of one we have found just now which is minus z d over d z of so d over d z of z over z
minus one ok now let us differentiate z over z minus
one with respect to z d over d z of z over z minus one we can write it as d over
d z of z minus one plus one divided by z minus one which is equal to d over d z of one plus
one upon z minus one so when we differentiate this we get minus one upon z minus one whole
square and thus z of the sequence u n equal n is minus z into minus one upon z square
which gives you z over z minus one whole square so that is how we find z of the sequence
u n equal to n then next you can find z of u n equal to n square in a similar manner
so z of n square now in the recurrence formula we take p equal to two so we shall have
in the recurrence formula take p equal to two so at z of n square is minus z d over
d z of z of n and z of n we have already found so minus z d over d z of z n is z over
z minus one whole square so while differentiating this result ok by
differentiating d over z over z minus one whole square we shall have
one derivative of z is one into z minus one whole square minus z into two times
z minus one divided by z minus one to the power four
so we can cancel z minus one and then what do we get minus z times z minus one minus
two z upon z minus to the power three so what do we get this is minus z minus one
so z square plus z divided by z minus one whole to the power three so we get the formula
for n square which is z square plus z over z minus one whole cube now we can find
we can find the value of z n cube in a singular manner by taking p equal to three in the difference
formula we will need the value of z n square which we have already found s by differentiating
the value of z n square and multiplying by minus z we shall arrive at the formula for
z n cube which is z cube plus four z square plus z over z minus one to the power four
and so on so this is how we find the z transforms of these sequences u n equal to one u n equal
to n u n equal to n square u n equal to n cube and then we can see that if we want to
find the inverse the inverse z transform then inverse z transform of z over z minus one
z inverse z inverse of z over z minus one will be equal to one so the inverse
z transform inverse z transform inverse z transform of z over z minus one will be
u n equal to one for all n equal to one two three and so on and zero for and less
than zero ok so now let us go to properties of the z
transform if we know that in the [la/laplace] laplace transform we have the linearity property
so here also we have the linearity property if n be are any complex constants real
or complex constants and u n and b n are two sequences defined on the discrete arguments
then z of a u n plus v v n equal to a z u n plus b z b v n this we can easily prove
by the definition of the z transform z of a u n plus b v n will be equal to summation
n equal to zero to infinity a u n plus b v n into z to the power minus n which can
be broken as summation n equal to zero to infinity a u n z to the power minus n plus
summation n equal to zero to infinity b v n into z to the power minus n and which is
equal to a times summation n equal to zero to infinity u n z to the power minus n
plus summation b times summation n equal to zero to infinity b n z to the power minus
and this is a times z u n plus b times z v n so z transforms satisfies the linearity
property and then we have the change of scale property if you remember we have the change
of the scale of the in the laplace transforms so here we also we have the change of scale
property if z of the sequence u n is given by u z then z of a to the power minus n u
n is equal to u a z this can be proved very easily
let us write z of a to the power minus n u n by definition this will be sigma n equal
to zero to infinity a to the power minus n u n into z to the power minus n and this
can be then written further as sigma n equal to zero to infinity u n into a z to the power
minus n now by our definition by our definition z of u n is summation n equal to
zero to infinity u n z to the power minus n which we have denoted by u z which we have
denoted by u z so here we have sigma n equal to zero to infinity u n a z to the power minus
in place of z we have a z so this implies sigma n equal zero to infinity u n a z to
the power minus n equal u a sorry u a z so that is how we get the change of scale
property when whenever we multiply u n sequence by a to the power minus n there is
a change of a scale in place of z in u z we get a z u z becomes u a z so now the corollary
of this result is z of a to the power n u n equal to u z by a so let us see a we have
z of in the dumping rule we have z of a to the power minus n u n equal to u u
a z ok so this if we replacing a by a by one by a ok we have z of a to the power n
u n equal to u z upon a on replacing a by one by a in the dumping rule so we have
z of a to the power n u n equal to u z by a the geometric sector let us see the geometric
sector u a to the power minus n which occurs here ok dumps the function u n when
mod of z is less than one and that is why we call it as the dumping rule so now using
this dumping rule z of u n equal to if z of u n equal to u z we have z of a to
the power minus n u n equal to u a z using this dumping rule we can find the z transforms
of several other sequences which will be important when we will solve the difference
equation we find the general solution of differential equation
so let us see what how we can use this dumping rule from the dumping rule it follows
that z of a to the po n a to the power n z of n a to the power n is equal to a z upon
z minus a whole square let us let us prove this we know that we know that
from the dumping rule z of a to the power n z of a to the power n u n is equal to u
z by a where u z is the z transform of the sequence u n ok so what we do is a in order
to prove z of n a to the power n equal to a z over n minus a whole square let us
take u n equal to n so n when u n equal to n we have a z over z minus a whole square
so this is how we find the z transform of the sequence n into a to the power n
by using the dumping rule similarly we can find the z transform of n square a to
the power n so we know that z transform of n square z transform of n square we had obtained
earlier it is z square plus z over n minus z minus one whole cube ok so we apply the
dumping rule again apply the we applying the dumping rule
z of a to the power n into u n equal to u z by a ok where u z is sigma u z is z
transform of u n sequence ok let u n be equal to n square then z transform
of z transform of n square then u z is equal to u z is equal to z square plus z divided
by z minus one whole cube ok so z transform of n square u n is n square n square a to
the power n will be replacing z by z over a so z by a whole square plus z by a divided
by z by a minus one whole cube this will give you a z square plus a square z over this
is a square here so a square when we multiply a cube we are multiplying ok a cube we
are multiplying so we get a z square over plus a square z over minus a whole cube
so that is how we find the z transform of n square into a to the power n similarly
you can find the z transform of n cube into a to the power n because we have found the
z transform of n cube so once z transform of n cube is known z transform of n cube into
a to the power n will be replaced z by z minus a in the z transform of n cube and you get
z transform of n cube a to the power n so these are some of the standard results
that we shall need in the sol in when we solve the difference equations so one should
memorize in fact these results because when we solve the difference equation there
we will have to inverse z transform so for taking the inverse z transform we need
to memorize some of these important results like when you take the lap inverse laplace
transform you use the known known results of laplace transforms laplace transforms
of some extended functions similarly here so we need to memorize some of these extended
results to take the inverse z transform while solving the difference equations
so with that i would like to conclude my lecture i
thank you very much for your attention