hello friends welcome to my lecture on applications of z transforms so this we will cover in two
lectures this is first lecture on application of z transforms we know that the difference
equations model those processes in which we know relationships between changes and
or differences rather than the rates of change which leads to the differential equations
now these difference equations arise in those situations where sequential relation exists
at various discrete values of the independent variable the difference relation that we
are obtained while solving the difference equation by power series or frobenius method
are also difference equations similarly numerical solution of differential equations also leads
to difference equations
now in control engineering often the input is in the form of discrete pulse of short
duration and the radar tracking devices receive such discrete pulse from the target
which is which it is being which is being tracked so difference equations is a relation
between the differences of an unknown function at one or more general values of the argument
for example let us consider delta yn plus one plus yn equal to two this delta is the
forward difference and then you can also consider delta yn plus one plus delta square yn minus
one equal to one both of these equations are difference equations these two equations can
be put in the following form we know that delta of fx the forward difference delta of
fx is fx plus h minus fx
so using this definition delta yn plus one will be equal to yn plus two minus
yn plus one and hence delta yn plus one plus yn equal to two gives us yn plus two
minus yn plus one plus yn equal to two which is so alternately the difference equation
one can be expressed like this and similarly the equations two the equations two is
delta yn plus one and plus delta square yn minus one yn minus one equal to one so here
we can write it as delta yn plus one is first order forward difference so we get yn plus
two minus yn plus one and delta square yn minus one here is delta square second order
forward difference so this will be delta of delta yn minus one equal to one now delta
yn minus one is equal to yn minus yn minus one and so delta of delta yn minus one
is equal to delta of yn minus yn minus one which is equal to delta yn minus delta
yn minus one but delta yn is yn plus one minus yn and here this is yn minus yn minus one
so we will get yn plus one minus two yn plus yn minus one so let us substitute the value
here so yn plus two minus yn plus one and then plus n plus one minus two yn and then
plus yn minus one equal to one so this cancels yn plus one cancels and we get yn plus two
minus two yn plus yn minus one equal to one so the equations one and two can be alternately
written in this in this forms so the equations one and two can be also written as yn plus
two minus yn plus one plus yn equal to two and yn plus two minus two yn plus yn minus
one equal to one now let us discuss how can we determine the order of a difference
equation
so the order of a difference equation defined as the difference between the largest and
the smallest argument occurring in the difference equation divided by the unit of increment
so let us see here this equation this equation which is alternate form of the first equation
number one and this one and this equation which is the alternate form of this equation
so here we see that the largest argument is n plus two ok so let me call it as equation
number this is equation number three and this is equation number four so in three largest
argument is n plus two and then the smallest argument is n divided by the unit of increment
unit of increment is one we have n n plus one n plus two so this is two so this is
of second order and here the largest argument is n plus two the smallest argument is n minus
one divided by the unit of increment is one so we get here three so equation one is of
second order while equation two is of third order or we can say equation three is of second
order and equation four is of order three now solution of difference equation so when
we have a difference equation the solution is defined as an expression for yn which satisfies
the given difference equation the general solution of a difference equation is that
in which the number of arbitrary constant is equals to the order of the difference equations
so suppose we want to call this difference equation three then its general solution must
involve two arbitrary constant because here there are where the order of the difference
equation is two similarly the equation four is of order three so the general solution
must involve three arbitrary constants so general solution will be that solution of
the difference equation in which the number of arbitrary constants is equal to the order
of the difference equation now particular solution of the difference equation will be
obtained from the general solution by assigning arbitrary by assigning the particular values
to the arbitrary constants let us now discuss how we can solve linear difference equation
with constant coefficients a linear difference equation is defined as an equation in which
yn plus one yn plus two etcetera that is all yn plus all the sequences arguments
yn plus one yn plus two and so on they occur in the first degree and they occur separately
like differ linear difference equation with constant coefficients in general will be of
this form yn plus k plus a one yn plus k minus one a to yn plus k minus two and so on a k
by n equal to fn where a one a two ak are constants
so here you can see this is k th order difference equation because n plus k s the highest or
the largest argument and n is the smallest argument so n plus k minus n divided by one
unit of increment is one so it is k th order difference equation and it is a linear difference
equation because yn plus yn plus k yn plus k minus one and so on yn they all occur in
first degree and separately now if u one u two u one if u one u two uk n are k independent
solutions of this equation yn plus k a one yn plus k minus one ak by n equal to zero
so here we say that k th k th k is the order of the difference equation k k k is the order
of the difference equation so let us consider the corresponding homogeneous part so yn plus
k plus a one yn plus k minus n plus ak yn equal to zero if u one u two u one uk and
vk independent solutions of this homogeneous equation then their linear combination c one
u one plus c two u two c k u k is called the complete solution of this homogeneous equation
let us denoted by say un if vn is a particular solution of equation one if vn is a particular
solution of this non homogeneous equation then the complete solution of equation
one will be given by yn equal to n plus vn where un is called the complementary function
and vn is called the particular integral now thus the general solution of the difference
equation one is the sum of the two that is complementary function and the particular
integral the solution of one equation one solution of one can be obtained by the classical
approach like we just has been solve the linear def linear differential equations
with constant coefficients so these linear difference equations with constant coefficient
can be solved in a analogous manner by the classical approach but here we will solve
them by using the z transform so we are solving these linear non homogeneous difference equations
with by using z transforms
let us see how we do this suppose we consider so let us consider four un minus un plus two
equal to zero with u naught equal to zero and u one equal to two so we are given this
difference equation you can see the largest argument here is an plus two the smallest
argument is n so and the unit of increment is one so it is the second order difference
equation second order difference equation and they are given two initial conditions
the conditions are u naught equal to zero so u one equal to zero so we will get with
these two initial conditions we shall get a particular solution of this differential
difference equation so what we do in order to solve this difference equation what we
do is we take the z transforms of z transform of both the sides of the difference equation
and then solve this solve the resulting equation for the value of uz where z is uz
is the z transform of this sequence un and then we divide uz by z and break uz by z into
partial fractions in such a way that when we write the corresponding expression
for a uz it comes in terms of the known z transforms so that while inverting uz we
get the sequence un easily
so let say v v let us take the z transform taking z transforms taking z transforms of
both the sides let me call it equation one so this equation one so taking z transform
of equation one we have four times z is a linear z is it is a linear property so
four times z of un minus z of un plus two equal to z of zero which is zero so now four
z of un is four of uz uz is the z transform this sequence un and here while writing z
transform of un plus two let us recall the shifting to the left shifting of un to
the left so we know that z transform of un plus k where k is greater than zero is equal
to z to the power k and then we have uz minus u naught minus u one by z and so on
minus uk minus one over z to the power k minus one
so let us apply this formula so z square into uz minus u naught minus u one by z equal to
zero let us make use of the given conditions u naught equal to zero un equal to two so
four uz minus z square and then we get uz minus zero minus two by z so the coefficient
of z is uz is four minus z square into uz and we get this term minus z square into
minus two y z so plus two z is equal to zero we can write it as uz equal to two z upon
z square minus four now let us write uz by z we break uz by z into partial fraction
so two over z square minus four which is two over z minus two into z plus two the partial
fractions of this expression is two times one over z minus two minus one over z plus
two divided by four ok so this is one by two one over z minus two minus one over z plus
two now we can write uz from here so uz now thus uz is equal to one by two z over
z minus two minus z over z plus two and hence un which is z inverse of uz is equal to one
by two z inverse of z over z minus two m z inverse of z over z plus two let us recall
that z transform of a to the power n is z over z minus a provided mod of z is greater
than mod of a so z inverse of z over z minus a will be a to the power n so here we have
two and therefore one by two two to the power n and for this mod of z must be greater than
two and then from here we get minus two raise to the power n
so here also mod of z must be greater than two so mod of z must be greater than two for
this two we true ok so one by two two raise to the power n minus minus two to the power
n we can simplify it further two to the power n minus one and then i can write it as plus
or minus two i can take from here so minus two to the power n minus one minus minus becomes
plus so the this is minus two to the power n minus one so this is the sequence un
equal to two to the power n minus one plus minus two to the power two n minus one
which is the solution of example one now let us take the example two so here we have this
sequence un plus two plus four un plus one plus three un equal to three to the
power n
again let us see n plus two is the largest argument and n is the smallest argument
here so n plus two minus n divided by the unit of increment is one this means the order
of this difference equation is n plus two minus n divided by one so that is two this
is the order of the difference equation so this is second order difference equation
let us solve this here we are given the initial conditions u naught equal to zero and u one
equal to one so the general solution of this difference equation will involve two or three
constants to determine two arbitrary constants we are given two conditions u naught equal
to one and u naught equal to zero and u one equal to one so this will give us a particular
solution of the difference equation
now let us take z transform here so z transform is a linear linear operation so z of un plus
two z of un plus two plus four times z of un plus one plus three times z of un equal
to z transform of three to the power now again let us apply this fifteen of un to the left
so z of un plus two will be z square uz where uz is z transform of the un sequence minus
u naught minus u one by z plus four times z of un plus one will be z times uz minus
u naught plus three times z transform of un is uz is equal to z to the power z of three
to the power n is z over z minus three provided mod of z is greater than three
now let us use the values of u naught and u one which are given to us u naught is given
to be zero so uz minus zero minus u one is one one over z plus four z times uz minus
zero plus three uz equal to z over z minus three let us collect the coefficient of uz
so z square plus four z plus three into uz we have and then we have minus z square by
z so minus z minus z when goes to the other side becomes plus z so z over z minus three
plus z and this will give you z square minus two z divided by z minus three now let
us find u z by z from here so then uz by z it will be equal to z minus two divided by
z minus three into z square plus four z plus three if you factorize you get z plus
one z plus three so z plus one and z plus three so now we have to break this uz by z
into partial fractions and then determine the inverse z transform so uz by z so let
us write it as a over z minus three b over z plus one c over z plus three the values
of abc can be obtained now a is equal to z minus three so z minus two divided by z
plus one z plus three evaluated at z equal to three
so this will be equal to three minus two that is one and three plus one is four three
into three three plus three six so twenty four so it is one by twenty four b will be
equal to z minus two divided by z minus three into z plus three evaluated at z equal
to one minus one so we get this yes minus three divided by minus four and here we have
two so three by eight and see we will have as z minus two over z plus z minus three in
to z plus one evaluated as z equal to minus three so this will be minus five and here
we will have minus three minus three minus six in to two so five by twelve and hence
u z by z is is equal to one by twenty four z minus three b is three by eight three by
eight z plus one and then we have five by twelve z plus three ok
now we can multiply by z so this gives u z is equal to one by twenty four z over
z minus three plus three by eight z over z plus one and five by twelve z over z plus
three now we can easily invert this so let us take the z trans inverse z transform ok
so this was u n plus two here so ok z then hence z inverse of u z is equal to one by
twenty four z inverse of z over z minus three plus three by eight z inverse of z over z
plus one and then we have five by twelve z inverse of z over z plus three again let us
let us recall that u z sorry z of u to the power a to the power n is z over z minus
a where mod of z is greater than mod of a let us apply this formula so z inverse of
z over z minus three so this is one by twenty four three to the power n where mod of z is
greater than three and then three by eight here minus one to the power a is minus
one and here mod of z is greater than one so mod of z is greater than three here here
mod of z is greater than one then by twelve minus three to the power n so here a is
minus three and therefore mod of z is greater than three so the common reason is mod
of z greater than three and z inverse of u z is u n ok so thus u n is one by twenty four
three to the power n three by eight minus one to the power n plus five by twelve minus
three to the power n where mod of z is greater than three
so this is the answer to this difference equation by taking z transform z transform takes
care of the initial conditions we do not have to use initial conditions after we have found
this solution like we do in this classical approach so here this how we apply the
transform techniques we take the z transform of the given difference equation collect the
coefficient of u z rest of the terms be tend to the right side divide by the coefficient
of u z and then take the then break then you write u z by z then these expression
of u z by z is then broken in to partial fractions in such a way that we while multiplying
by z we get the expressions in terms of the z transform of known z transforms so so
that while inverting we get easily the sequence u n so u n here can be obtained like
this we will in our next lecture we shall discuss some more problems where we will have
to be very careful while writing u z by z because there in the problems the the
expressions we will have to arrange of u z by z so that while inverting we get the
sequence even very easily so we will do that in the next lecture with that i would like
to conclude my lecture
thank you very much for your attention