hello friends welcome to my lecture on convolution theorem for z transforms we will begin
with the convolution theorem which says that if we know the z transforms of two sequences
say u n and v n z inverse u z is equal to un and z inverse vz is equal to vn then
the theorem says that z inverse of uz into vn vz is equal to the convolution of un and
vn and the convolution of un and vn is denoted by un star vn and this is defined as sigma
m equal to zero to n um vn minus m a similar theorem on convolution is there in the
laplace transform where we say that if we know the laplace transforms of two functions
then laplace transform of their convolution is equal to the a product of their laplace
transform so here also if we have two sequences un and vn then laplace transform z transform
f their convolution is equal to product of their z transform so to prove this let us
by definition of the z transform since uz is the transform of un z transform
of un sequence is equal to uz we are given and z transform of the sequence vn where given
as vz so we can see that this is equal to sigma n equal to zero to infinity un z
minus z to the power minus n and this is sigma n equal to zero to infinity vn z to the
power minus n
so when you multiply uz and vz what we will have then uz into vz will be equal to sigma
n equal to zero to infinity uz un z to the power minus n this is uz and vz over the
in vz we will take the summation over some other index let us say m equal to zero to
infinity vn z to the power minus n so this is uz into vz and then ok we can write it
in the expanded form so this is u naught plus u one by z plus u two by z square and so on
and here it is v naught plus v one by z plus v two by z square and so on now when you multiply
these two are infinite series so when you multiply we can collect the coefficient of
z to the power minus n so collecting the coefficient of z to the power minus n what we will have
is the coefficient of in in this product the coefficient of z to the power
minus n will be u naught vn plus u one vn minus one u two vn minus two and so on
un v naught this can be written as sigma m m equal to zero to m um vn minus m so this
is the coefficient of z to the power minus n when you multiply these two infinite
series and with this we can write it as so therefore uz into vz uz into capital vz
may be written as summation n equal to zero to infinity and then this expression sigma
m equal to zero to n u b um bn minus n into z to the power minus n so the coefficient
of z to the power minus n is this sum which is given by the summation notation sigma m
equal to zero n un bn minus m and definition this sum actually denotes the
convolution of the sequences un and vn so this is z transform of un and vn so this from
here we can say that z inverse of uz into vz is equal to un star vn and which is
equal to summation m equal to z to n um vn minus m so when we know the z transforms
of two sequences un and vn the z transformer their convolution is equal to product of
their z transforms so this is what this is the convolution theorem now let us apply
this theorem to find the z transforms
so let us see suppose we want to find the z transform of z inverse of z over z minus
a whole cube z inverse of z over z minus a whole cube this is our aim we have to find
the inverse z transform of z over z minus a whole cube so let us recall 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 we know z transform inverse z transform of z over z minus a is a to the
power n so by the convolution theorem z inverse of z over z minus a into z over z minus a
so this is uz this is vz z inverse of uz into vz is convolution of un and vn so a n is
star a n convolution of a n be the a n which is equal to sigma m equal to zero to n um
um is a to the power m into bn minus m so a n to the a to the power n minus m this is
um this is bm we have here um here we have bn minus m and this is equal to sigma m equal
to zero to n so a to the power n now a to the power n is independent of the index
m so a to the power n can be written outside this is sigma m equal to zero to n applied
to one ok so this is one summing one is summed n plus one times so this is and plus
one a to the power n now so this is inverse z transform of z over z minus a whole square
we have to again apply the convolution theorem now by the convolution theorem again z
inverse of z over z minus a whole square into z over z minus a it is convolution of n plus
one a to the power n and a to the power n n plus one a to the power n is the inverse
z transform of z over z minus a whole square and a to the power n is inverse z transform
of z over z minus a
so this is equal to summation m equal zero to n we can we can write it as this we
can write as a am and this we can write as a n minus we can choose any sequence to write
am ok so m plus one a to the power m and into a n minus m so this is um this is bn
minus m ok so a to the power m a to the power n m will cancel and we will have summation
m equal to zero to n m plus one a to the power n a to the power n is independent of m so
a to the power n can be written outside then summation m equal to zero to n plus one now
replacing m plus one by j i can write it to summation a to the this i can write as
a to the power n summation j equal to one to n plus one n sorry j so this is summation
of j when j runs from one to n plus one and we know the value of this a to the power n
sigma j equal to one to n j is n into n plus one by two so it is n plus one into n plus
two by two so that is the z inverse z transform of so thus z inverse of z over z minus a whole
cube this is equal to a to the power n and plus one plus two by two when mod of z is
greater than mod of a now in particular inverse z transform of z cube over z minus whole
cube can be obtained by taking a equal to one here so inverse z transform of z cube
over z minus one whole cube is one by two n plus one into n plus two provided mod of
z is greater than one let us take some more examples on convolution theorem suppose
we want to find inverse z transform of z square over z minus one z minus three so let us say
let uz be equal to z over z minus one and vz be equal to z over z minus three then we
know that inverse z transform of uz is equal to
one to the power n provided mod of z is greater than one because we know 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 taking a equal to one here we get that z inverse of z over z minus one is one to
the power n and here and similarly z inverse of vz will be equal to z inverse of vz
will be equal to this is z over z minus three so we will take a equal to three here so this
is three to the power n provided mod of z s greater than three ok so we know the inverse
z transforms of z over z minus one and z over z minus three so inverse z transform of hence
by the convolution theorem
inverse z transform of uz into vz will be equal to convolution of un with vn so the
required z trans inverse z transform that is z inverse of the z into vz is their convolution
here so sigma m equal to zero to n one to the power you can write three to the
power m this i can write as um three to the power m that i can write it as bn minus m
so one to the power n minus m so one to the power n minus m which is summation m equal
to zero to n three to the power now this is a geometric series and here the first term
is one so the total number of terms are n plus one so and the geometric ratio is three
so we apply the formula we know that a plus ar and so on ar to the power n minus one this
has some a times r r to the power n minus one divided by r minus one when r is greater
than one ok so here we have rs three so this is a a is equal to one so we have three to
the power n plus one minus one divided by three minus one so this is half of three to
the power n plus one minus one so that is inverse z transform of z square over z
minus one z minus three one by two z three on and the reason of convergence here will
be the region that is common to both the inverse z transforms inverse z transform
of uz has common the region of convergence has mod of z greater than one here have mod
of z is greater than three so the common region is mod of z greater than three so the region
should be mod of z greater than three now let us take up another question here we
have two sequences one sequence un is one by n factorial that there is sequence vn is
also one by n factorial and we have to show that the convolution of un with vn or you
can say un with vn is gives you two to the power n by y factorial so let us say let un
be equal to one by n factorial then we know that by def yz by definition of z transform
the z transform of un is equal to sigma n equal to zero to infinity un z to the power
minus n so this will give you a summation n equal to zero to infinity un is one by
one factorial
so one by n factorial z to the power minus n ok and this is nothing but e to the power
one by z provided mod of z is greater than zero so the region of convergence here is
mod of z greater than zero and e to the power one by z the exponential function e to the
power one by z is expanded into the series n equal to zero to infinity into the series
n equal to zero to infinity one by n factorial z to the power minus n now let us find
that z inverse of uz into vz so this is uz now z transform z inverse of uz nto vz we
know this is equal to un star vn so let us take vz to be same as uz ok so z inverse
of uz uz is e to the power one by z vz also is e to the power one by z this is equal
to un into star vn so one by n factorial star one by n factorial and now we see that z inverse
of e to the power two by z z inverse of e to the power two by z so
it is z of two to the power this is two to the power n by n factorial we have because
since z inverse z of two to the power n by n factorial
if we take this sequence this is summation n equal zero to infinity two to the
power n by factorial z to the power minus n so i can write it as summation n equal to
zero to infinity of one by n factorial two by z to the power n and therefore it is
e to the power two by z so z inverse e to the power two by z is nothing but two to the
power n by n factorial so we have we have seen here that since e to the power one
by z was equal to sigma n equal to zero to infinity one by n factorial z to the power
minus n from here one can guess that e to the power two by z will be sigma n equal to
zero to infinity one by n factorial two by z to the power minus n and therefore this
sequence that we will get and taking the inverse z transform of e to the power one by z will
be two to the power n by factorial so that gives us an idea about the inverse z transform
of e to the power two by z hat we can see from here also the by definition of convolution
un star vn is sigma m equal to zero to n um which is one by n factorial into vn minus
m so n minus m factorial by definition of convolution we can see his we know that
un star vn this is summation m equal zero to infinity um vn minus m so i have just put
the values here now i can multiply and divide by n factorial
so if we do that one by n factorial summation m equal to zero to n ncm we can write multiplying
and dividing by n factorial and this is nothing but
the sum of the binomial coefficients nc naught plus nc one plus nc two and so on ncn is given
by one plus one to the power n which is two to the power n by n factorial so the convolution
of one by n with one by n one by n factor one by n factorial will give you two to the
power n by n factorial from the inverse z transform when we apply inverse z transform
here we know that u u to the power one by z is i think i did not i should not have done
that rather we can see that here yeah z inverse un star i un star vn is one by n
factorial start one by n factorial and this we did we dont have to do this that is z inverse
of uz vz uz into vz and which is e to the power two by z and z inverse e to the
power two by z is nothing but two to the power n by n factorial so this is two to the power
n by n factorial we get so this is what we have and with that i would like to conclude
my lecture