WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on convolution
theorem for z transforms we will begin
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with the convolution theorem which says
that if we know the z transforms of two sequences
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say u n and v n z inverse u z is equal
to un and z inverse vz is equal to vn then
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the theorem says that z inverse of uz into
vn vz is equal to the convolution of un and
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vn and the convolution of un and vn is denoted
by un star vn and this is defined as sigma
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m equal to zero to n um vn minus m a similar
theorem on convolution is there in the
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laplace transform where we say that if we
know the laplace transforms of two functions
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then laplace transform of their convolution
is equal to the a product of their laplace
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transform so here also if we have two sequences
un and vn then laplace transform z transform
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f their convolution is equal to product of
their z transform so to prove this let us
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by definition of the z transform since
uz is the transform of un z transform
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of un sequence is equal to uz we are given
and z transform of the sequence vn where given
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as vz so we can see that this is equal
to sigma n equal to zero to infinity un z
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minus z to the power minus n and this is sigma
n equal to zero to infinity vn z to the
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power minus n
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so when you multiply uz and vz what we will
have then uz into vz will be equal to sigma
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n equal to zero to infinity uz un z to the
power minus n this is uz and vz over the
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in vz we will take the summation over some
other index let us say m equal to zero to
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infinity vn z to the power minus n so this
is uz into vz and then ok we can write it
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in the expanded form so this is u naught plus
u one by z plus u two by z square and so on
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and here it is v naught plus v one by z plus
v two by z square and so on now when you multiply
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these two are infinite series so when you
multiply we can collect the coefficient of
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z to the power minus n so collecting the coefficient
of z to the power minus n what we will have
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is the coefficient of in in this product
the coefficient of z to the power
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minus n will be u naught vn plus u one
vn minus one u two vn minus two and so on
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un v naught this can be written as sigma m
m equal to zero to m um vn minus m so this
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is the coefficient of z to the power minus
n when you multiply these two infinite
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series and with this we can write it as so
therefore uz into vz uz into capital vz
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may be written as summation n equal to zero
to infinity and then this expression sigma
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m equal to zero to n u b um bn minus n into
z to the power minus n so the coefficient
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of z to the power minus n is this sum which
is given by the summation notation sigma m
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equal to zero n un bn minus m and definition
this sum actually denotes the
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convolution of the sequences un and vn so
this is z transform of un and vn so this from
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here we can say that z inverse of uz into
vz is equal to un star vn and which is
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equal to summation m equal to z to n um vn
minus m so when we know the z transforms
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of two sequences un and vn the z transformer
their convolution is equal to product of
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their z transforms so this is what this
is the convolution theorem now let us apply
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this theorem to find the z transforms
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so let us see suppose we want to find the
z transform of z inverse of z over z minus
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a whole cube z inverse of z over z minus a
whole cube this is our aim we have to find
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the inverse z transform of z over z minus
a whole cube so let us recall that z inverse
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of z over z minus a this is equal to a to
the power n provided mod of z is greater than
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mod of a so we know z transform inverse z
transform of z over z minus a is a to the
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power n so by the convolution theorem z inverse
of z over z minus a into z over z minus a
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so this is uz this is vz z inverse of uz into
vz is convolution of un and vn so a n is
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star a n convolution of a n be the a n which
is equal to sigma m equal to zero to n um
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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
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um this is bm we have here um here we have
bn minus m and this is equal to sigma m equal
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to zero to n so a to the power n now a to
the power n is independent of the index
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m so a to the power n can be written outside
this is sigma m equal to zero to n applied
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to one ok so this is one summing one is
summed n plus one times so this is and plus
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one a to the power n now so this is inverse
z transform of z over z minus a whole square
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we have to again apply the convolution theorem
now by the convolution theorem again z
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inverse of z over z minus a whole square into
z over z minus a it is convolution of n plus
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one a to the power n and a to the power n
n plus one a to the power n is the inverse
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z transform of z over z minus a whole square
and a to the power n is inverse z transform
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of z over z minus a
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so this is equal to summation m equal zero
to n we can we can write it as this we
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can write as a am and this we can write as
a n minus we can choose any sequence to write
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am ok so m plus one a to the power m and
into a n minus m so this is um this is bn
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minus m ok so a to the power m a to the power
n m will cancel and we will have summation
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m equal to zero to n m plus one a to the power
n a to the power n is independent of m so
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a to the power n can be written outside then
summation m equal to zero to n plus one now
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replacing m plus one by j i can write it to
summation a to the this i can write as
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a to the power n summation j equal to one
to n plus one n sorry j so this is summation
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of j when j runs from one to n plus one and
we know the value of this a to the power n
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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
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two by two so that is the z inverse z transform
of so thus z inverse of z over z minus a whole
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cube this is equal to a to the power n and
plus one plus two by two when mod of z is
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greater than mod of a now in particular inverse
z transform of z cube over z minus whole
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cube can be obtained by taking a equal to
one here so inverse z transform of z cube
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over z minus one whole cube is one by two
n plus one into n plus two provided mod of
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z is greater than one let us take some
more examples on convolution theorem suppose
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we want to find inverse z transform of z square
over z minus one z minus three so let us say
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let uz be equal to z over z minus one and
vz be equal to z over z minus three then we
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know that inverse z transform of uz is equal
to
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one to the power n provided mod of z is greater
than one because we know that z transform
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of a to the power n is z over z minus a provided
mod of z is greater than mod of a
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so taking a equal to one here we get that
z inverse of z over z minus one is one to
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the power n and here and similarly z inverse
of vz will be equal to z inverse of vz
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will be equal to this is z over z minus three
so we will take a equal to three here so this
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is three to the power n provided mod of z
s greater than three ok so we know the inverse
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z transforms of z over z minus one and z over
z minus three so inverse z transform of hence
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by the convolution theorem
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inverse z transform of uz into vz will
be equal to convolution of un with vn so the
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required z trans inverse z transform that
is z inverse of the z into vz is their convolution
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here so sigma m equal to zero to n one
to the power you can write three to the
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power m this i can write as um three to the
power m that i can write it as bn minus m
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so one to the power n minus m so one to the
power n minus m which is summation m equal
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to zero to n three to the power now this is
a geometric series and here the first term
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is one so the total number of terms are n
plus one so and the geometric ratio is three
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so we apply the formula we know that a plus
ar and so on ar to the power n minus one this
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has some a times r r to the power n minus
one divided by r minus one when r is greater
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than one ok so here we have rs three so this
is a a is equal to one so we have three to
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the power n plus one minus one divided by
three minus one so this is half of three to
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the power n plus one minus one so that is
inverse z transform of z square over z
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minus one z minus three one by two z three
on and the reason of convergence here will
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be the region that is common to both the
inverse z transforms inverse z transform
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of uz has common the region of convergence
has mod of z greater than one here have mod
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of z is greater than three so the common region
is mod of z greater than three so the region
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should be mod of z greater than three now
let us take up another question here we
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have two sequences one sequence un is one
by n factorial that there is sequence vn is
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also one by n factorial and we have to show
that the convolution of un with vn or you
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can say un with vn is gives you two to the
power n by y factorial so let us say let un
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be equal to one by n factorial then we know
that by def yz by definition of z transform
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the z transform of un is equal to sigma n
equal to zero to infinity un z to the power
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minus n so this will give you a summation
n equal to zero to infinity un is one by
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one factorial
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so one by n factorial z to the power minus
n ok and this is nothing but e to the power
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one by z provided mod of z is greater than
zero so the region of convergence here is
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mod of z greater than zero and e to the power
one by z the exponential function e to the
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power one by z is expanded into the series
n equal to zero to infinity into the series
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n equal to zero to infinity one by n factorial
z to the power minus n now let us find
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that z inverse of uz into vz so this is uz
now z transform z inverse of uz nto vz we
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know this is equal to un star vn so let
us take vz to be same as uz ok so z inverse
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of uz uz is e to the power one by z vz
also is e to the power one by z this is equal
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to un into star vn so one by n factorial star
one by n factorial and now we see that z inverse
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of e to the power two by z z inverse of e
to the power two by z so
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it is z of two to the power this is two to
the power n by n factorial we have because
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since z inverse z of two to the power n
by n factorial
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if we take this sequence this is summation
n equal zero to infinity two to the
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power n by factorial z to the power minus
n so i can write it as summation n equal to
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zero to infinity of one by n factorial
two by z to the power n and therefore it is
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e to the power two by z so z inverse e to
the power two by z is nothing but two to the
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power n by n factorial so we have we
have seen here that since e to the power one
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by z was equal to sigma n equal to zero to
infinity one by n factorial z to the power
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minus n from here one can guess that e to
the power two by z will be sigma n equal to
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zero to infinity one by n factorial two by
z to the power minus n and therefore this
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sequence that we will get and taking the inverse
z transform of e to the power one by z will
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be two to the power n by factorial so that
gives us an idea about the inverse z transform
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of e to the power two by z hat we can see
from here also the by definition of convolution
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un star vn is sigma m equal to zero to n
um which is one by n factorial into vn minus
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m so n minus m factorial by definition of
convolution we can see his we know that
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un star vn this is summation m equal zero
to infinity um vn minus m so i have just put
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the values here now i can multiply and divide
by n factorial
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so if we do that one by n factorial summation
m equal to zero to n ncm we can write multiplying
00:19:57.460 --> 00:20:07.410
and dividing by n factorial and this is nothing
but
00:20:07.410 --> 00:20:13.600
the sum of the binomial coefficients nc naught
plus nc one plus nc two and so on ncn is given
00:20:13.600 --> 00:20:22.419
by one plus one to the power n which is two
to the power n by n factorial so the convolution
00:20:22.419 --> 00:20:27.580
of one by n with one by n one by n factor
one by n factorial will give you two to the
00:20:27.580 --> 00:20:34.040
power n by n factorial from the inverse
z transform when we apply inverse z transform
00:20:34.040 --> 00:20:41.340
here we know that u u to the power one by
z is i think i did not i should not have done
00:20:41.340 --> 00:20:51.610
that rather we can see that here yeah
z inverse un star i un star vn is one by n
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factorial start one by n factorial and this
we did we dont have to do this that is z inverse
00:20:59.559 --> 00:21:09.600
of uz vz uz into vz and which is e to
the power two by z and z inverse e to the
00:21:09.600 --> 00:21:15.179
power two by z is nothing but two to the power
n by n factorial so this is two to the power
00:21:15.179 --> 00:21:24.460
n by n factorial we get so this is what we
have and with that i would like to conclude
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my lecture