WEBVTT
Kind: captions
Language: en
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hello
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friends this is my first lecture on z transforms
so in this lecture we shall discuss
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the z transform and inverse z transform of
some elementary functions the z transforms
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are useful in solving difference equations
which arise in in the when in the
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discrete systems the development of the communication
branch of engineering is actually based on
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the theory of z transforms the z transforms
has the same role in discrete systems as
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the laplace transform in continuous systems
many properties of the z transforms are similar
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to the properties of the laplace transforms
z transform operates on a sequence of u n
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of discrete discrete integer valued arguments
n equal to zero plus minus one plus minus
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two and so on and like the laplace transform
which operates on continuous functions z transform
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is we can say is the discrete analog of the
laplace transform for every operational rule
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and application of laplace transform there
is a there is an operational rule and application
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of z transforms a discrete system can be expressed
as a difference equation and then its solution
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can be obtained by using z transforms so
let us see how we define the z transforms
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the z transform of the sequence u n defined
for discrete values n equal to zero one two
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three and so on and e one equal to zero for
n less than zero is denoted by z u n and is
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defined as z u n equal to sigma n equal
to zero to infinity u n z to the power minus
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n which is a function of z so we write
it as u z provided the infinite series convergence
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so whenever this infinite series converges
it will give us a function of z so it is u
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z the inverse z transform is written as z
inverse oh of u z z inverse of u z equal to
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u n so whenever we know the z transform
of a sequence its inverse z transform will
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give us back the sequence even like in the
case of laplace transform we have we define
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the inverse laplace transform here again
the inverse z transform is defined in a similar
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manner so z inverse u z equal to u n
now let us discuss z transforms of some
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a standard sequences so let us begin with
u n equal a to the power n so when u n is
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a to the power n then where a is a complex
number then z of a to the power n is sigma
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n equal to zero to infinity so u n here
is a to the n so n equal to zero to in between
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u n z to the power minus n so we have a at
to the power n and z to the power minus n
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this capital z this is small z now this
is summation n equal to zero infinity a by
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z raise to the power n now we know that
here z is a complex number a is also a a is
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a complex constant so a by z is a complex
number so the series of complex numbers sigma
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n equal to zero to infinity a by z to the
power n converges provided mod of z a by z
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is less than one and we have this is equal
to one over one minus a by z provided mod
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of a by z is less than one so this gives you
z over z minus a provided mod of a by z is
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less than one or mod of z is greater than
a so the condition for the convergence
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of the series here is that mod of z must be
greater than mod of a and now let us
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look look at the formula which is known
as the recurrence formula if you take u n
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to be to the power p let us say u n equal
to n to the power p where p is a positive
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integer then we have a z of n to the power
p this formula is very useful in determining
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the z transforms of other sequences so
minus z d over d z of z and to the power
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p minus one ok so here what do we let us
prove this formula ok
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so to prove this formula we have z of n
to the power p z of n to the power p by definition
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is sigma n equal to zero to infinity n to
the power p z to the power minus n ok so we
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have this we have given here the proof
this can be written as z times summation a
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equal to zero to infinity n to the power p
minus one into n into z to the power minus
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n plus one now since by definition z of n
to the power p minus one is equal to summation
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n equal to zero to infinity n to the power
p minus one z to the power minus n we have
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d over d z of d over d z of z n to the power
p minus one equal to summation n equal to
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zero to infinity n to the power p minus one
into minus n z to the power minus n plus one
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ok now let us combine this equation one and
two so when you combine one and two what you
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find is sigma n equal to zero to infinity
n to the power p minus one into n into z to
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the power minus n plus one that is z of n
to the power p so we get z of n to the power
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p ok equal to minus z into d over d z of n
to the power p minus one so combining one
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and two we have the [re/required] required
recurrence relation from here so z of n
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to the power p minus z d over d z z of n to
the power p minus one now this result is
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going to be very useful in finding the
z transforms of some other sequences
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so let us first look at the special cases
of the result z of a to the power
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n equal to z over z minus a here if you
take a equal to one we will have
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see if you take let us say we take let
a b equal to one then u n is equal to one
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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
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result so this is valid one mod of z is greater
than one so z of one z of one can be found
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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
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see how we find z of n equal to z over z minus
one whole square so let us use this result
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so let us take p equal to one in the recurrence
result in the recurrence formula in the
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recurrence formula so then z of n to the power
p that we are taking p equal to ones z of
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n equal to minus z d over d z of z to the
power n to the power zero which is one and
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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
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minus one ok
now let us differentiate z over z minus
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one with respect to z d over d z of z over
z minus one we can write it as d over
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d z of z minus one plus one divided by z minus
one which is equal to d over d z of one plus
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one upon z minus one so when we differentiate
this we get minus one upon z minus one whole
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square and thus z of the sequence u n equal
n is minus z into minus one upon z square
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which gives you z over z minus one whole square
so that is how we find z of the sequence
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u n equal to n then next you can find z of
u n equal to n square in a similar manner
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so z of n square now in the recurrence formula
we take p equal to two so we shall have
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in the recurrence formula take p equal to
two so at z of n square is minus z d over
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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
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z minus one whole square
so while differentiating this result ok by
00:11:31.240 --> 00:11:41.269
differentiating d over z over z minus one
whole square we shall have
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one derivative of z is one into z minus one
whole square minus z into two times
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z minus one divided by z minus one to the
power four
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so we can cancel z minus one and then what
do we get minus z times z minus one minus
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two z upon z minus to the power three so what
do we get this is minus z minus one
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so z square plus z divided by z minus one
whole to the power three so we get the formula
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for n square which is z square plus z over
z minus one whole cube now we can find
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we can find the value of z n cube in a singular
manner by taking p equal to three in the difference
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formula we will need the value of z n square
which we have already found s by differentiating
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the value of z n square and multiplying by
minus z we shall arrive at the formula for
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z n cube which is z cube plus four z square
plus z over z minus one to the power four
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and so on so this is how we find the z transforms
of these sequences u n equal to one u n equal
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to n u n equal to n square u n equal to n
cube and then we can see that if we want to
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find the inverse the inverse z transform then
inverse z transform of z over z minus one
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z inverse z inverse of z over z minus one
will be equal to one so the inverse
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z transform inverse z transform inverse z
transform of z over z minus one will be
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u n equal to one for all n equal to one
two three and so on and zero for and less
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than zero ok so
now let us go to properties of the z
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transform if we know that in the [la/laplace]
laplace transform we have the linearity property
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so here also we have the linearity property
if n be are any complex constants real
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or complex constants and u n and b n are two
sequences defined on the discrete arguments
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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
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by the definition of the z transform
z of a u n plus b v n will be equal to summation
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n equal to zero to infinity a u n plus
b v n into z to the power minus n which can
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be broken as summation n equal to zero to
infinity a u n z to the power minus n plus
00:15:09.600 --> 00:15:17.119
summation n equal to zero to infinity b v
n into z to the power minus n and which is
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equal to a times summation n equal to zero
to infinity u n z to the power minus n
00:15:24.639 --> 00:15:33.329
plus summation b times summation n equal
to zero to infinity b n z to the power minus
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and this is a times z u n plus b times z v
n so z transforms satisfies the linearity
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property and then we have the change of scale
property if you remember we have the change
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of the scale of the in the laplace transforms
so here we also we have the change of scale
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property if z of the sequence u n is given
by u z then z of a to the power minus n u
00:16:05.759 --> 00:16:09.230
n is equal to u a z this can be proved very
easily
00:16:09.230 --> 00:16:16.300
let us write z of a to the power minus n u
n by definition this will be sigma n equal
00:16:16.300 --> 00:16:22.579
to zero to infinity a to the power minus n
u n into z to the power minus n and this
00:16:22.579 --> 00:16:28.589
can be then written further as sigma n equal
to zero to infinity u n into a z to the power
00:16:28.589 --> 00:16:44.970
minus n now by our definition by our definition
z of u n is summation n equal to
00:16:44.970 --> 00:16:52.639
zero to infinity u n z to the power minus
n which we have denoted by u z which we have
00:16:52.639 --> 00:16:59.440
denoted by u z so here we have sigma n equal
to zero to infinity u n a z to the power minus
00:16:59.440 --> 00:17:07.981
in place of z we have a z so this implies
sigma n equal zero to infinity u n a z to
00:17:07.981 --> 00:17:22.579
the power minus n equal u a sorry u a z
so that is how we get the change of scale
00:17:22.579 --> 00:17:27.829
property when whenever we multiply u n
sequence by a to the power minus n there is
00:17:27.829 --> 00:17:37.899
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
00:17:37.899 --> 00:17:45.309
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
00:17:45.309 --> 00:17:57.990
z of in the dumping rule we have z of
a to the power minus n u n equal to u u
00:17:57.990 --> 00:18:09.340
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
00:18:09.340 --> 00:18:21.809
u n equal to u z upon a on replacing a by
one by a in the dumping rule so we have
00:18:21.809 --> 00:18:29.019
z of a to the power n u n equal to u z by
a the geometric sector let us see the geometric
00:18:29.019 --> 00:18:36.880
sector u a to the power minus n which occurs
here ok dumps the function u n when
00:18:36.880 --> 00:18:44.110
mod of z is less than one and that is why
we call it as the dumping rule so now using
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this dumping rule z of u n equal to
if z of u n equal to u z we have z of a to
00:18:51.070 --> 00:18:55.760
the power minus n u n equal to u a z using
this dumping rule we can find the z transforms
00:18:55.760 --> 00:19:02.669
of several other sequences which will be
important when we will solve the difference
00:19:02.669 --> 00:19:05.570
equation we find the general solution of differential
equation
00:19:05.570 --> 00:19:10.330
so let us see what how we can use this
dumping rule from the dumping rule it follows
00:19:10.330 --> 00:19:21.149
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
00:19:21.149 --> 00:19:38.930
z minus a whole square let us let us prove
this we know that we know that
00:19:38.930 --> 00:19:50.520
from the dumping rule z of a to the power
n z of a to the power n u n is equal to u
00:19:50.520 --> 00:20:06.169
z by a where u z is the z transform of the
sequence u n ok so what we do is a in order
00:20:06.169 --> 00:20:13.179
to prove z of n a to the power n equal
to a z over n minus a whole square let us
00:20:13.179 --> 00:21:37.160
take u n equal to n so n when u n equal to
n we have a z over z minus a whole square
00:21:37.160 --> 00:21:44.230
so this is how we find the z transform
of the sequence n into a to the power n
00:21:44.230 --> 00:21:51.240
by using the dumping rule similarly we
can find the z transform of n square a to
00:21:51.240 --> 00:22:00.940
the power n so we know that z transform of
n square z transform of n square we had obtained
00:22:00.940 --> 00:22:10.100
earlier it is z square plus z over n minus
z minus one whole cube ok so we apply the
00:22:10.100 --> 00:22:39.370
dumping rule again apply the we applying the
dumping rule
00:22:39.370 --> 00:22:57.679
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
00:22:57.679 --> 00:23:11.149
transform of u n sequence ok
let u n be equal to n square then z transform
00:23:11.149 --> 00:23:23.880
of z transform of n square then u z is
equal to u z is equal to z square plus z divided
00:23:23.880 --> 00:23:36.101
by z minus one whole cube ok so z transform
of n square u n is n square n square a to
00:23:36.101 --> 00:23:46.490
the power n will be replacing z by z over
a so z by a whole square plus z by a divided
00:23:46.490 --> 00:24:10.350
by z by a minus one whole cube this will
give you a z square plus a square z over this
00:24:10.350 --> 00:24:18.169
is a square here so a square when we multiply
a cube we are multiplying ok a cube we
00:24:18.169 --> 00:24:29.590
are multiplying so we get a z square over
plus a square z over minus a whole cube
00:24:29.590 --> 00:24:37.190
so that is how we find the z transform of
n square into a to the power n similarly
00:24:37.190 --> 00:24:41.860
you can find the z transform of n cube into
a to the power n because we have found the
00:24:41.860 --> 00:24:47.769
z transform of n cube so once z transform
of n cube is known z transform of n cube into
00:24:47.769 --> 00:24:54.380
a to the power n will be replaced z by z minus
a in the z transform of n cube and you get
00:24:54.380 --> 00:25:02.039
z transform of n cube a to the power n so
these are some of the standard results
00:25:02.039 --> 00:25:09.100
that we shall need in the sol in when we
solve the difference equations so one should
00:25:09.100 --> 00:25:15.370
memorize in fact these results because
when we solve the difference equation there
00:25:15.370 --> 00:25:21.679
we will have to inverse z transform so
for taking the inverse z transform we need
00:25:21.679 --> 00:25:29.640
to memorize some of these important results
like when you take the lap inverse laplace
00:25:29.640 --> 00:25:36.080
transform you use the known known results
of laplace transforms laplace transforms
00:25:36.080 --> 00:25:45.080
of some extended functions similarly here
so we need to memorize some of these extended
00:25:45.080 --> 00:25:51.539
results to take the inverse z transform
while solving the difference equations
00:25:51.539 --> 00:25:55.980
so with that i would like to conclude my
lecture i
00:25:55.980 --> 00:25:57.940
thank you very much for your attention