WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on initial
and final value theorems for z transforms
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in practical applications often we need
the values of u n for n equal to zero and
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tends to infinity without actually requiring
the complete knowledge of the sequence u n
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we know that the sequence u n is related
to the its z transform uz so by using that
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relation relationship we can find the values
of u n for n equal to zero and n tends
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to infinity without actually knowing the
sequence u n so suppose the z transform
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of u n sequence is known it is given by uz
then it tells out that the value of u n for
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n equal to zero is given by limit of uz as
z tends to infinity the proof is quite
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simple we consider the definition of the
z transform z is given by which is z transform
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of the sequence u n is given b y sigma n equal
to zero to infinity u n z to the power minus
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n which when expressed as u naught plus u
one z plus u two yz square and so on on taking
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the limit as z tends to infinity gives
us the value of naught as limit z tends to
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infinity uz
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so when we take the limit of this equality
as z tends to infinity we get limit z tends
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to infinity uz as u naught so this limit
of uz as z tends to infinity gives us the
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value of u naught now similarly we can see
that limit z tends to infinity z times z minus
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u naught is equal to u one so in order to
see this we have uz equal to z of u n sequence
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which is sigma n equal to zero to infinity
u n z to the power minus n which is u naught
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plus ui one by z plus u to by z square and
so on
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so from here we see that uz minus naught
multiplied by z this is equal to u one
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plus u two by z plus u three by z square and
so on and so taking the limit as z tends to
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infinity z
times uz minus u naught gives us the value
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of u one similarly from equation one
we can see that uz minus u naught minus
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u one by z uz minus u naught minus u one by
z when it is multiplied by z square gives
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us a u two plus u three by z plus u four by
z square and so on ok so taking the limit
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again as z tends to infinity ok we get limit
z tends to infinity z square times uz minus
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u naught minus u one by z equal to u two and
so on so this how we can get the initial value
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of the sequence u n by its z transform
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so let us now talk about the final value theorem
here we will get the value of u n as n
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tends to infinity so if the z transform of
the sequence u n is known say it is uz then
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this result tells us that the final value
of the sequence u n that is u n as n tends
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to infinity is given by the limit of z
minus one uz as z tends to one let us look
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at the proof of this final value theorem
so we can take the z trans[form] suppose
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we take the z transform of the sequence u
n plus one minus u n then by linearity property
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we can write it as z of u n plus one minus
z of u n and then we use the shifting property
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here shifting to the right shifting of
the sequence u n to left so z of u n plus
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k is z to the power k into uz minus u naught
minus u one by z minus u two by z square and
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so on minus uk minus one z to the power k
minus one
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if we use this shifting property then z of
u n plus one will be equal to z times u z
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minus u naught so we will get z of u n
plus one equal to z times uz plus u naught
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so let us substitute the value of z of u n
plus one here so z of u n plus one minus u
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n is z times uz minus u naught minus uz
and then we can take write this further as
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z times z minus one minus u naught z
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now by definition of z transform z transform
of the sequence u n plus one minus u n is
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equal to sigma n equal to zero to infinity
u n plus one minus u n into z to the power
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minus n as z tends to one then we will
have limit z tends to one z minus one uz
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minus u naught as limit z tends to one
n equal to zero to infinity u n plus one minus
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u n z to the power minus n because z
minus one into uz minus u naught z is equal
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to this so from this equation and this
equation we arrive here as z tends to one
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limit of this expression is equal to limit
z tends to one this so when z tends to one
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will it will tend to sigma n equal to zero
to infinity u n plus one minus u n which can
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be expressed as limit n tends to infinity
the its partial sum and partial sum so
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u one minus u naught minus u two minus u one
and plus u one u n min u n plus one minus
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u n and the this will reduce further to u
n plus one limit n tends to infinity u n plus
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one minus u naught which is equal to limit
of u n as n goes to infinity minus u naught
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because u naught is independent of n so
then we can cancel and the left hand side
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is limit z tends to one z minus one uz minus
u naught because u naught is independent of
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z so we can cancel u naught on both sides
and then we get limit of u n as limit z tends
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to one z minus one uz so this is the proof
of the final value theorem let us use this
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initial value theorem to find the
initial values of the sequence u n when
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u its z transform is known
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so let say uz is known to us and we want to
determine the value of u n for n equal to
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two and n equal to three so let us see how
we get this uz is equal to two z square
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plus five z plus fourteen divided by z
minus one to the power four ok and the initial
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value by initial value theorem initial value
theorem u naught the initial value of the
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sequence u n naught is limit z tends to infinity
z tends to infinity uz so this is equal
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to now uz can be written as in the numerator
we have polynomial of degree two in z and
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the denominator has a polynomial in z of
degree four so i can write it as one by z
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square two plus five by z plus fourteen
by z square divided by one minus one by z
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to the power four on taking z square common
from the numerator and z to the power four
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common from the denominator
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so this will get we will get this now as
z tends to infinity his expression tends to
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two this tends to one so his whole thing
tends to two and this goes to zero so the
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limit is zero so we get the value of u naught
as zero let us now find the value of u one
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u one is equal to limit z tends to infinity
z times uz minus u naught so this is limit
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z tends to infinity z times two z square
plus five z plus fourteen by z minus one to
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the power four minus zero so uz uz can be
expressed in this form when we multiply this
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form by z ok what we get is one by z so
this z cancels with one z here and we get
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one by z two plus five by z plus fourteen
by z square divided by one minus one by z
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to the power four again this expression tends
to two and this then goes to zero so we get
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the limit as zero
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now let us find u two ok so u two is limit
as z tends to infinity z square times uz
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minus u naught minus u one by z u naught and
u one both are zeros and when we multiply
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uz by z square ok what we get is limit z tends
to infinity two plus five by z plus fourteen
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by z square five by z divided by one minus
one by z to the power four and when z goes
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to infinity this limit clearly is two so we
get the value of u two as two we can similarly
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find u three
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so similarly u three three is equal to
limit z tends to infinity z cube times
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uz minus u naught minus u one by z minus u
two by z square now we have obtained u naught
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as zero u one as zero u two equal to two by
z u two equal to two so we can see find
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the limit so u three is equal to limit z tends
to infinity z cube times uz is one by z square
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two plus five by z five by z plus fourteen
by z square divided by one minus one by z
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to the power four minus this zero this
zero this two by z square so two by z square
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z cube can be multiplied so this is limit
z goes to infinity z cube when multiplied
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gives you two z plus five plus fourteen by
z divided by one minus one by z to the power
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four minus two and we can simplify it further
to find the limit
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so this will give you this z z cube
when multiplied by this is two z so two z
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times this ok now we can write it further
as
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one minus one by z to the power four can be
expanded by binomial expansion so we have
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one then four times minus one by z then
six times minus one by z square then we have
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four times minus one by z cube and then we
have one by z to the minus one by z to
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the power four so we can do this now we
can see when you multiply two z here and what
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does then happens
so two z minus two z two z cancel out we have
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then five
five ok so we two z two z cancel out we get
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five plus fourteen by z ok and here what we
get his z this z cancel out we get minus four
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into minus two so plus eight and then this
z when multiplied here to z square what we
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get is minus twelve into one by z and then
we get here minus one by z cube so we get
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plus eight times one by z square and then
we get hm minus two times one over z to
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the power three
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so when z goes to infinity this goes to
zero this goes to zero this this goes to zero
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and we what we get is thirteen so this is
equal to thirteen so this is how we get the
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value of u three now so there is
example one an example two we have take
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an example of inverse z transform let us find
inverse z transform of l n z over z plus one
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so we shall apply the power series method
to determine the inverse z transform let
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us see we ln z over z plus one is equal
to ln z plus one over z to the power minus
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one and this is ln or i can say this is minus
ln one plus one by z
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now if mod of z is greater than one then
i can write it as minus one over z minus
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one over two times z one by z square plus
one by three times one by z cube and so on
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provided mod of z is greater than one so
i can write it as minus one by z plus one
00:17:18.060 --> 00:17:27.819
by two z square minus one by three z cube
plus one by four z to the power four and so
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on
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now comparing this series in finite series
with sigma [vocalized noise] we note that
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here there is no term which is free from z
so we get z inverse ok so if so z inverse
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of ln z over z plus one this is equal to
u n sequence ok then u n is equal to zero
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when n is equal to zero and so here when n
is equal to one the coefficient is minus
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one when n is equal to two the coefficient
is one by two so we get minus one to the power
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n divided by n when n is equal to one two
three and so on
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this is the example two let us now go to
the next slide and see how we find the z trans[verse]
00:18:38.480 --> 00:19:01.340
inverse z transform of z over z plus one whole
square we know that z transform of n in to
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a to the power n is a z over z minus a
whole square provided mod of z is greater
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than mod of a so when mod of a is greater
than mod of a we know that z transform of
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n a to the power n is equal to a z over z
minus whole square so taking a equal to minus
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one we have z transform of z transform
of n times minus one to the power n equal
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to minus z divided by z plus one whole square
provided mod of z is greater than mod of minus
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one so which is one
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and this gives us inverse transform of
so using the linearity property of the z transform
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what we get is minus z transform of n times
minus one to the power n minus one is equal
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to z over z plus one whole square when mod
of z is greater than one or we can say that
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inverse z transform of z over z plus one whole
square this is equal to minus one to the
00:20:31.909 --> 00:20:35.519
power n minus one into n
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now this is the method where we have used
the known result to get the inverse transform
00:20:43.580 --> 00:20:48.299
there is a method which we call long division
method we can find the inverse transform by
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that using the using that long division
method also so let us see how we that method
00:20:52.659 --> 00:21:05.679
works so we shall apply the long division
method what we will do here is that we have
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z over z plus one whole square z over z plus
one whole square is z over z square plus two
00:21:15.590 --> 00:21:27.059
z plus one so let us divide z by z square
plus two z plus one we have to divide z
00:21:27.059 --> 00:21:32.259
by z square plus two z plus one in such a
way that it will lead us to an finite series
00:21:32.259 --> 00:21:34.620
in powers of one over z
00:21:34.620 --> 00:21:42.590
so z square plus z plus one by we multiply
this b y z to the power minus one so that
00:21:42.590 --> 00:21:47.480
here when we multiply by z to the power minus
one we get z here so that that z will cancel
00:21:47.480 --> 00:21:52.269
with this z here so z to the power minus one
we take then we multiply z to the power minus
00:21:52.269 --> 00:21:58.870
one we take here so we get z plus two times
z in to z minus one is two plus z to the power
00:21:58.870 --> 00:22:07.509
minus one and what we get is when we subtract
the sign changes so minus two minus z to the
00:22:07.509 --> 00:22:13.899
power minus one now our our aim will be to
eliminate minus two so we write here minus
00:22:13.899 --> 00:22:20.360
two times z to the power minus two so when
you multiply by minus two times z to the power
00:22:20.360 --> 00:22:27.350
minus two to z square you get minus two and
then we get minus four z to the power minus
00:22:27.350 --> 00:22:34.240
one and then minus two times z to the power
minus two again we subtract so [we/the] sign
00:22:34.240 --> 00:22:39.120
changes and this cancels and we get four z
to the power one minus one minus z to the
00:22:39.120 --> 00:22:46.640
power minus one so we get three z to the power
minus one plus two times z to the power minus
00:22:46.640 --> 00:22:47.640
two
00:22:47.640 --> 00:22:55.330
now we multiply by three times z to the power
minus three so that when we multiply it to
00:22:55.330 --> 00:22:59.590
z square it will give us three times z to
the power minus one so three times z to the
00:22:59.590 --> 00:23:08.070
power minus one and then we get six times
z to the power minus two and we will get
00:23:08.070 --> 00:23:16.549
three times z to the power minus three
so like this we continue so will get here
00:23:16.549 --> 00:23:24.910
four times z to the power minus two be the
negative sign ok minus three times z to
00:23:24.910 --> 00:23:26.269
the power minus three
00:23:26.269 --> 00:23:31.789
so now our aim will be to multiply z square
by such by an expression which will give us
00:23:31.789 --> 00:23:38.679
minus four z to the power minus two so minus
four z to the power minus four so minus four
00:23:38.679 --> 00:23:48.960
z to the power minus two and then we get minus
eight z to the power minus three and then
00:23:48.960 --> 00:23:58.549
we get minus four z to the power minus
four so this is plus plus this cancels and
00:23:58.549 --> 00:24:06.809
we get this minus this is five times z to
the power minus three plus four times z
00:24:06.809 --> 00:24:15.769
to the power minus four
so continuing this process ok so thus z
00:24:15.769 --> 00:24:23.210
over z plus one whole square it will come
out to be z to the power minus one then
00:24:23.210 --> 00:24:35.139
minus two times z to the power minus two and
then three times z to the power minus three
00:24:35.139 --> 00:24:42.320
minus four times z to the power minus four
and then next one will be plus five
00:24:42.320 --> 00:24:52.480
times z to the power minus five and so on
ok and when we compare it with that series
00:24:52.480 --> 00:25:02.250
sigma u one z to the power minus n so this
gives us then z inverse of z over z plus
00:25:02.250 --> 00:25:18.190
one whole square this equal to this equal
to u n equal to zero because there is no
00:25:18.190 --> 00:25:24.240
term contain[ed] which is free from z so zero
when n is equal to zero n minus one to the
00:25:24.240 --> 00:25:34.909
power n minus one when minus one to the
power n into n when n is equal to when
00:25:34.909 --> 00:25:51.610
n is equal to one we will write here minus
one
00:25:51.610 --> 00:25:58.639
so this how we get the inverse transform of
z over z plus one whole square with that
00:25:58.639 --> 00:26:01.970
i would like to conclude my lecture
00:26:01.970 --> 00:26:03.340
thank you very much for your attention