WEBVTT
Kind: captions
Language: en
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hello friends welcome to my second lecture
on properties of z transforms so first
00:00:26.240 --> 00:00:33.820
we will discuss what happens when we
know the z transform of a sequence un say
00:00:33.820 --> 00:00:42.690
uz and then the z transform of n into un ok
n into un that is un is multiplied by n hen
00:00:42.690 --> 00:00:50.290
we see that z transform of n un is minus z
into d over dz of uz so this result is also
00:00:50.290 --> 00:00:59.600
going to be very useful when we solve difference
equations so by the definition of z transform
00:00:59.600 --> 00:01:09.040
z and un z and un is sigma n equal to zero
to infinity n un into z to the power minus
00:01:09.040 --> 00:01:17.100
n and this we can write as minus z times sigma
n equal to zero to infinity un into minus
00:01:17.100 --> 00:01:23.681
z n minus n times z to the power minus n minus
one so applying the definition of z transform
00:01:23.681 --> 00:01:27.970
we will we are able to write z transform of
n un in this form
00:01:27.970 --> 00:01:34.659
now but when we see that this can be written
as minus z times sigma n equal to zero to
00:01:34.659 --> 00:01:41.420
infinity un d over dz of z to the power minus
n when you differentiate z to the power minus
00:01:41.420 --> 00:01:47.460
n with respect to z what you get is minus
n times z to the power minus n minus one so
00:01:47.460 --> 00:01:54.719
we express in this form in this form but
then this is equal to minus z d over dz of
00:01:54.719 --> 00:02:00.859
sigma n equal to zero to infinity even z to
the power minus n and therefore it is equal
00:02:00.859 --> 00:02:08.300
to minus z times d over dz of uz now if you
apply mathematical induction on p we can see
00:02:08.300 --> 00:02:15.530
that z transform of n to the power p un is
equal to minus z to the power p dp over dzp
00:02:15.530 --> 00:02:22.620
uz where p is the positive integer
now say suppose we have un equal to n cos
00:02:22.620 --> 00:02:30.290
n theta then n we want to find the z transform
of u one then we can use the known result
00:02:30.290 --> 00:02:39.000
for z transform of cos n theta and make use
of the theorem which we have just proved
00:02:39.000 --> 00:02:48.730
that is the z transform of n un is minus
z times d over dz of uz so let us see we
00:02:48.730 --> 00:03:00.510
let us recall that z transform of cos n
theta is equal to we we know that z transform
00:03:00.510 --> 00:03:08.549
of cos n theta this we have proved in the
previous lecture that it is z into z
00:03:08.549 --> 00:03:20.560
minus cos theta divided by z square minus
two z cos theta plus one so now let us apply
00:03:20.560 --> 00:03:28.830
the theorem multiplication by n so then
z transform of n cos n theta will be equal
00:03:28.830 --> 00:03:39.290
to minus z times d over dz of let us call
it uz so d over dz of uz
00:03:39.290 --> 00:03:50.340
so this is minus z times d over dz of z
square minus z cos theta divided by z square
00:03:50.340 --> 00:04:01.040
minus two z cos theta plus one ok so this
will be equal to minus z times the derivative
00:04:01.040 --> 00:04:10.510
of the numerator is two z minus cos theta
multiplied by z square minus two z cos
00:04:10.510 --> 00:04:29.590
theta plus one minus two z minus two cos
theta into z square minus z cos theta divided
00:04:29.590 --> 00:04:42.030
by z square minus two z cos theta plus
one whole square ok now when we simplify the
00:04:42.030 --> 00:04:51.830
numerator we will have this as minus
z times minus z times z square sorry
00:04:51.830 --> 00:04:59.680
so this will be if you simplify the numerator
you will get z cube cos theta minus two
00:04:59.680 --> 00:05:13.069
z square plus z cos theta divided by z square
minus two z cos theta plus one whole square
00:05:13.069 --> 00:05:21.660
so on simplifying we get the z transform
of un z cube cos theta minus two z square
00:05:21.660 --> 00:05:26.620
plus z cos theta over z square minus two z
plus cos z cos theta plus one whole square
00:05:26.620 --> 00:05:33.280
now suppose we have un equal to n plus one
square we want to find the z transform
00:05:33.280 --> 00:05:49.370
of un equal to n plus one square then here
we can use we know that z transform of n square
00:05:49.370 --> 00:06:02.389
z transform of n square is z square plus
z divided by z minus one raise to the power
00:06:02.389 --> 00:06:16.160
three ok so and we know we know this shifting
theorem if z of un is equal to uz then
00:06:16.160 --> 00:06:30.889
z transform of un plus one z transform of
un plus one is z times uz z times times
00:06:30.889 --> 00:06:44.009
uz minus
u naught so here we will get z transform of
00:06:44.009 --> 00:06:50.880
un plus one so here un you can take as n square
if you take un as n square then z transform
00:06:50.880 --> 00:06:58.340
of n square is z square plus uz is z square
plus z over z minus one whole cube and then
00:06:58.340 --> 00:07:09.130
z transform of n plus one whole square so
z transform of n plus one whole square will
00:07:09.130 --> 00:07:28.789
be z times uz uz is z square plus z my
over z minus one whole cube so this is minus
00:07:28.789 --> 00:07:34.430
u naught now here we need the value of u naught
ok
00:07:34.430 --> 00:07:43.349
so for that i need to know what is u naught
and this u naught we shall require we shall
00:07:43.349 --> 00:07:49.229
we shall do the initial value theorem there
we shall know that if we know z transform
00:07:49.229 --> 00:07:54.770
of a sequence un that is uz then by taking
the limit of uz z tends to infinity we get
00:07:54.770 --> 00:08:09.229
u naught so we know it it we we shall show
that
00:08:09.229 --> 00:08:23.080
u naught is equal to limit z tends to infinity
uz and here uz is z transform of n square
00:08:23.080 --> 00:08:31.320
so z square plus z over z minus one whole
cube so u naught is equal to limit z tends
00:08:31.320 --> 00:08:39.789
to infinity z square plus z divided by z minus
one whole cube and this is a polynomial in
00:08:39.789 --> 00:08:44.390
z of degree two here we have a polynomial
in z of degree three so the limit is call
00:08:44.390 --> 00:08:54.070
obviously zero ok so u naught is equal to
zero and hence z transform of n plus one whole
00:08:54.070 --> 00:09:05.709
square is equal to z cube plus z square
divided by z minus one whole cube
00:09:05.709 --> 00:09:12.310
now you may wonder how do you determine
how we prove that limit of uz as z tends to
00:09:12.310 --> 00:09:21.670
infinity is u naught so this is very simple
we know that since uz
00:09:21.670 --> 00:09:30.990
is equal to z transform of un is sigma n is
equal to zero to infinity un z to the power
00:09:30.990 --> 00:09:40.930
minus n which is u naught plus u one by z
plus u two by z square and so on so when you
00:09:40.930 --> 00:09:50.130
take the limit of this z tends to infinity
this gives you u naught so this implies so
00:09:50.130 --> 00:09:56.450
uz equal to naught plus u one by z plus u
two by z square and so on this implies limit
00:09:56.450 --> 00:10:04.560
as z tends to infinity uz is equal to u naught
this is called initial value theorem which
00:10:04.560 --> 00:10:11.100
we will prove in which we will prove in the
next lecture so we can use this that initial
00:10:11.100 --> 00:10:16.740
value theorem to arrive at the z transform
of n plus one whole square this is z cube
00:10:16.740 --> 00:10:22.820
plus z square over z minus one whole cube
so this is this result uses z transform of
00:10:22.820 --> 00:10:28.090
that the value of u naught which we get by
initial value theorem but if we do not want
00:10:28.090 --> 00:10:35.720
to use the initial value theorem then we can
write by by linearity property so we can
00:10:35.720 --> 00:10:53.880
do this result by applying in by initial
value by linearity property z transform of
00:10:53.880 --> 00:11:02.040
plus one whole square is equal to z transform
of n square plus two times z transform of
00:11:02.040 --> 00:11:10.649
n plus z transform of one and z transform
of n square is z square plus z divided by
00:11:10.649 --> 00:11:18.750
z minus one whole cube plus two times z transform
of n which is z over z minus one whole
00:11:18.750 --> 00:11:26.519
square and this is z transform of one is z
over z minus one
00:11:26.519 --> 00:11:36.710
so simplifying this we shall have z cube plus
z square divided by z minus one so by using
00:11:36.710 --> 00:11:42.610
the initial value property but by using
the linearity property we can obtain the z
00:11:42.610 --> 00:11:48.769
transform of n plus one whole square and if
you do like this that is you use the shifting
00:11:48.769 --> 00:11:54.610
property then you need to know the value of
u naught which we will get by using the
00:11:54.610 --> 00:12:01.829
definition of uz as sigma n equal to
zero to infinity un z to the power minus
00:12:01.829 --> 00:12:09.110
taking the limit as z tends to infinity to
get the value of u naught
00:12:09.110 --> 00:12:18.210
now let us go to unit impulse sequence
it is given by we define a unit impulse sequence
00:12:18.210 --> 00:12:25.779
by delta n equal to one when n is zero n zero
for all n different from zero so if you find
00:12:25.779 --> 00:12:31.171
z transform of delta n let us apply the definition
the definition sigma n equal to zero to
00:12:31.171 --> 00:12:37.620
infinity delta n z to the power minus n when
is equal to zero we are given delta n to be
00:12:37.620 --> 00:12:44.110
equal to one otherwise it is zero so all other
terms are zero except the term for n equal
00:12:44.110 --> 00:12:50.269
to zero where delta n is one so we get one
into z to the power zero which is one so zero
00:12:50.269 --> 00:12:56.470
transform of delta n is equal to one so whenever
we need the z the inverse z transform of
00:12:56.470 --> 00:13:05.019
one we will write there delta n then we
need the z transform unit step function
00:13:05.019 --> 00:13:10.920
so unity step function we defined as un equal
to zero when n is less than zero one for all
00:13:10.920 --> 00:13:16.930
n greater than or equal to z that is for all
non negative integers we define un to be equal
00:13:16.930 --> 00:13:23.370
to one and for all negative integers we
define it as zero so z transform of the unity
00:13:23.370 --> 00:13:28.660
step function un is then again by definition
sigma n equal to zero to infinity un zero
00:13:28.660 --> 00:13:33.070
to the power minus n which is sigma n equal
to zero to infinity z to the power minus n
00:13:33.070 --> 00:13:38.670
because un is equal to one for all n bigger
than or equal to zero now this is z to the
00:13:38.670 --> 00:13:46.520
power so this is one plus z plus z square
and so on unit step function this is sigma
00:13:46.520 --> 00:13:50.390
n equal to zero to infinity z to the power
minus n
00:13:50.390 --> 00:13:58.899
so this is one plus one by z plus one by z
square one by z cube and so on so this series
00:13:58.899 --> 00:14:05.860
converges when mod of z is greater than
one and the sum of the series is one over
00:14:05.860 --> 00:14:13.680
one minus one by z so this is true when this
is greater than this is less than one or this
00:14:13.680 --> 00:14:20.940
is z over z minus one when mod of z is greater
than one so there is enough convergence of
00:14:20.940 --> 00:14:29.040
this series is mod of z greater than
one the z transform of unity step function
00:14:29.040 --> 00:14:40.910
is z upon z minus one now let us consider
a an example of a rational function in z and
00:14:40.910 --> 00:14:53.329
see how we determine the inverse z transform
so here we know uz so let uz be equal to
00:14:53.329 --> 00:15:11.940
four z square minus two z divided by z cube
minus five z square plus eight z minus four
00:15:11.940 --> 00:15:22.790
ok
so we will now let us recall that z
00:15:22.790 --> 00:15:34.329
transform of n is z over z minus one whole
square z transform of n square is z square
00:15:34.329 --> 00:15:44.670
plus z divided by z minus one whole cube ok
now these tendered technique to determine
00:15:44.670 --> 00:15:55.410
the inverse z transform is that when uz is
a rational function of z we write first uz
00:15:55.410 --> 00:16:07.310
by z so uz by z is equal to four z minus
two divided by z cube minus five z square
00:16:07.310 --> 00:16:17.260
minus eight z minus four it will be clear
in when we do the solution why we are taking
00:16:17.260 --> 00:16:23.690
uz bt z so let us break uz by z into partial
fractions we see that when you put z equal
00:16:23.690 --> 00:16:29.519
to one here this denominator becomes zero
so z minus one is the factor of this so we
00:16:29.519 --> 00:16:44.240
have to find factors of this polynomial z
cube minus five z square plus eight z minus
00:16:44.240 --> 00:16:51.980
four so let us find the factors of this
z minus one is the factor and it is a polynomial
00:16:51.980 --> 00:16:57.480
so what we do is we write it as z minus one
z minus one we multiply by z square we get
00:16:57.480 --> 00:17:12.260
z cube minus z square then minus four z minus
four z square minus four z square then
00:17:12.260 --> 00:17:26.230
plus four z plus four
so we get z minus one into z square minus
00:17:26.230 --> 00:17:34.750
four z plus four which is z minus one into
z minus two whole square so we get here four
00:17:34.750 --> 00:17:45.430
z minus two divided by z minus one into z
minus two whole square and the factors then
00:17:45.430 --> 00:17:53.820
then the partial fractions we can write as
a over z minus one plus b over z minus two
00:17:53.820 --> 00:18:01.140
plus c over z minus two whole square corresponding
to z minus one and z minus two whole square
00:18:01.140 --> 00:18:07.980
now the values of a and c can be found directly
so a is equal to we remove z minus one
00:18:07.980 --> 00:18:21.030
from here so four z minus two over z minus
two whole square at z equal to one so this
00:18:21.030 --> 00:18:26.730
will be four minus two that is two when minus
two is minus one whole square there is one
00:18:26.730 --> 00:18:32.920
so we get the value yes two and then c can
also be obtained directly c is equal to z
00:18:32.920 --> 00:18:39.830
minus two whole square we remove from here
so four z minus two divided by z minus one
00:18:39.830 --> 00:18:44.650
and when we put z equal to two
so this will be four into two eight minus
00:18:44.650 --> 00:18:51.880
two six so c is equal to six we have to
determine now the value of b for that we write
00:18:51.880 --> 00:19:01.580
the lcm so lcm is z minus one z minus two
whole square and so then four z minus two
00:19:01.580 --> 00:19:15.600
is equal to a times z minus two whole square
plus b times z minus one z minus two and c
00:19:15.600 --> 00:19:23.940
times z minus one now on the right side we
get terms in z square but n the left side
00:19:23.940 --> 00:19:29.380
there is no term in z square so let us collecting
the terms in z square here the coefficient
00:19:29.380 --> 00:19:35.760
of z square is a here here the coefficient
of z square is b so a plus b is equal to zero
00:19:35.760 --> 00:19:42.430
so we get a plus b equal to zero and a we
have already found so this implies b is equal
00:19:42.430 --> 00:20:00.060
to minus two so thus we get the value of
uz by z so hence uz by z is equal to
00:20:00.060 --> 00:20:10.190
a a is two two over z minus one and then b
over z minus two so minus two over z minus
00:20:10.190 --> 00:20:27.640
two and then c is six so six over z minus
two whole square
00:20:27.640 --> 00:20:34.310
ok
now k now we multiply by z so we get r
00:20:34.310 --> 00:20:46.690
uz equal to two z upon z minus one then minus
two z upon z minus two and then six z upon
00:20:46.690 --> 00:20:58.290
z minus two whole square now let us see why
we see we bring we consider uz by z
00:20:58.290 --> 00:21:03.290
because then the partial fraction will be
some constant divided by z minus one and constant
00:21:03.290 --> 00:21:08.120
divided by z minus two square and what we
need when we find the inverse z transform
00:21:08.120 --> 00:21:16.890
of uz we need here the forms the standard
forms z over z minus one whole square or z
00:21:16.890 --> 00:21:23.130
minus z over because we know that z transform
of a to the power n is equal to z over z minus
00:21:23.130 --> 00:21:31.070
a provided mod of z greater than mod of a
so here we need z over z minus there if if
00:21:31.070 --> 00:21:36.870
we do not consider uz over there and instead
consider uz equal to uz four z square minus
00:21:36.870 --> 00:21:42.550
two z upon this and then we bracket into partial
fractions we shall not get the a standard
00:21:42.550 --> 00:21:49.440
forms where z should come in the numerator
so we consider uz by z now this is two z over
00:21:49.440 --> 00:21:54.660
z minus one so we can take the inverse transform
now and let us see when we take the inverse
00:21:54.660 --> 00:22:09.200
z transform so z inverse of uz is equal to
z inverse two times z inverse of z over z
00:22:09.200 --> 00:22:21.700
minus one then minus two times z inverse of
z over z minus two and then plus six times
00:22:21.700 --> 00:22:33.410
z inverse of z over z minus two whole square
z transform of a to the power n is z over
00:22:33.410 --> 00:22:39.560
z minus a so when mod of z is greater than
one this transform will be put a equal to
00:22:39.560 --> 00:22:50.120
one so two times one to the power n for this
and then minus two times z transform of
00:22:50.120 --> 00:22:58.980
n is z over z minus one whole square we
need the z transform z inverse z transform
00:22:58.980 --> 00:23:10.151
of z over z minus two so if we take z transform
of n is z so this implies so then z transform
00:23:10.151 --> 00:23:19.550
of n into a to the power n if we want this
then we multiply then we divide using the
00:23:19.550 --> 00:23:29.760
dumping rule we will get z over a divided
by z over a minus one whole square and
00:23:29.760 --> 00:23:41.210
so alright k ok so this will be ok his
is az divided by z minus a whole square ok
00:23:41.210 --> 00:23:46.740
now here here we can get it directly so minus
two times z over z minus taking mod of z greater
00:23:46.740 --> 00:23:53.110
than two ok mod of z get two to the power
n so here mod of z greater than we want ere
00:23:53.110 --> 00:24:02.610
we want mod of z greater than two and then
plus six times and then z inverse of ok
00:24:02.610 --> 00:24:16.040
so z inverse of az divided by z minus a whole
square this is equal to n into a to the power
00:24:16.040 --> 00:24:23.300
n so we can here we have a in place of a we
have two here so this will be n into a to
00:24:23.300 --> 00:24:31.040
the power n divided by a so six into
n into two to the power divided by two we
00:24:31.040 --> 00:24:37.470
can multiply and divide by two here six by
two and then we can bring two here like this
00:24:37.470 --> 00:24:47.060
ok so this will be so this is this first
inverse z transform is valid when mod of z
00:24:47.060 --> 00:24:52.410
is greater than one second transform is valid
when mod of z is greater than two so the common
00:24:52.410 --> 00:24:58.180
region we have to take so mod of z greater
than one and mod of z greater than two so
00:24:58.180 --> 00:25:02.920
mod of z greater than one if we have mod of
z greater than than two we have mod of z greater
00:25:02.920 --> 00:25:10.100
than two then both will be valid
so we have two minus two to the power n plus
00:25:10.100 --> 00:25:23.860
one plus three times n into two to the power
n so this is valid for mod of z greater than
00:25:23.860 --> 00:25:28.110
the two we have taken the common region of
convergence mod of z greater than one and
00:25:28.110 --> 00:25:34.920
mod of z greater than two so in so this is
how we find the inverse z transform un ok
00:25:34.920 --> 00:25:43.810
now we take up example two where suppose un
is equal to one over n plus one then z
00:25:43.810 --> 00:25:52.360
transform of one over n plus one we have to
find so we know that z transform of one by
00:25:52.360 --> 00:25:58.890
n if we find the z transform of one by n z
transform of one by n then z transform of
00:25:58.890 --> 00:26:08.840
one by n is given by sigma n equal to zero
to infinity one by n z to the power minus
00:26:08.840 --> 00:26:16.250
n and z transform of one by n plus one
for this we can use the shifting property
00:26:16.250 --> 00:26:22.180
if you take un equal to i by n then un plus
one can be considered as one by n plus one
00:26:22.180 --> 00:26:33.250
so this is z times uz minus u naught u
naught is the term in z n minus one which
00:26:33.250 --> 00:26:43.560
does not contain your your z does which
does not contain n your z term so here we
00:26:43.560 --> 00:26:50.620
will consider n to begin with i think i
should have taken z directly one by n plus
00:26:50.620 --> 00:26:59.680
one let me write it again z transform of one
by n plus one is sigma n equal to zero to
00:26:59.680 --> 00:27:08.670
infinity one by n plus one z to the power
minus n and if you expand this what we will
00:27:08.670 --> 00:27:17.390
get n equal to zero so one minus n equal to
one by two over z then n equal to two one
00:27:17.390 --> 00:27:26.230
by three here also plus one by three z to
the power two one by four z to the power three
00:27:26.230 --> 00:27:37.510
and so on
now what we will do is we can write it
00:27:37.510 --> 00:27:46.930
as see let us recall that as log ln one plus
z let us recall the extension of ln z plus
00:27:46.930 --> 00:27:56.970
one this is z minus z square by two plus
z cube by three and so on minus z cube z four
00:27:56.970 --> 00:28:07.830
by four and so on ok so ln one minus one by
z let us so this is mod of z less than one
00:28:07.830 --> 00:28:17.320
and ln one minus one by z if we write this
then what we have minus one by z minus
00:28:17.320 --> 00:28:28.990
one by two z square minus one by three
z q minus one by four z four and so on
00:28:28.990 --> 00:28:39.650
so this so this i can write as minus z times
if you identify with this minus one upon z
00:28:39.650 --> 00:28:52.970
minus one by two z square minus one by three
z q we can write like this so here mod
00:28:52.970 --> 00:28:57.350
z less than one here mod of one by z will
be less than one this means mod of z must
00:28:57.350 --> 00:29:10.280
be greater than one so mod of z must be greater
than one so this is minus z ln one minus one
00:29:10.280 --> 00:29:22.410
by z provided mod of z is greater than one
and this is raised to the power ln z minus
00:29:22.410 --> 00:29:32.110
ok so this is z times ln one minus one by
z raised to the power minus one or z times
00:29:32.110 --> 00:29:41.960
ln z over z minus one so this is the z transform
of one over n plus one provided mod of z is
00:29:41.960 --> 00:29:49.940
greater than one and then we can find the
z transform of one over in to n plus one z
00:29:49.940 --> 00:29:58.690
transform by using linearity property
so this is equal to z transform of one by
00:29:58.690 --> 00:30:11.010
n minus z transform of one by n plus one
if we know the z transform of un we can
00:30:11.010 --> 00:30:19.090
also find the z transform of one by n then
we will need that n equal to zero will term
00:30:19.090 --> 00:30:29.960
we will cannot take z equal to z transform
of we know that if z transform of un is equal
00:30:29.960 --> 00:30:43.200
to uz then z transform of un minus k z to
the power k into uz z to the power minus k
00:30:43.200 --> 00:30:51.720
into u z so z transform of we know z transform
of one over n plus one so z transform of
00:30:51.720 --> 00:31:03.300
one over n z transform of one over n plus
one we know this is z of ln z over z minus
00:31:03.300 --> 00:31:10.860
one so from here if we write z transform of
one by n using this result we have z to the
00:31:10.860 --> 00:31:23.330
power minus one into uz that is a z
in to ln z over z minus one so we have ln
00:31:23.330 --> 00:31:32.790
z over z minus one
so this will be ln z over z minus one by using
00:31:32.790 --> 00:31:37.600
the shifting theorem so this is ln one
minus z times ln z over z minus one provided
00:31:37.600 --> 00:31:46.340
mod z is greater than one so by using the
shifting theorem so this is ln one minus
00:31:46.340 --> 00:31:56.320
z times ln z over z minus one so y z mod z
is greater than one so by using shifting theorem
00:31:56.320 --> 00:32:04.130
we can find this now let us consider
this problem example three so example three
00:32:04.130 --> 00:32:16.620
we have to find inverse z transform of one
over z minus five whole to the power three
00:32:16.620 --> 00:32:27.880
inverse z transform of one over z minus five
to the power three let us write one over
00:32:27.880 --> 00:32:37.630
z minus five to the power three as one over
z cube times one minus five by z to the power
00:32:37.630 --> 00:32:48.300
three so then one by z q one minus five
by z to the power three minus three
00:32:48.300 --> 00:32:53.530
now we are given that mod of z is greater
than five so mod of five by z is less than
00:32:53.530 --> 00:33:01.690
one so we can apply the one angle theorem
and write it as one over z q and one plus
00:33:01.690 --> 00:33:13.210
three times five by z then three in to four
divided by one in to two two factorial five
00:33:13.210 --> 00:33:24.190
by z whole square then three into four into
five divided by one into two in to three five
00:33:24.190 --> 00:33:35.120
by z whole cube and so on three four five
six divided by one two three four five by
00:33:35.120 --> 00:33:47.280
z whole cube and so on ok i can write it as
one by two into one by z cube then two into
00:33:47.280 --> 00:34:00.820
five by z to the power zero then two into
three five by z to the power one then three
00:34:00.820 --> 00:34:12.300
into four five by a to the power two here
we shall have four into five five by z to
00:34:12.300 --> 00:34:22.639
the power three and so on five in to six
five into six five by z to the power
00:34:22.639 --> 00:34:29.530
here five by z to the power zero five by z
to the power one five by z to the power two
00:34:29.530 --> 00:34:34.740
five by z to the power three there five by
z to the power four so five by z to the power
00:34:34.740 --> 00:34:47.100
four and so on so this is this can be written
as one by two zq sigma we have two two
00:34:47.100 --> 00:35:02.000
into three so we have m plus two five by
z to the power
00:35:02.000 --> 00:35:12.860
let us say m here m plus three m here m m
equal to if i write it for say from zero to
00:35:12.860 --> 00:35:22.000
infinity then what do we get here m equal
to five m plus two m plus three we get here
00:35:22.000 --> 00:35:33.171
n is equal to m plus three so m plus three
means m plus two m plus one let me see m is
00:35:33.171 --> 00:35:41.820
equal to zero if i put here i get here two
into one m equal one i put here two into three
00:35:41.820 --> 00:35:48.840
m equal to two i put here so three into four
yeah so m plus i can write here m plus one
00:35:48.840 --> 00:35:59.500
and here m plus two m plus two and then five
by z to the power m now i think it is ok m
00:35:59.500 --> 00:36:04.740
equal to zero put two into five by a to the
power zero then two in to three five by a
00:36:04.740 --> 00:36:10.140
to the power one then three in to four five
by a to the power two and so on ok so we can
00:36:10.140 --> 00:36:21.130
write like this and then this will be equal
to one by two summation m equal to zero till
00:36:21.130 --> 00:36:33.300
infinity ok so m plus one in to m plus two
into five to the power m z to the power minus
00:36:33.300 --> 00:36:41.610
m minus three
z to the power three we have in the denominator
00:36:41.610 --> 00:36:50.430
so z to the power m minus minus m minus three
we get now let us put n equal to m plus three
00:36:50.430 --> 00:36:56.490
let us put n equal to m plus three so then
this will be one by two times summation n
00:36:56.490 --> 00:37:04.230
is equal to three to infinity and m is equal
to n minus three so we will get n minus two
00:37:04.230 --> 00:37:15.840
here and here we will get n minus three plus
two so n minus one and then we get five to
00:37:15.840 --> 00:37:29.290
the power n minus three five to the power
n minus three into z to the power z to the
00:37:29.290 --> 00:37:35.850
power minus n where we are putting n equal
to m plus three
00:37:35.850 --> 00:37:42.970
so from here we can see that now this is nothing
but this summation n equal to three to infinity
00:37:42.970 --> 00:37:52.020
n minus one n minus two divided by two five
to the power n minus three into z to the power
00:37:52.020 --> 00:37:57.410
minus n so when n is greater than or equal
to of this now let us compare it with the
00:37:57.410 --> 00:38:05.930
z uz is equal to sigma n is equal to zero
to infinity u n z to the power minus n
00:38:05.930 --> 00:38:12.240
this is z transform of un sequence so if we
compare with this then we can say that here
00:38:12.240 --> 00:38:22.160
un is equal to one by two n minus one and
n minus two five to the power n minus three
00:38:22.160 --> 00:38:28.840
when n is greater than or equal to three and
this is equal to zero when n is less than
00:38:28.840 --> 00:38:39.410
three so this is the inverse z transform of
z minus five over q with that i would like
00:38:39.410 --> 00:38:40.770
to conclude my lecture
00:38:40.770 --> 00:38:41.829
thank you for your attention