WEBVTT
Kind: captions
Language: en
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hello friends welcome to my lecture on convergence
of z transform before discussing the convergence
00:00:26.020 --> 00:00:32.951
of z transform let us look at some more
methods where we can find the z transform
00:00:32.951 --> 00:00:39.770
we can find the inverse z transform of
uz the partial diffraction method we have
00:00:39.770 --> 00:00:47.460
discussed earlier also here we have one more
case where we can discuss the we can find
00:00:47.460 --> 00:00:53.300
the inverse z transform of uz so in this method
as i have said in my previous lecture when
00:00:53.300 --> 00:01:00.329
we want to use the partial different diffraction
method we consider uz by z and break uz
00:01:00.329 --> 00:01:06.450
by z into partial fractions and then multiply
the resulting expansion by z and then we invert
00:01:06.450 --> 00:01:13.190
that so suppose we have the example of uz
equal to two z square plus three z over z
00:01:13.190 --> 00:01:20.020
plus two z minus four then and we are to find
the inverse z transform of this function of
00:01:20.020 --> 00:01:33.310
z so then what we will do is let us say let
uz equal to two z square plus three z divided
00:01:33.310 --> 00:01:47.740
by z plus two z minus four so then we consider
uz by z let us write uz by z his will be
00:01:47.740 --> 00:01:58.950
two z plus three divided by z plus two z minus
four we will break this two z plus three
00:01:58.950 --> 00:02:01.929
over z plus two z minus four into partial
fractions
00:02:01.929 --> 00:02:10.289
so let us write this as a over z plus two
and b over z minus four then the values of
00:02:10.289 --> 00:02:21.000
a and b can be found easily so then a will
be equal to two z plus three divided by
00:02:21.000 --> 00:02:28.850
z minus four evaluated at z equal to minus
two and this will be giving you minus four
00:02:28.850 --> 00:02:35.680
plus three so minus one divided by minus two
minus four so that is minus six so we will
00:02:35.680 --> 00:02:46.040
get one by six and b will be equal to similarly
two z plus three divided by z plus two
00:02:46.040 --> 00:02:52.480
and we will have to evaluate this at z equal
to four so when we put z equal to four here
00:02:52.480 --> 00:02:58.030
four into two eight plus three is eleven divided
by four plus two is six so b is six so thus
00:02:58.030 --> 00:03:14.050
uz by z is equal to one by six z plus two
and then eleven by six into one over z minus
00:03:14.050 --> 00:03:27.310
four so we will then multiply this equation
by z so multiplying by z we get uz as z over
00:03:27.310 --> 00:03:38.439
one over six into z over z plus two and then
eleven over six oh z over z minus four now
00:03:38.439 --> 00:03:52.099
let us recall that we know we know that z
inverse of z over z minus a this is equal
00:03:52.099 --> 00:04:05.120
to a to the power n provided mod of z is greater
than mod of a so so z inverse of z over z
00:04:05.120 --> 00:04:14.310
plus two will be equal to minus two to the
power n provided mod of z is greater than
00:04:14.310 --> 00:04:26.460
mod of minus two which is two and similarly
z inverse of z over z minus four will be equal
00:04:26.460 --> 00:04:43.930
to four to the power n provided mod of z is
greater than four and hence z inverse of uz
00:04:43.930 --> 00:04:51.120
will be equal to one by six z inverse of z
over z plus two which is minus two to the
00:04:51.120 --> 00:05:03.620
power n and plus eleven by six z over z
minus four inverse in z inverse of z over
00:05:03.620 --> 00:05:11.139
z minus four which is four to the power n
and the common region is mod of the in case
00:05:11.139 --> 00:05:15.900
of z inverse z over z plus two the region
of convergence was mod of z greater than two
00:05:15.900 --> 00:05:21.509
and here it is mod of z greater than four
so we will have this is valid when mod of
00:05:21.509 --> 00:05:22.979
z is greater than four
00:05:22.979 --> 00:05:33.410
so here un is equal to one by six two to
the power minus two to the power n and plus
00:05:33.410 --> 00:05:42.199
eleven by six four to the power n where mod
of z is greater than four so this is how we
00:05:42.199 --> 00:05:48.000
can find the inverse z transform in this case
let us now there is another method which
00:05:48.000 --> 00:05:55.879
is known as inversion integral method here
we will need um analysis of complex analysis
00:05:55.879 --> 00:06:01.080
knowledge of complex analysis that is the
contour integration so i think i hope that
00:06:01.080 --> 00:06:08.539
you are familiar with this contour integration
so if you know contour integration you
00:06:08.539 --> 00:06:17.699
can use this method also to find the inverse
z transform so the suppose we have sequence
00:06:17.699 --> 00:06:23.889
un the inverse z transform we we know the
z transform of a sequence un it is given as
00:06:23.889 --> 00:06:30.530
uz then the inverse z transform f uz is given
by un equal to one over two pi i integral
00:06:30.530 --> 00:06:42.169
over c ez uz into z to the power n minus
one dz where c is the contour which encloses
00:06:42.169 --> 00:06:46.940
the similarities of the function uz so
and we calc
00:06:46.940 --> 00:06:53.610
so to evaluate un we calculate the sum of
residuals of the analytical function uz into
00:06:53.610 --> 00:07:00.729
z to the power n minus one at the poles of
uz which lie inside the contour c according
00:07:00.729 --> 00:07:06.280
to the given region of convergence so by the
given region of convergence we decide how
00:07:06.280 --> 00:07:14.009
many poles of uz lie inside the contour
c and then calculate the residues at those
00:07:14.009 --> 00:07:23.250
poles of uz and take their sum to get the
sequence un so let us look at this method
00:07:23.250 --> 00:07:30.690
suppose we want to determine the inverse z
transform of three z over z minus one z
00:07:30.690 --> 00:07:43.740
minus two so z inverse of three z over z minus
one z minus two here let us say u un uz is
00:07:43.740 --> 00:07:59.310
equal to three z over z minus one z minus
two so let us find the poles of uz the
00:07:59.310 --> 00:08:20.379
similarities of uz are the simple poles
at z equal to one end z equal to two now
00:08:20.379 --> 00:08:26.599
we are not given the region of convergence
so we shall assume that the region of convergence
00:08:26.599 --> 00:08:32.979
is such that it encloses all the similarities
ok so z equal to one z equal to two are enclosed
00:08:32.979 --> 00:08:49.760
inside the contour c ok so lets see the contour
c enclose both the poles z equal to one z
00:08:49.760 --> 00:08:59.950
equal to two then we have to find the residue
at z equal to one so residue at z equal to
00:08:59.950 --> 00:09:14.110
one of the function uz into z to the power
n minus one is equal to limit z tends to one
00:09:14.110 --> 00:09:21.660
z minus one into uz into z to the power n
minus one
00:09:21.660 --> 00:09:27.340
we know that when he function fz in complex
analysis has a simple pole at the point z
00:09:27.340 --> 00:09:33.790
equal to a and we want to determine the residue
there then we multiply fz by z minus a and
00:09:33.790 --> 00:09:38.950
take the limit as z tends to a so here we
multiply uz into z to the power n minus one
00:09:38.950 --> 00:09:45.160
by the factor z minus one and find the limit
at z tends to one so this is equal to limit
00:09:45.160 --> 00:09:55.400
z tends to one z minus one and then uz is
three z into z to the power n minus one divided
00:09:55.400 --> 00:10:03.650
by z minus one z minus two
00:10:03.650 --> 00:10:12.820
so this z minus one we can cancel and when
z tends to one it is three into one to
00:10:12.820 --> 00:10:19.210
the power n divided by three minus two
so one this is three into one to the power
00:10:19.210 --> 00:10:35.840
m and similarly limit z tends to two similarly
the residue at z equal to two of the function
00:10:35.840 --> 00:10:56.700
uz into z to the power n minus one is equal
to limit z tends to two z minus two into
00:10:56.700 --> 00:11:07.820
uz into z to the power n minus one which
will be equal to limit z tends to two z minus
00:11:07.820 --> 00:11:19.140
two into three z divided by z minus one z
minus two into z to the power n minus one
00:11:19.140 --> 00:11:27.960
so this will cancel with this and when z tends
to two what we get is three into two
00:11:27.960 --> 00:11:41.520
to the power n ok so thus we get thus un
by the formula un equal to sum of residues
00:11:41.520 --> 00:11:48.780
at the poles z equal to one and z equal to
two so three into one to the power n plus
00:11:48.780 --> 00:11:56.280
three into two to the power here we are
getting one minus two minus one we are
00:11:56.280 --> 00:12:06.540
getting so this will be minus so this is minus
here minus three into one to the power n plus
00:12:06.540 --> 00:12:13.880
three into two to the power n so we can write
it three times two to the power n minus one
00:12:13.880 --> 00:12:20.360
so this is the this how we can find this
sequence un by using the invergent integral
00:12:20.360 --> 00:12:28.170
method ok so let us discuss one more example
z inverse of two z over z minus one
00:12:28.170 --> 00:12:39.700
z square plus one ok let us say let us say
uz be equal to two z over z minus one z
00:12:39.700 --> 00:12:49.170
square plus one so then we write uz by z so
then uz by z will be equal to two over z minus
00:12:49.170 --> 00:13:00.930
one z square plus one which we can write as
a over z minus one b over z minus i and c
00:13:00.930 --> 00:13:08.990
over z plus i because z square plus one can
be broken into linear factors z minus i
00:13:08.990 --> 00:13:18.730
z plus i now so then a will be equal to
put z equal to one and two over z square plus
00:13:18.730 --> 00:13:25.820
one so in two over z square plus one when
we put z equal to one we will get two over
00:13:25.820 --> 00:13:35.390
two we get one then we we can find so two
over z minus one and here z minus two we shall
00:13:35.390 --> 00:13:46.190
leave z plus i we will take
so this evaluated at z equal to i this will
00:13:46.190 --> 00:13:54.490
give you two over i minus one into two i so
we shall get these two will cancel with his
00:13:54.490 --> 00:13:59.730
two and we shall get i square which is
minus one so minus one minus i so minus one
00:13:59.730 --> 00:14:11.550
upon one plus i and similarly c will be equal
to two over z minus one into z minus i
00:14:11.550 --> 00:14:21.570
at z equal to minus i so this will be equal
to two over minus i minus one and then minus
00:14:21.570 --> 00:14:31.440
two i so two will cancel with two when
we multiply by minus i here so minus i into
00:14:31.440 --> 00:14:41.050
minus i is i square which is minus one so
minus one and then we get plus i so we get
00:14:41.050 --> 00:14:54.610
minus one upon one minus i and hence uz will
be equal to z we multiply in this equation
00:14:54.610 --> 00:15:01.870
and a is one so z over z minus one we have
and then we have v is equal to minus one over
00:15:01.870 --> 00:15:11.800
one plus i into z over z minus i and then
c is minus one over one minus i so minus one
00:15:11.800 --> 00:15:23.010
over one minus i into z over z plus i and
this implies u one the inverse z transform
00:15:23.010 --> 00:15:30.140
of uz will be z over z minus one means
one to the power n because inverse z transform
00:15:30.140 --> 00:15:38.810
of z over z minus i is a to the power n
and then minus one over one plus i inverse
00:15:38.810 --> 00:15:43.470
z transform here will be i to the power n
00:15:43.470 --> 00:15:47.360
so here mod of z is greater than one here
mod of z is greater than mod of i mod of i
00:15:47.360 --> 00:15:52.990
is also one so region of convergence is same
mod of z greater than one and here we have
00:15:52.990 --> 00:15:59.580
minus one over one minus i into minus i to
the power n and the region of convergence
00:15:59.580 --> 00:16:05.120
is mod of z greater than mod of minus i
which is again mod of z greater than one so
00:16:05.120 --> 00:16:13.530
we get mod of z greater than one so we have
one minus i to the power n over one plus
00:16:13.530 --> 00:16:19.850
i minus minus i to the power n over one minus
i where mod z is greater than one so this
00:16:19.850 --> 00:16:29.350
is the solution of example two now let us
go to a two sided z transform so far we consider
00:16:29.350 --> 00:16:37.080
the z transform of which was one sided z transform
where one sided z transform means we consider
00:16:37.080 --> 00:16:44.080
the values of n to be non negative n equal
to zero one two three and so on and we considered
00:16:44.080 --> 00:16:50.340
uz equal to sigma n equal to zero to infinity
un z to the power minus n but n can take negative
00:16:50.340 --> 00:16:56.010
values also so when n takes values from minus
infinity to plus infinity the z transform
00:16:56.010 --> 00:17:01.190
will be defined as sigma n equal to minus
infinity to un z to the power minus n and
00:17:01.190 --> 00:17:07.589
this transforms then called as two sided z
transform in the case of one sided z transform
00:17:07.589 --> 00:17:12.240
where we had taken n greater than or equal
to zero the region of convergence as we have
00:17:12.240 --> 00:17:18.380
seen is always of the form mod of z greater
than mod of a which is the region outside
00:17:18.380 --> 00:17:24.439
the circle mod of z equal to mod of a the
circle where the center is at the origin and
00:17:24.439 --> 00:17:39.759
a dash is mod of a so the region of convergence
will look like this
00:17:39.759 --> 00:17:55.870
if this is the circle with mod of with ds
mod of a in the z plane then the region
00:17:55.870 --> 00:18:01.880
outside the circle mod of z equal to mod of
a is this shaded region so the region of convergence
00:18:01.880 --> 00:18:07.179
is always the circle is mod of z equal to
mod of a
00:18:07.179 --> 00:18:13.120
so in the case of the values of n greater
than or equal to zero the region of convergence
00:18:13.120 --> 00:18:17.529
is always outside the circle it is of the
form mod z greater than mod of a but when
00:18:17.529 --> 00:18:23.159
n takes the negative values that is from
minus infinity to zero the region of convergence
00:18:23.159 --> 00:18:33.190
we shall see is of the form mod z less than
mod of b so it will be of this form suppose
00:18:33.190 --> 00:18:39.159
this is your mod z equal to mod of b circle
with center at origin and ready as mod of
00:18:39.159 --> 00:18:48.399
b then it is interior of this circle the region
of convergence lies to the interior of
00:18:48.399 --> 00:18:54.760
the circle so in the case of two sided
z transform for the negative values of
00:18:54.760 --> 00:19:01.100
the negative powers of z we get the
region of convergence which is outside the
00:19:01.100 --> 00:19:06.850
circle for the power of z which are positive
we the region of convergence will look like
00:19:06.850 --> 00:19:14.120
it is inside the circles so it so for the
z two sided z transform the region of convergence
00:19:14.120 --> 00:19:19.850
will be of the form mod of a less than
mod of z less than mod of b that is a the
00:19:19.850 --> 00:19:24.110
region of convergence lying between two concentric
circles so such a region of convergence is
00:19:24.110 --> 00:19:31.110
known as annular region and the inner circle
than inner circle is mod of z equal to a mod
00:19:31.110 --> 00:19:37.960
of z equal to mod of a so it bounce the comes
in negative powers of z and the outer circle
00:19:37.960 --> 00:19:43.250
which is mod of z equal to mod of b it bounds
the terms in the positive powers of z so for
00:19:43.250 --> 00:19:49.179
example let us consider this two less than
mod of z less than three so it is an anywhere
00:19:49.179 --> 00:19:57.700
region the region concentric to two circles
mod of z equal to mod of two mod of z equal
00:19:57.700 --> 00:20:05.860
to two and then mod of z equal to three
00:20:05.860 --> 00:20:20.870
so this region is the given region of convergence
so here two is less than mod of z less than
00:20:20.870 --> 00:20:26.919
three let us find the inverse z transform
of the function uz equal to here so uz
00:20:26.919 --> 00:20:38.470
is equal to two times z square minus five
z plus six point five divided by z minus two
00:20:38.470 --> 00:20:46.629
z minus three whole square so you can put
it as a over z minus two let us break it into
00:20:46.629 --> 00:20:56.000
partial fractions b over z minus three and
then c over z minus three whole square so
00:20:56.000 --> 00:21:03.679
we can determine the values of a b c very
easily a is equal to two times z square
00:21:03.679 --> 00:21:13.480
minus five z plus six point five divided by
z minus three whole square evaluated at z
00:21:13.480 --> 00:21:22.230
equal to two so let us put z equal to two
here so then two times we have four minus
00:21:22.230 --> 00:21:30.700
ten plus six point five divided by two minus
three whole square so this is one and here
00:21:30.700 --> 00:21:34.720
we have ten point five minus ten so we get
point five point five into two is equal to
00:21:34.720 --> 00:21:42.070
one and similarly we can find directly the
value of c c will be two times z square minus
00:21:42.070 --> 00:21:54.019
five z plus six point five divided by z minus
two and let us put z equal to three in this
00:21:54.019 --> 00:22:04.909
so that we will get two times nine minus
fifteen and then plus six point five divided
00:22:04.909 --> 00:22:06.649
by three minus two is one
00:22:06.649 --> 00:22:11.129
so we get here fifteen point five minus
fifteen is point five point five into two
00:22:11.129 --> 00:22:20.000
is equal to one now to determine b we have
to write the identity so two times let us
00:22:20.000 --> 00:22:27.750
take the lcm two times z square minus five
z plus six point five this is equal to i am
00:22:27.750 --> 00:22:36.090
taking lcm this is equal to a times z minus
three whole square plus b times z minus two
00:22:36.090 --> 00:22:47.240
z minus three plus c times z minus two
now let us equate the terms containing z square
00:22:47.240 --> 00:22:55.749
on both sides so here the term containing
z square is a and then here b so a plus b
00:22:55.749 --> 00:23:00.490
the coefficient of z square on the right side
is a plus b and the left side it is two so
00:23:00.490 --> 00:23:06.850
a plus b equal to two a we have got already
equal to one so a equal to one gives us b
00:23:06.850 --> 00:23:17.290
also equal to one so a b c here are all equal
to one ok so thus we have so thus uz is equal
00:23:17.290 --> 00:23:31.279
to z over a say sorry one over z minus
two and then one over z minus three and then
00:23:31.279 --> 00:23:38.759
one over z minus three square ok now we are
given the region of convergence as two or
00:23:38.759 --> 00:23:43.910
less than mod of z less than three so here
we can see that mod of z greater than two
00:23:43.910 --> 00:23:56.450
implies two over mod of z two over mod
of z is less than one so what we will do here
00:23:56.450 --> 00:24:03.220
two expand one over z minus two by using
the binomial theorem his i can write as one
00:24:03.220 --> 00:24:10.799
over z times one over one minus two over z
and mod of two over z is less than one so
00:24:10.799 --> 00:24:16.990
we can expand it in the binomial by binomial
theorem and here this is mod of z is less
00:24:16.990 --> 00:24:22.879
than three so mod of z divided by three is
less than one so what i will do is i will
00:24:22.879 --> 00:24:30.059
take minus one over three out and write it
as one minus z upon three
00:24:30.059 --> 00:24:36.769
so that mod of z less than three gives us
the infinite series expansion of one over
00:24:36.769 --> 00:24:44.840
minus z by three so here again here we shall
write one over minus three whole square and
00:24:44.840 --> 00:24:53.039
then i will write one over one minus z by
three whole square so now let us expand by
00:24:53.039 --> 00:25:00.370
binomial theorem so if we we do that i will
write it as one over z since mod of two by
00:25:00.370 --> 00:25:07.499
z is less than one i will write it sigma n
equal to zero to infinity two by z raise to
00:25:07.499 --> 00:25:15.149
the power m and then minus one by three again
this is my mod of z by three is less than
00:25:15.149 --> 00:25:16.149
one
00:25:16.149 --> 00:25:26.510
so sigma n equal to zero to infinity z by
three raise to the power n and here plus one
00:25:26.510 --> 00:25:35.320
by nine summation n equal to zero to infinity
i think i should write like first i would
00:25:35.320 --> 00:25:42.139
write it in expanded form so i should write
it as this is one minus z by three raise to
00:25:42.139 --> 00:25:52.649
the power minus two so i will get here one
plus two times z by three plus two into three
00:25:52.649 --> 00:26:01.309
by two factorial and then i will have z y
three m z by three whole square and then two
00:26:01.309 --> 00:26:12.710
three four divided by three factorial minus
z by three whole cube and so on ok here
00:26:12.710 --> 00:26:19.840
sorry minus minus minus and minus becomes
plus so i will get this here so ok so this
00:26:19.840 --> 00:26:34.610
will be equal to so this will be one plus
two times z by three and then two into three
00:26:34.610 --> 00:26:44.710
by two factorial z by three whole square
and then two into three into four divided
00:26:44.710 --> 00:26:56.220
by three factorial z by three to the power
three and so on ok now let us see at this
00:26:56.220 --> 00:27:03.460
this further let us expand this so one
by z this is what when n is equal to zero
00:27:03.460 --> 00:27:13.690
first term is one then two by z we have then
we have four by z two square by z square
00:27:13.690 --> 00:27:22.119
two cube by z cube and so on and so these
terms give us negative powers of z by these
00:27:22.119 --> 00:27:28.549
two series are in the positive powers of z
so minus one by three and then here we have
00:27:28.549 --> 00:27:41.230
one plus z by three plus z square by three
square z cube by three cube and so on this
00:27:41.230 --> 00:27:47.649
the series in the second term in the third
term we get one by nine i can multiply one
00:27:47.649 --> 00:28:00.549
by nine are ok one by nine one by nine
into plus two by nine into z by three and
00:28:00.549 --> 00:28:10.309
then we have two into three divided by nine
into two factorial z by three whole square
00:28:10.309 --> 00:28:15.730
and so on
00:28:15.730 --> 00:28:22.379
now let us determine the sequence un from
here so for n greater than or equal to zero
00:28:22.379 --> 00:28:30.730
we will get the value of un from this so because
sequence un we have to find un for the values
00:28:30.730 --> 00:28:36.980
of n greater than or equal to zero so when
n is z to the power minus n when n is equal
00:28:36.980 --> 00:28:46.179
to one and on wards so n greater than or equal
to one gives you because when n is equal to
00:28:46.179 --> 00:28:52.730
one we get one here when n is equal to two
here we get two to the power two minus one
00:28:52.730 --> 00:28:58.989
so we get two here the coefficient of z to
the power minus two is two and the coefficient
00:28:58.989 --> 00:29:05.279
of z to the power minus three is two square
so un is two to the power minus n and we
00:29:05.279 --> 00:29:15.549
get here minus n plus two into three to the
power n minus two so this is for n less than
00:29:15.549 --> 00:29:22.070
or equal to zero let us take for n equal to
zero what do we get when n is equal to zero
00:29:22.070 --> 00:29:28.159
we have here minus two over three square so
this is minus two over three square minus
00:29:28.159 --> 00:29:33.299
two over three square means minus two over
nine and here what do we get minus one over
00:29:33.299 --> 00:29:38.259
three minus one over three plus one over nine
00:29:38.259 --> 00:29:47.059
so this is nine minus three plus one so minus
two over three square so here also we get
00:29:47.059 --> 00:29:53.889
minus two over three square and the coefficient
of then z z here is minus one by three
00:29:53.889 --> 00:30:02.080
square and here it is two by three cube
so coefficient of z is two by twenty seven
00:30:02.080 --> 00:30:12.929
minus one by nine which is we can say it is
six minus one six minus three so we get
00:30:12.929 --> 00:30:21.230
three by twenty seven or one by nine and what
we get here i think minus one by nine and
00:30:21.230 --> 00:30:27.409
here we get two by twenty seven two by twenty
seven minus one by nine so we get here twenty
00:30:27.409 --> 00:30:33.919
seven this will be two here two here and here
we will get minus three so we will get minus
00:30:33.919 --> 00:30:42.380
one by twenty seven minus one by twenty seven
and here what we get so when n is equal to
00:30:42.380 --> 00:30:49.830
minus one it will be minus minus one plus
two and three to the power minus three so
00:30:49.830 --> 00:30:57.039
this will be minus one over three cube which
is minus one by twenty seven so the terms
00:30:57.039 --> 00:31:04.640
which contain positive powers of z there un
is minus n plus two three to the power n minus
00:31:04.640 --> 00:31:11.500
two and when the negative powers of z give
you un equal to two to the power n minus one
00:31:11.500 --> 00:31:13.159
where n is bigger than or equal to one
00:31:13.159 --> 00:31:22.539
so in the case of this annular region which
we were get the two sided z transform and
00:31:22.539 --> 00:31:30.310
the sequence un then is two sided for
n greater than or equal to one we get two
00:31:30.310 --> 00:31:34.919
to the power n minus one and for n less than
or equal to zero we get minus n plus two three
00:31:34.919 --> 00:31:41.330
to the power n minus two we shall be applying
the knowledge of z transform to the difference
00:31:41.330 --> 00:31:46.690
equations as we know that the laplace transforms
are very useful in solving linear differential
00:31:46.690 --> 00:31:53.950
equations we shall show in our next two lectures
the z transforms are also very useful to solve
00:31:53.950 --> 00:31:59.330
liner difference equations when we have a
discreet system the performance of the discreet
00:31:59.330 --> 00:32:05.470
system is expressed by a difference equation
and z transforms are very useful in the analysis
00:32:05.470 --> 00:32:11.929
and representation of discreet time systems
so the solution of the difference equation
00:32:11.929 --> 00:32:18.549
is required to determine the for frequency
response of such a discreet system so in our
00:32:18.549 --> 00:32:24.240
next two lecture we shall be discussing how
we can solve a difference equation by applying
00:32:24.240 --> 00:32:28.340
the z transforms with that i would like
to conclude my lecture
00:32:28.340 --> 00:32:29.659
thank you very much for your attention