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Kind: captions
Language: en
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hello friends welcome to my lecture on applications
of z transforms so this we will cover in two
00:00:25.539 --> 00:00:33.780
lectures this is first lecture on application
of z transforms we know that the difference
00:00:33.780 --> 00:00:39.190
equations model those processes in which
we know relationships between changes and
00:00:39.190 --> 00:00:45.170
or differences rather than the rates of
change which leads to the differential equations
00:00:45.170 --> 00:00:50.879
now these difference equations arise in those
situations where sequential relation exists
00:00:50.879 --> 00:00:56.020
at various discrete values of the independent
variable the difference relation that we
00:00:56.020 --> 00:01:01.840
are obtained while solving the difference
equation by power series or frobenius method
00:01:01.840 --> 00:01:07.520
are also difference equations similarly numerical
solution of differential equations also leads
00:01:07.520 --> 00:01:09.140
to difference equations
00:01:09.140 --> 00:01:15.560
now in control engineering often the input
is in the form of discrete pulse of short
00:01:15.560 --> 00:01:21.590
duration and the radar tracking devices
receive such discrete pulse from the target
00:01:21.590 --> 00:01:26.630
which is which it is being which is being
tracked so difference equations is a relation
00:01:26.630 --> 00:01:32.310
between the differences of an unknown function
at one or more general values of the argument
00:01:32.310 --> 00:01:37.909
for example let us consider delta yn plus
one plus yn equal to two this delta is the
00:01:37.909 --> 00:01:44.759
forward difference and then you can also consider
delta yn plus one plus delta square yn minus
00:01:44.759 --> 00:01:50.060
one equal to one both of these equations are
difference equations these two equations can
00:01:50.060 --> 00:01:58.479
be put in the following form we know that
delta of fx the forward difference delta of
00:01:58.479 --> 00:02:03.359
fx is fx plus h minus fx
00:02:03.359 --> 00:02:14.790
so using this definition delta yn plus
one will be equal to yn plus two minus
00:02:14.790 --> 00:02:32.920
yn plus one and hence delta yn plus one
plus yn equal to two gives us yn plus two
00:02:32.920 --> 00:02:42.250
minus yn plus one plus yn equal to two which
is so alternately the difference equation
00:02:42.250 --> 00:02:47.680
one can be expressed like this and similarly
the equations two the equations two is
00:02:47.680 --> 00:03:01.290
delta yn plus one and plus delta square yn
minus one yn minus one equal to one so here
00:03:01.290 --> 00:03:07.650
we can write it as delta yn plus one is first
order forward difference so we get yn plus
00:03:07.650 --> 00:03:15.090
two minus yn plus one and delta square yn
minus one here is delta square second order
00:03:15.090 --> 00:03:23.379
forward difference so this will be delta of
delta yn minus one equal to one now delta
00:03:23.379 --> 00:03:38.970
yn minus one is equal to yn minus yn
minus one and so delta of delta yn minus one
00:03:38.970 --> 00:03:50.250
is equal to delta of yn minus yn minus one
which is equal to delta yn minus delta
00:03:50.250 --> 00:04:04.830
yn minus one but delta yn is yn plus one minus
yn and here this is yn minus yn minus one
00:04:04.830 --> 00:04:16.040
so we will get yn plus one minus two yn plus
yn minus one so let us substitute the value
00:04:16.040 --> 00:04:30.729
here so yn plus two minus yn plus one and
then plus n plus one minus two yn and then
00:04:30.729 --> 00:04:41.789
plus yn minus one equal to one so this cancels
yn plus one cancels and we get yn plus two
00:04:41.789 --> 00:04:53.450
minus two yn plus yn minus one equal to one
so the equations one and two can be alternately
00:04:53.450 --> 00:04:59.150
written in this in this forms so the equations
one and two can be also written as yn plus
00:04:59.150 --> 00:05:05.110
two minus yn plus one plus yn equal to two
and yn plus two minus two yn plus yn minus
00:05:05.110 --> 00:05:11.340
one equal to one now let us discuss how can
we determine the order of a difference
00:05:11.340 --> 00:05:12.340
equation
00:05:12.340 --> 00:05:16.370
so the order of a difference equation defined
as the difference between the largest and
00:05:16.370 --> 00:05:22.120
the smallest argument occurring in the difference
equation divided by the unit of increment
00:05:22.120 --> 00:05:31.750
so let us see here this equation this equation
which is alternate form of the first equation
00:05:31.750 --> 00:05:39.410
number one and this one and this equation
which is the alternate form of this equation
00:05:39.410 --> 00:05:46.770
so here we see that the largest argument is
n plus two ok so let me call it as equation
00:05:46.770 --> 00:05:52.530
number this is equation number three and this
is equation number four so in three largest
00:05:52.530 --> 00:06:01.039
argument is n plus two and then the smallest
argument is n divided by the unit of increment
00:06:01.039 --> 00:06:07.970
unit of increment is one we have n n plus
one n plus two so this is two so this is
00:06:07.970 --> 00:06:15.509
of second order and here the largest argument
is n plus two the smallest argument is n minus
00:06:15.509 --> 00:06:26.530
one divided by the unit of increment is one
so we get here three so equation one is of
00:06:26.530 --> 00:06:33.110
second order while equation two is of third
order or we can say equation three is of second
00:06:33.110 --> 00:06:38.550
order and equation four is of order three
now solution of difference equation so when
00:06:38.550 --> 00:06:43.940
we have a difference equation the solution
is defined as an expression for yn which satisfies
00:06:43.940 --> 00:06:48.819
the given difference equation the general
solution of a difference equation is that
00:06:48.819 --> 00:06:53.990
in which the number of arbitrary constant
is equals to the order of the difference equations
00:06:53.990 --> 00:06:59.569
so suppose we want to call this difference
equation three then its general solution must
00:06:59.569 --> 00:07:04.099
involve two arbitrary constant because here
there are where the order of the difference
00:07:04.099 --> 00:07:09.120
equation is two similarly the equation four
is of order three so the general solution
00:07:09.120 --> 00:07:15.520
must involve three arbitrary constants so
general solution will be that solution of
00:07:15.520 --> 00:07:19.669
the difference equation in which the number
of arbitrary constants is equal to the order
00:07:19.669 --> 00:07:24.680
of the difference equation now particular
solution of the difference equation will be
00:07:24.680 --> 00:07:30.129
obtained from the general solution by assigning
arbitrary by assigning the particular values
00:07:30.129 --> 00:07:37.990
to the arbitrary constants let us now discuss
how we can solve linear difference equation
00:07:37.990 --> 00:07:42.680
with constant coefficients a linear difference
equation is defined as an equation in which
00:07:42.680 --> 00:07:53.069
yn plus one yn plus two etcetera that is all
yn plus all the sequences arguments
00:07:53.069 --> 00:07:58.310
yn plus one yn plus two and so on they occur
in the first degree and they occur separately
00:07:58.310 --> 00:08:03.759
like differ linear difference equation with
constant coefficients in general will be of
00:08:03.759 --> 00:08:11.199
this form yn plus k plus a one yn plus k minus
one a to yn plus k minus two and so on a k
00:08:11.199 --> 00:08:16.280
by n equal to fn where a one a two ak are
constants
00:08:16.280 --> 00:08:23.710
so here you can see this is k th order difference
equation because n plus k s the highest or
00:08:23.710 --> 00:08:29.949
the largest argument and n is the smallest
argument so n plus k minus n divided by one
00:08:29.949 --> 00:08:36.010
unit of increment is one so it is k th order
difference equation and it is a linear difference
00:08:36.010 --> 00:08:42.300
equation because yn plus yn plus k yn plus
k minus one and so on yn they all occur in
00:08:42.300 --> 00:08:49.699
first degree and separately now if u one u
two u one if u one u two uk n are k independent
00:08:49.699 --> 00:08:57.799
solutions of this equation yn plus k a one
yn plus k minus one ak by n equal to zero
00:08:57.799 --> 00:09:04.150
so here we say that k th k th k is the order
of the difference equation k k k is the order
00:09:04.150 --> 00:09:10.339
of the difference equation so let us consider
the corresponding homogeneous part so yn plus
00:09:10.339 --> 00:09:15.870
k plus a one yn plus k minus n plus ak yn
equal to zero if u one u two u one uk and
00:09:15.870 --> 00:09:21.730
vk independent solutions of this homogeneous
equation then their linear combination c one
00:09:21.730 --> 00:09:27.569
u one plus c two u two c k u k is called the
complete solution of this homogeneous equation
00:09:27.569 --> 00:09:34.959
let us denoted by say un if vn is a particular
solution of equation one if vn is a particular
00:09:34.959 --> 00:09:40.960
solution of this non homogeneous equation
then the complete solution of equation
00:09:40.960 --> 00:09:47.350
one will be given by yn equal to n plus vn
where un is called the complementary function
00:09:47.350 --> 00:09:54.510
and vn is called the particular integral now
thus the general solution of the difference
00:09:54.510 --> 00:09:59.990
equation one is the sum of the two that
is complementary function and the particular
00:09:59.990 --> 00:10:05.810
integral the solution of one equation one
solution of one can be obtained by the classical
00:10:05.810 --> 00:10:12.580
approach like we just has been solve the
linear def linear differential equations
00:10:12.580 --> 00:10:17.080
with constant coefficients so these linear
difference equations with constant coefficient
00:10:17.080 --> 00:10:22.610
can be solved in a analogous manner by the
classical approach but here we will solve
00:10:22.610 --> 00:10:30.179
them by using the z transform so we are solving
these linear non homogeneous difference equations
00:10:30.179 --> 00:10:33.130
with by using z transforms
00:10:33.130 --> 00:10:44.720
let us see how we do this suppose we consider
so let us consider four un minus un plus two
00:10:44.720 --> 00:11:01.420
equal to zero with u naught equal to zero
and u one equal to two so we are given this
00:11:01.420 --> 00:11:07.380
difference equation you can see the largest
argument here is an plus two the smallest
00:11:07.380 --> 00:11:13.380
argument is n so and the unit of increment
is one so it is the second order difference
00:11:13.380 --> 00:11:28.030
equation second order difference equation
and they are given two initial conditions
00:11:28.030 --> 00:11:32.490
the conditions are u naught equal to zero
so u one equal to zero so we will get with
00:11:32.490 --> 00:11:37.699
these two initial conditions we shall get
a particular solution of this differential
00:11:37.699 --> 00:11:42.230
difference equation so what we do in order
to solve this difference equation what we
00:11:42.230 --> 00:11:47.799
do is we take the z transforms of z transform
of both the sides of the difference equation
00:11:47.799 --> 00:11:55.050
and then solve this solve the resulting
equation for the value of uz where z is uz
00:11:55.050 --> 00:12:02.199
is the z transform of this sequence un and
then we divide uz by z and break uz by z into
00:12:02.199 --> 00:12:08.131
partial fractions in such a way that
when we write the corresponding expression
00:12:08.131 --> 00:12:14.760
for a uz it comes in terms of the known z
transforms so that while inverting uz we
00:12:14.760 --> 00:12:17.430
get the sequence un easily
00:12:17.430 --> 00:12:23.270
so let say v v let us take the z transform
taking z transforms taking z transforms of
00:12:23.270 --> 00:12:34.640
both the sides let me call it equation one
so this equation one so taking z transform
00:12:34.640 --> 00:12:46.580
of equation one we have four times z is
a linear z is it is a linear property so
00:12:46.580 --> 00:12:58.660
four times z of un minus z of un plus two
equal to z of zero which is zero so now four
00:12:58.660 --> 00:13:06.890
z of un is four of uz uz is the z transform
this sequence un and here while writing z
00:13:06.890 --> 00:13:11.770
transform of un plus two let us recall the
shifting to the left shifting of un to
00:13:11.770 --> 00:13:24.620
the left so we know that z transform of un
plus k where k is greater than zero is equal
00:13:24.620 --> 00:13:36.910
to z to the power k and then we have uz
minus u naught minus u one by z and so on
00:13:36.910 --> 00:13:43.030
minus uk minus one over z to the power k minus
one
00:13:43.030 --> 00:13:55.000
so let us apply this formula so z square into
uz minus u naught minus u one by z equal to
00:13:55.000 --> 00:14:00.190
zero let us make use of the given conditions
u naught equal to zero un equal to two so
00:14:00.190 --> 00:14:15.620
four uz minus z square and then we get uz
minus zero minus two by z so the coefficient
00:14:15.620 --> 00:14:26.190
of z is uz is four minus z square into uz
and we get this term minus z square into
00:14:26.190 --> 00:14:37.669
minus two y z so plus two z is equal to zero
we can write it as uz equal to two z upon
00:14:37.669 --> 00:14:49.030
z square minus four now let us write uz
by z we break uz by z into partial fraction
00:14:49.030 --> 00:15:01.490
so two over z square minus four which is
two over z minus two into z plus two the partial
00:15:01.490 --> 00:15:12.290
fractions of this expression is two times
one over z minus two minus one over z plus
00:15:12.290 --> 00:15:26.730
two divided by four ok so this is one by two
one over z minus two minus one over z plus
00:15:26.730 --> 00:15:42.830
two now we can write uz from here so uz
now thus uz is equal to one by two z over
00:15:42.830 --> 00:16:01.660
z minus two minus z over z plus two and hence
un which is z inverse of uz is equal to one
00:16:01.660 --> 00:16:17.150
by two z inverse of z over z minus two m z
inverse of z over z plus two let us recall
00:16:17.150 --> 00:16:28.689
that z transform of a to the power n is z
over z minus a provided mod of z is greater
00:16:28.689 --> 00:16:35.400
than mod of a so z inverse of z over z minus
a will be a to the power n so here we have
00:16:35.400 --> 00:16:43.770
two and therefore one by two two to the power
n and for this mod of z must be greater than
00:16:43.770 --> 00:16:49.150
two and then from here we get minus two raise
to the power n
00:16:49.150 --> 00:16:54.620
so here also mod of z must be greater than
two so mod of z must be greater than two for
00:16:54.620 --> 00:17:01.010
this two we true ok so one by two two raise
to the power n minus minus two to the power
00:17:01.010 --> 00:17:07.730
n we can simplify it further two to the power
n minus one and then i can write it as plus
00:17:07.730 --> 00:17:12.400
or minus two i can take from here so minus
two to the power n minus one minus minus becomes
00:17:12.400 --> 00:17:22.200
plus so the this is minus two to the power
n minus one so this is the sequence un
00:17:22.200 --> 00:17:27.860
equal to two to the power n minus one plus
minus two to the power two n minus one
00:17:27.860 --> 00:17:36.140
which is the solution of example one now let
us take the example two so here we have this
00:17:36.140 --> 00:17:51.320
sequence un plus two plus four un plus
one plus three un equal to three to the
00:17:51.320 --> 00:17:52.450
power n
00:17:52.450 --> 00:17:59.950
again let us see n plus two is the largest
argument and n is the smallest argument
00:17:59.950 --> 00:18:05.980
here so n plus two minus n divided by the
unit of increment is one this means the order
00:18:05.980 --> 00:18:12.420
of this difference equation is n plus two
minus n divided by one so that is two this
00:18:12.420 --> 00:18:18.360
is the order of the difference equation so
this is second order difference equation
00:18:18.360 --> 00:18:26.010
let us solve this here we are given the initial
conditions u naught equal to zero and u one
00:18:26.010 --> 00:18:31.340
equal to one so the general solution of this
difference equation will involve two or three
00:18:31.340 --> 00:18:36.620
constants to determine two arbitrary constants
we are given two conditions u naught equal
00:18:36.620 --> 00:18:46.650
to one and u naught equal to zero and u one
equal to one so this will give us a particular
00:18:46.650 --> 00:18:48.220
solution of the difference equation
00:18:48.220 --> 00:18:55.620
now let us take z transform here so z transform
is a linear linear operation so z of un plus
00:18:55.620 --> 00:19:07.770
two z of un plus two plus four times z of
un plus one plus three times z of un equal
00:19:07.770 --> 00:19:15.040
to z transform of three to the power now again
let us apply this fifteen of un to the left
00:19:15.040 --> 00:19:23.790
so z of un plus two will be z square uz where
uz is z transform of the un sequence minus
00:19:23.790 --> 00:19:35.480
u naught minus u one by z plus four times
z of un plus one will be z times uz minus
00:19:35.480 --> 00:19:44.560
u naught plus three times z transform of un
is uz is equal to z to the power z of three
00:19:44.560 --> 00:19:55.470
to the power n is z over z minus three
provided mod of z is greater than three
00:19:55.470 --> 00:20:01.130
now let us use the values of u naught and
u one which are given to us u naught is given
00:20:01.130 --> 00:20:13.920
to be zero so uz minus zero minus u one is
one one over z plus four z times uz minus
00:20:13.920 --> 00:20:26.480
zero plus three uz equal to z over z minus
three let us collect the coefficient of uz
00:20:26.480 --> 00:20:43.520
so z square plus four z plus three into uz
we have and then we have minus z square by
00:20:43.520 --> 00:20:51.170
z so minus z minus z when goes to the other
side becomes plus z so z over z minus three
00:20:51.170 --> 00:21:04.230
plus z and this will give you z square
minus two z divided by z minus three now let
00:21:04.230 --> 00:21:17.270
us find u z by z from here so then uz by z
it will be equal to z minus two divided by
00:21:17.270 --> 00:21:26.360
z minus three into z square plus four z
plus three if you factorize you get z plus
00:21:26.360 --> 00:21:40.810
one z plus three so z plus one and z plus
three so now we have to break this uz by z
00:21:40.810 --> 00:21:49.890
into partial fractions and then determine
the inverse z transform so uz by z so let
00:21:49.890 --> 00:22:06.660
us write it as a over z minus three b over
z plus one c over z plus three the values
00:22:06.660 --> 00:22:17.410
of abc can be obtained now a is equal to
z minus three so z minus two divided by z
00:22:17.410 --> 00:22:26.610
plus one z plus three evaluated at z equal
to three
00:22:26.610 --> 00:22:38.530
so this will be equal to three minus two that
is one and three plus one is four three
00:22:38.530 --> 00:22:46.330
into three three plus three six so twenty
four so it is one by twenty four b will be
00:22:46.330 --> 00:22:58.740
equal to z minus two divided by z minus
three into z plus three evaluated at z equal
00:22:58.740 --> 00:23:11.110
to one minus one so we get this yes minus
three divided by minus four and here we have
00:23:11.110 --> 00:23:28.070
two so three by eight and see we will have
as z minus two over z plus z minus three in
00:23:28.070 --> 00:23:40.330
to z plus one evaluated as z equal to minus
three so this will be minus five and here
00:23:40.330 --> 00:23:56.860
we will have minus three minus three minus
six in to two so five by twelve and hence
00:23:56.860 --> 00:24:13.140
u z by z is is equal to one by twenty four
z minus three b is three by eight three by
00:24:13.140 --> 00:24:30.920
eight z plus one and then we have five by
twelve z plus three ok
00:24:30.920 --> 00:24:40.270
now we can multiply by z so this gives
u z is equal to one by twenty four z over
00:24:40.270 --> 00:24:55.260
z minus three plus three by eight z over z
plus one and five by twelve z over z plus
00:24:55.260 --> 00:25:06.440
three now we can easily invert this so let
us take the z trans inverse z transform ok
00:25:06.440 --> 00:25:19.950
so this was u n plus two here so ok z then
hence z inverse of u z is equal to one by
00:25:19.950 --> 00:25:32.680
twenty four z inverse of z over z minus three
plus three by eight z inverse of z over z
00:25:32.680 --> 00:25:49.760
plus one and then we have five by twelve z
inverse of z over z plus three again let us
00:25:49.760 --> 00:25:59.500
let us recall that u z sorry z of u to
the power a to the power n is z over z minus
00:25:59.500 --> 00:26:07.070
a where mod of z is greater than mod of a
let us apply this formula so z inverse of
00:26:07.070 --> 00:26:14.900
z over z minus three so this is one by twenty
four three to the power n where mod of z is
00:26:14.900 --> 00:26:24.380
greater than three and then three by eight
here minus one to the power a is minus
00:26:24.380 --> 00:26:29.690
one and here mod of z is greater than one
so mod of z is greater than three here here
00:26:29.690 --> 00:26:39.700
mod of z is greater than one then by twelve
minus three to the power n so here a is
00:26:39.700 --> 00:26:46.320
minus three and therefore mod of z is greater
than three so the common reason is mod
00:26:46.320 --> 00:26:59.070
of z greater than three and z inverse of u
z is u n ok so thus u n is one by twenty four
00:26:59.070 --> 00:27:07.970
three to the power n three by eight minus
one to the power n plus five by twelve minus
00:27:07.970 --> 00:27:13.230
three to the power n where mod of z is greater
than three
00:27:13.230 --> 00:27:21.590
so this is the answer to this difference equation
by taking z transform z transform takes
00:27:21.590 --> 00:27:26.330
care of the initial conditions we do not have
to use initial conditions after we have found
00:27:26.330 --> 00:27:35.059
this solution like we do in this classical
approach so here this how we apply the
00:27:35.059 --> 00:27:39.730
transform techniques we take the z transform
of the given difference equation collect the
00:27:39.730 --> 00:27:45.860
coefficient of u z rest of the terms be tend
to the right side divide by the coefficient
00:27:45.860 --> 00:27:52.330
of u z and then take the then break then
you write u z by z then these expression
00:27:52.330 --> 00:27:59.840
of u z by z is then broken in to partial fractions
in such a way that we while multiplying
00:27:59.840 --> 00:28:06.620
by z we get the expressions in terms of
the z transform of known z transforms so so
00:28:06.620 --> 00:28:13.510
that while inverting we get easily the
sequence u n so u n here can be obtained like
00:28:13.510 --> 00:28:19.270
this we will in our next lecture we shall
discuss some more problems where we will have
00:28:19.270 --> 00:28:28.000
to be very careful while writing u z by z
because there in the problems the the
00:28:28.000 --> 00:28:34.580
expressions we will have to arrange of
u z by z so that while inverting we get the
00:28:34.580 --> 00:28:41.510
sequence even very easily so we will do that
in the next lecture with that i would like
00:28:41.510 --> 00:28:42.510
to conclude my lecture
00:28:42.510 --> 00:28:42.770
thank you very much for your attention