1
00:00:46,480 --> 00:00:55,070
So, in the last class we had seen that the
Hamilton’s approach provides a nice way
2
00:00:55,070 --> 00:00:57,379
of
writing the first order differential equations.
3
00:00:57,379 --> 00:00:58,379
.
4
00:00:58,379 --> 00:01:25,470
And, in that method the Hamiltonian method,
we write first express T and V in terms of,
5
00:01:25,470 --> 00:01:34,479
of what p and q. In the Lagrangian method,
we wrote it in terms of q and q dot, so there
6
00:01:34,479 --> 00:01:49,020
lies 1 difference, then you obtain the, partial
derivatives H is just T plus V, then write
7
00:01:49,020 --> 00:01:52,820
the
Hamiltonian function H as T plus V. Then,
8
00:01:52,820 --> 00:02:01,479
you obtain delta q and once you have done
so
9
00:02:01,479 --> 00:02:15,590
or because there will be different directions
do not talk, there will be different directions
10
00:02:15,590 --> 00:02:19,800
and so you will have to take the derivative
of the Hamiltonian function with respect to
11
00:02:19,800 --> 00:02:22,980
each of these.
And then finally, once you have done that,
12
00:02:22,980 --> 00:02:27,690
the equations would nicely be expressed as
q i
13
00:02:27,690 --> 00:03:00,151
dot is equal to and p i dot is equal to minus,
so this is the essential technique. Now, you
14
00:03:00,151 --> 00:03:05,950
have seen that in some cases, these equations
will yield dotted terms in the right hand
15
00:03:05,950 --> 00:03:09,659
side, which again you have to substitute from
here. in order to express in terms of
16
00:03:09,659 --> 00:03:15,019
.undotted terms, so that is the essential
technique. But, now so far we have, solved
17
00:03:15,019 --> 00:03:25,709
problems that are you know, more of the circuit
or sort of physics oriented.
18
00:03:25,709 --> 00:03:32,920
And that might, erroneously give you an idea,
that this method is these methods, that we
19
00:03:32,920 --> 00:03:37,590
have discussed so far, Lagrangian and Hamiltonian.
These are not all that much useful in
20
00:03:37,590 --> 00:03:43,390
engineering systems, so today let us, attack
an engineering system right away, most of
21
00:03:43,390 --> 00:03:49,990
you are electrical engineers, so let us do
a electrical engineering problem, a DC machine
22
00:03:49,990 --> 00:03:51,360
driving a load.
.
23
00:03:51,360 --> 00:04:06,069
So, what is the structure of the DC machine,
you have the armature, the armature is fed
24
00:04:06,069 --> 00:04:13,749
by means of a supply a DC supply. There will
of course, be some resistance and
25
00:04:13,749 --> 00:04:19,970
inductance of the armature winding, so, let
us take that out, so that we can, directly
26
00:04:19,970 --> 00:04:28,650
see
that they are in the circuit, so that is the
27
00:04:28,650 --> 00:04:36,389
armature circuit. Assume, that the field is
constant we will later assume that the field
28
00:04:36,389 --> 00:04:42,139
is also variable, but to begin with, let us
assume that the field is constant, which means
29
00:04:42,139 --> 00:04:47,650
it could be a permanent magnet machine
or a separately excited machine.
30
00:04:47,650 --> 00:04:54,580
And you have applied a voltage say E, to this
and this is what the inductance of the
31
00:04:54,580 --> 00:05:02,180
armature, resistance of the armature. Now
here, this fellow rotates and there has to
32
00:05:02,180 --> 00:05:05,720
be a
mechanical system to, so from here, there
33
00:05:05,720 --> 00:05:30,350
is a shaft and the shaft is moving something
say a disc, with some mass against this. Now,
34
00:05:30,350 --> 00:05:33,920
this armature has a mass or moment of
35
00:05:33,920 --> 00:05:42,490
.inertia, assume that to be I 1, here there
will be a bearing and bearing will have some
36
00:05:42,490 --> 00:05:50,920
friction, so assume that to be R 1.
Similarly here, this disc will have a mass
37
00:05:50,920 --> 00:05:57,030
or moment of inertia call it I 2, and these
will
38
00:05:57,030 --> 00:06:07,380
have another bearing friction called R 2.
There is another degree of freedom, you must
39
00:06:07,380 --> 00:06:16,880
notice, that these 2 are not moving exactly
in step, there has to be a some kind of a
40
00:06:16,880 --> 00:06:24,200
springiness of the shaft, so that is represented
by a spring constant K. If it is not there,
41
00:06:24,200 --> 00:06:26,340
if
it is rigid, then they will not, have their
42
00:06:26,340 --> 00:06:29,580
own independent degrees of freedom., so in
order
43
00:06:29,580 --> 00:06:36,110
for these 2 to have their independent degrees
of freedom, this fellow should be slightly
44
00:06:36,110 --> 00:06:39,680
tortionable, so there will be a tortional
spring here.
45
00:06:39,680 --> 00:06:46,560
So, suppose this is the system, how do we
represent it, in terms of a set of differential
46
00:06:46,560 --> 00:06:54,350
equations. First let us, identify what are
the variables, this the electromechanical
47
00:06:54,350 --> 00:06:56,900
system,
the main advantage of whatever we are doing
48
00:06:56,900 --> 00:07:02,510
is that, you know the methods of writing
the differential equations for electrical
49
00:07:02,510 --> 00:07:08,030
systems, you know it for the mechanical systems
individually, but when you attack a electromechanical
50
00:07:08,030 --> 00:07:12,381
system, then these methods are
very handy, because they treat them in the
51
00:07:12,381 --> 00:07:19,410
same way, so what are the state variables,
what are the the variables in the system.
52
00:07:19,410 --> 00:07:26,521
We are trying to identify the minimum set
of coordinates, tell me what are they here
53
00:07:26,521 --> 00:07:36,900
there would be a current flowing, in our notation
we will consider a charge flowing here.
54
00:07:36,900 --> 00:07:42,800
The electrical system, will it need any other
variable, no 1 will suffice, how, how about
55
00:07:42,800 --> 00:07:48,940
the mechanical system. We will need the position
of this armature and the position of
56
00:07:48,940 --> 00:07:56,530
this that uniquely identifies, so there will
be a position coordinate q, say let us equal
57
00:07:56,530 --> 00:08:03,020
q 1,
q 2 and q 3, so these are the position coordinates.
58
00:08:03,020 --> 00:08:11,240
Position means, angle essentially and q 2
dot will be the angular velocity, q 3 dot
59
00:08:11,240 --> 00:08:20,240
will be
its angular velocity. Now, notice that the
60
00:08:20,240 --> 00:08:25,140
electrical system and the mechanical system,
how are they coupled, I am not talking about
61
00:08:25,140 --> 00:08:29,970
the physical coupling, I am talking about
the mathematical coupling between them, how
62
00:08:29,970 --> 00:08:37,719
are they coupled. The electrical system
applies a torque on the mechanical system,
63
00:08:37,719 --> 00:08:43,370
and the mechanical system responds by
applying a back EMF on the electrical system,
64
00:08:43,370 --> 00:08:48,680
that is how they coupled.
And, what are there, these couplings, the
65
00:08:48,680 --> 00:09:01,900
back EMF E b is K phi I a is q 1 dot, and
the
66
00:09:01,900 --> 00:09:07,320
torque, we will not write it as T, because
T we are representing the kinetic energy by,
67
00:09:07,320 --> 00:09:16,029
so
we will write it as say F is
68
00:09:16,029 --> 00:09:26,889
so this is the torque. Sorry, no this is my
F is K phi I a is the
69
00:09:26,889 --> 00:09:42,930
.torque, back EMF E b is K phi the speed of
rotation of this one, so
70
00:09:42,930 --> 00:09:47,550
now these 2 K phi’s
are. in fact, the same, these 2 K’s are
71
00:09:47,550 --> 00:09:50,760
in fact, the same. That is what, if you assume
that
72
00:09:50,760 --> 00:09:58,339
the power input into the, motor is equal to
the power output into, out of the motor.
73
00:09:58,339 --> 00:10:02,550
If you ignore the losses, then these 2 are
actually equal, else there will slight difference,
74
00:10:02,550 --> 00:10:08,120
which you can ignore for now. So, these 2
are the relationship of the interaction between
75
00:10:08,120 --> 00:10:14,610
the mechanical subsystem and the electrical
subsystem, now one convenient way of
76
00:10:14,610 --> 00:10:19,300
handling these would be, to assume to handle
the electrical subsystem and the
77
00:10:19,300 --> 00:10:24,620
mechanical subsystem separately. Why because
then the electrical subsystem, will be
78
00:10:24,620 --> 00:10:30,930
simply a circuit with, which sees a voltage
which is the back EMF. You can do it that
79
00:10:30,930 --> 00:10:35,959
way, a mechanical subsystem is just a the
mechanical system, which sees a torque, you
80
00:10:35,959 --> 00:10:43,360
can do it that way, first let us do it that
way, and then we will handle it as integrated
81
00:10:43,360 --> 00:10:47,160
system, so for the electrical subsystem and
the mechanical subsystem.
82
00:10:47,160 --> 00:10:48,160
.
83
00:10:48,160 --> 00:11:00,540
So, electrical subsystem, what will be the
kinetic energy, what will the potential energy,
84
00:11:00,540 --> 00:11:17,779
so we will put the subscript e. . Half q 1
dot square L a, so this is
85
00:11:17,779 --> 00:11:29,829
half L a q 1 dot square, that is the kinetic
energy, the potential energy is q 1 is a
86
00:11:29,829 --> 00:11:35,600
direction, in which E is added E is applied,
but it is acted on in the opposite direction
87
00:11:35,600 --> 00:11:40,110
by
back EMF. So, the total voltage effectively
88
00:11:40,110 --> 00:11:49,509
applied on this, circuit is E minus E b and
that as acts in the direction of E q 1.
89
00:11:49,509 --> 00:12:05,459
.So, we will have to put a minus, E minus
E b sorry, E b into q 1, the Rayleigh term
90
00:12:05,459 --> 00:12:17,009
of the
electrical subsystem is half R a q 1 dot square,
91
00:12:17,009 --> 00:12:37,689
so done. From here, we can write the
momentum, because momentum is, is
92
00:12:37,689 --> 00:12:52,600
what Lagrangian is T minus V, yes, so no only
the
93
00:12:52,600 --> 00:13:03,689
Lagrangian. So, the p 1 is the derivative
of only the Lagrangian the Rayleigh does not
94
00:13:03,689 --> 00:13:08,509
appear, so this is nothing but, the momentum
see, L is equivalent to the mass q 1 is equal
95
00:13:08,509 --> 00:13:15,369
to the velocity, so it is the electrical momentum,
so this is the momentum.
96
00:13:15,369 --> 00:13:27,199
So, now we can write your Hamiltonian function
as T plus V is equal to, so what is T
97
00:13:27,199 --> 00:13:34,940
plus V, T we substitute it here, so that we
can express in terms of p, it will be, it
98
00:13:34,940 --> 00:13:40,540
will
come twice, so 1 by twice L a in the denominator
99
00:13:40,540 --> 00:13:54,829
p 1 square, this term minus V, so is
becomes plus
100
00:13:54,829 --> 00:14:03,490
E minus E b q 1.
Student: .
101
00:14:03,490 --> 00:14:16,880
Sorry, so T plus V it is minus, you are right,
so this is the Hamiltonian function, so if
102
00:14:16,880 --> 00:14:19,610
you
write the Hamiltonian function, then the rest
103
00:14:19,610 --> 00:14:22,800
falls in place immediately.
.
104
00:14:22,800 --> 00:14:30,019
q 1 dot is the derivative of the Hamiltonian
with respect to I will write e with respect
105
00:14:30,019 --> 00:14:36,550
to p
1, that is derivative of the Hamiltonian with
106
00:14:36,550 --> 00:14:39,470
respect to p 1 is…
Student: .
107
00:14:39,470 --> 00:14:54,360
.Yes, P 1 by L a, the other one, P 1 dot is
equal to . we have to use
108
00:14:54,360 --> 00:15:05,339
this, so take a derivative of this and this,
it is . with respect to q 1 it
109
00:15:05,339 --> 00:15:22,089
will be this term, minus of that term, so
E minus E b, because it was negative here,
110
00:15:22,089 --> 00:15:26,790
minus
Student: .
111
00:15:26,790 --> 00:15:35,499
R a q 1 dot, but dot I do not want in the
right hand side, so we can substitute, E minus
112
00:15:35,499 --> 00:15:52,749
E
b minus R a p 1 by equation done. Now, in
113
00:15:52,749 --> 00:16:07,779
the mechanical subsystem
114
00:16:07,779 --> 00:16:12,279
the T, Kinetic
energy of the mechanical side, will consists
115
00:16:12,279 --> 00:16:15,959
of the 2 bodies their own kinetic energies,
so
116
00:16:15,959 --> 00:16:41,720
it is half I 1 q, q 2 yes, q 2 dot square
plus half I 2 q 3 dot square. Now, the potential
117
00:16:41,720 --> 00:16:50,940
energy would be, potential energy would be
due to the K the K term here and there
118
00:16:50,940 --> 00:16:56,249
would be the torque applied, that will be
acting in the direction of q 2.
119
00:16:56,249 --> 00:17:08,690
So, we have to take him to that into account,
so it is half K, it is q 2 minus q 3, q 2
120
00:17:08,690 --> 00:17:13,679
minus
q 3 is the torsion in that spring, that is
121
00:17:13,679 --> 00:17:20,090
squared plus F acting in the direction of
q 2, so
122
00:17:20,090 --> 00:17:36,500
minus F q 2. And the Rayleigh term is the
two friction elements, so half R 1 q 2 dot
123
00:17:36,500 --> 00:17:48,930
square plus half R 2 q 3 dot square is that
right. So, the three terms we have written
124
00:17:48,930 --> 00:17:55,460
down, now we have to express T plus V, in
terms of p and q remember that, so we have
125
00:17:55,460 --> 00:18:03,100
to find out, what is p or p 2 and p 3.
.
126
00:18:03,100 --> 00:18:15,289
So, p 3, p 2 is basically the derivative of
the, this term, derivative of the Lagrangian
127
00:18:15,289 --> 00:18:17,429
we
had written, because the let derivative of
128
00:18:17,429 --> 00:18:21,350
the potential with respect to q dot was 0,
so we
129
00:18:21,350 --> 00:18:35,770
.can write directly like this q 2 dot is I
1 q 2 dot, that is p 2. Similarly, p 3 is
130
00:18:35,770 --> 00:18:43,080
the derivative
of the this thing in terms of q 3 dot is I
131
00:18:43,080 --> 00:18:47,620
2 q 3 dot, Obviously because these are the
two
132
00:18:47,620 --> 00:18:57,700
angular momenta, now we have to express this
in terms of that, so we will have to write,
133
00:18:57,700 --> 00:19:05,390
H of the mechanical side is . we have to express
these, which
134
00:19:05,390 --> 00:19:10,480
means, these terms we will have to write in
terms of the ., that we
135
00:19:10,480 --> 00:19:20,929
have just derived.
It will be 1 by twice I 1 p 2 square plus
136
00:19:20,929 --> 00:19:32,149
1 by twice I 2 p 3 square plus the potential
term,
137
00:19:32,149 --> 00:19:58,370
plus half K q 2 minus q 3 square, then minus
no F q 2, K is only one, there is only one.
138
00:19:58,370 --> 00:20:07,720
Student: .
No, wait, wait, wait, we will do that later,
139
00:20:07,720 --> 00:20:11,970
first let us, obtain the mechanical systems
stuff
140
00:20:11,970 --> 00:20:19,090
independently and then we will substitute.
So, this is the total Hamiltonian function,
141
00:20:19,090 --> 00:20:23,190
so
we can now write q 2 dot is derivative of
142
00:20:23,190 --> 00:20:38,610
the Hamiltonian with respect to p 2, what
remains only these, so p 2 by I 1. Similarly,
143
00:20:38,610 --> 00:20:53,220
q 3 dot will be trivial is simple p 3 by I
2
144
00:20:53,220 --> 00:21:01,830
right from here, so these three things are
relatively easier, let us go to the momentum
145
00:21:01,830 --> 00:21:03,409
terms.
.
146
00:21:03,409 --> 00:21:21,070
P 2 dot that will be, let us first write and
then evaluate H m by del q 2 minus del R plus
147
00:21:21,070 --> 00:21:28,200
what, the first term when differentiated with
respect to q 2, . yields
148
00:21:28,200 --> 00:21:51,050
two terms here. So, you write down them carefully,
K minus K q 2 minus q 3, so plus F
149
00:21:51,050 --> 00:22:06,799
.yes F and this term minus R 1 q 2 dot, and
then this has to be substituted, so we will
150
00:22:06,799 --> 00:22:22,519
write minus K q 2 minus q 3 plus F minus R
1 q 2 dot is p 2 by I 1 done. The next
151
00:22:22,519 --> 00:22:36,669
equation would be p 3 dot is, is what, again
just use this with 2 substituted by 3. So,
152
00:22:36,669 --> 00:22:44,780
you
have the first equation this remains, so this
153
00:22:44,780 --> 00:22:54,740
remains only it will be K q 2 minus q 3 and
then…
154
00:22:54,740 --> 00:22:59,410
Student: .
Yes, and this Rayleigh term will remain, that
155
00:22:59,410 --> 00:23:22,450
will be minus R 2 q 3 dot, which you can
now substitute as R 2 p 3 by done. So, at
156
00:23:22,450 --> 00:23:26,750
this stage, your certain things remain, where
are
157
00:23:26,750 --> 00:23:33,929
the equations, where are the equations? Finally,
here are equations E b remained, and we
158
00:23:33,929 --> 00:23:43,000
need to substitute that, F remained we need
to substitute that. So, just substitute them,
159
00:23:43,000 --> 00:23:50,909
from whatever we had written earlier and . put
them back, the
160
00:23:50,909 --> 00:23:55,279
moment you put them back, you will need to
substitute this, because there are dotted
161
00:23:55,279 --> 00:24:00,380
terms.
So, once you substitute, can you write down
162
00:24:00,380 --> 00:24:06,690
the equations now, this term will have
another term, this term will have to be substituted
163
00:24:06,690 --> 00:24:15,370
from there, you do that, I will leave.
So, this is how, the equations for the DC
164
00:24:15,370 --> 00:24:21,700
motor could be obtained, you might argue,
that
165
00:24:21,700 --> 00:24:29,600
now here we, took the two things separately,
what if you take them as one single
166
00:24:29,600 --> 00:24:34,960
electromechanical system, cannot you do that,
yes you can do that.
167
00:24:34,960 --> 00:24:40,460
But in doing so, in when you do this this
kind of problems in exams, you might as well
168
00:24:40,460 --> 00:24:44,460
argue that I will do, it as one single electromechanical
system, but there has to be some
169
00:24:44,460 --> 00:24:49,299
cautions exercised, let me illustrate that,
because otherwise, you might commit problem
170
00:24:49,299 --> 00:24:55,950
there. Let us do it, as when we treat the
whole thing . as a single
171
00:24:55,950 --> 00:25:04,769
electromechanical system, there the variables
remain the same, but there has to be, some
172
00:25:04,769 --> 00:25:08,559
problems regarding the interaction between
the electrical part and the mechanical part
173
00:25:08,559 --> 00:25:14,929
in
terms of these, let us see. Let us do it by,
174
00:25:14,929 --> 00:25:19,389
the Lagrangian method first not going by the
Hamiltonian rule, because that is another
175
00:25:19,389 --> 00:25:22,679
method I thought, you have to learn how to
use
176
00:25:22,679 --> 00:25:23,679
it.
177
00:25:23,679 --> 00:25:24,679
..
178
00:25:24,679 --> 00:25:44,740
So, in this case the T, I will write treated
as a single electromechanical system, then
179
00:25:44,740 --> 00:25:55,100
T is
it will be the same the three kinetic energies,
180
00:25:55,100 --> 00:26:09,049
half L a q 1 dot square plus half I 1 q 2
dot
181
00:26:09,049 --> 00:26:30,611
square plus half I 3 q 3 dot square done.
Potential energy V, here no it is, what are
182
00:26:30,611 --> 00:26:33,279
what
is the total potential energy, potential energy
183
00:26:33,279 --> 00:26:39,020
will include the back EMF, potential energy
will include the torque and that is what we
184
00:26:39,020 --> 00:26:49,110
need to treat carefully.
So, first the electrical part, it is minus
185
00:26:49,110 --> 00:27:03,039
E minus E b into q 1 it appeared, now plus
half k q
186
00:27:03,039 --> 00:27:14,520
2 minus q 3 of the spring minus F q 2 this
is the torque. So, notice there was this term.
187
00:27:14,520 --> 00:27:20,740
and there was this term, that naturally appeared,
they are inside the system still they
188
00:27:20,740 --> 00:27:24,870
appeared, now we have to do something about
that, so let us first substitute, what we
189
00:27:24,870 --> 00:27:44,870
know they are, substitute it, you get minus
E minus K phi, E b is q 2 dot q 1 plus half
190
00:27:44,870 --> 00:27:55,909
k q
2 minus q 3 square minus F is k phi q 1 dot,
191
00:27:55,909 --> 00:28:13,139
K phi I a q 2, leave it like that.
And the total Rayleigh function is half R
192
00:28:13,139 --> 00:28:24,590
a q 1 dot plus half R 1 q 2 dot square plus
half
193
00:28:24,590 --> 00:28:35,990
R 2 dot q 3 dot square. Now, notice one problem,
the problem is that in the whole
194
00:28:35,990 --> 00:28:40,850
derivation we had assumed, that the v is independent
of the velocities, and they are
195
00:28:40,850 --> 00:28:48,509
dependent, now V is now dependent on the velocities.
And therefore we cannot use
196
00:28:48,509 --> 00:28:56,059
exactly the same formulation. But still we
can use the similar formulation, because we
197
00:28:56,059 --> 00:29:01,409
had proceeded up to 1 point by without assuming
that, and at some point where we had
198
00:29:01,409 --> 00:29:04,019
introduced that assumption at what point.
199
00:29:04,019 --> 00:29:05,019
..
200
00:29:05,019 --> 00:29:13,590
When we had said that we know what the delta
of the Lagrangian of the q 1 dot is, we
201
00:29:13,590 --> 00:29:24,960
said that is equal to, that is what will not
happen here, where did we use it at one stage
202
00:29:24,960 --> 00:29:34,660
we had obtained the Lagrangian equation as
d d t of the derivative of T with respect
203
00:29:34,660 --> 00:29:42,279
to q,
q 1, q i dot minus derivative of the V with
204
00:29:42,279 --> 00:29:50,020
respect to q i, plus the derivative of the
Rayleigh with respect to q i dot equal to
205
00:29:50,020 --> 00:29:57,600
0. This we said that if we can write, this
as the
206
00:29:57,600 --> 00:30:02,269
no here it was the Lagrangian, here it was
Lagrangian.
207
00:30:02,269 --> 00:30:07,450
So, there was no problem here, the problem
was here, we said that since V is
208
00:30:07,450 --> 00:30:13,320
independent of q i dot we can substitute this
as the Lagrangian, this we cannot do now.
209
00:30:13,320 --> 00:30:18,760
So, we will have to use this equation, as
the basic equation on the base basis of which
210
00:30:18,760 --> 00:30:21,290
we
obtain the equation, everything will the same,
211
00:30:21,290 --> 00:30:28,679
will be the same, go ahead do it. In terms
of this we have, we will have the second order
212
00:30:28,679 --> 00:30:36,090
equations . as T first,
the in the direction of the q 1 will be this
213
00:30:36,090 --> 00:30:46,350
derivative, it will be L a q 1 double dot,
first
214
00:30:46,350 --> 00:30:49,250
term.
Second term the derivative of the Lagrangian
215
00:30:49,250 --> 00:30:58,279
T minus V with respect to q 1, it will be
this term only other terms vanish, so it will
216
00:30:58,279 --> 00:31:26,370
be minus, E minus K phi q 2 dot, ((Refer
Time: 31:14)) yes, plus this term plus R a
217
00:31:26,370 --> 00:31:36,519
q 1 dot is equal to 0, that is the first equation.
Second equation do the same thing where q
218
00:31:36,519 --> 00:31:50,009
2 is here, q 2 is here, q 2 is here, you will
have, I 1 q 2 double dot minus in terms of
219
00:31:50,009 --> 00:31:53,780
q 2, what remains here it remains, here it
220
00:31:53,780 --> 00:32:10,629
.remains. So, write it carefully, minus K
phi q 1 dot plus K q 2 minus q 3, plus R,
221
00:32:10,629 --> 00:32:17,510
R 1 q 2
dot equal to 0.
222
00:32:17,510 --> 00:32:29,259
The third equation will be, I similarly I
2, I can write almost by similarity with this,
223
00:32:29,259 --> 00:32:36,499
q 3
double dot minus, when you do it in terms
224
00:32:36,499 --> 00:32:42,070
of . three what remains,
this remains nothing else. So I will have
225
00:32:42,070 --> 00:32:48,320
to write it carefully, it will be minus K,
q 2
226
00:32:48,320 --> 00:32:59,629
minus q 3 that does the potential part plus
the Rayleigh part, it will be plus R 2 q 3
227
00:32:59,629 --> 00:33:04,639
dot
equal to 0. So, these three are the Lagrangian
228
00:33:04,639 --> 00:33:13,429
equations, but we want to derive it in the
first order, and we have already derived it,
229
00:33:13,429 --> 00:33:16,549
we want to see that we are doing it.
.
230
00:33:16,549 --> 00:33:31,230
So, we define the potentials, P 1 is equal
to what, you have already done that, similarly,
231
00:33:31,230 --> 00:33:39,580
p
2 is equal to I 1 q 2 dot and p 3 is equal
232
00:33:39,580 --> 00:33:53,330
to I 2 q 3 dot, in terms of this. Now, once
you
233
00:33:53,330 --> 00:34:03,309
have define that, we can write the Lagrangian
equation had taken the form p 1 dot is
234
00:34:03,309 --> 00:34:10,830
equal to derivative of the Lagrangian with
respect to q 1 minus derivative of the Rayleigh
235
00:34:10,830 --> 00:34:18,790
with respect to q 1 dot. So, p 1, this we
can write in the right hand side, it will
236
00:34:18,790 --> 00:34:22,730
be p 1 dot
is derivative of the Lagrangian with respect
237
00:34:22,730 --> 00:34:31,240
to q 1, Lagrangian with respect to q 1 this
goes T minus V, so this becomes plus and this
238
00:34:31,240 --> 00:34:48,460
remains nothing else.
So, it will be E minus K phi q 2 dot minus
239
00:34:48,460 --> 00:34:59,599
the Rayleigh term R a q 1 dot, both the dotted
terms in the right hand side, so substitute
240
00:34:59,599 --> 00:35:12,110
you get, E minus K phi q 2 dot is p 2 by I
1,
241
00:35:12,110 --> 00:35:24,640
and this is here, minus R a P 1 by L a, so
that is the equation. Notice that equation
242
00:35:24,640 --> 00:35:28,710
has
come out to the same as the one that we have
243
00:35:28,710 --> 00:35:33,160
derived, using the Hamiltonian method,
244
00:35:33,160 --> 00:35:42,609
.check Similarly, p 2 dot is the derivative
of the Lagrangian with respect to q 2 minus
245
00:35:42,609 --> 00:35:51,609
derivative of the Rayleigh with respect to
q 2 dot is… It will be derivative of Lagrangian
246
00:35:51,609 --> 00:35:59,240
with respect to q 2, this term will remain
this term will remain.
247
00:35:59,240 --> 00:36:25,130
So, you will have to write it as K phi q 1
dot minus K q 2 minus q 3 minus R 1 q 2 dot,
248
00:36:25,130 --> 00:36:35,250
and then substitute these two. So, you more
or less understand. how to do it, do you,
249
00:36:35,250 --> 00:36:38,060
and
you, I would request you to check, that this
250
00:36:38,060 --> 00:36:41,770
method is yielding the same set of
differential equations, that you obtain by
251
00:36:41,770 --> 00:36:48,809
the Hamiltonian method. So, is is the point
clear, here in this case, there were some
252
00:36:48,809 --> 00:36:54,550
forces acting inside the system, the torque
and
253
00:36:54,550 --> 00:37:00,359
the back EMF, both were inside the system.
If you take the whole thing as one electromechanical
254
00:37:00,359 --> 00:37:05,500
system, it was not that somebody
else is applying that, but then those things
255
00:37:05,500 --> 00:37:11,589
are also, properly accounted for the moment,
you use the Lagrangian equation properly.
256
00:37:11,589 --> 00:37:14,720
Only thing is that, that is the character
of any
257
00:37:14,720 --> 00:37:20,300
electromechanical device, that the torque
which is the force is proportional to the
258
00:37:20,300 --> 00:37:24,040
current,
which is a velocity. So, there will always
259
00:37:24,040 --> 00:37:30,000
be velocity dependent potentials in any
electromechanical system. And therefore, we
260
00:37:30,000 --> 00:37:41,030
have to be careful about such things. In the
last class, we had started handling some electrical
261
00:37:41,030 --> 00:37:49,859
circuits there you probably noticed a
problem, let us do one electrical problem,
262
00:37:49,859 --> 00:37:52,509
in order to illustrate that carefully.
.
263
00:37:52,509 --> 00:37:58,410
.
Here we consider say a voltage source, an
264
00:37:58,410 --> 00:38:16,950
inductor, a capacitor and a resistor, simple
circuit, how will you do it E, L, C, R will
265
00:38:16,950 --> 00:38:23,180
say simple q 1 and q 2 and we can simply
266
00:38:23,180 --> 00:38:30,640
.write, it fine lets write. You will have
the Lagrangian as the kinetic energy half
267
00:38:30,640 --> 00:38:39,110
L q 1 dot
square, minus the potential energy half 1
268
00:38:39,110 --> 00:38:49,380
by 2 c q 1 minus q 2, then it will become
plus,
269
00:38:49,380 --> 00:39:04,609
E q 1, that is the Lagrangian function. And
the Rayleigh function is half R q 2 dot square,
270
00:39:04,609 --> 00:39:23,520
no problem, so we can directly go ahead.
So, along q 1 the equation would be p 1 is
271
00:39:23,520 --> 00:39:29,130
the derivative of the Lagrangian with respect
to q 1 dot, then L q 1 dot, that is the p
272
00:39:29,130 --> 00:39:36,260
1, now, if you have any difficulty ask me,
if you
273
00:39:36,260 --> 00:39:46,530
have any difficulty ask me, I am the person
to ask. So, you can write p 1, then you can
274
00:39:46,530 --> 00:40:08,130
express P 1 is this and p 2 is a problem there,
it is 0, if you do it like this q 2 dot is
275
00:40:08,130 --> 00:40:16,050
0, so p
2 is 0, That is obvious, because this line,
276
00:40:16,050 --> 00:40:19,290
does not have this particular loop, does not
have
277
00:40:19,290 --> 00:40:21,609
inductance p 2 is 0.
.
278
00:40:21,609 --> 00:40:38,350
If p 2 is 0, then the Hamiltonian is expressed
as half L Hamiltonian would be the
279
00:40:38,350 --> 00:40:44,569
potential energy plus kinetic energy, Kinetic
energy would be half L q 1 dot square.
280
00:40:44,569 --> 00:40:53,400
Now, these will have to be substituted, so
we will write it as, it is easy to write 1
281
00:40:53,400 --> 00:41:05,350
by twice
L p 1 square, using this plus the potential
282
00:41:05,350 --> 00:41:14,750
function it is 1 by twice c q 1 minus q 2
square
283
00:41:14,750 --> 00:41:23,730
minus E q 1 done, that is the Hamiltonian
function. Then we can write the equations
284
00:41:23,730 --> 00:41:33,319
directly q 1 dot is the derivative of the
Hamiltonian with respect to p 1 is simple.
285
00:41:33,319 --> 00:41:53,440
Here p 1 by L, where . I have written the
Lagrangian I have not put.
286
00:41:53,440 --> 00:41:54,720
Student: .
287
00:41:54,720 --> 00:42:02,609
.You are right, so yes whenever i make such
careless mistakes please point out, because
288
00:42:02,609 --> 00:42:20,690
that goes into the recording. So, you have
q 1 dot is this and then p 1 dot is p 1 dot
289
00:42:20,690 --> 00:42:30,780
is
minus, minus delta R delta q 1 dot is you
290
00:42:30,780 --> 00:42:39,069
can easily write now,
Student: .
291
00:42:39,069 --> 00:42:49,720
Yes minus 1 by c q 1 minus q 2 and plus E,
done, right hand side does not have any
292
00:42:49,720 --> 00:43:04,020
dotted term, so we are happy. Then along the
p q 2 direction, we have q 2 dot is q sorry,
293
00:43:04,020 --> 00:43:15,990
p 2 dot is what?
Student: .
294
00:43:15,990 --> 00:43:30,450
Yes something, so 1 by c, q 1 minus q 2 only
the sign is different minus, in this direction
295
00:43:30,450 --> 00:43:39,180
there is the Rayleigh term, so it will be
R q 2 dot. So, what are the ultimate final
296
00:43:39,180 --> 00:43:48,930
equations, it is 1, 2, 3, these 3 are the
equations.
297
00:43:48,930 --> 00:43:58,710
Student: .
Wait, I will i have not yet talked about the
298
00:43:58,710 --> 00:44:03,930
q 2 dot thing, I have written it, but wait,
q 2
299
00:44:03,930 --> 00:44:13,440
dot is the derivative of the Hamiltonian with
respect to p 2, which has turned out to be
300
00:44:13,440 --> 00:44:17,260
0,
which has or in other words, it cannot be
301
00:44:17,260 --> 00:44:22,930
evaluated, it could not be evaluated. But
still,
302
00:44:22,930 --> 00:44:34,030
where is p 2,
Student: p 2 is 0.
303
00:44:34,030 --> 00:44:46,799
p 2 is 0, if p 2 is 0 p 2 is always 0, what
is p 2 dot obviously 0, so, if p 2 dot is
304
00:44:46,799 --> 00:44:52,480
0, actually
this is not correct, this is all that we can
305
00:44:52,480 --> 00:44:56,020
say, it we could not evaluate it. If p 2 dot
is 0, we
306
00:44:56,020 --> 00:45:09,359
now have q 2 dot, from here we will be able
to write, q 2 dot is this goes in the other
307
00:45:09,359 --> 00:45:18,790
side
1 by R C can you see, yes q 1 minus q 2, that
308
00:45:18,790 --> 00:45:26,481
is it. When you write this it was actually
wrong, you could not evaluate it, from what
309
00:45:26,481 --> 00:45:32,619
you have, so we have taken a different route
and that argument was sound, since you know
310
00:45:32,619 --> 00:45:35,960
that your p 2 is 0, p 2 dot also has to be
0
311
00:45:35,960 --> 00:45:40,309
there is no change in that.
And so this term is 0, and therefore from
312
00:45:40,309 --> 00:45:53,280
there you can extract, so that gives you the
equation 3, which this equation, we had obtain
313
00:45:53,280 --> 00:45:56,589
this, now in the left hand side it is p 2
dot,
314
00:45:56,589 --> 00:46:06,740
which you argued that it is 0, so if you put
0 here, then you can extract this. See, you
315
00:46:06,740 --> 00:46:12,859
needed 3 equations, but in the circuit there
are only two storage elements
316
00:46:12,859 --> 00:46:19,520
and the minimum number of equation, that you
really need is equal to the
317
00:46:19,520 --> 00:46:24,280
number of the storage elements.
So, we had done all this exercise, but ultimately
318
00:46:24,280 --> 00:46:29,040
ended up with 3 equations, which is not
very economical, they are all right, they
319
00:46:29,040 --> 00:46:33,650
are all correct, but they are not economical.
So,
320
00:46:33,650 --> 00:46:39,079
the conclusion is that this Lagrangian, Hamiltonian
formalism, that we are taking, this
321
00:46:39,079 --> 00:46:46,089
often leads to uneconomical set of differential
equations. But they are all right, they are
322
00:46:46,089 --> 00:46:52,490
not wrong and we will have to learn, how to
obtain the correct economical set of
323
00:46:52,490 --> 00:47:00,260
differential equations.
You might argue, that this problem happened
324
00:47:00,260 --> 00:47:07,950
because we sort of blindly took the q 1 and
q 2 as the coordinates . the two loops. And
325
00:47:07,950 --> 00:47:14,180
these two loops
independently may not be the minimum or correct
326
00:47:14,180 --> 00:47:21,390
set, let us follow that line of argument,
Let us see, what where we end up.
327
00:47:21,390 --> 00:47:22,390
.
328
00:47:22,390 --> 00:47:30,710
It is the capacitor here and the resistor
here, instead of defining the loops, this
329
00:47:30,710 --> 00:47:33,460
time what
we will say, we will argue that, the current
330
00:47:33,460 --> 00:47:35,980
or the charge through the inductor is one
state
331
00:47:35,980 --> 00:47:43,970
variable q 1, and the charge through the capacitor
is another state variable q 2, you can
332
00:47:43,970 --> 00:47:50,230
do that. So, we are not using the mesh current
kind of argument, we are using a different
333
00:47:50,230 --> 00:47:59,691
argument to see whether that helps, here is
your E, here is your L, C, R then can you
334
00:47:59,691 --> 00:48:09,869
write the Lagrangian and the Rayleigh. The
Lagrangian will be in this case, half L q
335
00:48:09,869 --> 00:48:10,869
1
336
00:48:10,869 --> 00:48:23,260
.dot square due to this, now minus due to
this, it is 1 by twice c q 2 square, now it
337
00:48:23,260 --> 00:48:35,390
will not
be q 1 minus q 2 q 2 square plus E q 1.
338
00:48:35,390 --> 00:48:40,549
And the Rayleigh equation would be simply
now this term, how much current flows
339
00:48:40,549 --> 00:48:56,460
through this, q 1 minus q 2, so here it will
be half R q 1 dot minus q 2 dot square. Now,
340
00:48:56,460 --> 00:49:08,849
therefore, p 1 is the derivative of the Lagrangian
with respect to q 1, it will be L q 1 dot,
341
00:49:08,849 --> 00:49:23,700
no problem. P 2 will be derivative of the
Lagrangian with respect to q 2 is 0, so in
342
00:49:23,700 --> 00:49:26,839
this
case, also we get it 0, there is no problem
343
00:49:26,839 --> 00:49:32,340
about it is still 0.
Student: .
344
00:49:32,340 --> 00:49:44,810
Sorry, q 2 dot is 0, q 2 dot does not appear
in this equation, so if p 2 is 0 and p 1 is
345
00:49:44,810 --> 00:49:47,230
this
and then we can, we can write the Hamiltonian
346
00:49:47,230 --> 00:50:01,990
as 1 by twice L p 1 square and the rest
remains it will be plus 1 by twice c q 2 square
347
00:50:01,990 --> 00:50:16,430
minus E q 1. So, now your you can either
write q 1 dot is equal to simply p 1 by L
348
00:50:16,430 --> 00:50:21,180
from here, or you can write q 1 dot is the
derivative of Hamiltonian with respect to
349
00:50:21,180 --> 00:50:29,410
p 1, which is same thing, so it is not really
necessary to evaluate these, this automatically
350
00:50:29,410 --> 00:50:34,410
follow from the definitions of p 1 p 2. q
1
351
00:50:34,410 --> 00:50:41,350
is this, and I will not bother about the q
2 at this stage, because here there is something
352
00:50:41,350 --> 00:50:43,440
wrong.
.
353
00:50:43,440 --> 00:50:55,771
.Let us do it for, p 1 dot is minus derivative
of the Hamiltonian with respect to q 1 minus
354
00:50:55,771 --> 00:51:05,260
derivative of the Rayleigh with respect to
q 1 dot is what, . when
355
00:51:05,260 --> 00:51:08,190
you write the Hamiltonian with respect to
q 1 only this survives.
356
00:51:08,190 --> 00:51:15,710
Student: .
Only this survives yes, . write E minus.
357
00:51:15,710 --> 00:51:21,900
Student: .
Yes, the Rayleigh term survives R times q
358
00:51:21,900 --> 00:51:35,549
1 dot minus q 2 dot, and p 2 dot is same thing
with respect to q 2, it will be Minus 1 by
359
00:51:35,549 --> 00:51:45,270
c q 1 by c q 2 plus R again this term will
remain
360
00:51:45,270 --> 00:52:04,520
q 1 dot minus q 2 dot. Now, you notice that
p 2 is 0, p 2 is 0, this implies p 2 dot is
361
00:52:04,520 --> 00:52:15,240
0, if p
2 is 0 p 2 dot is 0, so you have q 2 dot is
362
00:52:15,240 --> 00:52:26,569
what, q 2 dot is, there would be, there will
be
363
00:52:26,569 --> 00:52:33,200
something in the right hand side wait, we
could not extract it from here. q 2 dot is
364
00:52:33,200 --> 00:52:35,480
the
derivative of the Hamiltonian with respect
365
00:52:35,480 --> 00:52:45,710
to p 2, so how do you write it.
Student: .
366
00:52:45,710 --> 00:52:58,849
Yes, this will not give here, you have this
will be 0 we know this and therefore, this
367
00:52:58,849 --> 00:53:02,470
is
obtained. So, we will not use this, we will
368
00:53:02,470 --> 00:53:11,220
use this 0 is equal to minus q 2 by c plus
R q 1
369
00:53:11,220 --> 00:53:24,810
dot we already know is p 1 by L minus R q
2 dot, so you have q 2 dot is, you can now
370
00:53:24,810 --> 00:53:41,690
extract it, minus q 2 by R C plus p 1 by L,
so this is how. Ultimately, what are the
371
00:53:41,690 --> 00:53:51,250
equations you have, ultimately what are the
differential equations you have, tell me.
372
00:53:51,250 --> 00:54:07,799
This is this is definitely 1.
Student: .
373
00:54:07,799 --> 00:54:10,920
This is another.
Student: .
374
00:54:10,920 --> 00:54:19,789
Yes, so you still ended up with 3 equations,
so this actually, did not help and we need
375
00:54:19,789 --> 00:54:27,680
to
still refine this and in fact, the proper
376
00:54:27,680 --> 00:54:31,630
methods, you know, that the for electrical
circuits,
377
00:54:31,630 --> 00:54:37,089
which are pure electrical circuits, we have
the Kirchhoff’s laws for defining how the
378
00:54:37,089 --> 00:54:43,569
currents and voltages are distributed and
their relationship. We will show then that
379
00:54:43,569 --> 00:54:46,130
can
be used in order to formulate the minimum
380
00:54:46,130 --> 00:54:50,440
set of equations, but that will also tell
us how
381
00:54:50,440 --> 00:54:55,690
.to formulate the correct variables for this
formulation. So, that is what we will take
382
00:54:55,690 --> 00:54:58,349
up
from the next class, typically for electrical
383
00:54:58,349 --> 00:55:01,750
circuits, how do we, obtain the minimum set
of differential equations.
384
00:55:01,750 --> 00:55:04,170
That is all, thank you.
385
00:55:04,170 --> 00:55:04,170
.