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In the previous class, we discussed the analysis
of a simplified model of a power system. In
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fact, the model we chose was very simplified
one of our single machine connected to constant
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voltage source and our focus was on the analysis
of that system. We will continue that today.
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So, todays lecture is titled the analysis
of dynamical systems. We continue with our
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analysis.
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In the previous class if you recall, we discussed
the concept of equilibrium or equilibria of
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dynamical systems, states, small and large
disturbance stability and we began on this
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example which I just talked about, the single
machine connected to a voltage source. In
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today’s class, we will try to focus on large
disturbance behavior of this particular system
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and then, we will turn our attention to the
general analysis of linear system which is
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certain class of systems. We considered the
toy model last time of a power system. So,
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let us just recap quickly on what we did last
time.
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So, we had single machine connected to a voltage
source. This was a stiff voltage source or
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we call it an infinite bus. The synchronous
machine was modeled as a voltage source behind
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transient reactants. The phase angle of the
voltage source was in fact delta which is
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related to the position of the rotating machine.
So, the simplified model, we did not drive
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this model. I just gave it to you. This is
in fact a very simplified model as I mentioned
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in the previous lecture. It gives the motion
of the rotor angle and the rotor speed deviations
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from omega naught which is the frequency of
this infinite bus. This particular simplified
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model, what we did was first of all we tried
to study the small disturbance stability of
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this particular system.
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So, the first step which we took was finding
out the equilibria of the system. In fact,
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equilibrium is defined as a point or the values
of the states for which the derivative of
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the states become 0 and that yielded us two
equilibrium points. Both have omega e equal
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to omega. This comes out of this particular
equation and delta e was sine inverse of this
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quantity. This value of delta e results in
this derivative become equal to 0.
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Now, between 0 and 180 degree, they could
be two equilibria. So, if this is for example,
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P m mechanical power which you assume to be
constant, then there two equilibria delta
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1 and delta 2. One is less than 90 degrees
and the other one is greater than 90 degrees.
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Of course, you could have this continuing
and you could actually have many more equilibria,
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but you noticed that they are spaced 200,
that is 360 degrees away from these. So, actually
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they are indicative of the same position of
the rotors.
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So, we will not talk of equilibria beyond
180 degrees. We said upon ourselves to find
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out the stability of the system and we to
some extent did an analysis for small disturbance
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around these two equilibrium. We need to really
understand the large disturbance behavior.
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In fact, large disturbance behavior really
is something which is important in the sense,
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that in the first lecture I told you of the
phenomena of loss of synchronism. It is in
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fact a manifestation of the non-linear behavior
of the system. So, small disturbance stability
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around equilibria is actually easier to analyze
and easier to you know describe. So, that
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is what we did in the last class. So, just
again looking at what we did.
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We assumed that we had small deviations around
the equilibrium. So, we assume that delta
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omega and delta delta are very small, P m
is a constant. Of course, I mentioned it some
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time back. Then, we got this particular model,
where k was E s E cos delta e by x by delta
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e was an equilibrium value. So, if you are
studying deviations around an equilibrium
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delta e omega e, then k would be this.
So, after that we obtained the response for
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small disturbance in fact whether a system
is going to be stable or not after you given
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it a disturbance, that is, whether it is going
to come back to the equilibrium is determined
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by the response of the system. So, response
of this particular system was written down,
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in fact, it was guessed by analogy with a
spring mass system. We did not actually derive
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it. We will do that of course for generally
linear systems in the following lectures.
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So, response for small disturbance is given
by this. You notice it is an oscillatory response,
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where omega n is dependent on this value of
k and h. It is a natural frequency of the
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oscillation in delta and omega. So, for small
disturbance is around the equilibrium, we
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have got an oscillatory response, just one
small point. This assumes that k is greater
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than 0. So, this is true actually. This response
is true provided k is greater than 0. There
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is an oscillatory response if k is greater
than 0, that is, delta e less than 90 degrees.
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The values of A and phi incidentally are obtained
from the initially value of delta delta and
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delta omega.
Note that whenever we are talking of a stability
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of a system, we are really discussing the
situation when we are away from the equilibrium.
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We are not at the equilibrium. If you are
at the equilibrium, of course, you simple
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stay there. The point is that you are given
a push from the equilibrium due to some disturbance.
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So, this disturbance could be things like
step change in the voltage of the infinite
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bus or something small. So, we will not discuss
what the disturbance is actually, but we will
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assume that it has been given a disturbance
around equilibrium and of course, once you
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get disturbed, you will oscillate around the
equilibrium provided your original equilibrium
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was less than 90 degrees, that is delta e
was less than 90 degrees. This is true for
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the first equilibrium, that is, delta e is
delta 1 and omega e is omega 0.
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So, this is the first equilibrium. When I
say first equilibrium, I am talking of this
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particular equilibrium point, delta 1 which
is less than 90 degrees. Of course, point
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which may occur to you, would you call this
oscillatory behavior is stable or unstable.
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The thing is that if you look at the response
here, it just continues the system, just continues
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to oscillate around the equilibrium. So, would
you call it as stable or unstable response?
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Actually, we would like the system to come
back to the equilibrium. So, if you just oscillate
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around the equilibrium, you would call it
marginally stable system. It is not really
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coming back to the equilibrium nor it is going
away from the equilibrium with time.
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So, of course, if there is some damping in
the system, the oscillation will die on it.
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I am not proving this here, but you can really
imagine it that if there is some kind of viscous
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damping or something in the system, it will
be kind of oscillation will die down with
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time. In fact, in real system, in a real power
system where in you model everything including
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the damp providing the field winding, there
is usually some damping in the system. In
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fact, if you got control systems associated
with the control of field voltage, you even
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can make damping negative. That is the oscillation
which grows in time.
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So, this is an issue which will handle later
on. Right now, we have not considered any
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damping in our model. It is very simple model.
We are not going to get any damping. It is
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just going to be an oscillation around the
equilibrium. If you are near the equilibrium
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delta 1 delta is equal to delta 1, then that
is what you will expect. Later on, we will
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come across more detail models of synchronous
machines in which damping is present; some
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kind of damping will be there.
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The other equilibrium delta 2 which is greater
than 90 degrees, which is that unstable point
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k is greater less than 0. Why is it unstable?
That is because the response turns out to
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be this and one component, aware this omega
is real one of the component aware is going
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to be greater than 0, that is this or this
can be greater than 0. So, what will happen
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is that you will have a combination of growing
and decaying terms in general. Of course,
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there is an issue that will be k 1 or k 1
and k 2 are non zero. The answer is of course,
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k 1 and k 2 are dependent on the initial value
of deviations and for more situation, most
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initial conditions, k 1 and k 2 will be non
zero.
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So, what you will have is you will have a
combination of growing in decaying terms.
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The growing terms effectively mean that if
I give disturbance from the equilibrium, you
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are going to move away from the equilibrium
and this is what happens at this particular
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equilibrium point. So, this is what we discussed
in the previous lectures.
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One important point which you should note
when we are talking of operating it unstable
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equilibrium, you cannot actually operate an
unstable equilibrium part. If you give push,
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you are going to go away from the equilibrium
points, any small portion. So, it is small
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disturbance unstable system and if you got
small disturbance unstable system, you cannot
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operate at all. So, you have to be, you really
cannot operate at this equilibrium delta 2
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omega e is equal to omega naught. So, this
is an important point which you should keep
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in mind. For example, the earlier equilibrium
was to be unstable and that also would not
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be feasible to operate there either. So, this
is an important thing. A system should be
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small disturbance stable if you want to operate
it that point.
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So, just to summarize, if delta e is delta
1, omega is omega naught, you get oscillatory
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response for a small disturbance. So, this
is very important. The other equilibrium could
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have growing terms. So, it is an unstable
system at the other equilibrium delta is equal
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to delta 2. For large disturbances, we cannot
make any approximation based on omega or delta
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omega delta delta, delta being small. So,
we cannot assume that the disturbance is small
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if the disturbance is not, rather if the deviation,
initial deviation is small. If the deviation
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from the equilibrium is not small, you cannot
assume that sine of the deviation is equal
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to the deviation. That is what basically we
did in when we got the linear approximation.
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This particular approximation which we got
was based on the fact that sine delta delta
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was approximately equal to delta delta. This
was approximated as a key approximation and
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cos delta delta was approximately equal to
1. This is true only for small disturbance.
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For large disturbances, we cannot use this
model.
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So, we have used the full grown model in which
sine delta is retained. If you look at this
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particular term here, you can look at it as
some kind of accelerating torque or accelerating
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power. We are talking in terms of per unit
and will assume speed deviations from the
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equilibriums are not too large.
So, we can per unit torque and per unit power
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are in fact almost equivalent. So, we will
just take this approximation. So, if you take
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this accelerating power here, it is accelerating
power or approximately the accelerating torque.
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You will notice that as compared to this in
this particular system, this is a linearise
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system; we are assuming that the accelerating
torque is professional to the angular deviation
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here. If you do not assume small disturbance,
this expression does not tell you that, it
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tells you that. It is depended on the sign
of the angle.
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So, if you look at how this particular accelerating
power or accelerating torque looks like. If
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you look at the accelerating torque or accelerating
power, if you are at the equilibrium, that
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is delta delta is equal to 0, the act the
equilibrium delta 1, the accelerating torque
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is 0, but it becomes negative, that is P m
minus E S E sine delta by x becomes negative
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if delta delta goes away from the equilibrium
and delta delta is positive. If it is negative,
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this is positive. So, it goes negative.
So, in some way, if delta delta is greater
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than 0, you will find if there is some tendency
to get you back to the equilibrium because
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the torque is restorative in nature. That
is very important point like spring mass system.
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You take a spring mass system and you give
it push from the equilibrium. If it is a positive
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push, the spring is stretched and it has a
tendency to pull it back. So, nature of the
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forces is restorative. Convulsively, if you
push the spring, the spring is compressed
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and it has a tendency to push you back to
the equilibrium. So, we can say that the torques
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or the forces are restorative in nature in
this particular case, but look at this carefully.
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It is negative, all right, but after the point,
it becomes positive.
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So, if you look at this particular system,
it becomes positive after some time. For this
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spring mass system, if you assume that the
spring has got you know is ideal, then you
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can stretch it and force is always professional
to the stretch. This is not true here. You
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will find it accelerating torque upon becoming
negative. It is restorative till this point
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and thereafter, it becomes positive. So, for
small disturbances, we can take it as restorative,
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but for large disturbance, it is not correct
to assume it is restorative.
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So, in fact, when we did the linearise analysis,
what we effectively took was the slope of
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this particular curve at this point. So, this
slope was the k which we talked about sometime
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back. So, for small disturbances, you can
use this particular straight line with a slope
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of minus k, but for large disturbance, it
is not correct to assume that the torque are
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the accelerating torque may not be restoring
after some time. Now, the fact that the torque
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deviates from this linear or this line has
got important consequences.
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In fact, it goes; it kind of comes down after
some. It gives the maximum and comes down
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after sometime becomes negative. You will
find that if the disturbance is large enough,
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it is very interesting thing. You have got
for the small, for small deviations the force
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is almost propositional to the stretch, the
restorative force, but if you give the large
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enough disturbance, it will start to move,
it will move and then, the restorative force
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suddenly start coming down and becoming negative.
Now, the restoring force becomes negative,
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the mass will never come back to the equilibrium.
So, what we should do? You know if the spring
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stretches up to a point, if it stretches up
to a point at which this restoring force is
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become equal to 0, then you may not come back
to this old equilibrium again. So, that is
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the important thing.
So, if there is a large disturbance, you may
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or may not come back to the equilibrium. You
may actually just go away and you may just
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become unstable. So, that is an important
point. Interesting point is that the restoring
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torque in this particular system becomes negative
when the angular deviation reaches; angular
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deviation becomes delta 2 minus delta 1, that
is, delta becomes delta 2. So, this particular
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point, we should ensure that we can kind of
formulate kind of rule.
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So, if I give disturbance because of which
the angular deviation is greater than 0 and
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they are not small and if this speed deviation
from omega naught does not become 0 before
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delta 1 becomes equal to, sorry delta become
equal to delta 2 or delta delta equals delta
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2 minus delta 1, then you have reached a place
where torques are no longer restorative and
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you may become unstable. So, that is the basic
reason why for large disturbance is a synchronous
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machine connected to a voltage source or later
on, we will see connected to other synchronous
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machines could go unstable.
Now, the whole catch is after a disturbance
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are you going to be stable or not is something
which depends on the response of the system
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because the rule says that is the speed going
to be equal to 0 before the torques become
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non restorative. So, the thing is now you
have to actually see what the response of
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this speed is, when it will become 0 and so
on. So, that is something which is complicated.
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So, let us just take example of large disturbance.
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You have got synchronous machine connected
to another voltage source. It is connected
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by two lines. On one of the lines, there is
a three phase fault. So, you can have a three
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phase fault on a transmission line. If there
is a three phase fault, there is a fault current
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in fact, the voltage here if it is a three
phase bolted fault, then you will find at
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voltage at dips down to 0. There will be large
fault currents and typically, there will be
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release in the system and circuit breakers.
So, release will detect that there is a fault
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in this system in this part of the transmission
system and trip these circuit breakers. So,
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these lines get isolated. Suppose, fault you
just have this particular line remaining in
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the system, this line is gone off. So, this
is the typical stability question which you
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may come across for studying large disturbances,
that is there is a fault like large disturbances
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like a fault and because of that fault, there
is a significant deviation in these angles
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and the speeds from the equilibrium. Are we
going to come back to an acceptable equilibrium
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after the fault is clear?
So, this is the post fault system. So, are
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we going to come to the post fault equilibrium?
That is the question which we need to ask.
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So, just putting this let us try to formulate
this as a kind of a problem.
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At time before t is equal to 0 and t 1, the
system says in the pre-fault condition. So,
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this is in the pre-fault condition. In the
pre-fault condition, we are operating at delta
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1 with a superscript P. So, this is the equilibrium
point when you have fault. Let us say, this
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fault occurs at t 1, the electrical power
becomes 0. Why does it become 0? Because the
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terminal voltage is dropped down to 0. We
will assume up three phase fault. So, the
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voltage drops down to 0. So, electrical power,
this is in fact electrical power p e, it becomes
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0. So, there is suddenly mechanical power
become much greater than the electrical power
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and you will find the machines accelerate.
Delta and omega will suddenly start increasing.
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At time t is equal to 0, the fault is clear
by the fact that release detect the fault
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and trip the faulted component that is a transmission
line at t is equal to 0. So, at t is equal
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to 0, the electrical power will become this.
This is the electrical power p e, the equilibrium.
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Now, r delta 1 and delta 2. Remember that
P e, the electrical power after the fault
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is different from the pre-fault electrical
power because you have tripped one line. Now,
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your electrical power initially was E S E
by x pre-fault sine delta and now, it is E
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S E x sine delta. In the pre-fault system
x p superscript P is given by x 1 x dash plus
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x e by 2. After the fault is clear, it is
simply x dash plus x e.
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Now, of course, we have these equations. These
equations defined the movement of the rotor
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angle and we also see that due to a fault,
your angles and speeds have become greater
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than 0. They have deviated from the equilibrium.
So, there is a, there are substantial deviation,
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there is a substantial deviation from the
equilibrium.
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Now, how do you know whether the system is
going to be stable or not? Well, you could
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numerical integrate the system. So, if you
numerically integrate the system, you can
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actually find out. So, numerical integration
means this particular integration is performed
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by discretizing this set of equations. How
to do this is something we will discuss later
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in the course, but soon enough. If you do
a numerical integration of the system, I will
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just show you know numerical integration.
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If the clearing time of the fault that is
a fault duration, that is t 1 to t 0, that
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is the fault duration is small, the behavior
may mix that after small disturbance behavior.
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It is oscillatory. If the fault deviation
is increased, you will find that the oscillation
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magnitude increases and it still looks like
a sign wave, that is, what small disturbance
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is like, but if now the fault clearing time
and therefore, the deviation from the equilibrium
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increases beyond the point, you will find
that the nature of the curve is like this.
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It is no longer looking like a sign wave.
This is in fact numerical integrated solution
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of the system. So, actually use a numerical
integration technique to find out this. How
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00:24:25,601 --> 00:24:33,090
to divide it? Let me assure you we will learn
it sometime soon in this course.
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One important point is if the clearing time
becomes greater than the certain value, you
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become, you loss synchronism. You become unstable;
you are no longer going to come back. This
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00:24:44,110 --> 00:24:49,730
angle just grows with time. So, this is a
typical situation in which because of a large
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00:24:49,730 --> 00:24:53,860
disturbance, we are not going to reach equilibrium.
If the fault duration is large, then this
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00:24:53,860 --> 00:24:59,710
is what will happen. Remember the equilibrium
is stable. You know small disturbance stable,
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00:24:59,710 --> 00:25:05,460
but rather we can say that it involves oscillatory
behavior around the equilibrium, but for large
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00:25:05,460 --> 00:25:08,720
disturbances, it never reaches the equilibrium
again.
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So, the question is of course, this is why
do we need to do non-linear, why we need to
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do numerical integration in order to find
out this particular response? The answer is
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this. Whenever you have a non-linear system,
unfortunately the response cannot be written
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down in terms of simple function which you
know. So, this is not true for linear system.
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00:25:34,880 --> 00:25:39,980
So, if you take linear system, you can write
it in terms of it turns out that response
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00:25:39,980 --> 00:25:49,100
is in fact in all linear systems, you will
find that response contains terms like this,
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00:25:49,100 --> 00:26:01,850
e raise to lambda t, e raise to minus lambda
t, t e raise to lambda t sin. So, typically
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00:26:01,850 --> 00:26:06,960
linear systems are super position of terms
of this kind. You will find at this is how
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00:26:06,960 --> 00:26:12,890
most linear systems. So, you take out any
linear systems response and you find out this,
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00:26:12,890 --> 00:26:19,170
but for non-linear systems, we cannot write
down the response in terms of simple well
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00:26:19,170 --> 00:26:25,860
known functions and as a result of which you
may have in other than in very very special
253
00:26:25,860 --> 00:26:31,510
cases, you may actually have to, you can find
out the answer only if you numerically integrate
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00:26:31,510 --> 00:26:34,660
the systems.
So, the response in general for a non-linear
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00:26:34,660 --> 00:26:38,970
system for example, the one we are talking
of for large disturbance, we have to consider
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00:26:38,970 --> 00:26:44,900
the non-linear systems. We will have to obtain
from doing a numerical integrating. So, we
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00:26:44,900 --> 00:26:48,980
need to rely on a computer, typically if the
system is very large. So, that is one of the
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problem issue we will face in this particular
course that even if you get the model to get
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00:26:53,890 --> 00:26:59,050
to infer the responses is not always very
easy. In fact, when you talk of a system which
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00:26:59,050 --> 00:27:05,350
has got hundred of generators and tens of
thousands of lines and buses and it is non-linear
261
00:27:05,350 --> 00:27:11,460
system to infer the behavior of the system
becomes very tough because the response is
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00:27:11,460 --> 00:27:18,210
complicated function. It can only be obtained
by doing computer study of the system and
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00:27:18,210 --> 00:27:22,350
that is this little bit worry some actually,
philosophically worry some also. It basically
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says that unless you work out or really use
very high level of complicated computation,
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you cannot actually infer how the system behaves
and that is although you know the model.
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So, this is very interesting because you know
the physics of the system. The physics is
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00:27:37,840 --> 00:27:41,790
very well known, but still you do not know
how the system is going to behave unless you
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00:27:41,790 --> 00:27:46,890
actually do study which probably requires
a lot of computing resources. This is of course,
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00:27:46,890 --> 00:27:53,180
not a system like this. A single machine infinite
bus can be you know you can even do numerical
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00:27:53,180 --> 00:27:58,370
integration without the lot of computer resources,
but realistic system requires lot of computation.
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So, that is something which is philosophically
interesting about the behavior of system.
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00:28:03,240 --> 00:28:09,660
So, even if you know the physics of the problem,
you know the equation has described each particular
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00:28:09,660 --> 00:28:14,080
component. Once you integrate all of them,
thousands of such components, how the behavior
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00:28:14,080 --> 00:28:16,180
will require a lot of computation.
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Interestingly enough, this particular system,
this you know this particular system I mean
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00:28:26,100 --> 00:28:33,530
only this system, in fact only this system
model system are similar equations. Even though,
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00:28:33,530 --> 00:28:42,050
computation of response requires numerical
integration techniques stability evaluation,
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although it is not easy for this particular
problem, it can be obtained. For example,
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00:28:46,820 --> 00:28:54,480
look at this particular problem of a ball
in a valley. We have talked about this problem
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00:28:54,480 --> 00:29:02,570
in the previous section, previous class. I
know because of certain disturbance, this
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00:29:02,570 --> 00:29:07,110
particular ball has deviated from the equilibrium.
It has got displaced by equilibrium by certain
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00:29:07,110 --> 00:29:13,270
amount and it has also acquired some velocity
because of the disturbance. Then, the question
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00:29:13,270 --> 00:29:19,990
is, is this ball going to roll back or no?
So, one way of finding out you can make rule
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00:29:19,990 --> 00:29:26,030
that if this ball speed becomes 0 before it
reaches to this point, then it will roll back.
285
00:29:26,030 --> 00:29:32,380
So, at the limiting condition, the ball will
just go right up to this, just reach here.
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00:29:32,380 --> 00:29:37,030
It will just become, it will reach speed equal
to 0 just at this point. This is a kind of
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00:29:37,030 --> 00:29:43,240
limiting condition. Intuitively in this particular
case, if I tell energy is conserved is another
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00:29:43,240 --> 00:29:45,440
way of finding out whether this is going to
happen or not.
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00:29:45,440 --> 00:29:49,860
So, I am not going to evaluate the response,
but still from a certain criterion I can tell
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00:29:49,860 --> 00:29:55,770
you whether system is going to be stable or
not. If I compute kinetic in energy and potential
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00:29:55,770 --> 00:30:05,570
energy of this ball at this point at this
initial condition, if it is greater than just
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00:30:05,570 --> 00:30:10,760
the potential energy at this point. This you
know we assume kinetic energy is 0 at this
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00:30:10,760 --> 00:30:14,450
point, then it will just be stable, that is
the speed becomes 0. Just at this point it
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00:30:14,450 --> 00:30:19,220
will be stable. So, the criterion is with
a kinetic and potential energy here is equal
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00:30:19,220 --> 00:30:25,420
to the potential energy at this point. I will
call it delta 2. If it is greater, then it
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00:30:25,420 --> 00:30:31,880
is unstable, but if it is equal to you will
find the ball just goes and sits here. This
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00:30:31,880 --> 00:30:38,360
is of course true only if energy is conserved.
So, I am talking of very limited kind of situation,
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very specific situation. In fact, in our under
graduate courses we learned about a criteria
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called equal area criteria. In fact, it is
based on this particular feature.
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00:30:51,530 --> 00:30:58,070
So, the only difference is that will have
to define what is conserved actually. You
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00:30:58,070 --> 00:31:03,840
can actually show that w which is equal to
this particular quantity half of 2 H by omega
302
00:31:03,840 --> 00:31:11,930
B omega square minus P m into delta plus this
into cos delta is the quantity which is conserved.
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00:31:11,930 --> 00:31:20,620
How can you find it? Just try to evaluate
d w by d t. So, shall we do that? Yeah. So,
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00:31:20,620 --> 00:31:33,990
this is what it is. So, d w by d t is equal
to 2 H by omega B into omega minus omega naught
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00:31:33,990 --> 00:31:50,870
into d omega minus omega naught by d t minus
P m into d delta by d t plus, well minus of
306
00:31:50,870 --> 00:32:03,260
E s E sine delta or x b delta by d t and d
delta by d t of course, is equal to omega
307
00:32:03,260 --> 00:32:10,961
minus omega naught from our equations.
So, what we have is d omega by d w by d t
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00:32:10,961 --> 00:32:44,640
is nothing, but 2 H by omega B d omega minus
omega naught by d t minus P m minus this
309
00:32:44,640 --> 00:32:50,100
and this, there is some mistake in the signs.
So, obviously I have written something wrong
310
00:32:50,100 --> 00:33:01,901
here. This should be minus and this should
be plus, so minus. You look at this. So, if
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00:33:01,901 --> 00:33:11,990
I define my w to be this, this is equal to
0. Why it is equal to 0? It is because of
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00:33:11,990 --> 00:33:17,340
the second differential equation. This we
know that this is equal to this.
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00:33:17,340 --> 00:33:23,870
So, what we have is in this particular system?
This single machine connected to an infinite
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00:33:23,870 --> 00:33:33,850
bus. This is conserved. So, that is it is
a constant. So, if time t is equal to 0, your
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00:33:33,850 --> 00:33:41,290
deviations are delta in delta delta delta
omega, that is, the values of delta 0 and
316
00:33:41,290 --> 00:33:47,480
omega 0 is these. Then, the condition for
stability is that if the energy evaluated
317
00:33:47,480 --> 00:33:57,760
at the fault clearing time is greater than
the energy evaluated at delta 2 omega 0 because
318
00:33:57,760 --> 00:34:04,320
omega, this is omega 0. First term will become
0. So, basically we are using the same criteria
319
00:34:04,320 --> 00:34:11,419
as we are using here. So, in fact, this is
the basis for equal area criterion which we
320
00:34:11,419 --> 00:34:16,280
have learned in our under graduate years.
321
00:34:16,280 --> 00:34:25,200
Now, we will have quick recap. The aim of
this particular, this couple of lectures which
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00:34:25,200 --> 00:34:29,400
we are having is actually you talk about general
system. So, I have taken in example of single
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00:34:29,400 --> 00:34:35,750
machine infinite bus system because we will
get a kind of bird view of power system dynamics.
324
00:34:35,750 --> 00:34:40,490
We do not get lost in the detail of general
dynamical systems, but of course, study of
325
00:34:40,490 --> 00:34:45,950
that is interesting and important. So, just
a quick recap. A single machine connected
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00:34:45,950 --> 00:34:50,000
to an infinite bus with two equilibria. One
of the equilibria if you give small disturbance,
327
00:34:50,000 --> 00:34:55,300
you get oscillatory behavior. The other one
is unstable for small disturbance. You cannot
328
00:34:55,300 --> 00:35:00,250
operate there at all.
For the first equilibria of large disturbance,
329
00:35:00,250 --> 00:35:07,310
you can become unstable, but large disturbance
behavior requires you to consider the non-linear
330
00:35:07,310 --> 00:35:12,920
behavior of the system, that is the restoring
torques is non linear functions of the angular
331
00:35:12,920 --> 00:35:19,030
deviation. So, what you have is basically
large disturbance behavior is much more interesting
332
00:35:19,030 --> 00:35:22,880
in the sense that it shows that if you got
a large enough disturbance you know no longer,
333
00:35:22,880 --> 00:35:28,911
it is an oscillatory response, but something
like an unstable response and actually, this
334
00:35:28,911 --> 00:35:34,600
unstable response in this particular situation
is in fact the loss of synchronism phenomena
335
00:35:34,600 --> 00:35:40,010
which you have learned before. So, if you
recall in the first lecture, I said that machines
336
00:35:40,010 --> 00:35:43,541
which are connected to each other, synchronous
machine connected to another synchronous machine
337
00:35:43,541 --> 00:35:48,230
or synchronous machine connected to a good
voltage source, it tends to remain in synchronism
338
00:35:48,230 --> 00:35:55,030
with that with the rest of the system. So,
if you give a small push disturbance, it tends
339
00:35:55,030 --> 00:36:01,440
to oscillate and come back to the equilibrium,
but if you give the large disturbance, it
340
00:36:01,440 --> 00:36:06,410
may lose synchronism.
So, one important point is the inherent characteristic
341
00:36:06,410 --> 00:36:11,590
of the system is like this. The physics of
the system is such that it behaves in such
342
00:36:11,590 --> 00:36:15,780
a way. So, if you give push for small disturbance
and it will oscillate and come back to the
343
00:36:15,780 --> 00:36:20,200
equilibrium. If there was a large push and
it will lose synchronism. So, this is as much
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00:36:20,200 --> 00:36:27,950
true for this toy example as it is for large
system. Of course, before we get lost into
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00:36:27,950 --> 00:36:35,400
a lot of mathematical detail involving you
know variables like x and y, let us look at
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00:36:35,400 --> 00:36:39,320
you know what happen in the large system.
If you got very large system, you got many
347
00:36:39,320 --> 00:36:46,560
machines connected to each other. So, what
happens for large disturbance in large system?
348
00:36:46,560 --> 00:36:51,540
The thing is that you have this loss of synchronism
phenomena, but as I mentioned in the first
349
00:36:51,540 --> 00:36:57,470
lecture, you have got groups of machines,
their angles, they move together. Their angles
350
00:36:57,470 --> 00:37:03,461
move or increased relative to the angles of
another group of machines or many other. They
351
00:37:03,461 --> 00:37:10,270
can be three groups of machines also. You
know this particular phenomenon occur in realistic
352
00:37:10,270 --> 00:37:15,560
large power systems and you know it is not
easy to analyze. It is ok to analyze this
353
00:37:15,560 --> 00:37:20,030
small system by equal area criterion, but
it turns out to be a big challenge to analyze
354
00:37:20,030 --> 00:37:26,390
large non-linear system.
So, coming back again, linear systems we are
355
00:37:26,390 --> 00:37:32,310
not actually defined linear systems in a very
regressive fashion. In this course, in fact
356
00:37:32,310 --> 00:37:37,400
a lot of things are invited, you have not
defined. So, for example, linear systems are
357
00:37:37,400 --> 00:37:43,190
the equation of this kind. Wherever constant
coefficient, the rate of change of a state
358
00:37:43,190 --> 00:37:48,980
is equal to a constant coefficient into a
state, of course, this is single order system.
359
00:37:48,980 --> 00:37:53,560
This first order system you can have higher
order system as well which is of two sets
360
00:37:53,560 --> 00:38:01,140
of states. Non-linear systems, this function
can be a bit more complicated. It is not simply
361
00:38:01,140 --> 00:38:05,970
a constant coefficient into the state. It
can be a complicated function like sine x
362
00:38:05,970 --> 00:38:09,650
as we have shown in single machine infinite
bus system.
363
00:38:09,650 --> 00:38:15,360
So, for example, you can take higher order
system as well. For example, this is the higher
364
00:38:15,360 --> 00:38:22,150
order non-linear system rate of change of
x 1 is the coupled system. Rate of change
365
00:38:22,150 --> 00:38:30,300
of x 1 is minus x 1 into x 2 plus x 1 square
x 2 itself. The rate of change is again dependent
366
00:38:30,300 --> 00:38:36,250
on x and so, this is the example of non-linear
system. You can have similar coupled equations
367
00:38:36,250 --> 00:38:43,510
of order 3, 4, 5000, one million and so on.
A linear system, a coupled linear system can
368
00:38:43,510 --> 00:38:47,910
be of this form. In fact, when you are talking
of a linear system, this is how it is. In
369
00:38:47,910 --> 00:38:53,970
fact, in these two systems are not given an
input. You can have of course another input
370
00:38:53,970 --> 00:38:59,550
quantity plus something into u and this also
could be a function of u as well which is
371
00:38:59,550 --> 00:39:05,210
an input like P m is an input in our single
infinite system, infinite bus system. So,
372
00:39:05,210 --> 00:39:09,680
you could have input as well, but we do not
have an input, it is called an autonomous
373
00:39:09,680 --> 00:39:17,020
system. So, these two are autonomous system
without any input u.
374
00:39:17,020 --> 00:39:23,990
In general, linear system can also you know
if you are having a higher order linear system,
375
00:39:23,990 --> 00:39:33,380
good way of writing this is A into x. So,
A is a matrix, x is a vector, x is this, d
376
00:39:33,380 --> 00:39:38,450
x by d t is also a vector of the individual
derivative of the states. So, this is the
377
00:39:38,450 --> 00:39:50,370
general way of writing a higher order linear
system. A higher order non-linear system would
378
00:39:50,370 --> 00:39:58,280
be d x by d t or we call it x dot because
x dot is easier, saves less, saves of bit
379
00:39:58,280 --> 00:40:09,080
of space. A non-linear system could be f x
u, where x is a vector, f is also vector.
380
00:40:09,080 --> 00:40:19,890
So, you have got in fact like x 1 dot is equal
to f 1 x 1, x 2, u 1, u 2 and x 2. This is
381
00:40:19,890 --> 00:40:31,280
the possible system f 2, x 1, x 2, u 1, u
2. So, this is a coupled non-linear system
382
00:40:31,280 --> 00:40:36,850
of two states, having two states.
383
00:40:36,850 --> 00:40:43,600
Now, linear system when we are talking of
maybe, you know may come across physical system
384
00:40:43,600 --> 00:40:47,610
which is inherently linear. That can happen.
Most physical system is silently turned out
385
00:40:47,610 --> 00:40:52,580
to have some non-linearity or other, but we
can actually have some systems which are inherently
386
00:40:52,580 --> 00:40:59,600
linear. For example, I can design a system
which is for all particle purposes linear.
387
00:40:59,600 --> 00:41:05,760
Some systems are inherently linear, other
linear systems are obtained. So, when you
388
00:41:05,760 --> 00:41:11,040
really come across a linear system, they may
as I said it may inherently linear or it may
389
00:41:11,040 --> 00:41:17,740
arise to an approximation.
For example, original non-linear system when
390
00:41:17,740 --> 00:41:25,160
we consider small deviation from equilibria
of non-linear systems, we end up with linear
391
00:41:25,160 --> 00:41:32,950
system. So, this is what in fact we did for
our single machine infinite bus system. We
392
00:41:32,950 --> 00:41:43,860
started off with a non-linear system. This
is the non-linear system because you are analyzing
393
00:41:43,860 --> 00:41:50,380
small disturbances; we could get a small disturbance
model around an equilibrium point. So, if
394
00:41:50,380 --> 00:41:55,840
this is the linear system and that was the
non-linear system, so linear system are obtained
395
00:41:55,840 --> 00:42:02,640
as approximation, that is, small deviation
from the equilibria of non-linear systems.
396
00:42:02,640 --> 00:42:08,810
Of course, there is actually a formal procedure
to get a linear system, a linearised system
397
00:42:08,810 --> 00:42:14,340
of originally non-linear system. So, if you
have got x dot is equal to this is a non-linear
398
00:42:14,340 --> 00:42:23,460
system; you assumed as small deviation from
the equilibrium. You substitute this and this
399
00:42:23,460 --> 00:42:32,220
into this, get a Taylor series approximation.
This particular term is equal to 0 because
400
00:42:32,220 --> 00:42:39,350
at equilibrium by the definition of equilibrium
x dot should be equal to 0. So, at equilibria
401
00:42:39,350 --> 00:42:43,800
x dot should be equal to 0. Therefore, f of
x e u e should be equal to 0.
402
00:42:43,800 --> 00:42:54,550
So, what you get eventually is delta x is
the partial derivative of f evaluated at the
403
00:42:54,550 --> 00:43:01,130
equilibrium point. It is a very important
evaluating at this at equilibrium point plus
404
00:43:01,130 --> 00:43:09,200
d f by d u partial derivative of f, sorry
doe f by doe u evaluated at the equilibria.
405
00:43:09,200 --> 00:43:14,140
So, of course, you know this is an approximation
because in neglecting higher order terms in
406
00:43:14,140 --> 00:43:20,620
this Taylor series expansion. So, the higher
order terms like delta x square delta u square
407
00:43:20,620 --> 00:43:26,110
extra those kinds of terms we are actually
neglecting. So, what we end up is linearised
408
00:43:26,110 --> 00:43:33,490
system with the understanding that it will
probably give you good results or give you
409
00:43:33,490 --> 00:43:37,250
correct results near about the equilibrium
point.
410
00:43:37,250 --> 00:43:41,860
So, if disturbance is a small, this should
give you good idea. Actually, this should
411
00:43:41,860 --> 00:43:48,460
be delta x dot. This should give you a good
idea how the system is going to behave. In
412
00:43:48,460 --> 00:43:58,210
fact, there are you know kind of exceptions
to this particular rule.
413
00:43:58,210 --> 00:44:05,640
For example, if you try to apply this kind
of procedure to evaluate the small disturbance
414
00:44:05,640 --> 00:44:15,500
stability of the system x dot is equal to
minus x square around the equilibrium x is
415
00:44:15,500 --> 00:44:23,690
equal to 0. See actually this as an equilibrium
x e is equal to 0. That is obtained simple
416
00:44:23,690 --> 00:44:27,670
by putting the derivative here equal to 0.
So, how do you get the equilibrium z is equal
417
00:44:27,670 --> 00:44:34,150
to 0, x e becomes equal to 0. So, the equilibrium
of this is this. So, if you linearise the
418
00:44:34,150 --> 00:44:42,750
system, you will have x e plus delta x dot
which is nothing, but delta x dot because
419
00:44:42,750 --> 00:44:48,220
x e is value. So, derivative of it does not
make any sense. In fact, it is 0. This is
420
00:44:48,220 --> 00:44:58,870
equal to minus of x e plus delta x square.
So, that becomes equal to this is equal to
421
00:44:58,870 --> 00:45:14,730
x e minus of x e square plus 2x e delta x
plus delta x square. Now, we will neglect
422
00:45:14,730 --> 00:45:21,390
this term. I told you we will neglect higher
order terms of small disturbances when we
423
00:45:21,390 --> 00:45:29,740
evaluate small disturbance stability. X e
because, actually it should be like this x
424
00:45:29,740 --> 00:45:36,810
e is equal to 0. So, the small disturbance
model of this system is by the procedure I
425
00:45:36,810 --> 00:45:41,230
have told is in fact, since x is equal to
0, it is 0.
426
00:45:41,230 --> 00:45:47,490
So, it says that if you have, if you got small
deviation from the equilibrium, there is no
427
00:45:47,490 --> 00:45:53,160
movement. This is clearly not true. So, this
is an exception to what I said, the formal
428
00:45:53,160 --> 00:45:58,650
procedure which I told you before. So, there
are certain systems in which you cannot apply
429
00:45:58,650 --> 00:46:03,940
this formal procedure in order to tell how
the behaviour for small disturbance is going
430
00:46:03,940 --> 00:46:12,620
to be. For small disturbances in fact, this
particular system if x is negative, then you
431
00:46:12,620 --> 00:46:19,470
will get this x dot is positive. So, if you
got this is 0, the equilibrium if you are
432
00:46:19,470 --> 00:46:23,800
here, this system tends to move to this equilibrium.
So, in fact, if you start from here, you may
433
00:46:23,800 --> 00:46:30,940
actually come back to this equilibrium. If
you are here on the other hand, if x is positive,
434
00:46:30,940 --> 00:46:36,100
x square is positive minus x square is positive
is negative and you will come back to this,
435
00:46:36,100 --> 00:46:42,210
I am sorry I made a mistake here.
If x is negative, you tend to move away from
436
00:46:42,210 --> 00:46:49,320
the equilibrium. So, if x is negative since
minus x square is negative, you move further
437
00:46:49,320 --> 00:46:54,870
away from the equilibrium. So, actually the
non-linear behavior, the small disturbance
438
00:46:54,870 --> 00:47:01,290
behavior of this system does not seem to be
obtained from the simplified system which
439
00:47:01,290 --> 00:47:07,660
we obtain by using the procedure which are
described in the previous slide. So, obviously
440
00:47:07,660 --> 00:47:14,410
there are some exceptions to try to understand
the small disturbance behavior by neglecting
441
00:47:14,410 --> 00:47:20,280
higher order terms, but these exceptions are
very very rare and for all practical purpose,
442
00:47:20,280 --> 00:47:26,610
for almost of the system, you are going to
encounter in fact, this simplification will
443
00:47:26,610 --> 00:47:30,250
tell you everything about the small disturbance
behavior of this system.
444
00:47:30,250 --> 00:47:36,930
So, please do not get roughly too much by
the example which I gave, that is the pathological
445
00:47:36,930 --> 00:47:42,380
example. For most systems by this formal procedure,
you can get the small disturbance behavior
446
00:47:42,380 --> 00:47:43,960
of the system.
447
00:47:43,960 --> 00:47:47,770
One important point to get the equilibrium,
we need to solve this particular equation.
448
00:47:47,770 --> 00:47:53,800
If this is the non-linear equation, in that
case, you will have to use numerical technique
449
00:47:53,800 --> 00:48:00,230
like Newton-Raiffeisen or Gauss-Seidel method
to obtain x e and u v. So, even this is an
450
00:48:00,230 --> 00:48:06,070
interesting step when you are going to linearise
the system in order to obtain the small disturbance
451
00:48:06,070 --> 00:48:11,200
model.
So, just to recap, a linear system may be
452
00:48:11,200 --> 00:48:19,720
obtained from a non-linear system and that
linear system is obtained by neglecting the
453
00:48:19,720 --> 00:48:25,890
higher order terms. In most situations, the
system so obtained will give you the correct
454
00:48:25,890 --> 00:48:30,350
small disturbance behavior for most system
which is going to be that way.
455
00:48:30,350 --> 00:48:41,030
Now, coming to a very important topic The
analysis of general linear systems. So, we
456
00:48:41,030 --> 00:48:50,220
are going to talk about general linear systems.
Now, if you look at this particular example
457
00:48:50,220 --> 00:48:56,400
x dot is equal to a x, the solution of it
is very simple. You can verify that x of t
458
00:48:56,400 --> 00:49:03,920
is equal to e raise to a t x 0. You take the
derivative of this.
459
00:49:03,920 --> 00:49:12,420
It turns out to be a into x. So, obviously
this is a solution of this system. Now, another
460
00:49:12,420 --> 00:49:19,500
consistency check issue is that add x is t
is equal to 0 is x is equal to x. So, if I
461
00:49:19,500 --> 00:49:26,970
told you that at time t is equal to 0, x is
equal x 0, this also should be satisfied as
462
00:49:26,970 --> 00:49:32,220
far as the response is concerned. So, your
plugging t is equal to 0. You will find at
463
00:49:32,220 --> 00:49:38,500
x of t at t is equal to 0 is equal to x of
0. So, this is also consistent.
464
00:49:38,500 --> 00:49:46,970
So, this is the solution of this system for
this initial condition. The interesting thing
465
00:49:46,970 --> 00:49:52,000
about the linear system which I have shown
you here is that first of all, response is
466
00:49:52,000 --> 00:49:57,940
written nicely in terms of function we fairly
well, we understand fairly well that is if
467
00:49:57,940 --> 00:50:05,320
e is greater than 0, you will have growing
response is less than 0, your oscillation
468
00:50:05,320 --> 00:50:15,530
died out with time. So, if I start off from
system from an initial condition x of 0, if
469
00:50:15,530 --> 00:50:22,320
A is less than 0, you will find it is going
down to 0 which incidentally is the equilibrium
470
00:50:22,320 --> 00:50:25,490
value. The equilibrium value of the system
is x is equal to 0.
471
00:50:25,490 --> 00:50:33,440
So, if A is less than 0, this is how the response
is going to be. If A is greater than 0, this
472
00:50:33,440 --> 00:50:39,260
is how the response is going to be. Just from
the value of A, you can tell whether this
473
00:50:39,260 --> 00:50:43,400
system is stable or not. So, if I start from
initial condition which is not equal to the
474
00:50:43,400 --> 00:50:49,850
equilibrium and A is less than 0, you are
stable if system comes back to the equilibrium.
475
00:50:49,850 --> 00:50:53,890
If it is not true, of course A is greater
than 0, you are going to go away from the
476
00:50:53,890 --> 00:50:59,260
equilibrium. So, this is unstable, this is
stable. Now, so this is what I meant when
477
00:50:59,260 --> 00:51:03,960
I said that linear systems are a bit easy
to understand. The nature response is or in
478
00:51:03,960 --> 00:51:07,680
terms of well known functions. You look at
this system now.
479
00:51:07,680 --> 00:51:13,350
This is the second order system. The response
is very easy to evaluate because x 1 is dependent
480
00:51:13,350 --> 00:51:18,530
on x 1 and x 2 dependent just on x 2. So,
the solution of this system is very easy.
481
00:51:18,530 --> 00:51:24,670
You just have to apply this individually whatever
I did in the previous slide to the individual
482
00:51:24,670 --> 00:51:32,240
states, but what if I have system which is
coupled. This is second order coupled system.
483
00:51:32,240 --> 00:51:38,170
In that case, the solution no longer seems
easy in the sense that it does not, you know
484
00:51:38,170 --> 00:51:42,690
I cannot infer right away. This is certainly
not a solution of this. You can just plug
485
00:51:42,690 --> 00:51:47,790
in the value of x 1 and x 2 in this system
and you can find out that this is not going
486
00:51:47,790 --> 00:51:53,470
to be solution of this.
So, the solution of this requires you to do
487
00:51:53,470 --> 00:52:00,340
bit of algebra and actually, by trying to
understand this particular solution, we introduce
488
00:52:00,340 --> 00:52:07,670
ourselves to very very important tool of engineering
which is the idea of transformation. So, I
489
00:52:07,670 --> 00:52:15,650
will give you the kind of simple system in
which a transformation can be used in order
490
00:52:15,650 --> 00:52:21,280
to make it simplified. So, the idea of a transformation
is something which I will discuss more in
491
00:52:21,280 --> 00:52:26,190
detail in the next class more formally, but
just look at this simple system.
492
00:52:26,190 --> 00:52:34,590
X 1 dot is equal to this and x 2 is this some
system I want to analyze. This is a coupled
493
00:52:34,590 --> 00:52:39,119
system. Again the solution is not obvious.
Just by in fraction is not easy to find out.
494
00:52:39,119 --> 00:52:45,150
I can write this in this fashion is a matrix,
x 1 dot is equal to x 1 plus 0.5 x 2, x 2
495
00:52:45,150 --> 00:52:53,850
dot is equal to 0.5 x 1 plus x 2. So, this
is what I have written in this matrix form.
496
00:52:53,850 --> 00:52:59,340
Now, to analyze this system, let us just do
one thing. We do not look at how x 1 and x
497
00:52:59,340 --> 00:53:05,280
2 individually behave. We rather look at how
the difference and sum of these states behave.
498
00:53:05,280 --> 00:53:12,790
So, I just subtract these two equations in
a linear system and subtract this, simply
499
00:53:12,790 --> 00:53:18,240
subtract in this. What I get and when I add
them up, this what I get? So, for this particular
500
00:53:18,240 --> 00:53:22,910
system, I am not saying it looks neat, right
but this is only for this particular system.
501
00:53:22,910 --> 00:53:27,190
I have given this simplified system so is
to introduce to you the idea of a transformation.
502
00:53:27,190 --> 00:53:33,300
So, instead of looking at variables x 1 and
x 2, look at these variables. So, what does
503
00:53:33,300 --> 00:53:41,250
it lead you to? It leads you to x 1 minus
x 2 is 0.5 into this, x 1 plus x 2 dot, sorry
504
00:53:41,250 --> 00:53:51,690
x 1 minus x 2 dot is this, x 1 plus x 2 dot
is this. So, if I call this is y 1, this variable
505
00:53:51,690 --> 00:53:58,210
x 1 minus x 2 as y 1, this is what you get
and the solution in y 1 is very very simple,
506
00:53:58,210 --> 00:54:03,000
right. It is it has become decoupled system
like the one we have considered some time
507
00:54:03,000 --> 00:54:06,560
ago.
So, this leads you to a simple solution. Of
508
00:54:06,560 --> 00:54:11,370
course, I have not got the solution in terms
still of x 1 and x 2. We will do this formally
509
00:54:11,370 --> 00:54:20,890
in the next class. So, this is something we
will do. It will lead us to the general analysis
510
00:54:20,890 --> 00:54:23,080
of linear systems.
511
00:54:23,080 --> 00:54:30,890
One small point I wish to make here is that
if you got a system x ex dot is equal to ax,
512
00:54:30,890 --> 00:54:40,780
your solution is x of t is equal to e raise
to a t x of 0. If your system is x dot is
513
00:54:40,780 --> 00:54:48,090
equal to ax by bu, what is the general solution?
The answer is that you can write down the
514
00:54:48,090 --> 00:54:54,830
solution again very neatly in the sense that
you can easily infer the properties of the
515
00:54:54,830 --> 00:54:59,619
response, but we will do this in the next
class. So, we will leave two things for the
516
00:54:59,619 --> 00:55:08,080
next class. One is a more formal way of understanding
the linear transformations to analyze these
517
00:55:08,080 --> 00:55:14,690
linear differential equations and also, what
happens when you got you know some input,
518
00:55:14,690 --> 00:55:20,760
some forcing function like u. So, these two
things we will do in the next class.
519
00:55:20,760 --> 00:55:26,180
In the next few classes, we will kind of review
a lot of analysis and we will be doing more
520
00:55:26,180 --> 00:55:32,280
of analysis of general systems, but all these
things which you are going to do are going
521
00:55:32,280 --> 00:55:38,450
to be useful later when we apply them to realistic
power systems and that is something let us
522
00:55:38,450 --> 00:55:44,270
keep it at the back of our mind. Sometime,
we will also find out that some of the modeling
523
00:55:44,270 --> 00:55:49,580
principles also require you to do a bit of
analysis. That is why I have decided to tell
524
00:55:49,580 --> 00:55:56,960
you a bit about analysis first before going
into modeling. That is a reason why we have
525
00:55:56,960 --> 00:56:03,430
gone into the analysis of general systems.
With that we, this is as far as this particular
526
00:56:03,430 --> 00:56:07,569
lecture goes. We will meet next time. Thank
you.