WEBVTT
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This is lecture 4 of module 1. In this lecture,
we will discuss some concepts in nonlinear
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systems analysis, specifically, the Lyapunov
based approach, the reason being that we will
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use this concept to derive some new training
algorithm for neural network based control
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schemes. Hence, I thought that we must discuss
this subject of nonlinear system analysis.
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You should have some background before you
can actually appreciate what we teach in this
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course.
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In this course, we will introduce what is
nonlinear system, what is linearization, Lyapunov
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stability theory and some examples. What you
are seeing is.ů This is our general state-space
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model ľ nonlinear system model. x dot is
equal to f t x u. Hence, it is explicit in
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time as well as function of x as well as input
u. This is a state variable model and output
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is again a nonlinear function h. You see x
is a vector, f is a vector, y is a vector
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and h is also a vector. This is a multi-input,
multi-output system. When I say this is nonlinear,
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what does it mean? Of course, the first thing
that you already know is that a linear system
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means it follows the principle of superposition
and a nonlinear system is expected to not
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follow the principle of superposition. Let
me for your benefit compare. A linear system
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is different from a nonlinear system in terms
of superposition principle.
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This is a linear system and this is a nonlinear
system. Here, this is my system, which is
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linear. If I give u1, I get y1. In this case,
if I give alpha u1, then also I get alpha
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y1; same system, say some system G. Now, if
I give another input u2 to the same system
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G, I get y2 and finally, the superposition
principle is I give alpha1 u1 plus alpha2
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u2 and I find alpha1 y1 plus alpha2 y2. A
nonlinear system is just the opposite. For
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example, if I give u1, I get y1, same system
G nonlinear. When it is a nonlinear system
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and if I give an input alpha1 u1, the output
is y and y is not equal to alpha1 u1.
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Similarly, if I provide these two inputs alpha1
u1 plus alpha2 u2 to the same nonlinear system,
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the output y is not equal to alpha1 y1 plus
alpha2 y2. This is the major difference between
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a nonlinear system and a linear system as
far as differences are concerned, but a nonlinear
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system is more than justů. It does not follow
the superposition principle; it has an even
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broader perspective and some of these perspectives
are summarized here.
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We will just discuss how a nonlinear system
has multiple equilibrium points ľ we will
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shortly discuss this. It has limit cycles.
If you give a specific sinusoid signal of
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a specific frequency, then in a linear system,
the output will have the same frequency whereas
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a nonlinear system may produce subharmonic
oscillations of frequencies. Also, the nonlinear
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system can behave or can exhibit chaos. Chaos
means its steady state behavior is highly
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unpredictable and random. The nonlinear system
also executes multiple modes of behavior and
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that is called bifurcation. That is one of
the examples.
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This is the example of an autonomous system,
a nonlinear system x dot equal to f x u. Autonomous
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means it does not explicitly depend on time
and y is h x u. This is the generalized state-space
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model of a nonlinear system that is autonomous.
Some systems that have been very widely studied
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in nonlinear system are known as affine systems.
They have a specific structure. In our control
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system design, many times we will discuss
this kind of system while designing a controller,
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that is, x dot is equal to f x plus g x into
u. This is a nonlinear function and this is
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another nonlinear function, but you can see
that this has a specific structure. When you
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have no external input, then x dot is equal
to f x ľ this is an unforced system.
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What is the equilibrium point? At equilibrium
point, x dot, the dynamics, becomes 0 ľ the
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derivative of the states becomes 0; that is
the equilibrium point. If xe and ue correspond
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to a specific equilibrium point, you replace
and then equate to 0. You will find what the
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equilibrium point is.
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What is linearization? Linearization is the
process of replacing the nonlinear system
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model by its linear counterpart in a small
region about its equilibrium point. We linearize
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because we have well-established tools to
analyze and stabilize the linear system.
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Let us try to understand the basic process
of linearization. Let us write the general
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form of a nonlinear system x dot is equal
to f x u as this. We have x1, x2 ů xn ľ
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there are n states. So dx1 upon dt is f1 of
x1, x2 ů xn and u1 to um. We have m inputs
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and n outputs. This is the generalized way
to write, a very explicit way to write this
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compact notation x dot equal to f x u. f is
a vector. You can see f1, f2 ů fn.
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Let x1e, x2e ů xne be a point of equilibriumů
the point of equilibrium xe and ue, where
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ue is a vector ľ u1e, u2e ů ume and xe is
x1e, x2e ů xne. We have xe and ue. This is
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an equilibrium point and this holds true:
f xe ue is 0.
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Then, we perturb this equilibrium state by
allowing x equal to xe plus delta x ľ little
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perturbation from xe and similarly, a little
perturbation from ue by delta u. Then, Taylor's
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expansion yields a veryů. . We can write
dx upon dt is f and this is your x, this is
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your u due to perturbation and we expand this
using Taylor's expansion. First, this is your
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equilibrium point, the operating point and
then del f upon del x, you know f is a vector,
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so x is also a vector.
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If you differentiate a vector with respect
to a vector, you get a matrix. So, del f upon
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del x into delta x and this matrix is computed
at the value x equal to xe and x equal to
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ue. These are first-order terms and this is
the operating point or equilibrium point.
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This is del f upon del u and you compute this
ľ again, another Jacobian matrix; you have
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to compute at xe and ue and multiply by delta
u. This is the first-order expansion plus
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second order, third order and higher orders.
But we only go up to the first-order expansion
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of Taylor series because that is where we
can apply our linear system theory. If you
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go to second order, again the system becomes
nonlinear. It is not always true that we can
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always approximate a nonlinear system around
an equilibrium point using first-order Taylor
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series expansion ľ there is some limit on
that. Let us compute and see what the Jacobian
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matrix is.
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del f upon del x. As I told you, this is the
matrix. The easiest way to remember is that
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the first vector in f1 is differentiated with
respect to each element of the x vector and
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that is in the first row. Similarly, in the
last row, the last vector of f, is the n th
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vector of f, is differentiated with respect
to all the elements of x and that is the last
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row. You can easily remember this: del f1
by del x1 to del f1 upon del xn; similarly,
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del fn upon del x1 to del fn upon del xn.
These values have to be computed by replacing
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x as xe. So x has to be xe and u has to be
ue. This will be a constant matrix xe. Once
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you replace xe and ue, this becomes a constant
matrix. Once you replace xe and ue, this becomes
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a constant matrix. Similarly, you can compute
also del f upon del u ľ similar principle.
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Obviously, you can see that this is n into
n matrix and this is n rows and m columns,
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so n into m matrix. These are all Jacobian
matrices.
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Let us note here that dx upon dt is dxe upon
dt plus d delta x upon dt, because x is xe
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plus delta x. When I differentiate x with
respect to dt, I differentiate xe with respect
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to dt and delta x with respect to dt, but
you can see that xe is an equilibrium point,
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and at that point, x dot is 0, the derivative
is 0, so this is 0. You only get this term:
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d delta x upon dt. Furthermore, f xe ue is
also 0 because that is how you computed xe
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and ue ľ by making it 0. At the equilibrium
point, x dot (the derivative) is 0.
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We have already computed in the previous slide.
This matrix is actually matrix A and this
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matrix is matrix B. As we defined, A is del
f upon del x and B is del f upon del u. We
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have neglected the higher order terms and
we arrive at the linear approximation ľ d
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delta x upon dt is A delta x plus B delta
u. This gives sometimes we also say small
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signal models, that is, the behavior of a
nonlinear system around an equilibrium point,
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where the dynamics is assumed to be linear.
Hence, the dynamic is in the linear state-space
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model delta x dot equal to A delta x plus
B delta u.
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Similarly, we had P outputs. For P outputs,
the functions are h1, h2 ů hP. In vector
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notation, you can write y equal to h x u.
We can again use Taylor's series expansion:
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y equal to ye plus delta y. That is being
perturbed ľ equilibrium point ye with delta
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y. Then, we have the actual system is y equal
to h x u. Using Taylor's series, we can always
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write delta y is C delta x plus D delta u,
where C and D are again Jacobian matrices
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(computed). Can you guess now what is C and
what is D? As we have already said, C has
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to obviously be del h upon del x. This has
to be a P into n matrix. Similarly, D is del
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h upon del u and this has to be a P into m
matrix. That is how we also linearize the
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output y around the equilibrium point ye.
ye corresponds to xe and ue. At value xe,
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ue, the system response is ye. So, ye is the
system response when the system state is xe
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or system is at the equilibrium point. Let
us take an example.
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You can see here that this is a simple scalar
differential equation. x is a scalar ľ one-dimensional,
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simply one-dimensional; it is a scalar quantity.
Here is a scalar differential equation but
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nonlinear because of this term x square. We
can verify that this equation will not follow
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the principle of superposition. You linearize
it about the origin and you can easily see
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that if I write f xe as 0, then you will find
when you make minus x plus x square equal
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to 0, this will lead you to x is 0. So origin
equilibrium point; x = 0 is an equilibrium
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point. Linearize it about the origin and you
get x dot equal to minus x. The solution is
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x t equal to x0 e to the power of minus t.
It is very simple because this is a simple
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linear differential equation: x dot is equal
to minus x. I hope that you are very well
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aware on how to get the solution of this.
You can verify this now.
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You put x0 e to the power of minus t here,
differentiate it and at x of 0, when t equal
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to 0, put this value x0 and you see this equation
is fine. Whatever may be the initial state
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x0, the state will settle at x t = 0, because
whatever the initial condition, when you perturb
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the system from its initial position, the
equilibrium point x0 will always come to the
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origin. The state will settle at x t. This
is your equilibrium point, which is the only
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equilibrium point that this linearized system
has. Let me again clarify what we actually
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did.
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We had x dot equal to minus x plus x square.
First, we found out the equilibrium point,
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that is, f xe is 0, which is minus x plus
x square. So x = 0 is an equilibrium point.
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At the equilibrium point, the linearized equation
isů if you follow the linearization scheme,
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it is minus x. This is your linearized system
around the equilibrium point x = 0. The solution
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we found out here is x t is x0 e to the power
of minus t ľ that is the solution to the
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equilibrium point.
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Now, if I actually take the exact solution
of x t dot, which is minus x plus x square,
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if I find the actual solution of this nonlinear
system, then x t equal to x0 e to the power
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of minus t upon 1 minus x0 plus x0 e to the
power of minus t. You can try it. It is an
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exercise for you. This is a simple nonlinear
system, for which we find a closed loop solution
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and this is the closed loop solution for this
nonlinear solution. In general, we cannot
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express the solution of a nonlinear system
in a closed form but because this is a simple
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one, we can actually find the closed form
solution of this nonlinear system.
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Earlier, we showed that the origin was one
of the equilibrium points, but there is another
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equilibrium point, that is, when I solve minus
x plus x square is 0, you have x of x minus
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1 is 0. This implies x = 0 and x = 1. There
are two equilibrium points. We talked about
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linearizing the system around only one equilibrium
point x = 0, origin, but there is also another
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equilibrium point x = 1. Let us see the system
behavior around these equilibrium points.
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This is a plot. You can see this is one equilibrium
point. What we are showing is this is the
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time axis and this is your x t ľ we are showing
the response. The value of x t in the beginning
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t = 0ů. We can start either near the first
equilibrium point, that is, 0 ľ this particular
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point and another is 1 ľ this is another
equilibrium point. If I give the initial condition
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as ?0.5, you see they all settle. ?0.5 and
+0.5, wherever I give in this zone, they are
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all going towards 0.
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Finally, the system at t tends to infinity,
goes back to the equilibrium point 0, that
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is, if the system is disturbed slightly from
its equilibrium point 0, it is near the origin,
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it goes back to the origin, whereas if you
look here, this is a little interesting. One:
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if you go back here, it diverges from 1. Very
near it diverges, but surprisingly what happens
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is that after it goes to a very high value,
it comes back and then again goes back to
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0. Similarly, we can see this one. Just after
1, it goes and again from here to here, it
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jumps and then again goes to 0.
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Similarly, another point here, this particular
line and from here, it jumps here and again
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goes to 0. What you are seeing is that as
I said in the beginning, the equilibrium point
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1 is actually an unstable equilibrium point.
From here, the system diverges from this point,
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whereas if you start somewhere near the origin,
then you could disturb the system near the
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equilibrium point, which is the origin. Then,
it goes back to the origin. The conclusion
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is x = 0 is a stable equilibrium point, whereas
x = 1 is an unstable equilibrium point. At
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the end of the lecture, we will discuss a
little more about these equilibrium points
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ľ we will solve through examples but let
us discuss some notion of stability in a nonlinear
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system. What we talked about now is that a
nonlinear system can be linearized around
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an equilibrium point. Some equilibrium points
are stable equilibrium points and some equilibrium
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points are unstable equilibrium points. Now,
what is the concept of stability on a nonlinear
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system?
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In linear system, we had only one equilibrium
point. We never talked about locally stable,
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globally stable ľ you never heard these kind
of notions in a linear system, because it
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has only one equilibrium point. But when it
becomes a nonlinear system, we always talk
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about whether it is locally stable, globally
stable ľ all these, because an equilibrium
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point can be locally stable, it can also be
globally stable. This is not a course on nonlinear
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systems and we will not go into details, but
we will cover whatever minimal is necessary
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in this course. x dot is equal to f of x.
This is your nonlinear system and the equilibrium
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point is x bar. This is your equilibrium point.
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The equilibrium point x bar is stable in the
sense of Lyapunovů. This is the definition
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we are giving ľ what we are seeing on the
board is the definition. We say that x bar
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is the stable equilibrium point in the sense
of Lyapunov if there exists a positive quantity
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epsilon such that for every delta (this is
a function of epsilon), we have this particularů.
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and this is satisfied, that is, this is my
initial state. My system state has been disturbed
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from the equilibrium point x bar and so, there
is some disturbance. After you disturb the
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system, now I am relaxing the system, that
is, system dynamically evolves and after dynamically
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evolving, where does it go? What happens to
the system states?
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If this disturbance that I have given is less
than delta, I cannot infinitely disturb the
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system and then conclude that system still
remains stable. I assume some value delta,
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that is, I put an upper bound by which I give
a disturbance to the system. If this is true,
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if this is my disturbance, this is the way
I disturb the system, then, x t minus x bar,
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that is the future state of my system is notů.
If x t minus x bar is less than epsilon, then
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the systemů. What is the meaning of that?
When I say this statement that if this is
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my disturbance, finally, always the magnitude
or absolute magnitude of my future state minus
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the equilibrium point is always less than
epsilon, what does that mean?
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If I am giving a simple disturbance to the
system, the future states do not go far away
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from the equilibrium point at any point of
time ľ they remain very close to their equilibrium
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point. That is the concept of stability that
we defined. Then, we define a concept of asymptotically
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stable or asymptotic stability, where this
particular difference that we said in the
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beginning is less than epsilon, we are now
saying that if it goes to 0, that is, my future
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state converges to the equilibrium point as
t tends to infinity, then, this is called
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asymptotically stable and we call x bar unstable
if it is not stable. If it is neither this
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nor this (obviously, this is neither of these),
then the system is unstable. These are all
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definitions we are talking about. We defined
stability, that is, it is just stable and
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then we defined the notion of asymptotic stability.
Now, how do we determine this notion ľ whether
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the system is stable or asymptotically stable?
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We will discuss two methods in this class
today: one is the indirect method and another
32:46.559 --> 32:53.299
is the direct method. The indirect method
is that we linearize the system, that is,
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x dot is equal to f x ľthat is my nonlinear
system. Imagine that we are allů. In the
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beginning, we are only considering the unforced
system u = 0. So, x dot is equal to f x means
33:05.600 --> 33:12.600
my u is 0 and I have no external force on
the system. Now, what I do is that I linearize
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the system and the delta x that we wrote earlier,
I am again redefining that delta x again as
33:24.500 --> 33:31.500
x just to make our notations very simple.
Here, delta x is defined as x. That is here.
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Now, x dot is equal to A x plus g x. This
g x represents all higher order terms ľ second
33:47.409 --> 33:54.409
order, third order and so on. What is A? A
has to be del f upon del x and this is an
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n by n matrix. This is A.
34:06.230 --> 34:13.230
Now, g x is my higher order term and I assume
that this higher order term satisfies this
34:18.119 --> 34:24.929
condition. You can see that the absolute magnitude
of g x upon the absolute magnitude of x as
34:24.929 --> 34:31.929
x tends to 0 is 0. It means that this approximation
linearization is valid. Using first-order
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Taylor's series expansion, this linearization
scheme is valid when this higher order term
34:42.819 --> 34:49.819
at g x becomes smaller, it becomes very small
in comparison to x ľ then only, it can become
34:49.999 --> 34:56.999
0 and that is the meaning. The nonlinear system
x dot = f x is asymptotically stable if and
34:58.799 --> 35:05.799
only if the linear system x dot equal to A
x is stable. This is our theorem or definition
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of the indirect method.
35:18.849 --> 35:25.849
How to determine whether the system is stable
or not? This is the definition of how to determine
35:27.799 --> 35:34.799
whether a system is locally stable ľ stable
around an equilibrium point. If a system is
35:35.160 --> 35:42.039
stable around an equilibrium point, then the
linearized system around that equilibrium
35:42.039 --> 35:49.039
point, which is x dot equal to A x and if
we analyze the stability of the system, all
35:54.739 --> 36:01.739
the Eigen values of A must be in the left
half of this plane, that is, the real part
36:07.279 --> 36:14.279
of the Eigen values must be negative. If the
real part of the Eigen values of A are negative,
36:15.769 --> 36:22.769
then the system is asymptotically stable around
the equilibrium point and that means all Eigen
36:24.329 --> 36:31.329
values of A have negative real parts. What
is the advantage of this method?
36:31.549 --> 36:38.549
It is very simple, because if x dot equal
to f x is given, find out what is x dot equal
36:40.529 --> 36:47.529
to A x, that is, A is del f upon del x at
equilibrium point x is equal to x bar or xe
36:52.960 --> 36:59.960
or whatever is the equilibrium point. Then,
check if A has all the Eigen values whose
37:02.989 --> 37:09.989
real parts are all negative, but the problem
here is that if any of these Eigen values
37:14.519 --> 37:21.519
are 0, the real parts are 0 or if some of
the Eigen values are imaginary values, then
37:26.999 --> 37:33.999
we cannot conclude that the system is locally
stable. So, if some Eigen values of A are
37:40.739 --> 37:44.670
0, then we cannot draw any conclusion about
the stability of the nonlinear system and
37:44.670 --> 37:51.670
also if the Eigen values are simply pure imaginary
quantities ľ the 0 real part. It is valid
37:55.759 --> 38:01.739
only if initial conditions are close to the
equilibrium point x bar, because of the condition
38:01.739 --> 38:08.739
that the higher order term g of x upon x limit
x tends to 0 should be 0. These are two drawbacks.
38:13.719 --> 38:20.719
Although this is very simple, now we must
look for a very nice method of determining
38:22.369 --> 38:27.979
stability, which is known very popularly as
Lyapunov stability theory or Lyapunov's second
38:27.979 --> 38:29.420
or direct method.
38:29.420 --> 38:36.420
For this, x dot is equal to f x. We define
a Lyapunov function V x with the following
38:37.299 --> 38:43.940
properties. This Lyapunov function at the
origin or at the equilibrium point is 0; this
38:43.940 --> 38:50.940
Lyapunov function is a scalar function, this
scalar function is always greater than 0 and
38:54.170 --> 39:01.170
x is not in the equilibrium point ľ it is
always a positive definite; and the rate derivative
39:01.950 --> 39:08.950
of this Lyapunov function is negative definite
along the trajectory x dot = f of x. Then,
39:11.700 --> 39:14.299
we say the system is stable.
39:14.299 --> 39:20.309
Then, x is asymptotically stable. The method
hinges on the existence of a Lyapunov function,
39:20.309 --> 39:27.309
which is energy-like function. We defined
a function V x. If I want to differentiate
39:29.180 --> 39:36.180
V x with respect to dt, so d V x by dt is
the rate derivative. Then, del V by del x
39:40.680 --> 39:47.680
into dx by dt. This dx by dt is already my
f x, this is my system. So I replace this
39:53.059 --> 40:00.059
f x here. dow V by dow x into f x. What is
f x now? V is a scalar function and x is a
40:05.329 --> 40:12.329
vector and my f x is also a vector. You can
see that dow V by dow x is a row vector and
40:19.719 --> 40:26.719
f x is a column vector. This is my column
vector and this is a row vector. If I expand,
40:34.989 --> 40:41.989
we can easily see that V dot x is dow V by
dow x1 into f1 plus dow V by dow x2 into f2
40:45.140 --> 40:50.319
and so on plus dow V by dow xn into fn. What
is the advantage?
40:50.319 --> 40:56.609
This answers the stability of nonlinear system
without explicitly solving the dynamic equations.
40:56.609 --> 41:01.539
It can easily handle time varying systems.
It can determine asymptotic stability as well
41:01.539 --> 41:06.259
as plain stability. It can determine the region
of asymptotic stability or the domain of attraction
41:06.259 --> 41:10.619
of an equilibrium. We will see all these things
through examples.
41:10.619 --> 41:17.619
Here is an example. This is a second-order
system with cubic nonlinearity ľ y cube is
41:21.289 --> 41:28.289
there. If you linearize this system, you get
y double dot plus 3 y dot equal to 0. The
41:28.450 --> 41:34.539
characteristic equation of the linearized
system is s into s plus 3. You see that you
41:34.539 --> 41:41.539
cannot say that by using Lyapunov's indirect
method, the system is stable, because one
41:43.079 --> 41:50.079
of the poles s is 0. So, the indirect method
fails here. So, there is a 0.
41:58.200 --> 42:05.200
We found out y double dot plus 3 y dot is
0. If I convert this into Laplace domain,
42:11.269 --> 42:18.269
s square y s plus s 3 y s is 0. If I look
at the characteristic polynomial, s square
42:24.069 --> 42:31.069
plus 3 s into y s
is 0. The poles are at s = 0 and s = ?3 and
because one of the poles is at 0, the indirect
42:51.549 --> 42:58.549
method fails in this case. We saw that the
indirect method failed and so now, we will
43:09.190 --> 43:12.670
apply Lyapunov's direct method.
43:12.670 --> 43:19.670
We had the system, which is y double dot plus
3 y dot plus y cube is 0. I select the states
43:27.299 --> 43:34.299
as x1 is y and x2 is y dot. I select these
two states and then, this equation can be
43:38.069 --> 43:45.069
written as a state-space format. x1 dot is
y dot, which is x2 and x2 dot, which is y
43:46.369 --> 43:53.369
double dot, is ?3ů. This is y double dot
plus 3 y dot. So minus 3 x2 minus x1 cube.
43:59.200 --> 44:06.200
This is my state-space model. For this, the
equilibrium point is obviously the origin.
44:11.099 --> 44:17.239
The origin is the equilibrium point. Now,
the Lyapunov function I select for the system
44:17.239 --> 44:24.239
V x is 1 upon 4 into x1 to the power of 4
plus 1 upon 2 into x2 square. This is an energy-like
44:25.259 --> 44:25.709
function.
44:25.709 --> 44:32.709
I take the derivative of this. V dot x is.ů
I put the formula dow V by dow x1 into x1
44:33.009 --> 44:40.009
dot plus dow V by dow x2 into x2 dot and you
can check that x1 dot is x2 and x2 dot is
44:42.729 --> 44:49.729
minus 3 x2 minus x1 cube. This is your f1
and this is your f2. This is f1 and this is
44:58.319 --> 45:05.319
f2. dow V by dow f1 plus dow V by dow x2 f2.
So, f1 is x2 and f2 is minus 3 x2 minus x1
45:07.670 --> 45:14.670
cube. If you simplify it, V dot x is minus
3 x2 square, which is always less than 0 because
45:16.759 --> 45:23.759
x2 squareů other than origin; when x2 is
not in the origin or when x2 is not 0, it
45:26.049 --> 45:33.049
is always negative. It follows that the equilibrium
point is asymptotically stable, which we could
45:33.180 --> 45:40.180
not conclude using the indirect method. What
is the disadvantage of this Lyapunov based
45:42.130 --> 45:49.130
approach? Although we showed an example where
we could not solve or we could not determine
45:49.910 --> 45:55.619
the stability using the indirect method, we
could determine that stability using the direct
45:55.619 --> 45:56.369
method.
45:56.369 --> 46:03.369
The problem is that there is no systematic
way of obtaining Lyapunov functions. Finding
46:03.829 --> 46:10.829
a Lyapunov function is more of an art than
science. This is an art, not science ľ how
46:16.329 --> 46:23.329
to find out a Lyapunov function. The Lyapunov
stability criterion provides only sufficient
46:23.640 --> 46:30.640
condition for stability. To end the lecture,
there are two examples that we will solve
46:38.170 --> 46:38.690
today.
46:38.690 --> 46:45.690
The first is to make your concept very clear
of how to linearize a nonlinear system. We
46:49.200 --> 46:56.200
will always encounter these in our intelligent
control course. More importantly, the Lyapunov
46:56.759 --> 47:03.640
stability theory will be used to determine
or to derive some new algorithm ľ the training
47:03.640 --> 47:10.640
algorithm as well as current control algorithm.
So, you see here, please note the system dynamics.
47:15.359 --> 47:20.999
Obtain a nonlinear state-space representation
of this and linearize this system around its
47:20.999 --> 47:27.999
equilibrium point. This is your question.
Let us solve it.
47:29.499 --> 47:36.499
This is the solution. We have y double dot
equal to 2 y minus y square plus one dy upon
47:44.839 --> 47:51.839
dt plus 1 plus u plus du by dt. This is the
nonlinear dynamics. Now, we have to find the
48:05.279 --> 48:12.279
state-space model. In the state model, we
first define the states. Obviously, let us
48:15.279 --> 48:22.279
take x1 equal to y. From our experience looking
at the form of the structure of this equation,
48:25.890 --> 48:32.890
we define x2 as x1 dot minus 2 u, which is
y dot minus 2 u ľ this is our definition.
48:39.219 --> 48:46.219
With this definition, you can write down x1
dot is y dot and y dot is x2 plus 2 u.
48:56.650 --> 49:03.650
State-space means we always write the derivative
of the state in terms of other states and
49:05.130 --> 49:12.130
input. Here, x1 dot is a function of x2 and
u ľ in terms of states and u. Similarly,
49:14.680 --> 49:21.680
for x2 dot, you can do little algebraic manipulation
ľ x2 dot is obviously y double dot minus
49:24.479 --> 49:31.479
2 u dot. You can see that y double dot minus
2 u dot is this particular quantity. We can
49:38.930 --> 49:45.930
write thatů 2 y, so 2 x1 minus y is x1 square
plus 1 and dy upon dt, that is, x1 dot is
49:57.209 --> 50:04.209
x2 plus 2 u plus 1 plus u. On simplification,
this becomes 2 x1 minus x1 square plus 1 into
50:19.890 --> 50:26.890
x2 plus 2 u plus 1 plus u. Actually, we did
not simplify ľ it is the same.
50:36.279 --> 50:43.279
Finally, y is x1 ľ that is my output. How
do I linearize it? Set u = 0 and x1 dot = 0
50:57.680 --> 51:04.680
and so also x2 dot. x1 dot and x2 dot are
both made 0. If you make u = 0, the first
51:10.039 --> 51:17.039
equation is x2 = 0 and the second equation
is 2 x1 minus x1 square plus 1 into x2 plus
51:27.289 --> 51:34.289
1 is 0. The solution is x1 = 1 and x2 = 0
ľ this is the equilibrium point. At this
51:44.709 --> 51:48.719
equilibrium point, we have to linearize the
system.
51:48.719 --> 51:55.719
Linearize
around the equilibrium point, which is 1 and
0. Apply the formula that we derived earlier
52:12.819 --> 52:19.819
using Taylor's series expansion, that is,
delta x dot is del f upon del x at 1, 0 delta
52:27.799 --> 52:34.799
x plus del f by del u at 1, 0 into delta u.
We are not writing the input here because
52:56.269 --> 53:03.269
we have assumed u to be 0 and we are linearizing
around the equilibrium point 1, 0 and u is
53:03.789 --> 53:10.789
0. That is the question that is asked to you.
If you compute del f by del x, it isů please
53:20.119 --> 53:27.119
verify, this will be your A matrix. The B
matrix isů. .
53:27.900 --> 53:34.900
How do I find it? 0 is
del f1 by del x1 and f1 is actually del x2
by del x1. By definition, this is 0. Similarly,
53:53.559 --> 54:00.559
the
second one del f1 by del x2, which is this
value, is del x2 by del x2 and this is 1.
54:14.199 --> 54:21.199
You are differentiating x2 with respect to
x2. Like that, please verify this particular
54:22.440 --> 54:29.440
solution that we got. Finally, delta y is
obviously (1, 0) delta x ľ this is the answer.
54:32.259 --> 54:35.420
Now, we will take the second example.
54:35.420 --> 54:42.420
The second example is that we again have a
nonlinear system and you are asked to find
54:45.079 --> 54:52.079
out whether (1, 1) is an equilibrium state
ľ first question; the second is that is this
54:54.459 --> 55:01.459
the only equilibrium state or there are many;
the third is whether the first one that is
55:01.459 --> 55:08.239
given ľ xe ľ is asymptotically stable. Let
us try to solve this equation again.
55:08.239 --> 55:15.239
The solution is that you are given x1 dot
is minus x1 square plus x1 x2 and x2 dot is
55:28.479 --> 55:35.479
minus 2 x2 square plus x2 minus x1 x2 plus
2. How do you know that xe = (1, 1) is an
55:50.880 --> 55:57.400
equilibrium point? Just put these values here
and see whether they are 0. If you put (1,
55:57.400 --> 56:04.400
1) here, this is minus 1 plus 1, which is
0. In this case, when you replace x2 as 1
56:05.809 --> 56:12.809
and x1 as 1, again you have minus 2 plus 1
minus 1 plus 2 and again, it is 0; hence,
56:19.069 --> 56:26.069
xe is an equilibrium point. This is the first
part of the answer.
56:29.489 --> 56:35.559
The second part is to find whether there are
other equilibrium points. A simple way is
56:35.559 --> 56:42.559
there to solve it ľ put x1 dot equal to 0
and x2 dot is equal to 0 and solve. You will
56:45.619 --> 56:52.619
find these equilibrium points areů. We have
more equilibrium points: (minus 2 by 3, minus
56:53.910 --> 57:00.910
2 by 3) is one more, then (0, 1 plus root
17 by 4), and (0, 1 minus root 17 by 4). There
57:16.289 --> 57:23.289
are three more equilibrium points. You can
verify this by just putting x1 dot equal to
57:24.779 --> 57:30.529
0 and x2 dot equal to 0 ľ you will find this
solution.
57:30.529 --> 57:37.529
The next question, the third part, was whether
xe is asymptotically stable. How do we do
57:55.219 --> 58:02.219
it? We linearize the system around xe. If
you linearize around xe, linearize f x around
58:17.630 --> 58:24.630
xe, then you get x dot equal to A x, where
A is del f upon del x. Please verify that
58:32.339 --> 58:39.339
A is (?1, 1, ?1, 4) and the Eigen values are
?5 plus or minus root 5 divided by 2. You
58:57.049 --> 59:04.049
can check that both the Eigen values have
negative real part ?5 and hence, it is asymptotically
59:05.150 --> 59:12.150
stable. We will continue further discussion
on nonlinear system in the next class.
59:23.910 --> 59:30.439
Here is a little help for you to follow some
reference books. You can follow one of the
59:30.439 --> 59:37.439
books by Khalil ľ Nonlinear Systems, Prentice
Hall. Another one is Slotine and Li's Applied
59:37.650 --> 59:40.880
Nonlinear Control ľ a very nice book. I like
this book on Applied Nonlinear Control. M.
59:40.880 --> 59:47.880
Vidyasagar has also written a nice book ľ
Nonlinear System Analysis, Prentice Hall.
59:50.789 --> 59:56.849
Thank you again.
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