This is lecture 4 of module 1. In this lecture, we will discuss some concepts in nonlinear
systems analysis, specifically, the Lyapunov based approach, the reason being that we will
use this concept to derive some new training algorithm for neural network based control
schemes. Hence, I thought that we must discuss this subject of nonlinear system analysis.
You should have some background before you can actually appreciate what we teach in this
course.
In this course, we will introduce what is nonlinear system, what is linearization, Lyapunov
stability theory and some examples. What you are seeing is.ů This is our general state-space
model ľ nonlinear system model. x dot is equal to f t x u. Hence, it is explicit in
time as well as function of x as well as input u. This is a state variable model and output
is again a nonlinear function h. You see x is a vector, f is a vector, y is a vector
and h is also a vector. This is a multi-input, multi-output system. When I say this is nonlinear,
what does it mean? Of course, the first thing that you already know is that a linear system
means it follows the principle of superposition and a nonlinear system is expected to not
follow the principle of superposition. Let me for your benefit compare. A linear system
is different from a nonlinear system in terms of superposition principle.
This is a linear system and this is a nonlinear system. Here, this is my system, which is
linear. If I give u1, I get y1. In this case, if I give alpha u1, then also I get alpha
y1; same system, say some system G. Now, if I give another input u2 to the same system
G, I get y2 and finally, the superposition principle is I give alpha1 u1 plus alpha2
u2 and I find alpha1 y1 plus alpha2 y2. A nonlinear system is just the opposite. For
example, if I give u1, I get y1, same system G nonlinear. When it is a nonlinear system
and if I give an input alpha1 u1, the output is y and y is not equal to alpha1 u1.
Similarly, if I provide these two inputs alpha1 u1 plus alpha2 u2 to the same nonlinear system,
the output y is not equal to alpha1 y1 plus alpha2 y2. This is the major difference between
a nonlinear system and a linear system as far as differences are concerned, but a nonlinear
system is more than justů. It does not follow the superposition principle; it has an even
broader perspective and some of these perspectives are summarized here.
We will just discuss how a nonlinear system has multiple equilibrium points ľ we will
shortly discuss this. It has limit cycles. If you give a specific sinusoid signal of
a specific frequency, then in a linear system, the output will have the same frequency whereas
a nonlinear system may produce subharmonic oscillations of frequencies. Also, the nonlinear
system can behave or can exhibit chaos. Chaos means its steady state behavior is highly
unpredictable and random. The nonlinear system also executes multiple modes of behavior and
that is called bifurcation. That is one of the examples.
This is the example of an autonomous system, a nonlinear system x dot equal to f x u. Autonomous
means it does not explicitly depend on time and y is h x u. This is the generalized state-space
model of a nonlinear system that is autonomous. Some systems that have been very widely studied
in nonlinear system are known as affine systems. They have a specific structure. In our control
system design, many times we will discuss this kind of system while designing a controller,
that is, x dot is equal to f x plus g x into u. This is a nonlinear function and this is
another nonlinear function, but you can see that this has a specific structure. When you
have no external input, then x dot is equal to f x ľ this is an unforced system.
What is the equilibrium point? At equilibrium point, x dot, the dynamics, becomes 0 ľ the
derivative of the states becomes 0; that is the equilibrium point. If xe and ue correspond
to a specific equilibrium point, you replace and then equate to 0. You will find what the
equilibrium point is.
What is linearization? Linearization is the process of replacing the nonlinear system
model by its linear counterpart in a small region about its equilibrium point. We linearize
because we have well-established tools to analyze and stabilize the linear system.
Let us try to understand the basic process of linearization. Let us write the general
form of a nonlinear system x dot is equal to f x u as this. We have x1, x2 ů xn ľ
there are n states. So dx1 upon dt is f1 of x1, x2 ů xn and u1 to um. We have m inputs
and n outputs. This is the generalized way to write, a very explicit way to write this
compact notation x dot equal to f x u. f is a vector. You can see f1, f2 ů fn.
Let x1e, x2e ů xne be a point of equilibriumů the point of equilibrium xe and ue, where
ue is a vector ľ u1e, u2e ů ume and xe is x1e, x2e ů xne. We have xe and ue. This is
an equilibrium point and this holds true: f xe ue is 0.
Then, we perturb this equilibrium state by allowing x equal to xe plus delta x ľ little
perturbation from xe and similarly, a little perturbation from ue by delta u. Then, Taylor's
expansion yields a veryů. . We can write dx upon dt is f and this is your x, this is
your u due to perturbation and we expand this using Taylor's expansion. First, this is your
equilibrium point, the operating point and then del f upon del x, you know f is a vector,
so x is also a vector.
If you differentiate a vector with respect to a vector, you get a matrix. So, del f upon
del x into delta x and this matrix is computed at the value x equal to xe and x equal to
ue. These are first-order terms and this is the operating point or equilibrium point.
This is del f upon del u and you compute this ľ again, another Jacobian matrix; you have
to compute at xe and ue and multiply by delta u. This is the first-order expansion plus
second order, third order and higher orders. But we only go up to the first-order expansion
of Taylor series because that is where we can apply our linear system theory. If you
go to second order, again the system becomes nonlinear. It is not always true that we can
always approximate a nonlinear system around an equilibrium point using first-order Taylor
series expansion ľ there is some limit on that. Let us compute and see what the Jacobian
matrix is.
del f upon del x. As I told you, this is the matrix. The easiest way to remember is that
the first vector in f1 is differentiated with respect to each element of the x vector and
that is in the first row. Similarly, in the last row, the last vector of f, is the n th
vector of f, is differentiated with respect to all the elements of x and that is the last
row. You can easily remember this: del f1 by del x1 to del f1 upon del xn; similarly,
del fn upon del x1 to del fn upon del xn. These values have to be computed by replacing
x as xe. So x has to be xe and u has to be ue. This will be a constant matrix xe. Once
you replace xe and ue, this becomes a constant matrix. Once you replace xe and ue, this becomes
a constant matrix. Similarly, you can compute also del f upon del u ľ similar principle.
Obviously, you can see that this is n into n matrix and this is n rows and m columns,
so n into m matrix. These are all Jacobian matrices.
Let us note here that dx upon dt is dxe upon dt plus d delta x upon dt, because x is xe
plus delta x. When I differentiate x with respect to dt, I differentiate xe with respect
to dt and delta x with respect to dt, but you can see that xe is an equilibrium point,
and at that point, x dot is 0, the derivative is 0, so this is 0. You only get this term:
d delta x upon dt. Furthermore, f xe ue is also 0 because that is how you computed xe
and ue ľ by making it 0. At the equilibrium point, x dot (the derivative) is 0.
We have already computed in the previous slide. This matrix is actually matrix A and this
matrix is matrix B. As we defined, A is del f upon del x and B is del f upon del u. We
have neglected the higher order terms and we arrive at the linear approximation ľ d
delta x upon dt is A delta x plus B delta u. This gives sometimes we also say small
signal models, that is, the behavior of a nonlinear system around an equilibrium point,
where the dynamics is assumed to be linear. Hence, the dynamic is in the linear state-space
model delta x dot equal to A delta x plus B delta u.
Similarly, we had P outputs. For P outputs, the functions are h1, h2 ů hP. In vector
notation, you can write y equal to h x u. We can again use Taylor's series expansion:
y equal to ye plus delta y. That is being perturbed ľ equilibrium point ye with delta
y. Then, we have the actual system is y equal to h x u. Using Taylor's series, we can always
write delta y is C delta x plus D delta u, where C and D are again Jacobian matrices
(computed). Can you guess now what is C and what is D? As we have already said, C has
to obviously be del h upon del x. This has to be a P into n matrix. Similarly, D is del
h upon del u and this has to be a P into m matrix. That is how we also linearize the
output y around the equilibrium point ye. ye corresponds to xe and ue. At value xe,
ue, the system response is ye. So, ye is the system response when the system state is xe
or system is at the equilibrium point. Let us take an example.
You can see here that this is a simple scalar differential equation. x is a scalar ľ one-dimensional,
simply one-dimensional; it is a scalar quantity. Here is a scalar differential equation but
nonlinear because of this term x square. We can verify that this equation will not follow
the principle of superposition. You linearize it about the origin and you can easily see
that if I write f xe as 0, then you will find when you make minus x plus x square equal
to 0, this will lead you to x is 0. So origin equilibrium point; x = 0 is an equilibrium
point. Linearize it about the origin and you get x dot equal to minus x. The solution is
x t equal to x0 e to the power of minus t. It is very simple because this is a simple
linear differential equation: x dot is equal to minus x. I hope that you are very well
aware on how to get the solution of this. You can verify this now.
You put x0 e to the power of minus t here, differentiate it and at x of 0, when t equal
to 0, put this value x0 and you see this equation is fine. Whatever may be the initial state
x0, the state will settle at x t = 0, because whatever the initial condition, when you perturb
the system from its initial position, the equilibrium point x0 will always come to the
origin. The state will settle at x t. This is your equilibrium point, which is the only
equilibrium point that this linearized system has. Let me again clarify what we actually
did.
We had x dot equal to minus x plus x square. First, we found out the equilibrium point,
that is, f xe is 0, which is minus x plus x square. So x = 0 is an equilibrium point.
At the equilibrium point, the linearized equation isů if you follow the linearization scheme,
it is minus x. This is your linearized system around the equilibrium point x = 0. The solution
we found out here is x t is x0 e to the power of minus t ľ that is the solution to the
equilibrium point.
Now, if I actually take the exact solution of x t dot, which is minus x plus x square,
if I find the actual solution of this nonlinear system, then x t equal to x0 e to the power
of minus t upon 1 minus x0 plus x0 e to the power of minus t. You can try it. It is an
exercise for you. This is a simple nonlinear system, for which we find a closed loop solution
and this is the closed loop solution for this nonlinear solution. In general, we cannot
express the solution of a nonlinear system in a closed form but because this is a simple
one, we can actually find the closed form solution of this nonlinear system.
Earlier, we showed that the origin was one of the equilibrium points, but there is another
equilibrium point, that is, when I solve minus x plus x square is 0, you have x of x minus
1 is 0. This implies x = 0 and x = 1. There are two equilibrium points. We talked about
linearizing the system around only one equilibrium point x = 0, origin, but there is also another
equilibrium point x = 1. Let us see the system behavior around these equilibrium points.
This is a plot. You can see this is one equilibrium point. What we are showing is this is the
time axis and this is your x t ľ we are showing the response. The value of x t in the beginning
t = 0ů. We can start either near the first equilibrium point, that is, 0 ľ this particular
point and another is 1 ľ this is another equilibrium point. If I give the initial condition
as ?0.5, you see they all settle. ?0.5 and +0.5, wherever I give in this zone, they are
all going towards 0.
Finally, the system at t tends to infinity, goes back to the equilibrium point 0, that
is, if the system is disturbed slightly from its equilibrium point 0, it is near the origin,
it goes back to the origin, whereas if you look here, this is a little interesting. One:
if you go back here, it diverges from 1. Very near it diverges, but surprisingly what happens
is that after it goes to a very high value, it comes back and then again goes back to
0. Similarly, we can see this one. Just after 1, it goes and again from here to here, it
jumps and then again goes to 0.
Similarly, another point here, this particular line and from here, it jumps here and again
goes to 0. What you are seeing is that as I said in the beginning, the equilibrium point
1 is actually an unstable equilibrium point. From here, the system diverges from this point,
whereas if you start somewhere near the origin, then you could disturb the system near the
equilibrium point, which is the origin. Then, it goes back to the origin. The conclusion
is x = 0 is a stable equilibrium point, whereas x = 1 is an unstable equilibrium point. At
the end of the lecture, we will discuss a little more about these equilibrium points
ľ we will solve through examples but let us discuss some notion of stability in a nonlinear
system. What we talked about now is that a nonlinear system can be linearized around
an equilibrium point. Some equilibrium points are stable equilibrium points and some equilibrium
points are unstable equilibrium points. Now, what is the concept of stability on a nonlinear
system?
In linear system, we had only one equilibrium point. We never talked about locally stable,
globally stable ľ you never heard these kind of notions in a linear system, because it
has only one equilibrium point. But when it becomes a nonlinear system, we always talk
about whether it is locally stable, globally stable ľ all these, because an equilibrium
point can be locally stable, it can also be globally stable. This is not a course on nonlinear
systems and we will not go into details, but we will cover whatever minimal is necessary
in this course. x dot is equal to f of x. This is your nonlinear system and the equilibrium
point is x bar. This is your equilibrium point.
The equilibrium point x bar is stable in the sense of Lyapunovů. This is the definition
we are giving ľ what we are seeing on the board is the definition. We say that x bar
is the stable equilibrium point in the sense of Lyapunov if there exists a positive quantity
epsilon such that for every delta (this is a function of epsilon), we have this particularů.
and this is satisfied, that is, this is my initial state. My system state has been disturbed
from the equilibrium point x bar and so, there is some disturbance. After you disturb the
system, now I am relaxing the system, that is, system dynamically evolves and after dynamically
evolving, where does it go? What happens to the system states?
If this disturbance that I have given is less than delta, I cannot infinitely disturb the
system and then conclude that system still remains stable. I assume some value delta,
that is, I put an upper bound by which I give a disturbance to the system. If this is true,
if this is my disturbance, this is the way I disturb the system, then, x t minus x bar,
that is the future state of my system is notů. If x t minus x bar is less than epsilon, then
the systemů. What is the meaning of that? When I say this statement that if this is
my disturbance, finally, always the magnitude or absolute magnitude of my future state minus
the equilibrium point is always less than epsilon, what does that mean?
If I am giving a simple disturbance to the system, the future states do not go far away
from the equilibrium point at any point of time ľ they remain very close to their equilibrium
point. That is the concept of stability that we defined. Then, we define a concept of asymptotically
stable or asymptotic stability, where this particular difference that we said in the
beginning is less than epsilon, we are now saying that if it goes to 0, that is, my future
state converges to the equilibrium point as t tends to infinity, then, this is called
asymptotically stable and we call x bar unstable if it is not stable. If it is neither this
nor this (obviously, this is neither of these), then the system is unstable. These are all
definitions we are talking about. We defined stability, that is, it is just stable and
then we defined the notion of asymptotic stability. Now, how do we determine this notion ľ whether
the system is stable or asymptotically stable?
We will discuss two methods in this class today: one is the indirect method and another
is the direct method. The indirect method is that we linearize the system, that is,
x dot is equal to f x ľthat is my nonlinear system. Imagine that we are allů. In the
beginning, we are only considering the unforced system u = 0. So, x dot is equal to f x means
my u is 0 and I have no external force on the system. Now, what I do is that I linearize
the system and the delta x that we wrote earlier, I am again redefining that delta x again as
x just to make our notations very simple. Here, delta x is defined as x. That is here.
Now, x dot is equal to A x plus g x. This g x represents all higher order terms ľ second
order, third order and so on. What is A? A has to be del f upon del x and this is an
n by n matrix. This is A.
Now, g x is my higher order term and I assume that this higher order term satisfies this
condition. You can see that the absolute magnitude of g x upon the absolute magnitude of x as
x tends to 0 is 0. It means that this approximation linearization is valid. Using first-order
Taylor's series expansion, this linearization scheme is valid when this higher order term
at g x becomes smaller, it becomes very small in comparison to x ľ then only, it can become
0 and that is the meaning. The nonlinear system x dot = f x is asymptotically stable if and
only if the linear system x dot equal to A x is stable. This is our theorem or definition
of the indirect method.
How to determine whether the system is stable or not? This is the definition of how to determine
whether a system is locally stable ľ stable around an equilibrium point. If a system is
stable around an equilibrium point, then the linearized system around that equilibrium
point, which is x dot equal to A x and if we analyze the stability of the system, all
the Eigen values of A must be in the left half of this plane, that is, the real part
of the Eigen values must be negative. If the real part of the Eigen values of A are negative,
then the system is asymptotically stable around the equilibrium point and that means all Eigen
values of A have negative real parts. What is the advantage of this method?
It is very simple, because if x dot equal to f x is given, find out what is x dot equal
to A x, that is, A is del f upon del x at equilibrium point x is equal to x bar or xe
or whatever is the equilibrium point. Then, check if A has all the Eigen values whose
real parts are all negative, but the problem here is that if any of these Eigen values
are 0, the real parts are 0 or if some of the Eigen values are imaginary values, then
we cannot conclude that the system is locally stable. So, if some Eigen values of A are
0, then we cannot draw any conclusion about the stability of the nonlinear system and
also if the Eigen values are simply pure imaginary quantities ľ the 0 real part. It is valid
only if initial conditions are close to the equilibrium point x bar, because of the condition
that the higher order term g of x upon x limit x tends to 0 should be 0. These are two drawbacks.
Although this is very simple, now we must look for a very nice method of determining
stability, which is known very popularly as Lyapunov stability theory or Lyapunov's second
or direct method.
For this, x dot is equal to f x. We define a Lyapunov function V x with the following
properties. This Lyapunov function at the origin or at the equilibrium point is 0; this
Lyapunov function is a scalar function, this scalar function is always greater than 0 and
x is not in the equilibrium point ľ it is always a positive definite; and the rate derivative
of this Lyapunov function is negative definite along the trajectory x dot = f of x. Then,
we say the system is stable.
Then, x is asymptotically stable. The method hinges on the existence of a Lyapunov function,
which is energy-like function. We defined a function V x. If I want to differentiate
V x with respect to dt, so d V x by dt is the rate derivative. Then, del V by del x
into dx by dt. This dx by dt is already my f x, this is my system. So I replace this
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
vector and my f x is also a vector. You can see that dow V by dow x is a row vector and
f x is a column vector. This is my column vector and this is a row vector. If I expand,
we can easily see that V dot x is dow V by dow x1 into f1 plus dow V by dow x2 into f2
and so on plus dow V by dow xn into fn. What is the advantage?
This answers the stability of nonlinear system without explicitly solving the dynamic equations.
It can easily handle time varying systems. It can determine asymptotic stability as well
as plain stability. It can determine the region of asymptotic stability or the domain of attraction
of an equilibrium. We will see all these things through examples.
Here is an example. This is a second-order system with cubic nonlinearity ľ y cube is
there. If you linearize this system, you get y double dot plus 3 y dot equal to 0. The
characteristic equation of the linearized system is s into s plus 3. You see that you
cannot say that by using Lyapunov's indirect method, the system is stable, because one
of the poles s is 0. So, the indirect method fails here. So, there is a 0.
We found out y double dot plus 3 y dot is 0. If I convert this into Laplace domain,
s square y s plus s 3 y s is 0. If I look at the characteristic polynomial, s square
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
method fails in this case. We saw that the indirect method failed and so now, we will
apply Lyapunov's direct method.
We had the system, which is y double dot plus 3 y dot plus y cube is 0. I select the states
as x1 is y and x2 is y dot. I select these two states and then, this equation can be
written as a state-space format. x1 dot is y dot, which is x2 and x2 dot, which is y
double dot, is ?3ů. This is y double dot plus 3 y dot. So minus 3 x2 minus x1 cube.
This is my state-space model. For this, the equilibrium point is obviously the origin.
The origin is the equilibrium point. Now, the Lyapunov function I select for the system
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
function.
I take the derivative of this. V dot x is.ů I put the formula dow V by dow x1 into x1
dot plus dow V by dow x2 into x2 dot and you can check that x1 dot is x2 and x2 dot is
minus 3 x2 minus x1 cube. This is your f1 and this is your f2. This is f1 and this is
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
cube. If you simplify it, V dot x is minus 3 x2 square, which is always less than 0 because
x2 squareů other than origin; when x2 is not in the origin or when x2 is not 0, it
is always negative. It follows that the equilibrium point is asymptotically stable, which we could
not conclude using the indirect method. What is the disadvantage of this Lyapunov based
approach? Although we showed an example where we could not solve or we could not determine
the stability using the indirect method, we could determine that stability using the direct
method.
The problem is that there is no systematic way of obtaining Lyapunov functions. Finding
a Lyapunov function is more of an art than science. This is an art, not science ľ how
to find out a Lyapunov function. The Lyapunov stability criterion provides only sufficient
condition for stability. To end the lecture, there are two examples that we will solve
today.
The first is to make your concept very clear of how to linearize a nonlinear system. We
will always encounter these in our intelligent control course. More importantly, the Lyapunov
stability theory will be used to determine or to derive some new algorithm ľ the training
algorithm as well as current control algorithm. So, you see here, please note the system dynamics.
Obtain a nonlinear state-space representation of this and linearize this system around its
equilibrium point. This is your question. Let us solve it.
This is the solution. We have y double dot equal to 2 y minus y square plus one dy upon
dt plus 1 plus u plus du by dt. This is the nonlinear dynamics. Now, we have to find the
state-space model. In the state model, we first define the states. Obviously, let us
take x1 equal to y. From our experience looking at the form of the structure of this equation,
we define x2 as x1 dot minus 2 u, which is y dot minus 2 u ľ this is our definition.
With this definition, you can write down x1 dot is y dot and y dot is x2 plus 2 u.
State-space means we always write the derivative of the state in terms of other states and
input. Here, x1 dot is a function of x2 and u ľ in terms of states and u. Similarly,
for x2 dot, you can do little algebraic manipulation ľ x2 dot is obviously y double dot minus
2 u dot. You can see that y double dot minus 2 u dot is this particular quantity. We can
write thatů 2 y, so 2 x1 minus y is x1 square plus 1 and dy upon dt, that is, x1 dot is
x2 plus 2 u plus 1 plus u. On simplification, this becomes 2 x1 minus x1 square plus 1 into
x2 plus 2 u plus 1 plus u. Actually, we did not simplify ľ it is the same.
Finally, y is x1 ľ that is my output. How do I linearize it? Set u = 0 and x1 dot = 0
and so also x2 dot. x1 dot and x2 dot are both made 0. If you make u = 0, the first
equation is x2 = 0 and the second equation is 2 x1 minus x1 square plus 1 into x2 plus
1 is 0. The solution is x1 = 1 and x2 = 0 ľ this is the equilibrium point. At this
equilibrium point, we have to linearize the system.
Linearize around the equilibrium point, which is 1 and 0. Apply the formula that we derived earlier
using Taylor's series expansion, that is, delta x dot is del f upon del x at 1, 0 delta
x plus del f by del u at 1, 0 into delta u. We are not writing the input here because
we have assumed u to be 0 and we are linearizing around the equilibrium point 1, 0 and u is
0. That is the question that is asked to you. If you compute del f by del x, it isů please
verify, this will be your A matrix. The B matrix isů. .
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,
the second one del f1 by del x2, which is this value, is del x2 by del x2 and this is 1.
You are differentiating x2 with respect to x2. Like that, please verify this particular
solution that we got. Finally, delta y is obviously (1, 0) delta x ľ this is the answer.
Now, we will take the second example.
The second example is that we again have a nonlinear system and you are asked to find
out whether (1, 1) is an equilibrium state ľ first question; the second is that is this
the only equilibrium state or there are many; the third is whether the first one that is
given ľ xe ľ is asymptotically stable. Let us try to solve this equation again.
The solution is that you are given x1 dot is minus x1 square plus x1 x2 and x2 dot is
minus 2 x2 square plus x2 minus x1 x2 plus 2. How do you know that xe = (1, 1) is an
equilibrium point? Just put these values here and see whether they are 0. If you put (1,
1) here, this is minus 1 plus 1, which is 0. In this case, when you replace x2 as 1
and x1 as 1, again you have minus 2 plus 1 minus 1 plus 2 and again, it is 0; hence,
xe is an equilibrium point. This is the first part of the answer.
The second part is to find whether there are other equilibrium points. A simple way is
there to solve it ľ put x1 dot equal to 0 and x2 dot is equal to 0 and solve. You will
find these equilibrium points areů. We have more equilibrium points: (minus 2 by 3, minus
2 by 3) is one more, then (0, 1 plus root 17 by 4), and (0, 1 minus root 17 by 4). There
are three more equilibrium points. You can verify this by just putting x1 dot equal to
0 and x2 dot equal to 0 ľ you will find this solution.
The next question, the third part, was whether xe is asymptotically stable. How do we do
it? We linearize the system around xe. If you linearize around xe, linearize f x around
xe, then you get x dot equal to A x, where A is del f upon del x. Please verify that
A is (?1, 1, ?1, 4) and the Eigen values are ?5 plus or minus root 5 divided by 2. You
can check that both the Eigen values have negative real part ?5 and hence, it is asymptotically
stable. We will continue further discussion on nonlinear system in the next class.
Here is a little help for you to follow some reference books. You can follow one of the
books by Khalil ľ Nonlinear Systems, Prentice Hall. Another one is Slotine and Li's Applied
Nonlinear Control ľ a very nice book. I like this book on Applied Nonlinear Control. M.
Vidyasagar has also written a nice book ľ Nonlinear System Analysis, Prentice Hall.
Thank you again. 360