I want to talk about stability, because as you
probably recall when we did a control
design, first order of business is to
design controllers so that systems don't
blow up. If they blow up, there's nothing
we can do about it. The quad rotors just
fall out of the air. The robots drive off
to infinity. The cars smash into things.
We don't want them to blow up, because the
deciding objectives are almost always
layered in this sense. First order of
business is stability. Then we want to
track whatever reference character or
reference point we have. We also want it
to be robust to parameter uncertainties,
and possibly noise. And then we can wrap
other objectives around it, like when you
want to move as fast, quickly as you can,
or use as little energy when you're
moving, or things like this. But,
regardless of which, stability is always
the first order of business. So let's
start with scalar systems, no inputs. So
only the A matrix now, in this case x dot
is little ax, which means that it's
scalar. Well then the solution x of t is e
to the a, we said t minus t naught x of t
naught. Here I simply picked t naught to
be equal to 0. So this is the solution.
Okay, lets plot what this solution looks
like. If a is positive, then x of t it
starts nicely and then pabaah. Its, its
blowing up as far as I can tell. So if a
is positive this system blows up. Well, if
a is negative, then e to the at, this is a
decaying exponential. So we get x to just
go, , nice down to zero. What happens if a
is zero in between these 2? Well, then you
have e to the zero t, which is 1. So then,
x of t is simply equal to x naught. x
never changes.
So here, it didn't blow up, but it didn't
actually go down to zero. And in fact,
what we have its, its really a sitution
where three possible things can happen you
blow up, you go down to zero, or you stay
put. So let's talk about these three
cases. The first case is what is called
asymptotic, stability. So the system is
asymptotically stable if x goes to zero
For all initial conditions, so this fancy
upside down a, is known as the universal
quantifier. All we need to know is that
when we see and upside down a the way we
pronounce it is for all x nought. So
asymptotic stability means that we go to
zero and almost always what we want to
design our system so that x actually goes
to 0 no matter where we start, that's
asymptotically stability and as you
recall, in the scalar case, a strictly
negative corresponds to asymptotically
stability. And then we have unstability,
instability where the system being
unstable. What that means is there exists
an initial condition, so the flipped e,
and to speak for the existential
quantifier, which we read it as exists. So
it's unstable if there exists so many
extra conditions from which the system
actually blows up. In the scaler case, we
had A positive corresponding to
instability. and then we have something we
call critical stability, which is somehow
in between. The system doesn't blow up.
But it doesn't go to zero either, and in
fact, for the scalar system, this
corresponded, corresponded to the, a equal
to zero case. So if you summarize that, if
you have a scalar system then a positive
means the system is unstable. A negative
means that the system is asymptotically
stable, which is code for saying that the
state goes to zero. And a zero means
critically stable. Okay.
Let use this way of thinking now on the
matrix case. X. is ax, capital A. So this
is now, x is a vector, a is a matrix. What
do we do there? Well, we can't just say.
Oh, a is positive, or a is negative.
Because a is a matrix. It's not positive
or negative. But what we can do is we can
go for the next best thing, which is the
eigenva lues.
And, in fact, almost always, the intuition
you get from a scalar system translates
into the behavior of the eigenvalues of
these matrices. And for those of you who
don't know what eigenvalues are, these are
the special things that are associated
with matrices. So, if I have a matrix A; N
by N, and I multiply it by a vector an N
by 1 vector, if I can write it as the same
vector times a scalar, then what this
means is that the way that A acts on this
vector is basically scaling it. And the
scaling factor is given by lambda. If I
can, if I can find lambda of v to satisfy
this, then what I have is a lambda that is
called an eigenvalue. And it's actually
not a real number. It's typically a
complex number. So it's a, a slightly more
general object than just a real number,
but that's an eigenvalue. And v is known
as an eigenvector. And eigenvalues and
eigenvectors are really these fundamental
objects in, in when you're dealing with
matrices and when you want to understand
how they behave. And, whenever you think
scalar first, you can almost always
translate it into what do the eigenvalues,
eigenvalues do for your systems. And, the
eigenvalues actually would tell you how
the matrix a acts in the directions of eh
eigenvectors. So, you can almost think of
them as scalar systems in the directions
of the different eigenvectors. And, you
know, sometimes you may want to compute
eigenvalues. I don't.
So, if you use MATLAB. You would just
write, eig(A), and out pops the
eigenvalues. whatever software you, your
comfortable with, you want to use C, or
Python, or whatever, there is almost
always a library that allows you to
compute eigenvalues. And, the command is
typically something like eig(A).
So, this would give you what eigenvalues
are, given a particular matrix. Okay.
Let's see what this actually means. Let's
take a simple example here. Here's my a
system. 1, 0, o minus 1. if you take eig a
of this. you get 1 eigenvalue being 1. And
the other eigenvalue being negative 1. And
the correspo nding eigenvectors are 1, 0,
and 0, 1. Okay.
What does this mean? It actually means the
following. So let's say that this is x1,
and this is x2. Okay.
V2 was 0, 1. So this was this direction.
So here is what, v2 is. This is the
direction in which v2 is pointing. Well,
the eigenvalue there is negative 1, which
means that, if you recall the scalar
system, when a was negative, we had
stability. So if I start here, my
trajectory is going to pull me down to
zero. Nice and stable, and in fact, if I
start here, it's going to pull me up to
zero, nice and stable. Right.
So, if I'm starting. on the x2 axis, my
system is well behaved. If I start on the
x1 axis, I have lambda 1 being positive,
which corresponds to little a being
positive in the scalar case, which means
that the system actually blows up. So,
here, the system goes off to infinity.
And, in fact, if I start here, my x2
component is going to shrink but my x1
component is going to go off to infinity.
So what I have is this is what the system
actually looks like. So the eigen vectors
in this case will tell me what happens
along different dimensions of, of the
system. So after all of this, if I have x
dot as big AX, and I can find a solution,
then the system is asymptotically stable,
if and only if, for the scalar case, we
had that little a had to be negative. What
we need in the matrix case is that the
real part, remember that Lambda are
complex, the real part of Lambda is
strictly negative for all eigenvalues to
a. For all, this is what asymptotic
stability means for linear systems.
Unstable means that there is one or more,
but one single bad eigenvalue spoils the
whole bunch. So a single eigenvalue that
has positive real part. This is an, a
sufficient condition for instability. And
we have critical stability only if so this
is a, a necessary condition that says the
real part has to be less than or equal to
0 for all igon values. But where we are
going to be spending our time is Typically
up here in the asymptotically stabl e
domain, because what we want to do, is we
want to design our system or our
controllers in such a way that the closed
loop system is asymptotically stable. So
we're going to somehow make the eigen
values have negative real part That's
going to be one of the design objectives
that we're interested in. And I want to
point out something about critical
stability that if one eigenvalue is 0 and
the rest of the eigenvalues have negative
real part, or if you have two purely
imaginary eigenvalues So they have no real
part, and the rest have negative real
part, then you have critical stability.
and we will actually see that a little bit
later on, but the thing that I really want
to take, you to take with you based on
this slide, is, you look at the a matrix,
you compute the eigenvalues. If the real
part of the eigenvalues are all negative
You're free and clear, the system is
asymptotically stable. If one or more
eigenvalues have positive real part, you
toast, your system blows up. That is bad.
So, let's end with a tale of two pendula.
Here is the normal pendula, well if you
compute the of this, you get this matrix.
And the eigenvalues are j and negative j.
Well, I don't know if you remember, but on
the previous slide, there was a bullet
that said if you have 2 purely imaginary
eigenvalues, which we have here. We have 2
purely imaginary eigenvalues and then no
more, then we have critical stability.
What this actually means is that, this
pendulum, clearly, there is no friction or
grav-, or damping here. It's just going to
oscillate forever. It's not going to blow
up. And it's, , excuse me. And it's not
going to go down to zero. It's just going
to keep oscillating forever and ever. It's
critically stable system.
Now, let's look at the inverted pendulum
where I'm moving the base, but in that
case, a is 0110. We already know, this
things is going to fall over. Right? So,
if you compute the Eigen values you get
one Eigen value to be equal to negative 1
and 1 to be positive 1, which means that,
we have one Rothton eigenvalue. This
eigenvalue that's going to spoil the
system. So this in an unstable system. So
now that we understand that eigenvalues
really matter, and they really influence
the behavior of the system, let's see, ,
excuse me, how we can use this to our
advantage when we do control design.
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