Well, x0 was equal to 0, so this whole thing is simply equal to B times u0.
Well x2, that's A times x1 plus B times u1. Well, x1 is Bu0, right, so the whole
thing becomes ABu0+Bu1. x3, well if I plug things in, I get
A^2Bu0+ABu1+Bu2. In fact, if I keep going to xn, I get A
n-1 Bu0, blah, blah, blah, blah, right?
So, we seem to have a formula, really, for
for where we end up. And, in fact, what we want to do, of
course, is make xn equal to this desired point.
And what I can do is I can rewrite this thing, this equation,
which is really an equation in the u's in the following form.
I have u n-1 times B. Well, that's this term.
And then, all the way down to u0B an-1, which is this term right here.
So, what I'm doing is I'm just saying, this is what I would like to solve with
respect to my u's. Well, when can I do that? Well, first of
all, this matrix here, the B AB all the way to A n-1 B,
it's a fancy-looking matrix and, in fact, it's an important-looking matrix.
I'm going to call this matrix, gamma. And gamma is an n by m times n matrix,
where n is the dimension of the state and m is the dimension of the input.
Well, this solution, or sorry, this equation x* is gamma times this U u
vector has a solution in terms of u's, if and only if the rank of gamma is equal to
n. And rank is the number of linearly
independent rows or columns in gamma. Well, so that's what we want to do,
right? We want to somehow have a gamma that's rich enough.
And it turns out that this thing generalizes, this way of thinking about
the control problem, generalizes quite nicely to continuous time systems as
well. So, if I have a continuous time system,
x dot is Ax+Bu, x in Rn. Well, first of all, this system is
completely controllable, which we're going to call CC.
If it is possible to go from any initial state to any final state,
so, meaning, if I start here and I want to end up here,
there is a u that takes me between these 2 points,
for any such points, that's what it means for the system to be
completely controllable. And now, I'm going to define gamma again,
which is this B AB all the way to A n-1 B.
This is known as the controllability matrix.
And here, ladies and gentlemen, is the first theorem of this entire course.
Controllability theorem or complete controllability theorem 1.
The system is completely controllable. If and only if the rank,
which means the number of linearly independent columns of gamma or rows,
it's the same. if and only if the rank of gamma is equal
to n, where n is the dimension of the system.
So, this is the rank test as its known and its a way of checking controllability
of a linear time-invariance system. So, let's see what happens.
Here I have two different systems, 2 0 1 1 x plus 1 1 u is the first one and 0 1 0
0 and 0 1 u, fine, there should be an x here, by the
way. so the lower one is the point mass the
upper one was one where we actually saw that pole-placement was not possible.
In fact, pole-placement is possible for the lower system and not possible for the
upper system. This is, in fact, the system used when we
couldn't do it. Well, let's look at them. First of all,
this top system is a two-dimensional system,
n=2. And gamma, in this case, is B AB, all the
way A n-1 B. But n-1=1 so gamma is simply B times AB.
Well, the lower system also istwo- dimensional.
Gamma is equal B times AB. Well, let's compute AB then.
It turns out to be equal to 2 2 for the upper system and if I apply this into
gamma, I get B here and AB here. So, this is 1, 1 the first column and 2,
2 the second column. And if I multiply the first column by
two, I get the second column. So, this thing does not have two linearly
independent columns. In fact, it has one linearly independent
column. So, the rank of gamma is equal to one,
which means, it's not completely controllable.
Well, looking at the lower system. Well, this is AB, this is my gamma, 0 1 1
0. There is no way I can multiply this
column by anything to get this column. So, the lower gamma has rank equal to 2
because that's two linearly independant columns.
So, it is completely controlable. So here, pole-placement not possible,
not completely controllable. Here, pole-placement possible,
not completely controllable. It seems like we're ready for theorem
number 2. So, if I have u as -Kx, x dot then become
A-BKx. This is the close loop dynamics.
Then, controllability theorem number 2 says that pole-placement to arbitrary
eigenvalues is possible if and only if the system is completely controllable.
So, what this tells me is that we need to check controllability.
If we don't have complete controllability, chances are we're not
going to be able to control them. So, this is the obstruction to
pole-placement. And, in fact let's let's see how we would
actually compute something like this for real.
So, let's say that I have again, my point, my, my system.
Well, in MatLab, luckily for us, we don't have to compute things at all.
We just say, here is the controllability matrix.
So, gamma or g is controllability of AB, we can check the rank.
In this case, we get the answer out, 2. So, the rank of g is 2.
n is 2, in this case, so we have indeed, a completely
controllable system, which means, again I forgot an x there, I
apologize, which means that we can place the poles
wherever we want and we all, it also means that it's possible to go between
any two points. So, let's say that I start here and I
want to go to x*. Is it possible to go like this? Well, I
know it's possible to go from x0 to x* but it turns out that just because you
can go between any two points, doesn't mean you can follow any trajectory
because this system, the point mass, well, x1 here is
position, right? And x2 is velocity. Well, if I'm here, that means I have a
positive velocity and a positive possession.
And now, what I'm going to do if I start moving like this, I'm going backwards,
meaning, x1 is reduced with a positive velocity and there is no way I can go
backwards with a positive velocity, so this is not possible.
So, what do we need to do? Well, here I have a positive velocity so I'm going to
keep growing but I can reduce the velocity.
And here, I have zero velocity, right? Down here.
And that negative velocity, I start going backwards.
And [SOUND] I'm going backwards. And then, I can start going forward again
to get to x*. So, this is how you would have to go from
x0 to x*. So, you can't follow arbitrary
trajectories just because the system is completely controllable,
but you know that you can go between arbitrary points. So, that gives us the
tools that we really need to understand when we can control the system and that
tool is controllability.
