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.