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>> Hello and welcome to our third Glue lecture.

Â To sort of round out the third week of the course, today's Glue lecture is titled

Â Systems, and I'm going to talk kind of how we think about systems in this course.

Â And this will be helpful for quiz 3, we'll do 2 examples that are similar to

Â questions on the quiz. So the goal of this lecture is to think

Â about how inputs and outputs can define a system, and that's really How we are going

Â to think of systems in this course. And then, I hope to kind of get you used

Â to this matrix representsation, this states based form of systems by doing an

Â example of putting a second order system into states based form.

Â And to do a linearization of a non-linear second order system as well.

Â So first, let's just think for a few minutes about what a system is.

Â So, you know, in general, we might think of a satellite being a system, as a fancy

Â technological device that, that does something that we design it to do.

Â We might think of it as something that has this kind of confusing block diagram.

Â Description. We might think of it as something in our

Â car, right or a, or a interface which was, a technological interface that we, we

Â interact with.. Of course there's also the solar system,

Â that is somehow a system of planets that's working together.

Â That we think of, or systems of the body, different organs that work together to

Â perform basic functions. So the point of this slide is just the

Â idea that the word system a lot and we might all have different ideas about what

Â a system is and, you know, this course is really kind of all about systems.

Â And so I kind of want to clarify what we'll mean, and, when we think about them,

Â and the idea, really is that, in this course, we want to understand how inputs

Â really relates to outputs. And so any of these systems, we can be

Â considering different inputs, different parts of the system that we can control or

Â actuate on and different outputs of the system, different things that we might be

Â measuring for different goals. So you know, on, you know, a satellite,

Â you might be controlling orientation. And so you might think of different inputs

Â and outputs there, or you might be controlling altitude, and so now you have

Â different actuators, different goals for the altitude than the orientation.

Â So we could think of those really in this course, as, as distinct systems that have

Â different control objectives. And so the picture that we're thinking

Â about is really this idea that we have some input here that we're mapping to an

Â output defined by these A, B and C matrices.

Â And so getting into the first example, we're going to talk about how to move a

Â dynamical equation that describes some system into this form that looks like

Â this. And at first that can be a little bit

Â tricky and so I want to go over it and of course remembering that both of these

Â equations, have some initial condition. F of 0 equals f not.

Â X of 0 equals x not. Keeping that in mind, we're just going to

Â think right now about how to go between something that looks like this to

Â something that looks like this. And, really that's just a matter of

Â picking our state variables, de, deciding what our state is, and then to/g choosing

Â these inputs and outputs. So which of these variables is what we're

Â controlling and which of these is what we're measuring.

Â And then we're going to write this second order differential equation as a pair of

Â first order ones, so that our second order system is now represented In terms of two

Â first order equations. And then, put these in terms of our x, our

Â u, and our y, right? So first, for this example, we'll select

Â our state to be f and f dot That's a pretty common choice for, for a state at

Â not necessarily the only choice, but you'll see that it's convenient because

Â when we think of x dot, x dot is going to look like this right?

Â Oops, f double dot. And now we ha, we, we need to be able to

Â know something about f double dot right to for this choice of state, and so of course

Â we do, and so that's why you'll see that Often as the choice of the state.

Â And now we pick our input to be this function p here, so this is somehow,

Â someway that we're influencing what f, the, the various derivatives of f are

Â doing, and now we'll chose our output to be f, but again this could be chosen as f

Â dot, this could be chosen as both f and f dot, And that will determine the form of

Â these a, b, and C matrices. So those choices are important and usually

Â they are described in the problem setup. So now that we have chosen our state, our

Â input and our output we need to write this equation in terms of two first order

Â equations. So we can think of f-dot and f-double-dot.

Â Well now we're going to write them in terms of our state variables x 1 x 2 in

Â our input and our output. So, here we have f dot.

Â Well that was x2, right? So we can directly relate f dot in terms

Â of x2 and now, x2 of course is related to x1 through this, through this time

Â derivative. And then for F double dot, well F double

Â dot is x two dot, and this equation that we were given here.

Â So where I'm dividing by everything by m, and now I'm writing remember X two is F

Â dot, so I filled that in here, and here I'm filling in that X one is f.

Â And here I'm filling in that u is p. So just rewriting this to this.

Â And so now if we come back, this is exactly what I had on the last slide.

Â X dot in terms of the other state variables as we renamed them in the first

Â step. And so now it's just a matter of getting

Â it to look something like a x plus b u, right?

Â So I need to pull out, we need to, what I would like to do is write an empty matrix,

Â and I know that I want that to be times x, right?

Â So, I write x here, so that's the ax term. And, now it's time some other matrix, this

Â one's going to be tall and skinny, times u.

Â Okay, so here, I just need to kind of rearrange these terms.

Â And here we'll have, so what, remember that this is x 1 dot, and x 2 dot, that

Â we're trying to equal. So what does x 1 dot equal?

Â Well, it equals x 2, so we just need no x 1 contribution, and 1 x 2 contribution,

Â and then no u. >> So, now this first line, because, is

Â correct, right? This x1 gets multiplied by the first entry

Â of the matrix, x2 gets multiplied by the second.

Â They get added together, so this all adds up to x2.

Â Now, for x2 dot, well that equals this whole term.

Â So now we have a contribution from x1, x2 and u.

Â And that contribution for x one is beta over, and from x two is alpha over m.

Â And now from u, we need to multiply u by c, to get this o, over m, to get this term

Â tied up. So That ends up looking like this, right?

Â Where this is our a matrix, and this is our b matrix.

Â And we can do the same thing for y. Y equals c x, and since y was f, and f is

Â x one, you get just a one in the, in the entry of the matrix that x one will be

Â multiplied by. So this is how you work.

Â Something that looks like this and to stay space form, So now we have that our system

Â where we picked our state, our input, our output its represented by these three

Â matrices A, B and C. So to pick up to get into example two.

Â This is going to be an example of linearizing a non-linear system.

Â So this is a system where we have this z squared term.

Â Sorry let me fix my marker here. So now we have this z squared term, which

Â means we can't do what I just did where I Wrote this empty a matrix and pulled out x

Â and just filled in the indices right. We have this z squared term which means

Â it's not going to fit into that abc form. And so what we do is we linearize it

Â around some operating point, and here that point is going to be 0, 0.

Â And so first what we need to do, of course, is just write this guy, the second

Â order. Differential equation in terms of these 2

Â first order equations. So just I'm writing this kind so that you

Â remember that x1 dot because of this twice that we;re given in the problem that x and

Â z and z dot and our input u is tau. That this is how, that d dot and d double

Â dot relate to x1 dot and x2 dot. And so now we have that x1 dot is x2, x2

Â dot is this whole thing because that's what d double dot is.

Â Again writing it in terms of our state variables.

Â But now, again we have this x1 squared, so we can't just write this in abc form.

Â So the way we do it in this case is to just compute this A matrix, this

Â linearization. So here we have f1, f and f2 which are

Â going to be, here's our f1, here's our f2. And so we just have to compute these

Â derivatives in terms of our stay variables.

Â So this first, let's I'll write it out. For A, we have the derivative of this guy

Â with respect to x1. That's 0.

Â Now the derivative x2 with respect to x2. That's 1.

Â Now we look at f2 to compute the second row.

Â Derivative of x2 with respect to x1 is 2lx1, this term.

Â Both of these other two terms go to zero, and now f2 in terms of x2 is just gamma,

Â right, because here's the x2 term. And so now this is our a matrix evaluated

Â at our operating point, right? So this term is going to go away, and this

Â whole thing is going to equal zero, zero 0,1,0 gamma, for this choice of this

Â operating point. So we're linearizing around that, around

Â that point. In the case of the pendulum, when it's

Â hanging down. Right?

Â That we're just going to be able to control it around that hanging down

Â position. So, in nicer text, this is what we have.

Â And we do the same, a very similar process for the B matrix, where Now the b matrix

Â has a different size, right? But we have the derivative f one in terms

Â of u, and the derivative f two in terms of u.

Â So that's these two equations again. And we get this, that this is zero c, and

Â evaluated at any operating point this is actually going to be the same thing

Â because it's not And a function of any of our state variables anymore.

Â So now, the linearization of this system is given by these 2 matrices.

Â And our output is going to be a similar process, that I think Dr.

Â Iverson went over in lecture, as well. So with that, good luck on quiz 3, and I

Â hope you're enjoying the course.

Â