0:40

>> I'm going to say no.

Â >> [LAUGH] Yeah, the way I'm asking question, that's a good 50-50.

Â Right, but if you have the state space form,

Â we're going to have this quickly there.

Â It's typically written this way, right?

Â They always write these systems as first order equations,

Â I'm going to show that once.

Â But the states that go into the system, a rigid body, Casey,

Â how many degrees of freedom is it now?

Â >> Three or six.

Â >> Right, if you do translation and rotation.

Â If you only do attitude, it's three.

Â So the number of degrees of freedom,

Â that's the number of states required to perfectly describe the dynamical system.

Â So the quick way to remember that is just if you know the degrees of freedom on

Â the system, that's how many states you must have as a minimum representation.

Â Sometimes we use more, and we've seen that.

Â I could use quarternions, and all the sudden it's four.

Â You didn't make it a four dimensional problem.

Â I could use DCMs, it didn't make it a nine dimensional problem because I have all

Â these constraints I have to satisfy, right?

Â So this formulation is going to be, well, for the most part, unconstrained.

Â But it's just that's the number of states that we're throwing in.

Â That's what we've chosen to do.

Â 1:44

And you can have up to N.

Â Typically, N has to be at least the size of

Â the number of degrees of freedom of your system.

Â Then we have these differential equations where autonomous means

Â it doesn't explicitly depend on time.

Â Yes, these states' x varies.

Â Bring my sniper system, x goes up and down, up and down.

Â That's fine, but there's no driving force that says

Â u is equal to 3 times t just because, you know?

Â So the time isn't explicitly in there.

Â If it is explicitly in there,

Â it's a non-autonomous systems and then proofs get a lot harder, actually.

Â Then you have to talk about uniform stability and is it stable for today,

Â great, but what about tomorrow?

Â And this is classic problem in orbits because you want to intercept

Â an asteroid that's about to hit the Earth.

Â And yes, today you could be stably getting there, but

Â tomorrow it's not enough fuel, it's not going to work.

Â Now your position relative to that asteroid impacts your stability argument

Â and that makes it a non-autonomous system.

Â So we're doing primarily time-independent systems, kind of what we have.

Â Our masses are not deploying slowly with time, we have a rigid body,

Â all this kind of a thing.

Â So we can rewrite this.

Â This is our just natural dynamics, always x = to f.

Â But that's a first order form, so let's look at that.

Â So if we have theta + omega n sine theta = 0,

Â that's my natural dynamics.

Â How would I write this in first order form?

Â 4:19

Anyway, no excuse, I was wrong.

Â This would be basically what we just did.

Â It would be theta and theta dot.

Â Those are the two things.

Â If you have N degrees of freedom, we have three attitudes,

Â three positions, it's a second order differential equation.

Â I also have to keep track of their velocities.

Â So this becomes a 2 times N here would be 2 times the number of degrees

Â of freedom, yes.

Â That would make more sense, good.

Â So we've got that and that's basically so if we go here,

Â if we just wrap this up and you save this, then you're saying,

Â okay, x dot is going to be theta dot and theta double dot.

Â And theta dot, let's call this x1, x2.

Â Then theta dot, the x1 dot is going to be nothing but x2.

Â And x2 dot is going to be -omega and

Â sine of x1 that you would have, right?

Â So you'd just introduce these forms.

Â So you could always take a set of second, third, fourth order differential equations

Â and introduce, as you're talking about, these other derivatives.

Â That's what we're doing, right?

Â So, this can always be done, so this form here is very general.

Â And we have our first order differential equations.

Â Now we want to talk about control.

Â So we don't just have x double dot plus something x equal to 0,

Â but it's equal to some control effort.

Â There's some server, some thruster, something you're controlling, right?

Â So you have an authority over this.

Â How much thrust is produced?

Â How much torque are you creating with wheels?

Â And this has some function, so depending on the states that happen here,

Â there's some function that could be going in here.

Â This function is written as g(x) so it could be non-linear.

Â But you can also, as you will see in several problems we do,

Â we have some elegant controls where we could just throw in -k times sigma,

Â sigma being my attitude error.

Â And it's still going to be stable, but

Â we can also throw in non-linear feedback controls.

Â We'll see some of them. So very general.

Â It's just our control formulation is a non-linear, encompasses linear.

Â And then when you apply this into the dynamics,

Â you get what's called the Closed-Loop System.

Â So that's when you had the natural dynamics, which were unstable,

Â you apply some control, hey, I'm going to take this attitude angle and

Â feedbacks on it, and that's going to stabilize it.

Â And then once you apply that control, that's the Closed-Loop System.

Â So you just plug in that u into it, and

Â that's the stability that we're going to be looking at.

Â Not the original system, but

Â we really care about the new system that we've created this way.

Â Equilibrium states, we discussed this.

Â We've seen this with the rigidi body spinners,

Â we've seen it with dual spinners.

Â It's when, if you have always this form, x dot = f, for

Â what set of states x is this derivative across all the states going to go to zero?

Â That means in an example we had earlier with the theta double dot + omega

Â n sin theta, that vector was theta and theta dot.

Â So I have to figure out for what set of states is that all going to vanish?

Â And typically, you also need theta dots to be zero.

Â You're really just looking for the states in the end.

Â And when you find that, that's your equilibrium point.

Â 7:28

This does not imply stability.

Â This just means if I put a spin perfectly around that axis, I can't do it.

Â It's too hard.

Â It always flips. But if you did a perfect spin,

Â that would be an equilibrium.

Â It's just not a stable equilibrium, right?

Â The other ones are much easier to do.

Â Those will be the stable equilibriums.

Â So that's what we're going to be defining.

Â That's what nature gives you.

Â But what you're going to find too for the tracking problem,

Â there's some mathematical tricks.

Â And the next thing you do it looks very, very similar.

Â And essentially what we're doing is if this is what nature gives you,

Â I can make other points equilibrious in space with feedback.

Â I could make something hover and fight gravity and

Â that becomes a stabilizing thing, if I have enough fuel and thrust.

Â That's how your quad rotors work, right?

Â The little UAVs and cameras flying around.

Â There's no equilibria that should be hovering over your house at 15 meters.

Â But your control made that work, and that became all of the sudden,

Â equilibrium, it hovers, you know?

Â