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So, let's see.

Â If we want to do the complete six-degree-of-freedom, the six-state control.

Â It's three degrees of freedom, purely rotation, right?

Â If we want to do this.

Â Let's actually, part of the review,

Â let's go through the regulation problem.

Â In class, last time,

Â we derived the full-tracking problem which had extra complexity.

Â But let's go through the steps.

Â We can do that much quicker by just looking at this.

Â So, first question, as always,

Â if I want to bring this pen to rest pointing at a particular attitude,

Â how many states do I have to throw in there?

Â Or what state, specifically,

Â would I throw in there?

Â Brian, what do you think?

Â The attitude coordinates and rates.

Â Attitude and rate. So a lot MRPs,

Â we can use MRPs,

Â and we have the rate measure, right?

Â That's what we're gonna do because we want both of those things to come to rest.

Â Now for this system too,

Â we need a Lyapunov function.

Â It's positive definite in terms of all of these quantities.

Â And you could, in theory, come up with a complex function

Â that couples sigmas and omegas and so forth.

Â In practice, people tend to separate them,

Â but there's nothing that says it has to be true, right?

Â So we get to pick our prototypical function and we're gonna use kinetic energy.

Â For my rate measure,

Â and I need something positive definite in terms of MRPs,

Â so I think that one had a two here, k,

Â one plus sigma squared,

Â which is the normal sigma squared of this function.

Â So, is this function,

Â is this positive definite for all initial conditions?

Â Trevor?

Â For any omega and for any sigma,

Â it's only zero if you zero, and positive elsewhere.

Â So omega, for that first term it's not.

Â Why would it not be for this one?

Â Well, because... All I know the first term is positive definite because...

Â But is it globally? I mean,

Â is it positive definite for any initial conditions?

Â Yes.

Â Away from zero, right?

Â So if it's zero, it's zero.

Â That's what we need.

Â And, actually, any omega.

Â Great. Same thing here.

Â If sigma's attitude error is zero,

Â then natural log of one is zero. We have what we want.

Â If it's nonzero, this actually is always positive.

Â Is this function radially unbounded?

Â What do you guys think? Yes, in this it is.

Â I have one question, actually. If in the state,

Â we put only the attitude...

Â Yes.

Â If we set it to be zero, for example,

Â or relative to [inaudible] ,

Â why do we also need the rate?

Â Because if we say that if we control the attitude is in one position,

Â why do we also need the rates?

Â OK. Let me go to another page to do that.

Â So you're saying is, you know, let me just use this.

Â Natural log one plus sigma squared, right?

Â You take the derivative of this and you get k times omega transpose sigma, I believe.

Â Something like that, remembering correctly.

Â The thing you're controlling is a mechanical second order system.

Â If I don't have omegas up here and I differentiate this,

Â there's no way to apply in a differential form my dynamical system, right?

Â So right now, this is purely just numbers and kinematics.

Â There's nothing specific to the dynamics.

Â Is this a rocket body? Is this a multibody?

Â Is it a fuel, you know, who knows what the dynamics is of the system.

Â So that's where you would need that.

Â If you wanted to control this,

Â we could do it in a two-step loop which I discussed this morning.

Â Or if I treat omega as a control variable,

Â I could just do attitude and then control on the attitude,

Â but that gives me a steering law.

Â So I'm not giving a torque law with Newton meters and getting a steering law that says,

Â "Look, rotate to your left one degree per second."

Â And then, you know, and this is guaranteed to converge asymptotically.

Â And actually in fact, we do build laws like that as well.

Â I am developing here torque-level attitude control.

Â If you had a steering law, the several system,

Â you could write another Lyapunov function just for that,

Â get in a reference and make sure you track it precisely, right?

Â But now you have a two-separated system.

Â Individually, they're stable; coupled,

Â they're only stable if the separation principle holds,

Â that the outer loop functions at a much slower rate.

Â Its bandwidth is much slower than the inner loop.

Â The inner loops always have to be much tighter or run at a faster speed,

Â you know, update cycles.

Â So, different ways you could do that. No, good question.

Â So here, if we have this,

Â so what are the steps now?

Â Good. So we've got this.

Â This is good for any initial conditions.

Â It's radially unbounded.

Â We take the derivative.

Â This part, pretty quickly,

Â is gonna give me this,

Â and this part, I believe,

Â in the end is just gonna be plus k Sigma.

Â So, how to get from this math to here?

Â That's one of the homeworks you do.

Â You know, you kind of plug it all in,

Â the usual matrix math.

Â So now, at this point, this V dot,

Â nothing has been applied that's specific to this dynamical system.

Â So, we can actually plug in our dynamics in this term.

Â And let's do that next.

Â So, you're gonna have omega transposed minus omega tilde, I,

Â omega plus the control torque, plus k sigma.

Â What's the next step in this control development?

Â It's different ways we could go.

Â Daniel?

Â Notify the Vdot and just find the u in terms of everything else.

Â But if you just say you u is equal to -k Sigma,

Â plus omega tilde, I, omega, that will cancel this.

Â Then your V dot is zero.

Â You got a stable response, right?

Â But we know actually it's not gonna be asymptotically stable.

Â What else do we have to do, right?

Â You want this V dot.

Â If it's always zero, we never get rid of energy.

Â That means these will never get better.

Â Ever. Right? So somehow we want to at least get rid of

Â V. We want a V dot that's not just purely zero ideally.

Â And we can do that.

Â This is basically omega transpose something.

Â The trick we keep playing in mechanical systems is I'm gonna say, "Look,

Â let's set it equal to this negative semidefinite function."

Â In mechanical systems, kind of what you were asking earlier,

Â we tend to have a rate measure and a state measure.

Â And when we differentiate, the rates become accelerations,

Â plug in equations of motions.

Â But both terms will have rates that you can factor out and that

Â allows you to always do that trick on a second order dynamical system.

Â But this is good because then we can pick and putting in a fully matrix P,

Â it's symmetric positive definite,

Â gives me lots of knobs to tune.

Â And now we can say, 'well,

Â this bracketed part has to be equal to -P omega'.

Â And if we do that,

Â then basically we add an extra P omega to here,

Â right? And off we go.

Â And we're done.

Â Now we have a control,

Â but this V dot is only negative semidefinite.

Â It's negative definite in terms of omegas.

Â So compared to the earlier control,

Â I didn't know what happens to omega now.

Â Here, at least with this one,

Â I know right away my omegas have to go to zero.

Â That's what this V dot is gonna go to zero. So that's good.

Â I just don't know yet what happens to convergence.

Â That's a new thing we'll cover today.

Â But let's play with some modifications.

Â So here, we compensated for all this stuff.

Â Great. And we have this one in there.

Â Did we have to compensate for this one term here?

Â Do we need this?

Â Kevin, what do you think?

Â Not really, because when we carry out the multiplication,

Â we have omega transposed times omega tilde.

Â Exactly. So, here, kinetic energy is actually a physical kinetic energy,

Â it's not just some energy-like term.

Â And this is literally a non-working term.

Â So these gyroscopics don't- that's the stuff we saw with torque-free motion.

Â They're there, but with no external torque.

Â T was constant.

Â T dot was omega dotted with L. And if you don't have any external torque,

Â Tdot is definitely constant.

Â So, all these gyroscopics have to be nonworking.

Â In more complicated tracking problem,

Â this morning we saw del omega transpose del omega tilde.

Â It's not exactly energy,

Â mechanical energy, but it's the same mathematics.

Â It's a term that cancels out to zero. So, yeah.

Â So you could use in your Lyapunov analysis and say,

Â 'look, this is actually gonna cancel'.

Â And I can write a new control that is simply- I still need to compensate for this.

Â So that gives you your feedback on attitude,

Â and then you have feedback on omega.

Â This will also be globally stabilizing.

Â In fact, you can do the same steps you are about to do for

Â showing this asymptotically stabilizing.

Â So that's really slick.

Â This one is a nonlinear control because it's got a quadratic rate measure.

Â This one is a perfect linear control,

Â but it's globally stabilizing on a nonlinear system.

Â It's really nice and easy to implement.

Â So you can see we can often reuse the same Lyapunov analysis,

Â change some terms, add a term,

Â or find, well, I could compensate.

Â What's the difference between them?

Â Kind of like the multilink system.

Â We had two controls that are both guaranteed to be stable,

Â but the performance can be different.

Â So here you are feeding back on a quadratic measure.

Â For spacecraft, typically, these rates tend to be slow.

Â People declare emergency as if they're tumbling more than three degrees per second.

Â That's not much of a tumble,

Â they come from a ferry's wheel or a playground, you know?

Â So that's a small number squared,

Â it's actually quite small, not a significant part.

Â But what the nice thing is,

Â feeding back on this term,

Â that's where we get those closed loop dynamics we saw this morning,

Â where we had a nice linear form in omegas.

Â If we don't compensate for that,

Â your closed loop dynamics has a quadratic form.

Â And somebody this morning was asking about gain selection.

Â So when we linearize and you have a more nonlinear closed loop dynamics,

Â your linearized performance prediction won't be as

Â largely valid as one that that feedback linearizes more.

Â So there's performance tradeoffs,

Â what you want to look at, you know?

Â Are you dealing with large tip-off velocities?

Â Maybe you don't want to feedback on this because you will saturate quickly.

Â Or maybe it doesn't matter.

Â Maybe you can still prove stability.

Â But anyway, that's some quick ideas on how we can generate this.

Â So... So, good.

Â That kind of wraps up our review of all of this.

Â So we did the more complicated one where we have reference tracking.

Â We talked about global stability and, obviously,

Â we have to switch as well because radial unbounded meant sigma goes to infinity,

Â but sigma infinity means a 360 rotation,

Â so you can never do a revolution past the origin again,

Â and we want aircraft to be able to tumble and tumble and then recover whenever.

Â So with MRP switching,

Â we can break it up into discrete sections that we stabilize.

Â So we still get global stability,

Â and you just fight it as long as it makes sense.

Â And then you just let it finish up the long way around and

Â it will catch it there in the flip side.

Â Excellent. Good.

Â