0:18

The last week of classes we're actually going to look at,

Â what if the dynamics become more complicated?

Â But a common pattern you will find because the controlled development strategies

Â are very, very similar.

Â It's just as you do all the math, instead of just having these gyroscopics,

Â if you've got reaction wheels there is additional gyroscopics you

Â have to account for.

Â If you have CMG's different stuff.

Â Variable BCMG's even more stuff, so you can adds to it.

Â But the control development you'll find is very, very similar in that.

Â So we're going to focus here on rigid body.

Â If you want to have more on that then 6010, the follow on class to this,

Â actually delves much more into systems of reaction mills, and CMGs, and

Â variable speeds, and discernible laws, and everything that comes from them.

Â But in here too we've got a tracking control.

Â So we're bringing omega equal to the reference.

Â And the attitude is a body relative to reference.

Â And in the end, we developed this control law.

Â This one is for the unconstrained control.

Â So there's no limit in how big u could become, if your omegas go to infinity,

Â your u goes to infinity, which of course is not realistic.

Â because if we implement with thrusters, that's probably full-on is all you can do.

Â That's as much torque as you're going to get as you fire those things.

Â But also reaction mill, CMG's, and kind of attitude devices have limitations in how

Â strongly they can push, again, torque the spacecraft.

Â So this one you went through.

Â Now, let's see, what do we have to argue for stability?

Â 1:41

Kevin, let me start with you, here.

Â What did we have to do? What is the key part of this slide that

Â shows me this system is going to be stable?

Â >> We need to look at the the Lyapunov function.

Â >> Okay, so we have a Lyapunov function.

Â >> Need to make sure it's positive definites.

Â >> All right, which this form it would be for any non 0 states.

Â And we often break up the rates, energy like and then the states sometimes spring

Â like or here we get just natural logs that work well for the rigid parameters, good.

Â And what's the next step?

Â >> And make sure that [INAUDIBLE] >> Right,

Â which we actually imposed down here.

Â This is a common trick we do when we develop these controls,

Â that we take Lyapunov function, we take it's derivative,

Â we set it equal to something negative semi-definate.

Â But now, we note we set it equal to something in the rates, all right.

Â It's negative-definite, in terms of my rate error, not the state error.

Â And the reason for that, in a mechanical system,

Â you always need some rate to measure, which tends to be something squared.

Â State measures, take a derivative, rates appear on both sides,

Â we can factor them out and then that's how we equate them again, right.

Â It's the same process.

Â It doesn't matter if it's the spring mass system with cubics or rotational stuff.

Â The most mechanical systems has something like this.

Â So good, so this was actually the argument.

Â This one was negative semi-denifite only.

Â The attitudes are also important but don't appear in this expression.

Â So V dot can be 0 when we have 50 degrees error, in this mathematics, all right.

Â The question is, does that actually happen?

Â And this is a necessary condition, but

Â it doesn't guarantee, if this is only semi-definite, that means it is stable.

Â But it doesn't mean it is not asymptotically stable, right?

Â That's why we have to have additional arguments.

Â So what was the additional stuff we do, do you remember?

Â To look at asymptotic stability on such systems?

Â 3:47

>> Well, if we could make it, yes.

Â But unfortunately, for these kind of second order systems it's very tough.

Â I haven't seen one that actually does it successfully.

Â >> We're looking then at the second order derivative.

Â >> Exactly, so there's other methods.

Â If you do nominal control, you may have heard of invariance principle.

Â It's kind of related.

Â It's a different approach.

Â You find the largest invariant set where this thing manages and

Â then there's some extra arguments.

Â But the chain theorem I like very much,

Â because it's a nice approach especially for engineers.

Â You know how to take derivatives, you do it and

Â then we have to look at that, right?

Â So Daniel look at your chin, could you quickly outline that one for me?

Â >> Well, we keep on deriving to the Lyapunov function until we

Â get to something that's not 0.

Â >> Right, so

Â we take an extra time derivative of that second derivative, third derivative.

Â And you expect these things to be 0 for the even ones and you're looking for

Â an odd derivative that's non 0.

Â On which set, Kevin, do we actually evaluate these higher derivatives?

Â >> The set where the V dot function is going to 0.

Â >> Right, which in this kind of formulation is always tends to be

Â the rates of 0.

Â With integral feedback we had something where it was the measure of the rates and

Â integral term combined goes to 0.

Â But it's something of that form, exactly.

Â So good, we went through that.

Â Global stability we argued if you treat these as just going to infinity,

Â which we can do.

Â But of course there we exclude the case that the spacecraft tumbles 360 degrees,

Â because then our measure goes to infinity.

Â So you can talk about global stability, but

Â you still can't do a completely continuous tumbling body.

Â To do that, we kind of had to start talking about switch Lyapunov functions.

Â And what saves us there was that as we switch at sigma is equal to 1,

Â the Lyapunov function itself actually remains continuous.

Â Not smooth, but continuous.

Â And in those arguments you can do.

Â So we do have a globally stabilizing one and

Â here's the outline of what we had with asymptotic stability.

Â You take the second derivatives.

Â Why could we always assume that delta omega dots and so forth or

Â delta omega double dots are finite?

Â What was that argument?

Â >> [INAUDIBLE] >> Exactly, all right, sometimes students.

Â because you could plug in and in the spring mass damper system I show you,

Â you have an x double dot.

Â You can take the dynamics, plug it in, put it in x dot equal to 0, and

Â then show okay, if this is all that it vanishes.

Â But if you can argue that this is stable, this is finite.

Â 0 times finite has to be 0 and

Â you save yourself some steps in algebra so, that's kind of a nice approach.

Â 6:17

So now, we have asymptotic stability.

Â There we looked at robustness last time, so let's review that one quickly.

Â The way we write that mathematically, is we have now, three torques [LAUGH].

Â This is the torque we controlled u, due to thrusters.

Â Something's applying directly a pure coupled torque to the space craft.

Â We have no new torques L, which could be a gravity gradients, or pressures, or

Â radiation torques something we know, some outgassing on a thruster,

Â because the valve's stuck and it's not quite closed.

Â because that's a really common problem.

Â You can model that and compensate for it.

Â And then delta Ls are the unknown things that happen.

Â So, you can also think, if you're holding a certain attitude and

Â you're firing a thruster as an orbit correction and

Â one of the thrusters fires hotter than the other one.

Â This would give you a continuous disturbance, and

Â it actually would be fixed as in a body frame.

Â The following analysis, what you want to remember then is that we're assuming

Â if all the other stuff, that delta L is fixed.

Â And when I'm taking derivatives here,

Â these are actually technically body frame derivatives.

Â I put dots on there but my next book is going to fix that and make them primes.

Â They're really body frame derivatives.

Â So as seen by the body frame, we need these delta Ls to be fixed and

Â then we can make this assumption.

Â We did some other math.

Â The Lyapunov doesn't tell us what happens here, because this function is indefinite.

Â All I know is delta omega can't blow up to infinity, but that's not very comforting.

Â So we look at other steps and

Â we could turn it something like a spring mass tamper with a variable spring

Â stiffness and there's some stability guarantees, as we'll have steady states.

Â So my rate errors will go to 0, which is nice.

Â But my steady state errors,

Â if you have unmodeled torques actually go non 0, all right.

Â 7:53

So now, if we have unmodeled torques, we can predict what the error is and

Â as Daniel was talking last time, we get to pick K.

Â So that's what's called modeled disturbance rejection part of the gain

Â selection.

Â You may have gain selection criteria that specify how you over damped, under damped,

Â how rapid is your response, but

Â you may also have to consider external disturbances.

Â These gains help stabilize something when you have unknowns coming in.

Â Feedback is a fantastic way to get rid of ignorance, all right?

Â We think it's rigid but really not, we think it's point.

Â All the stuff that's going on, and to the feed back it's a way to clean things up

Â but you have to see to what degree.

Â It never does it perfectly.

Â And this gives you a nice analytic estimate.

Â So good, we're just jumping through these the way we fix that.

Â In linear control, you add integral but

Â with non linear motion it's not easy just to add integral feedback.

Â This will relate very closely to the last part of the lecture today,

Â where I'm going to show you perfectly linearized closed set dynamics.

Â And there it's actually easy to add classic linear proportional, derivative,

Â integral terms, but you pay the price somewhere else to get that kind of a form.

Â So here we're looking at how to modify the Lyapunov approach.

Â We still have our del omega, we still have a sigma.

Â And we're introducing a new variable z, which is the integral of attitude and

Â in integral of inertia times angular acceleration error.

Â This is what makes all this math work.

Â You're v dot, we found was still negative semi-definite.

Â And then to look at asymptotic conversions.

Â This is 0, if these two terms are summed to 0, right?

Â Not if del omega 0 is a del omega 0, individually.

Â So we apply this, if you only have known torques this control with integral

Â still converges, and everything should go to 0.

Â The more interesting part was when we included unmodeled errors, right,

Â then we get that extra term.

Â And at the end, we can argue that this has to be bounded.

Â So the integral term of something having debounded means these terms have to at

Â some point go to 0.

Â And this helped us argue that the states should go to 0, but

Â the z part with extra arguments doesn't go to 0.

Â But anyway, so that was integral which relates to today's lecture.

Â 10:25

It can read now as- >> If you say linear,

Â they're nearly linear, right?

Â I mean, fundamentally, they're non-linear.

Â If you double your angle, it's not exactly double the MRP, but

Â it's pretty darn close, right?

Â So we don't have a completely linear description in that sense, but

Â it's really, really close to a linear one.

Â In fact, for a very large range, typically well past 90 degrees.

Â 90, 120ish, depends on your tolerance, all right?

Â So in H180 nothing crazy happens, like you have with CRP.

Â CRPs has blow up to infinity at 180.

Â MRPs just start to, you get a little bit more non-linearity but it's not too bad.

Â So that's the key, with these things we can linearize.

Â And your MRP differential equation, the key thing to remember is this one.

Â This B matrix basically linearizes to identity.

Â So as MRPs are angles over 4,

Â MRP rates are rates over 4 essentially, and you can plug that in.

Â And if you wanted to, you could use this form with a complete P,

Â a complete inertia tensor and place roots and so forth.

Â But realistically we tend to have spacecraft where we've

Â picked the coordinate frame that is pretty darn close to principle.

Â May not be exact.

Â Maybe some astronaut put the suitcase on the left side instead of the right side.

Â Now we're slightly different.

Â But it's pretty diagonally dominated so that's a good assumption.

Â And the gains, that's a pretty common assumption too.

Â So you could pick your selections from here, right?

Â The next steps are not required for this step.

Â We've already done the linearization.

Â It was dead trivial because the kinetic, the closed loop side was already linear.

Â it was just the MRP dynamics that we had to linearize.

Â So that's kind of a nice thing of this, and then you do that.

Â So if you have a diagonal inertia, diagonal p matrix, you can come up with

Â this simple decoupled that one axis controls the attitude and rates of one.

Â Attitudes and rates of two, attitudes and rates of three, and

Â you can set your gains individually.

Â One of the challenges here, still, is we can have three different rate feedbacks,

Â but we still always with the MRP that we have only have a single gain on the MRP.

Â That gain had to be on the outside of the lateral log function.

Â It was k times natural log 1 plus sigma squared.

Â 12:34

If you'd move it inside as a matrix, the math doesn't work as nicely,

Â you have lots of muddle in your terms.

Â And then you can linearize them again, but it won't match up nearly as well.

Â So once you know where the roots are, you can look up on how to get natural

Â responses, damping, decay times, damped angle velocities.

Â It's kind of nice.

Â And so, we get now something that's for

Â the non-linear motion, globally asymptotically stabilizing.

Â With integral, we can make this also robust.

Â