0:06

Now, in our modeling so far, we've said, okay, these are equations of motion,

Â user control torque.

Â L is the known external torque.

Â The f gravity gradients, atmospheric drag, any other disturbances,

Â some tank is outgassing and causing a disturbance.

Â You could include that but you never know it perfectly.

Â So how sensitive is this to unmodeled errors?

Â So the Delta L here, treat that as the unknown, that's our ignorance.

Â L is the part we know, and then Delta L could be the part we don't know.

Â And if you think Apollo 13, it was leaking gas, right?

Â And it was twirling everywhere.

Â A lot of uncertainty exactly what the torques are when acting on that system.

Â So we want to investigate this.

Â If you plug the very same control into the development we did earlier,

Â at some point we had an I omega dot and you plug in the equations of motion, and

Â you use that exact same control that we developed, that feedback compensates on L.

Â If you didn't have this, you would end up with the same v dot again, right?

Â That's the same development.

Â But with this delta l, there's one extra term we don't get rid of.

Â We don't feed back compensate perfectly for the l, but

Â only the part that we know, then the part we don't know that's still there.

Â So you end up with this extra term, that's a delta omega transpose delta l.

Â And this is negative semi-definite.

Â This term itself is negative definite in terms of delta omega, but

Â this term could have either sign depending on the uncertainty of delta l.

Â I mean, what's the signs of delta l?

Â What's the signs of delta omega?

Â This could be positive, it could be negative.

Â This is an example of an indefinite component.

Â So once you add up negative semi definite plus an indefinite there's no guarantee

Â this is even negative in this case.

Â So we can't from here on just use classic

Â theory to argue stability of this kind of a problem.

Â 2:34

This term is only linearly dependent on delta omega.

Â So a common assumption you find in control papers they assume.

Â They will say assume bounded unknown external disturbance.

Â There is a bound that you can come up with.

Â I don't know if it's one or two but I know it's less than five newton meters, right?

Â So if you can come up with a bound, you can imagine if this gets large enough,

Â if the rates pump up large enough at some point this is going to compensate again.

Â And this stabilizing effect is going to be stronger than this potentially

Â destabilizing.

Â And we don't even know if it's destabilizing.

Â But there's no way that delta omegas could go off to infinity in this one here.

Â Because if delta omegas get too large,

Â this v dots always going to bring you back down again.

Â But it's not a very strong argument, it's a pretty loose argument, in fact.

Â But that's a little bit of insight that you can get out of this equation.

Â So we have to look at the close loop dynamics a little bit more carefully.

Â 3:25

Earlier we had these, if you have feedback compensated for

Â l perfectly, you would have all these equal to 0.

Â Now, with this delta l this is extra term we don't compensate for it.

Â So it's still left there and I put on the right hand side, that is a delta l.

Â And what I'm doing next is I'm making a reasonably common assumption that I'm

Â saying this delta l is actually constant over a short enough period of time.

Â Which is kind of, if you have out gassing or other disturbances,

Â drag effect on a nominal attitude.

Â That's a reasonably good assumption.

Â And that allows us to differentiate this equation once.

Â So delta omega dot becomes double dots, and so forth.

Â And instead of sigma dot, I'm plugging in the differential kinematic equation.

Â Now, I have something in terms of all delta omegas.

Â And this is going to be approximately zero, because I'm going to hold my

Â unmodeled torque reasonably constant for that time period.

Â So this looks actually like mx double dot + cx dot + kx but

Â our stiffness here varies with the states.

Â So it's a time varying thing.

Â So it's not quite the classics premass damper system but

Â something with a time varying stiffness.

Â 4:35

To argue stability of the system, you need to be able to argue that this matrix b

Â is actually positive definite, over the departures that we're looking at.

Â And to be positive definite, we always check, omega transpose b omega.

Â Is that bigger than 0?

Â If you plug in the v definition that we would have.

Â Which was one minus sigma squared identity plus two tilde plus two the outer product

Â and put this together.

Â And then rearrange because you can carry out just products just, you know identity

Â times omega is indentity times omega transpose, this gives you this.

Â Omega transpose Why is this term going to zero?

Â Omega transpose, sigma tilde, omega.

Â 5:32

So why does omega transposed sigma tilde omega.

Â Why does that go to zero?

Â >> The sigma tilde times omega is equivalent to sigma cross omega, so

Â we are perpendicular to omega.

Â >> Yep exactly, with this being a cross break operator you

Â could just put in a negative sign and make this as omega transposed omega tilde sigma

Â with a minus sign or recognize that this is a cross product.

Â Whatever the answer is has to be orthogonal to omega and to sigma.

Â 6:04

>> And that's good enough.

Â And that's actually makes it go to zero again.

Â Yeah, exactly.

Â So with that little trick, we'll like to still see the matrices.

Â You're all set.

Â So that's how it vanishes.

Â So now if you look at this, if sigma squared is not one and

Â sigma squared being equal to one is 180 degree attitude, right?

Â You're just completely upside down.

Â Anything else, you're back to something less than that.

Â This part is definitely pause indefinite in terms of omegas, so that's good.

Â And this term here is positive or 0.

Â It's positive semi definite terms of omega.

Â Maybe omegas and sigmas are orthogonal in which case that product could go to zero.

Â But this part would guarantee to be positive so the whole thing is positive.

Â Positive zero is still positive.

Â There's that one point, 180 degrees where this doesn't work.

Â Where you now have something where it's a zero times this.

Â And then you left a semi definiteness.

Â So if we exclude this, and really with this external torque,

Â what we'd use as announce is more like the steady state error at the end.

Â I'm trying to predict what that error is going to be.

Â 7:17

Why is that nice?

Â Because then if you have this dynamical system, the second order,

Â eventually creation of delta omega, this arguments you're making control theory,

Â this has to have a steady state solution.

Â Which means, these things will double dots and dots go to rest and

Â you're only left with this part that you have to deal with.

Â And, So

Â since B was nearly orthogonal, it's definitely full ranked.

Â Remember the inverse of B was the transpose divided by 1

Â plus sigma squared, squared.

Â Something like that, nearly orthogonal.

Â It's definitely a full rank matrix.

Â So the steady state part is actually going to be zero.

Â So with these arguments,

Â what we can make is even with external disturbance, I don't know yet

Â what happens to the attitude error but my rate errors will actually come to rest.

Â But a key assumption is also that your disturbance is constant.

Â A body fixed constant.

Â But you're not doing a time-varying disturbance,

Â otherwise this would not hold.

Â And you'd have, you'd really have to go to numerics, the analysis.

Â I haven't seen a nice analysis on that one.

Â So for the constant-disturbance torque, this steady-state condition holds.

Â If we go back to the original equation, if the steady-state

Â condition delta omega is zero, then it's derivative is also going to be zero.

Â All this drops out,

Â and then the steady-state attitude error, you can actually predict as being

Â this unmodeled torque which we said is constant unmodeled divided by that gain.

Â 9:14

>> That's not quite You can, but

Â you're changing the dynamical system to do that, right?

Â because this would be like arguing a spring mass damper system.

Â We said, hey, this thing,

Â if we tilt it upside down, now it's going to deflect one meter, great.

Â You're saying, hey, if I want to be ten centimeters closer, forget that spring.

Â I'm putting in a new spring that's stiffer, now I'm here, right?

Â So that's true from a dynamical point of view.

Â From a control perspective, changing the stiffness of the spring is not

Â one of the variables necessarily that you'd have.

Â This is a controlled gain variable, true, but so you can play with that.

Â If you have your controlled solution, I think that's what you have to look at it.

Â You pick the gain of 1, very popular gain.

Â And you will end up here regardless of how close you started out from, right.

Â So what do we call that kind of a stability

Â where we always- >> Lagrange.

Â >> Lagrange, boundedness, exactly.

Â So with unmodeled external torque, right now,

Â we just have bounded Lagrange stability.

Â But you can predict with this gain and you can have a bound on

Â hey it might be between point one and one, so let's just pick one as the worst case.

Â I can give you an upper bound on how big that error is.

Â Now as a control designer you can do exactly what you were saying,

Â you would pick a key going you know what, I need enough,

Â they call this disturbance rejection with this proportional feedback, right?

Â It makes it a little bit more robust and

Â had the impact of the ignorance won't be as bad on your attitude error.

Â So you can make K big.

Â 11:02

What else is an issue?

Â >> You have saturation >> Right?

Â You might hit saturation limits so that might be an issue.

Â So astronauts might really complain and fire you as a control designer and

Â you may be breaking the spacecraft.

Â The wings are dropping off because you're actuating so hard.

Â I mean, that can happen.

Â Space structures are very flimsy.

Â You know, there's great thrusters.

Â You better check the structural limits with the solar panels and when they snap.

Â When you are doing like we are thinking orbit insertions right now.

Â Big clusters firing.

Â It's a challenge right?

Â So you want to make sure things don't break.

Â But what else is happening?

Â When we get feedback on sigma.

Â Right? The control was minus K times sigma.

Â So if I have the big K, I'm being very aggressive.

Â But what's the other problem if you look at feedback and measurements?

Â How good is that measurement going to be?

Â 11:45

All right?

Â Especially noise, if you have noise in the measurements times a huge K, you've

Â added a huge issue because you're deriving it from not the real attitude error but

Â the sensed erroneous attitude error times a big gain and that might excite all

Â kinds of stuff and then you might also with the performance kind of what your

Â talking about to is your going to potentially excite unmodeled modes.

Â You thinking it's rigid but once accelerated that quickly this stuff is

Â going to bend and slosh and [INAUDIBLE] with polar bear emission to you know so

Â those things are considerations as well.

Â So we can't practically typically just make this infinitely large but

Â you do kind of pick at the ends this is the performance you get and

Â maybe your going to bump it up a little bit if you have disturbances.

Â In a linear system, how do we make it robust unmodel torques,

Â if you do your classic linear control?

Â 12:39

>> Integrator. >> Integrator, right,

Â if you have integral control, it helps make it much more robust,

Â and I'll be showing you how we can do integral control with all the stability,

Â with complete non-linear stability guaranties.

Â It gets complicated compared to this but it builds on the same mathematics, so

Â we're kind of adding layer and layer.

Â 12:58

So if we do an example here, that's my unmodeled torque in the body frame.

Â It's a fixed number.

Â Just pretend some valve didn't close properly,

Â you've got some outgassing out of that valve, that's a particular problem.

Â Now you've got this disturbance.

Â And we want to do a maneuver just bringing it to rest, regulation problem,

Â as we're predicting our status rate error should go to zero.

Â So my rate error should go to zero.

Â And my status state attitude error should be this divided by,

Â I think I have a gain of one.

Â Yep.

Â So it's easy to do the math.

Â This should be my attitude error in MRPs.

Â If you run it through the simulation then, you will see my rate errors,

Â I'm tumbling at the beginning getting things going with the control.

Â But it stabilizes and my write errors do go to 0 as predicted, so that's nice.

Â And the attitude errors of MRPs, they start out with here.

Â They try to drive into 0.

Â But if I have 0 added to my control basically turns off.

Â I have 0 rate, 0 attitude and

Â the unmodeled disturbance keeps moving you away from where you want to be.

Â Once you move far enough away, the feedback says no, move right and

Â the disturbance says no move left, and they find their balance.

Â That's the new equilibrium, and that's what's happening here.

Â It gets close to it, but you get precisely the predicted offsets.

Â If you're not happy with that performance,

Â you would have to increase K if you're keeping the same control structure.

Â Make it ten times bigger, this would shrink ten times.

Â But that's kind of nice, you can see now we've got nice analytic guarantees,

Â with at least constant disturbances.

Â And if you can bound them, you can get a feel for just how much of an offset you

Â can have, even if they're time-varying, and go in there.

Â >> This is assuming that you're constantly in control of feedback, right?

Â It's not- >> Yes, continuous control.

Â Yep, not impulsive implementations, or on and off.

Â If you turn it off again you would start to drift,

Â this would then, without the control this would accelerate infinitely.

Â Fact, it would go off to infinity.

Â In fact, if you look at asteroid stuff that's the york effect you have

Â a torque that's always there, and no control in an asteroid.

Â >> Fall apart.

Â >> Fall apart, exactly.

Â They'd become binary systems, and tertiary systems, and fun things happen.

Â