0:06

Good now we're actually going to make these things do something.

Â Lyapunov functions.

Â And how do you come up with these.

Â We talked about energy.

Â Energy's a big inspiration for this stuff.

Â But there's a modern mathematical structures especially for CRP, MRPs and

Â all the parameters that I want to show you.

Â Some of them you're going to prove the math in your homework.

Â It's pretty straightforward, I'm going to show you the results.

Â Coming up with them originally, that's the hard part.

Â But now that you see them, it's not that hard.

Â 0:38

Anybody remember that mission, what was it?

Â One of the Gemini missions, where there were two astronauts in the Gemini,

Â trying to dock with another capsule up there.

Â And then something went wrong and they were tumbling.

Â I mean, they were tumbling crazy rates.

Â Amazed they didn't just black out.

Â But they were able to figure out something was miswired in this firing strategy or

Â what they thought would be improving it, no,

Â they were moving off to infinity speed rate.

Â At least as long as they had fuel.

Â And they were going crazy revolutions, they finally got it to stabilize.

Â Do you think those astronauts cared when they stop the spin if they point

Â left or right?

Â 1:12

I wouldn't, right?

Â I just want to have the bloody spinning stop, right?

Â And that's kind of the strategy here.

Â Tip offs, the classic thing with spacecraft.

Â You get the little small sides that get released.

Â Basically, they get chunked out of the spacecraft, right?

Â And you know they're going to tumble.

Â It's not going to be elegant.

Â You're just a secondary, tertiary payload, so you're out, thank you.

Â So now we have to always think about untumbling stuff, right.

Â And how do we do this?

Â So, we're going to talk about the open off controls just for tumbles.

Â That makes it a first order system.

Â I don't care about my attitude.

Â Shayla earlier was talking about having attitude and rates.

Â Now, I just care about my omegas and

Â I've got the first order differential equation for my omega, right.

Â I want to bring my rates to zero, that's the goal.

Â That's everything that we're doing.

Â I'm using this q notation.

Â 1:57

So, we're going to be talking a lot about specific single rigid body

Â equations proving all these properties.

Â I'm going to show you the stuff now for much more general,

Â this is really any mechanical system can be written in this form.

Â So if you're familiar with Lagrangian dynamics, congratulations!

Â You'll know exactly where I'm going with this stuff.

Â If you're not familiar with it, no problem.

Â You're not asked to do it as an exam, or something.

Â But I'm going to highlight the key concepts.

Â This is how we get the equations.

Â What I'm hoping you get out of this is that I can show you,

Â some of these properties that we're using are actually really powerful.

Â And you can do these on any kind of a mechanical system.

Â And you can expand these controls.

Â So the goal is, while it's a second order mechanical system, we have states and

Â rates, I only care about my rates.

Â The kinetic energy you can write with a system mass matrix.

Â It looks something like mass over 2 times rate squared.

Â If you have a rigid body this would be omega tranpose i omega, right?

Â Before general mechanical system you may have multirobot links who knows.

Â We can always write in that form that looks very, very similar.

Â 3:00

And it says symmetric posi definite matrix that's M is equals to M

Â transpose it means it's symmetric, greater than 0 means it's posi definite.

Â Its always going to be true with the mechanical system, not just for

Â the inertia tensor.

Â So we can write it this way, so kinetic energy becomes a very convenient

Â positive definite measure of my errors, which are my rates.

Â If you go through like Rajon Dynamics or Kaine's equations, or

Â Dolombert's principle, you end up with these equations that are fairly long

Â system mass matrix times acceleration.

Â This would be something like inertia times I omega dot is equal to minus omega tilde,

Â I omega.

Â Which is kind of this term plus other stuff and the [INAUDIBLE].

Â You wouldn't have this in the rigid body dynamics.

Â Because [INAUDIBLE] is constant.

Â A simple body frame.

Â But you get this general description.

Â This thing is called the Christoffel operator.

Â If you're very curious it has a lot to do with matrix math.

Â But it's there, again, the detail is not important.

Â Q here is just a generalized force.

Â But you think attitude dynamics, that was the u.

Â That was the external torque that we had on our dynamics.

Â If this has translational forces and

Â rotational torques this would all be in that q vector in a very general way.

Â So we need a Lyapunov function.

Â Step one is always, what do you care about?

Â In this case, I only care about my q dots.

Â So I want a Lyapunov function that's positive definite only in q dots.

Â Q happens to appear in here, but it doesn't impact the definiteness of it.

Â In fact, n, regardless of q, is always symmetric definite.

Â So, no matter what, like a multi link system.

Â No matter what I throw in, I'm guaranteed this is a symmetric posi-definite matrix.

Â And this kinetic energy, essentially what I'm setting it to is equal to my v,

Â all right?

Â But over here, so while there's a q here, we don't care about the q, right?

Â We only care about stability of the rates.

Â Are the rates coming to zero?

Â I donâ€™t care if I point left or right.

Â So, good, weâ€™ve done this, you take a derivative,

Â this is where Iâ€™m going to wave my hands a lot, because there's a lot of math

Â goes into this, Iâ€™m just kind of showing you the pattern, right?

Â Before we get into the details of [INAUDIBLE] that's where you

Â have to own it.

Â That's where you'll be doing it yourself.

Â This stuff, chain rule, this kind of becomes this with the double dot.

Â But then M itself depends on Q, which can depend on time and

Â there's an extra term that you get.

Â So lots of math.

Â This takes a few pages putting it all together.

Â You can write this out, rearrange this.

Â 5:45

And with the minus half plus half, those two terms actually cancel.

Â And lo and behold, my V dot, despite all this nastiness and

Â implied craziness that happens, is extremely simple.

Â And this is V here happens to be true kinetic energy, so

Â that means this is my power equation.

Â What's my time derivative of kinetic energy, which we've solved already for

Â a rigid body?

Â For a rigid body, what was T dot, if you're only looking at attitude motion.

Â Andrea, do you remember?

Â 6:16

Anybody remember?

Â It was omega dotted with l, right, that was it.

Â And for force, it's force dotted with velocity.

Â All right, there's a clear pattern.

Â So you can see, no matter how complex, how weird the mechanical system, whatever

Â these generalized forces are, are mixtures of torques, and forces, and craziness.

Â In the end, it boils down to this very simple system.

Â This is actually often called the work energy principle that we will re-examine

Â when we get to motor torques and CMGs and so forth as well.

Â I can predict what the power input will be,

Â how it changes kinetic energy with an extremely simple term.

Â If you derive this, this takes a lot of effort.

Â And in six, two, ten you actually do that.

Â So just, it's fun.

Â So, this is my derivative.

Â This is v dot.

Â Now here, I'm looking for a control to arrest any q dots.

Â So, what could I make my big q?

Â That's my control.

Â I get to pick what forces, torques are acting on the system.

Â 8:23

So how can you control the closed loop of

Â performance if you're setting this to just be equal to minus q.

Â >> Wait, what was?

Â Repeat that question.

Â >> At this point, once you have rates,

Â I am tumbling at six radians per second about this axis.

Â That's it.

Â >> Just doing the opposite?

Â >> It's going to do the opposite and with the unit gain, essentially.

Â And then that's it.

Â There's no knob to tune to say, hey, I want to be aggressive or no, hey,

Â I'm flying.

Â >> Some K positive K, well that's a gain control I guess.

Â >> Yeah so again, you're picking, you're making this control.

Â You can pick a positive constant in there.

Â And then in that case, this gets multiplied times K.

Â It doesn't change the stability as long as your gain is positive.

Â Right, now you've imposed a constraint on your control selection.

Â But now I have a knob to tune, and I can make this basically just

Â a proportional feedback, which is a linear control, but

Â I can prove global asymptotic stability on the complex non linear system.

Â 9:26

Which is kind of cool.

Â So we're going to do the same for this spacecraft.

Â So, this is the one that we're going to use instead of a single gain.

Â I can make it more complex and give the control designer a lot more knobs to tune

Â by making it a fully populated, symmetric posi-definite p matrix.

Â because then, the response will simply be, I think v dot is minus q dot pq dot,

Â which is v is a symmetric posi-definite matrix.

Â Minus data is going to be negative definite.

Â And I have asymptotic stability.

Â So now you've given him his.

Â So your result with Q,

Â it would be K times identity and we often do that in real life.

Â It's just like okay same gain across all the axis.

Â If you want more knobs that's were it goes.

Â Does that make sense?

Â 10:12

Tracking Instead of driving q dot to 0 what we want to

Â drive is our rates minus our reference rates to 0.

Â So if you got a spinner and you're supposed to be scanning across the sky and

Â you have this constant rotation you should be doing you have a non zero rate.

Â Then you want to compare your actual rate, I'm going at 1 degree per second and

Â I'm supposed to be going at 10 degrees per second.

Â And you're going I've got 9 degree per second error right.

Â I want to make that match a non zero rate and

Â even can be time varying this makes it a tracking problem.

Â 10:45

What we tend to do you will find is we use the same algebraic formula if this,

Â if you drop the deltas this is kinetic energy on this mechanical system.

Â By having deltas, it's something energy like in the algebraic structure.

Â But it's not exactly kinetic energy anymore, right.

Â But it's still positive definite, because m is positive definite.

Â And this tends to be useful, because to plug in equations emotion,

Â we always end up have mass times m x double dot somewhere.

Â And with the derivatives of this, you're going to have some where delta Q double

Â dots times this mass is going to make it easy to substitute.

Â If you do this, plug it in, plug it in equations, the motion we have before,

Â things don't just drop out where you have Q dots transposed with big Q but

Â you have all these extra terms.

Â And some of them have rs on them.

Â So then I can solve for a Q here that will make this, there's a dot missing here.

Â There defiantly should be a dot on this one too.

Â So V dot would have to be negative definite,

Â which makes it again globally asymptotically tracking.

Â But I can track a reference rate not just a zero rate with this one.

Â And if you look at the control,

Â before we just had this part, that was the proportional feedback that made it work.

Â The tracking ones always are more complicated.

Â You often,

Â to get asymptotic conversions we have to feed forward the reference motion.

Â Hey, you should be moving at this rate ideally, and

Â then that kind of dictates what, if your reference motion is time varying,

Â you may have an open loop control.

Â Torque that says, I need to accelerate, slow down, accelerate,

Â slow down, that's the nominal one and then you super impose the rest of it.

Â So these terms without that's your feedback terms and the ones with r,

Â those are your class equals called feed-forward compensation.

Â We're giving the control knowledge of what the nominal motion should be but

Â we will go through these steps.

Â This is going to go quickly it probably doesn't all sink in.

Â I just wanted to show you this works on a very general dynamical system.

Â We can apply it to a variety of stuff and

Â then we're going to specialize the rest of the class on rigid body, rigid body,

Â rigid body going through all the different primitive issue.

Â