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Good, now how do we prove this?

Â We don't solve nonlinear equations.

Â So we use the Lyapunov functions, Lyapunov's direct method.

Â So Bender, tell me,

Â in brief words, what is Lyapunov's direct method? What are we doing?

Â You have to find a functional,

Â and then that function needs to be positive definite,

Â have continuous partial derivatives,

Â and then also, the derivatives and...

Â Negative definite? For what?

Â What, for what type of stability are you arguing?

Â The semi definite.

Â What type of stability do you have in this case?

Â This means it could be zero.

Â The rates could go to zero away from zero.

Â Origin, right? So now,

Â you've got this flat spot that you can get stuck in actually.

Â We talked about this. The pendulum has some of those.

Â You'd say that in the homework.

Â You could do this,

Â and that's a thing where Vdot could go to zero and you stay there.

Â Right? So this only gives you Lyapunov stability.

Â If you want convergence, asymptotic stability,

Â everything else is the same,

Â except for this argument has to be stronger to be

Â negative definite within some neighborhood.

Â Right? Now, what goes into these functions?

Â Matt? What variables do you throw in here?

Â State variables and time.

Â You could have time.

Â In this class, we're typically dealing with

Â time-independent systems so they're not explicitly time dependent.

Â But you're absolutely correct, time could be in there.

Â State variables, too.

Â But let me ask you then this, which state variables?

Â So, let me show you something. This is a system.

Â Alright.

Â And you have omega dot or I omega dot.

Â Plus some control u.

Â All this stuff.

Â That's a classic rigid body.

Â Which states do you have to put into here?

Â Ansel, what do you think?

Â Why?

Â I'm not sure.

Â Caley, what do you think?

Â Omega dot.

Â So omega dot?

Â That's my angular acceleration.

Â Sorry. It should be sigma dot.

Â So you want sigma dot in here?

Â I'm not sure.

Â Right.

Â But this is the key.

Â That doesn't let me erase it.

Â What is going on?

Â There we go.

Â Feels like Monday.

Â There we go.

Â Just took a little smacking.

Â Okay. Anybody? What states go in there? John?

Â Your attitude coordinates and your omegas.

Â Okay. Do you agree, Spencer?

Â Sounds reasonable.

Â Another trick question.

Â There's no real answer to that question because I'm not giving you the prompt statement.

Â Every time you have a control problem,

Â you should ask what is the control goal.

Â What are we trying to achieve?

Â Right? And that is the key especially in the current homework.

Â There will be some we do with

Â spring-mass-damper system or cubic spring in your homeworks,

Â and you do the same math as what we've been

Â doing and apply it to this kind of similar function. That's great.

Â But there's one of them where the equations of motion is just XDOT equal to u.

Â And, in fact, our rate equations are just Omega dot

Â equal to whatever the Omega equations are.

Â Right? That's a first order dynamical system.

Â So it depends on your control gain.

Â And what we've been discussing last time was just rate-based control.

Â We're looking at prototypical Lyapunov functions that are good measures,

Â that are positive definite,

Â all the stuff Bender needed here,

Â right, to get this going.

Â And we take derivatives and now we have to guarantee this is gonna be negative

Â semidefinite least or hopefully negative definite then we've proven convergence,

Â which is even stronger. All right?

Â But there, the goal is just to drive Omegas to zero.

Â You may have, and in that mechanical system we had,

Â the kinetic energy which is typically what we use as

Â a good prototypical rate-based function,

Â you don't have to use it.

Â You can be creative and use other ones,

Â but energy tends to be a pretty good one.

Â So for a general mechanical system,

Â we have the state space or we call these q's,

Â I believe. So now I can erase it.

Â q M which depends on q times qdot again,

Â and a transpose, right?

Â That's what you end up with. q's are in there,

Â but we don't care about the cues.

Â When you differentiate, they're there,

Â and you have to take derivatives of those q's and it impacts the math.

Â Fine. But from a control perspective, in that one,

Â we said V is equal to this and V was just a function of

Â q dots in the math that we just wrapped up with last time.

Â All right. Because there,

Â we only care about V states going to zero.

Â I don't care where I stand once I settle it.

Â I'm just trying to bring it to rest.

Â If you're trying to do a complete six off motion control,

Â the spacecraft has to point in this direction and be at rest.

Â Now you have to track what you guys were saying earlier,

Â you would have to have attitude and rates.

Â Just because my attitude is zero doesn't,

Â for a second order mechanical system,

Â doesn't guarantee you're gonna stay there.

Â You might just be moving to zero.

Â At this instant, you're zero but your rates are nonzero that means

Â next time step you're gonna be overshooting and so forth, right?

Â So for convergence, you have to make sure your attitudes and your rates go to zero.

Â And we will see that today.

Â Okay. But it's that simple little step.

Â What do you care about?

Â So at the beginning, I really recommend,

Â be anal about this and every time you have a V right out,

Â this is a V of Sigma and this,

Â 'cause some Vs will have Sigma and Omegas in it but you only care about Omegas.

Â And that's the only part,

Â is something negative definite or not is really only a statement about the Omegas.

Â For nonzero omegas, what is this statement or not?

Â That makes sense?

Â So, we can apply this today and go to some examples. So good.

Â That was a nice review.

Â Okay.

Â So Lyapunov functions -

Â We use a control goal.

Â For right now, it's the same.

Â We wrapped up for a general mechanical systems

Â last time so if you haven't seen these equations, nevermind.

Â It's just this is the structure and we go through...

Â Yes, sir, got a question?

Â Why don't we include also the control input,

Â the control function in the Lyapunov function?

Â Because if not, we are trying to prove that maybe the natural system is...

Â Right.

Â You're still looking at the response of,

Â let's say you have a spring-mass-damper system, cubic,

Â quartic springs, crazy stuff happening in there.

Â You still care about the stability of your states, X and Xdot.

Â The control impacts that,

Â so the control then we have to substitute in.

Â When I differentiate, if I have

Â a second order mechanical system and I care about these things,

Â I have to have an Xdot in my Lyapunov function such that when I differentiate there's

Â an Xdoubledot and you've seen here too, at some point,

Â we always plug in the equations of motion which is,

Â I've taken from here and I plugged it in here,

Â and that's where the control appears.

Â So, in this step, I can come up with, well, that's what the control,

Â this is how the control impact my Lyapunov functions and now I can design controls.

Â Well, what we also do sometimes is we have a control and we plug it in explicitly,

Â upfront, and then put these Lyapunov rates in there and we can see what the math says.

Â Is it gonna be stable or not?

Â Can we argue it is stable or not?

Â Remember, we also have this if and if and only if.

Â If a statement says if,

Â Thomas, what does that mean?

Â A is true if B is true.

Â How is that different from if and only if?

Â That means that B doesn't imply A.

Â Right.

Â So one implies the other,

Â but the other doesn't imply the first.

Â If it says if and only if, then it's like, hey,

Â A cannot be unless B is true and B cannot be true unless A is true.

Â Right? Then it goes both ways, that arrow.

Â Good. So this is what we did last time,

Â and we derived this and there's different controls.

Â This was one of them. We will see other ones in this class that we deal with.

Â And because this is a first order,

Â we only care about... See here, there are q's.

Â Your states appear in these general mechanical systems,

Â but I don't care about the states.

Â I don't care what the final configuration is of this multilink system we looked at.

Â I only care that the rates are zero.

Â So you only have to argue that this Vdot is negative definite in terms of the rates.

Â That's why writing it like this is really convenient,

Â and that one is clearly the case.

Â And for tracking, it's basically we use the same kinetic energy-like function.

Â But instead of using really velocities,

Â we use these velocity errors.

Â It's just a nice mathematical structure that gives you something as

Â positive definite in terms of these tracking errors.

Â Here we care about,

Â you know, I should be having these rates.

Â This is the rate I do have,

Â and this is my tracking error rate.

Â So I replaced q's with Delta q dot and do the same math.

Â Tracking typically always gives you more complex stuff.

Â You have feedforward compensation,

Â where the reference motion impacts our stability arguments, you know.

Â How much knowledge do you put into this control?

Â