0:05

Then we want to look at Asymptotic Stability.

Â Here it's basically ditto, you do the same stuff you just did for stability, right?

Â We kept with the definitions, began from boundedness to stable to asymptotic.

Â There's extra layers that you add on arguments.

Â So to go from stability to asymptotic stability you do everything you did before

Â to prove that it's stable, again when I say stable it implies Lyapunov stable.

Â But then there is an extra step and that extra step is you're talking about V dot

Â be negative semi definite, that's what guarantees stability.

Â Now you have to prove that neighbourhood V dot is negative definite.

Â It basically means that V dot function, that space,

Â there's no other flat spots where you can get stuck on.

Â And that will only converge, you'll always have negative energy rate,

Â essentially, until you reach the origin.

Â There's no other saddle points or flat spots or

Â equilibrias or whatever you want to describe them.

Â So that's the property that we have to prove to guarantee asymptotic stability.

Â Now we know it will get there.

Â It doesn't guarantee you'll get there quick or

Â slow, it just guarantees it'll get there if t goes to infinity all right?

Â That's the distinctions.

Â So lets look at our favorite system Spring Mass Damper it's a nice system.

Â It's a linear system but all of the things we do should hold for

Â linear systems as well.

Â So with the spring mass damper system, I still need a Lyapunov function.

Â Now here's a key thing when we design Lyapunov functions,

Â you have to consider first you're looking at the stability about which point.

Â If it's an equilibrium, then define your coordinates relative to that equilibrium.

Â If it's a reference, you'll have delta x's,

Â then we define the things we care about as a departure motion relative to that.

Â Here, I'm just looking at the origin, right?

Â Is the origin, which is an equilibrium point, is that one going to be,

Â if I have x in here and x dot double dot, a zero,

Â this whole thing is zero and it stays to x.

Â The origin is an equilibrium.

Â Now, when you write your v function,

Â you have to do it in terms of all the states you care about.

Â We talked earlier about the state vector, right?

Â And for spring mass damper system, I need both x and x-dot to be 0 to remain at 0.

Â If somebody gives you a system which is x = 0, You have no guarantee in a spring

Â mass damper resist that you will remain at 0 because you might have a rate.

Â You might just be crossing 0.

Â So you need x and x dot to be 0 in this case.

Â >> Daniel, your class notes the V, the V dot is missing the dot just FYI.

Â >> Where?

Â >> In the notes in the left side?

Â >> This one?

Â >> Yeah. >> I may have fixed that later, thanks.

Â If you don't see that definitely put that in your notes.

Â Thanks for pointing that out.

Â [COUGH] So, this is why here we have, we care about both, what happens to x and

Â x., this is the second order of differential system that we have.

Â In one of the homework you're doing now, actually you only care about

Â It's a first order differential equation, x dot equal to something.

Â In that case, if I know x goes to zero, my x dots have to be zero and you're done.

Â It's a first order system.

Â In that case you only care about one of the variables.

Â Here we're dealing with a second order system so

Â I care about both position and rates.

Â So we still need a Lyapunov Function.

Â And what you'll find is rate this again is a very true.

Â Once you have good Lyapunov functions don't throw it away, because you probably

Â can use can use it for lots of different dynamical systems or different set ups.

Â We will apply the same Lyapunov function and then we modify the control, and

Â it's the same Lyapunov function by giving us slightly different v dots but

Â we can still argue these properties.

Â And then you modify it again with the same Lyapunov function.

Â And hopefully if your life is good, you can still argue some other properties and

Â it's conclusions and so forth right?

Â So in that spirit, we're going to reuse total energy that we had.

Â Kinetic plus potential energy, that's our Lyapunov function.

Â It's 0 when x and x dot are 0 and everywhere else, it's non-zero.

Â Which is great, and let me see where that goes.

Â So, now if you plug it in, you get these equations of motion, this part,

Â if you plug in these terms, this plus this has to be minus this.

Â So I get minus c times x.squared.

Â Now, is this v.negative semi-definite, or negative definite.

Â What do you guys think?

Â 4:44

All right, this is a- >> Burn I mean,

Â x dot is zero, only not zero when x is zero, first of all.

Â If it's the spring system so that's not.

Â >> I'm confused, what are you arguing, yes or which one?

Â >> I'm not sure, I'm thinking about it.

Â X-dot is not equal to 0 when x equals 0.

Â >> Immediately, even a simple problem you go, wait a minute.

Â And I'm glad, because this is often like, wait a minute, what's going on here?

Â The beta definition means at the origin, when your states are 0,

Â is this function 0?

Â Well, If x and x dot are zero this function is zero so that's good.

Â If for any non zero state, is this function guaranteed to be negative?

Â If yes, it's negative definite.

Â Is that true here?

Â For any non zero state, is this function guaranteed to be negative?

Â 5:45

0?

Â So what does that imply?

Â All right?

Â That's the crux of it, really.

Â So if you look at this function.

Â See that's why it's really important when you write out your v's, especially when

Â you're just getting started with the Lyapunov functions, use this.

Â Every time you write them, write out all your states.

Â This V is a function of x and x dot.

Â That means you care about both of them.

Â We will talk today about untumbling.

Â Just despinning objects in space and there we don't care about the attitude.

Â I just want to stop tumbling.

Â So then we only care about x dot and there'll be differentiation.

Â Yes, sir Matt.

Â >> The point where we're evaluating v dot to determine whether it

Â is some sort of definite or semi-definite, we don't care at all what v was?

Â >> No, v was actually posit-definite.

Â In fact, it was globally posit-definite, it's good for any initial conditions.

Â This v-dot though is only negative semi-definite,

Â because x-dot has to be 0 for V dot to be zero.

Â In fact, from this,

Â you can argue with some electro math, that x dot must go to zero.

Â The system will come to rest.

Â It'll stop moving, but the question is where.

Â Now we know from a spring mass damper system,

Â it's going to come to rest at the origin, but this math doesn't prove it yet.

Â Because v dot could be zero if x is five, x doesn't appear here.

Â So we just don't know with these and this is the important stuff with this argument,

Â it's an if statement.

Â If you can prove this is negative definite then it is as particularly stable,

Â if you can't prove it's negative definite.

Â It doesn't mean it's not asymptotically stable.

Â We just don't know.

Â 7:32

Maybe a smarter Lyapunov function would actually prove asymptotic stability.

Â And so we have to use different arguments.

Â So is everybody good why this is negative semi-definite?

Â We care about x and x dot, but it's only negative definite in terms

Â of x dot x doesn't appear, so it's not definite in terms of x.

Â But the function is always zero, then, so it's, it's semi-definite.

Â That means, we at least, we've still proven stability, which is nice, but

Â we know the system asymptotically stable.

Â So, how do we now prove asymptotic stability?

Â If you've taken any classes on non-linear systems.

Â Anybody here heard of Lasal's invariance principle by chance?

Â Okay, a few of you, this a coronary I really like this extra theory that we're

Â going to talk about here, this one's developed by Mukherjee and

Â Cheng I'm giving you the original paper.

Â So if you do a lot of work on those,

Â I definitely recommend pulling it up and reading it.

Â It's a great paper but what they're looking at is, well,

Â I just don't know if it's asymptotically stable or not yet.

Â 8:33

And with this long dot so I need extra arguments to see this.

Â And sadly for mechanical systems this is typically what we end up with.

Â This is not an odd ball case, this is almost the dominant case you run into

Â the mechanical systems that we can prove stability with the dot but

Â we can't prove asymptotic stability.

Â because we need functions that are built, anyway,

Â just the way the mathematics work out.

Â So what does Mukherjee/Chen do?

Â They basically say, don't stop at the first derivative.

Â We don't just look at V dot, we also look at V double dot, and V triple dot and

Â V quad dot, and V five dot.

Â Just, you keep taking derivatives.

Â And you're looking for a very particular pattern that happens.

Â So this is the process, the first thing you have to in identify here is

Â the domain where the status, the set of states where v dot actually goes to zero.

Â Because v dot here in this mechanical systems they always tend to go to zero

Â when x dot is zero, that's great.

Â That means it's negative definite and x dot.

Â I know my rates are going to go to zero, I just don't know where I'm going to settle.

Â So, in this here, in this state vector,

Â would be basically the set of states where x dot is zero, that's my domain omega.

Â So, I only need to look at that sub-space.

Â And then I have to take higher order derivatives and you evaluate them.

Â You take these derivatives of these functions and

Â you evaluate them on the set of states here, when is x dot equal to zero?

Â We plug it in and look what happens to the math.

Â And what will happen is we ended up here with a first derivative.

Â You take a second derivative.

Â That's going to be zero typically and you keep on taking derivatives.

Â For second order mechanical systems it'll always be the third derivative all of

Â a sudden, that's non-zero.

Â And if you can show the third derivative Is negative definite for

Â all the states you care about here.

Â From here we know x dot goes to zero.

Â We don't know what happens to x.

Â We will have a third derivative in terms of x that's actually negative definite.

Â So this is like looking at higher order curvatures of these bowls.

Â You've now proven you will actually converge in the system.

Â If it's positive definite then you're in trouble.

Â It's probably going to diverge.

Â And there's whole instability theorems around this as well.

Â As we're designing controls, we're not trying to be evil,

Â we're trying to do good.

Â So we're going to try to design controls that give us asymptotic conversions not

Â asymptotic diversions out of the system.

Â So that's the math.

Â Let's go through that, we'll see when it get supplied,

Â it's actually easy to follow.

Â We got to this point, first thing we identify when is x.v., = 0 and

Â that's only going to happen when x dot is 0, right?

Â -c* d bowl c had to be a positive non 0 value then that's true.

Â So I'm going to evaluate this on that set,

Â here my V dot was simply minus cx., squared you take

Â the derivative of that with chain rule you end up with -2cx.*x..

Â Now let's see, we know x dot will be zero.

Â I don't know what happens to x.

Â 12:42

So just saying a mechanical system wouldn't necessarily fly.

Â What's the key, why do we know x is not going to go to infinity?

Â It is stable exactly.

Â Don't ignore our early arguments.

Â This already proved the system is stable, so you give it some finite response.

Â There's going to be a finite region that the staff is going to be in, right?

Â So we know we have local stability already.

Â We just don't know if we have some asymptotic so x would never be infinity.

Â And if x is stable actually you can also, this is another approach,

Â you could keep it in this form and argue well, because the system is stable, the x

Â double dots can't be infinity otherwise you would not have a stable system.

Â It would have jumped infinitely faster than infinity with

Â an acceleration like that.

Â So finite times 0 would be 0 as well.

Â So, as you do this in your homework and

Â try this yourself, you can do either of those arguments.

Â But you have to argue, hey, this is something stable, it's bounded times 0.

Â So as we expected, the second derivative evaluated on the set where V dot vanishes.

Â That second derivative also goes to zero.

Â That's good, that's what we expect.

Â Now let's take the next derivative.

Â So we have this function, and taking another derivative, so a lot of chain

Â rule from this function here, and then you plug in the equations of motion.

Â Any time you get the derivative of x dot there will be an x double dot.

Â You plug in the equations of motion.

Â And I'm just pulling out the constants.

Â This is what I get.

Â Now I have to evaluate this triple derivative on the set where v dot vanished

Â which means x dot is equal to zero.

Â That means this is going to vanish, this is going to vanish, this is vanishing,

Â this is vanishing.

Â And again because everything is stable, we know x is finite and all that good stuff.

Â What it leaves you is this one term, now that was vanished.

Â Here's my term, it's going to leave you with this term times this stuff.

Â So in the end my V triple dot is -2 which is a positive number, c,

Â a positive number, k squared, a positive number, m squared, a positive number.

Â So positive minus positive*x squred,

Â this function is now negative definite in terms of the unknown state x.

Â I know my rates go to 0,

Â I just don't know where else I was going to settle all right.

Â So this is the procedure that we use applied to a very simple spring mass

Â damper system.

Â And how we can now prove asymptotic stability.

Â So it takes extra steps.

Â And we'll do the same thing for attitude motion.

Â Yes sir. >> Would you prove it's not asymptoticly

Â statement?

Â >> There, I'd have to go back to this one.

Â Typically if you look in dynamical systems then thereâ€™s curvature results.

Â The first none zero one is one of these that could be, thatâ€™s a big red flag that

Â itâ€™s not going to work you're not going to have conversion.

Â We know we have stability because of this.

Â But we just don't know about convergence.

Â So I think that's where it manifests but I want to just I have to go look at this

Â paper again to see exactly where it manifests.

Â So I do recommend if you work in this area, I'm showing you the one result when

Â we're designing controls we have power of what we design.

Â We will aim for something that makes this work and

Â give us an asymptotically stable control.

Â Yes Debo.

Â >> So well first of all to rephrase this question then it's not an if and

Â only if this theorem or another- >> Yes, that's correct.

Â This is not an if and only if.

Â This is, if this holds, then we have it.

Â Not inverse.

Â >> And I was going to ask, can you, I mean, it's more of a hands-on way.

Â But can you just show that x dot, if x is zero, x dot will not be zero?

Â More like within by integrating and looking at your function more closely.

Â >> You could, but that's basically the following LaSalle's invariance.

Â You're looking for the largest in variance set.

Â That happens and that one is going to be is equal to 0.

Â >> It's a local asymptotic stability if you showed that.

Â >> Yeah, I have really discussed local and global but, in this case,

Â this argument is true for any x which is nice.

Â To argue global stability,

Â that's on another slide right now that I'll show you.

Â That's an extra thing is required of the v functions.

Â Otherwise people came up with some weird degenerate cases.

Â But this one would actually be globally asymptotically stable,

Â which we know a linear spring mass damper should be in this case.

Â V holds for everything.

Â This argument, the definiteness holds for any x.

Â But there's one extra statement that you really have to prove for

Â global stability, okay?

Â So if you think asymptotic on our mechanical systems,

Â we typically have to look at second and third.

Â Second should be 0, third should be negative def in the stuff that wasn't

Â appearing in v-dot itself, that's the pattern.

Â