0:04

Moving on to new material now.

Â So, so far, globally stabilizing, that's what we had.

Â This Vdot is just negative semidefinite because the sigmas don't appear.

Â So we don't know if Vdot,

Â what sigma would be when Vdot goes to zero.

Â What we want to look at next is how do we prove stability,

Â and this is what we had with the spring-mass-damper system.

Â In fact, here this was -c * xdot squared.

Â And we ended up looking at higher order derivatives of V to figure out the stability.

Â This was the Mukai-Chen Theorem.

Â And that's what we're looking at here.

Â To apply Mukai -Chen,

Â you have to figure out for what set of states does Vdot vanish.

Â And with these systems, we always tend to write

Â Lyapunov rates like this â€“ that's what we're enforcing.

Â So, it's simply when delta omega is zero my Vdot vanishes.

Â So now, I want to take higher-order derivatives and

Â evaluate it on the sets where Delta omega is zero.

Â Taking your first derivative,

Â if I do this from here,

Â with chain rule, you do this and it's a scalar, you can flip it.

Â You can quickly prove to yourself this becomes this.

Â So del omega times P,

Â posit-definite, symmetric, del omega dot.

Â And evaluate it when del omega is zero.

Â This whole thing is zero.

Â So here's my first argument to you then.

Â As you could plug in what is del omega dot,

Â put in the equations of motion,

Â and do a lot more complicated stuff.

Â This is kind of a faster route.

Â This is gonna be zero because,

Â you know, del omega is zero.

Â And what's the key argument about del omega dot?

Â Zero times something's gonna give me zero.

Â If that something is infinite,

Â finite.

Â Finite. Yes. Finite, exactly.

Â Only you don't really care what the true value is.

Â At this stage, zero times as long as you can say it's finite.

Â Why do we know it's finite?

Â Because we know that's stable. Exactly.

Â We've already proved the system is stable.

Â There's no way you'd have

Â infinite acceleration because that would move you all over the place.

Â Right? That can't happen.

Â So that's nice. This gives you a little bit of a shortcut.

Â I don't have to plug in all the equations,

Â which you can do, it's just this is kind of a,

Â but you have to argue in an exam, "Hey,

Â this is gonna be finite because of stability.

Â Therefore, this term is guaranteed zero."

Â Good. So Mukai -Chen,

Â we take higher-order derivatives.

Â The even derivatives evaluated on the set where Vdot vanished should be zero.

Â And for stability, it should be an odd derivative since we're dealing with

Â a second-order system and this kind of stuff we're not

Â controlling circuits and other kind of a thing.

Â It tends to always be the third derivative.

Â So if you differentiate this one again,

Â you're gonna get a del omega dot P del omega dot.

Â And the other one, if you differentiate this one,

Â it's del omega * P del omega double dot evaluated on the set where del omega's zero,

Â because of the stability,

Â we know that all the del omegas have to be bounded.

Â You can't have infinite jerks somewhere again.

Â This is gonna go to zero or you could plug in

Â the equations and prove that that would work too.

Â What we're left with, we have to figure out is what happened to del omega dot.

Â When we derived the equations and plugged in the conditions,

Â this was the linear closed-loop dynamics.

Â This was the condition we had to enforce to get our Vdots to

Â have this particular negative semidefinite form.

Â So, let me just show you where that was.

Â I see some frowns.

Â So when we did this,

Â we took the derivative, the body frame derivative of that.

Â All this had to be true, then this set, it equal to here.

Â This all comes over into one form.

Â Really, here.

Â So that's actually a body frame.

Â I think I have a dot in there,

Â but that should be a prime.

Â Yeah. Doesn't matter. It works.

Â So what we're gonna look at now is what happens when these things are zero,

Â [inaudible] are not zero.

Â Del omega dot, what is del omega dot when del omega is zero?

Â That's what we need from here.

Â And, so here, I'm just solving for del omega dot.

Â You invert the inertia tensor,

Â bring it to here, then set del omega equal to zero.

Â This is the expression. That's the angular acceleration you would have with

Â this control if there's

Â no rate error and there's still a state error. That's when it comes in.

Â And then the last step is you plug this part back up into here and simplify,

Â and you end up with this expression.

Â And this turns out to be negative definite in terms of sigmas.

Â And why is that? Well, minus,

Â that's good, K squared.

Â It's a positive number squared, definitely positive.

Â Then there's sigma transpose,

Â a big three-by-three matrix, times sigma.

Â So if this matrix is positive definite, we're good.

Â And the inertia tensor we know is posit-definite.

Â Therefore, the inverse of a posit-definite matrix is still posit-definite.

Â That's good. P is also a posit-definite matrix and symmetric.

Â So there's a theorem that says if you

Â multiply two symmetric posit-definite matrices with each other,

Â you end up with a symmetric posit-definite matrix again.

Â And that's what we're taking advantage of here.

Â So the product of I inverse P I is gonna be a posit-definite symmetric matrix,

Â and that's why now here the first nonzero higher-order derivative is

Â this negative definite quantity in terms of the sigma and this

Â essentially proves we would be converging.

Â This is not just stable, it's,

Â in fact, asymptoticly stable.

Â This result, as you can see,

Â is good for any sigmas.

Â This is also a globally asymptotically stabilizing result that you'd have here.

Â Any questions?

Â You kind of do these steps, you definitely should do this by hand.

Â I'm kind of showing the primary thing,

Â trying to plug in the right equations.

Â Go through the stuff. You will get there.

Â But that's Mukai-Chen that we can use to prove that.

Â