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So welcome back. today, I would like to investigate a

particular hybrid system that has to do with a bouncing ball.

So, that is, I take a ball and I drop it on a surface and I would first of all,

like to model it. And secondly, like to figure out how the

ball behaves. And the reason why I want to do this is

not necessarily because I really like balls but because this bouncing ball

example holds the key to another little peculiarity of hybrid systems that we

should be aware of. So, here is the set up.

I have a ball. It's h above the ground.

Now, what happens in between the bounces? Well, Newton tells me that force is mass

times acceleration and, and the acceleration is, in fact, -g, where g is

the gravitational constant. So, if I let x1=h and x2 be equal to h

dot, which is what we always do, then what I get is that x dot is this

thing, 0100 x, which is just the point mass, plus this gravitational force

that's pulling it down. Now, let's say that the thing that I'm

interested in also is, where the ball actually is, how high, how high up it is,

so y is going to be 01 x. So, that is my model of the ball.

What happens at the bounces? Well, you could try to model ball deformations and

whatnot, but that seems rather complicated so what we're simply going to

say is that these are what's called inelastic bouncers which mean, which

means that a little bit of energies sucked out of the ball every time it

bounces. And in fact, what we're going to say is

that velocity, this is the velocity, so if the ball is coming down here [SOUND]

hits the ground then the velocity instantaneously flips, it means, that's

why we have a minus sign there because the ball instantaneously flips but we're

also scaling it with this gamma constant, so we're making the velocity change

direction from downwards to upwards, but a little bit less, which means we lost

some energy in the bounce. Well, the position doesn't change in the

bounce, so this means that my reset condition is going to be this.

This is just encoding this fact that I'm flipping a little bit, I'm flipping my

velocity, and I'm loosing a little bit of energy at the time of the bounce.

Alright, so lets see what this bouncing ball hybrid automaton might look like.

Well, x dot is Ax+b, in this case, not Bu and b is 0-g so this is this forcing term

that is gravity, no big deal. Now, what's going to happen is I'm going

to switch from that mode to itself when I'm bouncing.

And we already saw that the reset condition is this inelastic bounce

condition that says that we're losing a little bit of the velocity at the time of

the bounce. So, that's the reset condition.

Now, the guard condition is what? Well, clearly, we're bouncing when we're

hitting the ground. So, we would like the height to be less

than or equal to 0, right? Well, let's see what happens.

Here comes the ball [SOUND] hits the ground, now the height is 0, we're

switching and now the first thing we do is we reverse the velocity but, meaning,

we're switching. But hey, wait a second, the height is

still less than or equal to 0. The height is still 0.

So, what we're going to do is we're just going to end up spinning like crazy if we

have this guard. So, this is not a good guard.

Let's get rid of it. Instead, what we need to do is we're,

we're going to switch when we come down to the ground and the velocity is

pointing downwards. Meaning that h needs to be negative or

less than or equal 0. And h dot has to be less than or equal 0,

meaning, we're facing downwards. Or since h and h dot constitute our

state, then the guard should really simply be that x is less than or equal to

0. That's the guard condition.

Well, it could be it's just a, the height is just 0, but this is really the

condition that tells us that it's time to, to switch. Okay, good.

So now, we have that. Is this any good?

Well, I mean clearly, it's good. It's pretty and we have guards and

resets. But does it model, first of all,

faithfully what balls do when they bounce? And secondly, are there reasons

why a model like this would cause us to be a little bit nervous? To answer that,

we need to figure out what the ball is actually doing if we have this model.

So, first of all, we have this system again now.

What I would like to know is what is the output of this system.

Well, fast forward, sorry, not forward, reverse two weeks and we go back to

linear time-invariant systems. We know what the solution to this system

is. We know that the output is C times this

bad boy here, that we call the state transition matrix, which was fancy speak

for this matrix exponential times the initial condition plus C times this

rather awkward-looking integral there. Now, all of these is, is, is not

particularly pleasant. And I remember vividly how we struggled

to get through this because there were so many integrals to keep track of here.

Now, in this particular case, we're a little bit lucky because this A matrix is

what's called nilpotent because if I take A to the power two, or three, or any high

there, higher exponential here, then I get 0 out.

And that turns out to be quite wonderful for us.

Why is that? Well, remember, that e^At is really this infinite sum.

A, there should be a parentheses here, I apologize.

0 At to the power of k divided by k!. Well, in our case, all the higher order

A's are equal to 0, so I get first k=0, which gives me the identity matrix.

Then I get k=1, which simply gives me A times t.

And then, I have all the others which are equal to 0.

So, e^At, in this case, is simply this matrix.

Here is the part that comes from the adjacent matrix and since A is simply

that times t, I get the t up here. So, this is my e^At or my state

transition matrix. So now, I should be able to plug this

back into the equation for y and get a rather nice looking expression.

So, let's do that. Well, here is C.

Here is the state transition matrix. Here's my initial condition.

Here's this rather awkward-looking differential equation sorry, integral.

now, we can try to solve that. and if we do that, instead of actually

doing the math, I did the math,

I encourage you to go home and do that. I'm just going to write down what the

solution is. It's going to be h0 times the initial

velocity, h dot 0 of t-t0 and that's then minus g/2

t-t0 squared. So, this is what the output looks like in

between bounces. Good.

Now, that we have that, let's figure out how long it takes in between bounces.

So, let's say that we start at zero o clock, at height zero.

This is when the bounce is about to start and then, [SOUND] we want to know how

long this, this took. We want to know how long it took us to

get back the ground. Well, that's simple.

Let's pick a big time, big T and say, well, y at that time t has to be equal to

0. Well, then we plug that in.

h0=0 not so we don't any turn there we just get h dot 0 times T minus this

thing. or another way of writing it is like

this. And you know what, this needs to be equal

to 0 because h is 0 there and h is 0 there,

which is why we have two solutions. So, one solution corresponds to T=0,

which is at the beginning of the bounce. And the other solution is when we return

down. So, this is how long it took me to go

from the ground, up into the air, and then back down to the ground.

This is the time in between bounces. Let's compute the accumulated bounce

times, just because we find it amusing.

Well, the velocity at the beginning, so at the first time, let's just say that

the velocity is v, right? So then, the time of the first

bounce, instead of h dot 0, I just plug in v there.

I get 2v/g, that's the time of the first bounce.

Well, the velocity after the first bounce,

well, the reset condition says that I have gamma times v, so okay, I have a new

v here. That's what I'm going to plug into my h

dot 0 here that immediately tells me that the time of the second bounce is, well,

it's the time of the first bounce plus the time it took for the second bounce

and instead of v, I have v divided by gamma there.

So, that's how long it took me to perform two bounces, and so forth.

This tells me that the time of the nth bounce is given by this 2v/g which is

this term that shows up over and over again.

And then, every bounce gets this gamma to the power of k accumulated.

Now, this is what's called a geometric series.

And if you look that up or you happen to know it, if gamma is less than one, which

means that you're not adding energy at the bounce times,

this thing will actually, okay, and we have gamma less than one,

this will actually converge in the sense that when n goes to infinity, this thing

becomes equal to this expression. Well now, maybe some of you are asking,

so what? [LAUGH] Well, you shouldn't be asking so what.

Well, what's going on there? This is the time it takes to perform infinitely many

bounces, infinitely many bounces, and this is less

than infinity. What that means is this bouncing ball is

going to go [SOUND] infinitely many. And there is some time t infinity here

where this system is just going to grind to a halt and our model is just going to

keep computing bounce after bounce after bounce after bounce faster and faster and

faster. And this is actually a serious problem

that causes our system to actually crash. So, the ball bounces in infinite numbers

of times in finite time. And like I said, this is not just a

mathematical curiosity that shows that we can compute geometric series.

It means that if I'm simulating it, then my simulation would crash.

In fact, I encourage you to write this exact model into your favorite simulator

and watch it crash. This also means that if I write down my

hybrid system like this, then it's undefined beyond T infinity because it's

not clear what's happening beyond there. Time is not allowed to go to infinity.

So, stability, which says what happens when T goes to infinity [SOUND] you can't

even ask that question because t doesn't go to infinity.

what's worse obviously is that real balls don't do this, real balls bounce and

after awhile they lay still, right? So, this is also an indication that our

model is not particularly good. And, in fact, this type of phenomenon

where you have infinitely many switches in finite time is known as the Zeno

Phenomenon, after the philosopher Zeno, that had the a tortoise race, a hare.

And the paradox that Zeno came up with was the fact that the hare never managed

to overtake the tortoise. And in the next lecture, we will see,

first of all, why the tortoise is able to beat the hare and what to do about it,

meaning how do you make the rabbit overtake the turtle? Well, that, my

friends, is the topic of the next lecture.