Lecture 14: Stability and Control II - PowerPoint Presentation

Slide1

Lecture 14: Stability and Control II

Reprise of stability from last time

The idea of feedback control

Remember that our analysis is limited to linear systems although we will apply linear control to nonlinear systems sometimes successfully

1Slide2

Reprise

We are dealing with holonomic systems and working with Hamilton’s equations

Look at equilibria such that

With an equilibrium force

2Slide3

3

Reprise

We can combine the coordinates and the momentum into a state vector

and write the system in terms of

x

Equilibrium in this setting requires

Last time we worked in

q

,

p

space —Slide4

Reprise

We ask what happens if we perturb

q and p, but NOT Q

4Slide5

Reprise

The coefficients on the right hand sides are constant matrices

and we can write the equations in a unified matrix notation

5

How does this work in state space?Slide6

6

Reprise

This is a mixed notation to indicate what goes where

it is not a meaningful notation in terms of the location of the indicesSlide7

7

Reprise

The matrix is constant, so the state vector has exponential solutionsSlide8

8

We can write this symbolically as

which is a polynomial in

s of the same degree as the number of variables — generally twice as many as there are generalized coordinates/degrees of freedom

(This is not fully general, but it will suit our current purposes.)

RepriseSlide9

9

If Re(s) < 0 for all

s, the system is asymptotically stable

If Re(s) > 0 for any s, the system is unstable

If

Re(

s

) = 0 for all

s

, the system is marginally stable

stable: if we move the system away from equilibrium, the system will go back

unstable: if we move the system away from equilibrium,

the error will grow (initially) exponentially

marginally stable: if we move the system away from equilibrium,

the error will oscillate about its reference position

And the real part of

s

tells us about stability

RepriseSlide10

10

??Slide11

11

OK, let’s take a break and go look at the stability of a three link robot

Note that I told you some wrong things last time

I was working too fast and let some stuff slip

GO TO MATHEMATICASlide12

12

We quit here; the remainder of this set will reappear in our next class.Slide13

13

Let’s think about control in the context of the simple inverted pendulum

q

add a small, variable torque at the pivotSlide14

14

There’s a change of sign from the simple pendulum from last time

because I have chosen a different definition of

qWe have equilibrium at

q

= 0, and

Q

= 0 there as well.

We know that this will be unstable if it is perturbed with

Q

remaining zero

Let’s see how this goes in a state space representationSlide15

15

(I’ve put in the

e

because Q is zero at equilibrium)Slide16

16

If

q starts to increase, we feel intuitively that we ought to add a torque to cancel it

We can expand the feedback termSlide17

17

multiply the column vector and the row vector

combine the forced system into a single homogeneous systemSlide18

18

The characteristic polynomial for this new problem can be solved for

and so I can make

s2 negative by applying some gain g

1

.

So this very simple feedback can make an unstable system marginally stable

We can do better . . . Slide19

19

Suppose we feedback the speed of the pendulum as well as the position?Slide20

20

And now the characteristic polynomial comes from

Combining everything again we getSlide21

21

We can adjust this to get any real and imaginary parts we want

If you are familiar with the idea of a natural frequency and a damping ratio

then you might like to set the control problem up in that language

The linear term is the key — the feedback from the derivativeSlide22

22

The real part is always negative. If

z

is less than unity, there is an imaginary part.If z

equals unity the system is said to be

critically damped

can be made the same as the one degree of freedom mass-spring equation

by setting

givingSlide23

23

This suggests a bunch of questions

Is this generalizable to more complicated systems?

Is there a nice ritual one can always employ?

Is this always possible?

Will the linear control control the nonlinear system?

How much of this does it make sense to include in this course?

YES

SOMETIMES

NO

SOMETIMES

??Slide24

24

The question of possibility is really important so I’m going to address that as soon as I can develop some more notation

The general perturbation problem for control will be

For a single input system like the one we just saw

B

will be a column vector and

Q

a scalar and the equation isSlide25

25

We want Q (or Q

for one input) to be proportional to x

the minus sign is conventional

We see that

G

has as many rows as there are inputs

and as many columns as there are state variables

G

is a row vector for single input systemsSlide26

26

Rename some dummy indices to make it possible to combine terms

We have

for the single input case

Our control characteristic polynomial will come from

and the question is:

is it always possible to find

G

such that the roots are where we want them?Slide27

27

There are always at least as many gains as there are roots, so you’d think so

But it isn’t.

The controllability criterion, which I will state without proof, is that the rank of

must be equal to the number of variables in the state

There are as many terms in

Q

as there are variables in the stateSlide28

28

Q has as many rows as there are variables.

The number of columns in Q

is equal to the number of variables times the number of inputsIn the single input case Q is a square matrix

AND there is a nice simple way to figure out what the gains must be for stability

We are not going to explore this — we haven’t the time —

and it is covered in most decent books on control theory

We can get by with

guided intuition.Slide29

29

??Slide30

30

Single input systems are much simpler than multi-input systems but we have need of multi-input systems frequently

I will outline the intuitive approach to multi-input systems

which works best (at least for me) through the Euler-Lagrange equationsThis may be a bit hard to follow; we’ll do an example next timeSlide31

31

Euler-Lagrange equations

which we can rewriteSlide32

32

For a steady equilibrium, which is what we are learning how to do

perturbation

We can drop this term because of the

e

2

.Slide33

33

and we need to perturb the gradient of the Lagrangian to finish the linearization

orSlide34

34

We can use our old method of converting to first order odes on this and decide controllability (before we knock ourselves out trying to control it)

The state vector is Slide35

35

and the B matrix is

The

A matrix isSlide36

36

??