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Lecture 29: Preliminaries to Formal Control. Quick review of last time. State transformations I: diagonalization. Introduction to the ideas of controllability and observability. What did we do last time?. We looked at outputs and found transfer functions.

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slide1

Lecture 29: Preliminaries to Formal Control

Quick review of last time

State transformations I: diagonalization

Introduction to the ideas of controllability and observability

slide2

What did we do last time?

We looked at outputs and found transfer functions

The we took the transfer function and used It to find a new equivalent system

The examples we did gave a simpler equivalent system

typically an A matrix in companion form

Let me review some of the results from last time: suspension and pendulum

slide3

Gillespie

Us

z2

m2

k3

c3

z1

m1

k1

zG

slide4

Put these numbers into our picture

z2

m2 = 2.479 lb-sec2/in

m2

k3

c3

k3 = 143 lb/in

c3 = 15.06 lb-sec/in

z1

m1 = 0.194 lb-sec2/in

m1

k1 = 1198 lb/in

k1

zG

slide6

Simplified block diagram

red denotes the upper pair

x3

x1

u = zG

x4

x2

blue denotes the lower pair

slide7

I can form the transfer function between the ground motion and the axle motion

The numerator is not constant, so we will want to use the method we discussed last time

slide8

This defines a state, but it is not the state we started with!

The left hand equation defines the dynamics in terms of a new state

slide9

We can make a block diagram of the system

and we can add the output

y

37515

356217

+

u

-

83.7

6970

37515

356217

slide10

Let me add a couple of comments

We can make a new dynamical system from the block diagram

I have not addressed the connection between the old state and the new state

I’ll deal with that systematically another day

For now, let’s look at the dynamics of the two systems

slide11

Let’s compare the A matrices for the two systems

The original matrix

The new matrix

They have the same eigenvalues

slide14

Now I want to put in the motor that drives the force

force on the cart

the motor torque

wheel radius

slide15

This will modify the matrix A and make the input u the voltage e

so that we have a modified A and b

slide17

Last time we looked at the angle as the output

Today I’d like to look at the cart position as my output

This leads me to

slide18

From which we get a new state block diagram — much simpler than the old one

+

u

-

1/25000

12005

-249900

-117649

slide20

We can build A from the block diagram

+

u

-

1/25000

12005

-249900

-117649

slide23

The story so far . . .

We have discovered that we can solve the problem

In terms of a transfer matrix

or in the common case of a SISO system, a scalar transfer function

slide24

We only looked at the single input-single output (SISO) case

For which the transfer matrix is a scalar, the transfer function

If the numerator of H is not constant, we had to split the transfer function

This defined a new state z, and we didn’t know how to connect z and x

slide25

Today I want to look at transformations in general,

and at two very special transformations: diagonalization and companion form

I’m going to start with diagonalization, which is a close relative of modal analysis

We’ll eventually settle on companion form,

which is what we’ve been getting informally from our transfer function games

we’ll get a formal approach to that on Thursday

What would we like to do?

The eigenvectors are the state space equivalent of the modal vectors of modal analysis

Can we make the A matrix diagonal — uncoupling the eigenvectors?

The answer is yes, sometimes . . .

slide26

We need to transform the state x to a new state z that will separate the eigenvectors

A linear transformation between two vectors can be written

T is a constant matrix, and we want transformations such that it has an inverse

general transformation

The equation becomes

and we can simplify

slide27

We would be happy if TAT-1 were diagonal

If A has distinct eigenvectors we can make that happen

we candiagonalizeA.

Let V be a matrix whose columns are the eigenvectors of A

then

slide28

Thus if we choose Tto be V-1 we can diagonalizeA

In that case we’ll have

or, in terms of the individual equations

Uncoupled first order ordinary differential equations, which we know how to solve

Block diagrams on the next slide

slide29

u

l1

u

l2

etc

slide30

What is x in terms of z?

So if I want the full state,

I can find z and then use T-1 (here V) to obtain the physical state in term of z

slide32

All the left hand multipliers have to be nonzero for the input to reach all the zs

All the right hand multipliers have to be nonzero for all the zs to contribute to the output

slide34

What does all this have to say about the examples we’ve worked?

Two masses and some springs

The auto suspension

Artificial example 5A

The inverted pendulum on a motor-driven cart

slide35

From Lecture 13

f1

f2

k1

k3

k2

m1

m2

y1

y2

slide36

Let the input be f1

f1

k1

k3

k2

m1

m2

y1

y2

slide37

Let the input be f1

f1

k1

k3

k2

m1

m2

y2

Let the output be y2

slide40

Put in numbers and generally simplify a bit: k1 = 1 = k2, k3 = k, m1 = 1 = m2

I can do this without loss of generality by appropriate scaling

we have one variable parameter — the strength of the coupling spring

slide41

The matrix of eigenvectors, V = T-1 is given by

It is important to note that I can multiply the eigenvectors by a constant

without changing anything.

If you don’t like complex displacements, multiply by j

slide42

The two coupling matrices are given by

Both are complete, so the input affects all the components of z

and the output contains all the components of z

We can draw the block diagram

slide43

u

u

u

u

y

l1

l2

l4

l3

slide45

What is the interpretation of z?

The first two lines involve y1 and y2 moving together (sum of x1 and x2)

the second two lines involve them moving in the opposite directions

The coefficients reflect the eigenvalues: ±J and ±j√(1+2k)

slide48

Gillespie

Us

z2

m2

k3

c3

z1

m1

k1

zG

slide49

The differential equations, supposing z positive up

Define a physically reasonable state

slide51

Block diagram of the original problem

red denotes the upper pair

x3

x1

u = zG

x4

x2

blue denotes the lower pair

slide52

Put numbers into the picture and the system

z2

m2 = 2.479 lb-sec2/in

m2

k3

c3

k3 = 143 lb/in

c3 = 15.06 lb-sec/in

z1

m1 = 0.194 lb-sec2/in

m1

k1 = 1198 lb/in

k1

zG

slide53

Let me define an output. I’ll choose body position: z2 = x2

We looked at this using transfer functions to find a new state

slide55

and we can add the output

y

The new state block diagram — much simpler than the old one

37515

356217

+

u

-

83.7

6970

37515

356217

slide56

How does this compare to diagonalization?

The eigenvector matrix V = T-1

slide57

Check the transformed input and output matrices

They are complete and we are OK:

the input goes everywhere and the output sees everything

Let’s draw the block diagram (it looks a lot like the one we just saw)

slide58

u

u

u

u

y

l1

l2

l4

l3

slide60

Diagonalization gives us y in terms of u without knowing what z represents

If we want x in terms of z or vice versa, we need to do a little more

It’s numerically tedious; we can look in Mathematica if we want

Right now I don’t want

slide62

Friedland gives an example of order reduction — 5A

and I’d like to take a look at how that goes — what happens

It’s not a real physical system, but nonetheless instructive

We started on this last time and managed to reduce the 4th order system

to a first order system — a really weird result

We can use diagonalization to understand how this happened

slide63

From Friedland

The eigenvalues of A are -4, -3, -2, -1

slide64

The eigenvectors are

They are independent and span the space. The inverse is

slide65

Let’s look at the input and output vectors for the z state

We see that the input does not reach every piece of z

and that the output does not see every part of z

We can draw a new block diagram for this transformed state

slide66

0

u

u

u

u

y

l1

l2

l4

l3

slide67

-4

u

-3

y

-2

u

-1

Eliminate the zero paths

slide68

The only path from u to y is that shown in red below

It corresponds to the one dimensional transfer function

y

-2

u

-1

slide69

What do the solutions to this look like?

Note that each z responds at a single eigenfrequency — they are eigenfunctions

slide70

The output

The input and the output are only connected through z4

which suggests an explanation of the first order system we got before

slide71

We wrote a transfer function

and found a reduced block diagram

y

+

u

the single eigenvalue

that contributes to the

output

-1

slide72

The diagonalized block diagram

-4

u

-3

y

Compare to the

transfer function

state

-2

u

-1

slide73

What can we say about the relations between x and z?

We need z in terms of x to get the initial conditions for z

We need x in terms of z to convert the solution to a physical state

slide74

If we start from rest, z1 and z3 remain zero, and then we’ll have

u reaches all the components of x, but we are missing two of the eigenfunctions

slide77

Before we go on to the next example,

let’s say a little about controllability and observability

The goal of control is to use u to make x do what we want: u has to reach the whole state

If it cannot, then the system is uncontrollable

We are going to do this using feedback: we may have to see the whole state

If we cannot, then the system is unobservable

Let’s talk about this in the context of Example 5A

slide78

u cannot affect z1 and z3

the system is not controllable

-4

z1 and z2 do not appear in the output

the system is not observable

u

-3

y

-2

u

-1

slide79

We will develop better criteria for controllability and observability going forward

These will involve the companion form

All I want to do for now is to indicate that not all systems can be controlled

I’ll do one last example for this afternoon

slide81

Now I want to put in the motor that drives the force

force on the cart

the motor torque

wheel radius

slide82

This will modify the matrix A and make the input u the voltage e

so that we have a modified A and b

slide84

The matrix of eigenvectors is

(This is only to three significant figures; the two 953s are not actually equal)

It’s invertible and we can find the diagonalized problem

slide85

The transformed b and c matrices (vectors, here) are given by

The input vector is complete; the output vector is not.

slide86

30.2

u

u

u

u

y

-3.17

3.16

0

-0.47

z4 is simply the integral of u

The eigenvalues are unscaled; the other blocks are shown 1000 times bigger than

they actually are

slide88

In all the cases where we have done both modifications —

transfer function to new state and diagonalization —

the resulting new states have been different, with very different block diagrams

The different block diagrams correspond to different A and B matrices

Recall the suspension example

slide89

Diagonalized

u

u

u

u

l1

l2

l4

l3

y

slide90

This happens to be a partial companion form —

full companion form also changes B, and we’ll get to this

y

37515

356217

+

u

-

83.7

6970

37515

356217

slide91

The original matrix

Let’s compare the A matrices for the two methods

The diagonalized matrix we just did

The companion form matrix