MULTIVARIABLE POLE PLACEMENT AND OBSERVERS

For multiple inputs/outputs, controller and observer design is complicated by several factors:

ã Feedback gain K is not unique (as it is for SISO)

ã Some modes controllable (observable) throughsome inputs (outputs) but not others.

ã Controllability (observability) matrix non-square

Consider a plant with p inputs. We will define a “multi-input controllable canonical form” and an n x nsimilarity transformation.

CONTROLLABILITY INDICES:

P B AB A Bnn pn

= −×

M M L M 1

controllability matrix:

If there are n linearly indep. cols. amongthe n x p total columns. Therefore, there will be many ways to construct an n x n similarity transformation.

nPrank =)(

Let B b b bp= 1 2 L

(separating B into its p columns).

Then

P b b Ab Ab A b A bp pn n

p= − −1 1

11

1L M L M L M L1 24 34 1 244 344 1 2444 3444

each block contains contributions from each input;i.e., each column of B.

block 0 block 1 block n-1

P b b Ab Ab A b A bp pn n

p= − −1 1

11

1L M L M L M L1 24 34 1 244 344 1 2444 3444

block 0 block 1 block n-1

Starting from the left in this matrix, check each column, keepingcount of the number of linearly independent columns you encounter.Of course, you may stop counting when you have reached n linearly independent columns.

Let the block in which you find the last (i.e., the nth) linearly independent column be denoted the st block. Then the first block in which there are no more independent columnswill be the th block.

( )µ − 1

µ

µ is the controllability index of the system.

Note that this implies that the controllability could have been checked with the test:

[ ]BAABBrank 1−µL

Where . (m=n only for a controllable system)µ ≤ n

Suppose m=n . Now we want to gather n linearlyindependent columns from this matrix in order to generate a similarity transformation that will give us a

multi-input controllable canonical form.

There are two common ways to gather these columns:

From the matrix:

b b A b A bp p11

11L M L M Lµ µ− −

METHOD 1:

Gather columns from left to right, i.e., consider

b b b Ab Abp1 2 1 2, , , , , ,K K

keeping only those columns that are linearly independent of previously chosen columns.

When finished (n columns gathered), re-arrangethem as:

here,

# =µi

{ }=µµµ p,,, 21 K (individual ) “controllability indices of (A, B)”

Note that there is an index for each input.

[

=−µ

−µ−µ

ppp

n

bAAbb

bAAbbbAAbbM

p 1

1221

111

21

LL

LLL

of linearly independent columns associated with the ith input.

Also note that:

{ }np

p

≤µ++µ+µ

µµµ=µ

K

K

21

21 and ,,,max

THEOREM: The set of controllability indices isinvariant under any equivalence transformation, and any ordering of the columns of B.

METHOD 2:

Consider only one input (i.e., one column of B)at a time. That is, search in the order:

b Ab A b1 11

11, , ,K µ −

until is lin. dep. on 11 bAµ { }1

111

1,,, bAAbb −µK

Then continue with

21

222,,, bAAbb −µK

etc., until all n lin. indep. columns are found.

If then the entire system can be controlled from the first input alone.

µ1 = n

SUMMARIZING:

method 1: Left to right, then re-arrange.Resulting nonsingular transformation is:

M b A b b A bp pp

1 11

111= − −L M L M Lµ µ

method 2: One input at a time.

[ ]pp bAbbAbM p 11

112

1 −µ−µ= LMLML

We will use METHOD 1 to find a matrix to use asa similarity transformation (recall the similaritytransformation used to get the SISO controllablecanonical form). Now we will get a

[ ]LLLL pbAbAbbb 2121

Method 1

Method 2

multivariable canonical form

MULTIVARIABLE STATE FEEDBACK:

DuCxyBuAxx

+=+=&

pnBnnA

××

::

To simplify the notation, we will use a specific example:Let

Use METHOD 1 to get

[ ]33

32

332212

11 bAbAAbbAbbbAAbbM =

4444 34444 21 42339 321 =µ=µ=µ== pn

controllability indices

Compute

=−

34

33

32

31

22

21

13

12

11

1

mmmmmmmmm

M

e1 1µ

e2 2µ

e3 3µ

Now because ,1 IMM =−

3,2,1,0for 0

1,0for 0

1

1,0for 0

313

213

12

13

113

==

==

=

==

kbAm

kbAm

bAm

kbAm

k

k

k

Use to form the square matrix:3,2,1=µ ieii

=

334

234

34

34

22

22

213

13

13

AmAmAm

mAm

mAmAm

m

TUse this matrix as the similaritytransformation.

The result will be

== −1TATA

×

==

×××××××××

×××××××××

×××××××××

=

10000000000001000001000000

100001000010

000010

100010

TBBA

where denotes an arbitrary number, and blank spaces denote zero blocks.

×

×

In the matrix, if , the 3 nonzero rows of would be

If , they would be

For this example, we have , so the 3 nonzero rows are

Induce a pattern from this.

BB 321 µ>µ>µ

×××

10010

1

321 µ≤µ≤µ

100010001

4,2,3 321 =µ=µ=µ

×

10001001

Now because the 3 nonzero rows of are lin.indep., and all other rows are all zero, the threerows containing x’s can be arbitrarily assigned, and all of the other rows will be unaffected.

B

For example, we might choose such thatK

=−

4321

21

321

000001000010000100000000

10000000

100010

cccc

bb

aaa

KBA

This is clearly a set of three distinct controllablecanonical forms, just like the single-variable case.

Or, we could have chosen such thatK

=−

987654321

100001000010000100000

10000001000

100010

ddddddddd

KBA

...which simply a single controllable form.

(We could also have formed two canonical blocks (in two different ways!))

In the previous system, the matrix would be p x n ; i.e., 3 x 9. Let

K

=

393231

292221

191211

kkkkkkkkk

KL

Then we will have a 9 x 9 matrix

=

393231

292221

191211

10000000000001000001000000

kkkkkkkkk

x

KBL

+++

=

393231

292221

291922122111

000000000

000

000000

kkk

kkk

xkkxkkxkk

KB

LLLLLLLLL

It is clear from the form of this matrix that any valuein the third, fifth, or ninth row of the A-matrix canbe arbitrarily chosen.

• Choose the 2nd and 3rd rows of to produce the desired fifth and ninth rows of .

• From the desired third row of and the second rowof as chosen above, select the first row of .

KA

Procedure:

KA

K

(If the nonzero rows of were ,

choose the third, second, then first rows of .)K

B

×××

10010

1

See example 10.2.1.1

uxx

−−

−−

+

−−−−

−−−−

=

711116141

3611163511425433111333112112

&

−−−−−−−−

−−−−−−−−

=

202246814236713313611120378111175245914206617316271211641

P

(linearly independent)

We wish to have closed-loop poles at }33,5,5{ j±−−−

−−−−−−

=−

32

31

611

21

31

319

34

1

013041

131

M

So the first four columns of P become the matrix M. Then

−−−−

−

=

=

34

31

34

31

32

31

617

611

61

61

611

21

31

22

22

12

12

01

1

Amm

Amm

T

and

==

−−−−

−−== −

10000100

521000

0010

314

32

67

38

37

34

1 TBBTATA

From this, the transformed system is:

=

24232221

14131211

kkkkkkkk

K

Let the state feedback matrix be denoted

Directly applying this feedback,

+−+−+−+−

+++−+−=

+

−−−−

−−=+

24314

232232

21

1467

1338

1237

1134

24232221

14131211

314

32

67

38

37

34

521000

0010

10000100

521000

0010

kkkk

kkkk

kkkkkkkk

KBA

α−α−α−α−

β−β−

α−α−

321010

10

100001000010

or

001000000010

We have two choices the for final form of our transformed system:

β+β+=++=−+++

α+α+=++=++

∆

∆

0122

0122

186)33)(33(

and

2510)5)(5(

ssssjsjs

ssssss

We can accomplish both of these, with either

or

012

23

34

234 45033010316)33)(33)(5)(5(

α+α+α+α+=

++++=−+++++

ssss

ssssjsjsss

−−

−−=+

6180010000010250010

KBA

Computing the first of these,

giving

−−−−−−

=3

43

26

73

83

23

2

1 132723

K

And the second choice (a single block)

−−−−

=+

16103330450100001000010

2KBA

or

−−−−−−

=3

343

16

13

83

73

4

2 98329448K

(then don’t forget to transform back to whatever the original basis was!!)

These examples considered only the placement of poles in a controller. By using the same principles of duality as we used in the design of observer gain for single-output systems, we can now find observer gains for multiple-output systems.

This will require observability indices, observer canonical forms, etc.