Separation Principle

Controllers & Observers

( ) ( ) ( )

ˆ

( )

ˆ( ) ( ) ( )

(

ˆ( ) ( ) ( ) ( )

)

) (

u t Kx t

x t A L

x t Ax t B

C x t Bu t

u t y t Cx

y t

t

L

r t

Plant

Controller

Observer

Picture

Composite state model

( ) ( ) ( ) ( ) ( )

ˆ ˆ( ) ( ) ( ) (

ˆ( ) ( )

) ( )

ˆ( )ˆ

(

( )(

)

0(

))

)(

x t A LC x t Bu t Ly t

L

A B x t B

C A LC BK x tx t

x t Ax t Bu t y t Cx t

x

u t Kx t r t

Kr t

t

Similarity Transformation

1

( )

( ) 0

ˆ ˆ( )

(ˆ( )ˆ( )

0 0

0

()

0

) Kr t

A B x t B

C B

x t I x xP

e t I I x x

IP P

I I

I A B I

I I C I I

A BK I

A LC A

L A LC BK x tx t

L A LC B

C

K

L I

K

I

x t

0

( ) ( )

( ) 0 ( ) 0

A BK BK

A LC

x t A BK BK x t Br

e t A LC e t

Controllability of Composite System

2 2 1

( ) ( )

( ) 0 ( ) 0

0 0 0 0

nc c c

Comp

x t A BK BK x t Br

e t A LC e t

B A B A A

Composite system cannot be controllable. (rows of zeros)•Look at transfer function of composite system

•Poles and zeros

Stability Discussion

• K stabilizes “control system”

• L stabilizes “observer”

• Values of K and L are largely independent of each other

• Observer must be “more stable” than controller

• Separation Principle

Satellite Example

More Stability

Energy Storage & State

component Energy/Power

Resistor I2R=V2/R

Capacitor, v V2/(2C)

Inductor, i LI2/2

Friction element Bv2/2

Mass, v Mv2/2

Spring, dx K(dx)2/2

Free Response and Energy

• Free response indicates how energy stored in the system “leaves” the system

( ) (0)Atx t e x• If x(t) goes to zero, then the energy initially

stored in the system (kinetic, potential, inductors, capacitors) is dissipated in friction elements, resistors, or radiates to the ambient world

Forced Response and Energy

• The forced response shows how energy from the input function (and bias sources) enters the states

Stability and Energy Dissipation

• The conceptual (not literal) connection between energy stored in the states and whether or not it dissipates to zero in the free response, leads to a powerful test of stability associated with the name of Lyapunov.

• See Chapter 8: pps. 193, 198, 200, 209

Definitions

• The origin is stable if, for any given value of ε > 0, there is a value of δ(ε) > 0 such that if ||x(0)|| < δ then ||x(t)|| < ε for all t > 0 (see Fig. 8.1 pg. 195).

• The origin is asymptotically stable if it is stable, and if there exists δ’ > 0 such that whenever ||x(0)|| < δ’ then limt||x(t)|| = 0.

• Diagonal systems

Energy Functions & Lyapunov stability

• Lyapunov functions energy functions• Consider a mechanical system with no

energy sources, consider any function built up from kinetic and potential energy of its components. This function does not increase as time passes.

• Consider an electrical system with no energy sources, consider any function built from the energy stored in the capacitors and inductors. This function does not increase as time passes.

2nd Method of Lyapunov

Along every solution x(t) contained in Ω

1. If v(x(t)) is decreasing except at the origin, then the origin is asymptotically stable, and all initial conditions inside Ω produce trajectories that approach the origin;

2. If v(x(t)) is increasing except at the origin, then the origin is unstable.

Energy Functions for LTI systems• v(x) = xTPx, P symmetric positive-definite.

( ) ( )

( )

( )

T T

T T

T T T

T T

T

T

x t Ax t

v x x Px x Px

Ax Px x PAx

x A Px x PAx

x A P PA x

x Qx

A P PA Q

• Choose real, symmetric, positive definite Q.

• Find, if possible, real, symmetric, positive definite P

• If successful, conclude stability.

• Approach generalizes to non-linear systems

Positive/Negative (semi-) definite matrices, x != 0

• xTQx > 0 positive definite• xTQx < 0 negative definite• xTQx >= 0 positive semi-definite• xTQx <= negative semi-definite

• Symmetric QT=Q• Real every element of Q is real• Q is positive definite if, for some orthogonal S, STQS is

– Diagonal,– Has non-zero, positive elements on the diagonal

• Diagonal matrix examples

Positive Definiteness

• A real symmetric matrix M is positive definite if and only if any one of the following conditions holds:

– All eigenvalues of M are positive

– There exists a nonsingular matrix N such that M = NTN

Connection between Q and P• Q is any positive definite matrix and let

( ) ( ) ( ), ( ) (0)T Atv t x t Qx t x t e x • Total energy dissipated is

0

0

0 0

( ( )) ( ) ( ) (0) (0)

(0) (0) (0) (0)

T

Tt

A t At

t

t t tT T A t At

t t t

T T

v t dt x t Qx t dt x e Qe x

e Qe

dt

x x x xdt P

• Now consider ATP+PA

ATP+PA=?

0

0 0

0 0

0

lim

0

T T

T T

T T

T

T

T T

t tT

t tA t At A t At

t t

A t At A t At A t A

A t At A t A

t t

t

t

t

t

t t

A t At

P P e Qe dt e Qe dt

e Qe e Qe

A A A A

dA e Qe

e Qe e Q

A dt dtdt

e Q e

Q Q

e

Solving Lyapunov Eqns

• ATP+PA=-Q– Given Q, A, find P– “vectorize” to get Mp=q– Solve for p– Reconstruct P from p

– Solutions exist “if and only if” λi+ λj ≠0 for eigenvalues of A, including when i = j.

Satellite Example

1,1 1,2 1,1 1,2 1,1 1,2

1,2 2,2 1,2 2,2 1,2 2,2

1,2 1

2

,1

2

,2

,2 1,2

1,2

2,2

0 1

0 0

0 1 0 0

0 0 1 0

2

0

T

A

AP PA Q

p p p p q q

p p p p q q

p q

p q

q

p q

Q cannot be totally arbitrary since q2,2 must be 0.Can Q and P be positive definite?

Satellite Example

1,1 1,2 1,1 1,1

1,2 1,1 1,2

2 21,1 1,1 1,22 2 2

1,1 1,2 1,1 1,2 , /

21,1 1,2 1,1 , /

2,

0 1

0 0

/ 2

0 / 2

4det( ) ( )

2

det( ) ( )( ) / 4

det( )

T

Q

P

A

A AP PA Q

q q p qQ P

q q q

q q qsI Q s s q q s q s q

sI P s p s q q

sI A s

, , , ,0 0A A A A

Q cannot be positive definite since it has a negative EV.

With Feedback

11 12 11 12 11 12

12 22 12 22 12 22

12 11

11 12 2

22 11 1

2 12 12 22 22

2 12

0 1

2 2

0 1 0 2

2 2 1 2

2

2 2 4 4

0 2 0

2

2 2

2 1

0 4 4

fb

Tfb fb

A A BK

A P PA Q

p p p p q q

p p p p q q

p q

p p p

p p p q

q p p q

11 111

12 12

22 22

11 11 12 22

12 11

22 11 22

6 4 11

4 0 084 0 2

6 4 /8

/ 2

2 / 4

p q

p q MP Q M

p q

p q q q

p q

p q q

Any Q produces a P.Does pos-def Q produce pos-def P?

With Feedback

11 12 11 12 11 12

12 22 12 22 12 22

11

12

22

211 22 12

0 1

2 2

0 1 0 2 1 0

2 2 1 2 0 2

1

1/ 2

1

det( ) ( )( ) ( 1)( 2)

det(

Tfb fb fbA A BK A P PA Q

p p p p q q

p p p p q q

p

p

p

sI Q s q s q q s s

s

2 2 2

11 22 11 22 12) ( ) 2 3/ 4

( 1.5)( .5)

I P s p p s p p p s s

s s

Any Q produces a P.Does pos-def Q produce pos-def P?

With Feedback and Observer

Uncancelled pole-zero pairs

• CCF– Observable?

• OCF– Controllable?

• Subspaces– CO– Not C, O– Not O, C– Not O, Not C

• Controllable if input affects all state components directly or indirectly

• Observable if all state components affects output directly or indirectly

Example

23 2

0

1( )

3 3 10 1 0 0

0 0 1 0 1 1 0

1 3 3 1

0

( )

( )

0 1 0

1 2 1 1 0

1

2

??

1

10 1

0

0

0

c

c

o

sH s

s s s

x x u y x

rank

rank

z z u y z

z Px

s

P

s

Kalman Canonical form (1)

0 0

00

c c cc c cc c

c c c

o c o oo

o oo c o o

x A A x Bu y C C x

x A x

x A x Bu y C x

x A A x B

Uncontrollable states can affect controllable states.Controllable states cannot affect uncontrollable states.

Observable states can affect unobservable states.Unobservable states cannot affect observable states.

Picture.

Compute Transfer functions.

Kalman Canonical Form (2)

?0

? ?

0 0 0 0

0 ? 0

0 0

0

?

0co co co co

co co co co

co co co

co co co

co

coco co

co

co

x A x B

x A x Bdu

x A xdt

x A x

x

xy C C

x

x

Compute Transfer Function.

Depends only on Observable and controllable portion.

Hence lower degree polynomials.

Design via Similarity Transformations -- Picture

Signal Energy

• Example 7, pg. 202

• Define signal energy

• Compute energy in free response due to x(0)

• P is solution to Lyapunov Eqn.

Optimal Control• See Example 8, Pg 203

• Introduce SVFB, need to choose K

• Define quadratic cost functional in terms of – Positive definite R– Positive semidefinite Q

• Substitute y=Cx, u=-Kx+u’

• Substitute free response, with initial state x(0)

• Compare integrand with signal energy integrand, solve resulting Lyapunov eq.