1

Modern Control System Engineering

The class deals with a trend of modern control system theory and its application,

including linear control system, nonlinear control system, optimal control system,

etc.

Week 1 : Definition of the control system and introduction to modern control systems

Week 2-4 : Linear control system and MATLAB

Week 5-7 : Optimal control system

Week 8: Mid-term Test

Week 9-10 : Optimal control system

Week 11-14 : Nonlinear control system

Week 15 : Nonlinear control system, adaptive control system, fuzzy control system,

etc.

Week 16 : Final Test

2

Control System (C/S) : system which has control inputs

control input : input which can be assigned arbitrarily (with

restrictions). = input = control

- Control Objectives

Ex: ”Magnetic-ball-suspension system”

3

Modeling

Md2y(t)

dt2= Mg − i(t)2

y(t)

Ldi(t)

dt+Ri(t) = e(t)

State Equation ; Output Equation

x(t) = F (x(t), u(t)) ; y(t) = h(x(t), u(t))

x1(t) = y(t), x2(t) = y(t), x3(t) = i(t)x1

x2

x3

=

x2

g − 1M

x23

x1

−RLx3 + 1Le

; y = x1

4

Classification of Control Systems

Single Input Single Output system (SISO)

Multi-Input Multi-Output system (MIMO)

Continuous-time C/S

Discrete-time C/S

Time-invariant C/S

Time-varying C/S

Linear C/S

Nonlinear C/S

F (x, u) = Ax+Bu ; h(x, u) = Cx+Du

5

Feedback

- --

-

6

Controllerx = F (x, u)

y = h(x, u)

ux

y

x OR y

- State FB

- Output FB

- Open-loop C/S

- Closed-loop C/S

6

Modern Control System Theory

- Linear Control System

- Nonlinear Control System

- Digital Control System

- Optimal Control System

- Adaptive Control System

- Stochastic Control System

- Intelligent Control System

- Robust Control System

7

PART 1. LINEAR CONTROL SYSTEM- Transfer Functions Representation

- State Space Representation

Ex:

y(3)(t) + 2y(t) + 3y(t) + 4y(t) = 5u(t)

Assume zero initial conditions

s3Y (s) + 2s2Y (s) + 3sY (s) + 4Y (s) = 5U(s)

G(s) =Y (s)

U(s)=

5

s3 + 2s2 + 3s+ 4

Define state variables x1, x2, x3 by

x1(t) = y(t) ; x2(t) = y(t) ; x3(t) = y(t)

8

Then

x1(t) = y(t) = x2(t)

x2(t) = y(t) = x3(t)

x3(t) = y(3)(t) = −2y(t)−3y(t)−4y(t)+5u(t) = −4x1(t)−3x2(t)−2x3(t)+5u(t)

x(t) = Ax(t) +Bu(t)

y(t) = Cx(t) +Du(t)

A =

0 1 0

0 0 1

−4 −3 −2

, B =

0

0

5

, C =[1 0 0

], D =

[0]

** Transfer function can not be defined for nonlinear systems.

9

Ex:

y(n)(t)+an−1y(n−1)(t)+· · ·+a0y(t) = bmu

(m)(t)+bm−1u(m−1)(t)+· · ·+b0u(t)

Assume zero initial conditions

snY (s)+an−1sn−1Y (s)+· · ·+a0Y (s) = bms

mU(s)+bm−1sm−1U(s)+· · ·+b0U(s)

G(s) =Y (s)

U(s)=bms

m + bm−1sm−1 + · · ·+ b1s+ b0

sn + an−1sn−1 + · · ·+ a1s+ a0

10

CLASSICAL CONTROL

Y (s) = G(s)U(s)

U(s) = R(s)−B(s)

B(s) = H(s)Y (s)

Y (s) = G(s)R(s)−G(s)H(s)Y (s)

M(s) =Y (s)

R(s)=

G(s)

1 +G(s)H(s)(closed-loop transfer function)

11

-� ��- - -

6

Gc(s) Gp(s)r +

−

u y

PID controller:

Gc(s) = KP +KI

s+KDs

MODERN CONTROL

state FB

12

dx(t)

dt= Ax(t) (∗)

Def.: State Transition Matrix of A

Φ(t) : n× n matrix which satisfies

d

dtΦ(t) = AΦ(t) and Φ(0) = In×n

FACT: x(t) = Φ(t)x(0) is the solution of (*) and Φ(t) = eAt.

sΦ(s)− Φ(0) = AΦ(s) ⇒ (sI −A)Φ(s) = Φ(0) = I

Φ(s) = (sI −A)−1 =1

s(I − 1

sA)−1 =

1

s(I +

1

sA+

1

s2A2 +

1

s3A3 + · · · )

=1

sI +

1

s2A+

1

s3A2 + · · ·

∴ Φ(t) = I + tA+t2

2!A2 + · · · , etA

13

Solution of State equation and Output equation

x = Ax+Bu

y = Cx+Du

sX(s)− x(0) = AX(s) +BU(s)

(sI −A)X(s) = x(0) +BU(s)

X(s) = (sI −A)−1x(0) + (sI −A)−1BU(s)

x(t) = eAtx(0) + eAt ∗Bu(t) = eAtx(0) +

∫ t

0

eA(t−τ)Bu(τ)dτ

14

y(t) = CeAtx(0) + C

∫ t

0

eA(t−τ)Bu(τ)dτ +Du(t)

= CeAtx(0) + CeAt∫ t

0

e−AτBu(τ)dτ +Du(t)

x(t) = eA(t−t0)x(t0) + eAt∫ t

t0

e−AτBu(τ)dτ

x(t) = AeA(t−t0)x(t0) +AeAt∫ t

t0

e−AτBu(τ)dτ + eAte−AtBu(t)

= A

{eA(t−t0)x(t) + eAt

∫ t

t0

e−AτBu(τ)dτ

}+Bu(t) = Ax(t) +Bu(t)

15

EX: Discrete-time Linear C/S

x(k + 1) = Ax(k) +Bu(k), k = 0, 1, 2, · · ·

y(k) = Cx(k) +Du(k)

x(k + 1) = Ax(k) ⇒ x(k) = Akx(0)

x(1) = Ax(0) +Bu(0)

x(2) = Ax(1) +Bu(1) = A2x(0) +ABu(0) +Bu(1)

· · ·

x(k) = Akx(0)+Ak−1Bu(0)+· · ·+Bu(k−1) = Akx(0)+

k−1∑`=0

Ak−1−`Bu(`)

16

Def: (characteristic equation)

∆A(s) , det(sI −A) = 0

- the roots of the characteristic equation are said to be eigenvalues of A.

∆A(s) = (s− λ1) · · · (s− λn)

Def: Eigenvector of A associated with the eigenvalue λi

Aei = λiei & ei 6= 0

Theorem: (Cayley-Hamilton Theorem)

∆A(s) = sn+αn−1sn−1+· · ·+α0 ⇒ An+αn−1A

n−1+· · ·+α0In = On×n

PF: exercise.

17

Functions of Square Matrix

A : n× n matrix

∆A(s) = sn + αn−1sn−1 + · · ·+ α1s+ α0 = (s− λ1)m1 · · · (s− λσ)mσ

m1 + · · ·+mσ = n

i 6= j ⇒ λi 6= λj

f(A) , c0I + c1A+ · · ·+ cn−1An−1 = g(A)

where

g(s) = c0 + c1s+ · · ·+ cn−1sn−1

f (`i)(λi) = g(`i)(λi), 1 ≤ i ≤ σ, 0 ≤ `i ≤ mi − 1

18

Ex:

∆A(s) = (s− λ1)3(s− λ2)

f(λ1) = g(λ1); f ′(λ1) = g′(λ1); f ′′(λ1) = g′′(λ1)

f(λ2) = g(λ2)

Ex:

A =

2 1

0 3

eA =?, eAt =?, Ak =?, A2 =?, lnA =?, sinAt =?, cosAt =?

∆A(s) = (s− 2)(s− 3)

f(s) = est ; g(s) = c0 + c1s

f(2) = g(2) ⇒ e2t = c0 + 2c1

f(3) = g(3) ⇒ e3t = c0 + 3c1

19

c1 = e3t − e2t, c0 = 3e2t − 2e3t

eAt = f(A) = c0I + c1A = c0

1 0

0 1

+ c1

2 1

0 3

=

c0 + 2c1 c1

0 c0 + 3c1

=

e2t e3t − e2t

0 e3t

d

dteAt = AeAt ??

sin2At+ cos2At = I ??

20

Exercise: Find eAt and Ak, where

A =

5 1 2

0 5 0

0 0 7

Theorem: (Spectral Mapping Theorem)

Eigenvalues ofA are λ1, · · · , λn ⇒ Eigenvalues of f(A)are f(λ1), · · · , f(λn)

Theorem: Functions of the same matrix commutes, i.e.,

f(A)g(A) = g(A)f(A)

21

Properties of the State-Transition Matrix

1. Φ(0) = I.

2. det(Φ(t)) 6= 0, for all t.

eigenvalues of A : λ1 , · · · , λn ⇒ eigenvalues of eAt : eλ1t , · · · , eλnt

det(eAt) = e(λ1+···+λn)t 6= 0

3. Φ(t)−1 = Φ(−t)

eAte−At = eAt−At = eO = I ⇒ (eAt)−1 = e−At

In general, eP eQ 6= eP+Q

4. Φ(t2 − t1)Φ(t1 − t0) = Φ(t2 − t0)

eA(t2−t1)eA(t1−t0) = eA(t2−t0)

22

Transfer function

x = Ax+Bu

y = Cx+Du

Assume zero initial condition:

sX(s)− x(0) = AX(s) +BU(s)

(sI −A)X(s) = BU(s)

X(s) = (sI −A)−1BU(s)

Y (s) = CX(s) +DU(s) ={C(sI −A)−1B +D

}U(s)

⇒ G(s) =Y (s)

U(s)= C(sI −A)−1B +D

23

Controllability

x = Ax+Bu

y = Cx+Du

Definition: (controllability)

The state x0 is said to be controllable if there exists a piecewise

continuous input u(t) which drives the state x(0) = x0 to the state

x(tf ) = 0 for a finite time tf (> 0). If every state x0 of the system is

controllable, then the system is said to be completely controllable.

Theorem :

c.c. ⇔ S ,[B AB A2B · · · An−1B

]has rank n

24

Ex : x1

x2

=

1 0

0 1

x1

x2

+

1

0

u

S = [B AB] =

1 1

0 0

⇒ rank(S) = 1 6= 2 ⇒ NOT c.c.

Ex : x1

x2

=

0 1

0 0

x1

x2

+

0

1

u

S = [B AB] =

0 1

1 0

⇒ rank(S) = 2 ⇒ c.c.

25

x(0)

u(t)→ x(tf ) ⇒ u(t) =?

x(t) = eAtx(0) +

∫ t

0

eA(t−τ)Bu(τ)dτ

x(tf ) = eAtfx(0) +

∫ tf

0

eA(tf−τ)Bu(τ)dτ

x(tf )− eAtfx(0) = eAtf∫ tf

0

e−AτBu(τ)dτ

e−Atfx(tf )− x(0) =

∫ tf

0

e−AτBu(τ)dτ

26

u(t) = BT (e−At)TW (0, tf )−1[e−Atfx(tf )− x(0)

]

W (0, tf ) ,∫ tf

0

e−AτBBT (e−Aτ )T dτ

FACT : (Control Input which steers x(0) = x0 to x(tf ) = xf )

Suppose that the system is completely controllable. Then

det(W (0, tf )) 6= 0

and

u(t) = −BT (e−At)TW (0, tf )−1[x0 − e−Atfxf

]will transfer x0 at time t = 0 to the state xf at time t = tf .

27

PF:

x(tf ) = eAtf x0 + eAtf∫ tf

0

e−AτBu(τ)dτ

= eAtf x0 − eAtf∫ tf

0

e−AτBBT (e−Aτ )TW (0, tf )−1[x0 − e−Atf xf

]dτ

= eAtf x0 − eAtf∫ tf

0

e−AτBBT (e−Aτ )T dτW (0, tf )−1[x0 − e−Atf xf

]= eAtf x0 − eAtfW (0, tf )W (0, tf )−1

[x0 − e−Atf xf

]= eAtf x0 − eAtf

[x0 − e−Atf xf

]= xf

Remark: The control used here is the minimum energy control: i.e.,∫ tf

0

u(t)2dt

is minimal among all the controls which transfer x0 at time t = 0 to the state

xf at time t = tf .

28

Observability

x = Ax+Bu ; y = Cx+Du

Definition : (observability)

The state x(0) = x0 is said to be observable if given any input u(t),

there exists a finite time tf such that the knowledge of u(t) for

0 ≤ t ≤ tf ; the matrices A, B, C, and D; and the output y(t) for

0 ≤ t ≤ tf are sufficient to determine x(0) = x0. If every state x0 of the

system is observable, then the system is said to be completely

observable.

x(t) = eAtx(0) +

∫ t

0

eA(t−τ)Bu(τ)dτ

y(t) = CeAtx(0) + C

∫ t

0

eA(t−τ)Bu(τ)dτ +Du(t)

29

Theorem :

c.o. ⇔ V ,

C

CA...

CAn−1

has rank n

PF:

CeAtx(0) = α(t) , y(t)− C∫ t

0

eA(t−τ)Bu(τ)dτ +Du(t)

where α(t) is a known function.

C(I +At+1

2!A2t2 + · · · )x(0) = α(t)

30

Ex 1 : x1

x2

=

1 0

0 2

x1

x2

+

3

4

u ; y =[1 0

]x1

x2

V =

C

CA

=

1 0

1 0

⇒ rank(V ) = 1 6= 2 = n ⇒ NOT c.o.

Ex 2 : x1

x2

=

0 1

0 0

x1

x2

+

2

3

u ; y =[1 0

]x1

x2

V =

1 0

0 1

⇒ rank(V ) = 2 = n ⇒ c.o.

31

Similarity Transformation

y + 2y + 3y = 4u

x1 , y

x2 , y⇒

x1 = x2

x2 = −3x1 − 2x2 + 4u

x1

x2

=

0 1

−3 −2

x1

x2

+

0

4

uy =

[1 0

]x1

x2

+[0]u

32

z1 , y

z2 , y⇒

z1 = −2z1 − 3z2 + 4u

z2 = z1z1

z2

=

−2 −3

1 0

z1

z2

+

4

0

uy =

[0 1

]z1

z2

+[0]u

z1

z2

=

0 1

1 0

x1

x2

(state coordinates transformation)

33

Σ1 :

x = Ax+Bu

y = Cx+Duz = P−1x

x = PzCoordinates Transformation

z = P−1x = P−1(Ax+Bu) = P−1Ax+P−1Bu = P−1APz+P−1Bu , Az+Bu

y = Cx+Du = CPz +Du , Cz + Du

A = P−1AP , B = P−1B , C = CP , D = D

Σ2 :

z = Az + Bu

y = Cz + Du

34

Definition: (system equivalence)

System Σ1 is equivalent to Σ2, if there exists P such that

A = P−1AP , B = P−1B , C = CP , D = D

Theorem 1 :

If Σ1 and Σ2 are system equivalent (Σ1 ∼ Σ2), then

det(sI − A) = det(sI −A)

PF:

sI − A = sI − P−1AP = P−1(sP −AP ) = P−1(sI −A)P

det(sI − A) = det{P−1(sI −A)P

}= det(P−1) det(sI −A) det(P )

= det(sI −A) det(P−1P ) = det(sI −A)

35

Theorem 2 :

GΣ1(s) = GΣ2

(s)

PF:

GΣ2(s) = C(sI − A)−1B + D = CP (sI − P−1AP )−1P−1B +D

= CP{P−1(sI −A)P

}−1P−1B +D

= CPP−1(sI −A)−1(P−1)−1P−1B +D

= C(sI −A)−1B +D = GΣ1(s)

Theorem 3 :

Σ1 is c.c. ⇔ Σ2 is c.c.

PF:

SΣ1=[B AB · · · An−1B

]

36

SΣ2 =[B AB · · · An−1B

]

B = P−1B , AB = (P−1AP )(P−1B) = P−1AB ,

A2B = P−1A2B , · · · , An−1B = P−1An−1B

SΣ2=[P−1B P−1AB · · · P−1An−1B

]= P−1

[B AB · · · An−1B

]= P−1SΣ1

rank(P−1) = n ⇒ rank(SΣ2) = rank(SΣ1)

37

Theorem 4 :

Σ1 is c.o. ⇔ Σ2 is c.o.

PF:

VΣ2 =

C

CA...

CAn−1

=

CP

CAP...

CAn−1P

=

C

CA...

CAn−1

P = VΣ1P

( ∵ CA = (CP )(P−1AP ) = CAP )

rank(P−1) = n ⇒ rank(VΣ2) = rank(VΣ1

)

38

FACT :

etA = P−1eAtP

PF:

(sI −A)−1 ={P−1(sI −A)P

}−1= P−1(sI −A)−1P

etA = L−1{

(sI − A)−1}

= L−1{P−1(sI −A)−1P

}= P−1L−1

{(sI −A)−1

}P = P−1eAtP

***** Find out a NICE system Σ2, which is equivalent to

system Σ1

39

Controllability Canonical Form

A =

0 1 0

0 0 1

−α0 −α1 −α2

and B =

0

0

1

Properties

1© c.c.

PF:

S =[B AB A2B

]=

0 0 1

0 1 ∗

1 ∗ ∗

2©

det(sI − A) = s3 + α2s2 + α1s+ α0

40

PF:

det(sI−A) = det

s −1 0

0 s −1

α0 α1 s+ α2

= s2(s+α2)+α1s+α0 = s3+α2s2+α1s+α0

3©

C =[β0 β1 β2

]⇒ G(s) =

β2s2 + β1s+ β0

s3 + α2s2 + α1s+ α0

PF:

(sI − A)−1 =1

∆A(s)

∗ ∗ 1

∗ ∗ s

∗ ∗ s2

, (sI − A)−1B =1

∆A(s)

1

s

s2

G(s) = C(sI − A)−1B =β2s

2 + β1s+ β0s3 + α2s2 + α1s+ α0

41

A =

0 1 0

0 0 1

−α0 −α1 −α2

and B =

0

0

1

P =

[p1 p2 p3

]

P−1AP = A and P−1B = B ⇒ AP = PA and B = PB

A[p1 p2 p3

]=[p1 p2 p3

]0 1 0

0 0 1

−α0 −α1 −α2

42

B =[p1 p2 p3

]0

0

1

p3 = B

Ap3 = p2 − α2p3 ⇒ p2 = Ap3 + α2p

3 = AB + α2B

Ap2 = p1 − α1p3 ⇒ p1 = Ap2 + α1p

3 = A2B + α2AB + α1B

[p1 p2 p3

]=[B AB A2B

]α1 α2 1

α2 1 0

1 0 0

, SM

Ap1 = −α0p3 ?

Ap1+α0p3 = A3B+α2A

2B+α1AB+α0B = (A3+α2A2+α1A+α0I)B = 0

(by Cayley Hamilton’s Theorem )

43

Observability Canonical Form

A =

0 0 −α0

1 0 −α1

0 1 −α2

and C =[0 0 1

]

Properties

1© c.o.

PF:

V =

C

CA

CA2

=

0 0 1

0 1 ∗1 ∗ ∗

: rank 3

2©det(sI − A) = s3 + α2s

2 + α1s+ α0

44

PF: det(sI − A) = det

s −1 0

0 s −1

α0 α1 s+ α2

T

= s3 + α2s2 + α1s+ α0

3©

B =

β0

β1

β2

⇒ G(s) =β2s

2 + β1s+ β0

s3 + α2s2 + α1s+ α0

PF: (sI−A)−1 =1

∆A(s)

∗ ∗ 1

∗ ∗ s

∗ ∗ s2

T

, C(sI−A)−1 =1

∆A(s)

[1 s s2

]

G(s) = C(sI − A)−1B =β2s

2 + β1s+ β0

s3 + α2s2 + α1s+ α0

45

P−1AP = A and CP = C

P = Q−1

QAQ−1 = A and CQ−1 = C ⇒ QA = AQ and C = CQ

Q ,

q1

q2

q3

⇒ Q = MV =

α1 α2 1

α2 1 0

1 0 0

C

CA

CA2

(∗)

Exercise : Show eq (*).

46

Diagonal Canonical Form

Find out P such that

A =

λ1 0 0

0 λ2 0

0 0 λ3

= P−1AP

where λi’s are the eigenvalues of A, if possible.

P =[p1 p2 p3

]

PA = AP ⇒[p1 p2 p3

]λ1 0 0

0 λ2 0

0 0 λ3

= A[p1 p2 p3

]

47

[λ1p

1 λ2p2 λ3p

3]

=[Ap1 Ap2 Ap3

]⇒ Ap1 = λ1p

1, Ap2 = λ2p2, Ap3 = λ3p

3

pi is an eigenvector of A associate with eigenvalue λi.

P =[e1 e2 e3

]P−1 must exist.

Diagonalization

- not always possible.

- possible if and only if A has n linearly independent eigenvectors.

- possible if A has distinct eigenvalues.

48

Ex 1 :

A =

0 1

−2 3

det(sI −A) = s2 − 3s+ 2 = (s− 1)(s− 2)

λ1 = 1 : A−λ1I = A−I =

−1 1

−2 2

,−1 1

−2 2

ab

= 0 ⇒ e1 =

1

1

λ2 = 2 : A− λ2I = A− 2I =

−2 1

−2 1

⇒ e2 =

1

2

P =

1 1

1 2

⇒ P−1AP =

1 0

0 2

49

P =

1 1

2 1

⇒ P−1AP =

2 0

0 1

Ex 2 :

A =

2 0

0 2

P = I

P−1AP =

2 0

0 2

50

Ex 3 :

A =

2 2

0 2

λ1 = 2, λ1 = 2

There do not exist 2 linearly independent eigenvectors.

⇒ Can not diagonalize.

51

FACT : Assume distinct eigenvalues.

A =

λ1 0 0

0 λ2 0

0 0 λ3

, B =

b1

b2

b3

C =

[c1 c2 c3

]Then

(a)

c.c. ⇔ b1 6= 0, b2 6= 0, and b3 6= 0

(b)

c.o. ⇔ c1 6= 0, c2 6= 0, and c3 6= 0

PF:

52

(a)

S =

b1 λ1b1 λ2

1b1

b2 λ2b2 λ22b2

b3 λ3b3 λ23b3

⇒ det(S) = b1b2b3(λ2 − λ1)(λ3 − λ1)(λ3 − λ2)

(b)

V =

C

CA

CA2

=

c1 c2 c3

λ1c1 λ2c2 λ3c3

λ21c1 λ2

2c2 λ23c3

det(V ) = c1c2c3(λ2 − λ1)(λ3 − λ1)(λ3 − λ2)

53

FACT :

G(s) = C(sI−A)−1B+D =[c1 c2 c3

]1

s−λ10 0

0 1s−λ2

0

0 0 1s−λ3

b1

b2

b3

+d

=c1b1s− λ1

+c2b2s− λ2

+c3b3s− λ3

+ d

54

Jordon Canonical Form (JCF)

Find out P such that

A = P−1AP = Λ

where Λ is a ”almost” diagonal matrix which satisfies the following

properties: (see Text)

(a)

A =

λ 1 0

0 λ 1

0 0 λ

, P =[p1 p2 p3

]

55

AP = PA ⇒ A[p1 p2 p3

]=[p1 p2 p3

]λ 1 0

0 λ 1

0 0 λ

[Ap1 Ap2 Ap3

]=[λp1 p1 + λp2 p2 + λp3

]Ap1 = λp1

Ap2 = p1 + λp2

Ap3 = p2 + λp3

⇒

(A− λI)p1 = 0

(A− λI)p2 = p1

(A− λI)p3 = p2

⇒ (A− λI)3p3 = 0 and (A− λI)2p3 6= 0

Def: Generalized eigenvector w of A associated with the eigenvalue λi

of degree k

(A− λI)kw = 0 & (A− λI)k−1w 6= 0

56

⇒ generalized eigenvector w of degree 3 ⇒

p3 = w

p2 = (A− λI)w

p1 = (A− λI)2w

{p1, p2, p3} are linearly independent.

c1p1 + c2p

2 + c3p3 = 0 ⇒ c1(A− λI)2w + c2(A− λI)w + c3w = 0

⇒ c1(A− λI)4w + c2(A− λI)3w + c3(A− λI)2w = 0

⇒ c3(A− λI)2w = 0 ⇒ c3 = 0

A = P−1AP

57

(b)

A =

λ 1 0

0 λ 0

0 0 λ

AP = PA

[Ap1 Ap2 Ap3

]=[λp1 p1 + λp2 λp3

]

(A− λI)p1 = 0

(A− λI)p2 = p1

(A− λI)p3 = 0

⇒

generalized eigenvector w of degree 2

generalized eigenvector v of degree 1

⇒

p1 = (A− λI)w

p2 = w

p3 = v

58

(c)

A =

λ 0 0

0 λ 0

0 0 λ

AP = PA

[Ap1 Ap2 Ap3

]=[λp1 λp2 λp3

]

(A− λI)p1 = 0

(A− λI)p2 = 0

(A− λI)p3 = 0

⇒

generalized eigenvector p1 of degree 1

generalized eigenvector p2 of degree 1

generalized eigenvector p3 of degree 1

59

A Procedure for Computing a Jordan Canonical Form

Representation

1. Find out the characteristic polynomial of A:

∆A(λ) = (λ− λ1)n1(λ− λ2)n2 · · · (λ− λσ)nσ

2. Compute n1 linearly independent generalized eigenvectors of A

associated with λ1 as follows: Compute (A− λ1I)i, for i = 1, 2, · · · ,until the rank of (A− λ1I)k is equal to the rank of (A− λ1I)k+1. Find

a generalized eigenvector of degree k, say w. Define

wi = (A− λ1I)k−iw, for i = 1, 2, · · · , k.

If k = n1, proceed to step 3. If k < n1, try to find another generalized

eigenvector of the largest possible degree; that is, try to find another

generalized eigenvector of degree k; if this is not possible, try k − 1, and

so forth, until n1 linearly independent generalized eigenvectors are

60

found. Note that if rank(A− λ1I) = α, then there are totally (n− α)

chains of generalized eigenvectors associated with λ1.

3. Repeat STEP 2 for the eigenvalues λ2, · · · , λα.

4. Let

P =[w1 w2 · · · wk · · ·

]5. A = P−1AP is a Jordan canonical form.

61

FACT : Let λ1 6= λ2 and

A =

λ1 1 0 0 0

0 λ1 1 0 0

0 0 λ1 0 0

0 0 0 λ2 1

0 0 0 0 λ2

B =

0

0

1

0

1

C =

[c1 c2 c3 c4 c5

]D = 0

Then

(a) c.c.

(b)

c.o. ⇔ c1 6= 0 and c4 6= 0

62

(c)

eAt =

eλ1t teλ1t 12! t

2eλ1t 0 0

0 eλ1t teλ1t 0 0

0 0 eλ1t 0 0

0 0 0 eλ2t teλ2t

0 0 0 0 eλ2t

(d)

G(s) =c1

(s− λ1)3+

c2(s− λ1)2

+c3

s− λ1+

c4(s− λ2)2

+c5

s− λ2

PF:

63

(c)

(sI − A)−1 =

1s−λ1

1(s−λ1)2

1(s−λ1)3 0 0

0 1s−λ1

1(s−λ1)2 0 0

0 0 1s−λ1

0 0

0 0 0 1s−λ2

1(s−λ2)2

0 0 0 0 1s−λ2

(d)

(sI − A)−1B =

1(s−λ1)3

1(s−λ1)2

1s−λ1

1(s−λ2)2

1s−λ2

64

G(s) = C(sI − A)−1B + D =[c1 c2 c3 c4 c5

]

1(s−λ1)3

1(s−λ1)2

1s−λ1

1(s−λ2)2

1s−λ2

65

Realization of G(s) (or Decomposition)

Given

G(s) =β2s

2 + β1s+ β0s3 + α2s2 + α1s+ α0

find

x = Ax+Bu

y = Cx

whose transfer function is G(s).

1. Direct Decomposition

1© C. C. F

x =

0 1 0

0 0 1

−α0 −α1 −α2

x+

0

0

1

u

66

y =[β0 β1 β2

]x+

[0]u

state diagram

2© O. C. F

x =

0 0 −α0

1 0 −α1

0 1 −α2

x+

β0

β1

β2

uy =

[0 0 1

]x+

[0]u

state diagram

2. Parallel Decomposition

3© D. C. F

G(s) =β2s

2 + β1s+ β0s3 + α2s2 + α1s+ α0

=k1

s− λ1+

k2s− λ2

+k3

s− λ3

67

where λ1, λ2, and λ3 are distinct.

x =

λ1 0 0

0 λ2 0

0 0 λ3

x+

1

1

1

uy =

[k1 k2 k3

]x+

[0]u

state diagram

4© J. C. F

G(s) =k1

s− λ1+

k2(s− λ1)2

+k3

s− λ2(λ1 6= λ2)

68

x =

λ1 1 0

0 λ1 0

0 0 λ2

x+

0

1

1

uy =

[k2 k1 k3

]x+

[0]u

state diagram

Ex :

G(s) =s+ 2

(s+ 1)(s+ 2)=

s+ 2

s2 + 3s+ 2

(a) C.C.F.

⇒

x1x2

=

0 1

−2 −3

x1x2

+

0

1

u ; y =[2 1

]x1x2

69

V =

2 1

−2 −1

⇒ det(V ) = 0 ⇒ NOT c.o.

(b) O.C.F.

⇒

x1x2

=

0 −2

1 −3

x1x2

+

2

1

u ; y =[0 1

]x1x2

S =

2 −2

1 −1

⇒ det(S) = 0 ⇒ NOT c.c.

FACT: G(s) = N(s)D(s)

N(s) and D(s) are coprime ⇔ c.c. and c.o.

70

Stability

x = Ax+Bu ; y = Cx+Du

Definition : (asymptotically stable)

The system is asymptotically stable , if Φ(t)x(0) (= eAtx(0)) goes to 0n×1 as

t goes to ∞ for all x(0).

Theorem :

a.s. ⇔ All the eigenvalues of A lie in open left-half plane

State Feedback x1 = x1 + u

y = x1unstable

u = −2x1 + r

71

x1 = −x1 + r

y = x1stable

x = Ax+Bu ; y = Cx+Du

u = −Kx+ r

x = (A−BK)x+Br = Acx+Br

y = (C −DK)x+Dr = Ccx+Dr

72

Arbitrary pole assignment via state feedback

possible if and only if c.c.

det(sI −Ac) = (s− λ1)(s− λ2)(s− λ3) = s3 + c2s2 + c1s+ c0

(1) C.C.F.

A =

0 1 0

0 0 1

−α0 −α1 −α2

and B =

0

0

1

u = −Kx+ r = −

[k1 k2 k3

]x+ r

Ac = A−BK =

0 1 0

0 0 1

−α0 − k1 −α1 − k2 −α2 − k3

73

⇒ det(sI −Ac) = s3 + (α2 + k3)s2 + (α1 + k2)s+ (α0 + k1)

⇒ k1 = c0 − α0, k2 = c1 − α1, k3 = c2 − α2

(2) NOT C.C.F.

Find P such that z = P−1x and

z = P−1APz + P−1Bu , Az + Bu

y = CPz +Du , Cz + Du

where

A =

0 1 0

0 0 1

−α0 −α1 −α2

and B =

0

0

1

u = −Kz + r

74

K =[k1 k2 k3

]By (1),

k1 = c0 − α0, k2 = c1 − α1, k3 = c2 − α2

⇒ det{sI − (A− BK)} , det(sI − Ac) = s3 + c2s2 + c1s+ c0

Note that

K = KP ⇒ A− BK = P−1AP − P−1BKP = P−1(A−BK)P

⇒ det(sI − Ac) = det(sI −Ac)

u = −Kz + r = −KPx+ r = −Kx+ r

where

K = KP−1

75

Ex 1 :

x =

0 1

−3 4

x+

0

1

u∆A(s) = s2 − 4s+ 3 = (s− 1)(s− 3) ⇒ unstable

Want

∆(A−BK)(s) = (s+ 2)(s+ 3) = s2 + 5s+ 6

A−BK =

0 1

−6 −5

=

0 1

−3 4

−0

1

[k1 k2

]=

0 1

−3− k1 4− k2

⇒ k1 = 3 and k2 = 9

76

EX 2 : NOT c.c.

x =

1 0

0 2

x−1

0

[k1 k2

]x+

1

0

v =

1− k1 −k20 2

x+

1

0

v

∆(A−BK) =

∣∣∣∣∣∣s− 1 + k1 k2

0 s− 2

∣∣∣∣∣∣ = (s− 2)(s− 1 + k1) = 0

⇒ s = 2, − k1 + 1 ⇒ unstable

Ex 3 :

A =

1 0

0 2

B =

1

1

A−BK =

1 0

0 2

−1

1

[k1 k2

]=

1− k1 −k2−k1 2− k2

77

[s−(1−k1)][s−(2−k2)]−k1k2 = s2+(k1+k2−3)s+(2−2k1−k2) = s2+2s+1

⇒ k1 = −4 , k2 = 9

Theorem : (Controllability of closed-loop system)

State feedback cannot alter the controllability of the system.

PF:

S =[B AB A2B · · ·

]

Sc =[B (A−BK)B (A−BK)2B · · ·

]=[B AB −BKB A2B −ABKB −BKAB +BKBKB · · ·

]

78

= S

1 −KB −K(A−BK)B · · ·

0 1 −KB · · ·

0 0 1 · · ·...

......

. . .

Theorem : (Observability of closed-loop system)

State feedback can alter the observability of the system.

Ex :

x =

0 1

0 0

x+

0

1

u ; y =[0 1

]x

V =

C

CA

=

0 1

0 0

⇒ NOT c.o.

79

u = −Kx+ r = −[−1 0

]x+ r

x =

0 1

1 0

x+

0

1

r = Acx+Br

y =[0 1

]x

Vc =

C

CAc

=

0 1

1 0

⇒ c.o.

80

Observer

Assume that x is not measurable. (Often, this is the case).

Can we estimate x(state) by using the knowledge of {A,B,C,D} and input

u(t) and the output y(t)?

x = Ax+Bu ; y = Cx

˙x = Ax+Bu

e(t) = x(t)− x(t)

e(t) = ˙x(t)− x(t) = Ax+Bu−Ax−Bu = A(x− x) = Ae(t)

A is asymptotically stable ⇒ limt→∞

e(t) = 0

81

Luenberger Observer

˙x = (A− LC)x+ Ly +Bu

e , x− x

e = ˙x−x = (A−LC)x+Ly+Bu−Ax−Bu = (A−LC)(x−x) = (A−LC)e

A− LC is asymptotically stable ⇒ limt→∞

e(t) = 0

FACT :

c.o. ⇒ CAN find L such that (A− LC) is asymptotically stable

82

Ex : Find a Luenberger observer for the following system:

x =

1 0

2 3

x+

0

4

u ; y =[0 1

]x

A− LC =

1 0

2 3

−`1`2

[0 1]

=

1 −`12 3− `2

∆A−LC = (s− 1)(s− 3 + `2) + 2`1 = s2 + (`2 − 4)s+ 2`1 − `2 + 3

∆A−LC = s2 + 3s+ 2 ⇒ L =

`1`2

=

3

7

Luenberger observer

˙x = (A− LC)x+ Ly +Bu =

1 −3

2 −4

x+

3

7

y +

0

4

u

83

DESIGN OF A FEEDBACK CONTROL SYSTEM

x =

0 1 0

0 0 1

−α0 −α1 −α2

x+

0

0

1

u ; y =[β0 β1 β2

]x

Want to find a FB control law

u = −Kx+ pr = −[k1 k2 k3

]x+ pr

such that y(t) −−−→t→∞ r (constant).

x =

0 1 0

0 0 1

−α0 − k1 −α1 − k2 −α2 − k3

x+

0

0

p

r ; y =[β0 β1 β2

]x

84

Y (s)

R(s)=

p(β2s2 + β1s+ β0)

s3 + (α2 + k3)s2 + (α1 + k2)s+ (α0 + k1)

R(s) =r

s⇒ Y (s) =

rp(β2s2 + β1s+ β0)

s(s3 + α2s2 + α1s+ α0)

limt→∞

y(t) = lims→0

sY (s) =rpβ0

α0= r ⇒ p =

α0

β0

s3 + α2s2 + α1s+ α0 = (s2 + 2s+ 2)(s+ 10) choose a set of new poles

α0, α1, α2 ⇒ k1, k2, k3