chap4_student.pdf

Chapter 4. The Design of State Feedback Control

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(�� Ç)

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012

Outline

1 4.1 Introduction

2 4.2 State feedback

3 4.3 Pole placement using state feedback

4 4.4 Design of state observers

5 4.5 Feedback from estimated states

6 4.6 Simulations with MATLAB

4.1 Introduction

Introduction

What is the fundamental idea of control theory?

◮ Feedback!!!

How to achieve feedback control?

◮ Classical control: Cascade compensator, output feedback controller,

...

◮ State space control: State feedback

How to achieve state feedback control

◮ State feedback

◮ State feedback + state observer = output feedback

Prof J Wang (Tongji Uni) Chap 4. State feedback control Spring 2012 4 / 42

Outline

1 4.1 Introduction






4.2 State feedback

State feedback scheme

Consider an nth-order SISO system

x(t) = Ax(t) + bu(t)

y(t) = cx(t)

In state feedback, the input u(t) is given by

u(t) = r(t) − kx(t) = r(t) − [k1 k2 · · · kn]

x1(t)

x2(t)...

xn(t)

Such a scheme is called state feedback and the vector k is called the

state feedback gain matrix.Prof J Wang (Tongji Uni) Chap 4. State feedback control Spring 2012 6 / 42

4.2 State feedback

ub

∫

c y

A

+

r

k

+

−

The system with the state feedback is

x(t) = (A − bk)x(t) + bu(t)

y(t) = cx(t)

What can we observe from the above model?


4.2 State feedback

Will the state feedback change controllability?

Let’s first study a 4-th order system, i.e. n = 4. The controllability

matrices of the open-loop and closed-loop systems are respectively

Qc =[

b Ab A2b A3b]

Qck =[

b (A − bk)b (A − bk)2b (A − bk)3b]

=[

b Ab A2b A3b]

1 −kb −k(A − bk)b −k(A − bk)2b

0 1 −kb −k(A − bk)b

0 0 1 −kb

0 0 0 1

rank Qc = rank Qck =⇒ Controllability is NOT changed.


We can conclude that the closed-loop system is controllable via state

feedback if and only in the open-loop system is controllable. In other

words, the state feedback does NOT change the controllability of the

system.

How to understand this feature of the state feedback? Let’s go back to

the definition of state controllability.u

b∫

c y

A

+r

k

+

−

Denote the open-loop and

closed-loop systems by Σ and Σcl.

Let’s prove in two steps:

Σ controllable −→ Σcl controllable

Suppose for Σ there exists an input u1(t): x(t0)u1(t)−−−→ x(t1).

Then, for Σcl, we have r1(t) = u1(t) + kx(t): x(t0)r1(t)−−−→ x(t1).

Σcl controllable −→ Σ controllable

If Σ is not controllable, that is u(t) cannot control x(t), then r(t)

cannot control it.

4.2 State feedback

Will the state feedback change observability?

Let’s study it by an example! Given a system described by

x1

x2

=

1 1

0 −1

x1

x2

+

0

1

u

y =[

1 1]

x1

x2

the observability of the system can be verified by

rank Qo = rank

c

cA

= rank

1 1

1 0

= 2


4.2 State feedback

Suppose the state feedback gain is k =[

k1 k2

]

. Then the closed-loop

system matrix is

A − bk =

1 1

0 −1

−

0

1

[

k1 k2

]

=

1 1

−k1 −1 − k2

and the observability matrix is changed into

Qo−cl = rank

1 1

1 − k1 −k2

It is clear that rank Qo−cl depends on the value of k1 and k2.

rank Qo−cl =

1, k1 − k2 = 1;

2, otherwise.

Yes, the state feedback can change the observability of the system!Prof J Wang (Tongji Uni) Chap 4. State feedback control Spring 2012 11 / 42

4.2 State feedback

Will the state feedback change stability?

The stability of an LTI system depends completely on its

eigenvalues.

The state feedback control can change the controllable eigenvalues

of the open-loop system.

If an open-loop system is completely state controllable, the state

feedback can definitely change the stability of the system.


Outline

1 4.1 Introduction






4.3 Pole placement using state feedback

Pole placement by state feedback

Definition

Given an nth-order open-loop system


y(t) = cx(t)

design an appropriate state feedback gain k such that the closed-loop

system constructed by u(t) = r − kx(t)

x(t) = (A − bk)x(t) + bu(t)

y(t) = cx(t)

has the expected poles{

λ∗

1 , λ∗2 , · · · , λ∗n}

.


Necessary and sufficient conditions for pole placement

If a system is not completelycontrollable, it can bedecomposed according tocontrollability as follows

˙xc

˙xnc

=

Ac A12

0 Anc

xc

xnc

+

Bc

0

u

y =[

Cc Cnc

]

xc

xnc

Suppose the feedback gain k

can be partition as [k1, k2].

u Bc

∫

Cc

Ac

A12

∫

Cnc

Anc

y

+

+y1

+

+y2

+

+u

k1

k2

−

−

It is clear that the state feedback cannot affect the uncontrollable modal.

The state controllability is the necessary condition for the pole

placement by state feedback.


If the system is completely controllable, we can transform it into the

following controllable canonical form (Ac, bc, cc), where

Ac = P−1c APc =

0 1...

. . .0 1

−a0 −a1 ··· −an−1

, bc = P−1c b =

0...01

, cc = cPc

The characteristic equation of the open-loop system is

det(sI − A) = sn + an−1sn−1 + · · · + a1s + a0 = 0

Let k = kPc =[

k1 k2 · · · kn

]

. Then we have

Ac − bck =

0 1...

. . .0 1

−a0 −a1 ··· −an−1

−

0...01

[ k1 k2 ··· kn ]

=

0 1...

. . .0 1

−a0−k1 −a1−k2 ··· −an−1−kn



The characteristic equation of the closed-loop system is

det(sI−Ac +bcK) = sn +(an−1 + kn)sn−1 + · · ·+(a1 + k2)s+(a0 + k1) = 0

Provided that the expected poles of the closed-loop system are

Γ ={

λ∗

1 , λ∗2 , · · · , λ∗n}

one can obtain the related characteristic equation is

∆∗(s) = sn + a∗n−1sn−1 + · · · + a∗1s + a∗0 = 0

Comparing the above two characteristic equations, we have

k1 = a∗0 − a0

k2 = a∗1 − a1

...

kn = a∗n−1 − an−1



Therefore, we have

k = kPc =[

a∗0 − a0 a∗1 − a1 · · · a∗n−1 − an−1

]

The state feedback gain k for the original system Σ is

k = kP−1c =

[

a∗0 − a0 a∗1 − a1 · · · a∗n−1 − an−1

]

P−1c

It is clear that the eigenvalues of the completely-controllable open-loop

system Σ can be arbitrarily placed to expected positions by using the

state feedback.

In other words, controllability of an open-loop system is a sufficient

condition for arbitrarily pole placement.

Conclusion

The state controllability is a necessary and sufficient condition for pole

placement.


4.3 Pole placement using state feedback 4.3.1 Feedback gain by canonical transformation

Determine k by canonical transformation

The proof of the sufficient condition also provides an approach to

designing the state feedback gain k, as summarized in the following

1 Check whether the open-loop system is completely controllable.

2 Determine the characteristic polynomial ∆(s) for matrix A and the

expected characteristic polynomial ∆∗(s)

∆(s) = det(sI − A) = sn + an−1sn−1 + · · · + a1s + a0

∆∗(s) = (s − λ

∗

1)(s − λ∗

2 ) · · · (s − λ∗

n) = sn + a∗n−1sn−1 + · · · + a∗1s + a∗0

3 Calculate the transformation matrix P−1c .

4 Calculate the required state feedback gain

k =[

a∗0 − a0 a∗1 − a1 · · · a∗n−1 − an−1

]

P−1c


Example

Consider an open-loop system defined by

x1

x2

=

−1 0

1 −2

x1

x2

+

1

2

u

y =[

0 1]

x1

x2

Try to find the state feedback gain matrix k such that the closed-loop

system poles are located at s = −1 ± j

Solutions

(1) Verify the controllability condition.

As rank(Qc) = rank[b Ab] = rank[

1 −12 −3

]

= 2, the system is completely

controllable.

4.3 Pole placement using state feedback 4.3.1 Feedback gain by canonical transformation

(2) Determine the characteristic polynomials.

∆(s) = det(sI − A) = det[

s+1 0−1 s+2

]

= s2 + 3s + 2

∆∗(s) = (s + 1 + j)(s + 1 − j) = s2 + 2s + 2

(3) Determine the transformation matrix P−1c .

pc1 = [ 0 1 ]Q−1c = [ 0 1 ]

[

1 −12 −3

]−1= [ 2 1 ]

P−1c =

[

pc1

pc1A

]

=[

2 −1−3 2

]

(4) The required state feedback gain matrix is

k = [ a∗0−a0 a∗1−a1 ]P−1c = [ 0 −1 ]

[

2 −1−3 2

]

= [ 3 −2 ]

�


4.3 Pole placement using state feedback 4.3.2 Feedback gain by direct substitution method

Determine k by direct substitution method

If a system is of low order (n 6 3), direct substitution of the feedback

gain k into the desired characteristic polynomial may be simpler.

For example, if n = 3, then write the state feedback gain as

k = [k1 k2 k3]

Suppose the expected eigenvalues are λ∗

1 , λ∗2 and λ∗

3 .

Hence, we have

det(sI − A + bK) = (s − λ∗

1)(s − λ∗

2 )(s − λ∗

3)

By equating the coefficients of the powers of s on both sides, we can

determine the values of k1, k2 and k3.


4.3 Pole placement using state feedback 4.3.3 Feedback gain by Ackermann’s formula

Determine k by Ackermann’s formula

Given an nth-order open-loop system


y(t) = cx(t)

one can calculate a state feedback gain k

k = [0 0 · · · 1]Q−1c ∆

∗(A)

such that the closed-loop system

x(t) = (A − bk)x(t) + bu(t)

y(t) = cx(t)

places its poles in{

λ∗

1 , λ∗2 , · · · , λ∗n}

.


4.3 Pole placement using state feedback 4.3.3 Feedback gain by Ackermann’s formula

Example

Consider the previous example Σ(A, b, c), where A =[

−1 01 −2

]

,

B =[

12

]

, c = [ 0 1 ].

Try to find the state feedback gain k by the Ackermann’s formula such

that the closed-loop system poles are −1 ± j.

Solutions

(1) Q−1c =

[

1 −12 −3

]−1=

[

3 −12 −1

]

.

(2) ∆∗(s) = (s + 1 + j)(s + 1 − j) = s2 + 2s + 2.

(3) ∆∗(A) =[

−1 01 −2

]2+ 2

[

−1 01 −2

]

+ 2[

1 00 1

]

=[

1 0−1 2

]

.

(4) k = [ 0 1 ]Q−1c ∆

∗(A) = [ 0 1 ][

3 −12 −1

] [

1 0−1 2

]

= [ 3 −2 ].

�


Outline

1 4.1 Introduction






4.4 Design of state observers

Introduction to state observers

What is a state observer?◮ A state observer estimates the state variables of a system based on

the measurements of the system’s output and control variables.

Why use?◮ In pole placement design, we assumed that ALL state variables are

available for feedback.

◮ In practice, not all state variables are measurable.

◮ Sometimes, even if some states are measurable by sensors, it may

be too expensive to implement in engineering.

What types?◮ Full-order observer: Observe all the state variables of the system,

regardless of whether some state variables are measurable or not.

◮ Minimum-order observer: Observe only the minimum number of

the state variables.Prof J Wang (Tongji Uni) Chap 4. State feedback control Spring 2012 26 / 42


How to construct a state observer

Given an SISO system


y(t) = cx(t)

the state observer can be

˙x(t) = Ax(t)+bu(t)+f[

y(t) − cx(t)]

where x is an estimation of x.

u b∫

c y

A

x

+

b∫

x

A

+

c

f−

+

+

Define e(t) = x(t) − x(t). We have

e = x − ˙x

= [Ax + bu] −[

Ax + bu + f (cx − cx)]

= Ae − fce = (A − fc)e


What can we observe from the observer equation

e(t) = (A − fc)e(t)

If the eigenvalues of (A − fc) can be assigned arbitrarily, the

behavior of the error e(t) can be controlled.

The design of a full-order state observer is to determine the

feedback gain f such that (A − fc) has the expected eigenvalues.

Matrices (A − fc) and (A − fc)T = (AT − cTfT) have the same

eigenvalues.

The design of a full-order state observer is to determine the

feedback gain fT such that (AT − cTfT) has the expected

eigenvalues.

The design of a full-order state observer is to equivalent to the

state-feedback pole placement problem for the system

Σd(AT, cT, bT), which is the dual system of Σ(A, b, c).


Procedures for designing a state observer

Given an SISO system Σ(A, b, c), design a full-order state observer

˙x(t) = (A − fc)x(t) + bu(t) + fy(t)

such that the poles of the observer, i.e. eig(A − fc), are in the desired

position Γ ={

λ1, λ2, · · · , λn

}

.

1 Transform the system Σ(A, b, c) into its dual system Σd(AT, cT, bT).

2 Design a state feedback gain k so that the closed-loop system

Σd(AT − cTk, cT, bT) has the expected poles.

3 Obtain the gain f for state observer, i.e. f = kT.



Example

Consider the system Σ(A, b, c), where

A =[

0 1−2 −1

]

, B =[

01

]

, c = [ 2 0 ]

Assuming that the desired eigenvalues of the observer are s1,2 = −8,

design a full-order state observer.

Solutions

(1) Verify the observability condition.

As rank(Qo) = rank[

ccA

]

= rank[

2 00 2

]

= 2, the system is completely

observable.

(2) Transform the full-order observer design problem for Σ(A, b, c) into

the pole placement problem for Σ(A, b, c), where

A =[

0 −21 −1

]

, B =[

20

]

, c = [ 0 1 ]


(3) Determine the characteristic polynomials.

∆(s) = det(sI − A) = det[

s 2−1 s+1

]

= s2 + s + 2

∆∗(s) = (s + 8)2 = s2 + 16s + 64

(4) Determine the transformation matrix P−1c .

pc1 = [ 0 1 ] Q−1c = [ 0 1 ]Q−1

o = [ 0 1 ][

2 00 2

]−1= [ 0 0.5 ]

P−1c =

[

pc1

pc1A

]

=[

0 0.50.5 −0.5

]

(5) The required state feedback gain matrix is

k = [ a∗0−a0 a∗1−a1 ] P−1c = [ 62 15 ]

[

0 0.50.5 −0.5

]

= [ 7.5 23.5 ]

Hence, f = kT =[

7.523.5

]

.

(6) The full-order state observer is given by

˙x(t) = (A − fc)x(t) + bu(t) + fy(t)

∴ ˙x(t) =[

−15 1−49 −1

]

x(t) +[

01

]

u(t) +[

7.523.5

]

y(t)

�

Outline

1 4.1 Introduction






4.5 Feedback from estimated states

How about feedback from estimated states?

We know how to design state feedback gain and how to design a state

observer? What will happen if we combine they together?

ub

∫

c y

A

+r

k

+

−

ub

∫

c y

A

x

+

b∫

x

A

+

c

f−

+

+

r

k

+

−



Consider a completely controllable and observable system


y(t) = cx(t)

ub

∫

c y

A

x

+

b∫

x

A

+

c

f−

+

+

r

k

+

−

The control input u(t) is

u(t) = r(t) − kx(t)

∴ x = Ax(t) − bkx(t) + br(t)


ub

∫

c y

A

x

+

b∫

x

A

+

c

f−

+

+

r

k

+

−

∴ x = Ax(t) − bkx(t) + br(t)

The state equation of the state observer can be written as

˙x(t) = (A − fc)x(t) + bu(t) + fy(t)

= (A − fc)x(t) + b[

r(t) − kx(t)]

+ fcx(t)

= Ax(t) + fc[

x(t) − x(t)]

− bkx(t) + br(t)

= Ax(t) + fce(t) − bkx(t) + br(t) (define e(t) = x(t) − x(t))

∴ e(t) = (A − fc)e(t)


We have the following differential equations

x(t) = Ax(t) − bkx(t) + br(t)

e(t) = (A − fc)e(t)

Therefore, we can build up the state-space model for the whole system

consisting of the state observer and state feedback control:

x(t)

e(t)

=

A − bk bk

0 A − fc

x(t)

e(t)

+

b

0

r(t)

y(t) =[

c 0]

x(t)

e(t)



The characteristic equation of the whole system is

∆(s) =

∣

∣

∣

∣

∣

∣

∣

sI −

A − bk bk

0 A − fc

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

sI − A + bk −bk

0 sI − A + fc

∣

∣

∣

∣

∣

∣

= 0

or

∆(s) = (sI − A + bk) (sI − A + fc) = 0

Remarks

The closed-loop poles of the whole system consists of two parts

◮ (sI − A + bk) is associated with feedback pole placement.

◮ (sI − A + fc) is associated with state observer.

The feedback controller and the state observer can be designed

separately and combined together to construct the feedback

controller with a state observer — The Separation Principle.



We now discuss the transfer function of the whole system. From the

state space model, we have

Y(s)

R(s)=

[

c 0]

sI − A + bk −bk

0 sI − A + fc

−1

b

0

= c(sI − A + bk)−1b

Remarks

This transfer function is exactly the same as the closed-loop

feedback control system directly from the state.

The observer is completely canceled and does not appear in the

transfer function from r to y.

The closed-loop control system with state observer is not

completely observable.


Outline

1 4.1 Introduction






4.6 Simulations with MATLAB

Matlab commands used in this chapter

Command Description

acker Calculate the feedback gain matrix by

Ackermann’s formula

place Calculate the feedback gain matrix

ctrb Compute the controllability matrix

obsv Compute the observability matrix

eig Compute eigenvalues and eigenvectors

initial Initial condition response of state-space

models


4.6 Simulations with MATLAB

A numerical example

Example

Consider the system Σ(A, B), where

A =

[

0 1 00 0 1−1 −5 −6

]

, B =[

001

]

Determine the state-feedback gain K such

that the closed-loop poles are

s1,2 = −2 ± j4, s3 = −10.

Solutions

MATLAB programs that generate the

matrix K are shown right by using the

commands acker and place.

MATLAB Programs 1

>> A = [0 1 0;0 0 1;-1 -5 -6];

>> B = [0;0;1];

>> J = [-2+j*4 -2-j*4 -10];

>> K = acker(A,B,J)

K =

199 55 8

MATLAB Programs 2

>> A = [0 1 0;0 0 1;-1 -5 -6];

>> B = [0;0;1];

>> J = [-2+j*4 -2-j*4 -10];

>> K = place(A,B,J)

K =

199.0000 55.0000 8.0000


Chapter 4. The Design of State Feedback Control

Modern Control Theory (Course Code: 10213403)

Professor Jun WANG

(�� Ç)

Department of Control Science & Engineering

School of Electronic & Information Engineering

Tongji University

Spring semester, 2012

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