Lecture 15: State Feedback Control: Part I

• Pole Placement for SISO Systems

• Illustrative Examples

Feedback Control Objective

In most applications the objective of a control system is to regulate or track the system output by adjusting its input subject to physical limitations. There are two distinct ways by which this objective can be achieved: open-loop and closed-loop.

Open-Loop vs. Closed-Loop Control

SystemOutput

ControllerDesired Output

Command Input

Open-Loop Control

Closed-Loop Control

-SystemController

OutputDesired Output

Sensor

Closed-Loop Control• Advantages

– automatically adjusts input– less sensitive to system variation and

disturbances– can stabilize unstable systems

• Disadvantages– Complexity– Instability

State Feedback Control Block Diagram

Z-1H

G

xC

y

-K

r u

rKxu

State-Feedback Control Objectives

• Regulation: Force state x to equilibrium state (usually 0) with a desirable dynamic response.

• Tracking: Force the output of the system y to tracks a given desired output yd with a desirable dynamic response.

Closed-loop System

Plant:)()(

)()()1(

kk

kkk

Cxy

uHGxx

Control: rKxu

Closed-loop System:

Cxy

)(rH)(xHKG)1(x

kkk

Pole Placement Problem

Choose the state feedback gain to place the poles of the closed-loop system, i.e.,

HKG:G of sEigenvalue

At specified locationsdesdes

n ,,

1

State Feedback Control of a System in CCF

Consider a SISO system in CCF:

State Feedback Control

11K,rKxu nkk

uHxGx cc kk )(ˆ)1(ˆ

nnnn

c

nn

c

azazazzIs

,

aaaa

11

1

121

G)(

1

0

0

0

H

1000

0100

0010

G

Closed-Loop CCF System

n

nn

kk

aaaa

1

121 1

0

0

0

1000

0100

0010

G

Closed loop A matrix:

nnnn kakakaka 112211

1000

0100

0010

G

Choosing the Gain-CCF

Closed-loop Characteristic Equation

1211

1)( kazkazkazz nnn

nn

Desired Characteristic Equation:

desn

desn

ndesnn

i

desi

des azazazzz

1

11

1

)(

Control Gains:

niaaK indes

ini ,,2,1,11

Transformation to CCFTransform system uHGxx To CCF

uxaxaxax

xx

xx

nnnn

cc

ˆˆˆˆ

ˆˆ

ˆˆ

ˆˆ

1211

32

21

uHxGx

First, find how new state z1 is related to x:

vectorrow p,pxˆ11 nppx

Where x+(k)=x(k+1) (for simplicity)

Transformed State Equations

ux

xux

xux

xux

nnnn

nnnn

n

HpGxpGxpG

HpGxpGxpG

pGHxpGpGx

pHpGxpx

11

2121

32

2

21

ˆ

ˆˆ

ˆˆ

ˆˆ

0

0

0

1

Necessary Conditions for p:

100HGGHHp 1 n

1Mep Tn

Vector p can be found if the system is controllable:

State Transformation Invertibility

x

pG

pG

p

Tx

ˆ

ˆ

ˆ

1

2

1

nnx

x

x

State transformation:

Matrix T is invertible since

HpGHpG1

HpG10

100

HGGHH

pG

pG

p

22

1

1 nn

nn

n

By the Cayley-Hamilton theorem.

Toeplitz Matrix

I

001

01

1

TM 2

11

n

n

a

aa

Matrix on the right is called Toeplitz matrix

The Cayley-Hamilton theorem can further be used to show that

State Transformation Formulas

1

1

Mep,

pG

pG

p

T

Tn

n

Formula 1:

Formula 2:

1

2

11

001

01

1

MT

n

n

a

aa

State Feedback Control Gain Selection

11K̂,rx̂K̂u aaaa desn

desn

1

11K

pG

pG

p

rTK̂u

n

desn

desn aaaaKx

By Cayley Hamilton: nnnn GGaGaIa 1

11

IaGaGaGpK11

1 desdesndesn

nn

or

GΦMeK desTn

1

Bass-Gura Formula

1

2

11

1

11

100

10

1

MK1

n

n

T

ndes

ndes

des

a

aa

aa

aa

aa

n

n

11K̂,rx̂K̂u aaaa desn

desn

1

2

11

11

11

001

01

1

MK

n

n

T

des

desnn

desnn

a

aa

aa

aa

aa

Double Integrator-Matlab Solution

T=0.5; lam=[0;0];G=[1 T;0 1]; H=[T^2/2;T]; C=[1 0];K=acker(G,H,lam);Gcl=G-H*K;clsys=ss(Gcl,H,C,0,T);step(clsys);

Flexible System ExampleConsider the linear mass-spring system shown below:

m1

um2

x2x1

kParameters:

m1=m2=1Kg. K=50 N/m

• Analyze PD controller based on a)x1, b)x2

• Design state feedback controller, place poles at j 125,20,20

Collocated Control

Transfer Function: 100

50

)(

)(22

21

ss

s

sU

sXGp

PD Control: 20, aasKGc

-50 -40 -30 -20 -10 0-25

-20

-15

-10

-5

0

5

10

15

20

25

Real Axis

Imag

Axi

s

Root-Locus

Non-Collocated Control

Transfer Function: 100

50

)(

)(22

2

sssU

sXGp

PD Control: 20, aasKGc

Root-Locus

-35 -30 -25 -20 -15 -10 -5 0 5-15

-10

-5

0

5

10

15

Real Axis

Imag

Axi

s

Unstable

State Model

u

x

x

x

x

050

050

1000

0100

x

x

x

x

4

3

2

1

4

3

2

1

0

1

0

0

050

050

Discretized Model: x(k+1)=Gx(k)+Hu(k)

0

01.0

0

0

,

9975.00025.04992.04992.0

0025.09975.04992.04992.0

01.00997500025.0

001.00025099750

H.

..

G

Open-Loop System Information

003.00002.0001.00

0097.00098.00099.001.0

0000

0003.00002.00001.00

M

HGGGHGGHHM 2

Controllability matrix:

Characteristic equation:

|zI-G|=(z-1)2(z2-1.99z+1)=z4-3.99z3+5.98z2-3.99z+6

State Feedback ControllerCharacteristic Equations:

5819.06675.25822.44963.3)(

0658.09294.08187.0)(234

222

zzzzs

zzsdes

des

1

99.3000

99.3100

98.599.310

99.398.599.31

M

65819.0

99.36675.2

98.55822.4

99.34963.3

K

T

rxxxxu

4321 75.10554.4517.144757

75.10554.4517.14400.757K

|zI-G|=(z-1)2(z2-1.99z+1)=z4-3.99z3+5.98z2-3.99z+6

Matlab Solution%System Matricesm1=1; m2=1; k=50; T=0.01;syst=ss(A,B,C,D);A=[0 0 1 0;0 0 0 1;-50 50 0 0;50 -50 0 0];B=[0; 0; 1; 0];C=[1 0 0 0;0 1 0 0]; D=zeros(2,1);cplant=ss(A,B,C,D);

%Discrete-Time Plantplant=c2d(cplant,T);[G,H,C,D]=ssdata(plant);

Matlab Solution

%Desired Close-Loop Polespc=[-20;-20;-5*sqrt(2)*(1+j);-5*sqrt(2)*(1-j)];pd=exp(T*pc);

% State Feedback ControllerK=acker(G,H,pd);

%Closed-Loop Systemclsys=ss(G-H*K,H,C,0,T);gridstep(clsys,1)

Time Respone

Time (sec.)

Am

plitu

deStep Response

0

0.5

1

1.5

2x 10-3 From: U(1)

To:

Y(1

)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2x 10-3

To:

Y(2

)

Steady-State GainClosed-loop system: x(k+1)=Gclx(k)+Hr(k), Y=Cx(k)

Y(z)=C(zI-Gcl)-1H R(z)

If r(k)=r.1(k) then yss=C(I-Gcl)-1H

Thus if the desired output is constant

r=yd/gain, gain= C(I-Gcl)-1H

Time Response

Time (sec.)

Am

plit

ude

Step Response

0

0.5

1

1.5From: U(1)

To: Y

(1)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

To: Y

(2)

Integral Control

g

yje d

k

jIs

1

0

)(KxKuControl law:

Z-1H

G

xC

y

-Ks

yd u Ki

Integral controller

plant

Automatically generates reference input r!

1/g

Closed-Loop Integral Control System

dIs y

r

k

k

k

k H

v

xKK

0

H

)(v

)(x

IC

0G

)1(v

)1(x

Plant:)(Cx)(y

)(uH)(Gx)1(x

kk

kkk

yyeKxKru dIs kv ),(Control:

Integral state: )()()1( kkk evv

Closed-loop system

Double Integrator-Matlab Solution

T=0.5; lam=[0;0;0];G=[1 T;0 1]; H=[T^2/2;T]; C=[1 0];Gbar=[G zeros(2,1);C 1];Hbar=[H;0];K=acker(Gbar,Hbar,lam);Gcl=Gbar-Hbar*K;yd=1; r=0; %unknown gainclsys=ss(Gcl,[H*r;-yd],[C 0;K],0,T);step(clsys);

Closed-Loop Step Response

Time (sec.)

Am

plitu

de

Step Response

0

0.2

0.4

0.6

0.8

1From: U(1)

To:

Y(1

)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-4

-2

0

2

4

To:

Y(2

)