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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
)