Download - Lecture 15: State Feedback Control: Part I Pole Placement for SISO Systems Illustrative Examples
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Lecture 15: State Feedback Control: Part I
• Pole Placement for SISO Systems
• Illustrative Examples
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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.
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Open-Loop vs. Closed-Loop Control
SystemOutput
ControllerDesired Output
Command Input
Open-Loop Control
Closed-Loop Control
-SystemController
OutputDesired Output
Sensor
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Closed-Loop Control• Advantages
– automatically adjusts input– less sensitive to system variation and
disturbances– can stabilize unstable systems
• Disadvantages– Complexity– Instability
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State Feedback Control Block Diagram
Z-1H
G
xC
y
-K
r u
rKxu
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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.
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Closed-loop System
Plant:)()(
)()()1(
kk
kkk
Cxy
uHGxx
Control: rKxu
Closed-loop System:
Cxy
)(rH)(xHKG)1(x
kkk
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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
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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
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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
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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
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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)
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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:
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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.
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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
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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
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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
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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
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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);
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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
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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
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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
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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
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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
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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
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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);
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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)
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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
)
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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
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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)
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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
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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
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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);
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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
)