Post on 31-Mar-2015
Modern Control Systems 1
Lecture 08 State Feedback Controller Design
8.1 State Feedback and Stabilization
8.2 Full-Order Observer Design
8.3 Separation Principle
8.4 Reduced-Order Observer
8.5 State Feedback Control Design with Integrator
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00)( , xtxBuAxx
nBFAI 21det
Fxu
xBFA
BFxAxBuAxx
Stabilization by State Feedback: Regulator Case
Given ,n,ii 1 ,
Controllable )(A,B
There exists a state feedback matrix, F, such that
Plant:
State Feedback Law:
Closed-Loop System:
Theorem
State Feedback and Stabilization
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State Feedback System (Regulator Case)
CC
AA
BB
DD
FF
xu x y
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nn
cc
fafafa
FBA
2211
000
010
1
0
0
cB
xfffFxu n 21
State Feedback Design in Controllable Form
111det fasfasFBAsI n
nnn
cc
n
nc
aaa
IA
21
1
0
0
(8.1)
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11
1
cn
cnn
n
ii asass
111
111
aaf
aaf
aaf
c
ncnn
ncnn
ncncc aaaaaaF 2211
Comparing (8.1) and (8.2), we have
Suppose the desired characteristic polynomial
(8.2)
(8.3)
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CC
AA
BB
DD
FF
State Feedback: General Case (Non-Zero Input Case)
rFxu nfffF 21
State Feedback Control System
r xu x y
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Cxy
BuAxx
n
ii
n
s1
1
: Poly.Char. Desired
,, :poles Desired
zCy
uBzAz
c
cc
Tc
n
nc
B
aaa
IA
100
0
0
21
1
zFu c Tncnccc aaaaaaF 2211
xTFzFu cc-1
121
1
det ccn
cnn
n
ii asasassBFAsI
CTC
BTB
ATTA
c
c
c
1
1
Tzx
State Feedback Design with Transformation to Controllable Form
Controllable From:
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n
nnc aaa
IA
21
110
1
0
0
cB
ncccc CCCC
21
Tc
Tc
Tc BCCBAA ooo
Cxy
RxBuAxx n
nvvvT 21
TzxxTz 1
11
1
1 where
V
l
l
L
Al
Al
l
TT
n
T
nTn
Tn
Tn
Transform to Controllable Form
xCy
RxuBxAx
c
ncc
CTC
BTBATTA
c
cc
1
11
lecontrollab is ),( ,)(
1-
BAnUrank
BAABBU n
Coordinate Transform Matrix
Controllable Form:
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uxx
1
0
22
10
18633 :l PolynomiaChar. Desired
33 :poles Desired222
sss
j
Example
41626218 F
(A, B) is in controllable from, we can derive the state feedback gain from eq. (8.3)
xFxu 416
1 1)(sU
2
-1s -1s 1x2x
2
164
y
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BuAxx 1 nRx RuCxy Ry
)()()( trtKxtu
BrxBKAx )(
0 BKAsI
nBABAABB n ] [rank 12
nRK 1
Plant:
State Feedback:
Closed Loop System:
Char. Equation:
Suppose that the system is controllable, i.e.
Obtain the State Feedback Matrix by Comparing Coefficients
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n ,,1
)()( 1 nss
BKAsI )()( 1 nss
Then, for any desired pole locations:
(8.4)
We can obtain the desired char. polynomial
By controllability, there exists a state feedback matrix K, such that
From (8.4), we can solve for the state feedback gain K.
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)10)(1(
8
)(
)(
ssssU
sY
122
33
1
8)810()811(
8
)(
)(
ksksks
k
sR
sY
Example
Plant:
r
10
1 y1x 2x
1s 1s8u1 1s3x
1
2k
1k
3k
1
Fig. State Feedback Design Example
)()( tKxtu State Feedback: 321 kkkK
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ploes)(dominant 882,1 js 403 s
Desired pole locations:
707.0,8nPercent Overshoot 5%, Settling Rise time 5 sec.
)40)(88)(88( 8)810()811( 12
23
3
sjsjsksksks
-6401 k 94.752 k 5.6253 k
By comparing coefficients on the both sides of 8.5), we obtain
From (8.4), we get
(8.5)
5.62594.75640-K
Spec. for Step Response:
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0 0.5 1 1.5 20
0.2
0.4
0.6
0.8
1
1.2
Simulation Results
Fig. Step response of above example
)(ty
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IaAaAaAAA ccn
cnn
n
iic 12
1
1
)(
100 K 112 BABAABB n )(Ac
n ,, :poles Desired 1
The Matrix Polynomial
BuAxx 1 nRx RuCxy Ry
)()()( trtKxtu nRK 1
Plant:
State Feedback:
Then the state feedback gain matrix is
121
1
: Poly.Char. Desired ccn
cnn
n
ii asasass
Ackermann Formula for SISO Systems
nR 1
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Steady State Error
)()()()()(
tCxtytBrtAxtx
From (3.6)
)()()( sYsRsE
])(-)[1(lim
)( lim)(lim1-
0
0
BsI-ACssR
ssEtee
s
stss
][1 )()( 1 BAsICsRsE
By Final Value Theorem
Lapalce Transform of the Error Variable
Error Variable )(-)()( tytrte
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CxyxtxBuAxx
00)( ,
LyBuxLCA
xCyLBuxAx
ˆ
ˆˆ
Full-Order Observer
Suppose x is the observer state
xxe ˆ
CxyeLCA
LyBuxLCABuAxxxe
ˆˆ
Estimation error:
Error Dynamics Equation:
Plant:
L: Observer gain
Full-Order Observer Design
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Hence if all the eigenvalues of (A-LC) lie in LHP, then the error system
te as ,0
eLCAe
is asy. stable and
Fig. Full-Order Observer
)(ˆ tx
)(tu)(ty
)(tx)(txC
A
B s
1
C
A
B s
1
+L
)(ˆ ty
)(ˆ tx
)(tey
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By duality between controllable from and obeservable form
Tc
Tc
Tc BCCBAA ooo
nLCAI 21det
Given ,n,ii 1 , Observable )(A,C
There exists a observer matrix, L, such that
Theorem
we have the following theorem.
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The eigenvalues of can be assigned arbitrarily by proper choice of K. Since
)( KCA TT
)( and ),( CKAKCA TTT
have same eigenvalues, if we choose
TKL
then the eigenvalues of (A-LC) can be arbitrarily assigned.
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Cxy
BuAxx
LyxLCBFA
xxLCxBFxAxCyLBuxAx
ˆ
ˆˆˆˆˆ
eLCAxxe
xxe
ˆ
ˆ
BFexBFA
BFeBFxAxxBFAxBuAxx
ˆ
Separation Principle
xFu ˆ
Plant:
State Feedback Law using estimated state:
Observer:
Error Dynamics:
State Equation:
(8.6)
(8.7)
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e
xA
e
x
LCA
BFBFA
e
x ~0
xI
C
y
yy
2
1
LCAIBFAI
LCAI
BFBFAIAI
0
~
Equation (8.9) tells us that the eigenvalues of the observer-based state feedback system is consisted of eigenvalues of (A-BF) and (A-LC).Hence, the design of state feedback and observer gain can be done independently.
From (8.6) and (8.7), we obtain the overall state equation
Eigenvalues of the overall state equation (7.17)
(8.8)
(8.9)
Separation Principle (Cont.)
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Observer-Based Control SystemObserver-Based Control System
)()(
)()()(
tCxty
tButAxtx
Plant:
BuLCxxLCABuxCyLxAx
ˆ)()ˆ(Observer :
State Feedback Law: )()(ˆ)( trtxKtu
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Fig. Observer-based control system
)(ˆ tx
)(tu)(ty
)(tx)(txC
A
B s
1
C
A
B s
1
L
)(ˆ ty
)(ˆ tx
)(te
)(tr
K
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)(ˆ tx
)(tu)(ty
)(tx)(txC
A
B s
1
C
A
B s
1
L
)(ˆ ty
)(ˆ tx
)(te
)(tr
K
ck
Fig. Observer-based control system with compensating gain
Modern Control Systems 26
)()(
)()()(
tCxty
tButAxtx
Consider the n-dimensional dynamical equation
nqpnnn RCRBRA ,,
(8.10a)
(8.10b)
Here we assume that C has full rank, that is, rank C =q. Then, there exists a coordinate transformation Pxx
xIxCQxCPy
PBuxPAPx
q 01
1
which can be partitioned as
1
2
1
2
1
2221
1211
2
1
0 xxIy
uB
B
x
x
AA
AA
x
x
q
Reduced-Order Observer Design
(8.11)
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Since 1xy , we have
uBxAxAx
uBxAyAy
22221212
121211
which become
uByAyw
uByAu
111
221
212
2222
xAw
uxAx
Observer
(8.12a)
(8.13a)
(8.12b)
(8.13b)
)()(
)()()(
tCxty
tButAxtx
BuLyxLCA
BuxCyLxAx
ˆ)(
)ˆ(Observer:
Plant:
uwLxALAx 212222 ˆ)(
where
Modern Control Systems 28
Note that and w are function of known signals u and y. Now if the dynamical equation above is observable, an estimator of can be constructed.
u2x
Theorem:The pair {A, C} in (8.10) or, equivalently, the pair in (8.12) is observable if and only if the pair in (8.13) is observable.
} ,{ CA} ,{ 1222 AA
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uwLxALAx 212222 ˆ)(
Such that the eigenvalues of can be arbitrarily assigned by a proper choices of . The substitution of w and into (8.143)
yields L u
1222 ALA
)()(
ˆ)(ˆ
221111
212222
uByAuByAyL
xALAx
To eliminate the term of the derivative of y, by defining
yLxz 2ˆ
(8.14)
(8.15)
(8.16)
Let the estimate of be2x
Modern Control Systems 30
Using (8.15), then the derivative of (8.16) becomes
yALALALA
uBLBzALA
uBLByALA
yLzALAz
)()(
)()(
)()(
)()(
11211222
121222
121121
1222
From (8.15), we see that
yLzx 2ˆ
is an estimate of .2xDefine the following matrices
yDzCL
y
zx
xx ˆˆ
y
0
ˆˆ
2
1
)()(ˆ
,)(ˆ ,)(ˆ
11211222
121222
ALALALAJ
BLBBALAA
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yJuBzAz ˆˆˆ
yDzCx ˆˆˆ
Reduced-Order Observer:
where
)()(ˆ
,)(ˆ ,)(ˆ
11211222
121222
ALALALAJ
BLBBALAA
yLzx 2ˆ,ˆˆy
0
ˆˆ
2
1 yDzCL
y
zx
xx
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Fig. Reduced-Order Observer
)(tz
)(tu)(ty
)(tx)(txC
A
B s
1
s
1 + )(ˆ tx)(tz
A
B C
DJ
+
Modern Control Systems 33
)(
ˆ
2
22
yLzx
xxe
then we have
eALA
xLzxALA
uBLxALxALuBLBxALA
xLzALAuBxAxA
xLzxyLzxe
)(
))((
)()(
))((
)()(
1222
121222
12121111211121
112222222121
122
Define Error Variable
Modern Control Systems 34
Since the eigenvalues of can be arbitrarily assigned, the r
ate of e(t) approaching zero or, equivalently, the rate of )( 1222 ALA
yLz
approaching can be determined by the designer. Now we combine with to form
2x1x yLzx 2ˆ
yLz
y
x
xx
2
1
ˆ
ˆˆ
xQPxx 1
We get
z
y
IL
IQQ
yLz
yQQxQx
qn
q 0
ˆ
21
21
Then from
Modern Control Systems 35
)()(
)()()(
tCxty
tButAxtx
Consider the n-dimensional dynamical equation
nqpnnn RCRBRA ,,
Here we assume that C has full rank, that is, rank C =q. Define
R
CP
where R is any (n-q)n real constant matrix so that P is nonsingular.
(8.17a)
(8.17b)
How to transform state equation to the form of (8.11)
Modern Control Systems 36
Compute the inverse of P as
211 QQPQ
where Q1 and Q2 are nq and n(n-q) matrices. Hence, we have
qn
q
n
I
I
RQRQ
CQCQ
QQR
CPQI
0
0
21
21
21
Modern Control Systems 37
Now we transform (8.17) into (8.11),
by the equivalence transformation Pxx
xIxCQxCPy
PBuxPAPx
q 01
1
which can be partitioned as
1
2
1
2
1
2221
1211
2
1
0 xxIy
uB
B
x
x
AA
AA
x
x
q
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BuAxx
z
xCy
ruB
z
x
C
A
z
x n
0
1
0
00
0 1
Cxy
Cxryrz
rCxz
nnnn RCRBRA 11 ,,
CxyBuAxx
RyRx n
Ru
SISO State Space System
Rz
Integral Control:
Augmented Plant:
Modern Control Systems 39
zKKxu e
z
xKK e
z
xCy
rz
x
C
BKBKA
z
x ne
0
1
0
01
Closed-Loop System:
State Feedback Control Design with Integrator
Modern Control Systems 40
)(tu )(ty)(tx)(txC
A
B s
1)(te)(tr
K
s
1
)(tz
Block diagram of the integral control system
Fig.Block diagram of the integral control system
eK
Modern Control Systems 41
xy
uxx
011
0
53
10
Example
0.598,nPercent Overshoot 10%, Settling time 0.5 sec.Spec. for Step Response:
State Feedback Design: )()()( trtKxtu 21 kkK
3.11816 :l PolynomiaChar. Desired
10.918 :poles Desired2
ss
j
Modern Control Systems 42
BrxBKAx rx
1
0
1683.11
10
xCxy 01
BBKAC 11
0.995
1
0
16183.1
10011
1
])(-)[1(lim
)( lim)(lim1-
0
0
BsI-ACssR
ssEtee
s
stss
)1
)( (s
sR
From the steady state analysis in Sec. 3.4
Modern Control Systems 43
State Feedback Design with Error Integrator:
z
x
x
y
r
z
x
x
Kkk
r
z
x
xKkk
z
x
x
e
e
2
1
2
1
21
2
1
212
1
001
1
0
0
001
)(5)(3
010
1
0
0
0011
0
1
0
53
10
(8.18)
Closed-Loop System:
100)3.1)(1816( :l PolynomiaChar. Desired100 10.91,8 :poles Desired
2
sssj
Modern Control Systems 44
1780.11 k 1112 k 18310eK
By comparing coefficients on left hand sides of (8.19) and (8.20), we obtain
From (8.18), we get the char. eq. of the closed-loop system is
(8.19)
111 1780.1K
0)3()5( 12
23 eKsksks
0)183.116)(100( 2 sss (8.20)The desired char. eq. of the closed-loop system is
18310eK
Modern Control Systems 45
xy
r
z
x
x
z
x
x
00011
0
0
001
831011161783.1
010
2
1
2
1
Closed-Loop System:
Final Value Theorem
)(
)(
183101783.1116
18310)(
23 sR
sY
ssssT
1183101783.1116
18310lim
1)(lim)(lim
2300
ssssssTty
sst
0sse
Steady State Error