Transcript 投影片 1 - nuu.edu.tw

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
Modern Control Systems
1
State Feedback and Stabilization
Stabilization by State Feedback: Regulator Case
Plant: x  Ax  Bu,
x(t0 )  x0
State Feedback Law:
u   Fx
Closed-Loop System:
x  Ax  Bu  Ax  BFx
  A  BF x
Theorem
Given
i , i  1,,n
( A,B)
Controllable

There exists a state feedback matrix, F, such that
detI   A  BF     1   2   n 
Modern Control Systems
2
D
u
B
 x


x
C
 
y
A
F
State Feedback System (Regulator Case)
Modern Control Systems
3
State Feedback Design in Controllable Form
 0
 
Ac  
 0

  a1
I n 1
 a2
u  Fx   f1






 an 
0
0
Bc   

 
1 
 f 2   f n x
0
1


0
0

Ac  Bc F 




 a1  f1   a 2  f 2 



0





  a n  f n 

0
detsI  Ac  Bc F   s n  an  f n s n1    a1  f1 
Modern Control Systems
(8.1)
4
Suppose the desired characteristic polynomial
n
n
n 1


s



s

a
s
   ac1

i
cn
(8.2)
i 1
Comparing (8.1) and (8.2), we have
f n  a cn  a n
f n 1  a cn1  a n 1

f 1  a c1  a1
F  ac1  a1
ac 2  a2  acn  an 
Modern Control Systems
(8.3)
5
State Feedback: General Case (Non-Zero Input Case)
F   f1
u   Fx  r
f2  fn 
D
r


u
B


x

x
C
 
y
A
F
State Feedback Control System
Modern Control Systems
6
State Feedback Design with Transformation to Controllable1Form
x  Ax  Bu
y  Cx
Desired poles: 1 ,, n
x  Tz
Controllable From:
i 1
z  Ac z  Bc u
 0
 
Ac  
 0

  a1





 an 
I n 1
 a2

Bc  0 0  1
T
y  Cc z
u  Fc z
Bc  T 1B
Cc  CT
n
Desired Char. Poly.:  s  i 
Ac  T AT
Fc  ac1  a1 ac 2  a2  acn  an 
T
u  Fc z  FcT -1x
n
det sI   A  BF    s  i   s n  acn s n 1    ac 2 s  ac1
i 1
Modern Control Systems
7
Transform to Controllable Form
 ln T 
 T 
l A 
1
T  n
  
 T n 1 
ln A 
 l1T 
 
where L      V 1
 ln T 
 
Coordinate Transform Matrix
1
z T x
x  Tz
xR


U  B AB  An-1B
rank(U )  n, ( A, B) is controllable
v2  vn 

Cc  Cc1
n
y  Cc x
Ac  T 1 AT
x  Rn
0
I n 1n1 

Ac  

 a1  a2    an 
Controllable Form:
x  Ac x  Bcu
T  v1
x  Ax  Bu
y  Cx
Ao  Ac
T
Cc2
 Ccn
Bo  Cc
T

0 
0 
Bc   

 
1 
Co  Bc
T
Bc  T 1B
Cc  T 1C
Modern Control Systems
8
Example
1  0
0
x  
x   u

 2  2 1
Desired poles:  3  3 j
2
Desired Char. Polynomia
l : s  3  32  s 2  6s  18
(A, B) is in controllable from, we can derive the state feedback
gain from eq. (8.3)
F  18  2 6  2  16 4
 16
4
U (s )
1
u   Fx  16 4x
s -1
x2 s -1
2
Modern Control Systems
x1 1
y
2
9
Obtain the State Feedback Matrix by Comparing Coefficients
Plant:
x  Ax  Bu
y  Cx
State Feedback:
x  R n1
yR
uR
u(t )   Kx(t )  r(t )
K  R1n
Closed Loop System: x  ( A  BK) x  Br
Char. Equation:
sI  A  BK  0
Suppose that the system is controllable, i.e.
rank[ B AB A2 B An1B]  n
Modern Control Systems
10
Then, for any desired pole locations:
1,, n
We can obtain the desired char. polynomial
( s  1 )( s  n )
By controllability, there exists a state feedback matrix K, such that
sI  A  BK  ( s  1 )( s  n )
(8.4)
From (8.4), we can solve for the state feedback gain K.
Modern Control Systems
11
Example
Y ( s)
8

U ( s ) s( s  1)(s  10)
Plant:
State Feedback:
K  k1 k2 k3 
u(t )   Kx(t )
 8k1
Y ( s)
 3
R( s ) s  (11  8k3 ) s 2  (10  8k 2 ) s  8k1
r
1
x1  k1 u 8
s 1 x3
1
s 1 x2
s 11
y
 10
 k3
 k2
1
Fig. State Feedback Design Example
Modern Control Systems
12
Spec. for Step Response: Percent Overshoot 5%, Settling Rise time 5 sec.
n  8,   0.707
Desired pole locations: s1,2  8  j8 (dominantploes)
s3  40
From (8.4), we get
s 3  (11  8k3 ) s 2  (10  8k2 ) s  8k1 
( s  8  j8)( s  8  j8)( s  40 )
(8.5)
By comparing coefficients on the both sides of 8.5), we obtain
k1  -640

k2  94.75
k3  5.625
K  - 640 94.75 5.625
Modern Control Systems
13
Simulation Results
y (t )
1.2
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
Fig. Step response of above example
Modern Control Systems
14
Ackermann Formula for SISO Systems
Plant:
x  Ax  Bu
y  Cx
State Feedback:
x  R n1
yR
uR
u(t )   Kx(t )  r(t )
K  R1n
Desired poles: 1,, n
n
Desired Char. Poly.:  s  i   s n  acn s n1    ac 2 s  ac1
i 1
The Matrix Polynomial
n
c ( A)    A  i   An  acn An1    ac 2 A  ac1I
i 1
Then the state feedback gain matrix is

K  0 0  1 B
2
n 1
AB A B  A B

1
 c ( A)
 R1n
Modern Control Systems
15
Steady State Error
x (t )  Ax(t )  Br(t )
y (t )  Cx(t )
Error Variable
e(t )  r(t ) - y (t )
Lapalce Transform of the Error Variable
E ( s )  R( s )  Y ( s )
From (3.6)
E( s)  R( s) [1  CsI  A B]
1
By Final Value Theorem
ess  lim e(t )  lim sE ( s)
t 
s 0
 lim sR ( s )[1- C ( sI-A)-1 B]
s 0
Modern Control Systems
16
Full-Order Observer Design
Full-Order Observer
Plant:
x  Ax  Bu,
y  Cx
Suppose
xˆ
x(t0 )  x0
is the observer state
xˆ  Axˆ  Bu  L y  Cxˆ 
  A  LC xˆ  Bu  Ly
L: Observer gain
Estimation error:
e  x  xˆ
Error Dynamics Equation:
e  x  xˆ  Ax  Bu   A  LC xˆ  Bu  Ly
  A  LC e
 y  Cx 
Modern Control Systems
17
Hence if all the eigenvalues of (A-LC) lie in LHP, then the error system
e   A  LC e
is asy. stable and
e  0, as t  
u(t )
B
x (t )


1
s
x(t )
y(t )
C
A
ey (t )
L
 xˆ(t )
B

1
s
+

xˆ (t )
C
yˆ (t )
A
Fig. Full-Order Observer
Modern Control Systems
18
By duality between controllable from and obeservable form
Ao  Ac
Bo  Cc
we have the following theorem.
T
T
Co  Bc
T
Theorem
Given
i , i  1,,n
( A,C) Observable

There exists a observer matrix, L, such that
detI   A  LC    1   2   n 
Modern Control Systems
19
The eigenvalues of (  AT  C T K ) can be assigned arbitrarily by
proper choice of K. Since
( AT  C T K ), and ( A  K T C )
have same eigenvalues, if we choose
L  KT
then the eigenvalues of (A-LC) can be arbitrarily assigned.
Modern Control Systems
20
Separation Principle
Plant:
x  Ax  Bu
y  Cx
State Feedback Law using estimated state: u   Fxˆ
State Equation:
Observer:
x  Ax  Bu  Ax  BFxˆ  Ax  BFx  BFe
(8.6)
  A  BF x  BFe
xˆ  Axˆ  Bu  L y  Cxˆ   Axˆ  BFxˆ  LC x  xˆ 
  A  BF  LC xˆ  Ly
Error Dynamics:
e  x  xˆ
e  x  xˆ   A  LC e
Modern Control Systems
(8.7)
21
Separation Principle (Cont.)
From (8.6) and (8.7), we obtain the overall state equation
 x   A  BF
 e    0
  
BF   x

A  LC  e 
~  x
A 
e
(8.8)
Eigenvalues of the overall state equation (7.17)
 BF
I   A  BF 

~
I  A  
0
I   A  LC 

 I   A  BF   I   A  LC 
(8.9)
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.
 y1  C 
y      x
 y2   I 
Modern Control Systems
22
Observer-Based Control System
Plant:
x (t )  Ax(t )  Bu(t )
y (t )  Cx(t )
Observer：
xˆ  Ax  L( y  Cxˆ )  Bu
 ( A  LC ) xˆ  LCx  Bu
State Feedback Law: u(t )   Kxˆ (t )  r(t )
Modern Control Systems
23
r (t )
 u(t )

 x (t )
B

1
s
x(t )
y(t )
C
A
e(t )
L
B


xˆ(t )
1
s


xˆ (t )
C
yˆ (t )
A
K
Fig. Observer-based control system
Modern Control Systems
24
 u(t )
r (t )
kc

 x (t )
B

1
s
x(t )
y(t )
C
A
e(t )
L
B


xˆ(t )
1
s


xˆ (t )
C
yˆ (t )
A
K
Fig. Observer-based control system with compensating gain
Modern Control Systems
25
Reduced-Order Observer Design
Consider the n-dimensional dynamical equation
x (t )  Ax(t )  Bu(t )
(8.10a)
y (t )  Cx(t )
(8.10b)
A  Rnn , B  Rn p , C  Rqn
Here we assume that C has full rank, that is, rank C =q. Then, there
exists a coordinate transformation x  Px
x  PAP1 x  PBu
y  CP1 x  CQx  I q
0x
which can be partitioned as
 x1   A11 A12   x1   B1 
   
  x      u
 x2   A21 A22   2   B2 
y  I q 0x  x1
Modern Control Systems
(8.11)
26
Since
y  x1 , we have
y  A11 y  A12 x2  B1u
x2  A21x1  A22 x2  B2u
(8.12a)
(8.12b)
which become
x2  A22 x2  u
w  A12 x2
where
u  A21 y  B2u
w  y  A11 y  B1u
Observer
xˆ2  ( A22  L A12 ) xˆ2  L w  u
Plant:
x (t )  Ax(t )  Bu(t )
(8.13a)
(8.13b)
y (t )  Cx(t )
Observer:
xˆ  Ax  L( y  Cxˆ )  Bu
 ( A  LC ) xˆ  Ly  Bu
Modern Control Systems
27
Note that u and w are function of known signals u and y. Now if the
dynamical equation above is observable, an estimator of x2 can be
constructed.
Theorem:
The pair {A, C} in (8.10) or, equivalently, the pair { A, C } in
(8.12) is observable if and only if the pair{ A22 , A12} in (8.13) is
observable.
Modern Control Systems
28
Let the estimate of x2 be
xˆ2  ( A22  L A12 ) xˆ2  L w  u
(8.14)
Such that the eigenvalues of A22  L A12 can be arbitrarily assigned
by a proper choices of L . The substitution of w and u into (8.143)
yields
xˆ2  ( A22  L A12 ) xˆ2
 L ( y  A11 y  B1u)  ( A21 y  B2u)
(8.15)
To eliminate the term of the derivative of y, by defining
z  xˆ2  L y
Modern Control Systems
(8.16)
29
Using (8.15), then the derivative of (8.16) becomes
z  ( A22  L A12 )  ( z  L y )
 ( A21  L A11 ) y  ( B2  L B1 )u
 ( A22  L A12 ) z  ( B2  L B1 )u
 ( A22  L A12 ) L  ( A21  L A11 )y
From (8.15), we see that
xˆ2  z  L y
is an estimate of x2 .
Define the following matrices
Aˆ  ( A22  L A12 ), Bˆ  ( B2  L B1 ),
Jˆ  ( A22  L A12 ) L  ( A21  L A11 )
 x1  0  y  ˆ
xˆ           Cz  Dˆ y
 xˆ2   z   L y
Modern Control Systems
30
Reduced-Order Observer:
z  Aˆ z  Bˆ u  Jˆy
xˆ  Cˆ z  Dˆ y
where
 x1  0  y  ˆ
xˆ           Cz  Dˆ y,
 xˆ2   z   L y
xˆ2  z  L y
Aˆ  ( A22  L A12 ), Bˆ  ( B2  L B1 ),
Jˆ  ( A  L A ) L  ( A  L A )
22
12
21
11
Modern Control Systems
31
u(t )
B
 x (t )
1
s

x(t )
y(t )
C
A
Jˆ
Bˆ
 z(t )


1
s
Dˆ
z(t )
Cˆ
+
+
xˆ (t )
Aˆ
Fig. Reduced-Order Observer
Modern Control Systems
32
Define Error Variable
e  x2  xˆ2
 x2  ( z  L y )
then we have
e  x2  ( z  L y )  x2  ( z  L x1 )
 A21x1  A22 x2  B2u  ( A22  L A12 )( z  L x1 )
 ( A21  L A11 ) x1  ( B2  L B1 )u  L A11x1  L A12 x2  L B1u
 ( A22  L A12 )( x2  z  L x1 )
 ( A22  L A12 )e
Modern Control Systems
33
Since the eigenvalues of ( A22  L A12 ) can be arbitrarily assigned, the
rate of e(t) approaching zero or, equivalently, the rate of
z  Ly
approaching x2 can be determined by the designer. Now we
combine x1 with xˆ2  z  L y to form
 xˆ1   y 
xˆ     

ˆ
x
z

L
y

 2 
Then from
We get
x  Px  Q 1 x
 y 
ˆ
x  Qx  Q1 Q2  

z

L
y


0   y
Iq
 Q1 Q2  
 

 L I nq   z 
Modern Control Systems
34
How to transform state equation to the form of (8.11)
Consider the n-dimensional dynamical equation
x (t )  Ax(t )  Bu(t )
(8.17a)
y (t )  Cx(t )
A R
nn
,BR
n p
,CR
qn
(8.17b)
Here we assume that C has full rank, that is, rank C =q. Define
C 
P 
 R
where R is any (n-q)n real constant matrix so that P is nonsingular.
Modern Control Systems
35
Compute the inverse of P as
Q  P 1  Q1 Q2 
where Q1 and Q2 are nq and n(n-q) matrices. Hence, we have
C 
I n  PQ     Q1 Q2 
 R
CQ1 CQ2   I q



 RQ1 RQ 2   0
Modern Control Systems
0 
I n q 
36
Now we transform (8.17) into (8.11),
by the equivalence transformation x
x  PAP1 x  PBu
 Px
y  CP1 x  CQx  I q
0x
which can be partitioned as
 x1   A11 A12   x1   B1 
   
  x      u
 x2   A21 A22   2   B2 
y  I q 0x  x1
Modern Control Systems
37
SISO State Space System
x  Ax  Bu
y  Cx
x  Rn
yR
uR
Integral Control:
z  r  y  r  Cx
Augmented Plant:
A  Rnn , B  Rn1, C  R1n
z R
x  Ax  Bu
z  Cx  r
y  Cx

 x   A 0  x   B  0n1 
 z    C 0  z    0 u   1  r
  
     
 x
y  C 0 
z
Modern Control Systems
38
State Feedback Control Design with Integrator
u   Kx  Ke z   K
 x
K e  
z
Closed-Loop System:
 x   A  BK
 z     C
  
 x
y  C 0 
z
BK e   x  0n1 
  r



0  z   1 
Modern Control Systems
39
Block diagram of the integral control system
r (t )
 e(t )

1
s
z(t )
Ke


u(t )
B
 x (t )

1
s
y(t )
x(t )
C
A
K
Fig.Block diagram of the integral control system
Modern Control Systems
40
Example
1
0
0
x  
x   u

 3  5
1
y  1 0x
Spec. for Step Response: Percent Overshoot 10%, Settling time 0.5 sec.
  0.59
n  8,
Desired poles:  8  j10.91
Desired Char. Polynomia
l : s 2  16s  183.1
State Feedback Design:
u(t )   Kx(t )  r(t )
Modern Control Systems
K  k1 k2 
41
1  0
 0
x   r
x   A  BK x  Br  

 183.1  16 1
y  Cx  1 0x
From the steady state analysis in Sec. 3.4
ess  lim e(t )  lim sE ( s)
t 
s 0
 lim sR ( s )[1- C ( sI-A)-1 B]
s 0
1
(

R
(
s
)

)
 1  C A  BK  B
s
1
0
1

 0
 1  1 0 
  
 183.1  16 1
 0.995
1
Modern Control Systems
42
State Feedback Design with Error Integrator:
Desired poles:  8  j10.91, 100
Desired Char. Polynomia
l : ( s 2  16s  183.1)(s  100)
Closed-Loop System:
 x1    0
1  0
 0   x1  0
  k1 k 2    K e     
 x    

x  0 r
 1   2   
 2     3  5 1
 z  
 1 0
0   z  1
0
1
0   x1  0

  (3  k1 )  (5  k2 ) K e   x2   0 r

   
0
0   z  1
  1
 x1 
y  1 0 0 x2 
 
 z 
Modern Control Systems
(8.18)
43
From (8.18), we get the char. eq. of the closed-loop system is
s3  (5  k2 )s2  (3  k1 )s  Ke  0
(8.19)
The desired char. eq. of the closed-loop system is
( s  100)(s2  16s  183.1)  0
(8.20)
By comparing coefficients on left hand sides of (8.19) and (8.20), we obtain
k1  1780.1

k2  111
K  1780.1 111
Ke  18310
Ke  18310
Modern Control Systems
44
Closed-Loop System:
0
1
0   x1  0
 x1  
 x    1783.1  116 18310  x   0 r
 2 
 2   
0
0   z  1
 z    1
y  1 0 0 0x
T ( s) 
18310
Y ( s)

s 3  116s 2  1783.1s  18310 R( s)
Final Value Theorem
1
18310
lim y (t )  lim sT ( s )   lim 3
1
2
t 
s 0
s

0
s
s  116 s  1783.1 s  18310
Steady State Error
ess  0
Modern Control Systems
45