Chapter 9 Controller Design u=-Kx - Computer Action...
Transcript of Chapter 9 Controller Design u=-Kx - Computer Action...
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Chapter 9
Controller Design Two Independent Steps:
1) Feedback Design – Control Law u=-Kx – assumes all states are accessible (a lot of sensors are necessary)
2) Design of Estimator – (also called an Observer) which estimates the entire state vector given the outputs and inputs
Control Law Design Assumed system for control law design u=-Kx
for an nth order system there are n feedback gains K1,…,Kn. By choice of K the roots can generally be placed anywhere
PLANT
BuAxx
x C y
-
+
ˆ x
r
ESTIMATOR
CONTROL LAW -K
matrix of constants
state vector estimate
COMPENSATION
BuAxx
C y
u=-Kx
x
-
2
xBKAx
Kxu
BuAxx
)(
characteristic equation is
I (ABK) 0
Placing Roots
Example: Undamped oscillator with freq. 0
y(s)
u(s)
1
s2 02
state space description
ux
x
x
x
1
0
0
10
2
1
2
02
1
let's place the roots both at -20
we want to double the natural frequency and increase
damping from =0 to =1
desired characteristic equation is
212
0
00
22
0
1
0
0
10
0
0det)(det
44)2()(
KKs
sBKAsI
ssssd
or
s2 K2s02 K1 0
equating same coefficients:
K2 40
02 K1 40
2
K1 302
K2 40 02021 43 KKK
n
jn 12
jn
1 2
-20 -j0
j0
j
cos
REVIEW
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Use of Canonical Forms A simple way of calculating the gains when order is greater than three is to use special
“canonical” forms of the state equations. The special structure of the system matrix is referred to as companion form. Example: Third order case: The characteristic equation is
s3 a1s2 a2s a3
Recall the phase variable form (a lower companion form)
G(s) b1s
n1 .... bn1s bn
sn a1sn1 .... an
11 ....
1
:
0...100
0...010
aaa
A
nn
1
0
:
0
B 11 ... bbbC nn 0D
The closed loop system matrix is
ABK Third order case:
312213
000
000
321
123
100
010
1
0
0
100
010
321
KaKaKa
KKK
aaa
BKA
KKK
Characteristic equation
sI (ABK) 0
s3 (a1 K3)s2 (a2 K2)s (a3 K1) 0 (1)
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if the desired pole locations result in the characteristic equation
d (s) s3 1s
2 2s3 0 (2) equating like coefficients of (1) and (2)
K3 a1 1
K2 a2 2
K1 a3 3
Example: Drill problem D9.1 page 635 of text
u
x
x
x
x
x
x
1
0
0
763
100
010
3
2
1
3
2
1
rx
x
x
kkku
3
2
1
321
1
2
3
2 0 1
x
y x
x
Find ki to place the closed-loop system poles at s= -3, -4, -5 ANS desired characteristic polynomial
d (s) (s 3)(s 4)(s 5)
s3 12s2 47s 60
1 12, 2 47, 3 60 we have
k3 a1 1
7 12 5
k2 a2 2
6 47 41
k1 a3 3
3 60 57
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Design Procedure
Given (A,B) and desired d(s), transform to (Ac, Bc) and solve for gains We then need to transform gain back to original state space note: the poles can only be placed arbitrarily if the system is fully controllable This procedure is encapsulated in Ackermann’s formula Ackermann’s Formula
)(10...0 1 AMk dC
where
BABAABBM nC 12 ... (controllability matrix) where n is the order of the system or the number of states and
d (A) is defined as
IAAAA nnnn
d ...)( 22
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where the si ' are the coefficients of the desired characteristic polynomial
d (s) sn 1s
n1 ...n Example Apply the formula for the undamped oscillator
2
00
2
2
0
44
)2()(
ss
ssd
01 4 , 2
02 4
recall
0
102
0A
1
0B
2
0
3
0
0
2
0
2
02
0
02
0
2
0
34
43
10
014
0
104
0
0)(
Ad
control canonical (companion) form i.e. phase variable form
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also
01
10ABBM c
01
101cM
2
0
3
0
0
2
021
34
43
01
1010
KKK
020 43 K which is the same as the result previously obtained.
Tracking Problems
For step input: Will find
N to ensure zero steady-state error to step inputs
rNKxu
rNBxBKA
BuAxx
)(
now
rNBBKACy
rNBBKAx
rNBxBKAx
ss
ss
ssss
1
1
)(
)(
)(00
A-BK is stable inverse exists now
rNBBKAC
rNBBKACryre ssss1
1
)(1
)(
to get zero steady state error
N 1
C(A BK)1B
steady-state error
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Integral Control (used to get zero steady state error) Integrator increases the system order by one, i.e. augment the plant model by an added state variable xi
Cxrx
dtCxrdtyredtx
i
i
)()(
Augmented systems becomes
ruB
x
x
C
A
x
x
ii
1
0
00
0
Control law is
i
iiix
xKKxKKxu
The design now proceeds as before. Example
Double Integrator
G(s) 1
s2
xy
uxx
01
1
0
00
10
Augment the plant
zero matrices (compatible dimensions)
integrator in the forward path
R(s) E(s)
1s
U(s) X(s) + +
+ -
Ki +
+ B
1s C
Y(s) Xi(s)
A
-K
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xCy
x
xCy
Cxy
i
0
i
ii
x
x
y
u
x
x
x
x
001
0
1
0
001
000
010
Select poles of the closed loop system to be at
1 j,5 N.B. 3 poles because of extra state
10712 K
Steady state output to a unit step input can be derived as follows
rCxx
BuAxx
i
ruB
x
x
C
A
x
x
dt
d
ru BB
ii
1
0
00
0
i
ix
xKKu
rBxKBAx
rBxKBxAx
xKu
rBuBxAx
ru
ru
ru
)(
in steady state 0ssx
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For the example
011
0
00
10
CBA
0011
0
0
0
1
0
001
000
010
CBBA ru
r
r
ry ss
1
1
0
0
001
10712
010
001
1
0
0
10712
0
1
0
001
000
010
001
1
1
det 112 10
1 0
10
adj(A)13 1 0
7 10
10
only term of interest
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Observer Design
We will estimate states rather than measure them Observer simulates the original system Original System
Cxy
BuAxx
x(0) x0
Observer
)ˆ(ˆˆ xCyLBuxAx 0ˆ)0(ˆ xx
(A,B)
x(t) C y(t)
-
+
ˆ x (t)
(A,B)
C
ˆ y (t)
L
u(t)
+
+ + -
+
+ B C
A
B
L
A
C
1s
1s
U(s)
Y(s)
ˆ X (s)
PLANT
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Error between states and their estimates
˜ x x ˆ x
0~)0(~~)(
)ˆ(ˆ
ˆ~
xxxLCA
xCyLBuxABuAx
xxx
Observer error will go to zero asymptotically A-LC is stable (i.e. eigenvalues are in the LHP)
note: the eigenvalues of A-LC are the observer poles, which can be placed arbitrarily if the system is observable
this can be done by choice of the observer gains L (a column vector for single output
systems)
Definition: 1) a system is detectable if the unstable modes are observable
2) a system is stablizable if the unstable modes are controllable Duality
Control Estimation Control Estimation
A AT MC Mo
T B CT K LT C BT
Example
MC B AB .... An1B
duality
11
1
1
::
)(....)(
....
n
o
T
n
TnTT
TnTTTTTo
CA
CA
C
M
CA
CA
C
CACAC
CACACM
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Ackermann’s formula to find observer gains
1
:
0
0
)( 1oe MAL
DERIVATION USING DUALITY Ackermann’s Formula for control problem
)(1.....0 1 AMK CC
Duality
T
oe
T
e
T
o
T
MA
AML
1
:
0
)(
)(1.....0
1
1
AT BTCT (CBA)T
1
:
0
)(1
oe MAL
Example Design an observer for
xy
uxx
ssG
01
1
0
00
10
1)(
2
Note the system is completely observable
Design the observer with poles at
2 j2 Actual characteristic equation:
21
2
2
101
00
10
10
01)(
lsls
l
lsLCAsI
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Desired characteristic equation:
(s 2 j2)(s 2 j2) s2 4s 8
Equating coefficients
l1 4
l2 8
8
4L
The observer equations are
uxyx
xyxx
)ˆ(8ˆ
)ˆ(4ˆˆ
12
121
STRUCTURE OF OBSERVER
+ - +
+ 4
1s
1s
u y
ˆ x 1
PLANT
1s
+
+ 8
1s
ˆ x 2
2x 12 xx
x1
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Control Using Observers
Plant: Cxy
xxBuAxx
0)0(
Observer: 0ˆ)0(ˆ)ˆ(ˆˆ xxxCyLBuxAx
Estimated state feedback:
uKˆ x closing the loop
x
x
LCBKALC
BKA
x
x
LCxxBKLCA
LyxBKxLCAx
xBKAxx
ˆˆ
ˆ)(
ˆˆ)(ˆ
ˆ
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Separation Principle Introduce transformation
1
0
ˆ
0
ˆ ˆ
N N
N N
N N
N N
Ix zP where P
I Ix w
note P P
Iz x x x
I Iw x x x x
Therefore using this transformation the old augmented state vector comprising
x (the plant states) and
ˆ x (the estimator sates) now becomes
x and
˜ x (the estimator error). The new
system matrix
˜ A P1AP
x
x
LCA
BKBKA
x
x~0~
note
˜ A is block-triangular Eigenvalues of a block-triangular matrix are equal to the eigenvalues of the diagonal blocks. So
the eigenvalues of the full system comprise the eigenvalues of the plant (i.e. eigenvalues of A-BK) and the observer (i.e. eigenvalues of A-LC).
Alternative Approach
xLCAx
xLCAxLCA
xLCBKALCxxBKAxxx
xLCBKALCxx
xBKAxx
~)(~
ˆ)()(
ˆ)(ˆˆ
ˆ)(ˆ
ˆ
xBKAxx ˆ
now
˜ x x ˆ x ˆ x x ˜ x
xBKxBKA
xxBKAxx~)(
)~(
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Compensator Transfer Function
H(s) U(s)
Y(s)
we have
)()()(
)()()(ˆ
ˆ)(
ˆˆˆ)ˆ(ˆˆ
ˆ
)(
1
1
sYLLCBKAsIKsU
sLYLCBKAsIsX
LyxLCBKA
xLCLyxBKxAxCyLBuxAx
xKu
sH
Design Issues
1) Problem with pole placement is that there is no control over compenstor poles and zeros 2) Optimum choice for observer initial conditions is
3) Choice of observer poles:
i. Choose them to be faster than controller poles ii. Alternatively, choose them to be at plant zeros (if the sytem has RHP zeros,
use their LHP images).
Compensator output, which is the plant input
Compensator input, which is the plant output
u y PLANT
OBSERVER -K
ˆ x
Compensator Input
COMPENSATOR
Compensator Output
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Reduced-Order Observer Design If system has n states and m measurements, then an observer of order (n-m) will be sufficient. If
y Cx we will assume C has the structure:
( )
11 12 1
21 22 2
22 21 2
0 , 0
0
( )
mxm mx n m
m
m
u
m m
u u
u u m
known input
C I I R R
xy I x
x
x xA A Bu
x xA A B
x A x A x B u
also u
tmeasuremenknown
um
xAuByAy
uBxAyAyx
12111
11211
Summarizing:
)1(
)1(
)1()(
12111
22122
bxAuByAy
auBxAxAx
u
tmeasuremenknown
inputknown
muu
Recall for full-order observer:
Plant: )2()0( 0
Cxy
xxBuAxx
Observer: )3(ˆ)0(ˆ)ˆ(ˆˆ 0xxxCyLBuxAx comparing (1) and (2) shows the correspondence
measured states
unmeasured states
dynamics of the unmeasured state
output relationship
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22
21 2
11 1
12
(4)
u
m
x x
A A
Bu A x B u
y y A y B u
C A
substituting (4) into (3) we get the reduced order equation
12
12
22 21 2 11 1 12ˆ ˆ ˆ( ) (5)
u
u
A x
u u m u
A x
x A x A x B u L y A y B u A x
let us now define the estimator error
˜ x u xu ˆ x u
therefore subtracting (5) from (1a) and using (1b) (i.e. uxAuByAy 12111 ) we get
)6(~)(~ 1222 uu xLAAx Design proceeds by, given an A22 and A12 we choose an L to place estimator poles. Rewriting (5) we have
yLuLBByLAAxLAAx uu )()(ˆ)(ˆ 1211211222 (7)
The presence of the derivative of the measurement (i.e. y ) is not good since this amplifies the noise. To get around this we introduce a new state z where
Lyxz u ˆ (8)
ˆ x u z Ly (9)
substituting (9) into (7) leads to the final form of the reduced-order observer
Lyzx
GuFyDzz
u
ˆ
where
D A22 LA12
F DL A21 LA11
G B2 LB1
the error dynamics are given by this
equation
z is the state of the estimator
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The block diagram of the reduced order observer is shown below
Example
Double integrator
G(s) 1
s2
Lyzx
GuFyDzz
u
ˆ
where
D A22 LA12
F DL A21 LA11
G B2 LB1
Dynamics of reduced order observer uu xLAAx~)(~ 1222
^
1
s
1
s
u
x2
x1 y
1
s2 ^
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Require observer pole at
s2
I A22 LA12 0 where 2
I L 0
L 2
D 2
F 4
G 1
Ackerman’s formula for reduced order observer gains
1
:
0
0
0
)(1
022 MAL e
for example:
2
)1)(1)(2(
1
20)(
2)(
120
22
L
AM
A
ss
e
e
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Reduced-Order Transfer Function
Lyz
u
y
m
u
mxKxK
x
xKKu
ˆ
ˆ2121
now
yLKKzKu
LyzKyKu
GuFyDzz
)(
)(
212
21
also
yLGKGKFzGKD
yLKKzKGFyDzz
)()(
)(
212
212
Transfer function:
U(s)
Y(s)C'(sI A')1B'D'
where
A' DGK2
B' F GK1 GK2L
C' K2
D' (K1 K2L)
When C is not of the form 0I ?
BuQAQzQzBuAQzzQQzx
DuCxy
BuAxx
11
y CQzDu
so find a transformation Q so that CQ is of form 0I
let
Q Q1 Q2
so 1 2 0CQ CQ CQ I
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let
T
CP be nonsingular
arbitrary matrix
if
PQ I
1 21 21 2
1
0
0
CQ CQC IPQ Q Q
TQ TQT I
Q P