Post on 13-Dec-2015
1
Chapter 11
Compensator Design
When the full state is not available for feedback, we utilize an observer. The observer design process is
described and the applicability of Ackermann’s formula is established. The state variable compensator is
obtained by connecting the full-state feedback law to the observer.
We consider optimal control system design and then describe the use of internal model design to achieve prescribed steady-state response to selected input
commands.
2
State Variable Compensator Employing Full-State Feedback in Series with a
Full State ObserveruBAxx
.
yu ˆˆˆ.
LBxAx
C
-K
C
System Model
Observer
xu
+
-
Control Law
y
x̂
Compensator
3
Compensator DesignIntegrated Full-State Feedback and Observer
LCBKx ˆ
LCBKA
ControlGain -K
ObserverGain L+
+
u y
LCBKA
x̂
4
e
x
LC- A 0
BK BKA
e
x
BKeBK)x(Ax
exxxBKAxBAxx
CxBAxx
eLCAe
xLCLyBuxABuAxxxe
xK
LxLCBKAx
xCLBxAx
BKAI
x
xK
ˆ ;ˆ
y ;
)(
error) estimation of derivative (time ˆˆˆ
ˆ
ˆ)(ˆ
lecture) previos thesee observer, the(From )ˆ(ˆˆ
stability retain the welaw controlfeedback theusingthat when
verify toneed Wefeedback. state-full From 0))((Det
) x(of placein controlfeedback in the )(ˆ estimate stateEmploy
LawFeedback )(ˆ)(
.
.
.
.
.
.
...
.
.
u
u
u
y
yu
s
tt
ttu
5
The goal is to verify that, with the value of u(t) we retain the stability of the closed-loop system and the observer. The
characteristic equation associated with the previous equation is
)(ˆ-)( using lawfeedback state-full theoobserver t eConnect th 3.
e.performancobserver achieve toroots theplace and plane halfleft in the
roots has 0))(Det(such that Determine 2.
criteria.design meet the oproperly t roots theplace and plane halfleft in the
roots has 0))Det(such that Determine 1.
stable. is system overall then the
observer, theofdesign the tosecond and lawfeedback state-full
torelated(first equation above theof partsboth of roots theif So
))(Det( ))Det(
ttu xK
LCAsIL
BK(AsIK
Procedure Design
LCAsIBK(AsIΔ(s)
6
The Performance of Feedback Control Systems
• Because control systems are dynamics, their performance is usually specified in terms of both the transient response which is the response that disappears with time and the steady-state response which exists a long time following any input signal initiation.
• Any physical system inherently suffers steady-state error in response to certain types of inputs. A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to a ramp input.
• Whether a given system will exhibit steady-state error for a given type of input depends on the type of open-loop transfer function of the system.
7
The System Performance• Modern control theory assumes that the systems engineer can
specify quantitatively the required system performance. Then the performance index can be calculated or measured and used to evaluate the system’s performance. A quantitative measure of the performance of a system is necessary for the operation of modern adaptive control systems and the design of optimum systems.
• Whether the aim is to improve the design of a system or to design a control system, a performance index must be chosen and measured.
• A performance index is a quantitative measure of the performance of a system and is chosen so that emphasis is given to the important system specifications.
8
Optimal Control Systems
• A system is considered an optimum control system when the system parameters are adjusted so that the index reaches an extreme value, commonly a minimum value.
• A performance index, to be useful, must be a number that is always positive or zero. Then the best system is defined as the system that minimizes this index.
• A suitable performance index is the integral of the square of the error, ISE. The time T is chosen so that the integral approaches a steady-state value. You may choose T as the settling time Ts
T
dtte0
2 )( ISE
9
The Performance of a Control System in Terms of State Variables
• The performance of a control system may be represented by integral performance measures [Section 5.9]. The design of a system must be based on minimizing a performance index such as the integral of the squared error (ISE). Systems that are adjusted to provide a minimum performance index are called optimal control systems.
• The performance of a control system, written in terms of the state variables of a system, can be expressed in general as
• Where x equals the state vector, u equals the control vector, and tf equals the final time.
• We are interested in minimizing the error of the system; therefore when the desired state vector is represented as xd = 0, we are able to consider the error as identically equal to the value of the state vector. That is, we desire the system to be at equilibrium, x = xd= 0, and any deviation from equilibrium is considered an error.
ft
dttgJ0
),ux,(
10
Design of Optimal Systems using State Variable Feedback and Error-Squared Performance Indices: Consider the following control system in terms of x and u
ControlSystem
u1
u2
um
Control signals State variablesx1
x2
x3
elements. above theofaddition thefrom resultingmatrix theis ;
Kxu that sofunction linear a ofunction tfeedbak Limit the ....
..
.....................
..
...
;.. ;
as vector control thechoosemay We]. ; ;[
thereforeand x variablestate measured theoffunction some is that so controllerfeedback aSelect
:equation aldifferenti vector by the drepresente becan system The
n
2
1
mnm1
1n11
m
2
1
32222111
222111
.
nn
x
x
x
kk
kk
u
u
u
xxkuxxku
xkuxkuxku mmm
HHxBKx-Axx
k(x)u
u
BuAxx
.
.
11
Now Return to the Error-Squared Performance Index
] :used be willP symmetric[A xPxPxxx-xPxx ; xx
matrix x theof transpose theis X ; ... ...
,...,,,xx
operationmatrix use
squared, variablesstate theof sum theof integralan of in termsindex ePerformanc
operationmatrix the Utilize]. variablesstate [Two
variable]state [Single )(
.T
T.TT
0
T
T232
22
21
2
1
321T
0
22
21
2
01
pjipijdt
ddtJ
xxxx
xn
x
x
xxxx
dtxxJ
dttxJ
f
f
f
t
nn
t
t
12
To minimize the performance index J, we consider two equationsThe design steps are: first to determine the matrix P that satisfies the second
equation when H is known. Second minimize J by determining the minimum of the first equation by adjusting one or more unspecified system parameters.
1Equation of minimum thegdetermininby of value
theminimize and 2)(Equation Pmatrix theDetermine :CriteriaDesign The
2)(Equation
1)(Equation (0) (0)x
(0)Px(0)xPx)x(-
x-xPH)xP(Hx
(Hx)xPx(Hx)xPxPxxPxx
0
TT
TT
0
TTT
TT.
TT.
T
J
dtJ
dtdx
dJ
dt
d
-IPHPH
Pxxx
T
13
State Variable Feedback: State Variables x1 and x2
u1 1/s 1/s
11
x1x2
x1(0)/sx2(0)/s
1- 0
0 1
- 1-
1 0
- 1
1- 0
minimized isindex eperformanc that theso find and 1Let
. - -
1 0 have weformmatrix In
feedback negative provide tonegativesign ;)( that so system controlfeedback
1
0 ;
0 0
1 0
)( 1
0
0 0
1 0
222 12
1211
22 12
1211
2
21
21
22112.
21.
2211
2
1
2
1
kpp
pp
pp
pp
k
kk
kk
xkxkxxx
xkxktu
tux
x
x
x
dx
d
-IPHPH
xHx x
Choose
BA
T
.
14
121 1
1- DetDet ;
2- 1-
1 0 3;
minimum is when 2 ;2
)42(2)22(2
2
4211
2
2
21
1
1 1
mequilibriu fromunit one displaced is stateeach Assume 1 1(0)x
index) eperformanc integral (The (0)(0)
2
2 ;
1 ;
2
1
1
0
1
2min
222
22222
2
2
222
22
22
2212112212
1211
T
T
2
22
112
2212
2221222212
2212211
1212
H-λIH
Pxx
J
Jkk
kkkk
dk
dj
k
kk
kk
kJ
ppppp
ppJ
J
k
kp
kpp
pkppkp
ppkp
pp
15
107.02/5.0
3/08.0
/
/, Assume;
/
/
is system optimalan ofy sensitivit The
system) (optimum 1.0 y Accordingl .2
form theof isequation sticcharacteri The system.order second a is This
222
22
2
kk
JJSkk
kk
JJS
nss
optk
optk
n
R(s)=0 Y(s)U(s)x2 x1
k2
k1
--
1/s 1/s
16
Compensated Control System
1 1/s 1/s
11
x1x2
u
x1(0)/sx2(0)/s
-k1= -1
-k2 = -2
17
Design Example: Automatic Test SystemA automatic test system uses a DC motor to move a set of test probes as shown in Figure 11.23 in the textbook. The system uses a DC motor with an encoded disk to
measure position and velocity. The parameters of the system are shown with K representing the required power amplifier.
5
1
s 1
1
s
2.5)(Section )/)(/(
)(ff LRsJbss
KsG
u
K 1/s
x1
x2x3
Amplifier Field Motor
x1 = ; x2 = d / dt; x3 = If
State variables
Vf
5/;1/ ff LRJb
33
2323
1)(;
)5)(1()(
Ks
K
KKsKsH
sss
KsG
18
State Variable FeedbackThe goal is to select the gains so that the response to a step command
has a settling time (with a 2%criterion) of less than 2 seconds andan overshoot of less than 4%
5
1
s 1
1
s
rK
1/s
x1
x2x3
Amplifier Field Motor
k3
k2
k1
-
+
rxkxkxkrxkkku 332211321
19
00.369.3;62.10 are roots The
240;76.0;12When
05.0;35.0;1
120 ;8
plane.- in theleft the tolocus thepull order toin
j2-4sat zerosget will we20 and 8set weIf
1 ;
0)5)(1(
)(1)()(1
3
321
33
32
33
32
23
jss
kkk
kkk
kk
kk
s
ba
kb
k
kka
sss
basskksHsG
20
To achieve an accurate output position, we let k1 = 1 and determine k, k2 and k3. The aim is to find the characteristic equation
056
)(5s -
1- 1s 0
0 1-
Det
0
0
)(5- - -
1 1- 0
0 1 0
0 0 1
0
0
5- 0 0
1 1- 0
0 1 0
232
323
32
32
.
.
kskkkskksksss
kkkkk
s
r
kkkkkk
y
u
k
xBuAxx
x
xBuAxx
21
3.003.69- and -10.62
240at roots The
240 ;05.0 ;35.0 ;1
201
8
24at zeros place We
20 ;8 ;1Let
1 ; ;0
)5)(1(
)(1
321
33
32
1
33
322
3
jss
k
kkkk
kk
kk
js
bak
kb
k
kka
sss
basskk
22
P11.1:
1/30 is index eperformanc theof value then the31, Select
realizablenot is this,Physically minimum. is them, If
1
1)0(
yields 1 assuming of value theCalculate
1. if stable is System )0()(
1
0
2)1(2
)1(
.
.
Jk
Jk
kdtxeJ
kJ
kxetx
xkkxxx
kxu
uxx
tk
tk
23
P11.3: An unstable robot system is described below by the vector differential equation. Design gain k so that the performance index is minimized.
unstable. is system The 2; and -19 where
03817A]-[SIDet by determined are roots system loop-closed The
Ax x8- 11-
10- 9-x ;
19
1 10,Select
feedback without unstable is system The 0. togoes , togoes As .1] [1When x(0)
12
12 (0) (0):be tocomputed isindex eperformanc The
)384(4
)362( ;
)384(4
122 ;
)384(4
762
1)2(22;0)1()23(;1)1(2)1(2
)-(2 )(1-
- )-(1
is system theapplied,feedback with Then,
1]x 1[)( );( 1
1
2 1-
0 1
21
2
.
T
221211T
2
2
222
2
122
2
11
22122211121211
212
1
2
1
ss
ss
Jk
Jk
kpppJ
kk
kkp
kk
kkp
kk
kkp
kpkpkpkpkpkpkp
kk
kk
kxxktutux
x
x
x
dt
d
Pxx
-IPHPH
xHxx
T
.