robust control systems (chapter 12) - university of ottawarhabash/elg4152l8.pdf · robust control...
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1
Robust Control Systems (Chapter 12)Feedback control systems are widely used in manufacturing,
mining, automobile and other hardware applications. In response to increased demands for increased efficiency and reliability, these control systems are being required to delivermore accurate and better overall performance in the face of
difficult and changing operating conditions. In order to design control systems to meet the needs of improved performance and robustness when controlling
complicated processes, control engineers will require new design tools and better control theory. A standard technique of improving the performance of a control system is to add extra sensors and actuators. This necessarily leads to a multi-input
multi-output (MIMO) control system. Accordingly, it is a requirement for any modern feedback control system design
methodology that it be able to handle the case of multiple actuators and sensors.
Robust means durable, hardy, and resilient
2
Why Robust?• When we design a control system, our ultimate goal is to control a
particular system in a real environment.
• When we design the control system we make numerous assumptions about the system and then we describe the system with some sort of mathematical model.
• Using a mathematical model permits us to make predictions about how the system will behave, and we can use any number of simulation tools and analytical techniques to make those predictions.
• Any model incorporates two important problems that are often encountered: a disturbance signal is added to the control input to the plant. That can account for wind gusts in airplanes, changes in ambient temperature in ovens, etc., and noise that is added to the sensor output.
3
A robust control system exhibits the desired performance despitethe presence of significant plant (process) uncertainty
The goal of robust design is to retain assurance of system performance in spite of model inaccuracies and changes. A system is robust when it has acceptable
changes in performance due to model changes or inaccuracies.
D(s) Disturbance
R(s) PrefilterGP(s)
ControllerGC(s)
PlantG(s)
Sensor1
+
-
+
+
++
Y(s)Output
N(s)Noise
4
Why Feedback Control Systems?
• Decrease in the sensitivity of the system to variation in the parameters of the process G(s).
• Ease of control and adjustment of the transient response of the system.
• Improvement in the rejection of the disturbance and noise signals within the system.
• Improvement in the reduction of the steady-state errorof the system
5
Sensitivity of Control Systems to Parameter Variations
• A process, represented by G(s), whatever its nature, is subject to a changing environment, aging, ignorance of the exact values of the process parameters, and the natural factors that affect a control process.
• The sensitivity of a control system to parameter variations is very important. A main advantage of a closed-loop feedback system is its ability to reduce the system’s sensitivity.
• The system sensitivity is defined as the ratio of the percentage change in the system transfer function to the percentage change of the process transfer function.
( ) ( )
( )GHSsHsG
S
GHGG
GHTG
GTS
sGHsGsT
GGTT
sGsGsTsTS
sRsYsT
GT
GT
larger for Less)()(1
1
1/.
11. ;
)(1)()(
//
)(/)()(/)(ty)(sensitivi ;
)()()(
2
+=
++=
∂∂
=+
=
∂∂
=∆∆
==
6
The sensitivity of the feedback system to changes in the feedback element H(s) is
( )
y)sensitivit(Root /
y)Sensitivit (System //
rulechain the Use.block a offunction transfer within the a is where, determine toneed Often we
11/.
1.
2
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αα
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αα
∂∂
=
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==
+−
=+−
+
=∂∂
=
ir
GTG
T
Tα
TH
rS
TTSSS
GαS
GHGH
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TH
HTS
i
parameter
7
Robust Control Systems and System SensitivityA control system is robust when: it has low sensitivities, (2) it is stable
over the range of parameter variations, and (3) the performance continues to meet the specifications in the presence of a set of changes
in the system parameters.
( )
( )( )
( )11 ;
1 isroot The1
1.
parameter theoft independen are )( of zeros/
:isy SensitivitRoot
function) tranfer theis parameter; theis ( // :isy sensitivit System
1
1
++−=−−=−
++=++
−=
+−=
=
=
∑=
αα
ααα
ααα
ααα
ααα
αα
α
α
sSSS
rsrs
SS
sTd
drS
Td
TdTS
i
i
i
rTT
i
n
i
rT
ir
T
1/s+α+
-R(s) Y(s)
8
Let us examine the sensitivity of the following second-order system
( )Kss
sssGH
S
kssKsT
TK
+++
=+
=
++=
2
2
1)(1
1 that(4.12) Eq. from know We
)(
K/s(s+1)+
-R(s) Y(s)
9
Example 12.1: Sensitivity of a Controlled System
1.get to2Then . and 1condition normal heConsider t
)(
)(11
n21n
122
12
122
2
====
++
+=
++=
+=
ξbb
bsbsbsbsT
bsbss
sGGS
C
TG
ωωξ
diagram Bode aon Tlog 20 and Slog 20Plot
Controllerb1+b2s
Plant1/s2
R(s) +
-
Y(s)G(s)GC(s)
controller (PD) derivative-alproportion a is )(sGC
10
Bode PlotFrequency response plots of linear systems are often displayed in the form of logarithmic plots, called Bode plots, where the horizontal axis represents the
frequency on a logarithmic scale (base 10) and the vertical axis represents the amplitude ratio or phase of the frequency response function.
11
Disturbance Signals in a Feedback Control System
• Another important effect of feedback in a control system is the control and partial elimination of the effect of disturbance signal.
• A disturbance signal is an unwanted input signal that affects the system output signal. Electronic amplifiers have inherent noise generated within the integrated circuits or transistors; radar systems are subjected to wind gusts; and many systems generate all kinds of unwanted signals due to nonlinear elements.
• Feedback systems have the beneficial aspects that the effect of distortion, noise, and unwanted disturbances can be effectively reduced.
12
The Steady-State Error of a Unity Feedback Control System (5.7)
• One of the advantages of the feedback system is the reduction of the steady-state error of the system.
• The steady-state error of the closed loop system is usually several orders of magnitude smaller than the error of the open-loop system.
• The system actuating signal, which is a measure of the system error, is denoted as Ea(s).
G(s)
H(s)
Y(s)R(s)Ea(s)
1)( When )()(1
1)(1
)()()()()( =+
=+
−=−= sHsRsGsGH
sGsRsYsRsE
13
Compensator• A feedback control system that provides an optimum performance
without any necessary adjustments is rare. Usually it is important to compromise among the many conflicting and demanding specifications and to adjust the system parameters to provide suitable and acceptable performance when it is not possible to obtain all the desired specifications.
• The alteration or adjustments of a control system in order to provide a suitable performance is called compensation.
• A compensator is an additional component or circuit that is inserted into control system to compensate for a deficient performance.
• The transfer function of a compensator is designated as GC(s) and the compensator may be placed in a suitable location within the structure of the system.
14
Root Locus Method
• The root locus is a powerful tool for designing and analyzing feedback control systems.
• It is possible to use root locus methods for design when two or three parameters vary. This provides us with the opportunity to designfeedback systems with two or three adjustable parameters. For example the PID controller has three adjustable parameters.
• The root locus is the path of the roots of the characteristic equation traced out in the s-plane as a system parameter is changed.
• Read Table 7.2 to understand steps of the root locus procedure.• The design by the root locus method is based on reshaping the root
locus of the system by adding poles and zeros to the system openloop transfer function and forcing the root loci to pass throughdesired closed-loop poles in the s-plane.
15
The root Locus Procedure
( )( )
zero. aat arrival of angle theand pole a from locus theof departure of angle theDetermine :10 Stepaxis. real on thepoint breakway theDetermine :9 Step
axis.imaginary thecrosses locus heat which tpoint theDetermine :8 Step. angle with and at centered asymptotes alonginfinity at zeros the toproceed loci The :7 Step
axis. real horizontal therespect to with lsymmetrica bemust lociroot The :6 Steppoles). ofnumber the to(equal loci separate ofnumber theFind :5 Step
zeros. at the ends and poles at the begins locus Thedegrees. zero is 4- sat pole and zero thefrom angle theand ,180 isorigin at the
p1 angle thebecause 2,- and 0 points ebetween th axis real on the satisfied iscriterion angle The :4 Stepaxis. real on the zeros and poles thelocate we
(Step3) 0gain for the roots of locus thedetermine To .2 isparameter gain tivemultiplica The
04221 :zeros and poles of in terms written is )(function transfer The :2 Step
0141
121
1)(1equation sticcharacteri The :1 Step
AA
2o
φσ
=
∞≤≤
=++
+
=
+
+
+=+
p
KKsssKsGH
ss
sKsGH
16
Example
( )( )321)(
++=
sssG
ControllerGC(s)
PlantG(s)
R(s) Y(s)
-+
z1=-3+j1
-1-2
-z1
-z1
( )( )( )( )( )112
11ˆ
ˆ3)()(1
)()()(
rsrsrszszsK
sGsGsGsG
STC
C
+++++
=+
=
j1
j2
17
Analysis of Robustness
PrefilterGP(s)
ControllerGC(s)
PlantG(s)
Sensor1
+-
+
+
++
Y(s)Output
N(s)Noise
small. )(Better ;1)()( then 1,)( When ;)()(1
)()()(
function transfer loop-closed The .)]()(1[)(
r so )( tosmall bemust )( noiseSensor
).( edisturbanc afor small )(output thekeep and )(input an for )]()()([error trackingSmall
1
sSsTsSsGsGsG
sGsGsT
sGsGsS
ntrtn
tdtytrtytrte
PC
C
C
=+=+
=
+=
⟩⟩
−=
−
:goals System
18
The Design of Robust Control Systems• The design of robust control systems is based on two tasks:
determining the structure of the controller and adjusting the controller’s parameters to give an optimal system performance. This design process is done with complete knowledge of the plant. Thestructure of the controller is chosen such that the system’s response can meet certain performance criteria.
• One possible objective in the design of a control system is that the controlled system’s output should exactly reproduce its input. That is the system’s transfer function should be unity. It means the system should be presentable on a Bode gain versus frequency diagram with a 0-dB gain of infinite bandwidth and zero phase shift. Practically, this is not possible!
• Setting the design of robust system requires us to find a propercompensator, GC(s) such that the closed-loop sensitivity is less than some tolerance value.
19
PID ControllersPID stands for Proportional, Integral, Derivative. One form of controller
widely used in industrial process is called a three term, or PID controller. This controller has a transfer function:
A proportional controller (Kp) will have the effect of reducing the rise time and will reduce, but never eliminate, the steady state error. An integral
control (KI) will have the effect of eliminating the steady-state error, but it may make the transient response worse. A derivative control (KD) will have the effect of increasing the stability of the system, reducing the
overshoot, and improving the transient response.
∫ ++=
++=
dttdeKdtteKteKtu
sKs
KKtG
DIp
DI
pC
)()()()(
termderivative a and n term,integratioan term,alproportion a provides controller The
)(
20
Proportional-Integral-Derivative (PID) Controller
kp
ki/s
kis+
+
+
e(t) u(t)
sKsKsK
sEsUsG
sEsKs
KKsU
KKKtytrte
tseKsteKteKtu
IPDPID
DI
p
DIP
DIp
++==
++=
−=
++=
2
)()()(
)()()(
ly.respective gains,feedback derivative and integral,al,proportion theare and ,, output; system theand
signal reference ebetween therror theis )()()(
)()()()(
21
Time- and s-domain block diagram of closed loop system
PIDController System
r(t) e(t) u(t)
-+R(s) E(s) U(s)
y(t)
Y(s)
)()(1)()(
)()()(
sGsGsGsG
sRsYsG
PIDsys
PIDsys
+==
22
PID and Operational AmplifiersA large number of transfer functions may be implemented using
operational amplifiers and passive elements in the input and feedback paths. Operational amplifiers are widely used in control systems to
implement PID-type control algorithms needed.
25
Figure 8.35
Op-amp DifferentiatorThe operational differentiator performs the differentiation of the input signal.
The current through the input capacitor is CS dvs(t)/dt. That is the output voltage is proportional to the derivative of the input voltage with respect to time, and
Vo(t) = _RFCS dvs(t)/dt
sCRsZsZ
sVsV
sG SFS
o _)()(
)()(
)(1
2 =−==
)(2 sZ
)(1 sZ
26
Linear PID Controller
vo(t)
R1
R2
vs(t)
C1 C2
Z2(s)
Z1(s)
( )( )
122121
22112
2121
2211212
21
2211
;1;;)(
111
)()(
)(
CRKCR
KCR
CRCRKPs
KsKsKsG
sCR
sCR
CRCRsCR
sCRsCRsCR
sVsV
sG
DIIPD
PID
S
o
−=−=+
−=++
=
++
+=
++−==
27
Tips for Designing a PID ControllerWhen you are designing a PID controller for a given system, follow the
following steps in order to obtain a desired response.
• Obtain an open-loop response and determine what needs to be improved
• Add a proportional control to improve the rise time• Add a derivative control to improve the overshoot• Add an integral control to eliminate the steady-state error• Adjust each of Kp, KI, and KD until you obtain a desired overall
response.
• It is not necessary to implement all three controllers (proportional, derivative, and integral) into a single system, if not needed. For example, if a PI controller gives a good enough response, then you do not need to implement derivative controller to the system.
28
The popularity of PID controllers may be attributed partly to their robust performance in a wide range of operation conditions and partly to their
functional simplicity, which allows engineers to operate them in a simple manner.
( ) ( )( )
plane- hand-left in the anywhere locatedbecan that zeros twoandorigin at the pole oneith function w transfer a
introduces PID y theAccordingl ./ and ;/ Where
)(
3231
2132
3
212
33
21
s
KKbKKas
zszsKs
bassKs
KsKsKsK
sKKtGC
==
++=
++=
++=++=
29
Root LocusRoot locus begins at the poles and ends at the zeros.
( )( )
( )( )( )
error. statesteady zero and responsefast a have willsystem thelarge, is If ;)(
have we1,)( Setting equal.ely approximat are rootscomplex theand zeros theBecause)()(
)()()(
ˆ)()(1
)()()()(
isfunction transfer loop closed The zero. theapproaches rootscomplex theincreases, controller theof K3 As locus.root plot thecan wezeros,complex with controller PID a use weAssume
521)(
33
3
2
3
2
3
112
113
KKs
Krs
KsT
sGrs
sGKsG
rsrsrszszsK
sGsGsGsGsG
sT
sssG
P
PP
C
PC
+≈
+=
=+
=+++
++=
+=
++=
-2
r2
j 2
j 4K3 increasing
z1
z1
r1
r1
30
Design of Robust PID-Controlled SystemsThe selection of the three coefficients of PID controllers is basically a search
problem in a three-dimensional space. Points in the search space correspond to different selections of a PID controller’s three parameters. By choosing different points of the parameter space, we can produce different step responses for a
step input.The first design method uses the (integral of time multiplied by absolute error
(ITAE) performance index in Section 5.9 and the optimum coefficients of Table 5.6 for a step input or Table 5.7 for a ramp input. Hence we select the three PID
coefficients to minimize the ITAE performance index, which produces an excellent transient response to a step (see Figure 5.30c). The design
procedure consists of the following three steps.
dttetT
∫=0
)(ITAE
31
The Three Design Steps of Robust PID-Controlled System
• Step 1: Select the ωn of the closed-loop system by specifying the settling time.
• Step 2: Determine the three coefficients using the appropriate optimum equation (Table 5.6) and the ωn of step 1 to obtain GC(s).
• Step 3: Determine a prefilter GP(s) so that the closed-loop system transfer function, T(s), does not have any zero, as required by Eq. (5.47)
on
nn
o
bsbsbsb
sRsYsT
++++==
−− 1
11 ...)(
)()(
32
Input Signals; Overshoot; Rise Time; Settling Time• Step: r(t) = A R(s) = A/s• Ramp: r(t) = At R(s) = A/s2
• The performance of a system is measured usually in terms of step response. The swiftness of the response is measured by the rise time, Tr, and the peak time, Tp.
• The settling time, Ts, is defined as the time required for the system to settle within a certain percentage of the input amplitude.
• For a second-order system with a closed-loop damping constant, we seek to determine the time, Ts, for which the response remains within 2% of the final value. This occurs approximately when
2
2
1/
1/
2
100 (PO)Overshoot Percentage
Response)(Peak 1 Time);(Peak 1
ratio) damping : frequency; natural undamped :(44 ;4 ;02.0
ξξπ
ξξπ
ξω
ξω
πξω
τξω
−−
−−
−
=
+=−
=
==≅⟨
e
eMT
ξωTTe
pn
p
nn
ssnTsn
33
Example 12.8: Robust Control of Temperature Using PID Controller employing ITAE performance for a step input and a settling time of less than 0.5 seconds.
( )
( ) ( )( )3223
212
33
212
31
212
3
02222
15271 :5.6) (Table ITAEfor equation sticcharacteri theof tscoefficien optimum The
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seconds. 0.5 than less of timesettling afor input step afor eperformanc ITAE optimiman obtain todesire Weinput. step afor seconds 3.2 is criterion) (2% timesettling theand 50%, iserror state-steady the1,)( If
122%;50
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1 ;
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121
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nnn
C
C
PC
C
sss
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ssnn
ωsω.sω.s
KsKsKsKsKsK
sGGsGG
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sGs
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sG
esGKKAe
ssssssG
+++
+++++
++=
+==
=++
=
=
======+
=++
=++
=+
=→
ξωξω
R(s)GP(s) GC(s) G(s) Y(s)
D(s)
+++
-
U(s)E(s)
34
Cont. 12.8
( ) ( )
( )1000. tonumerator overall thebring andequation previous in the zeros
theeliminate order toin 5.648.13
5.64)( require weTherefore
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2
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++=
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sGsss
jsjssss
sssT
KKK
T
P
c
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n
nsnωξ
ξωω
35
Results for Example 12.8
0.4%0.4%52%y(t)/d(t)maximum
0.0%0.0%50.1%Steady-state error
0.450.203.2Settling time (s)
1.9%31.7%0Percent overshoot
PID with GP(s) Prefilter
PID & GP(s)=1GC(s)=1Controller
36
E12.1: Using the ITAE performance method for step input, determine the required GC(t). Assume ωn = 20 for Table 5.6. Determine the step response with and without a prefilter GP(s)
prefilter with theand without response step theDraw ;8.14
8.14)( Where
40028400
)(1)()(
)()( isgain loop-closed theprefilter, aWith
4002840027
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400 and 27 have then we20When 4.1 :isequation sticcharacteri ITAE The
)1(it Find )(1
)()(
by given controller PI a Use;1
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2
2
21
22
212
21
+=
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+=
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===++
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=
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=
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sssGGsGGpsG
sRsY
sss
sRsY
KKss
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sGGsT
sKKG
ssG
P
C
C
n
nn
C
C
C
ωωω
GP(s) GC(s) G(s) Y(s)
D(s)
+++
-
U(s)E(s)R(s)
37
E12.3 A closed-loop unity feedback system has
( )
iprelationsh plot theThen 9
by determined is in changes to ofy sensitivit The9
)( isfunction y sensitivit The
99)( isfunction transfer loop-closed The
plot Bode aon )S( and )T(plot and Find ;3;9)(
2
2
2
2
++−==
++
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=+
=
pssps
Tp
dpdTS
pTpss
psssS
psssT
jωjωSppss
sG
TP
TP
38
E12.5
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( )
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40000K and 200 200;2 20%. P.O.for 0.5Let 200
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2001
100
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2nnn
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21
+====
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=×=++
+
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++
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+
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+
+
=
ssGT
kssKsT
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KKKKsssKKsK
sssKKsK
GG
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ssssG
cn
S
c
c
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ξω
ωωξωξ
39
DP12.10
6.1862.706.186)( ;/6.1862.70)(
6.186;2.70;7.5 :ITAE Use;10
)(
)( :controller PI the )c(
1004100)(;1004)(
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)(
)( :controller PD (b)73.50)( ;73.50 ;702.0 4.5%; For 10)(1
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2121
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21
2
+=+=
===+++
+=
+=+=
+=+=
==+++
+=
+====⟨
++=
+==
+=
ssGssG
KKKsKss
KsKsT
sKsK
sKKsG
ssGssG
KKKsKs
sKKsT
sKKsGsGcKPO
KssK
sGGcsGGcsTKsG
sssG
pc
n
c
pc
c
c
ω
ξ
Consider
Consider
Consider
40
5.14) (Eq. 1
;5.15) (Eq 100
peak time and time,settling overshoot, means ePerformancPID and PI, PD, K, :casesfour for the eperformanc theofsummary a Find
10002155.71000)(
10002155.7)( ;1000 ;215 ;5.7
10 with ITAE Use;10
)(
)( :controller PID (d)
21
2
2321
322
123
322
1
322
1
2
ξω
π
ω
ξξπ
−==
++=
++====
=++++
++=
++=
−−
np
l
c
n
c
tePO
sssGp
ssssGKKK
KsKsKssKsKsKsT
sKsKsKsGConsider