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Chapter 6Chapter 6Analysis of Feedback Analysis of Feedback Control SystemsControl Systems
Prof. Shi-Shang JangChemical Engineering DepartmentNational Tsing-Hua UniversityHsin Chu, TaiwanJune, 2015
6-1 Introduction6-1 IntroductionGiven a plant with a transfer function of
y(s)/m(s)=Gp(s), a controller is implemented with a transfer function of Gc(s), the system hence becomes a closed loop function.
It is very convenient to analyze the closed loop system in s-domain since only algebraic equations are involved.
In a closed loop system, there are two inputs for the system, the set point (servo) of the output and the disturbances (regulation) that affect the output.
6-1 Introduction 6-1 Introduction – Cont– Cont..
Variations in outlet temperature are sensed by the sensor-transmitter andsent to the controller causing the controller output signal to vary. This isin turn causes the control valve position and consequently the steam flowTo vary. The variations in steam flow cause the outlet temperature to vary, thus completing the loop.
Example: The Heat Exchanger Example: The Heat Exchanger System System
6-1.1 Closed-Loop Transfer Function6-1.1 Closed-Loop Transfer Function
(1)Set Point Change
0
1
setsp o
c
s v
o s s w
o
sp s v coseto s v c
E s K T C s
M s G s E s
W s G s M s
T s G s W s G s W s
C s H s T s
W s
K G s G s G sT s
T s H s G s G s G s
6-1.1 Closed-Loop Transfer Function - 6-1.1 Closed-Loop Transfer Function - ContinuedContinued
(2)Load Changes
0
( )
1
(3)General Case
% / %
/ % %
dimensionless
( )
1 1
(4)Closed
seto
o w
s v c
s v c
sp s v c setwo o
s v c s v c
T
T s G s
W s H s G s G s G s
TO C kg s COH s G s G s G s
C kg s CO TO
K G s G s G sG sT s W s T s
H s G s G s G s H s G s G s G s
Loop Transfer Functions
1f
i l
Z
Z
Z=OutputZi=Inputf=product in forward path from Zi to Zl=product of every transfer function infeedback loop
6-1.1 Closed-Loop Transfer Function - 6-1.1 Closed-Loop Transfer Function - ContinuedContinued
sFsGsG
sGsR
sGsG
sGsGsC
sHsGsG
sHsGsGsG
cc
c
w
sv
1
2
1
1
2
1
11
ExampleExample
Example – Example – Cont.Cont.
sWs
Ks
ss
ss
KsF
s
Ks
Laplacing
tUAtUAtWdt
dC
tUAcftUAtFTTcdt
dcV
ionLinearizat
tTtTUAtwdt
dTC
tTctftTtTUAtTctfdt
dTcV
c
w
cs
ssF
ss
M
psipp
ss
M
psipp
11
111
Example – Example – Cont.Cont.
1100200
0100;
1
100
ln;
1
Dynamics Valve Control
TT
TT
vv
vv
Ks
KsG
rTransmitteSensor
wK
s
KsG
Example – Example – Cont.Cont.
Example – Example – Cont.Cont.
Example – Example – Cont.Cont.
sGsWsGsF
s
K
ss
s
KsW
s
KsF
s
K
s
KsFs
sF
s
cc
wF
ss
F
11
1
11
11
6-1.1 Closed-Loop Transfer Function - 6-1.1 Closed-Loop Transfer Function - ContinuedContinued
sFsGsG
sGsR
sGsG
sGsGsC
sHsGsG
sHsGsGsG
cc
c
w
sv
1
2
1
1
2
1
11
Example – Example – Cont.Cont.
1
2
1 2
1 1
1.652 1.183 1.0
0.2 1 8.34 1 0.502 1 0.75 1
3.34 0.524 1 1.0
8.34 1 0.502 1 0.75 1
1 1
v s
F
c
c c
G s G s G s H ss s s s
sG s G s H s
s s s
G s G s G sC s R s F s
G s G s G s G s
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop SystemsSystems
OL
w
svcT
w
svc
wo
svc
wo
K
K
KKKK
K
GGGH
G
W
T
sGsGsGsH
sG
sW
sT
11
00001
0
0
0
1
OL
svcsp
svcT
svcsp
svc
svcsp
seto
o
svc
svcsp
seto
o
K
KKKK
KKKK
KKKK
GGGH
GGGK
T
T
sGsGsGsH
sGsGsGK
sT
sT
11
00001
000
0
0
1
ExampleExample
Example: New steady state of the Example: New steady state of the heating tankheating tank
1. Servo control with Kc=1;Tset=1
2 1.652 1.183 3.90860.7963
1 1 2 1.652 1.183 4.9086
1
150 1 150.7963
c v s
set c v s
set
K K KT
T K K K
T
T T
2. Regulation Control with F=1
3.34 -3.34-0.6804
1 1 2 1.652 1.183 4.9086
150 149.3196
F
c v s
KT
F K K K
T T
Example: Flow rate control Example: Flow rate control systemsystem
Example: Flow rate control Example: Flow rate control system – system – Cont.Cont.
1
1
11
1
11
1
;1
sKKKs
KKK
sF
sF
Set
sKKKss
sKKK
sF
sF
sKsG
sGsGK
sGsGK
sF
sF
KsHs
K
sM
sFsG
FCcvTv
cvTset
vI
IcvTvI
IcvTset
Icc
cvT
cvsp
set
Tv
vv
6.1.2 Characteristic Equation of 6.1.2 Characteristic Equation of the Loopthe Loop
11 0
1 2
1 2
1 2
1 2
1 ( ) 0
0
0
( )1 ( )
numerical terms( )
input terms
input terms
Unforced terms Forced terms
s v c
n nn n
n n
wo
s v c
on n
no
n
H s G s G s G s
a s a s a
a s r s r s r
G sT s W s
H s G s G s G s
T sa s r s r s r
bb bT t
s r s r s r
Example : Servo Problem for a First Order Example : Servo Problem for a First Order SystemSystem
11
1
11
11
1
11
11
11
1
PI)( ;1
1 : 2 Case
1'
'
11
1
11
1
1
; : 1 Case
0;11
sssKK
sKK
sK
ssK
sK
ssK
sK
sK
sK
sK
sR
sC
sKG
s
K
KKs
KKKK
KKs
KK
sK
K
sK
K
sR
sC
KG
sFs
KsG
IIc
Ic
I
Ic
I
Ic
Ic
Ic
Icc
c
c
c
c
c
c
c
cc
Proportional only
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop SystemsSystems
Proposition 6-1 : The closed loop dynamics of a first order system with a P-only controller behaves the same with another first order system, with a gain of KOL/(1+ KOL) and time constant of /(1+ KOL) for the servo problem.
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.
Corollary 6-1 : A P-only controller applied to a first order system can yield perfect control if Kc for a servo problem.
Proof: Consider (6-1), in case , we have K11, and 10. Corollary 6-2 : Assume that a step change with a
magnitude of A at the set point Tset(s) is implemented to a P-only controller applied to a first order system, then the offset of the controller is A/(1+KOL) .
Proof: Let Tset=A/s , then To(s)=(A/s)[K1/ (1s+1)] , we have:
offset =
OL
OL
so
so
t K
AKAK
s
AKssTtT
11limlimlim 1
1
1
00
OLKAAKA
1
11
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.
Kc 1 K1
1 1.67 0.667
2 1 0.8
5 5/11 10/11
5s+1
2
Transfer Fcn
simout
To Workspace
Scope
2
Gain
1
Constant
Add
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.
Regulation problem (disturbance rejection) using a P-only control for a first order system:
01
10
)0(1
1
1
1
)(
0;1
;1
1
2
1
2
2
2
1
2
2
221
FKK
KOffset
KK
K
F
CsK
K
sK
sGsG
sG
sF
sC
sRs
KG
s
KG
c
c
cc
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.
Corollary 6-3 : A P-only controller applied to a first order system can yield perfect control if Kc for a regulation problem.
Corollary 6-4 : Assume that a step change with a magnitude of A at the load F(s) is implemented to a P-only controller applied to a first order system, then the offset of the controller is .
01 1
2 FKK
K
c
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.
Kc Offset
1 -0.124
2 -0.077
5 -0.036
time
Kc=1
Kc=2
Kc=5
6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.
Proposition 6-2 : The offset of closed loop dynamics of a first order system with a PI controller results zero offset.
Proof:
10
0
11
1
c
c
IIc
Ic
KK
KK
R
C
sssKK
sKK
sR
sC
6-2 Stability of the Control 6-2 Stability of the Control LoopLoop
1 2
1 2 1
1 2
1 2
1 2
1 22 2
1
Let , , , be n roots of 1 0
input terms
= particular solution
At the cross over point:
other terms
'sin
n
n c
n
n
r tr t r tn
u
u
r r r G s G s
bb bC t
s r s r s r
b e b e b e
b s bC s
s
C t b t
6-2.1 Stability Criterion6-2.1 Stability CriterionProposition 6-3: For a feedback
control loop to be stable, all of the roots of its characteristic equation must be either negative real number or complex numbers with negative real parts.
Corollary 6-4: For a feedback control loop to be stable , all of the roots of its characteristic equation must fall on the left-hand half of the s plane, also known as the “left-hand plane” (LHP).
6-2.1 Stability 6-2.1 Stability Criterion Criterion - Example- Example
Kc p1 p2 p3
0 -3 -2 -1
0.23 -3.1 -1.75 -1.15
0.39 -3.16 -1.42* -1.42*
1.58 -3.45 -1.28-0.75j -1.28+0.75j
6.6 -4.11 -0.95-1.75j -0.95+1.75j
26.5 -5.1 -0.45-2.5j -0.45+2.5j
60* -6.0 0-3.32j* 0+3.32j*
100 -6.72 0.35-4j 0.35+4j
1
1
1 2 3P vG s G s G s H ss s s
11 2 3
11
1 2 3
C
C
KC s s s s
R s Ks s s
1 2 3
C
C
C s K
R s s s s K
Or
0321 CKsss
Characteristic equation can be derived:
6-2.1 Stability Criterion 6-2.1 Stability Criterion - - ExampleExample
6-2.1 Stability Criterion – 6-2.1 Stability Criterion – Cont.Cont.Consider the previous example, at Kc=0.39, the
system becomes oscillatory, at Kc=60, the system becomes unstable. At this point, we term the controller gain that makes the closed loop system unstable the ultimate gain (Ku=60) of the system, the frequency (u=3.32) that the system oscillates is termed the ultimate frequency.
Problem: Given the characteristic equation C(s,
Kc)=0, what is the ultimate gain and ultimate frequency of the system?
Answer: Direct substitution, Routh array, Root Locus
6-2.2 Method of Direct Substitution 6-2.2 Method of Direct Substitution – – Cont.Cont.Basic idea: Consider the characteristic
equation C(Kc,s)=0, in case of ultimate gain, i.e. Kc=Ku, then the root of the equation should be located at the imaginary axis, i.e. s=j. The problem of solving the ultimate condition, hence changed into C(Kc,s)=R(Kc, )+jI(Kc, )=0, i.e. R(Kc, )=0, I(Kc, )=0 (two unkowns, two equations)
Problem: many algebraic work invloved.
6-2.2 Method of Direct Substitution – 6-2.2 Method of Direct Substitution – Cont.Cont.
60;11
066 ;01111
0
6611
6116
6116321
23
23
23
23
uu
uuuu
uuuu
uuuu
uu
K
K
Kjjj
jjjK
sssKsssK
6-2.2 Method of Direct Substitution – 6-2.2 Method of Direct Substitution – Cont.Cont.
7.282186.0
2
8.23;/2186.0
0439008.01420
08.0143420900
08.013110130
013
016.0
110
1
130
501
%
/
13
016.0;
%
110
1;
/130
50
3.16 .
32
23
u
cuu
uucu
u
c
c
ccvs
vs
T
Ksrad
jK
jsLet
Ksss
Ksss
Ksss
sGsGsHsG
CO
skg
ssG
CssH
skg
C
ssG
Fig
6-2.3 Effect of Parameters6-2.3 Effect of Parameters
3 2
Let the temperature sensor/transmiter is re-calibrated (range reduced to
75-125 such that
2 %
10 150 2 0.016
1 1 030 110 1 3 1
900 420 43 1 1.6 0
20.2186 / ; 11.9;
0.
s v c c
c
u cu u
C
H ss C
G s H s G s G s Ks s s
s s s K
rad s K T
28.72186
Let the temperature sensor/transmiter is replaced to a faster one such that:
1 %
5 150 1 0.016
1 030 1 5 1 3 1
20.2345 / ; 18.7; 26.8
0.2186
s v c c
u cu u
H ss C
G s H s G s G s Ks s s
rad s K T
6-2.3 Effect of Parameters – 6-2.3 Effect of Parameters – Cont.Cont.
6-2.4 Effect of Dead Time – Pade 6-2.4 Effect of Dead Time – Pade ApproximationApproximation
...42
1
21
21
...62
1
330
220
00
0
330
220
00
ststst
st
st
ststste st
6-2.4 Effect of Dead Time – Pade 6-2.4 Effect of Dead Time – Pade Approximation Approximation – Cont.– Cont.
0
0
0
0020
0
0
1
1
21
12
01222
0121
211
01
11
10
0
tKK
t
t
KKst
KKt
st
sst
stKK
s
KeKsGK
s
KesG
cu
u
cc
c
stc
c
st
HomeworkHomework
Text , page 2206-1, 6-3, 6-5, 6-10, 6-12, 6-18, 6-
20, 6-25
ExampleExampleThe temperature of a tank is controlled by adjusting
the steam flow to the jacket of the tank. The temperature transmitter has a span of 100°F and is set between 100 and 200°F. The proportional controller is set at a proportional band of 20. The normal pressure to the valve is 8 psig and the valve is air to open type. The normal temperature of the tank is 170°F and the normal feed temperature is 65°F. The set point is moved from 170°F to 175°F. And the tank eventually comes to a steady state of 174.1°F. ◦ (i) What is the offset?◦ (ii) What is the pressure to the valve in the final steady state?◦ (iii) What is the process gain? (Assume the valve gain is unity)◦ (iv) What will be the offset if the proportional band was
changed to 10?
SolutionSolution
( )175 174.1 0.9
0.9( ) 12 0.108; 100 / 20 5;
1005 0.108 8 8.54
174.1 170( ) 7.59
5 0.10810 7.59
( ) 0.91 10 7.59 0.12
170 5 0.9 174.5
p
set
i F
ii e Kc
M psig
iii K
YivY
T F