2. Single-loop PID Control System
ysp u(t)
+
_
ym(t)
+
+ y(t)
MV
DVs
Disturbance
Path
Sensor &
Transmitter
Final Control
Element
Control
Path
PID
Controller
Extended
Controlled Process
（广义对象）
Problem: For an unknown extended controlled process,
how to design and tune our PID controller ?
3. Proportional-Integral-Derivative
(PID) Controller
0
0
1 ( )
( ) ( ( ) ( ) ) ,
t
c d
i
de t
u t K e t e d T u
T dt
1
( ) (1 )
c c d
i
G s K T s
T s
Td is the derivative time.
Ideal PID Controller
Industrial PID Controller (design and realization ?)
1 1
( ) 1
1
d
c c
d i
d
T s
G s K
T T s
s
A
The derivative
gain Ad = 10.
4. Problem Discussion
Explain the function of PID controller for a
stable controlled process.
Analyze the effect of PID parameter
changes on control performances
How can we realize the industrial PID
controller in Simulink ?
PID tuning example (See ../PIDControl
/PIDLoop.mdl )
5. Contents
Selection of PID Controller Types （PID控制
器类型选择）
Tuning of PID Controller Parameters （控制
器参数整定）
Flow Control （流量控制）
Level Control （液位控制）
Reset Windup and Its Prevention （积分饱和
与防止）
Summary
6. Type Selection of PID Controllers
*1: For some slow
processes with long time
constants, the derivative
action is suggested to use.
However, if there exists
strong measurement noises,
a first-order or average filter
should be added.
Please analyze the rule
of type selection ?
Controlled
Variable
Controller
Type
Temperature /
Composition
PID*1
Flow /
Pressure
/Liquid-Level
PI
Liquid-Level P
8. Offline Tuning Based on
Process Parameters: K, T,τ
Step 1: switch the controller to manual mode,
change the output of controller in step form,
and record input/output data of controller.
Step 2: obtain process characteristics: K, T,τ,
from the step response data.
Step 3: set the PID parameters Kc, Ti , Td, and
switch the controller to automatic mode.
Step 4: increase or decrease the gain Kc until
obtaining the satisfactory response.
10. 0 10 20 30 40 50 60 70 80 90 100
66
68
70
72
74
76
78
80
%
min
Transmitter Output
Step 2: Obtain Process Para.
T
t O
632
.
0
O
O t
t
T
283
.
0
632
.
0
5
.
1
,%
,%
final initial
final initial
TO TO
TO
K
CO CO CO
11. Step 3: Obtain Initial PID Para.
(Ziegler-Nichols Method)
Controller Kc Ti Td
P
PI
PID
1 T
K
0.9 T
K
1.2 T
K
3.33
2.0 0.5
Note: the above method was developed for 0 T
12. Initial Value
Step 3: Obtain Initial PID Para.
(Lambda Tuning Method)
Controller Kc Ti Td
P
PI T
PID T τ/2
1 T
K
Note: the above method is not limited by the value of /T
1 T
K
1 T
K
0
0.2
13. 0 50 100 150
59
59.5
60
60.5
61
61.5
62
62.5
63
Time, min
%
Output of Transmitter
Ziegler-Nichols method
Lambda tuning method
set point
Simulation Example #1
K = 1.75
T = 6.5,τ= 3.3 min
For PI Controller,
Z-N tuning: Kc = 1.0,
Ti = 11 min
Lambda tuning: Kc =
0.56, Ti = 6.5 min
14. 0 50 100 150 200
59
59.5
60
60.5
61
61.5
62
62.5
63
Time, min
%
Output of Transmitter
Z-N tuning
Lambda tuning
set point
Simulation Example #2
K = 1.75
T = 6.5,τ= 6.3 min
For PI Controller,
Z-N tuning: Kc =
0.53, Ti = 20.8 min
Lambda tuning: Kc =
0.30, Ti = 6.5 min
15. Procedure of Online Tuning:
Ziegler-Nichols Technique
Step 1: with the controller online (in
automatic mode), remove all the reset (Ti =
maximum) and derivative (Td = 0) modes.
Start with a small Kc value.
Step 2: make a small set point or load
change and observe the response of CV.
Step 3: if the response is not continuously
oscillatory, increase Kc, or decrease PB,
repeat step 2.
Step 4: Repeat step 3 until a continuous
oscillatory response is obtained.
16. 0 10 20 30 40 50
58
59
60
61
62
63
64
65
66
Time, min
%
Output of Transmitter
set point
Kc = 0.5
Kc = 1
Kc = 2
Kc = 4
Kc = 3.5
Tu
Ti = 6000 min, Td = 0 min
Example of Online Tuning
See ../PIDControl/PIDLoop.mdl
Process
Fluid
Fuel Oil
T(t)
u(t)
y(t) %, TO
%, CO ysp(t)
Ti (t)
TC
27
TT
27
Furnace
17. Online Tuning:
Ziegler-Nichols Technique
Controller Kc Ti Td
P 0.5Kcu
PI 0.45Kcu Tu /1.2
PID 0.65Kcu Tu /2 Tu /8
The gain that gives these continuous oscillations is the ultimate
gain (临界增益), Kcu. The period of the oscillations is called
the ultimate period (临界周期), Tu. the ultimate gain and the
ultimate period are the characteristics of the process being
tuned. The following formulas are then applied:
18. 0 20 40 60 80 100
59
60
61
62
63
%
Output of Transmitter
0 20 40 60 80 100
40
60
80
100
Time, min
% Output of Controller
set point
Inlet temp. drops 5 Cent.
Kcu = 3.4, Tu = 11 min
PID: Kc = 2.2, Ti = 5.5 min, Td = 1.4 min
Online Tuning Result
See ../PIDControl/PIDLoop.mdl
Process
Fluid
Fuel Oil
T(t)
u(t)
y(t) %, TO
%, CO ysp(t)
Ti (t)
TC
27
TT
27
Furnace
19. 0 10 20 30 40 50
58
59
60
61
62
63
64
65
66
Time, min
%
Output of Transmitter
set point
Kc = 0.5
Kc = 1
Kc = 2
Kc = 4
Kc = 3.5
Tu
Ti = 6000 min, Td = 0 min
Limitation of Online Tuning
Process
Fluid
Fuel Oil
T(t)
u(t)
y(t) %, TO
%, CO ysp(t)
Ti (t)
TC
27
TT
27
Furnace
20. Auto-tuning Based on Relay
Feedback (基于继电反馈的参数自整定)
PID
Auto
Tune
u
ysp
Controlled
Process
+ _
ym
e(t)
Relay
（+）
Here we suppose the process gain > 0
22. Response of Relay Feedback
Oscillation period
TU & amplitude AY
（振荡周期与幅
度）?
0 5 10 15 20 25 30 35 40 45 50
58.5
59
59.5
60
60.5
61
61.5
Ym
0 5 10 15 20 25 30 35 40 45 50
47
48
49
50
51
52
53
U
min
See the detailed results:
../ PIDLoopAutoTuning.mdl
23. The Ultimate Gain (临界增益)
Kcu Calculation
/
4d
a
Y
CU A
d
K
4
0 5 10 15 20 25 30 35 40 45 50
47
48
49
50
51
52
53
U
min
经FT变换可知，控制输出的一次谐波幅度为
而对应的控制器临界增益为
/
4d
24. Online Z-N Tuning Parameters
Controller Kc Ti Td
P 0.5Kcu
PI 0.45Kcu Tu /1.2
PID 0.65Kcu Tu /2 Tu /8
If we use a PID controller, then we select the
following parameters ……
25. Closed-loop Response of PID
Feedback System
0 20 40 60 80 100 120 140
60
65
70
75
Ym
0 20 40 60 80 100 120 140
40
60
80
100
U
min
Above auto-
tuning method
can be applied
to other
controlled
processes ?
26. Characteristics of Flow Loops
Fast dynamic response
Zero dead time, which results in an infinite
controller gain in every tuning equation
Large measurement noise
To decrease the change of control valve, a PI
controller is common used with very small
proportional action and a large integral action
to approximate an integral controller. (Why?)
27. 0 20 40 60 80 100
45
50
55
60
65
%
Output of Transmitter
0 20 40 60 80 100
0
20
40
60
Time, min
%
Output of Controller
Kc = 4, Ti =2 min
Kc = 1, Ti = 0.5 min
Tuning Example of Flow Loops
u(t)
y(t)
% CO
% TO
ysp(t)
F(t)
FC
See ../PIDControl/FlowLoop.mdl
Please compare the proportional
gain with the integral gain
29. Characteristics of Level Loops
Very often levels are integrating
processes
There are two types of possible control
objectives when the input flow varies:
(1) Tight Level Control;
(2) Average Level Control
(“液位均匀控制”)
30. Tight Level Control
The objective is to control the level tightly at
set point, and the output flow can be allowed
to vary without limitation
If a level process happens to be self-regulated,
and it is possible to obtain K, T andτ, the
above tuning techniques can be used directly
If a level process is integrating, a PI controller
is common used with large proportional action
and a very small integral action
31. Average Level Control
The objective is to smooth the output flow
from the tank, which feeds the downstream
unit, the level in the tank must be allowed to
“float” between a high and a low level
A P controller is common used in Average Level
Control with a small proportional gain
Tuning: the gain should be set to be as small
as possible, as long as the level changes
between a high and a low level for the
expected flow deviation from the average flow.
32. Example of Level Control
See ../PIDControl/
LevelLoop.mdl
u(t) % CO
% TO
h(t)
Fi(t)
Fo(t)
A
ysp
y(t)
LC
41
LT
41
33. Analysis of Average Level
Control Systems
Dynamic equation of the controlled
process:
where A is the area of the tank.
Suppose
LC
u(t)
y(t)
% CO
% TO
ysp(t)
h(t)
Fi(t)
Fo(t)
A
)
(
)
(
)
(
0 t
F
t
F
dt
t
dh
A i
)
(
)
(
0 t
u
K
t
F V
,
)
(
)
(
max
h
t
h
t
y
34. Analysis of Average Level
Control Systems (cont.)
For a proportional
controller, Gc = －Kc,
＋
－
＋
Gc
Fi (s)
－
Fo (s) h(s)
ysp
KV
u(s)
s
A
1
max
1
h
y(s)
1
1
1
)
(
)
(
max
max
max
s
K
K
Ah
s
Ah
K
K
s
Ah
K
K
s
F
s
F
V
C
V
C
V
C
i
o
1
1
1
1
1
)
(
)
(
max
max
max
s
K
K
Ah
K
K
s
Ah
K
K
s
Ah
s
F
s
y
V
C
V
C
V
C
i
Please analyze the
above models.
36. Simulation Results of
P-type Average Level Control
0 10 20 30 40 50 60 70 80 90 100
9
9.5
10
10.5
11
m3/hr
Fin, Fout
0 10 20 30 40 50 60 70 80 90 100
46
48
50
52
54
time, min
%
Ysp, Ym Kc = -0.5
Fin
Kc = -2.0
37. Reset Windup Problem
Please see the following simulation example
…/PIDControl/PidLoopwithLimit.mdl
＋
－
＋
＋
d(t)
Controlled
Process ym(t)
ysp(t) e(t) uc
1
1
C
I
K
T s
up
38. Simulation Result with Reset
Windup in a Single-Loop System
Discussion:
Which difference
exists between
reset windup and
the open or
closed status of
the control valve
completely ?
39. The Principle of Preventing
Reset Windup
Principle: remove the reset or
integral action if the control
output is beyond the normal
operation range.
＋
－
＋
＋
d(t)
A Controlled
Process ym(t)
ysp(t)
KC
＋
＋
1
1
s
TI
e(t) uc up
max
max
max
min
min
min
)
(
,
)
(
),
(
)
(
,
)
(
u
t
u
if
u
u
t
u
u
if
t
u
u
t
u
if
u
t
u
c
c
c
c
p
40. 0 20 40 60 80 100 120 140 160 180 200
55
60
65
70
75
%
Ysp, Ym
0 20 40 60 80 100 120 140 160 180 200
0
50
100
150
%
time, min
Uc, Up
Anti-reset Windup Example
41. Industrial PID Controller
＋
－
＋
＋
d(t)
A Controlled
Process ym(t)
ysp(t)
＋
＋
1
1
s
TI
1
1
s
A
T
s
T
K
D
D
D
C
PID Controller
e(t) uc up
KC
＋
－
ysp(t)
＋
＋
1
1
s
TI
Derivative Action First
PID Controller
1
1
s
A
T
s
T
D
D
D
e(t) uc up ＋
＋
d(t)
A Controlled
Process ym(t)
PID1
PID2
42. Summary
Selection of PID Controller Types
Tuning of PID Controller Parameters
Tuning of PID Controller for Flow Loops
Tight / Average Level Control
Reset Windup and Its Prevention
43. Problem Discussion
For an unknown stable temperature control system,
can you determine PID parameters in Offline and
Online tuning methods ?
Please realize the industrial PID controller in
Simulink ?
For the fast flow control loop, show me your tuning
principle and explain why.
For the AVERAGE level control loop, show me your
tuning principle and explain why.
Explain the existing reason of reset windup and
show me your prevention schemes
44. Exercise 3.1
A controlled process is shown in the Problem 2-1 (p.34) in
Automated Continuous Process Control.
(1) calculate its characteristics parameters K, T and τ;
(2) decide on the action of the valve and the controller;
(3) tune your PID controller.
