Process Control: Designing Process and Control
Systems for Dynamic Performance
Chapter 8. The PID Controller
Copyright © Thomas Marlin 2013
The copyright holder provides a royalty-free license for use of this material at non-profit
educational institutions
CHAPTER 8: THE PID CONTROLLER
When I complete this chapter, I want to be
able to do the following.
• Understand the strengths and weaknesses
of the three modes of the PID
• Determine the model of a feedback system
using block diagram algebra
• Establish general properties of PID
feedback from the closed-loop model
CHAPTER 8: THE PID CONTROLLER
Outline of the lesson.
• General Features and history of PID
• Model of the Process and controller - the
Block Diagram
• The Three Modes with features
- Proportional
- Integral
- Derivative
• Typical dynamic behavior
CHAPTER 8: THE PID CONTROLLER
PROPERTIES THAT WE SEEK IN A CONTROLLER
• Good Performance - feedback
measures from Chapter 7
• Wide applicability - adjustable
parameters
• Timely calculations - avoid
convergence loops
• Switch to/from manual bumplessly
• Extensible - enhanced easily
v1
TC
v2
CHAPTER 8: THE PID CONTROLLER
SOME BACKGROUND IN THE CONTROLLER
• Developed in the 1930-40’s,
remains workhorse of practice
• Not “optimal”, based on good
properties of each mode
• Programmed in digital control
equipment
• ONE controlled variable (CV) and
ONE manipulated variable (MV).
Many PID’s used in a plant.
v1
TC
v2
CHAPTER 8: THE PID CONTROLLER
MV =
controller
output
Proportional
E
+
Integral
Derivative
Note: Error = E  SP - CV
Final
element
SP = Set
point
+
CV =
Controlled
variable
sensor
Process
variable
PROCESS
Three “modes”: Three ways of using the time-varying
behavior of the measured variable
CHAPTER 8: THE PID CONTROLLER
Closed-Loop Model: Before we learn about each
calculation, we need to develop a general dynamic model
for a closed-loop system - that is the process and the
controller working as an integrated system.
v1
TC
v2
This is an example; how
can we generalize?
• What if we measured
pressure, or flow, or …?
• What if the process
were different?
• What if the valve were
different?
CHAPTER 8: THE PID CONTROLLER
GENERAL CLOSED-LOOP MODEL BASED ON BLOCK DIAGRAM
D(s)
SP(s)
+
-
E(s)
Gd(s)
MV(s)
GC(s)
Gv(s)
CVm(s)
GP(s)
+
+
CV(s)
GS(s)
Transfer functions
Variables
GC(s) = controller
Gv(s) = valve
GP(s) = feedback process
GS(s) = sensor
Gd(s) = disturbance process
CV(s) = controlled variable
CVm(s) = measured value of CV(s)
D(s) = disturbance
E(s) = error
MV(s) = manipulated variable
SP(s) = set point
CHAPTER 8: THE PID CONTROLLER
D(s)
SP(s)
+
-
E(s)
MV(s)
GC(s)
Gv(s)
CVm(s)
Let’s audit
our
understanding
Gd(s)
GP(s)
+
+
CV(s)
GS(s)
• Where are the models for the transmission, and signal
conversion?
• What is the difference between CV(s) and CVm(s)?
• What is the difference between GP(s) and Gd(s)?
• How do we measure the variable whose line is
indicated by the red circle?
• Which variables are determined by a person, which by
computer?
CHAPTER 8: THE PID CONTROLLER
D(s)
SP(s)
E(s)
+
Gd(s)
CV(s)
MV(s)
GC(s)
Gv(s)
GP(s)
+
+
CVm(s)
GS(s)
Set point response
G p ( s )Gv ( s )Gc ( s )
CV ( s )

SP( s ) 1  G p ( s )Gv ( s )Gc ( s )GS ( s )
Disturbance Response
Gd ( s )
CV ( s )

D( s ) 1  G p ( s )Gv ( s )Gc ( s )GS ( s )
• Which elements in the control system affect
system stability?
• Which elements affect dynamic response?
Proportional
CHAPTER 8:
MV
E
SP
-
+CV
Integral
+
Derivative
THE PID CONTROLLER,
Note: Error = E  SP - CV
The Proportional Mode
PROCESS
“correction proportional to
error.”
Time domain : MV (t )  K c E(t )  I p
MV ( s )
Transfer function : GC ( s ) 
 KC
E( s )
KC = controller gain
How does this differ from
the process gain, Kp?
Proportional
MV
CHAPTER 8:
E
SP
-
+CV
Integral
+
Derivative
THE PID CONTROLLER,
Note: Error = E  SP - CV
The Proportional Mode
“correction proportional to
error.”
PROCESS
Time domain : MV (t )  K c E(t )  I p
Kp depends upon the process (e.g.,
reactor volume, flows, temperatures,
etc.)
KC = controller gain
How does this differ from
the process gain, Kp?
KC is a number we select; it is used
in the computer each time the
controller equation is calculated
Proportional
CHAPTER 8:
THE PID CONTROLLER,
MV
+
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Proportional Mode
PROCESS
Time domain : MV (t )  K c E(t )  I p
Proportional
CHAPTER 8:
THE PID CONTROLLER,
MV
+
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Proportional Mode
PROCESS
Time domain : MV (t )  K c E(t )  I p
Physical Device:
Raising and lowering the
gate affects the flow in
Location of the fulcrum
determines the
gate/level
Float measures
the liquid level
Proportional
CHAPTER 8:
THE PID CONTROLLER,
MV
+
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Proportional Mode
PROCESS
Key feature of closed-loop performance with P-only
Final value
Kd
D K d
D
CV ' (t ) t   lim s

0
after
s 0
s 1  Kc K p 1  Kc K p
disturbance:
• We do not achieve zero offset; don’t return to set point!
• How can we get very close by changing a controller
parameter?
• Any possible problems with suggestion?
Proportional
MV
CHAPTER 8:
+
THE PID CONTROLLER,
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Proportional Mode
PROCESS
Disturbance in concentration of A in the solvent
CV = concentration of A in effluent
MV = valve % open of pure A stream
FS
solvent
FA
CV
pure A
From person
AC
MV
SP
E (t )  SP (t )  CV (t )
MV (t )  K c E (t )  I
THE PID CONTROLLER,The Proportional Mode
0.8
Controlled Variable
Controlled Variable
0.8
0.6
No control
0.4
0.2
0.6
0.4
0
0
20
40
60
80
100
Time
120
140
160
180
0
200
0
0.5
valve
20
40
60
80
100
Time
120
140
160
180
200
0
Manipulated Variable
Manipulated Variable
1
Kc = 0
0
-0.5
-1
0
20
40
60
80
100
Time
120
140
160
180
Kc =10
-2
-4
-6
200
0
Note change of scale!
20
40
20
40
60
80
100
Time
120
140
160
180
200
60
80
100
Time
120
140
160
180
200
80
100
Time
120
140
160
180
200
0.3
Controlled Variable
0.25
Controlled Variable
Offset (bad)
0.2
0.2
0.15
Less Offset, better
(but not good)
0.1
0.05
0.2
0.1
0
-0.1
0
0
0
20
40
60
80
100
Time
120
140
160
180
200
Unstable, very bad!
20
-5
Manipulated Variable
valve
Manipulated Variable
0
Kc = 100
-10
-15
-20
0
-20
-40
-60
-25
0
0
20
40
60
80
100
Time
120
140
160
180
200
Kc = 220
20
40
60
Proportional
CHAPTER 8:
MV
+
SP
-
+CV
Integral
Derivative
THE PID CONTROLLER,
E
Note: Error = E  SP - CV
The Integral Mode
PROCESS
“The persistent mode”
t
Kc
Time domain : MV (t ) 
E (t ' )dt '  I I

TI 0
MV ( s ) KC 1
Transfer function : GC ( s ) 

E( s )
TI s
TI = controller integral time (in denominator)
Proportional
MV
CHAPTER 8:
E
SP
-
+CV
Integral
+
Derivative
THE PID CONTROLLER,
Note: Error = E  SP - CV
The Integral Mode
PROCESS
t
Kc
Time domain : MV (t ) 
E (t ' )dt '  I I

TI 0
MV(t)
Slope = KC E/TI
time
Behavior when E(t) = constant
Proportional
MV
CHAPTER 8:
+
THE PID CONTROLLER,
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Integral Mode
PROCESS
Key feature of closed-loop performance with I mode
Final value
after
disturbance:
CV ' (t ) t 
Kd
D
 lim s
s 0
s 1  Kc K p
0
sTI
• We achieve zero offset for a step disturbance;
return to set point!
• Are there other scenarios where we do not?
Proportional
CHAPTER 8:
MV
E
SP
-
+CV
Integral
+
Derivative
THE PID CONTROLLER,
Note: Error = E  SP - CV
The Derivative Mode
PROCESS
“The predictive mode”
dE(t )
Time domain : MV (t )  K cTD
 ID
dt
MV ( s )
Transfer function : GC ( s ) 
 K cTd s
E( s )
TD = controller derivative time
Proportional
MV
CHAPTER 8:
+
THE PID CONTROLLER,
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Derivative Mode
PROCESS
Key features using closed-loop dynamic model
Final value
after
disturbance:
CV ' (t ) t 
Kd
D
 lim s
 D K d
s 0
s 1  K cTd s
We do not achieve zero offset; do not return to
set point!
Proportional
CHAPTER 8:
THE PID CONTROLLER,
MV
+
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Derivative Mode
PROCESS
dE(t )
Time domain : MV (t )  K cTD
 ID
dt
• What would be the behavior of the manipulated
variable when we enter a step change to the set point?
• How can we modify the algorithm to improve the
performance?
Proportional
MV
CHAPTER 8:
+
THE PID CONTROLLER,
E
SP
-
+CV
Integral
Derivative
Note: Error = E  SP - CV
The Derivative Mode
PROCESS
dE(t )
Time domain : MV (t )  K cTD
 ID
dt
X
We do not want to take the derivative of the set
point; therefore, we use only the CV when
calculating the derivative mode.
d CV (t )
Time domain : MV (t )   K cTD
 ID
dt
Proportional
MV
CHAPTER 8:
+
SP
-
+CV
Integral
Derivative
THE PID CONTROLLER
E
Note: Error = E  SP - CV
PROCESS
Let’s combine the modes to formulate the PID Controller!
E (t )  SP(t )  CV (t )

1
MV (t )  K c  E (t ) 
TI

d CV 
0 E (t ' )dt 'Td dt   I
t
Please explain every term and symbol.
Proportional
MV
CHAPTER 8:
E
SP
-
+CV
Integral
+
Derivative
THE PID CONTROLLER
Note: Error = E  SP - CV
PROCESS
Let’s combine the modes to formulate the PID Controller!
E (t )  SP(t )  CV (t )

1
MV (t )  K c  E (t ) 
TI

proportional
Error from set point
d CV 
0 E (t ' )dt 'Td dt   I
t
integral
derivative
Constant (bias) for bumpless transfer
CHAPTER 8: THE PID CONTROLLER
Let’s apply the controller to the three-tank mixer (no rxn).
Notes:
1) tanks are well mixed
2) liquid volumes are constant
3) sensor and valve dynamics are negligible
4) FA = Kv (v), with v = % opening
5) FS >> FA
CV = concentration of A in effluent
MV = valve % open of pure A
stream
FS
solvent
FA
CV
pure A
From person
AC
MV
SP
E (t )  SP(t )  CV (t )
t

1
d CV 
MV (t )  K c  E (t )   E (t ' )dt 'Td
I
TI 0
dt 

CHAPTER 8: THE PID CONTROLLER
S-LOOP plots deviation variables (IAE = 12.2869)
Controlled Variable
1.5
1
0.5
0
0
20
40
60
Time
80
• Is this good
performance?
• How do we
determine:
100
120
Kc, TI and Td?
Manipulated Variable
40
30
20
10
Kc = 30, TI = 11, Td = 0.8
0
0
20
40
60
80
100
120
CHAPTER 8: THE PID CONTROLLER
S-LOOP plots deviation variables (IAE = 20.5246)
Controlled Variable
2
1.5
1
0.5
0
0
20
40
60
Time
80
• Is this good
performance?
• How do we
determine:
100
120
Kc, TI and Td?
Manipulated Variable
150
100
50
0
Kc = 120, TI = 11, Td = 0.8
-50
0
20
40
60
Time
80
100
120
CHAPTER 8: THE PID CONTROLLER
Lookahead: We can apply many PID controllers when
we have many variables to be controlled!
E (t )  SP(t )  CV (t )

1
MV (t )  K c  E (t ) 
T
I

t
 E (t ' )dt 'T
d
0
d CV 
I
dt 
T6
C
Vapor
product
P1
C
E (t )  SP(t )  CV (t )
T1

1
MV (t )  K c  E (t ) 
T
I

T5
T2
Feed
t
 E (t ' )dt 'T
d
0
d CV 
I
dt 
E (t )  SP(t )  CV (t )
F1
T4
F2
Process
fluid
T3
L1
C

1
MV (t )  K c  E (t ) 
TI

F3
Steam
A1
L. Key
Liquid
product
t
 E (t ' )dt 'T
d
0
d CV 
I
dt 
CHAPTER 8: THE PID CONTROLLER
HOW DO WE EVALUATE THE DYNAMIC
RESPONSE OF THE CLOSED-LOOP SYSTEM?
• In a few cases, we can do this analytically
(See Example 8.5)
• In most cases, we must solve the equations
numerically. At each time step, we integrate
- The differential equations for the process
- The differential equation for the controller
- Any associated algebraic equations
• Many numerical methods are available
• “S_LOOP” does this from menu-driven input
CHAPTER 8: THE PID, WORKSHOP 1
• Model formulation: Develop the equations that
describe the dynamic behavior of the three-tank
mixer and PID controller.
• Numerical solution: Develop the equations that are
solved at each time step for the simulated control
system (process and controller).
FS
solvent
FA
pure A
AC
CHAPTER 8: THE PID, WORKSHOP 2
• The PID controller is applied to the three-tank mixer.
Prove that the PID controller with provide zero
steady-state offset when the set point is changed in a
step, SP.
• The three-tank process is stable. If we add a
controller, could the closed-loop system become
unstable?
FS
solvent
FA
pure A
AC
CHAPTER 8: THE PID, WORKSHOP 3
• Determine the engineering units for the controller
tuning parameters in the system below.
• Explain how the initialization constant (I) is
calculated. This is sometimes called the “bias”.
FS
solvent
CV
FA
pure A
From person
SP
AC
MV
E (t )  SP(t )  CV (t )
t

1
d CV 
MV (t )  K c  E (t )   E (t ' )dt 'Td
I
TI 0
dt 

CHAPTER 8: THE PID, WORKSHOP 4
The PID controller must be displayed on a computer
console for the plant operator. Design a console
display and define values that
• The operator needs to see to monitor the plant
• The operator can change to “run” the plant
• The engineer can change
FS
solvent
FA
pure A
AC
CHAPTER 8: THE PID CONTROLLER
When I complete this chapter, I want to be
able to do the following.
•
Understand the strengths and weaknesses of the
three modes of the PID
•
Determine the model of a feedback system using
block diagram algebra
•
Establish general properties of PID feedback from
the closed-loop model
Lot’s of improvement, but we need some more study!
• Read the textbook
• Review the notes, especially learning goals and workshop
• Try out the self-study suggestions
• Naturally, we’ll have an assignment!
CHAPTER 8: LEARNING RESOURCES
•
SITE PC-EDUCATION WEB
- Instrumentation Notes
- Interactive Learning Module (Chapter 8)
- Tutorials (Chapter 8)
CHAPTER 8: SUGGESTIONS FOR SELF-STUDY
1. In your own words, explain each of the PID modes.
Give at least one advantage and disadvantage for each.
2. Repeat the simulations for the three-tank mixer with
PID control that are reported in these notes. You may
use the MATLAB program “S_LOOP”.
3. Select one of the processes modelled in Chapters 3 or 4.
Add a PID controller to the numerical solution of the
dynamic response in the MATLAB m-file.
4. Derive the transfer function for the PID controller.