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This lecture was delivered by Dr. Sakal Japendu for Process Control course at Ambedkar University, Delhi. It includes: Strength, Weakness, Proportional, Integral, Differential, Controller, Feedback, Block, Diagram, Model
Typology: Slides
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CHAPTER 8: THE PID CONTROLLER
GENERAL CLOSED-LOOP MODEL BASED ON BLOCK DIAGRAM
G
(s)
G
(s)
G
(s)
G
(s)
G
(s)
D(s)
CV(s)
CV
(s)
SP(s)
E(s)
MV(s)
Transfer functionsG
(s) = controller
G
(s) = valve +
G
(s) = feedback process
G
(s) = sensor
G
(s) = disturbance process
VariablesCV(s) = controlled variableCV
(s) = measured value of CV(s)
D(s) = disturbanceE(s) = errorMV(s) = manipulated variableSP(s) = set point
CHAPTER 8: THE PID CONTROLLER
G
(s)
G
(s)
G
(s)
G
(s)
G
(s)
D(s)
CV(s)
CV
(s)
SP(s)
E(s)
MV(s)
Where are the models for the transmission, and signalconversion?
What is the difference between CV(s) and CV
(s)?
What is the difference between G
(s) and G
(s)?
How do we measure the variable whose line isindicated by the red circle?
Which variables are determined by a person, which bycomputer?
Let’s audit
our
understanding
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Proportional Mode
p
c
I
t
E
K
t
MV
=
)
(
)
(
:
domain
Time
C
C
K
s
E
s
MV
s
G
=
=
)
(
)
(
)
(
:
function
Transfer
K
C
= controller gain
“correction proportional to
error.”
How does this differ from
the process gain, K
?
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Proportional Mode
p
c
I
t
E
K
t
MV
=
)
(
)
(
:
domain
Time
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Proportional Mode
Key features using closed-loop dynamic model
Final valueafterdisturbance:
0
1
1
0
≠
∆
=
∆
=
→
∞
→
p
c
d
p
d c
s
t
K
K
K
D
K
K K
s D
s
t
CV
lim
)
(
We do not achieve zero offset; don’t return to set point!
How can we get very close by changing a controllerparameter?
Any possible problems with suggestion?
THE PID CONTROLLER,The Proportional Mode
0
20
40
60
80
100
120
140
160
180
200
-0.
0 0.3 0.2 0.
Tim e
Controlled V ariable
0
20
40
60
80
100
120
140
160
180
200
-20 -40 -
0 20
Tim e
M anipulated V ariable
0
20
40
60
80
100
120
140
160
180
200
0 0.8 0.6 0.4 0.
Tim e
Controlled V ariable
0
20
40
60
80
100
120
140
160
180
200
(^0) -2 -4 -
Tim e
M anipulated V ariable
0
20
40
60
80
100
120
140
160
180
200
0
Tim e
Controlled V ariable
0
20
40
60
80
100
120
140
160
180
200
-10 -15 -20 -
(^0) -
Tim e
M anipulated V ariable
0
20
40
60
80
100
120
140
160
180
200
0 0.8 0.6 0.4 0.
Tim e
Controlled V ariable
0
20
40
60
80
100
120
140
160
180
200
-0.
0
1
Tim e
M anipulated V ariable
Kc = 0
Kc =
Kc = 100
Kc = 220
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Integral Mode
I
I
c
I
dt
t
E
K T
t
MV
∫
=
∞^0
' ) ' ( ) ( :
domain
Time
Slope = K
E/T
MV(t)
time
Behavior when E(t) = constant
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Integral Mode
Key features using closed-loop dynamic model
Final valueafterdisturbance:
0
1
0
=
∆
=
→
∞
→
I
p
c
d
s
t
sT
K
K
K
s
D
s
t
CV
lim
)
(
We achieve zero offset; return to set point!
Are there other scenarios where we do not?
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Derivative Mode
Key features using closed-loop dynamic model
Final valueafterdisturbance:
d
d
d c
s
t
K
D
s
T
K K
s D
s
t
CV
lim
)
(
∆
=
∆
=
→
∞
→
1
0
We do not achieve zero offset; do not return to
set point!
Proportional
Integral Derivative
Note
Error = E
CHAPTER 8:
THE PID CONTROLLER,
The Derivative Mode
D
D
c
I
dt
t
dE
T
K
t
MV
=
)
(
)
(
:
domain
Time
What would be the behavior of the manipulatedvariable when we enter a step change to the set point?
How can we modify the algorithm to improve theperformance?
