Feedback Control
PID Control
Hadiyanto
Overall Course Objectives
To develop skill on feedback control and PID
design
Feedback Concept
FB v FF
FB vs FF
FB vs FF
Case of HE
Ratio Control
Method
General Feedback Control Loop
D(s)
G d (s)
Y sp (s) E(s) C(s) U(s) + Y(s)
+- G c (s) G a (s) G p (s) +
Y s (s)
G s (s)
Closed Loop Transfer Functions
From the general feedback control loop and using
the properties of transfer functions, the following
expressions can be derived:
Y (s) G p ( s ) Ga ( s ) Gc ( s )
Ysp ( s ) G p ( s ) Ga ( s ) Gc ( s ) Gs ( s ) 1
Y (s) Gd ( s )
D( s ) G p ( s ) Ga ( s ) Gc ( s ) Gs ( s ) 1
Servo , d=0, Regulator , ysp=0,
Characteristic Equation
Since setpoint tracking and disturbance rejection have
the same denominator for their closed loop transfer
functions, this indicates that both setpoint tracking
and disturbance rejection have the same general
dynamic behavior.
The roots of the denominator determine the dynamic
characteristics of the closed loop process.
The characteristic equation is given by:
G p ( s) Ga ( s) Gc ( s) Gs ( s) 1 0
Feedback Control Analysis
The loop gain (KcKaKpKs) should be
positive for stable feedback control.
An open-loop unstable process can be
made stable by applying the proper level of
feedback control.
Characteristic Equation Example
Consider the dynamic behavior of a P-only controller
applied to a CST thermal mixer (Kp=1; p=60 sec)
where the temperature sensor has a s=20 sec and a is
assumed small. Note that Gc(s)=Kc.
Substituting into the characteristic equation
1 1
Kc 1 0
60s 1 20s 1
After rearranging into the standard form,
1200 1.15
p
1 Kc 1 Kc
Example Continued- Analysis of the
Closed Loop Poles
When Kc =0, poles are -0.05 and -0.0167
which correspond to the inverse of p and s.
As K is increased from zero, the values of the
c
poles begin to approach one another.
Critically damped behavior occurs when the
poles are equal.
Underdamped behavior results when K is
c
increased further due to the imaginary
components in the poles.
In-Class Exercise
Determine the dynamic behavior of a P-
only controller with Kc equal to 1 applied to
a first-order process in which the process
gain is equal to 2 and the time constant is
equal to 22. Assume that Gs(s) is equal to
one and Ga(s) behaves as a first-order
process with a time constant of 5.
PID Control Algorithm
Proportional Integral Derivative
1
t
de(t )
c(t ) c0 K c e(t ) e(t )dt D
I 0 dt
where e(t ) ysp ys (t )
C (s) 1
Gc ( s ) K c 1 D s
E (s) Is
Definition of Terms
e(t)- the error from setpoint [e(t)=ysp-ys].
K - the controller gain is a tuning parameter
c
and largely determines the controller
aggressiveness.
- the reset time is a tuning parameter and
I
determines the amount of integral action.
- the derivative time is a tuning parameter
D
and determines the amount of derivative
action.
Proportional Control
P Control c(t ) c0 K c e(t )
Gc ( s ) K c
Example,
Properties of Proportional Action
c(t ) c0 K c e(t )
Gc ( s ) K c
Kc K p Closed loop transfer function
Kc K p 1 base on a P-only controller
Y (s)
applied to a first order
Ysp ( s ) p process.
s 1
Kc K p 1 Properties of P control
◦ Does not change order of
process
◦ Closed loop time constant is
smaller than open loop p
◦ Does not eliminate offset.
Effect P to first order
Effect of controller gain . kc
Figure 8.8. Process response with proportional control.
Increasing the controller gain.
less sluggish process response.
Too large controller gain.
undesirable degree of oscillation or even unstable response.
An intermediate value of the controller gain
best control result.
Effect P on 2nd order
Integral Control
Properties of Integral Action
Kc t
c(t ) c0
I
0
e(t ) dt
Based on applying an I-
Y (s) 1 only controller to a first
Ysp ( s ) I p I order process
s2 s 1 Properties of I control
Kc K p Kc K p
I p ◦ Offset is eliminated
p ◦ Increases the order by 1
Kc K p
◦ As integral action is
1 I increased, the process
becomes faster, but at the
2 p Kc K p
expense of more
sustained oscillations
Integral Action
The primary benefit of integral action is
that it removes offset from setpoint.
In addition, for a PI controller all the
steady-state change in the controller output
results from integral action.
Proportional Action for the
Response of a PI Controller
Example for a First Order Process
with a PI Controller
Kc 2 I 10 Kp 1 p 5
Characteristic Equation :
1 2
5s 1 2 10 s 1 0
Rearranging
25s 2 15s 1 0
p 5 1.5
Example of a PI Controller Applied to a
Second Order Process
K c 1; I 1; K p 1; p 5; 2
Characteristic Equation :
1 1
1 1 0
25s 20 s 1
2
s
Rearranging
25s 20 s 2 s 1 0
3 2
p1 0.764 and a second order
response with p 4.37 and 0.08
Effect of integral time . I
Figure 8.9. PI control: (a) effect of integral time (b) effect of controller
gain.
Increasing the integral time.
more conservative(sluggish) process response.
Too large integral time.
too long time to reach to the set point after load upset or set-
point change occurs.
Theoretically, offset will be eliminated for all values of .
I
Derivative Control
Properties of Derivative Action
de(t )
c(t ) c0 K c D
dt
Y (s) K c K p D s
2 2
Ysp ( s ) p s 2 p K c K p D s 1
Closed loop transfer function for derivative-only
control applied to a second order process.
Properties of derivative control:
◦ Does not change the order of the process
◦ Does not eliminate offset
◦ Reduces the oscillatory nature of the
feedback response
Derivative Action for the Response of
a PID Controller
ysp
ys
cder
Time
Derivative Action
The primary benefit of derivative action is
that it reduces the oscillatory nature of the
closed-loop response.
Effect of derivative time . D
Figure 8.10. PID control: effect of derivative time.
• Increasing the derivative time.
improved response by reducing the maximum deviation,
response time and the degree of oscillation.
• Too large derivative time.
measurement noise tends to be amplified and the response may
be oscillatory.
• Intermediate value of D is desirable.
Composite Control
PI
PD
PID
Effect PI
Typical Response of Feedback Control Systems
Figure 8.7. Typical process response with feedback control.
C is the deviation from the initial steady-state.
No feedback control make the process slowly reach a new
steady-state.
Proportional control speeds up the process response and reduces
the offset.
Integral control eliminates offset but tends to make the response
oscillatory.
Derivative control reduces both the degree of oscillation and
response time.
The Characteristics of P, I, and D controllers
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.
Proportional Control
By only employing proportional control, a steady state error
occurs.
Proportional and Integral Control
The response becomes more oscillatory and needs longer to
settle, the error disappears.
Proportional, Integral and Derivative Control
All design specifications can be reached.
The Characteristics of P, I, and D controllers
CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR
Decrease Increase Small Change Decrease
Kc
(++) (--) (+/-) (+)
Decrease Increase Increase Eliminate
Ki
(++) (--) (--) (++)
Small Change Decrease Decrease Small Change
Kd
(+/-) (++) (++) (+/-)
Tips for Designing a PID Controller
1. Obtain an open-loop response and determine what needs to be improved
2. Add a proportional control to improve the rise time
3. Add a derivative control to improve the overshoot
4. Add an integral control to eliminate the steady-state error
5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall
response.
Lastly, please keep in mind that you do not need to implement all three controllers
(proportional, derivative, and integral) into a single system, if not necessary. For
example, if a PI controller gives a good enough response (like the above
example), then you don't need to implement derivative controller to the system.
Keep the controller as simple as possible.
Proportional Band
100%
PB
Kc
Another way to express the controller gain.
Kc in this formula is dimensionless. That is, the
controller output is scaled 0-100% and the error
from setpoint is scaled 0-100%.
In more frequent use 10-15 years ago, but it still
appears as an option on DCS’s.
Conversion from PB to Kc
Proportional band is equal to 200%.
The range of the error from setpoint is 200 psi.
The controller output range is 0 to 100%.
100% 100%
K
D
c 0.5
PB 200%
100%
K c 0. 5 0.25 % / psi
200 psi
Conversion from Kc to PB
Controllergain is equal to 15 %/ºF
The range of the error from setpoint is 25 ºF.
The controller output range is 0 to 100%.
15% 25º F
K
D
c 3.75
º F 100%
100%
PB 26.7%
3.75
Open-Loop Control - Example
1
G( s )
2
s 10s 20
num=1;
den=[1 10 20];
step(num,den)
Proportional Control - Example
The proportional controller (Kp) reduces the rise time, increases
the overshoot, and reduces the steady-state error.
Kp
T( s )
MATLAB Example 2
s 10 s ( 20 Kp )
Step Response
From: U(1)
1.4
Kp=300; 1.2 Step Response
From: U(1)
1
num=[Kp]; 1
0.9
Amplitude
0.8 0.8
den=[1 10 20+Kp];
To: Y(1)
0.7
0.6
0.6
t=0:0.01:2;
Amplitude
0.4
To: Y(1)
0.5
K=300 0.4 K=100
step(num,den,t)
0.2
0.3
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0.2
Time (sec.)
0.1
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (sec.)
Proportional - Derivative - Example
The derivative controller (Kd) reduces both the overshoot and the
settling time.
Kd s Kp
T( s )
MATLAB Example 2
s ( 10 Kd ) s ( 20 Kp )
Step Response
From: U(1)
1.4
1.2
Kp=300; 1 Step Response
From: U(1)
1
Kd=10;
Amplitude
0.8
To: Y(1)
0.9
0.6 0.8
num=[Kd Kp]; 0.7
Kd=10
0.4
0.6
Amplitude
den=[1 10+Kd 20+Kp];
To: Y(1)
0.2 0.5
0.4
0
t=0:0.01:2;
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0.3
Time (sec.)
0.2
Kd=20
step(num,den,t) 0.1
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time (sec.)
Proportional - Integral - Example
The integral controller (Ki) decreases the rise time, increases both
the overshoot and the settling time, and eliminates the steady-state
error Kp s Ki
T( s )
3 2
MATLAB Example s 10 s ( 20 Kp ) s Ki
Step Response
From: U(1)
1.4
Kp=30;
1.2
Step Response
From: U(1)
1 1.4
Ki=70;
Amplitude
1.2
0.8
To: Y(1)
1
0.6
num=[Kp Ki];
Ki=70
Amplitude
0.8
To: Y(1)
0.4
den=[1 10 20+Kp Ki]; 0.2
0.6
0.4
0
t=0:0.01:2; 0 0.2 0.4 0.6 0.8 1
Time (sec.)
1.2 1.4 1.6 1.8
0.2
2
Ki=100
0
step(num,den,t) 0 0.2 0.4 0.6 0.8 1
Time (sec.)
1.2 1.4 1.6 1.8 2
Assignment
