Okay, let's solve this step-by-step: * Set point (Io) = 12 rpm * Range = 15 - 10 = 5 rpm * Initial controller output = 22% * KI = -0.15%/s/% error * Error = Actual - Set point = ? * Given: Initial output is 22% * To find: What is the actual speed? Using the integral control equation: Iout = Io - KI * ∫edt 22% = 12rpm - 0.15%/s/% * ∫e dt ∫e dt = (22% - 12rpm)/0.15%/s/% = 40%*

1. Closed Loop Systems
Professor CHARLTON S. INAO
Control System Design, PE -3032
Defence Engineering College
Bishoftu, Ethiopia
Controllers
1

2. Topics
• Proportional Mode
• Derivative Mode
• Integral Mode
• PID Controller
2

3. Continuous controllers
Now in the continuous controller’s theory,
there are three basic modes on which the
whole control action takes place and these
modes are written below. We will use the
combination of these modes in order to have
a desired and accurate output.
• Proportional controllers.
• Integral controllers.
• Derivative controllers.

4. Combinations of these three controllers are
written below:
• Proportional and integral controllers.(PI)
• Proportional and derivative
controllers.(PD)
• Proportional ,integral and derivative
controllers(PID)

5. Proportional control
Proportional control, the controller output is proportional to
the error signal,Proportional controllers there are two
conditions and these are written below:
• Deviation should not be large, it means there should be less
deviation between the input and output.
• Deviation should not be sudden.
Now we are in a condition to discuss proportional controllers,
as the name suggests in a proportional controller the output
(also called the actuating signal) is directly proportional to the
error signal.
This can be analyzed (proportional controller) mathematically.
As we know in proportional controller output is directly
proportional to error signal, writing this mathematically we
have,

6. Proportional control
Removing the sign of proportionality we have,
Where Kp is proportional constant also known as
controller gain. It is recommended that Kp should be
kept greater than unity. If the value of Kp is greater
than unity, then it will amplify the error signal and
thus the amplified error signal can be detected easily.

7. Advantages of Proportional Controller
 Now let us discuss some advantages of proportional
controller. Proportional controller helps in reducing the
steady state error, thus makes the system more stable.
 Slow response of the over damped system can be
made faster with the help of these controllers.
Disadvantages of Proportional Controller
 Now there are some serious disadvantages of these
controllers and these are written as follows: Due to
presence of these controllers we some offsets in the
system.
 Proportional controllers also increase the maximum
overshoot of the system.
Proportional control

8. Steady-state error

9. Integral Controllers
As the name suggests in integral controllers
the output (also called the actuating signal) is
directly proportional to the integral of the
error signal. Now let us analyze integral
controller mathematically. As we know in an
integral controller output is directly
proportional to the integration of the error
signal, writing this mathematically we have,

10. Integral Controllers
Removing the sign of proportionality we have,
Where Ki is integral constant also known as controller gain.
Integral controller is also known as reset controller.
Advantages of Integral Controller
Due to their unique ability they can return the
controlled variable back to the exact set point following
a disturbance that’s why these are known as reset
controllers.
Disadvantages of Integral Controller
It tends to make the system unstable because it
responds slowly towards the produced error.

11. Derivative Controllers
We never use derivative controllers alone. It
should be used in combinations with other modes
of controllers because of its few disadvantages
which are written below: It never improves the
steady state error.
It produces saturation effects and also amplifies
the noise signals produced in the system
Now, as the name suggests in a derivative
controller the output (also called the actuating
signal) is directly proportional to the derivative of
the error signal. Now let us analyze derivative
controller mathematically. As we know in a
derivative controller output is directly
proportional to the derivative of the error signal,
writing this mathematically we have,

12. Derivative Controllers
Removing the sign of proportionality we have,
Where Kd is proportional constant also known as controller
gain. Derivative controller is also known as rate controller.
Advantages of Derivative Controller
The major advantage of derivative controller is that it
improves the transient response of the system

13. Proportional and Integral Controller
As the name suggests it is a combination of
proportional and an integral controller the output
(also called the actuating signal) is equal to the
summation of proportional and integral of the
error signal. Now let us analyze proportional and
integral controller mathematically. As we know in
a proportional and integral controller output is
directly proportional to the summation of
proportional of error and integration of the error
signal, writing this mathematically we have,

14. Proportional and Derivative Controller
As the name suggests it is a combination of
proportional and a derivative controller the
output (also called the actuating signal) is equals
to the summation of proportional and derivative
of the error signal. Now let us analyze
proportional and derivative controller
mathematically. As we know in a proportional
and derivative controller output is directly
proportional to summation of proportional of
error and differentiation of the error signal,
writing this mathematically we have,

15. PID controller
The PID controller is the most widely used
type of process controller. When combined
into a single control loop the proportional,
integral and derivative modes complement
each other to reduce the system error to zero
faster than any other controller.

16. PID control

17. PID control
• PID control is a feedback control method that
combines proportional, integral, and
derivative actions.
• The proportional action provides smooth
control without hunting.
• The integral action automatically corrects
offset.
• The derivative action responds quickly to
large external disturbances.

18. Hunting

19. PID Action

20. PID Output Graph

21. What is tuning?
• Tuning is adjustment of control parameters to
the optimum values for the desired control
response. Stability is a basic requirement.
However, different systems have different
behaviour, different applications have
different requirements, and requirements
may conflict with one another.

22. Manual Tuning
• The operator estimates the tuning
parameters required to give the desired
controller response.
• The proportional, integral, and
derivative terms must be adjusted, or
tuned, individually to a particular
system using a trial-and-error method

23. Semiautomatic or Autotune
• The controller takes care of calculating and
setting PID parameters.
• Measures sensor output
• Calculates error, sum of error, rate of change
of error
• Calculates desired power with PID equations
• Updates control output

24. PID Tuning Software
• PID Tuning Software: There is some prepared
software that they can easily calculate the
gain parameter. Any kind of theoretical
methods can be selected in some these
methods.
Some Examples:
• MATLAB Simulink PID Controller Tuning,
• BESTune, Exper Tune etc.

25. Ziegler-Nichols Rules for Tuning PID Controllers
-Ziegler and Nichols proposed rules for determining values of the
proportional gain Kp, integral time Ti, and derivative time Td
based on the transient response characteristics of a given plant.
-Such determination of the parameters of PID controllers or
tuning of PID controllers can be made by engineers on-site by
experiments on the plant.
-Such rules suggest a set of values of Kp, Ti, and Td that will give a
stable operation of the system. However, the resulting system
may exhibit a large maximum overshoot in the step response,
which is unacceptable.
- We need series of fine tunings until an acceptable result is
obtained.

26. Ziegler-Nichols 1st Method of Tuning Rule
- We obtain experimentally the response of the plant to a
unit-step input, as shown in Figure 10-2.
-The plant involves neither integrator(s) nor dominant complex-
conjugate poles.
-This method applies if the response to a step input exhibits an S-
shaped curve.
- Such step-response curves may be generated
experimentally or from a dynamic simulation of the plant.
Figure 10-2 Unit-step response of a plant.

27. Figure 10-3 S-shaped response curve.
L = delay time
T = time constant

28. Transfer function:
Table 8-1 Ziegler–Nichols Tuning Rule Based on Step Response of Plant (First
Method)

29. Instructional Objectives
• At the end of the chapter, the students should
be able to
1) Understand the parameters of P, I,D and PID
control mode.
2) Explain t he concept of proportional band
and proportional gain.
3) Solve practical process and control
problems related to the closed cloop control
modes.
29

30. Proportional Control
Error
0
ControllerOutput
Proportional Band
Proportional Gain=Kp=100/PB
Output=Kp e
Transfer function for the controller
Gc(s)=Kp
30

31. System with Proportional controlControllerOutput
Time
0
Error
Time
0
Kp=100/PB
Iout= Kpe + Io
Change in output(s) =KpE(s)
Transfer Function= change
in output(s)/E(s)= Kp
31

32. Proportional Control
KpeSetpoifromOutputControllerinChange nt
errorinchangeKpSetpoifromOutputControllerinoChange 0
0
0
0 nt
BandoportionalKp Pr/100
)(
)(
sE
KpsoutputinChangeFunctionTransfer
IoKpeIout


32

33. Proportional Control Problems(Process)
• A proportional controller
has Kp of 20 and a set
point of 50% output. Its
output is to have a which
at the set point allows a
flow of 2.0 m3/s. the valves
change its output in direct
proportion to the controller
output. What will be the
controller output and the
offset error when the flow
has to be changed to
2.5m3/sec?
50%
62.5%
I0
2.0 m3/s
Solution:
By proportion:
a) Controller Output at the new
flow rate
50%/2.0m3/s =x%/2.5m3/s = 62.5%
Iout = 62.5%
2.5 m3/sIout
b) The offset error
Kp= change in output/E
E= change in output/Kp
= (62.5%-50%)/20
= 0.625%=0.63%
New Output
33

34. Proportional Control Problems
• A pneumatic proportional
controller is used to control
temperature within the range of
60 °C to 100 °C . The controller is
adjusted so that the output
pressure goes from 3 psi(valve to
fully open) to 15 psi(valve to fully
closed) as the measured
temperature goes from 71 °C to
75 °C with the set point held
constant. Find the gain and the
proportional band.
• Answer a) Gain= 3psi/ °C
b) PB= 10 %
Solution:
)s(E
Kp)s(outputinChangeFunctionTransfer 
b)PB=75-71/ 100-60=0.1x 100%=10%
a)15psi-3psi/75-71C
=3psi/C
34

35. Proportional Control Problem
• .Consider the proportional mode
level controller system in the
figure below. Valve A is a linear
with a flow scale factor of 10
m3/hr per percent controller
output.
• The controller output is nominally
50% with a proportional gain
Kp=10%.A load change occurs
when flow through valve B
changes from 500 m3/hr to 600
m3/hr .Calculate the new
controller output and the offset
error.
Figure
35

36. Solution
36

37. Derivative Controllers
• Derivative Control- the change in controller output
from the set point value is proportional to the rate of
change with time of the error signal.
dt
de
KdIoIout 
Transfer Function by
Laplace Transform
)())(( ssEKsIoIout D
Where Io = the setpoint output value
Iout=the output value that will occur when the error e is
changing at the rate de/dt
KD = the constant of proportionality and is commonly referred to
as the derivative time since it has units of time
KDs = transfer function
37

38. Derivative ControllersErrorControllerOutput
Time
Time
0
0
With derivative control, as soon as the error
signal begins to change there can be quite a
large controller output since it is proportional
to the rate of change of the error signal and
not its value. The controller output is
constant because the rate of change is
constant and occurs immediately as the
deviation occurs.
Derivative controllers do not respond to
steady state error signals, since with steady
error the rate of change of error with time is
zero. Because of this the derivative control is
often combined with proportional control.
Constant rate
of change of
error signal
with time
38

39. Sample 1: Derivative Control-Problem
A derivative controller
has a set point of
50% and derivative
constant KD= 0f 0.4
s. What will be the
controller output
when the error a)
changes at 1%/s, b)
is constant at 4%?
Solution:
dt
de
KdIoIout 
Iout= Kd de/dt + Io
= 0.4s (1%/s) + 50%
= 50.4%
a)
b) With de/dt = zero, then Iout=Io
that is 50%. The output only
differs from the setpoint value
when the error is changing.
Error is constant at 4%.
Derivative of a constant is
zero(0). Iout=50% 39

40. Integral Control
dt
I
K
t
Io)(Iout
0
e
Integral control – is one where the rate of change of the control
output is proportional to the input error signal e.
d /dt=KIeI
I
I
K is the constant of proportionality and, when the controller
output is expressed as a percentage and the error as a percentage,
has units of s -1 (per second). The reciprocal of KI is called the
integral time TI and is in seconds. Integrating the above equation
gives:
teI d
I
K
t
d
0
Iout
Io

Io is the controller output at
zero time
Iout= is the output at time t
40

41. • The transfer function is obtained by taking
Laplace transform. Thus
Integral Control
(s)EK
s
1
s)Io)(Iout I (
Transfer function = IK
s
1
41

42. Integral ControlControllerOutput
Time
0
The figure illustrates the action of an
integral controller when there is a
constant error input to the controller. We
can consider the graph in two ways.
When the controller output is constant,
that is , dP/dt is zero, the error is zero.
When the controller output varies at
constant rate, that is dP/dt is constant,
the error must have a constant value
Error
Time
0
Constant
percent error
value
Io
Area under the error
graph between t=0
and t
 edt
t
0
42

43. Thus up to the time when the error occurs,
the value of the integral is zero. Hence
Iout=Io. When the error occurs it
maintains a constant value.
Thus the area under the graph is increasing
as time increases. Since the area
increases at a constant rate, the
controller output increases at a constant
rate.
Integral Control
43

44. Integral Control-Problem
Consider an integral controller
with a value of KI =0.1/s and
an output of 40% at the set
point. What will be the
output after times a)1 s b)2
s, if there is a sudden
change to a constant error
of 20%.
Given
KI = 0.1/s
Io=40%
E=error(constant)=20%
Required:
a) Iout @ t=1s
b) Iout @ t=2s
edtKIoIout I
t

0
)(
When the error does not vary with
time, the equation becomes
IoetKIout
etKIoIout
I
I

 )(
a)t=1s; Iout=0.1x20x1 + 40= 42%
b)t=2s; Iout=0.1x20x2 + 40= 44%44

45. Integral Control Problem
An integral controller is used for speed control
with a set point of 12 rpm with a range of 10
to 15 rpm. The controller output is 22%
initially. The integral constant KI=-0.15%
controler output per second per percentage
error. If the speed jumps to 13.5% rpm,
calculate the out put after a) 2 seconds
b)4 seconds c)6 seconds for a constant ep.
45

46. Solution
Iout=p=-0.15/s x -30 %x 6 sec + 22= 49%
Iout=p=-0.15/s x -30 %x 4 sec + 22= 40%
46

47. PID Controller
Combining all three modes of
control(proportional, integral and derivative)
enables a controller to be produced which has
no offset error and reduces the tendency for
oscillations. Such a controller is known as
three mode controller or PID controller. The
equation describing its action is:
Io)
dt
de
KedtKKp(eIout DI
t
0
 
47

48. PID Controller
)(sEsK(s)EKpK
s
1
(s)KpE(s)Io)-(Iout DI 
Transfer function by Laplace
Transform
)DI sKK
s
1
(1KpFunctionTransfer 
Problem:
What will the controller output of a three
mode controller having Kp as 4, KI as
0.6/s, Kd as 0.5s, a set point output of
50% and subject to error change show in
the figure below, a) immediately the
change starts to occur from t=0 sec,
and b) 2 s after it starts.
1Error(%)
1 2Time ,s
48

49. PID Controller Problem
Io)
dt
de
KedtKKp(eIout DI
t
0
 
Solution:
a) For (a) e=0; de/dt= 1/s and
 edt
t
0
0
Thus: Iout= 4(0 + 0 + 0.5x 1) + 50 = 52 %
b) For (b) we have , at 2 sec, e=1%
1
Error(%)
1 2
Figure
Time ,s
 edt
t
0
1.5 s , since the integral is the area under the error-time graph up to
2 s, only 1.5 sec because of the half area covered in square 1
and de/dt= 0, for constant error rate change of 1 %
Iout=4( 1 + 0.6x 1.5 + 0) + 50= 57.6%
49

50. Summary of the Characteristics of the Most Commonly Used Controller
Modes
1. Two Position:
Inexpensive.
Extremely simple.
2. Proportional:
Simple.
Inherently stable when properly tuned.
Easy to tune.
Experiences offset at steady state. (OK for level control)
3. Proportional plus integral:
No offset.
Better dynamic response than reset alone.
Possibilities exist for instability due to lag
introduced.
4. Proportional plus derivative:
Stable.
Less offset than proportional alone (use of
higher gain possible).
Reduces lags, i.e., more rapid response.
5. Proportional plus integral plus derivative:
Most complex
Rapid response
No offset.
Best control if properly tuned.
50

51. PID Controllers Summary
• Proportional gain increases response speed, to
much gain causes system to ring.
• Integral gain kills steady-state error, wind-up
and/or too much gain can cause system to go
unstable.
• Derivative gain adds damping and stability, but
is sensitive to jitter and noise.
51

52. 52
PID Controllers Summary
• Proportional term responds immediately to the
current tracking error; it cannot achieve the desired
set point accuracy without an unacceptably large
gain. Needs the other terms.
• Derivative action reduces transient errors.
• Integral term yields zero steady-state error in
tracking a constant setpoint. It also rejects constant
disturbances.
Proportional-Integral-Derivative (PID)
control provides an efficient solution to
many real-world control problems


53. Assignment/Homework to be graded as
Project Report
PROBLEMS:
1)A proportional controller has Kp of 20 and a set point of 50% allows a flow of 8.0 m3/s . The valve changes its
output in direct proportion to the controller output. What will be the controller output and the offset
error when the flow has to be changed to 10 m3/s.
2) A derivative controller has a set point of 50% and a derivative constant Kd of 0.5s. The error starts at zero
and then changes at 2%/s for 3 s before becoming constant for 2 s, after which it decreases at 1%/sec to
zero. What will be the controller output at
a) 0 s b) 1 s c) 4 s d) 6 s
3) An integral controller has a set point of 50% and a value of KI of 0.1/s. the error starts at zero and changes
at 4%/s for 2 s before becoming constant for 3 s. What will be the output after times times of a) 1 s b) 3
s?
4) A three mode (PID) controller has Kp as 2, KI as 0.1/s, Kd as 1.0s, and a set point output of 50%. The error
starts at zero and changes at 5% for 2 s before becoming constant for 3s. It then decreases at 2%/s to zero
and remains at zero. What will be the controller output at
a) 0 s b) 3 s c) 7 s d) 11 s?
53

54. Assignment-Project No. 3
• A proportional controller is used to control the height of water
in a tank where the water level can vary from zero(0) to 4.0
meters. The required height of water is 3.5m and the
controller is fully close a valve when the water rises to 3.9 m
and fully open it when the water falls to 3.1m. What
proportional band and transfer function will be required? Ans.
a) PB=20% b) Kp=Transfer function=5
54

55. Assignment 2: Derivative Control-Problem
• A derivative controller has a set point of 50%
and derivative constant of Kd 0.5s. The error
starts at zero and then changes at 2%/s for 3
seconds before becoming constant for 2
seconds, after which it decreases at 1%/s to
zero. What will be the controller output a t
a) 0 s and b)1s c)4s d)6s
55

56. Assignment 3 Problem Integral
Control
• An integral controller has a set point of 50%
and a value of Ki of 0.1/s. The error starts at
zero and changes at 4%/s for 2 s before
becoming costant at 3 s, what will be the
ouput after times of a) 1 s b) 5s
56

57. Assignment 4. Problem PID Controller
A three mode (PID) controller has Kp as 2,
KI as 0.1/s, Kd as 1.0s, and a set point
output of 50%. The error starts at zero
and changes at 5% for 2 s before
becoming constant for 3s. It then
decreases at 2%/s to zero and remains at
zero. What will be the controller output
at
a) 0 s b) 3 s c) 7 s d) 10 s?
57