1. The document describes the objectives and topics of an introductory control systems engineering course. It will introduce modeling, analysis, and design tools for control systems including digital control systems. 2. It provides an overview of what constitutes a control system including open and closed loop examples. The goal is to provide a desired system response by interconnecting system components. 3. Feedback control systems provide advantages like greater accuracy, less sensitivity to disturbances, and improved transient response and steady-state error which can be controlled by adjusting loop gain.

1. 1
EEE 333
Control System I

2. 2
COURSE OBJECTIVES
 To introduce students to the basics of control system
engineering.
 To introduce students to the different modeling, analysis
and design tools of control systems.
 To introduce students to the basics of digital control systems

3. 3
TEXTS AND REFERENCES
TEXT BOOK:
‘Control Systems Engineering’, by Norman S. Nise

4. 4
What is Control System?
A control system is an interconnection of components forming a system
configuration that will provide a desired system response.
Fig. 1: A system
Fig. 2: An open loop control system

5. 5
Fig. 3: A Closed-loop control system

6. 6
WELCOME TO THE ENDEAVOR

7. 1
Chapter 1: Introduction

8. 2
Control System Definition
 System: is anything/a process/a plant having specific output for a
specific input
 A control system consists of subsystems/processes/plants assembled for
the purpose of obtaining a desired output with desired performance,
given the desired output as the input to the control system.
Fig. 2: Simplified description of a control system
Fig. 1: A system/process

9. Example: An Elevator Movement
 Input: desired output; the push of the 4th-floor button; shown as a step
function in the figure
 Performance: can be seen from the response curve
 Transient response: Passenger comfort and patience are dependent
on it. Too fast ฀ uncomfortable; too slow ฀ patience is sacrificed
 Steady-state response: minimum possible steady-state error is
desirable 3

10. Advantages of Control System
4
1. Power amplification: high power needed to move a large object can be
obtained by a power amplifier built-in to the control system.
2. Remote Control: Control systems are useful in remote (e.g. space) or
dangerous (e.g. nuclear reactor) locations or inconvenient applications
(e.g. robots)
3. Convenience of input form: in the previous example, just a push of a
button (4 for 4th floor). Otherwise, ………
4. Compensation for disturbances: consider an antenna system that is to
point in a desired direction. Disturbances, e.g. wind, may force antenna
away from its correct position. An antenna control system shall
compensate for this kind of disturbances/noise. So precision and
repeatability will be improved.

11. 5
Two major configurations of control systems:
 Open loop
 Closed loop
 Advantages of Closed-loop Systems:
 Greater accuracy
 Less sensitive to noise, disturbances and changes in environment
 Transient response and steady-state error can be controlled more
conveniently, often by a simple adjustment of gain in the loop.
 Disadvantages:
 More complex and expensive than open-loop systems

12. 8
Feedback Control Systems: Some Examples
 Egyptian Water Clocks 1200 BC:
Time left is given by the amount of water left in the pot
Problem: Measurement is limited to TIME LEFT and by amount of water left in
pot
Solution: Measure the amount of water that comes out of the pot

13. Solution: Maintain a constant water level in the top pot
9
Measuring Time passed
Time passed is the amount of water in the lower pot
Problem: Water flow varies by amount of water in the top pot

14. 10
Fig: Level control system. A sight tube and operator’s eye form a sensor.
Problem: Manually relling the top pot is labor intensive and inaccurate.
Solution: Design an automatic control system

15. 11
Ctesibius 220-285 BC
• Father of pneumatics
• Created most accurate clock until Huygens (1657 AD)

16. 12
 Car Cruise Control:

17. 15
Control System Design
Case study: Antenna Azimuth Position Control System
This is an example of a system with
position control system

18. 16
Attempt 1: Manually through gears:
 Impossible; too slow and imprecise

19. 17
Attempt 2: Use a motor to turn the heavy antenna
Question: How to convert an angle command into an electrical signal?
Answer: Use a potentiometer to convert angle into voltage

20. 18
Use a POT to convert angle command into voltage
This is still an unrealistic arrangement
A large motor requires a lot of power. A POT cannot supply that.

21. 19
Use power amplifier to magnify POT output voltage
Still something missing! What?
Still there will be inaccuracies and disturbances (e.g. wind)
Answer: Error correction

22. The central theme: Feedback
• Use another gear/pot to get voltage proportional to output angle θo(t)
• Compare it with θi(t)
• Use the error [θi(t) - θo(t)] to drive the system, 20

23. 21
Functional block diagram:

24. 22
Analysis and Design Objectives
 Producing the desired transient response
 Reducing steady-state error
 Achieving stability:
 Output should not grow without bound
 Other considerations:
- Finances
- Robust design: System parameters may change over time. So
system performances also change over time. The engineer must
create a robust design so that the performances will not be sensitive
to parameter changes.

25. Summary of the Previous Lecture
1
 Advantages of control systems
 Power amplification
 Remote Control
 Convenience of input form
 Compensation for disturbances
 Control systems usually use feedback
 Control system design involves a number of steps
 Design objectives:
 Better transient response
 Decrease steady-state error
 Stability

26. 2
Homework (Chapter 1)
Book: ‘Control Systems Engineering’, by Norman S. Nise (6th Edition)
 Review Questions:
1 - 16
 Problems:
2, 5, 7, 17, 18

27. 3
Chapter 2
Modeling in the Frequency Domain

28. 4
 Recall the steps of control design:
Specification Functional block diagram Schematic 
mathematical modeling Reduction analysis/Design
 In this lecture, we will focus on:
Schematic Mathematical Modeling for transfer function

29. 5
 Finding a transfer function can consist of 3 steps:
1. Apply physical laws (Newton’ Laws, Hook’s Law, KCL, KVL etc.)
- Get differential equation
2. Linearize the equations (if not already linear)
3. Get transfer function
 Q: What’s wrong with differential equations? Or, why do we need
transfer function?
 Ans:
- System parameters (m, b, k), output x and input F are intermixed. Not
easy to find a ratio of output to input.
- so, not easy to get a block like this:

30. 6
Secondly, we would like to interconnect mathematical models easily.
- Difficult to do this with differential equation description
we need a convenient tool to solve differential equations.
- all these problems can be solved by using Laplace Transform.
 Laplace Transform (LT): We already know, right?

31. 7
 Inverse Laplace Transform (ILT)
u(t) is the unit step function.
 We will not use these formulas in this course. Instead, we will use
(i) ready-made table of standard functions (Table 2.1)
(ii) Laplace transform theorems (Table 2.2).
Exercise: Memorize Table 2.1 and Table 2.2

32. 8

33. 9

34. 11
Solution of Linear ODE’s Using LT
Example:
Take LT by using the following formula:
Also, assume the initial conditions are zero: y(0-)=0 and dy/dt(0-)=0

35. 12
Clearly, we can solve for y(t) by taking ILT
Taking ILT:
So,
yt12e4t
e8
t
ut

36. 13
Transfer Function
Form the ratio of the output transform C(s) to the input transform R(s)
Note that, the above equation separates the output, C(s), the input, R(s), and
the system, G(s)
G(s) Transfer Function
This multiplication can be represented by a system block diagram that we
initially aimed for:

37. 14
Note: Such a separation only exists if the original differential equation was
linear, constant coefficient (time-invariant) and ordinary differential
equation (ODE).
So, transfer function (TF) exists only for Linear Time-Invariant (LTI) system

38. 15
 Given G(s) and R(s), one can ILT to find the time response.
Example: Let G(s) 
1
s 2
, and let input, r(t)=u(t). Find c(t).
Sol:

39. Transfer Functions of Physical Systems
 Since we already know how to derive TF from ODEs, we only need to
learn how to write down ODE for a given system.
 TF of Electrical Network:
Example:
Applying KVL around the loop:
(1)
Putting these in equation (1): 16
We want an ODE involving v(t) and vc(t)

40. Now, take LT on both sides:
17

41.  Transfer functions of electrical circuit elements:
 Now define the ‘Impedance’ transfer function:
18

42. Eliminating I1(s), Find G(s) 
I2 (s)
V(s)
 Example: Inverting Operation amplifier
TF of the inverting OP-AMP is:

43. 2 1 2 1
HW:
We get:
G(s) (R 

1
)(C s 

1
) [
R2

C1
]

1
.
1
R C s
C2s R1 R1 C2 C2R1 s

44. After putting the parameter values:

45. TFs of Electromechanical Systems
From the book of Dorf, 8th Edition: Chapter 2, Example 2.4
Transfer Function of a DC Motor:
 Field-Controlled DC Motor:
Neglect: spring constant, K (no spring action)
Vf If ϕ Tm TL θ (or ω)
1

46. 1. Vf If
2. If ϕ
3. ϕ Tm
To have a linear system, one current must be held constant. For field-
controlled, ia(t) is constant (approx.).
In LT domain:
where, ia=Ia is a constant armature current, and Km is defined as the motor
constant=K1 Kf Ia
2

47. 3
4. Tm TL
Load Torque (TL) and Motor torque (Tm) are related by:
5. TL
θ
Disturbance torque
The Equation of Motion for the mechanical load:

48. 4
TF of the field-controlled motor-load combination is:

49. 5
The TF may be written alternatively:
where, field-circuit time-constant:
and, load time-constatn:
Typically, τ f << τ L, and often it may be neglected.

50. 6
 Example: Armature-Controlled DC Motor:
Uses the armature current, ia as the control variable and the field is kept
constant (e.g. permanent magnet)
Here, the transformation is:
VaIa Tm TL θ (or ω)
We got the following equations:
1. VaIa
here, Vb is the back emf as follows:

51. 7
2. Ia Tm
3. Tm TL
4. TL θ

52. 7
Chapter 5
Reduction of Multiple Subsystems

53. 8
Reduction of Block Diagrams (BD)
 Components of a block diagram:

54. 9
Three familiar forms:
(1) Cascade form
So, the equivalent TF is
Fig: Equivalent system

55. (2) Parallel form
So, the equivalent TF is
Fig: Equivalent system 10

56. 11
(3) Feedback Form
So, the equivalent TF is
Fig: Equivalent system
The product GH is called the open-loop TF or loop gain

57. 12
Now get the transfer function of the armature-controlled DC motor:
Final result:

58. Moving Blocks to Create Familiar Forms:
1. Moving a block to the LEFT past a summing junction:
2. Moving a block to the RIGHT past a summing unction:
13

59. 14
3. Moving a block to the LEFT past a pickoff point:
4. Moving a block to the RIGHT past a pickoff point:

60. 15
Block Diagram Reduction Via Familiar Forms
Example:
1. Start with the innermost easiest loop
2. Move the block/pickoff point/summing junction to create any of the 3
familiar forms
3. Find the equivalent TF of each loop/subsystem
4. Finally, find the equivalent TF of the whole system

61. 16
Moving the pickoff point to the RIGHT of the G4 block
Finding equivalent TF of the innermost loop
Sequentially reduce the blocks:

62. 6