1. Introduction to Control Systems
Historical perspective
Introduction to Feedback Control Systems
Closed loop system examples

2. Historical Perspective
13.7B BC Big Bang
13.4B Stars and galaxies form
5B Birth of our sun
3.8B Early life begins
700M First animals
200M Mammals evolve
65M Dinosaurs extinct
600K First Trace of humans

3. Feedback Control Systems emerge rather recently
1600 Drebbel Temperature regulator
1781 Pressure regulator for steam boilers
1765 Polzunov water level float regulator

4. Closed loop example: Polzunov’s Water level float regulator

5. Feedback Control Systems emerge rather recently
1600 Drebbel Temperature regulator
1681 Pressure regulator for steam boilers
1765 Polzunov water level float regulator
James Watt’s Steam Engine and Governor

6. Closed loop example: James Watt’s flyball governor

7. Open loop and closed loop control systems
Open Loop System

8. Open loop and closed loop control system models
Open Loop System
Closed Loop System

9. Example: Feedback in everyday life

10. Multivariable Control System Model

11. Multivariable Control System

12. Robotics
A robot is a programmable computer integrated with a machine

13. Example: Disk Drive

14. Example: Automatic Parking Control

15. Feedback Control: Benefits and cost

16. Feedback Control: Benefits and cost
Reduction of sensitivity to process parameters
Disturbance rejection
More precise control of process at lower cost
Performance and robustness not otherwise achievable

17. Feedback Control: Benefits and cost
Reduction of sensitivity to process parameters
Disturbance rejection
More precise control of process at lower cost
Performance and robustness not otherwise achievable
More mathematical sophistication
Large loop gain to provide substantial closed loop gain
Stabilizing closed loop system
Achieving proper transient and steady-state response

18. Ideal Control System Elements
Element Type
Physical Element
Describing Equation
Energy (E) or power (Þ)
Inductive storage
Electrical Inductance
v = L di/dt
E = (1/2) L i 2
Translational spring
dx/dt = (1/k) dF/dt
E = (1/2) F 2 / k
Rotational spring
ω = (1/k) dT/dt
E = (1/2) T 2 / k
Fluid inertia
P = I dQ/dt
E = (1/2) I Q 2
Capacitive storage
Electrical capacitance
i = C dv/dt
E = (1/2) C v 2
Translational mass
F = M d 2x/dt 2
E = (1/2) M ( dx/dt ) 2
Rotational mass
T = J dω /dt
E = (1/2) J ω 2
Fluid capacitance
Q = Cf d P/dt
E = (1/2) Cf P 2
Thermal capacitance
q = Ct dŦ/dt
E = Ct Ŧ 2
Energy dissipators
Electrical resistance
v = iR
Þ = i 2R = v 2 / R
Translational damper
F = b dx/dt
Þ = b ( dx/dt ) 2
Rotational damper
T = bω
Þ = bω 2
Fluid resistance
Q = ( 1/ Rf ) P
Þ = ( 1/ Rf ) P 2
Thermal resistance
q = ( 1/ Rt ) Ŧ
Þ = ( 1/ Rf ) Ŧ

19. Example: Mechanical system
Determine y(t)

20. Example: Mechanical system
Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0
r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2
r(t) = M d 2y(t)/dt 2 + b dy(t)/dt + k y(t) L2CCDE

21. Example: Mechanical system
Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0
r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2
r(t) = M d 2y(t)/dt 2 + b dy(t)/dt + k y(t)
Homogenous solution: yH (t) r(t) = 0
Particular solution: yp(t) = f [r(t)]
Total solution: y(t) = yH (t) + yp(t)

22. Example: Mechanical system
Assume the system is initially at rest: y(0-) = dy/dt (0-) = 0
r(t) – k y(t) – b dy(t) /dt = M d 2y(t)/dt 2
r(t) = M d 2y(t)/dt 2 + b dy(t)/dt + k y(t)

23. Example: Mechanical system
Assume the system is initially at rest: y(0-) = v(0-) = 0
r(t) – k ∫0-tv(τ)d τ – b v(t) = M dv(t) /dt
r(t) = M dv(t) / dt + b v(t) + k ∫0-tv(τ)d τ

24. Example: Electrical system
Determine v(t)
Assume circuit initially at rest. What does this mean?

25. Example: Electrical system
Write node equation
r(t) = (1/L) C dv(t)/dt + G v(t)
Homogenous solution vH (t) : r(t) = 0
Particular solution: vp(t) = f [r(t)]
Total solution: v(t) = vH (t) + vP (t) t ò
v(τ)d τ 0-

26. How to describe a plant/system mathematically?
if a dynamic system
Differential Equation
KVL

27. Consider a plant that can be described by an n-order differential equation
are constant coefficients, and
independent variable
dependent variable
There are 2 cases to be considered
(1) If q(t)=0  Homogeneous Case
(2) If q(t) ≠0  In-Homogeneous Case

28. Homogeneous Case Let a 2nd order differential equation be:
Define a differential operator D

29. Homogeneous Case (cont.)

30. Homogeneous Case (cont.)
Re-write
In general
If one can solve a first order D.E, then it can be used to solve nth order D.E

31. Homogeneous Case (cont.)
Solve a 1st order differential equation
General eqn.
Notice that:
Multiply both sides by eat

32. Homogeneous Case (cont)
Integrate both sides of equation
Zero Input Response
Zero State Response
due to initial condition
due to input

33. Homogeneous Case (cont.)
Consider
Initial condition

34. Homogeneous Case (cont.)

35. Homogeneous Case (cont.)
What are C1 & C2???
Based on the initial condition:

36. Homogenous (cont.)
Thus

37. Homogeneous Case (cont.)
Easiest method:
Find the roots of the characteristic equations
For second order system:

38. Homogeneous Case (cont.)
Therefore,
So how to find these values?
Based on Initial Condition

39. Homogeneous Case (cont.)
Example:
Consider again
We’ll have
Thus,
So,
By considering the initial conditions:
and
We’ll get:
Therefore:

40. 4 cases to be considered Case 1: Distinct real roots
Case 2: Equal roots & real
GENERAL SOLUTION
Case 3: Imaginary roots
Case 4: Complex conjugate roots

41. In-Homogeneous Case The solution: Particular solution
Homogeneous solution
Determine the particular solution is
the most challenging!!!
Homogeneous solution can be found by setting q(t)=0,
and then find the solution as in homogeneous case

42. Finding the particular solution, xP(t)
Undetermined Coefficient Method
Assumption:
Various derivative of q(t) have a finite of functional form
Functional forms:
(1)exponential (eat)
(2)sinosoidal (sin ωt, cos ωt)
(3) polynomial (tm)
(4)combinational of these function and their products (teat, eatsinwt)
Then guess:

43. In-Homogeneous Case (cont.)
Example:
Initial condition

44. Finding Xp(t)

45. In-Homogenous case (cont.)
So, we’ll have
Now, we need to find c1,c2, A & B
First, how to find A & B? Let us consider xp(t) again and substitute it into the original differential equation

46. Re-arrange:
A = ½
B = -3/2
A-3B = 5
B+3A = 0
So,

47. In-Homogenous case (cont.)
So,
What about c1 & c2?
Again, you must find them by considering the given INITIAL CONDITION

48. In Class exercise
In a group of two students, please solve the following problem within 10 minutes. Please submit them after right after 10 minutes ends. Intellectual discussion is highly encouraged.
Find the solution the given differential equation
provided that

49. Using Laplace Approach
Consider the previous example again:
Apply the Laplace transform to given diff. eqn
Simplify it:
X1(s)
X2(s)

50. Using Laplace Approach (cont.)
Partial fraction expansion:
and
Determine the values for A,B,C,D,E & F
Then,

51. Using Laplace Approach (cont.)
Finally, x(t) can be found by applying the inverse Laplace transform of X(s)

52. Laplace Transforms
Def:
Inverse:
Linearity:
Shifting Theorom:

53. The Laplace Transform The Laplace Transform of a unit step is:
Laplace Transform of the unit step.
The unit step is called a singularity function because it does not possess a derivative at t = 0. We also say that it is not defined at t = 0. Along this line, you will note that many text books define the Laplace by using a lower limit of 0- rather than absolute zero. One must ask, if you integrate from 0 to infinity for a unit step, how are you handling the lower limit when the step is not defined at t = 0? We will not get worked up over this. You can take this apart later after you become experts in transforms.
The Laplace Transform of a unit step is:
*notes

54. The Laplace Transform (t – t0) = 0 t
The Laplace transform of a unit impulse:
Pictorially, the unit impulse appears as follows:
f(t)
(t – t0) t0
I remember when I was first introduced to the unit impulse function, in an electrical engineering course at Auburn University. Essentially, we were told that the unit impulse function was a function that was infinitely high and had zero width but the area under the function was 1. Such an explanation would probably always disturb a mathematician because I expect the proper way to define the function is in a limiting process as a function that is 1/delta in height and delta wide and we take the limit as delta goes to zero.
At this point, the best thing to do is not to worry about it. Accept it as we have defined it here and go ahead and use it. Probably one day in a higher level math course you may give more attention to its attributes.
Mathematically:
(t – t0) = 0 t
*note

55. The Laplace Transform The Laplace transform of a unit impulse:
An important property of the unit impulse is a sifting
or sampling property. The following is an important.

56. The Laplace Transform The Laplace transform of a unit impulse:
In particular, if we let f(t) = (t) and take the Laplace

57. The Laplace Transform An important point to remember:
The above is a statement that f(t) and F(s) are
transform pairs. What this means is that for
each f(t) there is a unique F(s) and for each F(s)
there is a unique f(t). If we can remember the
Pair relationships between approximately 10 of the
Laplace transform pairs we can go a long way.

58. The Laplace Transform
Building transform pairs:
A transform
pair

59. The Laplace Transform Building transform pairs: u = t dv = e-stdt
A transform
pair

60. The Laplace Transform
Building transform pairs:
A transform
pair

61. The Laplace Transform
Time Shift

62. The Laplace Transform
Frequency Shift

63. The Laplace Transform Example: Using Frequency Shift
Find the L[e-atcos(wt)]
In this case, f(t) = cos(wt) so,

64. The Laplace Transform
Time Integration:
The property is:

65. The Laplace Transform Time Integration:
Making these substitutions and carrying out
The integration shows that

66. The Laplace Transform Time Differentiation:
If the L[f(t)] = F(s), we want to show:
Integrate by parts:
As noted earlier, some text will use 0- on the lower part of the defining integral
of the Laplace transform. When this is done, f(0) above will become f(0-).
*note

67. The Laplace Transform Time Differentiation:
Making the previous substitutions gives,
So we have shown:

68. The Laplace Transform Time Differentiation:
We can extend the previous to show;

69. The Laplace Transform
Transform Pairs:
f(t) F(s)

70. The Laplace Transform
Transform Pairs:
f(t) F(s)

71. The Laplace Transform
Transform Pairs:
f(t) F(s)
Yes !

72. The Laplace Transform
Common Transform Properties:
f(t)
F(s)

73. The Laplace Transform Using Matlab with Laplace transform: syms t,s
Example
Use Matlab to find the transform of
The following is written in italic to indicate Matlab code
syms t,s
laplace(t*exp(-4*t),t,s)
ans =
1/(s+4)^2

74. The Laplace Transform Using Matlab with Laplace transform: syms s t
Example
Use Matlab to find the inverse transform of
syms s t
ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))
ans =
-exp(-3*t)+2*exp(-3*t)*cos(3*t)

75. The Laplace Transform Theorem: Initial Value Theorem: Initial Value
If the function f(t) and its first derivative are Laplace transformable and f(t)
Has the Laplace transform F(s), and the exists, then
Initial Value
Theorem
The utility of this theorem lies in not having to take the inverse of F(s)
in order to find out the initial condition in the time domain. This is
particularly useful in circuits and systems.

76. The Laplace Transform
Example:
Initial Value
Theorem:
Given;
Find f(0)

77. The Laplace Transform Theorem: Final Value Theorem: Final Value
If the function f(t) and its first derivative are Laplace transformable and f(t)
has the Laplace transform F(s), and the exists, then
Final Value
Theorem
Again, the utility of this theorem lies in not having to take the inverse
of F(s) in order to find out the final value of f(t) in the time domain.
This is particularly useful in circuits and systems.

78. The Laplace Transform
Example:
Final Value
Theorem:
Given:
Find .

79. List of Laplace Transforms
f(t)
L(f) 1
1/s 7
cos t 2 t
1/s2 8
sin t 3 t2
2!/s3 9
cosh at 4 tn
(n=0, 1,…) 10
sinh at 5 ta
(a positive) 11
eat cos t 6
eat 12
eat sin t

80. Review: Partial Fractions
Case I: unrepeated real factor
Case II: repeated real factor
Case III: unrepeated complex factor
Case IV: repeated complex factor

81. Transform of Derivatives
THEORM 1
Laplace of f(t) exists
f’(t) exists and piecewise continuous for t>=0
THEOREM 2

82. Differential Equations
1st step
2nd step
3rd step

83. Transform of Integrals

84. Unit Step Function

85. t-Shifting

86. Applications – t-shifting
THEOREM

87. Dirac Delta (unit impulse) Function
Impulse of f(t), for
“Generalized function”
Laplace transform

88. Convolution 1. Solve 2. Calculate integral Definition Property
Application
1. Solve
2. Calculate integral

89. Solution of Partial Fraction Expansion
The solution of each distinct (non-multiple) root, real or complex uses a two step process.
The first step in evaluating the constant is to multiply both sides of the equation by the factor in the denominator of the constant you wish to find.
The second step is to replace s on both sides of the equation by the root of the factor by which you multiplied in step 1

91. The partial fraction expansion is:

92. The inverse Laplace transform is found from the functional table pairs to be:

93. Repeated Roots Any unrepeated roots are found as before.
The constants of the repeated roots (s-a)m are found by first breaking the quotient into a partial fraction expansion with descending powers from m to 0:

94. The constants are found using one of the following:

96. The partial fraction expansion yields:

97. The inverse Laplace transform derived from the functional
table pairs yields:

98. A Second Method for Repeated Roots
Equating like terms:

99. Thus

100. Another Method for Repeated Roots
As before, we can solve for K2 in the usual manner.

102. Unrepeated Complex Roots
Unrepeated complex roots are solved similar to the process for unrepeated real roots. That is you multiply by one of the denominator terms in the partial fraction and solve for the appropriate constant.
Once you have found one of the constants, the other constant is simply the complex conjugate.

103. Complex Unrepeated Roots