Modern Control System
EKT 308
• General Introduction
• Introduction to Control System
• Brief Review
- Differential Equation
- Laplace Transform
Course Assessment
• Lecture
Number of units
3 hours per week
3
•
•
•
•
•
50 marks
10 marks
10 marks
15 marks
15 marks
Final Examination
Class Test 1
Class Test 2
Mini Project
Assignment/Quiz
Course Outcomes
• CO1: : The ability to obtain the mathematical
model for electrical and mechanical systems and
solve state equations.
• CO2: : The ability to perform time domain
analysis with response to test inputs and to
determine the stability of the system.
• CO3: The ability to perform frequency domain
analysis of linear system and to evaluate its
stability using frequency domain methods.
• CO4: The ability to design lag, lead , lead-lag
compensators for linear control systems.
Lecturer
Dr. Md. Mijanur Rahman
mijanur@unimap.edu.my
016 6781633
Text Book References
• Dorf, Richard C., Bishop, Robert H., “Modern
Control Systems”, Pearson, Twelfth Edition, 2011
• Nise , Norman S. , “Control Systems Engineering”,
John Wiley and Sons , Fourth Edition, 2004.
• Kuo B.C., "Automatic Control Systems", Prentice
Hall, 8th Edition, 1995
• Ogata, K, "Modern Control Engineering"Prentice
Hall, 1999
• Stanley M. Shinners, “Advanced Modern Control
System Theory and Design”, John Wiley and Sons,
2nd Edition. 1998
What is a Control System ?
• A device or a set of devices
• Manages, commands, directs or
regulates the behavior of other
devices or systems.
What is a Control System ?
Process (Plant) to be controlled
Process with a controller
(contd….)
Examples
Examples (contd…)
Human
Control
System Control
Classification of Control Systems
Control systems are often classified as
• Open-loop Control System
• Closed-Loop Control Systems
Also called Feedback or
Automatic Control System
Open-Loop Control System
Day-to-day Examples
• Microwave oven set to operate for fixed time
• Washing machine set to operate on fixed
timed sequence.
No Feedback
Open-Loop Speed Control of Rotating Disk
For example, ceiling or table fan control
What is Feedback?
Feedback is a process whereby some
proportion of the output signal of a
system is passed (fed back) to the input.
This is often used to control the dynamic
behavior of the System
Closed-Loop Control System
• Utilizes feedback signal (measure of the
output)
• Forms closed loop
Example of Closed-Loop Control
System
Controller:
Driver
Actuator:
Steering
Mechanism
 The driver uses the difference
between the actual and the desired
direction to generate a controlled
adjustment of the steering wheel
Closed-Loop Speed Control of Rotating Disk
GPS Control
Satellite Control
Satellite Control (Contd…)
Servo Control
Introduction to Scilab
• Scilab
• Xcos
Differential Equation
N-th order ordinary differential equation
dny
d n 1 y
dy
an n  an 1 n 1  .....  a1  a0  0
dx
dx
dx
Often required to describe physical system
 Higher order equations are difficult to
solve directly.
 However, quite easy to solve through
Laplace transform.

Example of Diff. Equation
di
1
L  Ri   i dt  ei
dt
C
1
i dt  eo

C
Example of Diff. Equation (Contd…)
Newton’s second law:
F  ma
dv
F m
dt
2
d  ds 
d s
F m  m 2
dt  dt 
dt
Table 2.2 (continued) Summary of Governing Differential Equations for Ideal Elements
Laplace Transform
• A transformation from time (t) domain to
complex frequency (s) domain
 Laplace Transform is given by

F ( s) 
 f (t )e
 st
dt L{ f (t )}
0
Where, s    j is a complex frequency.
Laplace Transform (contd…)
• Example: Consider the step function.
u(t)
u(t) = 1 for t >= 0
u(t) = 0 for t < 0
1
0
t
-1

L{u (t )}   u (t )e  st dt
0


e 
1
1
  e dt  
   0  1 
s
s
  s 0
0
 st
 st
Inverse Laplace Transform
• Transformation from s-domain back to t-domain
Inverse Laplace Transform is defined as:
1
f (t )  L {F ( s)} 
1
  j
F ( s )e

2 j 
 j
Where,

is a constant
 st
ds
Laplace Transform Pairs
• Laplace transform and its inverse are seldom
calculated through equations.
• Almost always they are calculated using lookup tables.
Laplace Transform’s table for common functions
Function, f (t )
Laplace Transform
Unit Impulse,  (t )
1
Unit step,
1
s
u (t )
1
s2
Unit ramp, t
Exponential,
e at

s2   2
Sine, sin  t
Cosain,
s
s 2
cos  t
Damped sine,
Damped cosain,
Damped ramp,
1
sa
2
e  at sin  t
e  at cos  t
t e  at

( s  a) 2   2
sa
( s  a) 2   2
1
(s  a) 2
Characteristic of Laplace Transform
(1) Linear
If
a1 and
a2
are constant and F1 ( s)
and
F2 ( s)
are Laplace Transforms
La1 f1 (t )  a2 f 2 (t )  a1 F1 (s)  a2 F2 (s)
Characteristic of Laplace Transform (contd…)
(2) Differential Theorem
For higher order systems

 df (t ) 
 df (t )  st
L

e dt
 dt  0  dt 

Let
 u dv  u v   v du

where f  df dt
u  e st
and
dv  df dt dt
du  s.e  st dt
v  f (t )
 df (t ) 
L
 f (t )e  st

 dt 

0
( n2)
( n 1)

 d n f (t )  n
L n   s F (s)  s n1 f (0)  s n2 f (0) .....s f (0) f (0)
 dt 


  se  st f (t )dt
0
  f (0)  sF ( s)
Characteristic of Laplace Transform (contd…)
(3) Integration Theorem

Let

g (t )  f ( x) dx  F ( s)
0
f (t ) 
L
where
f ( 0)
dg
dt
 f (t )dt   F s(s)  f (s0)
is the initial value of the function.
(4) Initial value Theorem

 df (t )  st
 dt e dt  sF (s)  f (0)
0

Initial value means
s   , thus
t  0 and as the frequency is inversed of time, this implies that
0  lim sF (s)  f (0)
s 
Characteristic of Laplace Transform (contd…)
(5) Final value Theorem
In this respect
t   as
s  0 , gives
lim f (t )  lim sF ( s)
t 
s 0
Example1
Consider a second order
d2y
 4 y   (t )
dt 2
Using differential property and assume intial condition is zero
( s 2  4)Y (s)  1
Rearrangge
Y ( s) 
1 2
2 s 2  22
Inverse Lapalce
y(t )  0.5 sin 2 t
Example 2
d2y
dy
2 2  4  3 y   (t )
dt
dt
 (t) is the impulse function
Assume, 0 initial conditions.
Taking Laplace transform, we obtain
2s Y ( s)  4sY ( s)  3Y ( s)  1
2
1
So, Y ( s)  2
2s  4s  3
Example 2 (contd…)
1
Y (s) 
2
2( s  2 s  3 / 2)
1
1

2 ( s  1) 2  0.5
1
1

2 ( s  1) 2  (0.7071) 2
1
0.7071

2  0.7071 ( s  1) 2  (0.7071) 2
Example 2 (contd…)

This resembles
( s  a) 2   2
Where, a  1 and   0.7071
From table, inverse Laplace transform is
e  at sin  t
Thus the solution of the differential equation
y(t )  e t sin( 0.7071t )
Example 3
d2y
dy
 4  3y  2
2
dt
dx
Non zero initial condition
dy
y (0)  1,
(0)  0
dt
Taking Laplace Transform
[ s 2Y ( s )  sY (0)]  4[ sY ( s)  y (0)]  3Y ( s )  2 / s
s 3Y ( s )  s 2  4s 2Y ( s)  4s  3sY ( s)  2
2  s 2  4s
2  s 2  4s
s 2  4s  2
Y (s)  3


2
2
s  4s  3s s( s  4s  3) s ( s  1)( s  3)
Example 3 (contd…)
Further simplifica tion gives,
s ( s  4)  2
s4
2
Y ( s) 


s ( s  1)( s  3) ( s  1)( s  3) s( s  1)( s  3)
Through partial fraction expansion, we obtain
 3 / 2 1/ 2   1 1/ 3  2 / 3
Y ( s)  






s
 s 1 s  3   s 1 s  3
Taking inverse Laplace transform
 3  t 1  3 t    t 1  3t  2
y (t )   e  e    e  e  
2
3
2
 
 3
Example 4
(a)
Show that sin( t ) is a solution to
the following differential equation
2
d y (t )
 y (t )  0
2
dt
(b)
Find solution to the above equation using
Laplace transform with the following initial
condition.
dy
y (0)  0 and
(0)  1
dt
Solution
(a)
y  sin( t )
dy
 cos(t )
dt
d2y
  sin( t )   y
2
dt
d2y
y0
2
dt
Solution
(b)
Taking Laplace, we obtain,
dy
2
[ s Y ( s )  sy (0)  (0)]  Y ( s )  0
dt
s 2Y ( s )  1  Y ( s )  0
Y ( s )[ s 2  1]  1
1
Y (s)  2
s 1
y (t )  sin( 1.t )  sin( t )