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Modern Control System EKT 318
• General Introduction • Introduction to Control System • Brief Review - Differential Equation - Laplace Transform
Course Assessment
• Lecture 3 hours per week
Number of units 3
• Final Examination 50 marks
• Class Test 1 10 marks
• Class Test 2 10 marks
• Mini Project 15 marks
• Assignment/Quiz 15 marks
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
1. Dr. Saidatul Norlyana Azemi
[email protected] 012 – 5104805
(8 – 5pm only!!)
2. Dr. Mijanur, [email protected] 3. Dr. Ali Amer
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 ? (contd….)
Process (Plant) to be controlled
Process with a controller
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
Differential Equation
0..... 012
2
21
1
1
adx
dya
dx
yda
dx
yda
dx
yda
n
n
nn
n
n
N-th order ordinary differential equation
Often required to describe physical system - represent their rates of change in the system
Higher order equations are difficult to solve directly.
But, can be solve through Laplace transform.
Example of Diff. Equation
o
i
edtiC
edtiC
Ridt
diL
1
1
Example of Diff. Equation (Contd…)
Newton’s second law:
2
2
dt
sdm
dt
ds
dt
dmF
dt
dvmF
maF
m = mass a = acceleration
a = acceleration a = changes of velocity over timedv/dt
velocity = speed / time v = ds/dt
Table 2.2 (continued) Summary of Governing Differential Equations for Ideal Elements
Laplace Transform
• A method for solving differential equations. • Convert differential equations in time (t) domain to
complex frequency (s) domain
Laplace Transform is given by:
frequency.complex a is Where, js
0
)()()]([ dtetfsFtfLst
Laplace Transform (contd…)
• Example: Consider the step function.
t
u(t)
1 0
-1
00
0
0
)()}({
)()}({
s
edte
dtetutuL
dtetftuL
stst
st
st
u(t) = 1 for t >= 0 u(t) = 0 for t < 0
s
s
1
101
(1)
(2)
(3)
(4)
(final answer)
Laplace Transform’s Pair table for common functions
Function, )(tf Laplace Transform
Unit Impulse, )(t 1
Unit step, )(tus
1
Unit ramp, t
Exponential, ate
Sine, tsin
Cosain, tcos
Damped sine, te at sin
Damped cosain, te at cos
Damped ramp, atet
2
1
s
as
1
22
s
22 s
s
22)(
as
22)(
as
as
2)(
1
as
Inverse Laplace Transform
• Transformation from s-domain back to t-domain
Inverse Laplace Transform is defined as:
j
j
stdsesFj
sFLtf
)(2
1)}({)( 1
Where, is a constant
The Laplace Transform of a function, f(t), is defined as;
0
)()()]([ dtetfsFtfLst
The Inverse Laplace Transform is defined by
j
j
tsdsesF
jtfsFL
)(2
1)()]([1
Eq A
Eq B
The Laplace Transform
Laplace Transform Pairs
• Laplace transform and its inverse are seldom calculated through equations.
• Almost always they are calculated using look-up tables.
http://lpsa.swarthmore.edu/LaplaceXform/FwdLaplace/LaplaceAll.html
Characteristic of Laplace Transform
Table: Properties of Laplace Transforms
Example 1
Consider a second order: )(42
2
tfdt
fd
Using differential property (from Laplace table)
1)](4[)]0(')0()()[( 2 sFfsfsFs
Rearrange : 22 2
2
2
1)(
ssF
Inverse Laplace : ttf 2sin5.0)(
Assume, initial conditions is 0.
1)( impulseunit t
Using linearity property (from Laplace table)
1)()4[( 2 sFs
)(44 sFf
Example 2
function impulse theis (t)
)(3422
2
tydt
dy
dt
yd
Assume, initial conditions is 0.
1]342)[(
1)(3)(4)(2
)(342
2
2
2
2
sssY
sYssYsYs
tydt
dy
dt
yd
Using differential property (from Laplace table)
342
1)( ,
2
sssYSo(1)
(2)
(3)
(4)
Example 2 (contd…)
22
22
2
2
0.7071)()1(
0.7071
0.70712
1
0.7071)()1(
1
2
1
5.0)1(
1
2
1
)2/32(2
1)(
s
s
s
sssY
Inverse Laplace step: 342
1)( ,
2
sssYSo
How to get this?? Use calculator!
Why we do this?? Because to resemble: tableLaplace from
)(
22
as
(5)
Example 2 (contd…)
0.7071 and 1 Where,
)( resembles This
22
a
as
From table, inverse Laplace transform is: te at sin
Thus the solution of the differential equation
)7071.0sin(7071.0)( )1( tety t
22 0.7071)()1(
0.70717071.0)(
ssY
Example 3
2342
2
ydt
dy
dt
yd Non zero initial condition
ssssssY
ssYssYsssYs
ssYssYsYsYs
ssYyssYdt
dysYsYs
dt
dyy
4234)(
2)(34)(4)(
/2)(3]4)(4[)]0()([
/2)(3)]0()([4])0()([
Transform Laplace Taking
0)0(,1)0(
223
223
2
2
(1)
Example 3 (contd…)
)3(
6/1
)1(
2/13/2:
6/1
2/1
3/2
)3()1(24
fraction partial :stepNext
2
sssso
C
B
A
s
C
s
B
s
Ass
)3)(1(
24
)34(
42)(
:simplify is stepnext
2
2
2
sss
ss
sss
sssY
Example 3 (contd…)
tt eety
ssssY
3
6
1
2
1
3
2)(
transformLaplace inverse:stepNext
)3(
6/1
)1(
2/13/2)(
obtain weexpansion,fraction partialThrough
Example 4
Show that )sin(t is a solution to
the following differential equation:
0)()(
2
2
tydt
tyd
(a)
Find solution to the above equation using Laplace transform with the following initial condition.
1(0) and 0)0( dt
dyy
Solution
)sin().1sin()(
1
1)(
1]1)[(
0)(1)(
0)()]0()0()([
obtain, weLaplace, Taking
2
2
2
2
ttty
ssY
ssY
sYsYs
sYdt
dysysYs
0)()(
2
2
tydt
tyd1(0) and 0)0(
dt
dyy
(1)
