Modern Control System EKT 308 General Introduction Introduction to Control System Brief Review - Differential Equation - Laplace Transform.

Modern Control SystemEKT 308

• 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

Dr. Md. Mijanur Rahman

[email protected]

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 ? (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 someproportion of the output signal of asystem is passed (fed back) to the input.This is often used to control the dynamicbehavior of the System

Closed-Loop Control System

• Utilizes feedback signal (measure of the output)

• Forms closed loop

Example of Closed-Loop Control System

Controller: DriverActuator: 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

0..... 011

1

1

adx

dya

dx

yda

dx

yda

n

n

nn

n

n

N-th order ordinary differential equation

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

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

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

)}({)()(0

tfLdtetfsF st

frequency.complex a is Where, js

Laplace Transform (contd…)

• Example: Consider the step function.

t

u(t)

1

0

-1

sss

edte

dtetutuL

stst

st

110

1

)()}({

00

0

u(t) = 1 for t >= 0u(t) = 0 for t < 0

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

Laplace Transform Pairs

• Laplace transform and its inverse are seldom calculated through equations.

• Almost always they are calculated using look-up tables.

Laplace Transform’s 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

Characteristic of Laplace Transform

(1) Linear

1a 2a )(1 sF )(2 sFIf and are constant and and

are Laplace Transforms

)()()()( 22112211 sFasFatfatfaL

Characteristic of Laplace Transform (contd…)

(2) Differential Theorem

0

)()(dte

dt

tdf

dt

tdfL st

Let duvvudvu

steu

and

dtdtdfdv

dtesdu st .

)(tfv

)()0(

)()()(

00

ssFf

dttfseetfdt

tdfL stst

For higher order systems

)1()2(21 )0()0(.....)0()0()(

)(

nnnnn

n

n

ffsfsfssFsdt

tfdL

where dtdff

Characteristic of Laplace Transform (contd…)

(3) Integration Theorem

Let)()()(

0

sFdxxftg

dt

dgtf )(

s

f

s

sFdttfL

)0()()(

where )0(f is the initial value of the function.

(4) Initial value Theorem

)0()()(

0

fssFdtedt

tdf st

Initial value means 0t

and as the frequency is inversed of time, this implies that

s , thus)0()(lim0 fssF

s

Characteristic of Laplace Transform (contd…)

(5) Final value Theorem

t 0sIn this respect as , gives

)(lim)(lim0

ssFtfst

Example1Consider a second order

)(42

2

tydt

yd

Using differential property and assume intial condition is zero

1)()4( 2 sYsRearrangge

22 2

2

2

1)(

ssY

Inverse Lapalce

tty 2sin5.0)(

Example 2

function impulse theis (t)

)(3422

2

tydt

dy

dt

yd

Assume, 0 initial conditions.

Taking Laplace transform, we obtain

342

1)( ,

1)(3)(4)(2

2

2

sssYSo

sYssYsYs

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

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()( tety t

Example 3234

2

2

ydx

dy

dt

yd

Non zero initial condition

)3)(1(

24

)34(

42

34

42)(

2)(34)(4)(

/2)(3)]0()([4)]0()([

Transform Laplace Taking

0)0(,1)0(

2

2

2

23

2

223

2

sss

ss

sss

ss

sss

sssY

ssYssYsssYs

ssYyssYsYsYs

dt

dyy

Example 3 (contd…)

3

2

3

1

2

1

2

3)(

transformLaplace inverse Taking

3/2

3

3/1

1

1

3

2/1

1

2/3)(

obtain weexpansion,fraction partialThrough

)3)(1(

2

)3)(1(

4

)3)(1(

2)4()(

gives,tion simplificaFurther

33

tttt eeeety

ssssssY

sssss

s

sss

sssY

Example 4

Show that )sin(t is a solution to

the following differential equation

0)()(

2

2

tydt

tyd

(a)

(b) Find solution to the above equation using Laplace transform with the following initial condition.

1(0) and 0)0( dt

dyy

Solution

(a)

0

)sin(

)cos(

)sin(

2

2

2

2

ydt

yd

ytdt

yd

tdt

dy

ty

Solution

(b)

)sin().1sin()( 1

1)(

1]1)[(

0)(1)(

0)()]0()0()([

obtain, weLaplace, Taking

2

2

2

2

tttys

sY

ssY

sYsYs

sYdt

dysysYs