Control System Engineering - BETR 3393

Chapter 1 : Introduction of Control Design

Mr. Mohd Hanif Che Hasanhanif.hasan@utem.edu.my

Learning Outcome

At the end of this lecture, you should be able to:

• Briefly know how control system engineering involves in daily life application.

• Recap fundamental control theory• Explain control system design step by step

procedure

2

Contents

1.1 INTRODUCTION

1.2 REVIEW ON CONTROL FUNDAMENTAL

1.3 EXAMPLES

1.4 DESIGN METHOD

What do these two have in Common?

o Highly nonlinear, complicated dynamics!o Both are capable of transporting goods and people over long distances

BUT

o One is controlled, and the other is not.o Control is “the hidden technology that you meet every day”o It heavily relies on the notion of “feedback”

CONTROL consists ofSensing , Computation , Actuation

5

SenseVehicle Speed

ComputeControl “Law”

ActuateGas Pedal

In Feedback “Loop”

Example of Control System Block Diagram in Vehicle Auto Cruise Control

Goals in Vehicle Speed Control

Stability: system maintains and hold steadily on desired speed operating point.

Performance: system have fast responds and rapidly changes once desired speed is set.

Robustness: system tolerates various disturbances dynamics (mass, drag, etc)

1.2 REVIEW ON CONTROL FUNDAMENTAL

1. Various cases of poles location for second order/approx. system

2. Four (4) cases of damping ratio and relation to poles position for second order/approx. system

3. Step responses for second-order system damping cases

4. Underdamped second order system characteristics

5. Steady state error6. 2nd order approximation

Various cases of poles location for 2nd order system

Damping ratio, ζ Poles location Step Response

Overdamped, ζ>1Matlab Command:>>y=zpk([],[-1 -3],3)>> step(y)

Critically damped, ζ=1Matlab Command:>>y=zpk([],[-2 -2],4)>> step(y)

Under damped, 1>ζ>0Matlab Command:>>y=zpk([],[-2+3i -2-3i],13)>> step(y)

Undamped, ζ=0Matlab Command:>>y=zpk([],[5i -5i],25)>> step(y)

Re

Im

Re

Im

Re

Im

Re

Im

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Step Response

Time (seconds)

Am

plit

ude

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Step Response

Time (seconds)

Am

plit

ude

0 1 2 3 4 5 6 7 80

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Step Response

Time (seconds)

Am

plit

ude

0 1 2 3 4 5 6 7 8-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

Step Response

Time (seconds)

Am

plit

ude

Four (4) cases of damping ratio and relation to poles position for 2nd order system

Damping ratio, ζ Case Location of poles

ζ>1 Overdamped Different locations on the real axis

ζ=1 Critically damped Overlapping on the real axis

1>ζ>0 Under damped Conjugated and complex

ζ=0 Undamped On the imaginary axis, with same magnitude

2

21 1, nn jss

1, 2

21 nnss

nss 21 ,

njss 21 ,

Step responses for second-order system damping cases

0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Step Response

Time (seconds)

Am

plit

ude

Undamped, ζ=0

Under damped, 1>ζ>0

Critically damped, ζ=1

Overdamped, ζ>1

2nd order systems specification

n

sT

4

21

n

PT

100)1/( 2

%X

eOS

)100/(%ln

)100/ln(%

22 OS

OS

General Formula Constants

With Position Constant,

With Velocity Constant,

With Acceleration Constant,

Steady state errorR(s) C(s)

+

-)(sG

E(s)

Closed loop negative feedback system

P

stepK

ee

1

1)(lim

0sGK

sp

v

rampK

ee1

)(lim0

ssGKs

v

a

parabolaK

ee1

)(lim 2

0sGsK

sa

2nd order approx. system

• System Response with Additional Poles

0 0.5 1 1.5 2 2.5 3 3.50

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (seconds)

Am

plit

ude

c1

c2

c3

Step response of system T1(s), system T2(s),and system T3(s)

)32)(32)(5(

65

)134)(5(

65

)32)(32)(10(

130

)134)(10(

130

)32)(32(

13

134

13

23

22

21

jsjsssssT

jsjsssssT

jsjsssT

1.3 EXAMPLES

14

1- DC MOTOR

2- HEAT TRANSFER

3- WATER TANK

0 0.05 0.1 0.15 0.2 0.25 0.3 0.350

5

10

15

20

25

Step Response

Time (seconds)

Am

plit

ude

0 0.5 1 1.5 2 2.5 3

x 104

0

1

2

3

4

5

6x 10

5 Step Response

Time (seconds)

Am

plit

ude

DC MOTOR

Motor Speed

Motor Position

25.179687.112

54.364212

ss

G

ss

sG

8.24

0546455.162191.4632

Example of Matlab command :ms_sys= tf(36421.54,[1 112.87 1796.25])step(ms_sys)

HEAT TRANSFER

Transfer function157

5.1

sG

0 50 100 150 200 250 300 3500

0.5

1

1.5

Step Response

Time (seconds)

Am

plit

ude

WATER TANK

Transfer functions

G39315.2491

004323.10

0 500 1000 15000

1

2

3

4

5

6

7

8

9

10

Step Response

Time (seconds)

Am

plit

ude

1.4 DESIGN METHOD

18

1- Design in the s-plane. (Classical method)

2- Design in the frequency domain. (Classical method)

3- Design in the State Variable Feedback Systems

Design in

the s-plane (Classical)

CONTROLLER DESIGN

19

Design in

the

frequency

domain

(Classical)

Design in

state

variables

Classical design: mainly use transfer function as a system model, design of simple controller using intuitive technique

Modern design: based on state space description, for complex system (MIMO, nonlinear), analytical design

• Root locus displayed both transient response and stability information • Changes in gain resulted changes in performance• Adding poles and/or zeros change the root locus and also the performance

Design of Control Systems

1. Modeling of System - model and behavior

2. Determine control goals (requirements)

3. Selection on control architecture (cascaded, feedback, etc.)

4. Set control gains (have various methods)

5. Are goals (requirements) are fulfilled?

6. Investigate sensitivity of performance to changes in system parameters

7. modify gains or architecture if needed

that’s all … TQNext Lecture : Chapter 2 Root Locus Technique

21

References

1] Norman S. Nise, Control Systems Engineering, 6th Edition, John Wiley & Sons Inc., 2011.

2] Richard C. Dort, Robert H. Bishop, Modern Control Systems, 12th Edition, Pearson, 2011.

3] Katsuhiko Ogata, Modern Control Engineering, 5th Edition, Pearson, 2010.