Automatic control theory.ppt

7/27/2019 Automatic control theory.ppt

1/53

Automatic control theory

A Courseused for analyzing and

designing a automatic control system


2/53

Chapter 1 I ntroduction

Figure 1.1

* Operating principle

* Feedback control

1) A water-level control system

21 centuryinformation age, cybernetics(control theory), system

approach and information theory , three science theory mainstay(supports)in 21 century.

1.1 Automatic control

A machine(or system) work by machine-self, not by manual operation.

1.2 Automatic control systems

1.2.1 examples


3/53


Wat er exi t

wat er

ent rance

f l oat

l ever

Fi gure 1. 2

* Operating

principle* Feedback

control

2) A temperature Control system(shown in Fig.1.3)

M

+

e

ua=k( u

r- u

f)

ur

uf

ampl i f i er

t hermo

met er

Gear

assembl y

cont ai ner

Fi gure 1. 3

* Operating principle

* Feedback

control(error)

Another example of the water-level

control is shown in figure 1.2.


4/53


3) A DC-Motor control system

M

M

+

-

+

r egul at or

t r i gger

rect i f i er

DC

mot or

t echomet er

l oad

e

Uf( Feedback)

ur

Fi g. 1. 4

ua

Uk=k( u

r- u

f)

* Principle

* Feedback control(error)


5/53


4) A servo (following) control system

ser vopot ent i omet er

M

+

-

I nput

Tr

out put

Tc

ser vomechani sm

ser vo mot or

ser vomodul at or

l oad

* principle

* feedback(error)

Fig. 1.5


6/53


* principle

* feedback(error)

gover nment

( Fami l y pl anni ng commi t t ee)

census

soci et y

excess

procreat e

Desi re

popul at i on popul at i on+

- Pol i cy orstatutes

Fig. 1.6

5) A feedback control system model of the family planning

(similar to the social, economic, and political realm(sphere or field))


7/53


x2

x3

Si gnal

( var i abl e)xxx

Component s

( devi ces)

+

-

+x1 eAdders ( compar i son)

e=x1+x

3- x

2

x

Fig. 1.7

Example:

1.2.2 block diagram of control systems

The block diagram description for a control system :Convenience


8/53


amplifier Motor Gearing Valve

Actuator

Water

container

Processcontroller

Float

measurement

(Sensor)

Error

Feedback

signal

resistance comparator

Desiredwater level

Input

Actual

water level

Output

Fig. 1.8

For the Fig.1.1, The

water level control

system:

Figure 1.1


9/53


For the Fig. 1.4, The DC-Motor control system

Desi red

rot ate speedn

Regul at or Tri gger Rect i f i er DCmot or

Techomet er

Actuator

Processcont rol l er

measurement ( Sensor)

comparat or

Act ual

rot ate speedn

Error

Feedback si gnal

Reference

i nput ur

Output n

Fig. 1.9

auk

ua

uf

e


10/53


1.2.3 Fundamental structure of control systems

1) Open loop control systems

Cont rol l er Act uat or Process

Di st urbance

( Noi se)

I nput r(t)

Ref erence

desi red out put

Out put c( t )

( act ual out put )

Cont rol

si gnal

Act uat i ng

si gnal

uk

uact

Fi g. 1. 10

Features: Only there is a forward action from the input to the

output.


11/53


2) Closed loop (feedback) control systems

Cont rol l er Act uat or Process

Di sturbance

( Noi se)

I nput r( t )

Ref erence

desi red out put

Out put c( t )

( act ual out put )

Cont rol

si gnal

Actuat i ng

si gnal

uk

uact

Fi g. 1. 11

measur ementFeedback si gnal b( t )

+-

( +)

e( t ) =

r ( t ) - b( t )

Features:

1) measuring the output (controlled variable) . 2) Feedback.

not only there is a forward action , also a backward actionbetween the output and the input (measuring the output and

comparing it with the input).


12/53


Notes: 1) Positive feedback; 2) Negative feedbackFeedback.

1.3 types of control systems

1) linear systems versus Nonlinear systems.

2) Time-invariant systems vs. Time-varying systems.

3) Continuous systems vs. Discrete (data) systems.

4) Constant input modulation vs. Servo control systems.

1.4 Basic performance requirements of control systems

1) Stability.

2) Accuracy (steady state performance).

3) Rapidness (instantaneous characteristic).


13/53


1.5 An outline of this text

1) Three parts:mathematical modeling; performance analysis;

compensation (design).

2) Three types of systems:l inear continuous; nonlinear continuous; l inear discrete.

3) three performances:stability, accuracy, rapidness.

in all: to discuss the theoretical approaches of the control

system analysis and design.

1.6 Control system design process

shown in F ig.1.12


14/53


1. Establish control goals

2. Identify the variables to control

3. Write the specifications

for the variables

4. Establish the system configuration

Identify the actuator

5.Obtain a model of the process,

the actuator and the sensor

6.Describe a controller and select

key parameters to be adjusted

7. Optimize the parameters and

analyze the performance

Performance does not

Meet the specifications

Finalize the design

Performance

meet the

specifications

Fig.1.12


15/53


1.7 Sequential design example: disk drive read system

Actuator

motor

Arm

SpindleTrack a

Track b

Head slider

Rotation

of arm Disk

Fig.1.13 A disk drive read system

A disk drive read system Shown in F ig.1.13

Configuration Principle


16/53


Sequential design:

here we are concerned with the design steps 1,2,3, and 4 of F ig.1.12.

(1) Identify the control goal:

(2) Identify the variables to control:

Position the reader head to read the date stored on a track on the disk.

the position of the read head.

(3) Write the initial specification for the variables:

The disk rotates at a speed of between 1800 and 7200 rpm and the read head

flies above the disk at a distance of less than 100 nm.

The initial specification for the position accuracy to be controlled:

1 m (leas than 1 m ) and to be able to move the head from track a to track b

within 50 ms, if possible.


17/53


(4) Establish an initial system configuration:It is obvious : we should propose a closed loop system, not

a open loop system.

An initial system configuration can be shown as in Fig.1.13.

Control

device

Actuator

motor

Read

arm

sensor

Desiredhead

position

error Actualhead

position

Fig.1.13 system configuration for disk drive

We will consider the design of the disk dr ive fur ther in the after -

mentioned chapters.


18/53


Exercise: E1.6, P1.3, P1.13


19/53

Chapter 2 mathematical models of systems

2.1 Introduction

Controller Actuator Process

Disturbance

Input r(t)

desired output

temperature

Output T(t)

actual

output

temperature

Control

signal

Actuating

signal

uk uac

Fig. 2.1

temperature

measurement

Feedback signalb(t)

+

-()

e(t)=

r(t)-b(t)

1) Easy to discuss the full possible types of the control systemsin terms of thesystems mathematical characteristics.

2) The basisanalyzing or designing the control systems.

For example, we design a temperature Control system :

The keydesigning the controller how produce uk.

2.1.1 Why


20/53


2.1.3 How get1) theoretical approaches 2) experimental approaches

3) discrimination learning

2.1.2 What is Mathematical models of the control systemsthe mathematical

relationships between the systems variables.

Different characteristic of the processdifferentuk:

T(t)

uk

T1

T2

uk12uk11

uk21

For T1

12

11

k

k

u

u

For T1

22

21

k

k

u

u


21/53


2.2.1 Examples

2.2 Input-output description of the physical systemsdifferential

equations

2.1.4 types

1) Differential equations2) Transfer function

3) Block diagramsignal flow graph

4) State variables(modern control theory)

The input-output descriptiondescription of the mathematical

relationship between the output variable and the input variable of the

physical systems.


22/53


ur

uc

R L

C

i

define: input ur output ucwe have

rccc

crc

uu

dt

duRC

dt

udLC

dt

duCiuudt

diLRi

2

2

rccc uu

dt

duT

dt

udTTT

R

LTRCmake 12

2

2121:

Example 2.1 : A passive circuit


23/53


Example 2.2 : A mechanism

y

k

f

F

m

Define: input Foutput y. We have:

Fkydt

dyf

dt

ydm

td

ydm

dt

dyfkyF

2

2

2

2

Fk

ydt

dyT

dt

ydTThavewe

Tf

m

Tk

f

makeweIf

1:

:

12

2

21

2,1

Compare with example 2.1: ucy; urF analogous systems


24/53


Example 2.3 : An operational amplifier (Op-amp) circuit

ur uc

R1

C

R2

R4

R1

R3

i3

i1

i2

+-

Input ur output uc

)3........(....................).........(1

)2...(........................................

)1)......(()(1

223

3

112

2342333

iRuR

i

R

uii

iiRdtiiC

iRu

c

r

c

(2)(3); (2)(1); (3)(1)

r

r

CRRR

RR

R

RR

c

c

CR udt

du

udt

du

)( 432

32

4 1

32

)(:

)(;;: 432

32

1

324

r

r

c

c

udt

du

kudt

du

Thavewe

CRRR

RRk

R

RRTCRmake


25/53

Chapter 2 mathematical models of systemsExample 2.4 : A DC motor

ua

w1

Ra La

ia

M

w3

w2 ( J

3, f

3)

( J1

, f1

)

( J2

, f2

)

Mf

i 1i

2Input ua output 1

)4.....(

)3.....(....................)2.....(....................

)1....(

11

1

fdt

dJMM

CEiCM

uEiRdt

diL

ea

am

aaaaa

a

(4)(2)(1) and (3)(1):

MCC

RM

CC

Lu

C

CCfR

CCJR

CCfL

CCJL

me

a

me

aa

e

mea

mea

mea

mea

1

)1()( 111


26/53


):(

..........................

......

......

:

321211

21

22

21

321

21

2221

3

21

21

iiifromderivedbecan

torqueequivalent

ii

MM

ntcoefficiefrictionequivalentii

f

i

fff

inertiaofmomentequivalent

ii

J

i

JJJ

here

f

Make:

constant-timeelectricfriction

CC

fRT

constant-timeelectric-mechanicalCC

JRT

constant-timemagnetic-electricR

LT

me

af

me

am

a

ae

-.......

.......

............


27/53


a

e

mme uCdt

dT

dt

dTT 12

2

Assume the motor idle: Mf= 0, and neglect the friction: f= 0,

we have:

)(11

)1()( 111

MTMTT

J

u

C

TTTTTT

mmeae

fmfeme

The differential equation description of the DC motor is:


28/53


Example 2.5 : A DC-Motor control system

+

t r i ggerUf

ur

-M

M

+

-rect i f i er

DC

mot or

t echomet er

l oad

ua-

uk

R3

R1

R1

R2 R3

w

Input urOutput ; neglect the friction:

(4)MTMTT

J

u

Cdt

dT

dt

dTT

(3)uku(2)u

(1)uukuuR

Ru

mmeae

mme

kaf

frfrk

)......(11

...........................................

..................................)......()(

2

2

2

112


29/53


2134we have

)(1

)1( 211

212

2

MMTJ

Tu

Ckkkk

dt

dT

dt

dTT e

mr

eCmme e

2.2.2 steps to obtain the input-output description (differential

equation) of control systems

1) Determine the output and input variables of the control systems.

2) Write the differential equations of each systems components in

terms of the physical laws of the components.* necessary assumption and neglect.

* proper approximation.


30/53


2.2.3 General form of the input-output equation of the linear

control systemsA nth-order differential equation:

mnrbrbrbrbrb

yayayayay

mmmmm

nnnnn

.........)1(1)2(

2)1(

1)(

0

)1(1

)2(2

)1(1

)(

3) dispel the intermediate(across) variables to get the input-output

description which only contains the output and input variables.

4) Formalize the input-output equation to be the standard form:

Input variableon the right of the input-output equation .

Output variableon the left of the input-output equation.Writing polynomialaccording to the falling-power order.

Suppose: input routput y


31/53


2.3 Linearization of the nonlinear components2.3.1 what is nonlinearity

The output is not linearly vary with the linear variation of the

systems (or components) input nonlinear systems (or

components).2.3.2 How do the linearizationSuppose: y = f(r)

The Taylor series expansion about the operating point r0 is:

))(()(

)(!3

)()(!2

)())(()()(

00)1(

0

30

0)3(20

0)2(00

)1(0

rrrfrf

rrrfrrrfrrrfrfrf

00 :)()(: rrrandrfrfymake

equationionlinearizatrrfywehave ............)(: 0'


32/53


Examples:

Example 2.6 : Elasticity equation kxxF )(

25.0;1.1;65.12:suppose 0 xpointoperatingk

11.1225.01.165.12)()( 1.00'1' xFxkxF

equationionlinearizatxF

xxxFxF

..............11.12:isthat

)(11.12)()(:havewe 00

Example 2.7 : Fluxograph equation

pkpQ )(

QFlux; ppressure difference


33/53


equationionlinearizatpp

kQ

p

kpQbecause

...........2

:thus

2)(':

0

2.4 Transfer function

Another form of the input-output(external) description of control

systems, different from the differential equations.

2.4.1 definitionTransfer function:The ratio of the Laplace transform of the

output variable to the Laplace transform of the input variable,with

all initial condition assumed to be zero and for the linear systems,

that is:


34/53


)()()(

sR

sCsG

C(s)Laplace transform of the output variable

R(s)Laplace transform of the input variable

G(s)transfer function

* Only for the linear and stationary(constant parameter) systems.

* Zero initial conditions.

* Dependent on the configuration and the coefficients of thesystems, independent on the input and output variables.

2.4.2 How to obtain the transfer function of a system

1) If the impulse responseg(t) is known

Notes:


35/53


)()( tgLsG

1)()()(if,)(

)()( sRttr

sR

sCsG

Because:

We have:

Then:

Example 2.8 :)2(

)5(2

2

35)(35)(

2

ss

s

sssGetg

t

2) If the output response c(t) and the input r(t) are known

We have: )(

)()(

trL

tcLsG

)()()( tgLsCsG


36/53


Example 2.9:

responseUnit step

sssssCetc

functionUnit stepsttr

t

.........

)3(

3

3

11)(1)(

........

1

R(s))(1)(

3

Then:

3

3

1

)3(3

)(

)()(

ss

ss

sR

sCsG

3) If the input-output differential equation is knownAssume: zero initial conditions;

Make: Laplace transform of the differential equation;

Deduce: G(s)=C(s)/R(s).


37/53


Example 2.10:

432

65

)(6)(5)(4)(3)(2

)(6)(5)(4)(3)(2

2

2

ss

s

R(s)

C(s)G(s)

sRssRsCssCsCs

trtrtctctc

4) For a circuit

* Transform a circuit into a operator circuit.

*Deduce the C(s)/R(s) in terms of thecircuits theory.


38/53


Example 2.11: For a electric circuit:

ucur C1 C2

R1

R2

uc( s )

1/ C1s 1/ C2s

R1

R2

ur( s )

2112222111

r

c

r

rc

CR; TCR; TCRT

sTTTsTTsU

sUsG

sUsTTTsTT

sCR

sCsU

sCR

sCR

sCR

sCsU

:here

1)(

1

)(

)()(

)(1)(

1

1

1

)(

)1

(//1

)1(//1

)(

12212

21

12212

21

22

2

22

11

22

1


39/53


Example 2.12: For a op-amp circuit

ur u

c

R1

R2

R1

+

-

C R2 1/ Cs

ur u

c

R1

R1

+

-

......;:here

.................)1

1(

11

)(

)()(

21

2

1

2

1

2

ntime constaIntegral tCR

R

Rk

ller.PI-Contros

k

CsR

CsR

R

sCR

sU

sUsG

r

c


40/53


5) For a control system

Write the differential equations of the control system, and Assume

zero initial conditions;

Make Laplace transformation, transform the differential equations

into the relevant algebraic equations;

Deduce: G(s)=C(s)/R(s).Example 2.13

+

t r i gger

Uf

ur -

M

M

+

-rect i f i er

DC

mot or

t echomet er

l oad

ua-

uk

R3

R1

R1

R2 R

3

w

the DC-Motor control system in Example 2.5


41/53


In Example 2.5, we have written down the differential equations

as:

(4)MMTJ

Tu

Cdt

dT

dt

dTT

(3)uku(2)u

(1)uukuuR

Ru

em

ae

mme

kaf

frfrk

)......(1

.......................................

.........................).........()(

2

2

2

11

2

Make Laplace transformation, we have:

(4)sMJ

TsTTsU

CessTsTT

(3)sUksU(2)ssU

(1)sUsUksU

mmeamme

kaf

frk

)......()(1

)()1(

.....).........()(......).........()(

...........................................)]........()([)(

2

2

1


42/53


(2)(1)(3)(4), we have:

)()(1

)()]1

1([ 21212 sM

J

TsTTsU

Ckks

CkksTsTT mmer

eemme

-......

-...........:

constanttimeelectricmechanicalCC

JRT

constanttimemagneticelectricR

L

There

me

am

a

ae

)1

1(

1

)(

)()(

212

21

emme

e

r

CkksTsTT

Ckk

sU

ssG


43/53


2.5 Transfer function of the typical elements of linear systems

A linear system can be regarded as the composing of several

typical elements, which are:

2.5.1 Proportioning element

Relationship between the input and output variables:)()( tkrtc

Transfer function: ksR

sCsG

)(

)()(

Block diagram representation and unit step response:R( s) C( s)

k

1k

t

r ( t ) C( t )

t

Examples:

amplifier, gear train,

tachometer


44/53


2.5.2 Integrating element

Relationship between the input and output variables:

constanttimeintegralTdttrT

tc I

t

I

:..........)(1

)(

0

Transfer function:sTsR

sCsGI1

)()()(

Block diagram representation and unit step response:

1

R( s) C( s)

1

t

r ( t ) C( t )

t

sTI

1

TI

Examples:

Integrating circuit, integrating

motor, integrating wheel


45/53


2.5.3 Differentiating element


dt

tdrTtc D

)()(

Transfer function: sTsRsCsG D )()()(


Examples:differentiating amplifier, differential

valve, differential condenser

R( s) C( s)T

Ds

1 TD

t

r ( t ) C( t )

t


46/53

2.5.4 Inertial element



)()()(

tkrtcdt

tdcT

Transfer function:1)(

)()( Tsk

sRsCsG


Examples:inertia wheel, inertial load (such as

temperature system)1

R( s) C( s)

k

t

r ( t ) C( t )

t

T

1Tsk


47/53


2.5.5 Oscillating element


10)()()(

2)(

2

22 tkrtc

dt

tdcT

dt

tcdT

Transfer function: 1012)(

)(

)( 22

TssT

k

sR

sC

sG


Examples:

oscillator, oscillating table,

oscillating circuit

R( s) C( s)

12

1

22 TssT C( t )

k

t

1

t

r ( t )


48/53

2.5.6 Delay element



)()( tkrtc

Transfer function: skesRsCsG )()()(


Examples:

gap effect of gear mechanism,

threshold voltage of transistors

R( s) C( s)

1

t

r ( t )

ske

kC( t )

t


49/53

2.6 block diagram models (dynamic)

Portray the control systems by the block diagram models moreintuitively than the transfer function or differential equation models.

2.6.1 Block diagram representation of the control systems


Examples:

Si gnal( var i abl e)

G( s)Component( devi ce)

Adder ( compari son)E( s) =x

1( s)+x

3( s)- x

2( s)

X( s)

X3( s)

X2( s)

+

-

+X1( s) E( s)


50/53

Example 2.14


For the DC motor in Example 2.4

In Example 2.4, we have written down the differential equations as:

)4.....()3.....(....................

)2.....(....................)1....(

fdt

dJMMCE

iCMuEiRdt

diL

ea

amaaaaa

a

Make Laplace transformation, we have:

(8)sMsMfsJ

ssfssJsMsM

(7)sCsE(6)sICsM

(5)RsL

sEsUsIsUsEsIRssIL

ea

am

aa

aaaaaaaaa

)]......()([1

)()()()()(

..............................................................................).........()(.............................................................................).........()(

.............)()(

)()()()()(


51/53


Draw block diagram in terms of the equations (5)(8):

Ua

( s )

aa RsL

1C

m

Ia

( s ) M( s)

Ea( s ) Ce

)(s

fsJ

1

)(sM

-

-

Consider the Motor as a whole:

1)(

1

2 ffemme

e

TsTTTsTT

C

1)(

)(1

2

ffemme

mme

TsTTTsTT

TsTTJ

Ua

( s ) )(s

)(sM

-


52/53

Chapter 2 mathematical models of systemsExample 2.15 The water level control system in Fig 1.8:

Desi r ed

wat er l evel

ampl i f i er Motor Gear i ng Val veWat er

cont ai ner

Fl oat

Act ual

wat er l evel

Feedback si gnal hf

I nput hi

Out put h

-

e ua Q

1k 1

1

2 sTsTT

C

mme

e

s

ek s211

3

sTk

12

4

sT

k

)(

1

)1(

2sM

sTsTT

sTJ

T

mme

em


53/53


The block diagram model is:

Documents

Automatic control theory.ppt