3 Model Matematik Sistem

Model System Pengaturan

Dr. Ir. M. Sabri, MT

Komponen Sistem Kontrol

System atau proses (harus dikendalikan)

Actuator (mengubah sinyal kontrol untuk sinyal daya)

Sensors (menyediakan pengukuran keluaran sistem)

Input Referensi (menggambarkan output yang diinginkan)

Kesalahan deteksi (membentuk kesalahan kontrol)

Kontroller (beroperasi pada kesalahan kontrol untuk membentuk sinyal kontrol, Kadang-kadang disebut kompensator)

Karakteristik Sistem Umpan balik

Perhatikan sistem kontrol kecepatan berikut

Beban

Kl Motor

Km

Amp

Ka

Sensor kecepatan Ks

referensi

kecepatan

u +

_

gangguan

Momen puntir

wo

system loop terbuka

Lajur umpan balik

wr +

+

Td

Tm

Karakteristik System Loop terbuka

Keakuratan dari sistem loop terbuka tergantung pada kalibrasi gain dan pengetahuan gangguan sebelumnya (pilih kontrol u untuk memberikan diinginkan wo).

Masalah:

nonlinear atau berbagai gain waktu

Berbagai gangguan yang tidak diketahui

Beban

Kl Motor

Km

Amp

Ka

u

Gangguan

momen puntir

wo

system loop terbuka

+

+

Td

Tm

dllma

dmlo

TKuKKK

TTK

)(w

Karakteristik Loop Tertutup Perhatikan kes dengan umpan balik

d

slma

lr

slma

lmao

dlosrlma

dmlo

TKKKK

K

KKKK

KKK

or

TKKKKK

TTK

11

)(

)(

ww

ww

w

Beban

Kl Motor

Km

Amp

Ka

Sensor kecepatan Ks

Kecepatan rujukan

u +

_

Gangguan

momen puntir

wo

system loop terbuka

Feedback Path

wr +

+

Td

Tm

Karakteristik Loop tertutup

jika Ka sangat besar sehingga seperti ,

kemudian,

Ks adalah gain sensor dalam satuan volts per rad/s.

Hubungan input/output tidak terlalu sensitive terhadap gangguan atau perubahan dalam gain sistem

slmaslma KKKKKKKK 1

d

sma

r

s

o TKKKK

11 ww

rad/s volts 0

Karakteristik Loop tertutup

Sistem salah

Kesalahan kontrol adalah

lagi, jika gain loop, Ka Km Kl Ks besar, kemudian kesalahannya kecil.

d

slma

slr

slma

d

slma

slr

slma

slma

osr

TKKKK

KK

KKKK

TKKKK

KK

KKKK

KKKK

Ke

11

1

111

)(

w

w

ww

Catatan: Definisi Gain

Gain depan: Ka Km Kl

Gain umpan balik: Ks

Gain loop: Ka Km Kl Ks

Gain loop tertutup : gain depan

1 + gain loop

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

u k u a c

Fig. 2.1

temperature

measurement

Feedback signal b(t)

+

- ()

e(t)=

r(t)-b(t)

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

2) The basis analyzing or designing the control systems.

For example, we design a temperature Control system :

The key designing the controller how produce uk.

2.1.1 Why


2.1.3 How get

1) theoretical approaches 2) experimental approaches

3) discrimination learning

2.1.2 What is

Mathematical models of the control systems the mathematical

relationships between the systems variables.

Different characteristic of the process different uk:

T(t)

uk

T1

T2

uk12 uk11 uk21

For T1

12

11

k

k

u

u

For T1

22

21

k

k

u

u


2.2.1 Examples

2.2 Input-output description of the physical systems differential

equations

2.1.4 types

1) Differential equations

2) 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.


ur

uc

R L

Ci

define: input ur output uc we have

rccc

crc

uudt

duRC

dt

udLC

dt

duCiuu

dt

diLRi

2

2

rccc uu

dt

duT

dt

udTTT

R

LTRCmake 12

2

2121:

Example 2.1 : A passive circuit


Example 2.2 : A mechanism

y

k

f

F

m

Define: input F output y. We have:

Fkydt

dyf

dt

ydm

td

ydm

dt

dyfkyF

2

2

2

2

Fk

ydt

dyT

dt

ydTThavewe

Tf

mT

k

fmakeweIf

1:

:

12

2

21

2,1

Compare with example 2.1: ucy; urF analogous systems


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

rCRRR

RR

R

RRc

cCR udt

duu

dt

du)( 4

32

324

1

32

)(:

)(;;: 432

32

1

324

rr

cc u

dt

duku

dt

duThavewe

CRRR

RRk

R

RRTCRmake

Chapter 2 mathematical models of systems Example 2.4 : A DC motor

ua

w1

Ra

La

ia

M

w3

w2 ( J

3, f

3)

( J1, f

1)

( J2, f

2)

Mf

i1

i2

Input ua output 1

)4.....(

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

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

)1....(

11

1

ww

w

fdt

dJMM

CE

iCM

uEiRdt

diL

ea

am

aaaaa

a

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

MCC

RM

CC

Lu

C

CC

fR

CC

JR

CC

fL

CC

JL

me

a

me

aa

e

me

a

me

a

me

a

me

a

1

)1()( 111 www


): (

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

......

......

:

321211

21

22

21

321

21

22

21

321

21

www iiifromderivedbecan

torqueequivalentii

MM

nt coefficie frictionequivalentii

f

i

fff

inertia of momentequivalentii

J

i

JJJ

here

f

Make:

constant-timeelectricfrictionCC

fRT

constant-timeelectric-mechanicalCC

JRT

constanttimemagnetic-electricR

LT

me

af

me

am

a

ae

-.......

.......

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


ae

mme uCdt

dT

dt

dTT

12

2

www

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

we have:

)(11

)1()( 111

MTMTTJ

uC

TTTTTT

mmeae

fmfeme

www

The differential equation description of the DC motor is:


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 R

3

w

Input urOutput ; neglect the friction:

(4)MTMTTJ

uCdt

dT

dt

dTT

(3)uku(2)u

(1)uukuuR

Ru

mmeae

mme

kaf

frfrk

)......(11

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

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

2

2

2

11

2

www

w


2134we have

)(1

)1( 211

212

2

MMTJ

Tu

Ckkkk

dt

dT

dt

dTT e

mr

eCmme e

www

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.


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 variable on the right of the input-output equation .

Output variable on the left of the input-output equation.

Writing polynomial according to the falling-power order.

Suppose: input r output y


2.3 Linearization of the nonlinear components

2.3.1 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 linearization

Suppose: 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

rrrf

rrrf

rrrfrfrf

00 :)()(: rrrandrfrfymake

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


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 :is that

)(11.12)()( :have we 00

Example 2.7 : Fluxograph equation

pkpQ )(

Q Flux; p pressure difference


equationionlinearizatp

p

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 definition

Transfer 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:


)(

)()(

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 the

systems, independent on the input and output variables.

2.4.2 How to obtain the transfer function of a system

1) If the impulse response g(t) is known

Notes:


)()( 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


Example 2.9:

responseUnit step

sssssCetc

functionUnit step s

ttr

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 known

Assume: zero initial conditions;

Make: Laplace transform of the differential equation;

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


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 the circuits theory.


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


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 tCRR

Rk

ller.PI-Contros

k

CsR

CsR

R

sCR

sU

sUsG

r

c


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 ggerUf

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


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

www

w

Make Laplace transformation, we have:

(4)sMJ

TsTTsU

CessTsTT

(3)sUksU(2)ssU

(1)sUsUksU

mmeamme

kaf

frk

)......()(1

)()1(

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

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

2

2

1


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

)()(1

)()]1

1([ 21212 sM

J

TsTTsU

Ckks

CkksTsTT mmer

eemme

- ......

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

constanttimeelectricmechanicalCC

JRT

constanttimemagnetic electricR

LThere

me

am

a

ae

)1

1(

1

)(

)()(

212

21

emme

e

r

CkksTsTT

Ckk

sU

ssG


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

Fungsi Transfer

Dr. Ir. M. Sabri, MT


2.5.2 Integrating element


constant timeintegralTdttrT

tc I

t

I

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

)(

0

Transfer function: sTsR

sCsG

I

1

)(

)()(


1

R( s) C( s)

1

t

r ( t ) C( t )

t

sTI

1

TI

Examples:

Integrating circuit, integrating

motor, integrating wheel


2.5.3 Differentiating element


dt

tdrTtc D

)()(

Transfer function: sTsR

sCsG D

)(

)()(


Examples:

differentiating amplifier, differential

valve, differential condenser

R( s) C( s)TDs

1 TD

t

r ( t ) C( t )

t

2.5.4 Inertial element



)()()(

tkrtcdt

tdcT

Transfer function: 1)(

)()(

Ts

k

sR

sCsG


Examples:

inertia wheel, inertial load (such as

temperature system) 1

R( s) C( s)

k

t

r ( t ) C( t )

tT

1Ts

k


2.5.5 Oscillating element


10 )()()(

2)(

2

22 tkrtc

dt

tdcT

dt

tcdT

Transfer function: 10 12)(

)()(

22

TssT

k

sR

sCsG


Examples:

oscillator, oscillating table,

oscillating circuit

R( s) C( s)

12

122 TssT C( t )

k

t

1

t

r ( t )

2.5.6 Delay element



)()( tkrtc

Transfer function: skesR

sCsG

)(

)()(


Examples:

gap effect of gear mechanism,

threshold voltage of transistors

R( s) C( s)

1

t

r ( t )

ske

kC( t )

t

2.6 block diagram models (dynamic)

Portray the control systems by the block diagram models more

intuitively 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 ( compar i son)

E( s) =x1( s)+x

3( s) - x

2( s)

X( s)

X3( s)

X2( s)

+

-

+X1( s) E( s)

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....(

ww

w 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

)( )()()()(

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

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

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

)( )()()()(


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

Ua( s)

aa RsL

1Cm

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

-

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

Desi red

wat er l evel

ampl i f i er Mot or 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 w Q

1k 1

1

2 sTsTT

C

mme

e

s

ek s211

3

sT

k

12

4

sT

k

)(1

)1(

2sM

sTsTT

sTJ

T

mme

em

End

?

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