Control Systems [1 ed.]

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スの 子奮 ル〃 響

CONTROL SYSTEMS ABHFAtt H uttAIN HARロ ロ N ABHFAロ

3。

Mo Singh sc_■ こ

ゝ 識

y 露 09826266161

DHANPAT RA:a CO. /

CONTROL SYSTE AS Formerly,

and Technology,

J7tirerf″ POlytechniご

DR. HARtrON AEiHFAQ .{ssstant Ptofessor Deprtment of Elrtrial Engineering,

Faaityp! Engtne*ing &

TechnobgY,

Jamia Millia Islamia, NEW

DEIfrI @tDU)

DHANPAT RAI&CO.{PvI。 )Lld ￨

EDUCATIONAL&丁 ECHNICAL PUBttSHERS

CONTROL SYSrEMS Published

fr Sales

by,

GAGAN KAPUR

Dhanpat Rai & Co. (P) Ltd.

Oftce :1682, Nai Saiak Delhi-110m6 Wrorc :2325 5367, 2325 O25l Offce : 4576|15, Agarwal Road Darya Ganl, New Delhi-110@2

Regd.

Phone : 2324

7/,

37, 38, [email protected]

@

aurxon

All Trademarks Acknowledged

Disclaimer Every olfort haa bccn madc to Evoil orrora or ombsions in thb publicatbn. In spite ot thi8, some errora might havc crcpt i1. Any miatalc, crror or dbcrcpency notod may ba browht to our noticc whi*t shall bc taken carc of h thc ncxt cditbn. It ie notifbd tlat ncidlcr thc publishcr nor th€ author or selbr will bo reaponsiblg lor any damag! or losg o, actbn to any on9, of gny khd, h any m6nner, thorefrom.

2oll Pf{,,intia.ld%lt

Firct Edit*n :

2Ol4

nBpnnq jqols I

Repfint : 2016

P(?trlinhnl74.

P/c:

マ3∞

llo

Tryesetting by : North Delhi Computer Convention, Delhi-110009, [email protected] inted at I Nafiaj Offs€t, Delhi

P This book is intended lo serves os o textbook for ihe subied of 'Control Systems'

{or B.E./B.Tech Degreo students for Eleclricol, Elechonics, lnslrumentolion, Communicolion, Computer Science, Mechonicol, Chemicol ond Biomedicol Engineering. h will olso be useful lo condidoles oppeoring for AMIE, IETE, GATE, UPSC Engineering Services Enlronce Exominoiions. lt would be equolly

helpful

to

prociising engieers

to

understond theorelicol ospects

of

their

profession. The subiect motler in eoch chopter hos been developed in o systemotic monner with emphosis on bosic concepls. ll is writlen in o very simple longuoge so thol

sludents moy eosily grosp the subiecl. The illu$rotlons of principles ond concepts hove been done by meons of o lorge number of corefully selected wo*ed exomples. Step-by-slep procedures {or solving problems ore provided' Exomples hove been wo*ed out in detoil so thot lhe reoder moy {ollow eoch exomple wilh confidence from beginning to the end. Most simplified methods hove been provided for solving problems. At the end of eoch chopler o lorge number of Reviow Questions ond Unsolved Problems wilh onswers hove been included for prociice.

We hope this book will be useful to studenls ond teochers olike. We will be groleful to reoden for their suggestions for the improvement of the book'

ASHFAQ HUSAIN

HAROON ASHFAQ

3.

M.singh Sengar secimen Copy NOt For Sale

o982626161

I oppreciote the potience, underslronding ond support of my wife, Dr. Nigor Minhoi, fusociote Professor in Electronics, University Wopren's Polylechnic, Foculty of Engineering ond Technology, Aligorh Muslim Unive6ity, Aligorh. olso oppreciote the concern of *y rontAhmod fuhfoq, M.Tech. (Environmenlol Engineering), fusistont Professor in Civil Engineering, University Polytechnic, I

Foculty of Engineering ond Technology, Aligorh Muslim University, Aligorh. He

hos helped me

o lot in the preporotion of the monuscript ond proof reoding.

I om thonHul to Mr. Gogon Kopur, publisher, for bringing out this book in o scheduled time.

ASHFAQ HUSAIN

W -1-10 CHAPTER 1

lnlroduction To Control SYstems ■

1.1 Inhoduction 1.2 Basic Definitions

1

2

1.3 Control System Configuratiors 1.3,7 ,Pen-lnoP Estem

L3.2

3

qosed-LooP Systefi 3 5

1.4 Feedback Control 1.5 Effects of Feedback

6 7

1.6 Servomedranism 1.7 Regulating Systems (Regulators) 1.8 Types of Feedback Control Systems 7.8.7 Linear Confiol Estem

cHAPTER 2

7 7

I

7.8.2

Nonlineor Control SYstrms 9

L83

fime'Inva'iant

7,8.4

C.ontinuous'Data Confiol Ester,ns 10

7.8.5

Ditcrete-Data Control Estems 10

Versus frme'Varying

Syttenj

9

1\ -17

Trons{er Function

2.1 Introduction 2.2 Procedure for Finding Transfer Functions of Electric Networks

CHAPTER 3

3.1 Introduction 3.2 Block Diagrams 3.3 Simple or Canonical Form of a Closed-Loop System 3.4 Block Diagram Reduction Techniques 3.4.7 &scade Connedion ol Bb&s 21 Parullel bnnection

of Blocks

_‖ 酬‖

11

r8-55

Block Diogrom RePresentotion

3.4.2

11

22

18 18 19

A

Summing Point Before a Bl * 2j 3.4.4 gtifting a Summing Point &r a M. 24 3.4.5 Shifring o flakgf,fr Point After a Bbd.. 25 3.a.6 Sltifiing ol fakeofr Point Wrc a Bl,r*. 26 3.4.7 nmnangement ol fumming PoinB 2t 3-5 Proedure for Reduction of Block Diagran 3.6 Superposition of Multiple Inputs

3.4.3 Stijting a

4

56-88

Signol Flqw Grophs

4.1 tnhoduction

66

4.2 /Signal Flow Graph Terrninology

56

“ お

89-139 ” ９ ０ ” ” ”

5.1 Induction 5.2 Modelling of Mechanical Systems 5.3 Translational Motion 5.4 Network Elemmts of a Medranical Translational System 5.5 Mass (M

“

Mothemoticol Models of Physicol Systems

”

4.8 Comparison of Signal Flow Graphs with Block Dagrams 4.9 Signal Flow Graphs for Eleclric Networks

”

4.5 Signal-Flow Graph Developed from System Equations 4.6 Signal-Flow Graph from Block Diagram 4.7 Mason's Gain Formula

ω

“

4.3 Rules for Signal-Flow Graph 4.3.7 Additbn ,I/Je 60 4.3.2 TraIlf ission h/le 60 4.3.3 Cdscade &nne(i;ion 60 4.3.4 Parallel Connection 61 4.4 Properties of Signal-Flow Graphs

CHAPTER 5

”  ”

CHAFTER

” ９ ．

5.6 Linear Spring 5.7 Friction 5.8 Damper Element

颯 ” ７ ９

‖耐 ‖

”

5.9 Rotational Medranical Systems 5.10 Gear Trains 5.11 Free-Body Diagram

６７ ５・ ４ ４・ ・ 瑯・ ・ ・・ ・ ・Ш・

512 Nodal Metlnd of Developing MedEnical Equivalent Network

513 Analogous CircuitE 514 Series RLC Circuit 515 Parallel RLC Cirortt 5.16

Force-Voltage AnalogY

5.17

ForceCurrcnt AnalogY

522

９鄭瑯″””脱 ・ ・

518 Procedure lor Solving Analogous Systems for Drawing The /-z Analogous / Electric Network 5.19 Mule frcm a Medanical S),stem 'i*a"r" for Drawing the TorqueCurrent (T-, Analogous 5" Electric Network ftom Medranical Rotational Systert 521 Procedure for Drawing the Torque-Voltags (T-u) Analogous Electric Network from a Nledranical Rotational System Thermal Systems

5.23 Mathesratical Model of a Thermal System 5.24 Mathematical Model of a Simple Merorry Thermometer System

5.5

Liquid-Level SYsEms

5,% Mathematical Model of Liquid-Level

System

5.27 Ma0Nematical Model of Pneumatk Systent CHAPTER 6

140-204

Time-Response Anolysis

6.1 Time-ResPque o[ Cor*rol SYsmu

140

6.7.7 l:ransient ,4ryp.nse ,4O 6.1.2 Slf,dy-St,.b R€5,,.tt,P 14o 141

6.2 Standad Test InPuts

' 53

W

Futlrtion 111 6.2.2 nanl4. tundjion U2 6.2.7

6.2.3 tu'a,tr/lic F./J/,(irion 112 6.2-4 htl8,/dl€ Arndion 143 TimeResponse for a First Order Sysbm 6.3.1 Iesporce of the First Met SJf[em vith hit W htput ,45 6.3.2 Re5?onse ol the Fitst-@ater System wifi ttnit Rn',,'p Punf,ian 146 6.3.3 Resrr/nse oJ the Fitst'&iter qlr/em with Unit-Inpulse Pun(fion 148

6.4 Re8pql8e of a Secondorder System 6.5 Time Reeponse of Secondorder System with Unit Ramp Function

‖七‖

144

1{8

1ss

5.11 Analysis of Steady-State Error 5.12 Static Error Constants 6.13 Static Position Error Constant Kp 6.14 Static Velocity Error Constant tr(,

6.18 Generalized Error Coefficients or D),namic Error Coefficients 6.78.7 Cotrelation betuteen Static and Wnamic Enor Coelficients 173 6.19 Elfect of Adding Poles and Zeros to Transfer Functions

6.N.f

″  ”

6.15 Static Acceleration Error Constant K. 6.16 Steady-State Erors for Type O 1 and 2 Systems 6.76.7 Enors Jor Step Input 169 6.76.2 Enors for Ro'mp Input 170 6.76.3 Enots fot Parobolic Input 171 6.17 Disadvantages of Static Error Coefficient Method

７卸蜘ｕ螂協協”畑６ ９ 螂・ ５ ・

6.5 Impulse Response of Second Order Systems 6.7 Effect of (on Pole Location in Second-Order Systems 6.8 Time Response Specifications 6.9 Derivation of Time Response Specifications 5.10 Ty?es of Control Systems

174

Addition of a Pole to the Forword Path'llonsfer Fundion 174

6.79.2 Addition of o

to the Aosed-Loop Tlansrer Function 124 6.79.3 Ad.dition of a Zero to the Aosed-Loop Ttansfer Function 174 6.79.4 Addition of d Zerc to the Fotword. Path fransfer Function 174

CHAPTER 7

Pole

6.20 Dominant Poles of Transfer Function

774

6.21 Methods to Improve Time response

t?5

5.22 Proportional Controller

175

5.23 Proportional Plus Derivative (PD) Control 5.24 Proportional Plus Integral (PI) Control

176

7n

6.25 Proportional-Integral-Derivative (PID) Control

779

205-243

Stobiliiy of Control Systems

7.1 Introduction 7.2 Location of Poles and Stability 7.3 Routh's Array 7.4 Division or Multiplication of a Row by a Positive, Non-Zero Constant

205 lX'5

xt6 208

7.5 Routh's Stability Criterion

‖″‖

2(E

76 SPedJ Cases of Rou■ Hurwiセ 甲 7 7 APpiaton of Rouh Stabihty C五

nOn

“ bttm

7 8 Reltte Stability Analysls

CHAPTER 8

210 211

244-307

BOde Plol

81F“ quency

209

墾

ResPon∞

244

82 3ode Plot 8.2.l βOde Magnitude plot 2``

822 30d●

P力 as●

4ngle plo` 2`5

8 3 Standard Fonn of G(プ

o

245

8 4 3asic Factors of C(701

246

8 5 Bode Plot of Constant Gah K

247

8 6 Bode Plot of lntegral Factor

248

8 7 Bode PIot of FirstOrder Factors in the DenommatOr (Simple POle on Real Axis)

CHAPTER 9

型

8 8 First Order Factors in the Numerator

258

8 9 Quadratc comPleX POles

262

8 10 Step‐ By― Step Procedure to Draw Bode Plot

266

8 1l Relaive Stab■ ity

268

8 12 Gam Adlustment h Bode Plot

272

8 13 Detennhaion of Statlc Error Ccleffldents from Bode Plot

272

814 Al卜 Pass

and― mum Phase Syttems

274

8 15 Correlatlon beneen Phase Margm and DalnPmg Raio

277

8 16 Advantages of Bode Plots

278

8 17 Calculaton of Transfer Funcion from 3ode Magitude plot

273

308-360

Rooi Locus Anolysis

308

9.1 Introduction 9.2 Root tocus Concept 9.3 9.4 9.5 9.5

308

Angle and Magnitude Conditions

鰤

Symmetry of Root l-oci

310

Starting Points of Root toci

3■ 0

Terminating Points of Root Loci

311 311

9.7 Number of Root Loci

‖χ `￨￨

f.e f4sencr

of Root Loci on the Real Axis (Real Axis Segn€nb)

9.9 Asymptotes and Centroid 9-10 Breakaway and Br€ak-in Points 9.10.7 Pro(Pdure ol Detemining heakavay PoinB 9.11 Angle of Departure at a Corrplex Pole

313 314 315 3■ 6

9.12 Angle of Arrival at a Complex Zero 9.13 Imaginary-Axis Crossing Point

316 316

9.14 Value of K for a Givm Damping Ratio ( hom tre Root Locr.rs Plot

3■ 7

9.15 Value oI Gain Margin from Root Loc,us 9.16 Summary of General Rules for Constructing Root Loci 9.17 Efffi of Adding Poles to Qs)H(s) 9:18

CHAPTER

l0

Effuc.ts

317 317 321

of Adding Zeros to (s)H(s)

320

361 -399

Frequency Response Anolysis 10.1 Introduction

361

10.2 Advantages oI Frequency-Response Analysis

361

10.3 Limitations of Frequency-Response Analysis 10.4 Frequency Response Plots

362 362

10.5 Frequency Domain Specifications

362

10.6 Resonant Peak Magnitude (M,) and Resonant Frequency

10.7 Correlation between Time and Frequency Responses 10.8 Polar Plots

CHAPTER ll

312

(-l9t

1

y2っ

y4 毛

a.l ｌ

二

一Ｓ

θ

1

―亀 (rl

I

ng. +.ro Step 4

There are three forward paths between R and C. The forward path gains are as

follows

:

Forward path y,

-

!z- !s- !erh

Pt =

Forward

"1.

!.b..

"l

="1

s5 path y, -Az -!s -a a -! P,

11 =1 ' '- 'br'

Forward path y, -Uz

ss

-At

-A

+

Az, path gain

e

-Az

| -;b2 s-

-As -A t -U z

z=tl11 " s s s t,.t ' -\s' Ft.p 5 There are three individual loops

as shown

The loop gains are

-al , L. l-at)-\= ss

in Fig. 4.16(c).

S:GNAL FLOW GRAPHS 75 α

2 12=1・ 1・ ←α 2)=三 s s s2

ち

=:・ :・

:く

一 %)=子

Step 6 An the l∞ Ps touCh that is there are no nontouching looPs・ The graph deteminant is therefore′ △=1-(1+12+13)

―

―

=1-(千

争

争

)

=1+÷ +多 +争 △1=1-0=1 △2=1 △3=1

0=1 0=1

Step 7 APplying Mason′ s mlQ dle trarsfer fmction is

器

=:(Pl△ 1+P2△ 2+P3△ 3)

_:卜

・ 1+:争・ 1+:}・

1

1+`生 +:争 +:計

_ Ls2+b2S tt b3 s3+.s2+α 2S+α 3

Ex_PI`14191

,t"*

in the systatr shoum in Fig. 4.77(a). funcfir"W r Y;(s)

the cloxil-loop transfer r r

)L(S) y6

はFlg.4.17

76 coNTROL SYSTEMS sOh■ On.

カーy2 y3 ゝ r

Step l There b omy one brwa」 Pa■ Path gain

1 1..t. 1-.1 sq & sCz szRr q C2

P,=1.

'

!gp_l

y5 y6・

＼

There are three loops as shown in Fig. 4.17(b). The loop gains are

一 一面 つ L=岩 士〈 缶 12=毒 ( 寺 )=―

走

=寺 一面 つ ち ・ 毒← お l_

1 SCl y3

Rl

y2ヮ

SC2 y4y5、

y6 こ :ァ -1

R2

-1

(bl

I

rig. r.rz Step 3

Loop

I,

does not toudr loop

t,

while

I.

toudres Ig

*d

12 toudres

I".

Therefore, A=

1-(

11

+

I, + I")+ \1,

I - t . 1 1 =1-s&q = sR, C, sRrC, s'R,RrC,C, Step 4

Since all three loops touch the forward path,

and I" from determinant

A

to determine

A,

it is nec€ssary to remove I1,

that is, Ar =1

By Mason's gain fomrula, the overall transfer function is givm by

=iPlヘ 器

L

一・ 市 ¨ ・ 卜 一 “ ︲ 市 ¨

SiGNAL FLOW GRAPHS 77

＋

=( 1+

×

s2RI R2 CIC2 )

1)

(裁

s2RI R2 Ct C2+(RICl tt R2C2+R2q)s.1 E轟 轟ile‐ 4.lol

0″ 肩 胞

グ ル SⅢ ″ル

ル 耐 ゎ"器

押 みS彎 中

偽

め

1°

カ

▽ ー ´1

―ら

0)

0) Flg。

“

一 Ｓ

S

Ю

カ

一 Ｓ

1

1

80

１

ｌ

1

磁ψ

4.18

sOlu6on. Step l

There is only one forward Path

/r -

lz-

Vz- At-

As.

Path gain

_' n =t'! -

s

Step 2

L'u =

s

b s2

There are two feedback loops shown

ir

Fig. 4.18(D) with loop gains

1

-41 I.'_ss = (- o.,)-: =-

4 =!.!.por1=-1, ss s2 Step 3

There are no possible combinations of two nontouching ioops. Therefore,

A=l-(L +lr)=t+1 + Ste,4

,L

The forward path touches both the loops. Therefore, A, =1The overall transfer function is given by

y(r)

1p a. X(s)= A "

=爾

7=膏

78 CONTROL SYSTEMS SIGNAT FLOW GRAPHS FOR ELECTRIC NETWORXS The following procedure is used to draw signal flow graphs for electric networks

:

1. Draw the givea network in the s-domain. 2. Write the equations for different branch currents and note voltages. Care should be taken to write separate equation for separate branch and each element must be considered at least once.

3. Identi{y dilferent

variables and represent each variable by a separate node. The input and output nodes are to be shown separately. 4. Draw signal flow graph for eadr equation in step 2. 5. Combine all signal flow graphs to get total signal flow graph for the given network.

6.

Use Mason's gain formula to find the kansfer function of the given network.

涸 鰺 鰈 塚 餃

月π

tt rtlasPル C勧

′凛"″ レ

R

げ 滋 渤 岬 κttmに 加 Flg.4.19(α

)“ 竹 MaS● ″t』Ⅲ

ι

●)

警Fig.4.19 SOL■ 。n The neLvork OfFig 4 19(α

)iS dra￢m

m hes_domttn asshOν m m Fig 4 19(b)Let

r(s)be tt mesh current From Fig 4 19(b)′

Ю

∝ Alsc

=扮

Ю {轟 → 轟) ← =竃

o=Iけ 尭 ‰

■■ e node va五 ables are%(S〉 dra、

)+啄

・M‐ つ … 041l υ

I(S)and路 (S)USmg Eq(E4111)′ the dgnal■ow graph is

m asshown h Fig 419(c) ∼

Ushg Eq(E4112)′ he signal now graph iS drawn as sh。

■ ν n h Fig 419(′ )

COmbttg thetwo graphs ofFig 4 19(c)and Fig 4 19(′ )the tOtal slgnal graph is shO_in Fig 4 19(a)

S:GNAL FLOW GRAPHS 79

1

R+sL Its) 1

R+sL

警

Fig。

sc

LIS)

√゛ )R+sL

(C)

‰

IIS)

↑

1

１一 ￡

(の 1

1

4.19 LIsι

o/Mα Sο η Ъνj"/0箭

:α

“

ne signal■ Ow graPh Of Fig.4.19(`)has Only One foward Path as shown in Fig.4.19∽ Thttrefore′

the padl gain is

Pl=(轟

)(1島 ,)

The signal flow graph of Fig.4.19(`)has Only One looP. TLerefore′ the looP gain is

L=15(責 武1) △=1-1 =1十

1

sC(R+s Ll

一

l+sRC+s2LC sC(R+s L) Since the loop is toudring the only forward path, therefore, A, =1

Transfer function

t(S)_Pl△ 1

4(S)

△ (R+s L)sC

1+sRC+s2LC s2LC+sRC+1 sC(R+s L)

.

80 CONTROL SYSTEMS E=mple 4 12

Find the transfer function of the nehtork slown in Fig. 4.20(a) using Mason's gain formula. 1

1

Rl

０  ４

(0 ● ち

●

r2(S) SL ン ls) Ъ

40-― 一

Ｃ ︲ 一 ｓ

-

1

R2

0 4(S)

rll)

120)

1 L(S)

Rl

1 sc

略(S)● ― ― ●― ―・

上で０ ︲下型 一︺︶  ・

!

R2

'L

-0-OL(S)

ュ シ ・ 一 ︺ 下 ︺ ︶

1

Rl

1

ng. n.zo sOL“ n Figure 4120(b)shOWS the s― domah neれ vork of Fig 4 20(′ )From Fig 4 20(b)we can write

軍→=Ψ

(E4121)

寮 →=7

(E4122)

壕 →=7

,(E41213)

70(s)=ち (S)(SD

".(E4.12.4)

｀ ヽ ヽ

S:GNAL FLOW GRAPHS 81 The signal■ ow graphs representing Eq。

(E4.12.1)to Eq.(E4.12.4)are shOWn in Fig.4.20(c)

to Fig.4.20o respectively.The total signal■ ow graph for the network is shown in Fig.4.201g).

We sh」 l use Mason′ s gahお mJa b find tt harlsttr hncuon yOISl. 4(S) Step l h Fig.4.201g)′ there is only one foⅣ vard Pa■ aS ShOWn in Fig.4.20(り The gain ofthe fottard Path iS given by

L= L l=寺・ 尭・づ

寺 2 RI R2 C

Step 2 1heindi宙 dual feedback looPs are shown in Fig.4.20(り 。 ■ e looP galllS are

L=(1,)(―

「)=― ホ 武

12=(寺 )(― 尭 )=― 寺 13=(S Ll(―

￨

武r)=一 難

Step 3 Thel∞ Ps t and 13 are nOntouching。 ■Frefore′ L13=( 丁

[)=云 tて )(号 ご

市亀て

Step 4 There are no higher order nontouching looPs. Step 5 Since all the looPS tOuぬ the forward Paれ

Step 6 . II

△1=1

△=1-(L+12+13)+(L13)

=1+1+1十 sRIC sR2C R2 RIR2C

二

十

二

Step 7 By Mason′ s gain fomulみ halllsfer ttction

4(S)_Pl△ 1 4(S) △ L

_

RIR2 C l+ 1 + 1_+三 + L sRIC sR2C R2 RI R2 C sL s2L&C+s (L+ & R,C)+(R, + R, )

.

82 coNTROL SYSTEMS 働 ″赫

勧

ψ ル耐 勧

SⅢ ″ ル

器

…

絆

みS励

21ω

聴 “

“

I ng...rt Solrrtiort. Step l There are hvo fonrard PaJ随 aS ShoWn il Fig 4 21 wih Pa■ gtts as ′

Pl=1::b2=多 P2‐

XO‐

l

T

1

T

1:bl=:

XKS)。

1

)

― ‐ ― ‐ ― → "‐ ‐ … Ⅲ … Ⅲ … → ― ‐ Ylg)

XIS)。

& ng.

。

`ヽ

ち

r.zr Step 2 There arぃ 、 feedback looPs as shOWn h Fig 4 21(ε )with 10oP gai“

‐ ム :l al)=÷ ら=::C)=子 !g3-,]

There are no nontouching loops. Tlrerefore, △ =1-(1+12)

=1+=+争

O Yls)

:

SiCNAL FLOW GRAPHS 83 Step 4 3o■

■ e foward Paths tOuch the indi宙 dual looPs.Therefore/ △1=1′ △2=1

The transfer ttction is given by

1+鳥 △ 器=i蝿 △ 2)

=義 惨喜 与 →

耳m・nple141i41

月 グル 磁 ψ “

ル c施

"妥 ル

滋

s"Jル

押 力S励 滋 鞠 。422● リ

化 「 ▼ G5

c

場

亀 小

I・

Loop Ll

ffi

ffi

2

LooP L2

Loop

L3

ti1.4.22 s01ution.

Step l hreares破 forward

Paths Wial f。

11。

wing Pah gains:

Path R― χl― χ2 χ3 χ 4

Cf° rWard Path gahし 1=1・ G2・ G4・ G6・ 1=G2G4G6

Path R― χl― χ6 χ5 χ 4 P2=1・

Cf° Ward Path gai島 G3・ G5・ G7・ 1=G3G5G7

Path R― χl― χ2 χ 5 χ 4

Cf° nrard Path gふ,

P3=1・ G2・ q.G7・

Path R― χl

Tχ 6

1=G2qG7

Cf° n″ ard Path ga五 し P4=1・ G3・ G8・ G6・ 1=G3G8G6

χ3 χ 4

、   一 ・ ・一 ヽ

84 CONTROL SYSTEMS Pa■ R― ,1-χ 2 χ 5 χ6 χ 3 χ 4 Cf° nVard Path gam7

P5=l G2 q(H2)G8Q・ = G2q H2G8G6 Pah R― χl― χ6 χ3 χ 2 χ5 χ4 Cf° コ″ard Pa■ l gaれ P6‐ l G3 G8( rfl)q G71‐ G3G8HlqG7 Step 2 There are three mdi宙 du」 looPs cig 4 22(b))宙 hl∞ P ga血

L=G4(Hl) Q■ 12=G5(H2)= QH2 13=q(比 )G8(― Hl)=qH2QHl Step 3 There is only one Possble combma■ on oftwo nontou(hhgl∞ Ps wi■ 100P

gah product

L12=(G4rfl)(_QH2)=G4HIG5 H2 Step 4 hre are no combhaions of tt nontouttdng l∞

Ps′

four nontoung

looPs etc Therefore′ the detennmant of tt graph is

△=1(1+12+13)+Lら

=1+G4Hl+QH2 q H2G8■

+G4HIQH2

Step 5 The cofactor△ たof the dete...lmanta10ng a Pathκ iS evaluated by remo―g■e 100PS that touch Pa■ ■た■om Δ

The Arst fomard Path is not h tcluch輌 ■ one looP(wi■ galn― G5H2)

hrd∝

′ △1=1-(G5H2)

=1+G5H2 ■e second fomard Pah iS¨ t h tmch市 h melooP(轟 th

gain― G4■

)

Therefore/ △2‐ ■

1 ( G4HI)‐ 1+G4Hl

e ottrfomard Pa■

fourt■ 饉 and s破 h are h touch

'namely7 hird′ wlth a11 looPs idi宙 duallyJherefore △3‐

1′

The overall trarlsfer im“

景

△4=1′ △5=1′ △6=1

on

=i(Pl△ 1+P2△ 2+P3△ 3+P4△ 4+P5△ 5+P6△ 6)

IG2G4G6(1+G5H2)+QQら (1+G4Hl) +G2qG7+QQG6 G2CiH2G8G6 QG8HlqG7】 1+G4Hl+G5H2 q H2Q■ +G4rfl G5亀

S:GNAL FLOW GRAPHS 85

EXERCiSES l. What is a node ? Name different types of nodes. 2. Define loop. Name different types of loops. What is meant by the term loop gain ? 3. Explain the following terms : (a) Path (b) Forward path (c) Forward path gain (d) Feedback path (e) Loop gain

4. What do you mean by a signal flow graph ? State the properties of signal flow graphs. 5. State the Mason's gain formula. 6. Compare block diagram and signal flow graph methods. 7. A system having input and output represented by u and : is described by the following equations

:

X=xt+a3u

:￨￨￨￨

、ト 「

xt=brxr+xr+aru xr=-brxr+a.ru Draw the signal flow diagram. Obtain the transfer function

{R

of signal flow graph shown in Fig. P4.2 using Mason's gain

formula..

警Hgo P4.2 ‐LR

l+QHlt G5H2+9'S%+qG2G49+

い o

86 CONTROL SYSTEMS 9。

ObtainthetransfermctimfdthesirlnoWgraphsho―

h Fig.R。 3 using陥

師ヽ

gain

fomula..

一 一 枠 一 ３． ． ４ Ｐ

籐磯 雛

助

Ｃ一 ■

ヽ１

， ，

10。

Convert」 隆 block diagram of Figo P4.4(α )to a Sip」 flow graph and detemine the dosed loop

mula 響SttmCtimttbyttdMasorsg血 お

鱗Flgo P4.4

L二

１ １ ︲ ヨＪ ︲︲

今￨￨:[詈

S:GNAL FLOW GRAPHS 87

lL ttteminettdosedlooptransfermdimtth腱

tm shO― mFI.R.ユ 警

C(S)

凩Figo P4.5

Dettmine the ttalsた r hctiOn IΩ from the dgnd■ ow graph sho― R(s)

hF七 .P4.6

`l G2 G3

鷺Fig.P4.6

.■ _

■ ■・ ′ ‐け 摯タ イー t'‐

‐ ■

_

_

‐ ‐

‐

‐

‐

_‐

'3U3「

13. Draw the signal flow

G,(1- GH" - G H,

66ダ

%早

￨￨_. `‐::1‐

+02・12%夕 6+%ギ 27.ャ グ lゃ ￨￨

‐

‐.￨￨

‐

‐

‐ .

￨￨

‐

)l

-l

やや lや や71子 ‐‐ ￨サ

1」

graph and evaluate the closed-loop transfer function of a system whose block diagram is shown in Fig. P4.7(a).

88 CONTROL SYSTEMS

= Figo P4.7

闊打 ５

MAttH EMAttICAL MODELS OF PHYSiCAL SYSTEMS メ

:NTRODU釘 :ON For he analysis and design ofcontrolsystemtt we fomulate a ma■

ematical descriPtton Of

the system.The process of obtaining the desired mathemadcal descriPiOn ofthe system is known as ELOdening.A rna■ ematical model is not uluque to a glven systern.A system maical may have many ma山 〕 may be represented in many different ways at thereforら

models depending on reqlllrement. The dynamcs of many systems may be described in te...ls of differential equatiors obtained by using PhysiCal laws governmg a pard(‐ lar system.Once a mathematical model of a system is obtainea Various analy● cal and computer metk)ds can be used for analysis and syntheds Purposes.

議饉

MODEL‖ NG OF MECHANiCAL SYSTEMS The mouon of mettanicalsystems may be trarlsla饉 岨

′Юtadonal or combinauon of bo山

.

The motion of trarslation is defined as a motion ttat takes place along a straight line.Ihe rotational motion of a body is defined as a lnotion about a fixed axis.

釉

TRANSLAT10NAL MO■ ON In trarlslational motion′

the systems are charactenzed by linear disPlacement linear

veloclty and linear acceleraion.¶ 隆 equatiollls of motion are fo.uLulated from Newton′

laws of mouon.Newm′ slaw Ofmouonstates thattt algebraic ttm ofthe forces acting on a五 gid body in a given direcion is equalto h Product ofthe mass ofthe body and its accebradon in the salne d士 ection.恥 sl,w.響 ."‐ lXPrSSed as、

whereM= mass of the body 4 = acceleration of the body in the direction considered r = linear velocity of the body x= linear displacement of the body.

89

s

90 CONTROL SYSTEMS ￨,事 F「

NETWORK ELEMENTS OF A MECHAN:CAL TRANSLAT:ONAL SYSTEM There are′ 乃rec nttndamental PassiVe elements or system parameters which characterize a linear mechanical translational systern.These are:

(i)

mass

(i)

(iii) friction

spring

i編 ÷ MASS IM〕 L■

■ e physical model of a masttitis assumed thatits massis concentrated atits centre.恥

function of lnass in linear lnotion is to store kinetic energy.Mass calu■ ot store potential

eneigy.SPbolic鞠 ′mass

is represented by a block as shOwn in Fig.5.1.

Suppose that a force f (t)is applied to mass. The applied force produces a displacementx(f)in the

direction

￨一

of the applied force. The relation

between force

/(f)

and Mis given by

お

――――→レスの

[II]―

=Hg.5.l A Mass.

颯

L:NEAR SPRING

おstretchet it● eS tO Conttact and r itis compressed it tries to exPand tO itS no...lallengh.It stores Potential energy during h variation of its A sPring iS a medanical device.I■

ShaPeduetOdasicdeformationresultingfromtheapplicadonOfaforceoWhenafOrce riS apphed to the sprln,a reStOrlng force/K iS ProduCed.

Consider a spring having negligible mass and connected to a rigid support as shown in Fig. 5.2. Let the applied force in the spring.

/

produce a displacement

/

r

一

s

$ Eig.

The restoring force is given by

fx=k

5.2

A spring fixed at one end.

...(s.5.1)

where K = stiffness constant of the spring By Newton′ s

∴

law′

ノ=/K ノ=豫

¨.(5.6.2) ¨.(5.6.3)

卜

χ2

卜

Suppose that the spring has displacement at both

@-+f K

ends as shown in Fig. 5.3.

fig.

5.3

,`1

A spring with displacement at both ends.

MATHEMATICAL MODELS OF PHYSICAL SYSTEMS 91 Net displacemmt in the spring is (r, to the net displacement.

-r,

)

and opposing force by the spring is proportional

/K∝ lrl― '2) 2) 良 =K(■ ―χ By Newton's law, Therefore,

1517

…66o (565)

′=ノ K ′=K(■ ―■ 2)

(566)

FRiCT:ON Whenever there is a motion

or tendency of motion between two physical

elements,

frictional forces exist. The frictional forces encountered in physical systems are usually of non-linear nature. There are three types of friction in practical cystems namely, viscous frictiory static ftiction and Coulomb friction. I Coulomb Friction Force, This is the force of sliding friction between dry surfaces. O Static Friction Force. This is the force required to initiate motion between two surfaces in contact. t Viecous Friction Force. This is the force of friction between moving surfaces separated by viscous fluid or the force between a solid body and a fluid medium. Lr most physical systems, the viscous friction dominates and the other two are neglected.

●・81

DAMPER ELEMENT The damping or viscous friction B characterizes the mechanical element that absorbs energy. The symbolic network rePresentation of damping action is a viscous damper or dashpot as shown in Fig. 5.4. A dashpot is used to damP unwanted mechanical vibrations. Sometimes, it is necessary to introduce damping to improve the dlmamics of the system. In a dashpot viscous damping occurs between a piston and a rylinder.

I t r. t.,

X rig. s.+

A damper

witl

one end fired.

An ideal dashpot with one end fixed is shown in Fig. 5.5. When a force / is applied the dashpot produces a reaction damping force proportional to the velocity. t'u

n4 dt

t',

=

B+ dt

where B is the damping coefficient.

/,

which is

...(5.8.1)

92 CONTROL SYSTEMS By Newton's law

ノ=ノ 3

¨。 (5.8.2)

Therefore,

ノ=B争

¨。 (5.8.3)

When the dashpot has displacement at both ends as shown in Fig. 5.6, the reaction damping force is proportional to the relative velocity of the two ends. That is, /B∝

ー ζ子)

(ζ ト

ん

=B(争

― 争)… 68→

著Fig.5.6 A damping with both ends iee. ¨.(5.8.5)

一

騨讐

I珪∃ 車二 B

れ丁 名丁

一 一    〓 ノ   ノ

Therefore,

ん 了入

By Newton′ s law

/―

.¨

(5.8.6)

ROTA■ ONAL MECHANICAL SYSTEMS

︲ １︲ Ｔ

Rotational mechanical systems are handled by the s,une way as translational mechanical systems, except that torque replaces force, and angular displacement replaces translational displacement. The mechanical components for rotational systems are the same as those

for

translational systems, except that the components undergo rotation instead of

translation.

Torque equations are written in rotational systems. The applied torque is equal to the sum of the reaction torques.

輩 鳳 甘 〔i7 Ⅷ詰 H■

syStem。

lnertial torque (77 )

The torque applied to a body having a moment of inertia / produces an angular acceleration. The reaction torque T, is equal to the product of moment of inertia and acceleration. The torque equation is

T,

where

= I.

=l#=l#

0 = angular displacement ; or= angular velocity ; cr =

angular acceleration.

"。

(5.9。

1)

MATHEMATICAL MODELS OF PHYSICAL SYSTEMS 93

Reaction Spring Torque

I"

When a torque is applied to a spring, the spring is twisted by a. angle 0. The applied torque is transmitted through the spring and appears at the other end. The reaction sPring torque T" that is produced is equal to the product of the spring constant Kand the angle of twist. The reaction torque is given by Tx = K(0, -0a )

...(5.e.2)

where cand, dare the ends of the spring. The angles 0, and 0, denote the positions of the two ends of the spring measured from the neutral position. Reaction Damplng Torque

(Tr)

a body moves through a fluid, which may be either a liquid or produce motion of the body a torque must be applied to overcome the agas such as air. To reaction damping torque. The damping is represented by a dashpot with a viscous friction coefficient B The damping torque is equal to the relative angular velocity of the ends eand / of the dashpot. The reaction torque of a damper is

Damping (rccurs whenever

T"=B(a,_r.f)=

rt aa_ "l) dt (

"

...(s.e.3)

)

Table 5.1 shows analogous quantities in translational and rotational motions.

faUte 5.!

Analogous Quantities

in

Translational and Rotational Motions

TFa"`Ltio,,α Z ttO“ θ

IROtttiο α″鷲醸

Mass (M)

Inertia

Damper coefficient (B)

Damping constant (B)

Spri.g constant (K)

Sp.i.g constant (K)

Force

(/)

Displacemmt Velocity

ffiffi

GEAR

0"

“

“ (

/)

Torque (Q

(r)

(")=#

A.gul*

displacement (0)

Ar,gola. velocity

,=#

rRAlNs

Gear trains are used in control systems to attain mechanical matching of the motor to the load. Usually, a servomotor operates at a high spoed but at a low torque. To drive a load with a high torque and low speed by such a motor, the torque magnification and speed

reduction are obtained by a gear train. Thus, in medranical systems, gear trains act as matddng devices like transformers in electrical systems.

94 CoNTROL SYSttEMS Consider a gear train shown in Fig. 5.8.

F

Gear l

(Primary gear)

目 脳

NI teeth

T1 01

Input torque from motor TM

Gear 2 (Secondary gear)

,■ 9.5.8 Gear train.

Let

Nr = number of teeth on the first gear Nz = number of teeth on the second gear Tu = torque developed by the motor

Tr = load torque 4 = inPut torque on the first gear Tz = input torque on the second gear 0r = angular displacement of the first gear 0z = angular displacement of the second gear /r = radius of gear 1 rz = radius of gear 2 .lr = moment of inertia of gear

/z = moment of inertia

1

of. gear 2

Since the number of teeth on the gear wheel is proportional to its radius

rcc

rr

N

€Nr,

t2aN2 …(5.10.1)

     

一 一       〓

   

２ １ り ０ 一 〇

４

１ ０   ■ 一４

Assuming that no slip takes place between the gear teeth that is the gears are perfectly meshing then the linear distance travelled by both the wheels is the same. Mathematically,

¨.(5.10.2)

  ・

MATHEMATICAL MODELS OF PHYSICAL SYSTEMS 95

Assuming ideal gear conditions with no power loss, work done on the motor and load side is same 10。 3α ) ¨。 (5。

¨.(5.10.3b)

111警 ￨ Combination of Eqs。

(5.10.2)′

(5。

10.1)and(5。 10.3)gives ¨。 (5.10.4)

=ql=0272 “

‐ 寺￨:￨ For an ideal traFtSfO■

¨。 (5.10.5)

..ler2

υl_■ υ2

and

'1 _“ ■00狙

2

…(5.10,71

■

'2 I we∞ mptt Eq&●

¨.(5。 10。 6)

42

dO■ 00 We ObSe配

価

they aК

dentid.I芳

お

considered as the tums ratio ttL for an ideal transformeL torque is analogous to voltage. "2 Thほ gears have a teeth ratio and can be treated as torque transfb...lers.

Companson of Eqs.(5.10.5)and(5。 10.71 showS that anmar velodty is analogous to current.罰 s forms the basis of the force― voltage￠ η andOgy as sho― in Fig。 5.9(α

)・

If

r2:rl iS taken as h turrls ratio■ :“ 2′ We have the analogous ideal transfo.uler based on

forc― rrent(ノ ー:)ana10gy sho― m brackets.

in Fig.5.9(b)・ The corresPonding medanical quantities are

i2(ω リ

ち(01)

'2(T2)

'1(TI)

″1:″ 2

(rl:r2)

(r2:″ 1)

(α )

(b)

f an■ Ogy. S Fig.5。 9 1deal transfOrllner as eLct五 cal analog of a gea train● )′ v anttOgy(b)′

96 cONTROL SYSTEMS Thc differential equalon of the rnotor― side shaft is

+.aOl+.

生

L=ム

(5108)

The d■ fferential equatlon of the load― side shaft is

T2=J2':+32劣

(5_109)

￨+TL

=1与

We have′ 与

r 。

.=N2Tl

COmb― g Eqs(5109)and(51010)

・ 持1=争 +ら 十 ・ Differentahg Eq(5103α

)ル Йth

…(51011)

resPed tO`

・等 争

¨(51012)

=Ъ

Differentaing Eq(51012)■・lth

respect to t

η 争 争 =Ъ

Combinaton of Eqs(51010)′

¨(51013)

(51011)′

(51012)and(51013)Jves +321[う

TI=:[[J21与 争

°r

Tl=ア 2(:半

￨+TLI +(:学

+Iち

:)21il‐

(:単 :)21;‐

:)TL

…(51014)

Subsitutt the value of Tl from Eq(5104)In Eq(5108)′ we get

TM=IJl+ノ 2(持

Or

● ・ here

■〈+(1:)TL “ ′ 「

TM=/1“ /1“

十

)21J:;=+IBl+32(ミ 与)21`;卜 〔特 )TL

1;;￨十

1 +(:キ

BI`,=■

_15.10161

;卜

2

:ヶ

…15.10.15)

…15.10.171

)2

+(1:)232

¨(510.18)

MATHttMAT:CAL MODELS OF PHYS!CAL SYSTEMS 97 The terms

ur" known as equivalent moment of inertia and equivalent viscous (motor side).

^d 4r, friction coefficients respectively referred to primary

The term

]rr,

(ru. )

[^,, ]r.

is the load torque referred to primary (motor side).

Similarly, we can write

(」

‖ ￨)TM=J2“

+TL

1:l争 +32“

…(5.10.19)

`;卜

where 12,,

= equivalent moment of inertia referred to secondary (load side)

(w \2 Iz,q=lz.[*], \rq

= viscous

friction coefficient referred to secondary (load side) ■

︲︲︲︲′ ヽ１ ノ

２鴫 Ｎ一

ら

イーーーヽ ＋

“

〓

ら

…(5。 10.21)

Ｌ

ヽ︱ ︱ ︱ ︱ ノ

イー ヽ ︲︲︱

２鴫 Ｎ一

11 FREE¨ BODY

"'(s'10'20)

¨.(5.10.22)

= motor torque referred to load side (secondary)

D:AGRAM

｀

=:二

h order to obtain the matheinaticalinodel of a rnechanical systein′ the first step is to draw the free body diagram。 ■■a free body diagraln eaこ h rnass is isolated from the rest of the

systern.The forces acting on the inass that causes disPlacement and forces oPPosing the diSPlacement are indicated. ■■ e following procedure is used to draw the free body diagram: 1. Assume the direction of disPlaCement of the rnass as PoSitiVe direction.

2.Find all the forces with their directions acting on the mass.h aPPhed fOrCe isin the positive direction of disPlacement.■ ℃forces due to spring/dash Pot and mass

are opposite to the direction of disPlacement.

￨

3.APPly Newtors laws of mouOn to exPresS all the forces in terms of disPlacement or velocity of mass. Once the force body diagram is obtained′

hc differential equauons can be written easily

by equating the sunl of apphed forces to the sunl of oPPOSing forces.

98 coNTROL SYSTEMS ￨11雲

曇 贔 島 出 二 :二

働 ″′ κル

11

Fな

リ ″ 制 “

α′ わηs “

go爾 ″jtt Йθ 燿 I systt sk切 れ `山 "滋

。 卸ω ■ル磁 わ 滋り勧 ・ ・ ψ為 "器 ル “ “ rル

"グ

Eig.

5.10.

(a) Mass, spring and damper rystem.

Solution. We first draw the free-body diagram, placing on the body all the forces that act on the body either in the direction of motion or opposite to the direction of motion. The following forces act on the system assuming that mass is moving towards right

:

1. Applied force /(f) acting towards right. 2. The inertia force, damping force and spring force impede the motion and act opposite to the direction of motion.

卜 わ

ん0=M争

Inertia b鳴

o=B争 ん Spring forcer fK(′ )=I狐 ′ Damping for鴫

′′

ん M=

)

$ ng.

t.ro

(b) Free-botly diagram

The free― body diagram is shown in Fig.5.10(b).

By Newton′ s la"the Jgebraic sum of all the forces must be equal to zero.

ノ(′ )一 /M(:)一 ノ3(′ )一 /K(′ )=0

0r

ん )+ん )+良 (′

M禦

(′

(′

)=/(サ )

+BΨ +Ц →=知

Assuming zero lrutial conditions we take the Laplace transfo...1 0f both Eq。 (E5.1.1).

r 。

¨。 (E5.1.1)

搬 ′

‐

sides of

Ms2xls)十 ,SX(s)十 KXls)=F(3)

(Ms2+3stt

ЮX(χ )=F(S)

.,.(E5。

1.2)

MATHEMATICAL MODELS OF PHYSiCAL SYSTEMS 99 The transformed free-body diagram is shown in Fig. 5.10(c). 卜

XIS)

豫 lS

BsX(s)

us2x1s1

*

ng. S.rO

(c) Transformed free-body diagram (d) Btock diagram.

Solving for the transfer functiori gives

-t G(s)=x(:)-\-' F(s)

¨。 (E5.1.3)

Msz + Bs+ K

The block diagram is shown in Fig. 5.10(d).

:趙

淵鳳

i

g夕88-b。 ゥ

′″ α″ α ′ ωγj″ 働θα髪 力 ′ ″J`7“ ″ れS滅 ゎcrル ,電 ′ ` `わ ““ “ 『 Sbω れ Fな 。5110 αη′θ″″れ ル 磁 解騰 施 SyStt ″わ ・ ウ げ "α "ウ ル "奪 書 “

D″

"働

卜

χl

卜

χ2

Eig. 5.11 (a) Mechanicat system

Solution. The free-body diagrams for masses fuf, *d

respectively K,(r,-r") I-r Ml I f$)+l arr. I 1+- M,---i(b)

M,

are

shown in Figs. 5.11(b) and

5.11(c)

Fig.

5.11

r

J-l+ K(x, -r)-+l M2 I I l.+-

Kg,

M,-ar*-

(c)

(b) Fee-body diagram for masslf, (c) Free-body diagram for massj[r.

Force equation for mass lz1, is

スの=Mli:￨十 く01-ち

)

¨。 (E5.2.1)

Force equa■ on for mass M2

ヽ

角 摯Tり =亀 12+1争 ‐

¨.(E5.2.2)

一 ●

ヽ ヽ  一 ゛

．

100 CONTROL SYSTEMS Assuming zero lllitial conditions and taking Laplace transforms Of bOth the sides of

equations(E5.2.1)and(E5.2.2)we getthe system equauOns as

F(s)=s2MIxl(S)+KIXl(S)一 KI KIXl(S)一

X2(S)

¨ ・(E5.2.3)

KIX2(S)=K2X2(S)+S2M2X2(S)

Frorn Eq.(E5.2.4)′

¨。 (E5.2.4)

0≪ 軌+え 十 )T 為 絶

I誦

■

2つ

Subsituing the value of Xl(s)frOm Eq。 (E5.2.5)in Eq.(E5.2.3)′ we get

24+ぇ

=聟

駒

十K211S24+角 卜

●

KIX20

Therefore′ X2(χ )_

F←

職黎

儡

Dα

Kl

)●

2Mltt K)● 2M2+え

″″ 竹 ″ 物 働α′′ ル ″θ′ わ is′

"χ

力 θθ″タナ “

4ψ γル

十K2)Кf

肋 “

′θ:4′ :α ′

″ “器

“ Pis′ 力″ 4′ 夕 ′fO

Vψ 燿 働ω 肩

r力 θsysr",α

"タ

Fな “

夕″ 昭

"2鰯

ルθ″θ′ :ο ηχ Oげ 働

`♭

.

s01ution.The equation of lnotiOn for the systelln showlll in

丁﹁

Fig.5。 12 is

M争

+く

争

号

Mll,キ

)+KIXO― 十

B年

十 燈

わ

=0

0=B争

¨ .(5.3.1) K巨 +燈

J

...(5.3.2)

Taking Laplace transform of Eq. (5.3.2), assuming zero initial conditions′ we get

墨Fig.5。 12

Ms2xO(s)十 BSXO(S)十 KXO(S)=BSXJ(S)十 KXf(S) 。r

(Ms2+3stt

Ю

XO(S)=(3stt Ю XJ(S)

Hence the trarlsfer function is given by

XO(S) 3s+K X:(S)Ms2+3s+K

[量 」

B

θグノ

MAttHEMAT:CAL MODELS OF PHYS!CAL SYSTEMS 101 恥

mメ e二 4

動 励

17a rr24ψ

rル

"ゎ

れ 器

/Or rル

V疵 ″ 盛

脇

"4F機

s01ution.The equation of motion fOr the system is

九 十/均 =/K2

・ 0r

(1;￨一

χ O)=亀 χ 〈 ;:)+Kは 十 =・ 1;F+K嶋 ・ ,一

0

4ふ :十

(え

K21χ 0 ...(E5.4.1)

Taking LaPlace transform of Eq.(5.4.1)′ aSSuming zero u■

itial

conditions′ we get BIsXJ(s)十 KIXJ(S)=BISXO(S)+(K+K2)XO(S)

or

(4s+&)X,(s)=(4s+&+Kr)\(s)

¨.(E5.4.2) I Fig.5。 13

Hence the transfer function is given by XO(ζ )_

BIStt Kl

Xj(S)■ S+Kl

tt K2

I

-.trxarnple 5.5

姥 励

ン “

IH…

ψ

ル

肋 “

“器

ル

滋 ツ 勧

Fを

説耐

地

“

14

S01utiOn.The equation of rnotion for the system showll ln Fig.5.14 is

/31=/M+/32

・際号)=M争 争 ・争=Mtt・ )争 +ち

∝

l・

+亀

¨。 (E5.5.1)

Taking the Laplace transform of Eq. (85.5.1), assuming zero initial conditions, we get ■ SX,(S)=MS2xO(s)十 (■ 十 ■ )SXO(S) Or

(31s)X,(S)=[MS2+(Bl+32)SIXO(S)

＼

102 coNTROL SYSTEMS Hence the transfer function is given by

4s _ X:(S)Ms2+(■ +ら )S X.(S)_ ■ X。 (S)MS+Bl+32 xo(s)

Ettmplc 5 6

(E552)

Obtain the transfer function of the mechanical system shown in Fig. 5.75(a). Also obtain the transfet function of Fig. 5.15(b). Shmt that the tuansfer functions of the tuo systefis arc of identical form and thus they are analogous systerns.

一

′

i

］ 

(a)

(.)

*rs. s.rs

Solufion. No extemal force is acting on the system and the displacements are due to initial condition only. The equations of motion of the mechanical systems are

-!)

-.xn;= c(o': s1,4a-4rnl.^,(x, "' '( dr dt) '[df dt)

s(!+-9)=** '\ d, dt)

...(Es.6.r)

...(Es.6.2)

By taking the Laplace transforms of Eqs. (E5.6.1) and (E5.6.2), assuming zero initial conditions, we get l ISX,(S)― SX。 (S)]+Kl【 Xi(S)― XO(5)]=B21SXO(S)― SYls)1

32卜 XO(S)― SYls】 =K2γ (S)

(E563) ¨(E56.4)

MATHEMATICAL MODELS OF PHYSICAL SYSTEMS 103 subsu仙 血碍 the Value of 噂

γ(S)frOm Eq(E5.6.4)h Eq.c5.6.31′ we get

ナ SXOISll■ 組

倒 →

SXOISl昨 裂

いOISl l・ 針 い うSttp …

lE56つ

Hence the transfer function is givm by

XO(S)_

31s+Kl

和 (昨 .KI.L}昨

乱

(■ S+Kt)(32Stt Kち (■ Stt Kl)(ら

)

S+K2)+32く 32S+K2)《

S2

_ (31s+KI)(32S+K2) (lS■ Kl)(32S+K2)+32SK2

ち

Klス (f:S+1)(fttS+1)

_

KI K2(f}S+1)(fiS・ 1)+:き S

〓

和一 和

Or

1)(1:S+1) (1:S‐

(E566)

(f)S+1)(fiS+1)+flS The transfo...led circuit of F七

515(b)is Sl10-h Fig 515(σ )

40=凡 +ホ Z21SI R2峰

=機

_

EOo_ 4o _ ・ +ホ

Ю知名0ぃ ホ +機 。r

EOISl= (RICIs■

1)(R2C2S・ 1)

Ei(s)(RI CIs■ 1)(R2C2S・ 1)+R2qs

…(E56の

By comparing the transfer functions of Eqs. (E5.6.6) and (E5.6.7) we see that the systems shown in Fig. 5.15(a) and 5.15(b) are analogous.

104 coNttROL SYSTEMS D″ υ ι夕θabθ ウ グ g″ 協 α″′ωγj″ ′ 力θヵ γ θ われSげ 働じ″ι αれ,“I ィタα′ j′

s力

r$

ffi Fig.

5.16

θ ″ “

`力 j“

“

Fな .5.16級 ).

“

sys″

“

(a) Mechanicat system.

Solution. The free-body diagram of &1, is shown in Fig. where

5.1,6(b),

ん=■ 争 ん 1=М 争

".(E5.7.1)

¨。 (E5.7.2)

ノn=Kl(χ l― χ 2)

¨。 (E5。 7.3)

By Newton's second law of motior; the force equation for mass ,i\( is

Im+fn+fru=0

¨。 (E5.7.4)

Therefore,

oq#*.q*.

K1(x1-xr)=s

卜

・ ん

/Kl

ん  島

/X3

ん

5。

χ2

力)

ふ ふ 魚

量Hg。

¨。 (E5。 7.5)

160)Free― bodyこ agraln of4(o Free― bodyこ agraln of″ 2(d)・ ee‐ bOdyぬ agraln of″ 3・ ■ le

free― body

diagram Of M2 iS Sh° wn in Fig.5.16(c)′ where

/K3=K3χ 2

¨。 (E5,7.6)

/Kl=Kl(χ 2 χ l)

.¨

(E5.7.η

MAttHEMAT:CAL MODELS OF PHYSiCAL SYSttEMS 105

7助 ・誦 C'Z幼 …

ん2=場 争 ん=場 (争 ― 争)

By Newton′ s second law of moion′ the force equauon fOr mass M2 iS given by

/M2+/32+/Kl+/K3=ノ

¨。 (E5.7.10)

(ナ )

Therefore

L争 +・ (争 号 く

02 Xl l・

)十

The free‐ body

=バ め ち 穐

diagram of M3 iS Sh° wll in Fig.5.16(′

)′

...(E5.7.11)

where

rM3=M31::￨

..。

(E5.7.12)

/32=B2(ζ 暑 1等 L)

¨。 (EI.7,13)

/K2=K2χ 3

(E5.7.14) “。

Dy Newton′ s second law of mOtion′ the force equaion for mass M3 iS

ん 3+/32+/k2=0 Therefore′

M3瓦

戸 扇 「

十B2(1;i 〈

;「 )+K2χ

3=0

¨,(E5.7.15) ¨。 (E5。 7.16)

Equauons(E5.7.5)′ (E5.7.11)and(E5.7.16)are the fOrce equa■ ons ofthe meⅢ aniCalsystem shown ln Fig.5.16(α ).

111121 NODAL METHOD OF DEVELOP:NG MECHAN:CAL EQUiVALENT NETWORK The following steps are used to develoP an equivalent mechanical network from a mechanical system omode basis :

L. Identify all the displacements in the medranical system. 2. Represent eactr displacement by a separate node. 3. Choobe arbitrarily a reference node. 4. Identify the elements whidr are under the influence of different displacements. 5. Connect all the elements in parallel under the respective nodes which are under 5.

the influmce of respective displacements. Elements causing change in displacement (either friction or spring) are connected between the respective nodes.

The force equation for each node can be written by equating the algebraic sum of all the forces at each node to zero.

106 coNttROL SYSTEMS

E_ple 5`

For the part of a physical system shown

in

Fig. s.17(a) draw the mechanical

equiaalent network.

馘H多 ■17

o腱

chaniC■ System.

(b) Equivalent mechanical network.

Solution. Lr this system all the elements

M B arrtd, K are under

the influence of r and all

displacement r. Hence, its mechanical equivalent network will contain one node the elements will be connected in parallel between node r and reference nod.e. The mechanical equivalent network of Fig. s.17(a) is shown in Fig. s.l7(b).

For the physical system shown in Fig. sJ3(a) draw the mechanical equioalent nefiaork. Write the system equntions in time domain and s domain. S01utiOn.

丁為

■

2

I

xg.

s.ta

(a) Physicat slrstem

Step"

1

(D) Mechanical equivatent network.

Since there are two linear displacements

r, and rr, choose two nodes r, and

x2 corresPonding to these displacements. Ctroose an additional reference node as shown in Fig. 5.18(b).

MATHEMATICAL MODELS OF PHYSICAL SYSTEMS 107 St p

2

lhe elemmts iV{, 4 and K, are cormected to a rigid support.

These elements

rr. connected in parallel betwem node .r, and

are under the inlluence of displacemmt

Therefore,

rq, q

and

I(

are

reference node.

St"p

3

There is a drange in displacement &om rr to x2 due to the presence of elements B, and Kr. These elements are under the inlluence of displacement (r1 -r2)

Therefore, \ nd K, are connected in parallel between nodes r, and rr. A mass cannot have two displacements, therefore, no mass can be connected between two nodes.

St p

4

Mass iVI, and force

/(l)

are under the influence of node

rr.

Therefore Al" and | (t) are connected in parallel between node refermce node.

r,

and the

The medunical equivalent network of the givm system is shown in Fig. 5.18(D).

At each node the algebraic sum of the forces must be equal to zero. At node 11,

1争 ■争:こ

(E591)

Assuming initial conditions as zero and taking Laplace transform of Eq. (E5.9.1), we get

lqs2xr1sl+ 4s4 。r

1s)+

I{,x,(s)+ K,t{,(4-xzG)l=0

[MIs2+(1+32)S+(Kl+K2)]Xl●

At node″

)=(32S+K2)X2(S)

(E592)

2′

¨(E593)

As興 颯

血

ial CO述 Em as zero and taking Laplaceむ ansfo...l of bo■ sides of

Eq(E593)we get M212x2(S)+32 1SX2(S)iSXl(S)卜 (4s2+32S+K2)X2(S)(ら

K2【 X2(S)

Xl(S)]=F(s)

S+K2)Xl(S)=F(s)

Equations (E5.9.1) and (E5.9.3) describe the mechanical system of Fit. domain.

...(E594) 518(′ )h

ime

Equations (E5.9.2) and (E5.9.4) describe the medranical system ofFig. 5.18(a) in s domain.

108 coNTROL SYSTEMS Draw the mechnnical equioalent nehuork for the system shown in Fig. 5.19(a). Wite the system equations and determine the transfer function relating ire displacements

i,

xr.

and.

χ2

卜

恥

.5。 19(a)Mechamcal

system.

SOlutiδ h.

S“ p l chOOse ttO nodes,l and

χ2 C° rresPOndmg tO dlsPlaCements■ ChoOse an addlt10nal iference nOde

Step 2 恥 the sPnng ls underthe duence of nodes tt and χ 2′ dement K be■veen tt and χ

md

C° nnect

χ2

the sPnng

2

Step 3 Theelements Mand Bare underthe m■ uence ofdisPlacement″ 山ereb瀾 ゥM 2′

and B"咀 ¨ cOnne¨ d h Pa● llel bemeen nOde χ 2 and rettrence node

Step 4 COnnectfOrceノ (r)be"eentt andreferencenodebecause f(r)iS COnnected to

spung The mechanical equivalent network is shown in Fig. 5.19(b).

At

each node the algebraic sum of the forces must be equal to zero.

flql

At node 11, ¨(E5101)

At node 12,

5.19 (b) Uechanical equivatent network.

良=M.生 +B争

¨.(E5102)

Eqs(E5101)and(L102)are he syStem equatons h temts Of山

貯 lacemen`朽

md

χ2

Combhmg Eqs(E5101)and(E5102)′ we get KIXl一

M争

+3年

つ=M争 +3年

+K■ 2=ス ■1

¨(E5103)

MATHttMA丁 :CAL MODELS OF PHYSiCAL SYSTEMS 109 Assullning zero initial conditions and taking Laplace Eq。 (E5.10.3)′ we get

transfom of both sides of

ん4s2x2(S)十 BSX2(S)十 KX2(S)=KXl(S)

(Ms2+Bs+K)X2(S)=K/Xl(S) The trallsfer function is

X2(S)_ K Xl(S) MS2+Bs+K J轟 轟

α′ :θ κ S力 γルθSysナ ακεθ `“ `7夕 “

事 曜￨1町

Wri′`′力θ′`ψ γ

茄 η 協 γ η ψ ル 器 “

ｏυ属留 聖屁

Fig.

5.20

¨。 (E5.10.4)

ω s力 θ

"物

Fな .5.2θ (α

)′

α グル ′働ι “

K2

(a) Mechanical system

Sohrtion. Step

1

Since there are two linear displacements

r,

and 12 choose two nodes

x,

arrd

x,

corresponding to these displacements. Take an additiorral referenee node as shown in Fig. 5.20(b)-

Step2 The elements Iv\,4 *d K,

are under the influence of displacement 11. Therefore, Mr, 4 at d & are connected in parallel between node r, and the reference node. Since, the force f (t)is applied to 1:;1.., it is connected between r, and reference node.

烹Fig.5。 20。

(b)Mechanical equvalent network.

110 CONTROL SYSttEMS

Stf"3

to xrdue to the presence of damper element Br. This element is under the influence of displacement (xr-xz\ Therefore { is connected between nodes r, and rr.

4

The elements Mr, B, and K, are under the influence of displacement xr. Therefore, M2, 4 and K, are connected in parallel between node r, and the

St..p

There is a change in displacement from x,

reference node. The medranical equivalent network of the given system is shown in Fig. s.zo(b). At each node the algebraic sum of the forces must be equal to zero.

At node

r, ￨‐

/Ml+L+名 ￨九 一ん3=ノ

ル 争￨》 二 十,11●争‐ At node

r,

￨￨ 半争‐

■￨1

1‐

¨。 (E5.11.1)

=′

¨。 (E5。 11.2)

￨￨￨￨￨:￨1111111111・

￨‐

・ 「 「■￨￨￨￨￨￨￨￨￨:￨11111111111:￨￨￨￨￨「 「 11111111111111 ￨「

Equauons(E5.11.1)and(E5.11.2)are the system equauons in dme dOmain.

Assuming zero initial conditions and taking Laplace transforms of both sides of Eqs. (E5.11.1)and(E5.11.2)′ we getthe system equauorls in,domain.

s2MIxl(S)+■ SXl(S)+33SXl(S)+KXl(S)一 B3SX2(S)=F(s) and s2M2X2(S)+32SX2(S)+33SX2(S)+K2X2(S) LSX10)=0

¨。 (E5.11.3) "。

(E5.11.4)

Equations(E5.11.3)and(E5.11.4)can be Written in the fo...1

(A4s2+BIs+亀 and

― ら

s+Kl)Xl(S)一 B3SX2(S)=F(s)

SXl(S)十 (M2S2+ら

s+33Stt K2)X2(S)=F(S)

¨。 (E5.11.5) ¨。 (E5。 11.6)

Equauons(E5。 11.5)and(E5.11.6)call alSO be w五 世en in matrix fo..ll aS )Stt KI

M2S2+(:∫

lL)s+K2][][￨:￨]=[FlS]

IMIs2+(::11lЪ

LetransferfunctiOnl:lilCanbefomedasfo1lows: From Eq.(E5.11.6)′ 亀

SXl(S)=(M2S2+32S+亀

猟 キ

Stt K2)X2(S)

琢 ⇒

¨。 (E5.11.7)

MATHEMATICAL MODELS OF PHYSiCAL SYSTEMS lll Subsituting tte value of Xl(s)frOm Eq.(E5.11.の in Eq.(E5.11.5)′ we get Щ

♂

I BISI場

―

印

LSX21キ

的

∝ 器

=

¨。 (E5.11.8)

Dγ 硼″ 働θπ

ari″ 働θ "ο

$ ng.

t.z,

α :“ J`″ i“ ″ ′ θ勧 戒 ル 7働 θSys佗 ″ s力 οω f“ Frig.5.210,α “ “ “ “

`働 “ ′αιθ 7“ α

"′

`あ "S・

(a) Mechanical sptem

sOlutiOn.

Step l

take tfuee nodes r, r, and x, corresponding to these displacements. Take an additional reference node as shown in Fig. 5.21,(b).

Step 2

The elements fuI, and xr. Therefore I{4, K1 reference node.

Step 3

There is a drange in displacemerrt frorrrr, to r, due to the presence of elements 4z md Krr. These elements are under the influence of displacement (r, -r, )

Since there are three linear displacements

Therefore, 4z md

x, x, arrdr,

& *d force f are under the influence of displacement {rd f are connected in parallel between node x, and

{rz

are connected in parallel between nodes

x,

and xr.

112 CONTROL SYSTEMS Ftqn..* The elements IvIr, K, and B, are under the influence of displacement rr. Therefore, M2, K2 and 4 are connected in parallel between node r, and reference node.

Ft"e".5

There is a change in displacement from xrto

x,

due to the presence of element

Krr. The spring element K* is under the influence of displacement Therefore, K* is connected between nodes x, and xr. Ftpp

6

(xz-xs\

The elements IvIs, K, and B, are under the influence of displacement xr. Therefore, Ms, K3 and B, are connected in parallel between node r, and reference node. The mechanical equivalent network of the system in Fig. 5.21.(a) is shown in Fig.5.21(b).

二Fig。

5。

21 (b)MeChanical equlvalent network.

Step 7

APPly fOrce equation at node

χl to get

― 12チ lXlリ =М の ス 争+ヽ けKメ 為 X21・

¨.(E5。 12.1)

By taking the Laplace transform of Eq.(E5.12.1)and assuming zero iniual conditions/we get F(s)=MIS2xI(s)+KIXl(s)+K12[Xl(S)一

X2(S)卜

■ 2SIXl(S)一

X2(S)卜 °(E5.12.2)

Step 8 The force equation at node χ2 iS given by ｀ メ■ ‐

>■ 2チ lXI X21 L争

+ら

十 K2X2

1昴

2助 ・

年

Assulning zero initial conditions and taking LaPlace transforin Of Eq.(E5.12.3)′ we get

K12[る

(S)一

X2(S']+B12SIXl(S)一 X2(S)]=M2S2x2(S)+32SX2(S)+K2X2(S) ¨。 (E5.12.4)

MATHEMAT!CAL MODELS OF PHYS:CAL SYSTEMS l13 Step 9 The fOrce equatiOn at node χ3 iS given by

K2302 χ 3)=M3生

+亀 ∠二二十K3χ 3

.:。

(E5.12.5)

By taking Laplace transform of Eq. (85.12.5) and assuming zero initial conditions, we get K23[X2(S) X3(S)]=AttS2x3(S)+亀 ::￨

=xample●

Wr′

SX3(S)十 K3X3(S)

″′ ためη ′ われ Sげ 働 協 α J sysセ sλ θ ″ ωれ′Fな 。 5.22o).ob協 れ″ `r。 `わ “ ``9“ ` “ “ “

彎密″ "器・ C島

O “ I

Fig。

5.22

¨。 (E5.12.6)

(a)Rotadonal system

T

@) Free-body diagram

Soludon. The free-body diagram of the mechanical system of Fig. 5.22(a) is shown in Fig,.5.22(b).

According to the Newton's law of -oUoL the applied torque is equal to the sum of the restraining torques. T

=7, +Tu

+T*

...(E5.13.1)

r=l**BS+Ko ' dtz dt

...(8s.13.2)

Assuming zero initial conditions and taking the Laplace transforrr of both sides of Eq. (E5.13.2), we get T(s) = /s20(s)+ Bs(s)+ KO(s) T(S)=(IS2+Bstt K)0(s)

¨。 (E5.13.3)

Thereforer the trmsfer hnction of he system is

‐‐

.

ALTERNATiVE M ETH口

‐

0(S)

1

T(s)Js2+Bs+K

¨.(E5。 13.4)

D

First we draw the mechanical equivalent network of the given rotational system. In it, there is only one node angular displacement 0. Therefore, choose one node corresponding to angular displacement e. The elements I, B K and torque T are under theinfluence of 0. Therefore, they are connected between node 0 and reference node. The mechanical equivalent network of the rotational system of Fig. 5.22(a) is shown in Fig. 5.22(c).

.

r(')

Reference node

凸 Fig.5.22 繁

(c) Mechanicat equivalent

network.

l14 CONTROL SYSTEMS 151131 ANALOGOuS C:RCU:TS The circuits which represent the systems for which differential equations are similar in form are called analogous circuito. The corresponding variables and parameters in two circuits repres€nted by equations of the same form are called analogous. Mechanical systems can be represented by electric networks, force-voltage analogy or force-current analogy.

or vice.versa either by

51141 SER:ES RLC CiRCUIT Fig. 5.23 shows a series RLC circuit. By KVL υ(′ )=υ R+υ 二十υc

∝

州=R+Lチ

q

+そ

・

R

●L

%

…15■ 41)

量Flg.5.23

^.

dq SmCe , = j

dt

υ (1)FRラ

F+L字

+1そ

∝[字 +R響 +:9=υ 0 15:■ 5‐

¨(5142)

PARALLEL RLC CiRCU:T Figure 5.24 shows a parallel RLC circuit. By KCL

jR+'L+'C=J(1)

景

+t∫ υ′

'+C:子

¨(5151)

=(`)

…(51521

If v is the nux鵬 ge ′′

V=∫

′ υ′

空 =血 ′ ,2 へ ′

■       Ｒ

υ =ニ ェ

¨(5153) ¨(5154)

¨(5155)

はFig.5.24

MATHEMATiCAL MODELS OF PHYS:CAL SYSTEMS l15 Subsituting the values from Eqs.(5。

∝

15。 3)′ (5。

15.4)and(5.15.5)in Eq.(5。 15.2)′ we get

十 +C争 り Ψ 七 +iv=く り Cう 十 1争

ISl161 FORCE‐ VOLTACE

=く

iγ

…50 0■

i半

ANALOGY

Figure 5.25 shows a translational spring-mass-damper system. The force /(l) on mass M produces a displacement r(f ) The inertial force f y, the damping force f , and spring force /* oppose the motion of the body. By Newton's second law of motion

fu+ fs+ f*= f (t)

...(s.16.1)

M#. ,#+ Kx = f (t)

...(s.15.2)

F,ro

.:$

$ Ftg. s.25 For a series RLC circuit, we have from Eqs. (5.1,4.2),

- -''1\ L4*nfu*1 dt il=tt(t)

...(5.16'.3)

dt,

If we compare Eqs. (5.1.6.2) and (5.15.3), it is observed that these equations are analogous and the following analogies are established :

1. Applied force /(f) is analogous to applied voltage. 2. Mass M is analogous to inductance L. 3. Coefficient of viscous friction B is analogous to resistance R 4. Spring constant K is analogous to the reciprocal of capacitance 5. Displacement I is analogous to electric charge. Since the quantities

C.

M4:+, B4l, K*, f (t)areforces *rd Lq ,*#,!aqa(t)arevoltageq dt' dt dt'' dt' c "

the above mentioned analogy is called force-voltage analogy.

ヽヽ

＼

116 CONTROL SYSttEMS 1駐

171 FORCE‐ CURRENT

ANALOGY

The force equation for a lnass/sPring′ damper systern is

M争 +B争 +隆 =/0 For a Parallel RLC drcuit we haVe from Eq。

…(5.17.1)

(5.15.6)′

C争 十 1+liv=く り

…(5.17.2)

By comparing Eq. (5.17.1) and Eq. (5.17.2), we see that they are analogous. The following analogies are observed :

1. Applied force f (t) is analogous to source current i(f) 2. Mass M is analogous to capacitance. 3. Coefficient of viscous friction B is analogous to reciprocal of resistance R a. Spring constant K is analogous to reciprocal of inductance. 5. Displacement r is analogous to flux linkages y.

s^r"

M*

,

B#,

Kr, f (t)are forcesan o

r# ,+# ,lv, i(t)are currents, the above

mentioned analogy is called force-current analogy. Table 5.2 and Table 5.3 show force

fable

5.2

curent (f -D t d force voltage (/-o) analogies respectively.

Analogous Quantities (Force-Current

(f-i)

Anatogy)

Medrnnicol Systems

Electric*l

￨

/

Torque

T

Current

M

Moment of inertia

J

Capacitance

Viscous friction coefficient

B

Viscous friction coefficient

B

Reciprocal of resistance

Spring constant

K

Torsional spring

K

Reciprocal of inductance

constant

Angular displacement

Flux linkage 〓

船一 凌

Angular velocity

ｎｖ      ω

〓

″ 歳一

Velocify

χ   ′

Displacement

Voltage

Srsstems

︲ ︲︲ ．． ︲ 肝 卜一 ．計 融″ ′Ｃ

■■■●■■

￨￨￨￨￨￨￨■ tr(o

MAttHEMAT:CAL MODELS OF PHYS:CAL SYSTEMS l17

fabte

5.3

Analogous Quantities (Force-Voltage

(f-v)

Analogy)

Mc`ル ″ α″

Mechanical Systems

EI′ r`I・ Ic′ J

“ `ε

Torque

Voltage

Mass

Moment of inertia

Inductance

Viscous friction coefficient

Viscous coefficient

Resistance

Spring stiffness

Torsional spring

Reciprocal of

stiffness

capacitance

Displacement

Angular displacement

Charge

Velocity

Angular veloclty

Current

friction

The analogies are

:

Force-voltage

(

f -tt analogy

nt (f-il analogy

Force-curre

f i

M0 and proceed to comPlete the rest of the array. We then apply the stability criterion by taking the limit as s +0. This method is called epsilon methoil. ￨

￨

210 coNTROL SYSTEMS Method 2 Put

s=

1 in the originar characteristic polynomial. Rearrange the polynomiar in descending

powers of r. Develop new Routh array. Examine the terms of the first column for the number of changes in sign. The method is called rmerce coeficiant methoil. Case

2

All elements of one Row are Zero (An Entire Row ls Zero! This situation occurs whaneoer the preceding two rows are proportional ta each other. Here we use the fbllowing procedure :

we form the aux iary polynomiaf using the coefficimts of the rast non-zero row. That iq the polynomial where coefficients are the elemmts of the row above the row of zeros iust in Routh array is called the auriliary porymomiar. This a,xiriary porynomial will be all-even or all-odd and its highest power is detemrined by the marginal notation (left margin). We then replace the row of zeros by the coefficients of the derivative of the auxiliary polynomial. The array is then completed. The roots of the auxiliary equation are also the roots of the original characteristic equation because auxiliary equation is the part of the auxiliary eguation

These roots occur in pairs and are the mirror images of eactr other with respect to the imaginary axis. Therefore, these roots may be imaginary (complex coniugates) or real (one positive and one negative), may lie in quadruplets (two pairs of complex conjugate roob), etc.

APPUCANOil OF ROUIH STAEIUW CRTTERIOI{

'

','

The Routh stability criterion is frequently used for the determination of stability of linear feedback systems. The dosed loop fansfer function is

,

C(s)

_

R(s)

xG(s) 1+ xc(s) Ir(s)

Hmce the duracbristic equation is 1+ KG(s) H(s)= 0 Therefore, the roots of the above equation depe'rd upon the proper selection of the varue

ofK

It is seen that unlcnorm K appears in the crraracteristic equation For zucrr a case, Routhrs array is constructed. Thm the range of values of r( are obtained in sudr a way that it will

I STAB:L:TY OF CONTROL SYSTEMS 211 l

not Produce any sign drange in the first column of the Routlfs array. Hence it is possible to obtain the range of values of I(for stability of the system. Such a system in which stability depends on the condition of parameter K is called conilitionally stable system.

一 Ｔ

RELATIVE STAB:LiTY ANALYS:S IWell■ hⅢ

Ⅲs:lhe、 yste轟

The Routh criterion ascertains

the of a system by determining whether any of the roots of the draracteristic equation lies in the right hall of the splane. In order to know how close

absolute stability

the system is to instability, we should know how far from the ioaxis is the pole closest to it. This can be obtained from the

Routh criterion by shifting the vertical axis in the eplane to obtain the pplane as shown in Fig. 7.1.

麟Hg。

,。

l shift of the axis tO the Ltt by

α

rT峨 鷲 憮茸∬:導電 鯖簿ふ蹴訛 :肥 鷺′ 性 『 Hence′ in the polynomi』

We aPply■ e Routh stab■

△(の haS

△(s)we

Al,Now

replace s by(′ ―α )we get a new POlynOmi」 ty crite五 〇n tO仕 is new POlyno血 ial in′ tO know hOw many Ю ots

in the五 ght half of theル Plane・ h number ofchanges ofbign tt he■ rst c01mul

d h aray devdord br h PdpOm」

1.Tttl∴ 肥T席 恵胤 鳳∬

located to h五 ght of the vertical line s=二

P01ynOmid sausies dle Routh cri健 五〇n7it implies that an the r∞ ぉ 。f山 ,品 ginal characte五 stic P01ynOnual are l■ Ore negative than― α.hs is alsO the number Of r00ts Of △(S)10Cated towards the五 ght of the line s=_α in the,Plane. ‐ h correct magnitude to shift h〕

vertical axis is Obtained by trial and eHOr basis.

‐ e shifting of the,Plane axis tO ascertain the relaive stability of a system is Paracularly useful fOr higher order systems with several Pairs Of 910sed‐

roots.

￨

100P cOmplex coniugate

212 coNTROL SYSTEMS

ryre

一

Dewtinc the

stobility

q

a systent

urto* choracterbtie equatian b

sa +5s3 +20s2 +40s+ 50

=0

Sol"U-r. The first two tows of Routh aray ar€ obtained from the coefficients of the draracteristic polynomial as'

5

η   Ｏ  ち

1

4 The s2 rOw elemen“ h andち are Ob● med iOm s4 and Se rows bl=―

:￨: ::=:(5×

20-lx 4Kl)=12

b2= :￨: ]l=:(5x50-lxO)=50 Therefore the s2 rOW iS

s2

The sl row element

q

1

12

is obtained from s3 and s2 rows. 5

411

12

5Cl

■

q=

lili l:￨=分 (12x411-5×

Therefore dle sl row is

l

魃鰈ギ h sO row element 4 L Obttd frOm s2 and Sl rows

50)=翌

署

STAB:L:TY OF CONTROL SYSTEMS 213 ２ ・

(115/6)

５ １ 一６ ・

-1

TI

50-12× 0)=50 =I13(1:塁 ×

merefore sO row is ”   ”

２

節 “ ”

５

一 一 一一 一 一 一一 一一一一一一 一 ・ 一一一 一 一 一 一一 一・一 一一 一 ・一 一 ・一 一 一 一 ・ ・一 ・ ・ ・ ・  ・ ・

Ｓ   ｌ ︲ ・

ｙ ａ

The complete

ｒ ｒ ａ

Ｒ

   

５ １一 ６ ・

一 一 ・ ．

０ ５

♂一 ｈ ｔ   一一 青 一 警 一 青   型 一 ｕ ・ ｏ

躙

In the first column there is no drange of the sign. Therefore the system is stable. rrayfo7′ 働 ′ ″ε ″お ′′ ο ‖ 欄瀾翻 1篤 護 Fo御 ル Ro“ 助α り"ο ねι “ :ε

Q(s)=s5+2s4+4s3+4s2+3s+8

SOlutiOn.Ihe first two rows of Routh array are obtained from the coefficients of Q(s)aS ３

２

  ８

    ４

４   ４   ち

１

■ e s3 rOW elements are obtained froln s5 and S4 rOWS.

2f量堅 二〒 2 4=一 :￨: :￨=土 益

b2= :￨: :￨=∵ ■■Юrefore the s3 rOW iS

曇

ヨ

2 -1

=1

The s2 row is o[tained from sa and s3 rows. 4 -1

ら

q=―

θ 2=

:￨:

4-2(-1)]=5 」 ￨￨=l12×

:￨: :￨=l12× 8-2xO]二 8

■ erefore s2 rOW iS

5

M輻

The sl row is obtained from s3 and S2 rOWS.

1l i: ・―半」=卜 04X81晉

1圭

dz =o Therefore the s1 row is

￨‐

The so row is

」￨せ

STAB:L:TY OF CONTROL SYSTEMS 215 Ｓ ・ ・  

The complete Routh array

ｌ

4

２

4

２

-1

５

8 ２． 一５ ８

There are two changes of sign in the first column: one from +5 to

-4

-45

and the other from

to +4. According to Routh's criterioru this means that the ctraracteristic polynomial has

two roots with positive real parts. Hence the system is unstable. We may conveniently multiply or divide a row by a positive, non-zero constant without affecting the results of the Routh's criterion. Thus the above aray can be modified to read : ４ ２

３   ８  ４

４ ．一 ８

１   ２   １   ２   ５

divided by 2

21

muldPly by 5

5

-21 8

蟷

‖榊〔

助′ 物 れθ″ル ′ レ αη O/ル

θお 0/ル カ :わ ω:暉 ″ り ο″滋ια″ れ 働θRHP: “ “

Q(s)=s6+455+3s4+2s3+s2+4s+4

２

４

ち

ｑ

ら ち

１   ４   ＬＪ  ａｂ

３

４ ４

１

一 ダ 一 一一占 一一一一み ． ・

sOlutiOn。 ■ ℃ POlynOIrual saisfies the necessary condi饉 on fOr stabttty because an theL] are positive and non― zero.The Routh array is

q /1

＼

｀

216 CONTROL SYSTEMS ■■ e■ rst two rows are obtained from the coefficients of Q(s) s6andS5 rOWS as fouows:

■ e s4 rOW iS obtained fron■

3二 1×

2)=:

b2= :￨: :￨=:(4×

1-1×

4)=0

b3= :li ll=:(4×

4-1× 0)=4

佐 =―

:￨: :￨=:(4×

Therefore,s4rOW iS

:

II督

0

4

1he s3 rOW lS obtained from s5 and S4 rOWS.

:ii

I鵞

４   ４   ０   ０

５ 一２

０

    ●︱・ Ｗ    ． ．

Therefore

２ 一５ ・ 一                                                                               ´ ︻ 一 ︵ ´ ． 一   一                                                                                                   一   一 ︲ ︲     ４ ０ ×         × ４           ４             一 一

 

４

%= 金

２           ２ ×         × ５ 一２     ５ 一２ １     １ ２ 一５     ２ 一５     ０ 一        〓 一 一          一

 

４ ５ 一２

 

1

５ 一２

62= 雨

２   ０

４

1 q= 5/2

s3

亀  

 

０

２

２ ５ ら  ・ 一

ｑ

STABIL:TY OF CONTROL SYSTEMS 217 The s2 rOW iS Ob五

¨

d frclm s4 and SO rclws

4=― :￨: _!IF:1210-:(-1111=3 ′2=:￨:

:￨=i12x4-:xO]=4

nerefore the s2 rOW iS

3 4

躙

Ｏ

ll3

Ｗ

２ 一５   ４ Ｓ  ・ 一 ， ・¨               ” ¨  ｎ ａ

Ｏ

'=

２  ３   角 ｍ  

■ ︱︱﹁■■■１１ ｄ   一     一・

め

Ｏ

Ｗ

ｅ

■

ｍ  ・ 一 ヤ ︸ 一 ﹁ ・

鼈

-?l=+t'"t

t)*l=-*^

Therefore the sl row is

W*^

o

The s0 row is obtained from s2 and s1 rows.

,ffi34

ffi*" gTE n=

o

-ffil * ;l=

-,.u[-*e)"a-o]

=+

218 coNttROL SYSTEMS Ｗ  ず

０     一

so

・ ｒ  ・一●一

Therefore the

The complete Routh array is

5 2 2

４   ０

34

  ４

4

１   ４

s5

一

５ ３   ％ 一１   ４

２ ５ ３   ２   ０     １ 一   ４ 一

1

３Ｓ    ′   ♂   ´

s6.

There are two changes of sign in the first column : one from +3to

-pand 15

the other from

-!15 to + +This indicates that there are two roots in the right half of the *plane. is the characteristic polynomial of an unstable system. Examine the stability of thc system haoing cluractoistic equation :

′ 青 一 青 一 青 一 一 ｍ   ・ ・ ・ ・ ・

Example 7:4

3sa +10s3 +5s2 +5s+3

ｙ ａ

ａ

sOlutiOn.Ihe

=0

using Routh's criterion.

lS

3

5

10

5

2

1

L

ち

3

ai*,ia"a Uy s

q

4 ３   ２    ３   ２

１ 一２ 一

〓

Ｌ

b2= :

:￨=lox5-,xl)=: :￨=:こ

x3-OxO)二 3

Hence, this

〓

１ ・ ％

２ ・ 牲

■ １

一 〓

ｑ

・ 囲 一 ・

STAB!LITY OF CONTROL SYSTEMS 219

1-2× 3)=1-1争 =―

: :￨=;(:×

4矧 :七 ￨=為

;・

:ll― :×

￨_:

3)-0]=3

il=―

Therefore the complete Routh array is

3

5

2 7

1 3

3

2 5 7、

3

There are two ctranges of sign in the first column : one from

*1,, -9 and the other from 27

-!7 to +a. Thus, there are two roots of the cfraracteristic polynomial

in the right half of the

rplane. Hence the system is unstable. 助 ′

"“

s。 lutiOn.Method

れ

αdθ S″ 10op ω″わ

s″ bル

リげ `施 55+3s4+2s3+6s2+6s+9=0

Isrtt

ωLOSθ 滋 ″ε″ た鹿 `9“

l:

Ъ e Routh array is

♂ ♂ ず●ず ♂ ず︱ ♂ 一ず一

q=

1

2

6

3

6

9

1

2

3

0

3

0

8

3

8

2ε

-3 3

ε

4=(3-幾 ■ =3

)

0

0 or K0

Combining these two requirements we conclude that the system is stable

if and only if

0> 1)

1roT

t

the

262 coNttROL SYSTEMS ￨よ 01

QUADRATiC COMPLEX POLES Consider the second― Order transfer functiOn

的 =

…(8.9,1)

昨 碗

Or

For steady state put

s=

".(8。 9.2)

fro

Therefore, G(iro)

=

---=J--

(8.9.3) “。

'.r,[f ).[,f;)' The response of the system to any frequenry rois determined by

4

(and

cor.

lf

C,>L the quadratic factor in the denominator of Eq. (8.9.3) can be expressed as a product of truo first order factors with real poles in whictr this case can be heated

like the case of real poles.

4 If 0 m

9,8‐

EXTSTENCE OF ROOT

rOCt ON THE REAr AX|S (REAL AX|S SEGMENTS)

Some of root loci may lie on the real axis. t real axis lies on the root loans poles and zeras, on the real axis, to the

the open-loop

concerned is an odd

This property can be proved by applying the angle condition at any point on the real axis of the s plane.

Consider the open-loop and zero configuration as shown in Fig. 9.2.

/s+zr=g

I

fig.

S.Z

Angl.e

/s+2, =rc0" /t*pz=tto"

/s+p.=g

/s+pr=1g4

contribution for a point on the real axis

Let so be the test point. To check whether the test point so is on the root locus or no! join all the poles and zeros to this point. At this point the angles made by the lines joinin g pt , pz, ps arrd p+ are

fso

+ pr), Z(so + pr) l(so + p, ) and 4so + pn ) respectively.

Similarly, the angles, made by the lines joining \,z2,zs and l(so + zr) and ls, + z! ) respectively. From Fig. 9.2,

itis

seen that the angles made by

h,p2 arrd4

z\

are Z(so

+4\

Z(so +zr),

are

4so + h) = Z(so + pz) = Z(so + 4) =180' The angles made by ps

md 22 ate

Z(so + Ps) = The angles made by

z,

l(so + 2,2) =0o

and,zi, are equal and opposite. The sum of the angles contributed

complex conjugate zetos 23 andzi, is zero [-0e + 0s =0]

by

ROOT LOCUS ANALYSIS

313

Therefore,

(a) the net angle contribution of a complex conjugate

Pole or coniugate zero is always

zeto.

(b) the angle contribution of all the poles and

zeros, on the real axis, to the riSht of the

test point is 180"

(c) the angle contribution of all the poles and zeros on the real axis to the left of the test point is 0'. For the root locus to exist at any point in the s plane , the phase-angle condition, that is,

zQs)H(s)= lQq+1)180" must be satisfied. This condition can be satisfied at any Point on the real axis only if the sum of the poles and zeros to the riSht of that Point is an odd number' ｕ ト ロ Ｚ ＩＩ

This rule should not be applied to the actual open'loop pol.es and zeros as these are always on the root locus. It should be apptied to sections between them' The cornplex poles and complex zeros do not affect the existence properties of root loci on the real axis.

19:91 ASYMPTOTES AND CENTRO:D If the number of finite zeros, m, is less than the number of finite poles n, th.en (n-tn) branches of the root locus must end at zeros at infinity. The branches which are approaching infinity, travel along the straight lines called asymptotes of the root locus.

Thrs, the number of asymptotes is (n-m\ intersect the real axis asymPtotes or centroid.

All the asymptotes The centroid is at

s

at a common poiit called the centre of

- - oA where [sum of real parts of poles of G(s) H(s) σ

A

-

zum of real parts of zeros of G(s) H(s)l

number of poles

number of poles

-

number of zeros

- number of

zeros

Centroid is always real. It may be located on negative or positive real axis. It may or may not be a part of the root locus.

314 coNTROL SYSttEMS Angles of Asymptotes The angles between the asymptotes and the positive real axis are given by

tF:1禦 ￨ ￨

4■

f.or q

- 0,L,2,...(n - m-l)

“

These angles are called the asymptotic angles.

9.10

BREAKAWAY AND BREAK-IN POTNTS On the root locus between two open-loop poles the roots move towards each other as K is increased till they are coincident. At the coincident point the value of K is maximum as far as the root locus between the

two open-loop poles is concerned. Any further increase in the value of K breaks the root locus in two parts. The point at which the root locus breaks into two parts is called the breakaway point. A

br$,t pqint o, * a point

on the real rods where two ot mare branches af the root

loas depart from or arrive at the real oris. Breakaway Point

Brea way point is defined as the po** at which root loans comes out of the real ucis and mwes into the compla< plane. Break-in Point or Re-entry point Break-ia point is defined as a point

ィ 中￨■IISaⅢ

at which root tocus enters the real aris. Break-in

ⅢⅢⅢ91■ p中 ￨

“

The breakaway or break-in points are the points on the root locus at which multiple roots

of the characteristic equation occur

:

e

If there are two adjacently placed poles on the real axis and the section of the real axis in between them is a part of the root locus, then there exists minimum one breakaway point in between adjacently placed poles.

e

If there are two adjacently placed zeros on the real axis and the section of the real axis in between them is a part of the root locus, then there exists minimum one breakaway point in between adjacently placed zeros.

1

If there is a zero on the real axis and to the left of that zero there is no pole or zero existing on the real axis, and complete real axis to the left of this zero is a part of this root locus, then there exists minimum one breakaway point to the left of thatzero.

R00T LOCuS ANALYSIS 315 Break Angles The root locus branches must aPProach or leave the break point on the real axis at an angle 180" of + , where r is the number of branches approaching or leaving the break point. f 9.10.1

Procedure of Determining Breakaway Points The following steps are adoPted to determine coordinates of breakaway points

SteUl

Write the characteristic equation 1+ Qs) H(s) =

Step

?

0 of the

sYstem.

From the characteristic equation, seParate terms involving K and terms involving s and write the value of K in terms of s. That is, the characteristic equation is rearranged

St.l9 4

as

K=

/(s)

Differentiate above equation with respect to s and equate it to zero. That

S!.p

:

is,

dK ds

-o

The roots of the equatio., the equation

'ds

4{

{ds

=g

rit ",

the breakaway Points.

All the roots of

=0 need not be break points, only those which lie on the root

locus are break points. The breakaway point or break-in point may be real or complex. The above method gives a1l the break points for the range of K from -"o to + co. In order to determine the actual break point, substitute the breal point value in the equation of K to get the value of K. If the value of ( is positive, that break point is the _valid breakaway or break-in point for the root locus. ln other words, if at

a

point at which

4{ =0 the value ds

of K

tales a real positive value, then that point is an actual breakaway or breal-in Point. The break points for which the values of K are negative are invalid for direct root locus but are valid for inverse root locus. In order to decide brealaway point or break-in point w" the breal ooir,1 uu1.r" ir', 4{

.

ds'

t d'{ ds'

.o.then that point

For break-in ooin

is a breakaway point.

t d'{ ,o ds'

6",".

i.t"

4{ -d

we substitute

316 CONttROL SYSTEMS 9.11

ANGLE OF DEPARTURE AT A COMPLEX POLE It is desired to determine the direction in which the locus leaves a complex pole or enters a complex zero. This information is needed to know whether the root locus originating from a complex pole moves towards the real axis or extends towards the asymptote. The angle at which a root locus branch leaves a complex pole in the s-plane is called the angle of departure. It is denoted by ba. The angle of departure of the root locus from a

complex pole is given be 0a

=180"-0

where 0=I0p -I0z I{p = sum of angles subtended (at

whidr {,

by the phasors drawn to this pole is to be calculated) from other poles

ZQz = sum of angles subtended by the phasors drawn to this pole

from all the zeros 9:12

ANGLE OF ARRIVAT AT A COMPLEX ZERO The angle of arrival of root locus branch at a complex zero is given by

by

fsum of all the angles zubtended

)

(s.,m of the angles

subtended )

0, =180'-l phasors drawn to this zero (at which 0, | *l by the phasors drawn to this zero [is to be calculated) from other zeros ,1 [from all the poles ) I

Both of these properties follow directly from the application of the angle condition at a point on the root locus very close to the pole or zero under consideration. The angle of departure and the angle of arriaal need to be calculated only when there are complex poles and zeros. The angle of departure

from

a real open-loop pole and the angle of

arrival of

a real open-loop zero is always equal to 0" or 1801

O;T3.. IUEGTNARY-AXIS CROSSING POINT

If the locus

crosses the iro axis, the point of intersection of the root locus with the ico (imaginary) axis and the critical value of K can be determined by applying Routh-Hurwitz criterion to the characteristic equation of the system. This allows knowing the stable operating condition of the system.

￨￨￨￨IⅢ

I喘1疇￨￨￨ Ⅲ

‐

POmt/thatュ ●the‐ 興aFttnal Value Of K.

■

‐ 1‐

・

・ ■ ■

.・

R00丁 LOCUS ANALYSiS 317

11二

義41

vALUE OF κ FOR A GIVEN DAMPING RAT:O ζFROM THE R00T LOCUS PLOT

To determine the value of K for a

given draw plot, locus root damping ratio (from the a line making an angle of 0 = ro-1 q from the

origin such that 0 is measured from negative real axis shown in Fig. 9.3.

in

the

clockwise direction

as

At the point of intersection P of this line with the root locus plot, apply the magnitude condition and determine the value of K.

膊H“ 飢3

1躙 VALUE OF GA:N MARGIN FROM R00T LOCUS ■■ ratio

of the value of K at■ F polnt ofintersection of the root locus withせ Fノ O axis to

the deslgn value of K ofthe system is called J℃ value ofthe gain margln for the rootlocus of the system.

(ValueOf Katthepont ofintersectionofroot

Gain margin (GM) =

locus with the imaginary axis)

designvalueof K

If the root locus does not intersect the frrlaxis, the gain margin is infinite.

瑳霧I

SUMMARY OF GENERAL RULES FOR CONSTRUCT:NG R00T L00 The following mles are used as guidelines for sketching the rootlod using Poles and zeros

of qs)Hls)aS K is va五 ed from zero to infinity. Rule l

The root loci is symmetrical about the horizontal real axis of the s plane.

Rule 2

As the open-loop gain K is varied from zero to infinity, each branch of the root

locus originates from an open-loop pole where K=0 and terminates on an open-loop zero or zero at infinity where K=oo.

R■de

3

The number of brandres of root locus terminating on infinity equals the number of open-loop poles minus number of zeros. A point on the real axis lies on the root locus, if and only if the sum of the open-loop poles and the open-loop zeros, on the real axis, to the right-hand side of this point is an odd number.

318 CoNTROL SYSTEMS

RJ.4 the (n-rz) branches

of the root locus which go to infinity travel along straight line asymptotes. The asymptotes arc inclined to the real axis at angles

givm by

i^ =Qq::L:' , RJe

5

where q=0,162,...(n-m-t)

The asymptotes cross the real axis at a common point known as centroid

givm by (sum of real parts of poles)-(sum of real parts ofzeros) _ ",__

RJ.

6

The breakaway points and break-in points (point at which multiple roots of the draracteristic equation occur) of the root locus ale found from the relation dK tls

RJe

7

=o.

The angle of departure of the root locus from a complex pole is givm by 6a =

rQq+l)180'+

I,

q=0,1u2,...

where $is the net angle contribution to this pole by all other open-loop poles and zeros. Similarly, the angle of arrival at an openJoop zero is givm by

0,

=

t(2q+1[8,0'-

i,

q=0,1y2,...

wher€ 0 is the net angle contribution at this openloop zero by all other open-loop poles and zeros

RJ"

8

The point of intersection of the root locus branches with the lo axis and the critical value of I( can be determined by applying Routtr-HurwiC criterion to the draracteristic equation of the system.

Rule

9

The value ofthe openJoop gain

(

at any point so on the root locus is given by

(product of phasor lengths drawn ftom the point s, to all theopen-loop poles) a_ (product of phasor lengths drawn ftom the point so to all the open- loop zeros)

This result can be obtained by using the magnitude criterion. The draracteristic equation is given by

1+Qs)H(s)=0

(s)H(s)=-r lQs)H(s)l=1

ROOT LOCUS ANALYSIS

319

That is, K(s+ z, )(s+2, )...

G-p,Xr-pr)t,

|

,

l-'

,,,

(s+prFs+ pzr. (s+ zr )(s+22)...

[{'*',) firr*r '

t

i=1

_ (product of phasor lentths ftom s, to oPm- looP Poles) (product of phasor lmgths from so toopen loopzeros) Thus, in Fig. 9.4 the value of K at so is

$ ng.

e.l

calculation of-tr

Procedure for Solving Problems on Root Locus The following steps are followed in construction of root loci

:

the characteristic equation in the pole-zero form so that the Parameter of interest ( appears as

1. Wdte

1+

KQ$H($=0

Factorize Qs) H(s) if necessary, and write the polynomial in the form of poles and zeros as follows :

fI(s+2,)

l+K'-t

- =0 ff1r*p,)

l=r

,

Identify the poles and zeros of Qs) H(s) and plot them on the s plane. 3. Locate the portions (segments) of the root loci on the real axis. Show the real axis loci by thick lines. 4. Determine the number

of separate root loci.

5. Find the number of asymptotes and their angles 6. Determine the centre

with the real axis.

of asymptotet and draw the asymptotes on the plot.

320 coNTROL SYSTEMS

7. 8.

Determine the breakaway/break-in points on the real axis (if any).

9.

Determine the angle of departure of the root locus from complex poles and the angle of arrival of root locus of complex zeros,

By utilizing the Routh-Hurwitz criteriory determine the point at which the locus crosses the imaginary (lro) axis, if the locus crosses the fol axis.

It is to be noted that for solving any particular problem of root loci, all the rules of construction are not needed.

S〕 191171 EFFEttS OF ADD:NG POLES TO G{S〕 〃〔

The effects of addition of poles to

1.

(s)

H(s) are as follows

:

There is a change in shape of the root locus and shifts towards the right half of the s plane.

2. The angles of asymptotes reduce. 3. The centroid is shifted to the left. 4. The relative stability of the system is decreased. 5. There is a reduction in the range of K. Even a system which was perfectly

stable

may become unstable as K increases.

菫轟落I

EFFEgs OF ADD:NG ZEROS TO G{s}″ 〔S〕 勁 e effects of addidon of zeros to αs)H(S)are as follows: 1. Therё is a change in shape ofthe rootlocus and it shifts towards the left half of the S Plane.

2. The relative stability of the system is improved.

E―plll,￨=￨

スタリルθ ′ bα ε たsys″ た α s α 4 θ θ 4-Iθ θ ″ S/arル 4θ ナ ′ ′ナ " “ “ “ :θ

的=判

′ 力θγθθ′Jθ ε夕sノ θ′ω:ル

Sセ

K

ε θ 燿ノα70ο お′ γ θ′ α げαεε `“

Isナ ル sysセ ″

r′

s′

:γ

Jι

αS α υαγjα bル ′α″ 解 ιr′ ′ s力 θω ルαけ た `′ “ j夕 s=V7. 7tti′ (1′ 0)α S Cι 4′ rθ 4′ rα

′braル γαJ:υ α:“ θs

力

α

o/K21/η

グ

θち グθ″γ解′ 4ι ′ λθ′ α

"rO/K力

Iθ

θ′げ

γS′ ′blθ

jθ F:4′ ′Isθ ′ θ′ ′ 4・ ル ″′暉 れ′:υ αル υ力′ 夕Sθ s s夕 s″ :4`グ `胡 `0/К θscil,α“ナ わ4s α4′ ′ ル タ θ?夕 ッ げ ′ 力 θ θSε :II′ ′ われs.Fro燿 “ ルθ“ roθ ′ :θ ε夕s′ Jθ ち `“ ル ″ ″:4θ ルθ υα:夕 Ks夕 c力 助αナ助“ s夕 ′ナ sys′ θ″ みαs αs`′ ′::4g′ ′ z`o/4 `げ `rθ '電

sys′ θ

S`θ

ナα θ ″ 魔.い物α ″ルι∞γ ″s′ ο 4カ 電 υ ガタ

`so/助

θ θγ θ お7

R00T LOCuS ANALYS:S 321 Solution.

1.

qs) H(,

K(s+1)

-j----= s(s l) -

The open loop poles are at s =0 and The open-loop zero is at s = Therefore

s

=L Therefore, r =2.

-1

m=l

The number of as lm|ptoles

=n-fl=z -7 =1

2. The open-loop poles and zero are on the real axis only. Therefore, the root locus is s),mmetrical about the real axis.

two branches of the root locus start at the oPen-loop poles s =0 and They terminate on the open-loop zeros at s = -1 and s = co, where K = co. 3. The

s=

l, ur'here K =0.

4. One branch of the root locus travels to zero at infinity along a shaight line asymPtote at an angle of OA=(29+1)x180° 7=0

-

(2'0

+1)

2-l

* 180. -180.

5. The centroid is given by σ

=

sum of poles - sum of zeros number of poles -number of zeros

_(0+1)-(-1)_2

2-7

Mark the centroid on the real

a;0

and

K-1

That is,

>0

K>l-

Therefore, the system is not stable for all values of K. The marginal value of K for stable system is K, =1 The range of values of K for stable system operation is

LO

Therefore

That is,

6′

and

8.66x8-6K>0 6

Kく 8。 66x8

K 0 K0. 11● mple●

Using Nyquist criterion inoestigate the stability of a closeil-loop control system

=41

whose open-loop transfer function is

K

G(s)H(S)=― S(1+STl)(1+ST2) s01utiOn.The given transfer hnction in sinusoidal foェ

α→Щ到.ω =α OЩ O=

..liS Obtained by putting s=ノ

=M4 K∠ 0°

1-1

ぎ )(Vl+02■

2 z tan l oTI

Therefore

lC17rola1io1l=

∠G(ノ ol

M=

K 1+。 2ィ

可

H(ノ Ol=φ =-90°

一tan

vl+。

2呼

l oTl― tan l oT2

There is one pole at the origm. The Nyquist path is shown in Fig,11.28(α sections Cl′ C2′

).It has four

C3and C4・

We map eadh section of the Nyquist pa■

.

〔 ]Mapping of sect:on cl h this section the frequency o varies from O to∞

When o=QM=∞ and When and

φ=-90°

.

麟

Fig。

-0° -0°

=-90°

ω=∝レM=0

φ=-90° -90° -90° =-270°

11.28

ω

440 CONTROL SYSTEMS

αノ OH(JOll① =0=∞ ∠―

Thus′

"°

G(ノ

OH(ル 01。 _∞ =0∠

-270°

■ erettt h maPPhg of"dCIn Cl i s plane is rePresented by CD h GH Plane Where

POmt c is at∞ ∠-90° and D is atO∠ -270° as sho― h Fig.11.28(b)

The mter∞ cton of the"la p10t Of Fig ll.28(b)宙 h the real axL can be obtamed by

equaing helmaFary Part of G(ノ OH(ノ COl

tO ZerO

薔Fig.11.28 {η

Mapping of Section C2

Sechon C2 h S Plane is a semidrcle of― te radius ■ is se“ on can be maPPed h C(S)H(S)Plane by subsitutng

s=hm Rσ R→ ∞

'0

hC(S)H(S)and Vaっ 積 g O from+9o° to-90■

鳳 GoHo[_ル ρ=鳳 As R→

鳴 (1+R Tl′

0)and(1+RT2´ 0)are aPPrOXimateけ equal to RTl″ O and RT2′ ′ '°

reSPeCtl■7ely

Therefore/

lim

cG)HG)l

R+o

l"=n

Iim

= P x-= lirn

R-+o

Reρ (R Tl″ ρ)(RT2′

X: --Rr44e,:'o

jO)

=gs-Fe

Thus, irrespective of the phase drange the amplitude is zero, so the entire semicircle C, is mapped to the origin in the GH plane.

NYQUIST STABILITY CRITER10N 441

(ii

Mapplng of Section

Ca

In this section ovaries from - o to0-. The plot of G(i(D) H (io) will be symmetrical about the real axis of GH plane. It will be the mirror image of Fig. 11.28(D) as shown in Fig' 11.28(c)'

(ivl Mapping of Section Cr The semicircle of section Cn around the pole at origin in s plane can be mapped in GH plane by substituting S=ε

`ρ

hC(5)H(3) G(s)H(S)=

。 ε′プ (1+Tlε θρ)(1+T2ε °ρ)

As O va五 es from―

from∞

r

レ

Tl tt T2＼ ｀ :

闘 GoHo=闘 論 =。 。´

KTIT2 -1

0 ′

。

f tof′

G(s)H(S)Va五 eS

∠f ∠ f to∞

tt Hg・

11・

28

Therefore this part of GH plot corresponds to infinite semicirde CDA. The complete Nyquist plot of G(s) H(s) is shown in Fig. 11.28(c).

Determinatlon of Stability To determine stability, we have to linow whether the Nyquist Plot encirdes the (-1, l0) point or not. For this, we determine the point where the plot crosses the negative real axis. The following steps are used

(i)

:

Rationalize G(lo) H (lor) by multiplying the numerator and the denominator by the complex coniugate of the denominator.

(i4 (iii)

Separate real and

(ia)

The magnitude of the real part of G(iro) H(jo) is found by zubstituting the value of o found in steP (irr.

inaginary parts'

Equate ttre imagirury Part to zero to determine the frequmcy Nyquist plot crosses the negative real axis.

ro

at which the

442 coNTROL SYSTEMS Rationalizing G(iro)

H(io|

we get K(― ノ 0(1-ノ OTl)(1-ノ OT2)

G(ノ ol H(ノ 0')=

0(1+ノ ① Tl)(1+ノ ① T2)( ノ

ノOl(1-ノ ① Tl)(1-ブ OT2)

_― K[0(Tl+T2)+バ 1 02 TIT2)]

2ィ 。 (1_ω )(1_o2プ

)

The imagnary partis zero whe巧 1_。 2TIT2=0 Let ① r be the frequency at which the Nyquist plot crosses the negative real axis.Thereforeノ ′

for o=①

r ′

l― ω

;rTIT2=0 ① c― ± ′ 嵩

Since o′ c is PositiVe′

At this frequency

野=嵩

①′ c the

一た(Tl

magnitude ofぬ e

tt T2)

real Part Of G(プ ol Ff(プ ol iS

K(TI+T2)

+弔 。 Xl+① π)ぃ ィXl+士・ 晴 `イ 希・ )

_― KTI T2

rl+T2 ■℃ open‐ 1∞ p transfer hnctiOn has no open-looP POle on he五 gh卜 hand side of the s Plane′ that is P=0. For stab」 ity′ the dical POint(-1′ ノ0)ShOuld nOt be endrded by the C(プ olH(ノ OI P10t・

This is 6btained from the relation

Z=N tt P=N+0=N For closed-looP stab五 ty Z J■ Ould be zero.hs can only happen if N=0 That is/Polnt(-1′

0)muSt ブ

not be endrded.For thisr the condition to be satisfied is

KTI T2 12

The range of values of K fOr whch the systern is stable is g市

ExantPle ll.6.

The

opmloop bansfer function of a control system

G(s)H(s)=

en by Oく Kく 12

is gioen by

K

s(sT -1)

Delermine the stability using Nyuist ffiterion.

sOLtiOn The given transfer function in sinusoidal form is obtained by putting s=lroin C(S)H(S)

C(slH(s) . , ._ /. -

K

lto(jtoT_l)

448 CONTROL SYSTEMS Raional■ zhg and separaing real and lmattary Parts

OЩ わ =品

■

満

“

Thereis one Pole at h O五 gh The Nyquist PathiSShOwn m Fig l1 30(α )It has four sectons Cl′

C2'C3and C4 We maP

each sectlon of the Nyq‖ lSt Pah

K∠ 0°

C(プ OH(ノ Ol=

10Z90°

J

ll… ぼ

Therefore,

G(Jol H00=M=

∠G(pHoO=0-f―

and

(π

tan lω

D (4)

= +90'+ tan-1

r' 6^ u111 vo -\''

...(13.16.1)

m

Let us define the oPerator s as

,r = dr,

d{

Equation

("''u') "*o

k=1,2,...,n

(;J":;:l"r'. 1.,,.;; rlT,i,u'u'

...('1g.16.2)

(13.163)

The characteristic equation of the system is defined as s"

+anrs'-r +...+a1s+40

=0

which is obtained by setting the homogeneous part of Eq (13 16'3) to zero'

"'(1316'4)

STATE ANALYSIS OF CONTROL SYSTEMS 533 ‐ Ex8mplc

13110

D′ "r″

i″ 働θ油 ″/″ ε″″ is″ r釘 ″α ::ο ″ oFfll′ ′ 滋!θ ぅ 確 ″″′

禦 巧禦 +7■

ヽ

2ylrl判

o

′ ″ `″

`わ

(E13101)

S01ution Def― g he operator s as St=::卜

″

=1′ 2″

″ ゥ

Eq(E13101)can be writtell as

(s375s2+s+2)y(′

¨(13102)

)=″ (ι )

The draracteristic equation is obtained by setting the homogeneous part of Eq. (813.10.2) to zero

+5s2 +s+2 =0

s3

13.15.2 Characterlstic Equation from a Trarcfer functlon Consider the transfer function G(s)=

s^l +...+&rs+bo s'+ [*, s*l +...+ a7s+ ao

b^sn +bo-,

¨(13163)

The characteristic equation is obtained by equating the denominator polynomial of the transfer lunction to zero.

Thus, the characteristic equation of the system described by Eq. (13.16.3) is sn

ExEraple

1.3111

+ an_,s' 1 +...+ars+ao = 0

The transfer function of a system is described by

Y(s) _

U(s) What is tlrc characteristic equation

__

r

s3 +5s2

+s +2

?

Solution. The characteristic equation is obtained by setting the denominator pollmomial oI the transfer function to zero. Therefore, the characteristic equation is given by s3 +5s2

+s +2 =0.

13.16.3 Characteristic Equation from State Equation The characteristic equation from state equation is obtained from the relation

Or

det lsI―

A]=0

sI―

AI=0

534 CONttROL SYSTEMS 1森轟

轟

￨11■ 1オ

Asソ

勧

お 暉 7gsι

０ ｕ

＋ Ｘ

２ ２     ・ ・

二 J"

ｘ

２

ロ

一 一      〓

・Ｘ         ｙ

日 ﹂ Ｆ  

わ S: グ by ttθ ヵ 〃οωれ gS″ ″ ′ ′ ο ″ ′θ9“ α′ “ “ “ “ 巨Ｌ Ｆ

"″

Find the poles of the systan.

Solution. The characteristic equation is given by

χ 「 [￨]― ‖

:]￨=`

13 λi21下

lλ

0

+2)::il[:

(λ +3)(λ

燿 +5λ +4=0 (λ

+4)(λ +1)=0

λl=

L

λ2=

4

Therefore/the poles are-l and-4.

軍扇再面編可l131

″θ ′ ″is′ 0b滋 れc力 α

jε

′ r放 ι ′ げ助`解 ′ 7ッ α “ :θ

A=￨_1: _' _￨￨ S01utiOn.The charactenstic equation of A is det lsI―

1sI一

Al=0 Al=￨￨ ￨ ￨￨―

￨_1: _' =[:

i sIII

det BI一 Al=lsI― AI

二 ￨￨

SttA丁 巨ANALYSiS

﹂Ｉじ

=IIi

OF CONTROL SYSTEMS 535

i sIII

=S{S(S+6)― (-2×

7)}― (-1){-3(s+6)―

=(S+1)(S+2)(s+3)

=s3+6s2+1ls+6 Thereforer the characte五 stic equations is det lsI― Al=0

。r

s3+6s2+1ls+6=0

Lmple ll:141 A sySr`″

is ttrasι 翅 by ttι /o″ θωi昭 々 ″α解:θ η 夕α′ :θ

“

女=[_:

_:][1:]+ll]“

y=11 0][1:]

笙 雀r懲 辮 1lnぉ

givenけ

薔￨::=CISI一

Al l B tt D

=CIsI― Al lB+0 1sI―

Al=s[: ￨]― [_: _:]

=[ 真

lsI― A】

も

]

adiISI一 Al 1_det lsI_Al

adi BI― A]=lcofactor matr破 of BI― A]]T ｀

=[1: &:]7

=ISl: :]

=[S+:

[]T

Sf “

(-2)(12)}+0

536 CONTROL SYSttEMS

置Ы 刊JS」 ns lllla 卜s2+3s+2 l Therefore,

1s I一 Al

1=!::: :]

+3s+2 C=[1 0]

B=[l]′

Al lB ３  ２ ＋   一

０

s2

:嚇

Юい

TraFISfer function=CIs I一

1 =-s2 +3s+2

+3s+2

ル α υ れgs協 ″解θ ′ われ/Or αsrた た 131161 0b″ j“ 働 ″ s/grル θ `′ “ “ “ :

[1:]=[ i

:][I:]+[]“

y=[1 1][:]′

D=0

sOlutiOn.The trarlsfer hnction is given by

・ Go=器 =CЫ ― 劇B+D Al lB+0

=C IsI一 1sI―

A] l adiISI Al lsI一

IsI一

AI

A]=s[: ￨]―

=[: :]―

[ i

:]

[ i

:]=[Sli _:]

adiBI― A】 =lcofactor matrix ofBI― A】 ]T

=[::l :重 =[S i sI:]

]T=ISI: s+:]T

STATE ANALYSiS OF CONTROL SYSttEMS 537 sI―

AI=(s-2)(s+2)― (4)(-3)

=s2_4+12=s2+8 ３ ・授

２  ４ 一

Therefore

IsI―

Al 1= s2

B=[￨]′

C=[1 1] 聰ＬＰ

３ ・授

２  ４ 一

陽︱ ﹂

Trttfer hnctiOn=CIsI― A] lB

+8

s2+8

+4 8sll _ll lltt s2+8 f::][枷 s2+8 高

:

L3.I7

STATE.SPACE REPRESENTATTON FROM TRANSFER FUNCTION We shall determine the state variable model when the transfer function is given. Since the state variable representation is not unicpe, there are, theoretically an infinite number of ways of writing the state equations. We shall present here some methods foi deriving a set

of state variable representations from the transfer function.

13.18

STATE.SPACE REPRESENTATTON USING PHASE VARIABLES

Ihe one

Strr* variabb are ihfind 6 those tute yariahte* ihich are ob,tafuted ' -" ' :'from af the systamvarir#ra ffid f& dettv@.,

Usually, rhe system .ru.iuUt" ,r"d'i, then the derivatives of the output.

;;*r*;;r;;;

;;;"

;;;;

'.rrru6*,

*"

Phase variables provide a powerful method of state variable formulation.

A disadvantage of phase variable formulation is that the phase variables in general are not physical variables of the system and therefore, not available for measurement and control PurPoses. However, a link is provided between transfer function approach and time domain approach while designing a control system. Thug the design implementation of the control system becomes easy and straight forward.

538 CONTROL SYSTEMS 13.18.l

Phase Variabie Form

The phase vanable forln of state mode can be eas"ObttFd if the sysぬ m

moddお

av』 able m the dbmialequah ortransferf面don fo.uL th order drttrentlalequaion for a har¨ vanant system

Letus consider a general″

め 0+亀 一%メ め 手メ ルメ ,r't-7

=bo where z is an integer

and.

i;u(t\+ \fi*t

u(t)+"'+b'u(tl

¨(13181)

q,a2,...,a, andbo,br,...,b" are constants'

The kansfur function relating outPut and input in s-domain is

+ "+b,

Y(s)

(13182)

-bos'+4s'-r "'"' ",.,, u(t) sn + qs*r +...+ a,

Thisisthetransferfunctionrepresentationofthedifferentialequation(13.18.1).

13.18.2 State'Space Representation of an nth€rder System in which the Forcing Funstlon does not lnvolve Derivative Terms

\

Ilthehansfurfunctiorrofthesystemhasnozeros(thatis,thediffer€ntialeqrratiorrofthesystern be obtained easily as follows has no derivatives of the input), the model of the system can

:

Consider the transfer function G(s)=

¨(13183)

s" + qs"-l +...+ in4 s+an

corresponding to the differential equation

物 t計 ′ の 訳→“ の ズ 間争 め井ズ チメ =ち

+亀

(13184)

Thisisanrrth-ordersystelr!sonstatevariablesarerequired.Letthefirststatevariablebe

derivative equal to the outPut variable y. Let the second state variable 12 be equal to the of the output variable and so on. That is,

rt

χl=1/

χ 2=響 =' χ 3‐

lF子

='

L=弊 =〃

STAttE ANALYSiS OF CONttROL SYSttEMS 539 These equations can be reduced to a set of first-order differential equations as follows

;

一 一  一   ・   一 一 一 一 一 一 一 一 一 一 ・ 一 一 一 ・

…(13.18.5) ︶  、

・ η」 1レ

為 0'

,■ =二

=五

The last equa● on is obtained by equaing the highest order

the differential equation to all other terms.

derivative term of the output in

0

1

・…

0 0

為 χ2

・ ︰

0 -α

0 ″

0

α

1

.…

α″ _2 -・

"_l

亀

χ "_1 χ″

彙=Ax+Bu

Or

夕

.¨ (13.18.6)

0 b0

…(13.18.71

0 0

(nx n)

…(13.18.8)

1

切

一

﹁

(nx n)

０ １ １ ０

１ ０ ⋮ ０

A=

〇 ０ ⋮ ０

where

 

・¨

0

０

0

 

1

０

0 〓

降降卜Ｌ

The above set of state equations Eq. (13.18.5) can be written in vector-matrix form as

― ′ _1 ″

.¨

(13.18.9)

The matrix A has a very special form. Ithas all 1s in the upper off diagonal. The last row of

A contains the negative values of the coefficients of the homogenous part of the different equation in ascending order, except for the coefficient of the high-order term, which is unity. This form of matrix A is known as the bush fonn or the companion forur. The matrix B is a column matrix with the last row equal to one, and the rest of the elements are all zeros. The state equations in Eq. (13.18.7) with A and B given in uqs. (13.18.8) and (13.18.9) are as the

loourn

phase-variable canonical fonri (PVCF) orthe controllable canonical fonn (CCF).

Hence, in view of these observations, A and B and, therefore, the state equations can be written directly by inspection of the differential equation.

540 coNTROL SYSTEMS The output y is equal to

.r,

and therefore the outPut equation is given by

l;: I y=r1 o orl .'I 1,, That

is,

...(13.18.10)

] ...(13.18.1r)

y(t) = Cx(l)

If the transfer function is known instead of the differential equation, we can easily obtain a differential equation fomr as shown.

Trst= ''

If

Y(s)

u?

U(s)=s" +qsn'+...+a,

...(13.r8.12)

(s' + qs*1 +...+ a, ) Y(s) = bo U(s) yln) * ory@ ') + ...*ony =bo,

or where

...(13.18.13)

d'! !.,r,1 I-dt

...(13.18.14)

Equation (13.18.13) is same as Eq. (13.18.4) and hence the state-sPace model is again given

by eqs. (13.18.6) and (13.18.11). Block oiagram Representatlon (state Diagram for phase variable forml

A block diagram representation of the state model of Eq. (13.18.5) is shown in Fig'13'18' Each block in the forward path represents an integration and the output of each integrator is taken a state variable. Fig. 13.18 represents the state diagram for phase variable form.

'm i

fiq. 13.18

Block diagram representation of the state model for

i=Ax+ Bu ardy =Cx

STATE ANALYSiS OF CONTROL SYSTttMS Signa卜 Flow Craph Representat:on

l

おWn m聴‐" ° ∬ l植 ￨∬ 鮮躙 軍窯翼雅1織鼻 Sh°

=Ax+Bu and y=Cx

13.19 Signal■ owgraphrepresentationofltatel.odel fbr文 ' Fig。

lEmmple■

‐ ■

1● Cθ

S″ 夕dル

“

s″ セ

θル “

崚

γαSrtt

grizθ ′ θたαrα θ′ り 働θ′髪

わ4 ″″ θ7夕 αオ ““

i+5Y+6Y=11 Solution. The given system is of second order and, therefore, two state variables

are

required. Rearranging the given differential equation so that highest order derivative term

*Llt' *

ヽ

equated to the rest of the terms, we have

4dtz=-rL-6v+u dt The state variables are defined as

¨.(E13.16.1) ヽ

¨.(E13.16.2)

xr(t) = y (t)

x^(il=dy(t) z' dt

_。

(E13.16.3)

The state equations are given bY ¨.(E13.16.4)

xr=x2

In matrix form,

iz ='Sxz-6x, +u *ritten the state equatio* "utt'b"

¨.(E13.16.5) as

¨.(E13.16.6) [ち ]=[_: _:I『

:]十

lli“

542 CONTROL SYSTEMS The output equation is

t=xt

or

y

=l't

of

['' Ij

"'(813

L',

rc'n

The state model is given by Eqs. (E13.16.6) and (E13.16.7). The block diagram representation of the state model is shown in Fig. 13.20.

d

ffi Eig.

13。

13.20

Block diagram representation of the state modet for Example 13.16

19 CONTROLLABLE CANONICAL FORM{CCF,OF THE ttATE SPACE REPRESENTAT:ON Consider the fouowing transfer mction system 1+・ γ壺 十bz_ls十 し ・ 2=boS"十 LS… ・

U(S) S"十

亀s711+.… +α p1

l s+α 0

The given trarlsfer hction can be di宙 ded in twO Parts.

Y(s)_Y(s)Xl(S)

U(S)Xl(S)U(S)

-([s' where and

x,

r'\

+

(s)

* qs*l *...* o-i :,,.fi)'ttt* ht'-'

U(s)=sn +qsd

#=

bosn

+...+bn_, s+bn)

1 .¨

+...+an_rs+ao

+b's*r +"'+b'*'s+bn

(13.19.2)

...(13.19.3)

Consider the function .¨

Equation (13.19.2) is same as Eq. (13.18.3) with

Do

=t-

(13.19.2)

STATE ANALYSiS OF CONTROL SYSTEMS 543 Hence its state‐ sPace representa饉 on is given by Eq.(13.18.6).

十

:

χ

lt_1

︰Ｆ ０﹂ Ｆ Ю Ｉ＾ ⅣＩ １

χl χ2

.¨

(13.19.4)

.¨

(13.19.3)

χ

"

Now conslder the function ′

器

=%♂ +摯

d∵ 唱針 し

By cross muldPlying we get γ

(S)=(bOS″

十

blS" 1+.…

¨。 (13.19.5)

十 ♭ _lS+b")xl(S) ″

Thtt differential equation corresPonding to this is

y=b01:年 The first state variable is

十 十 +し 4争 +L・ … 錢 1[[年

rr. The second

state variable

r,

¨。 (13.19.6)

is the derivative of x, and so on.

That is,

…(13.19.η

Therefore, Eq. (13.19.5) becomes V

= bo

*, + bt x n +...+b*_, x, + b, x,

subsututing力 "from(13.19.4)′ 1/=bO(二

y=χ l

…(13.19.8)

we get

α l― α +・ ¨ 十 _l χ _2χ 2 ・ ¨α 2χ 摯 ″ 4 ■χ l+“ )+Lχ ″ 2+b″ χ b″

"χ

b″ ― bo α +χ _1)]+・ … )+χ 2(b"1 %′ ″ ″ К )十 ♭ 0夕 "(4-bO■

y=К b“

一bO α "):(b″

l

...(13.19。 9)

_1-bOα ″ .)i

Equations(13.19.4)and(13.19。 10)are Said to be in the controllable canonical fonIL.

544 CONTROL SYSTEMS

13.20 0BSERVABLE CANON:CAL FORM 10CF,OF THE STATE‐ SPACE REPRESEttAT10N Corsider the followng trarlsfer hction system

Y(s)=bOS"十

blSl1 1+...+bl1 l s+b″

LI(s) s"十 亀SP1

1+...+α pl l S+′

¨。 (13.20.1)

"

Equauon(13.20.1)can be modified in dF folbwing forEIl: s″

Y(s)一 〔

・ %LI(s)1+S″ 十

D市 iding

Eq。

[亀

Y(s)一 bl U(S)]十

...

SIα ″ lγ (S)一 b′ l LI(s)]+α

″

γ

(S)一

♭ ″

lI(S)=0

・・ 。(13.20.2)

(13.20.2)by sr and rearranging/we get

Ylsl=%Щ

七 切 … lЦ 朝 ぺLЩ ⇒ UO― %γ 働 +升 レ lu。 _%4γ 側+ル ″ ‐ ¨。 (13.20.3)

Now define state variables as follows

:

丸6=lLu6‐ .,0キ 塩 側 +恥 側 七欧→ 琢 →=伊 LЩ ⇒ ¨。 (13.20.4) 141′

′、

1

言:]iれ(聯Fす T I■

■ en Eq。

■■￨

(13.20。

■

￨

3)can be― tten

as

¨。 (13.20.5)

Y(s)=祐 LI(s)+X"(S)

By substituting Eq.(13.20.5) into Eq. (13.20.4) and multiplying both sides of the equation by

球,長 ″14ホ ι l

■to=x潤 ‐

￨″ 4(S)= X"2(')￨121挙

`Xな

'あ

"(S)十

(b2‐ 12♭ 0)LI(3)

(b"1-′ ・ υ ヒ 「 場 : サ ご ,;0■ 高 =Lit(昇 ご ￨.:て β X2(S)=Xl(S)一

α″二iX″ (S)十

…

lb0

¨.(13.20.6)

STATE ANALYS:S OF CONttROL SYSTEMS 545 Taking the inverse Laplace transforms of n equations (13.20.6) and writing them in the reverse order, we get

‐ ″￨″ 十 ￨￨′ 4‐ =… ′ (レ

1‐

3o'￨￨‐

■ ￨￨￨

力 2=χ l `″ _lχ (b"_1■ 11_1%￨)“ "■ ‐■ ‐ ■■￨.￨￨‐ ￨￨‐ ■￨: ￨‐ 受 十(b2 α 2♭ 0)“ Ⅲ マ2χ ″

¨。 (13.20。 7)

":r=χ ‐ 二 1,二 ″ 二 :14 χ ll(ち 411:'￨￨￨‐ J―

Taking the inverse Laplace transform of Eq. (13.20.5), we get ¨。 (13.20.8)

y = xn+bou

ち島 ％

ｔ４  一

﹄年⋮ Ｌ

＋

降階Ｌ

丸４  ■

０  ０       １

０  ０       ０

〓

ｌⅣ ЮＬド﹂１

″ Ｌ ︰・・ ■・ χ ・

Writing the state and output equations (L3.20.n and (13.20.8) in standard vector-matrix form, we get

¨.(13.20.10)

Equations (13.20.9) and (13.20.10) are said to be in obsen able canonical form (OCF). 躙 閂 鱚 颯 趙 鱚

働

"燒 "′

α減カル s″ ″御 ル Iル タッ 勧

加

山 醐

リ ル ′抑

禦 巧禦 +742ylll諷 → “

"′

制 η αli9" “

sOhtion。 ■℃ given system is of third order and thereforら ables are "=3.hee state va五 requred.Remanging the given differential equatiOn So that h highest― order de五 vative

r

te...lis equated to■ hに restof h temε

we have

禦 -5毛響-7-2ytrluり The state variables are defined as

0=写 ろ 彎禦

l

546 CONTROL SYSTEMS Then the state equations are represented by the vector matrix equations

dx-(t) dt

-Ax(r)+ Bu(r) tt=3, ar =1, ar =L

12

l::￨=li1 11

as

=2,

bo =1

“

0r

ー ー ト

￨::￨=￨_13

υ ｍⅣＬ１﹂    タ 隔ｌ １ ｌ

ト 性

The state model is given by

﹁︱ ︱ ︱ ︱ ︱ ︱ Ｉ Ｊ

劇 難

For the given differential equatiort,

The output equation is ノ(サ )=χ l(′ )

=[1 0 0111:￨ The state diagram is shown in Fig.13.21.

蓋H■

1■ 21

E*"rrrpl.

13.18

A system is described by the following differential equation n *dt' *z *. dt' 4dt

+ 4x =

\

(t) +3 u, (t) + 4 u, (t)

The outputs are

y, = a Obtain the state

sPace

#

+3u.,

;

v,

representation of the system.

=#+Aur+u,

:

STATE ANALYS:S OF CONTROL SYSttEMS 547 Solution. The given system is of third order and therefore, n =3. Three state variables are required. Rearranging the given differential equation so that the highest-order derivative term is equated to the rest of the terms, we have 卜し

争=-3争 -4争

-4χ

+均

o+3%o+4“ 30

¨.(E13.18.1)

、

The state variables are defined as

From Eq.(13.18.1)′

χl=χ

(E13.18.2) …。

χ2=χ l

¨.(E13.18.3)

χ3=χ 2

¨.(E13.18.4)

we have

受 3= 4χ -4#-3争 十 0+3夕 20+4夕 30 ■

… G131鉤

Therefore/Eqs.(E13.18.3)′ (E13.18.4)and(E13.18.5)foming the set of state equa● ons are

written

as χl=χ 2

¨。 (E13.18.6)

χ2=χ 3

¨.(E13:18.7)

カ3=

(′

"。

均 ％

０

...(E13:18.9)

ち

４

３

ｒ

ＦＩ Ｉ Ｉ ｌ ｌｌ ｌ ｌ Ｌ

０

ド

十

﹁︱ ︱ ︱ ︱ ︱ ︱ ︱ ︱ ヨ

０

Ｌ

０

陶

一

﹁︱ ︱ ︱ ︱ ︱ １ １ ︰ Ｊ

０ ４ 一

４

ト

ＬＬＬ

１

０

ｌ

〓

劇劉 明 ︱・

０

ｌ

﹁︱ ︱ ︱ ︱ ︱︱ ︱ ︱ コ

ｒ

﹄ＬＬ

4χ l-4χ 2 3χ 3+均 )+3ッ 2(′ )+4夕 3(′ ) (E13.18,7) In matrix form, the set of Eqs. (813.18.5), (E13.18.7) and (E13.1g.g) can be written as

The outputs are given by

y, =4x,+3u-, Uz=xt+ 4ur+3u, 均 ％ ％

４

﹁︱︱︱︱コ ０ １

０

十

聰 ︱＾ Ⅳ

ｌ ｌ ｌ ｌ ｌ ｌ Ｊ

Ｏ︶ １

０

χ ．に Ｌ Ｆ

４

〓

Ю Ｉ＾ Ｐ

降Ｌ

In matrix form

¨。 (E13.18.10)

The state inodelis given by Eqs.(E13.18.9)and(E13.18.10).

=轟

島畠11=:,

勁r ttθ

g滋 4磁

ε ′ わ 4 ψア ″ “ ル

_

T(S)= s3 Obナ α れル `s`′

+arsz +a1s+ao

＼

″ 解θι レJ.

ヽミ ヽド

548 CONTROL SYSTEMS Solution. We can obtain the differential equation form for the given transfer function as follows :

TISl=器 =「

,.

,tt

+

ars2

頭特葛

¨(E13191) …(E13192)

+qs+ao)YG)=boU(s)

The differential equation form of Eq. (E13.19.2) is

d3v d2u du i*or#*o.t-a+aoY=bou

(E13193)

鴨 s is a dlird― order system′ so three state vaHables are required Letthe ttst State va五 able be χl equalto tt output vanables y Let he second state va五

able

vanabkり and dird state vanable x3 be equalto

χ2be equalto the deHvaive ofthe outpl■

dle second de五 vaive of dle output va五 able Thati勁

(E13194)

″ 1=y χ2=π

E13195) 〈

=′

'3=:F子

='

(E13196)

These equations can be reduced to a set of first-order differential equations as follows

:

均 =χ 2

(E1319η

χ2='3

(E13198)

Suppose that the highest order derivative terrr, that i+ the third-order

derivati* te""

4

is equated to all other terms.

da dzv , v '#=-oot - a1i - a2 .-+bou d.3

x, -a, x, +bou equation is givm by

= -40 \ Therefore, the third state

'3= ′

O

-\

¨(E13199)

χl― ■ χ2 ′ 2χ 3+bO″

ツ

％

﹁︱ ︱ ︱ ︱︱ ︱ コ

＋

０

ＬＬ Ｌ

０

﹁︱ ︱ ︱ ︱︱ ︱ コ

降Ｌ Ｌ

〇   一

０ １ “

１

ｒｌ ｌ ｌ ｌ ｌ ｌ Ｌ

〓

〇 ０ コ

Ｉ

﹄Ｌ Ｌ

in matrix form as

…(131910)

′     ヽ

The set of state equations given by Eqs. (E13.19.7), (E13'19'8) and (813'19'9) can be written

  ︲

ヽ

STATE ANALYS:S OF CONTROL SYSTEMS 549 The output y is equal to

r,

and therefore, the output equation is given by y = rr "l

or

A=[1

[,,

o 0]lxrl

...(E13.19.11)

L"-l Figure 13.22 shows the block diagram representation of the state model.

ffi

ffi Fig.

13.22

Block diagram representation of the-state mode[ given byEqs. (E13.19.10) and (E13.19.11)

′ ι 0虎 :ル r ttβ 秘盤曇塁鐵￨″ 01 0b″ ′ α “ :“

sサ

l

Tlsr=-

"

Solution.

Y(s)

uG)

-

F;7

Sr″ 協ル sCribθ ′ り

+los+s

Y(s) =

U(s)

(s3 +5s2

s3 +6s2

¨。 (E13.20.1)

+10s+5

+10s+5)Y(s)= U(s)

¨。 (E13.20.2)

Taking the inverse Laplace transform on both sides we get

#.r#.lvff+sy

葬模

= u(t)

¨。 (E13.20.3) that the highest order derivatiサ

etem

II[よ }icll:11:]lil:]IIisl:I:・ lf)S°

ター6字

-10争

-5ノ +ッ

」 二5y-10」2_6望 生 1∠ 二 ∠+“

o t

¨.(E13.20.4)

550 CONTROL SYSttEMS therefore′ three state variables are required.

The given systeln is of third order′

The state variables are defined as χl=y χ2=y=χ l χ3=y=χ 2 Therefore′ the state equations can be w五 tten as

1:Lttl」

雛牙 }1

In matrix form, these equations can be written

¨.(E13.20.5)

as

1 01 [x,l tol [i,t Io o 1ll,,l.lol, l;;l=1. L,,l l-t -10 -.] [;] L,]

..。

(E13.20.6)

The output equation is ¨。 (E13.20,7)

y = xr=Lx, +0-xr+0.x,

lrr-l O Ollrrl A=11

or

1,,

"。

(E13.20.8)

]

The state model is given by Eqs. (E13.20.6) and (E13.20.8).

翌轟甲紹饉簿

物

tra“ s/arル 4C'あ

げαω t701 “ “

Sys,mお

Y(s)

s2+3s+4

u(s)‐

s3+2s2+3s+2

初り ♂て

0b″ j“ αs協 ″ 観0′ θ:. s。 lutiOn.恥 given trallsfer hmction can be di宙 ded in two PartS・

Y(s)=Y(s).&(s) u(s) & (s) u(s)

where

and

&(s)_ U(s)

1

s3 +2s2 +3

Y(s) = s2 +zs+4 xr (s)

s

+2

+3s+4)

"。

(E13.21.1)

¨。 (E13.21.2)

¨。 (E13.21.3)

ノ   ヽ”

=s"^-J-.(s2 +2s'+3s+2

SttAttE ANALYS:S OF CONTROL SYSTEMS 551 Consider the function

&(s)= U(r)

1

s3

+2s\gs+2

By cross multiplication, (s3 +2s2

+3s+2)X,

(s) = 1- U(s)

The differential equation corresponding to this is

rt,!.r*.r*+Zx,=u dtr

dt'

¨.(E13.21.4)

dt

This is a third-order svstem, so three siate variables are required. Let the first state variable be equal to rr. The other state variables are r, and r, defined as follows :

xz=ir xg=iz Therefore, the state equations can be written as

¨.(E13.21.5)

The last equation is obtained by equating the highestorder derivative term in Eq. (E13.21.a) to all other terms. That is,

#

=

-z x,

-lff -r#

*

u=

-2 x, -3 x, -2 x, + u

The above set of state equations (E13.21.5) be written in vector-matrix form as

¨.(E13.21.6)

￨::￨=￨_l _! _llll:￨+￨￨￨“ The secOnd exPresSiOn is γ(S)=s2+3s+4

Xl(S) By cross multiPlying/we get γ(S)=(S2+3s+4)Xl(s) The differential equation corresPOnding tO dus is

y=争 +3年 +生 Or

y=4・ χl+3・ χ2+1・ χ3

1

=xu+3xr+4x,

552 CONTROL SYSTEMS h vector matrix ftt this output equation is wntten as y〓

釉e

...(E13.21.つ

[4 3 1111:￨

state model of the system is given by Eqs.(13.21.6)and(E13.21.り

.

￨

′ ″4s/arル 島 お♂ たり 励ι ルι Sysた s"″ θ 1轟 轟酵 /o7ルθ ￨ユ Ob″ 加α “ “ “ “ “ =轟 ‐ 253+7s2+12s+8 4ε

G(s)=

s3 +6s2 + l1s

+9

Solution. The given transfer function is

Y(s) 2s3 +7s2 +12s+8 G\"'' = :=:-: = --;-----i------

U(s)

¨.(E13.22.1)

s' +5s' +11.s+9

This can be written as

Yo=型 .型

LI(s) U(S)Xl(S)

=

.(2s3 +7s2 + 12s +8) s3 +5s2

+1ls+9

器 =冨計而 ' and

Y

(t)-

,,

trl

=2rt

…(E13.22.3)

¨.(E13.22.4)

+7s2 +12s+8

From Eq. (873.22.3), by cross multiplication (s3 +5s2 +11s+9)X, (s) =

1'U(s)

The differential equation corresponding to this is

#*#+n+ ■isお

¨。 (E13.22.5)

*ex,=v

a ttd‐ Order systtL so three state vanables are requlred.

χ2 and χ3 defined as Let the first state varlable be equal to為 .The other state vanables are

fonows: χ2=χ l χ3=χ

2

.

Therefore′ the state equations can be w五 tten as 141‐

and

￨lχ 」 〒￨・ 1111:￨￨‐ 19111

tlこ

tl11革 雪タ 1‐

…。 (E13.22.6)

STATE ANALYS:S OF CONTROL SYSTEMS 553 The last equation is obtained by eqgating the highest order derivative term in Eq. (813.22.5) to all other terrrs. That is,

dsxr

F-

*u

=-gr, 'nl -11tt-6d'*: " dt " dt'

.

- -9\-llxr-6xr+u The above set of state equations (813.22.6) can be written in vector-matrix fonn as

...(E13.22.つ

￨::￨=￨_￨ _11 _￨￨￨::￨+llタ The second exPresSiOn

Y(s)=2s3+7s2+12s+8 Xl(S)

Y(s)=(2s3+7s2+12s+8)Xl(s) The differential equation coHesPonding to this is

y=21111+7争

十 ソ 争

+蹟 1

=213+7受 2+12カ 1+8 χl =2(-9χ l-1レ 2 6χ 3+夕 )+7χ 3+12χ 2+8χ l =-10χ l-10χ 2 5χ 3+2“ h vector matrix fo...1′ 」iS Output equation can be w五 tten as

y=[-10 -10 -5]￨￨:￨+2“

...(E13.22.8)

The state model of the system is represented by Eqs. (813.22.n and (E13.22.8). 魃 鱚 鶉 欲 鰈 蟷

A“ ""I syS勧

お 燃 ″ b`′ リ ル ル IJ00iFg

Go=←

0レ ″′

"蘭

sOlutiOn。

s+3 +っ ← +2ア

“

"`liO"f

・

励 SO/′ ″Sys物 物 α “ “ “

gs″ ″α′0“ ″

ttψ ル

.

口隆 given transfer hction is

Glsl=器

葛爺号=

s+3 s3 +5s2 +8s+4

¨.(E13.23.1)

554 coNTROL SYSTEMS This can be divided in two parts as

Y(.s)_ Y(s)

&(4

¨(E13232)

uG) xl (, u(0 x,(s) i.l(s) s'

.

wnere

Y (s)

and

----j-:

&G)

1

r-

¨(E13233)

5sr +8s*4

= 0 is possible provided that Q(t) is known.

，  ¨ メ  ・

562 CONTROL SYSTEMS 13.27 PROPERTIES OF STATE‐ TRANS:T:ON MATR:X The state― transiton matrix o(′ )POSSesses

the following properties:

Proof.We know that

十 rた 。 十 対+ま ″′ =´ 到 ば り ま Putting I =0 gives ¨。 (13.27.1)

φ10)=I

￨

‐ ‐ ■響響? Proof.We know that 女(`)=Ax(サ

)

X(1)=0(サ )XO)

Therefore,

xo Ψ xo=AⅨ う T=AⅨ の

Or

1

3:φ

1(′

)=o(一

P“ げ We

¨.(13.27.2)

′ )

know hat 。 (r)=θ

MuliPlying both sides by

θ

φ )`―

A′

Ar giveS

A,=θ Ar`― Ar

(′

Or

A′

φ(r)θ

=I 1(う

Pre muldPlying both sides by φ 1(′

φ

or

)0(r)` Ar=φ

`―

Ar=φ

φ

Or

φ

1(オ

-1(`)I l(′ )

A,

1(′

or

We get

)=θ ) )=φ (― ナ

¨.(13。 27.3) ¨。 (13.27.4)

SttAttE ANALYSiS OF CONTROL SYSTEMS 563

4・

XO卜 千0← ￨)不 (す )

Proof.We know that

x(f)= se or

a1g;

ee x(0) = x(1)

Pre rnultiPly■

ng both sides by` A, ―Ar`A,x(0)=` Ar x(ナ ι

Using Eq。

(13.27.3)′

)

we get X(0)=φ (― ′ )X(′ )

¨.(13.27.5)

Proof.We know that

φ )=`At (′

Therefore

φ 2 亀 )=`A(,2 (′

and Hence

0(′ 1-′ 0)=ι 0(′ 2

rl) A(4-10)

′ 1)0(ll― ′ 0) =`A(,2 ll)`A(ll― =θ

A(,2 rl tt rl_′

′ b)

0)

=`A(t2 '0) =0(′ 2 ′ 0)

¨。 (13.27.6)

This property of state-transition matrix is called the hansition property. It implies that a state-transition Process can be divided into a number of sequential transitions. Figure 13.23 shows that the transition for t = ts to t = f, is equal to the transition from fo to f, and then from t, to ty ln general, the state-transition process can be divided into any number of parts.

0(r2 10)

卜●ュこｒＦｔ 卜Ｌｒ﹂トーーー ，

〔

Hg。

13.23■ o「

対ofよ ieitransitilnma轟

￨

564 CONTROL SYSTEMS

￨ ￨

‐

￨

‐ ‐ ￨￨￨￨:.(13.2夕 :ろ ‐

Proof.

t0(r)lk = eA' eAt..'k times

= rrx =61k ty

13.28 COMPUTATION OF 1. Laplace

ee

Method 0(f) = etu = €-1[s

I-A]-1

^-, Ieaikr-A1l =L bI-l'lJ

td"t

2. Power Series Method eAt

=l+

N*t_* *A, t, *... 21 3!

In the power series method, we have to calculate A,A2,A3,..- This because it is not practicable to calculate higher powers of A

is

time consuming

Hence, this method is rarely used to compute eAt. Laplace method is very convenient and hence 1;enerally used to compute eA'.

13.29

EIGENVALUES The roots of the determinant det Eigmoalues of ari zx n matrix characteristic equation

A

det

lsI-Alare known as eigenvalues. also referred to as characteristic roots, ate roots of the

[II-A]=

0

ll.r-Al=

o

■録■轟11e satte. The location of eigenvalues in the s plane indicertes the system instability. For a system to be stable, eigenvalues cannot be located in the ::ight hand side of the s plane. Eigenvalues can be directly obtained from state equations without converting the state equations into transfer functions.

STATE ANALYS:S OF CONTROL SYSTEMS 565 13。

30 EIGENVECrORS If A is an“ x“

ma臨

there are“ elgenvalues.

Any non zero vector x,which saisies the ma歯 敵 equation (λ :I―

Al x:〒 0

l,

2, ..., n, denotes the ith eigenvalues of associated with the eigenvalue 1.,. where Li =

…(13.30.1)

d

is called the eigenvalues of

4

If A has distinct eigenvalues, the eigenvectors

can be obtained directly from Eq. (13.30.1) by the solution of the set of z homogeneous equations written for different values of 1.,.

No eigenvector is unique. The fact at the x, are not unique follows from the fact Fi"α

Й7θ

|

),

I-A

|=

0.

″ α′ θ ε ra“ げル ′ 鷹χ “ “ “ `rgaυ “ A=l_: :“

『

.

』 S01uti。 )n (α

)Eigenvalue〔

)

A=[_: :]

︱

Ｊ

lλ

I―

AI=lλ

   〓

１  ３

＝

一

０  ２ 一 ｒｌｌｌＬ

ｑ司

〓

ｐ ｐ

[ I]一 [_:

￨+l

λI:￨

ne characte五 sicequauonOfAislλ I― A=0. λ(λ -3)― (-1)(+2)=0

メー3λ +2=0 (λ

-1)(λ

-2)=0

Therefore′ theれvo eigenvalues are λl=l

and λ2=2

(One COuld also write λl=2 and λ2=⊃・We Shall find eigenvectors for λl=land λ2=2.

(b)Eigenvectors Let the eigenvectors be written as

Xl=Li]and x2=[lil

〆 566 CONTROL SYSTEMS Substituting L=lⅢ Xl Weget lll l]一 [1 :]][il]=0

[ ]:][il]=0 χll χ21=0

This gives and

2

χll-2χ 21=0

That is,

If

xt

χll=χ 21

=L thm rr,

=t

r.

Therefore, one droicE of eigenvector is

t

nt

=Lr.l

Similarly, for L, =2

t't I t;lll

[x;]='

f, -l[x;]=' or and

-xu Zxrr-xo

2 xr,

is, x, =2 xt2 = 0, ttrat is, xu=Zxt2 = 0, that

Therefore, eigenvector for 1", =2 is Xl=[1重 ]=[]・ 1彗 豪 製 螺 11編 議

οb″ :4漬 ιSTM/Orル ιs協 ″ １  ３ 一

０  ２

〓

ｒ ト

Ａ

１

・３ 一

  ｓ

２ 陶 ２ 匡︲ ・  ０  ﹂ ・

０

︱ ・         ・ １ ０  １     １

Ｆ

一      〓 一

一

劇

弔

Solution.

ア ″ A,s gJυ θ′θ J ωた θs`“ ′′ ″ `“ “

_:]=lf sI:￨

STATE ANALYSiS OF CONTROL SYSTEMS 567

帽 adj

[s

I

J刊 4=愕

- A] = [cofactor matrix of

[s

I-

AI]r

―

=[::l :,IT=IS十

:

[]T

t =ISI::

lsI― AI=ISI:

:]

:￨=(S+3)s― (-2)(1)

=s2+3s+2=(s+1)(S+2) Therefore′

は胤 ■ le state transidon matrix(SnИ )iS giVen by

s+3

2

(S+1)(S+2) s+l

1 s+2

4島 =2〆 ― 2 ￡請議5=￡ 4島 一 ￡ 〆 ・ 1 =二 +⊥

(S+1)(S+2) s+l

s+2

2 ￡面れ5=￡・■・島=″ ― ・ 〆 +￡

-2

=三 2十 三生

(S+1)(S+2) s+l

s+2

I

568

CONTROL SYSTEMS

-z1',' =i 1+ f-t L=_Ze | +2 ezl (s+1)(s+2) s+l s+2 .-1 ') (s+1)(s+2) s+1 s+2 f,-1 s _-L t -1 +f, t 2 - .r-t *2rzt (s+1)(s+2) s+1 s+2 El

一  十

´π

一

一   ＋

う４  ●乙 一

13.28

〓

ＡＶ

Exarrrple

′π

Therefore,

A linear time-inttqr;ant sVstefi is characterized by the state equation

l.l=l'oll', fi,l [t t.] fx, l.lf I lr where u is a unit step function. The

, l

initial condition

is

x(o) - Itl Lol Llsing inuerse ktplace transt'onn methnd, obtain the solution of the state equdtion.

r=lll soturion. "-[iLl 9], 1l ', Lll For unit step functioir

U1s1 = 1

The state equation of a dynamic system is

i=Ax+Bu(t) The solution of this equation is

x(t)=

{

tlO(Ox(O)l

+f

zero_inPutresPonse

0(O=tsI-Al

1

:l -Lo 9l 1sr-e;=sl1 1l '1 L1 1l [s

tt

-L

-1 0 I

-,

,-tl

lt$(s) BU(s)l

zero-stateresPonse

一 Ｔ               Ｓ ￢ 川 句            ＜

  ０

   

   

   

   

   

   

   

   

  り

●

〓 一 ｌｓ       ↓ １ 一 

劃

劃

巨 ︱＾ Ⅳ

Ｈ ・ 市 ・ 一 ］ ＪＬ 判

             〓 一 一

ｒ    ｒｌｌ﹄Ｆ﹂

¨¶

The zero― inPut reSPOnse

           一

釧 目

￨

Ｆ

釧 日 一

Ａ

一           〓 一 一           一

・

降 φ(S)X10)=

一        千 Ｔ一

ｕ Ｒ

︱

…甲

Al]r Its

- Al = [cofactor matrix of I ts

adj

愕

lsI Alた

Therefore,

=降

=降

569 SttATE ANALYSIS OF CONTROL SYSTEMS

崎→ IR]=￨ム BЧ

TherefOre7礎

鋼 ０・一

〓

０  １

〓 Ｂ

ＡＶ

０ ・一 門

降

570 CONTROL SYSTEMS

=￨ム

zerO― state

￨

resPOnse 1[0(S)BU(S)〕

=￡

=￡

-11-:1島

-11π

l:51=￡

￨

=[_llθ `]

The total resPonse=Zero― mput resPOnse+zerO― state resPOnSe ′  ＋ ″ “Ｖ ＋

ｒｌト 〓

﹁ ︱ Ｉ Ｊ

ｎ︶  ＋

＋

Ｆト

〓

′ ″

Ｘ

CONTROLLAB:LiTY AND OBSERVABIL:lγ ■ F cOncep“ Of cOntrO■ ab■ ity and Observab■ ty were irst htroduced by Kalman h 19611 ■ ese cOncePお are necesm7 tO establish ttF vttdty Of iF dassical"ansfer imdiln

methods Deflnitio● ●f contro‖ abi:itv

The definition does not restrict the choice of input

move■e

u(f)

The idea is that

it is possible to

system state to any desired des歯 ■atOn Thus′ contrO■ ab■ tyis h ab■ ty of樋 control inPut to affect each state vanable lfevery ttate χ(,0)。 fttt System iscontollable h a hte hme m“ Ⅳ t

hesystm issaidto

be completely controllable or simply cOntrOnable The contrOllab■ ty defhed above is assoclated With he states and should nOt be cottused つ vi■ 」ね 。utput cOntro■ abi■ ty7

whchis h Propeゥ

An uncclntro■ able system is a system wh■ ch

discOrmected from the nput

Of the lnPu■

outPut relatothP

has a subsystem htお

PhysICally

STATE ANALYSIS OF CONTROL SYSTEMS 571

13.32 XAIT'AN'S

TEST FOR CONTROTIAEILIW

A linear time-invariant system described by the state equation

To=6,14*ro14 is completely state contollable, it is necessary and sufficient that the following controllability matrix has a rank of z, the order of the system

n

x /rP

Q"=lB : AB : A2B :...: A"1BI The

cnnfioltability ,natria. consists of the nlumns ofBfollowedby the columrc of AB, and so on,

Since the matrices A and B are involved, sometimes we say that the pair controllable which implies that Q" is of rank I'

[AB]

is

For a single-input system Qc is an nx z square matrix' The matrix Q" is non singular, that is, its determinant is non zero. Therefore, in the SISO case, the system is conhollable if and only if the determinant of Q" is tuxl zero. IF● r"響

iT聟 午 午

llabⅢ

●lr■ eヴv“

“

=:ヤ

13.33

Sげ Q・ ,11if“

I彎 夕:1,警

ndell‐

RANK OF A MATRIX

A square matrix of order nx n is said to have a rank r, if and only it it has at least one non singular sub-matrix of order r but has no non singular sub-matrix oI order more than r. Therefore, we find out the largest square matrix in the given makix whose determinant is not zeto.

″

１  ２ 一

十

レービ

〓

″  χ

珊

Test the follrusing system for controllability using Kalman's test :

ｐ レー´

,.$x.a|4pl.a:i1i.29i,

sOL■ On

A=[ 」 :]′ A is2×

2′

B=[三:]

so the contro■ ab■ ty matnx is

Qc=【 B

AB=[

: AB]

」 i][_:]=[ ξ

]

572 CONttROL SYSTEMS

hrefore/

Qc=[_:

:]

which is not of rank 2(order Of→ ′Since

︲︲︲１′ｌ ｊ ︰ 二´ ｒ 帯﹁ ，

二謝4引 →0却

Therefore, the system is not completely controllable.

‐ ■::mpl111:301 Forナ ルル θi4g s1/.sた 7」

J′

￨::￨=￨:i

SOlutiOn

ルθθ θ

777/′

ll′

α bjIJry:

]・

ll ll [:￨十

A=￨ i _i

ror′

"′

`Sナ

1-￨ ￨￨ [ち

B=￨一

￨ ￨￨

８ ・ ６ ・

〓

４

５ 卜”﹂

Ｏ５０ ・

１

１   ４  ０  ０ Ｂ

ｒ

４

２  ３  ０

２

Ｆ   トー ー ﹂

一 一        〓

ｃ Ｂ Ｑ     Ａ

０２ “２ ・ Ｏ ５ Ｆ︲ ・ ﹂Ｐ ｒ ，

〓

  ｎυ ら乙  う０一      ﹁︲コ

Ｂ

Ａ

︲ｊ ﹁ ８ ・引 ４ ５ 卜”﹂

郎 ・０・ ・ 郎 １０． ・ ２ ドμ ﹂ 国 Ｆ ・ 一 一           〓

A is 3× 3 so the cOntrollability IFnatrix is given by

-820 Therefore, Q" = [04-s -5 0 0 16 -5 |

4ol

-Z0l

Ioo s 4-s-24j]

To be of

full rank, the controllability matrix must have three linearly independent equations.

We have

lo4-51 o l-so o I

o

l=o-+1sy1-5)-o sl

=1oo

*o

Since the determinant is non zero, the system is completely controllable.

SttAttE ANALYS:S OF CONttROL SYSTEMS 573

A system is described by the dynamic equations

[1:]=[ l _￨][I:]+[ ￨]夕 G″ 解θ ″θ ナ ルθ θ′ γ θ bi::ry O/′ ルSr′ θ 解 ・ “ “

,4グ

・

[ ￨]χ

J:α

トエ

Ais2x2,

so the controllability matrix is

B=[: I]

．一

卜

０

_￨]

司

口

Ｂ ０ ＾ ・ Ａ   伸︱ ︱ 押    ．

ｐ＝

A=[ : _:]′

ｃ Ｂ ｃ Ｑ   Ａ     Ｑ

sOhtiOn.

Determinant of sub-matrix

It-1 1llisl -1 1l L0 -1.1 lo -11 r

I

l-1 ^ |0

1l

Nowl

- | =1, that is, the determinant is not equal to zero, and the rank of Q. is 2 (order

-11

of the system). Hence, the system is state controllable.

OBSERVAB:LiTγ DennltiOn

The concept of observability relates to the condition of observing or estimating the state variables from the output variables, which are generally measurable.

Observability is concemed with the ability of eadr state variable to influence the output of the system. There are no restrictions placed on the output. The definition indicates that any earlier value of the state vector x(f) is determinable by watching the output y(f)

574

CONTROL SYSTEMS

13.35

KALMA]T'S TEST FOR OBSERVABILITY

A MIMO linear time-invariant system described by the dynamic equations d*lt) = ar(r)+ n, (r) dt

Y(t) =

cx(t)

it is necessary and sufficient that the following nxzq obeervability matrix Qo has a rar* of n, the order of the system.

to be completely observablg

c CA

Qo=

=[Cr

;;,

Ar

Cr

A2r

Cr ...

A,-1 Cr ]

;

cA*r This condition is also referred to as the pair [A, Q being observable.

In particular, if the system has only one output Cis a 1xnrow matrix. Q. is an(nxr) square matrix. The system is completely observable if Q" is non singular, that is, determinant is not zero.l

13.36 DUAUW

PROPERTY (DUAUTY THEOREM)

The principle of duality was introduced by Kalman. We have seen the conditions of controllability and observability. The pair [A,B] is controllable if the rank of Qc = The pair

[A,9

[B AB ... A'i

is obsewable Qo =

B] is a.

if the rank of

[C?

Ar

cr

... 1a-1 1r cr

1

is z.

Thus, there is a link between the two concepts. The duality theorem states that if the dynamics of a system is described in state sPace with the help of matrices (d B, C D, ) then r, r the system is completely state controllable if and only if the dual system 1Ar , Cr , B D ,1 is completely observable. The duality principle specifies that

4 6

Pair [A,B! is controllable also means that the palr [Ar,Br] is observable.. Pair [A,9 is observable also means that the pair [Ar , Cr ] is controllable.

STATE ANALYSIS OF CONTROL SYSTEMS 575 Wi■ the use of ms theOrem state controllabihty of a glven wstem can be checked by per姉

g ObSewabiliけ test OfitS dual Similarlル the ObServability of a glven system can

be checked by perfor― g

13.37

state contrOllab■ ty test of its dual

CONTROLTABITITY, OBERVABITITY, AND POLE.ZERO CANCETTATION The concepts of controllability and observability are closely related to the properties of the

transfer function.

Let the input-output relationship of an nth-order system having distinct eigenvalues (roots) be represented as

y(s)

_ k(s_zr )(s_221...(s_z^) . m0.

I

20. A linear

_e-2t

o(t)=l ^ +2e-2' l-2r-'

e-2r I - e-' +2e-2')

f ?e-t -e-2t '(o=L-;; '*2"'

)'

Ze-t

e-t

x (t)

for I

>0

when the input is

_

|

e-t -e-2t

) -

[0.5-r-'+0.5e-2'l

*i ''.]'(o)+ l'" ;'-';;"

time-invariant system is characterized by

[r,]

[r

a'l

Lr,l=1, -rl

[,,] [o'1. 1,,1.

L,l'

v=rr ot['''l

lxz)

where a is a unit step funCion. The initial conditions

obtah (a) zero-input respons€ (c) total response Ans.

(a\ (c) 2I.

Lxz)

L

1

@\

v=rl orll:l=,,=i

l

it=-\+u i2=xr-Zxr+u State whether the system is controllable or not.

[Not completely controllable] is described by

[t,'l

-

0

*2,+J ,., [,, l-[1-,,-' (r) 1," l=llr,-_, i..,,1 _-_ g,. 3.

The state-space representation of a system is described by

22. A system

= 1, rr(0)

(d) output lesponse

.1

L3 3

xl0)

(b) zero-state response

7x,l lze-t --"1, I 't=t . +Je'r |

l-e _ _ l4 1._:r'l li:l=l ', .i",,

are

11 -11 [,,

L;,1=1, -,1

I

lol

L,,l-L,l'

y=tl 01[''l Lxz

l

Check the controllability and observability of the system.

[Completely conhollable and completely observablej

.l

-i, "

.l

''o

586 CONTROL SYSTEMS

23.

comment on ttre conhollability of the system described by the state equation

[Not controllable]

L]=F llL]+l]“

Comment on the complete state observability of the system described by

Io 1

*=l o

o

ol

rlx

l_-r -1r - 6l Y = [1 -1 UX

[Observable]

Comment on the complete state observability of the system described by

ｑ

Ｘ   Ｘ

  ．一

〇

︲ ︲１ ２ 劇劉

２

ｌ

ｌ ｌ

一 一        〓

・ Ｘ   Ｙ

巨 ｌｎに ｐ

[Not observable]

lNDEX Cenhoid

313

Acceleration error 169

Chain node 57

Acceleration function 143

Characteristic equation 149, 532

Advantages of Nyquist method 400

Characteristic polynomial 149

All-p""s system

Closed-loop stability 401

274

Analogous circuits 114

Closed-loop system 3

Analytic function r()1

Companion form matrix 539

Angle condition 309

Compensation 457

Angle criterion 309

Compmsator 457

Angle of arrival 314

Conditionally stable system

Angle of deparhrre 316

Constant magnitude

Angles of asymptotes 314

Constant-M circles rt20

Asymptotes 373,2ffi

Constant-N cirdes 42i!

Asymptotic angles 314

Costant phase

Asymptotic plot 245

Contour map 401

B Bandwidth

orr*

363

loci

Control system

1

Controllability

570

loci

211

rO0

42;)

Conholler 2

BIBO stability 205

Conhollable canonical forur (CCF)

Bode diagram 245

Comer frequency 255

Branch 57

plot 245 Critical point 409

Break angles 315

Cutoff frequmcy

Break frequency 255

Cutoff rate 353

Bode

plot

Breakaway

Comer

244

point

o.

point

353

314

p.■ ￨￨￨￨￨￨._‐ ‐ ￨￨￨￨￨￨■ ‐ ￨

Break-in point 314 Break

.539

314

Bush form matrix 539

Damping factor Damping ratio

(

(

149 149

Diagonalization matrix 529 Decade 249 Cascade Compensation 458

Decomposition 522

Cauchy's theorem 408

Delay time

Q

159

588 CONttROL SYSttEMS Delta (6) tunction 143 Diagonal canonical form (DCF) 525

Gain adjustrnent 272

Diagonalization 529

Gain crossover frequenry cor. 258,419

Disturbance 2

Gain margin (GM) 269,377,363, 477

Dominant poles 174

-

Generalized error coefficients 173

Dominant roots 478 Dominating pole 478

Dominating zero 478

Impulse function 143

Duality theorem 574

Input node 57

Dudity Woperty 574 Dummynode 57

Insignificant poles 174

Dynamic equations 509 Dynamic error coefficients 173

|ordan blocks 526

jordan canonical form flCF) 525 Effectof addingpoles and zeros to transferfunctions 774

￨

・ ￨‐ ‐K‐

Kalman's test for controllability 571,574

Effects offeedback 5

Eigenvalues 564

Eigmvectors 565 Lag-lead compensator 488

Encirdemmts 403 Epsilon

(e)

Liquid-level systems

method 209

Logarithmic

r Features of feedback 7

plot

129

245

Loop gain 59 Loop path 58

Feedback compensation 458 Feedback control system 4

Magnitude condition 309

Feedback path 58

First-order system

1t14

Mapping

401

Force-current analogy 116

Mapping theorem 405

Force-voltage analogy 115

Mason's gain formula 63

Forcing function 559

Maximum overshoot

Forward path 58

MIMO system

Forward path gain 58

Minimum phase system 274

Free-body diagram 97

Mixed node 57

Frequency response 2U, 367

159

8

TNDEX 589

Principle of argument

N

Process

Nidrols chart 428

408

1

Properties of signal flow graphs 52

Nodal method 105

Node

lmt

Proportional controller 175

57

Proportional plus derivative (PD) controller 176

3

Nonleedback system

,

Non-minimum-phase transfer function 275 Nontouching looPs 59

Proportional plus integral (PI) controller 17 Proportional-integral-derivative (PID) controller 779

Nyquist stability criterion

411'

_ Ramp function 142

ObservabilitY 573

Re-entry

,

\

q

Observabititfmatrix

574

pgint

Regulator

31'4

7

﹁︲ ，      ︲ ＼

Observable canonical form (OCF) 5tt4

Relative stabilitY 211

Octaves 249

Resolvent matrix 532

Open-loop stabilitY 400

Resonant peak magnitude

Open-loop sYstem 3

Resonant frequencY 363

Output node 57

Reverse coefficient method 210 Rise

time

M,

363

159

Root locus 308 Parabolic function

1t12

Routh stabilitY criterion 208

Parallel Compmsation 468

Routh's arraY 206

Path 58 Peak overshoot 159 Phase

condition 3@

''

Phase crossover frequency Phase margin

(PM)

ro*

269,417

364

Second-order sYstem 148 Selection of a comPensatot 469

Selfloop

59

Phase variable canonical form (PVCF) 539

Semilog paper 245

Phase variables 537

Sensor 5

Phase'lead comPensator 470

Series compensation 458

Phase-lag comPensator 480

Series-parallel compensation 459

Pickoff point 19

Servomechanism 7

Plant

Setting

1

Pneumatic sYst6m 132 Polar plots

time

4

159

Similarity transformation 529

373 1

Pole-zero cancellation 575

Single-valued function 401

590

coNTRoL SYSTEMS

Singularity

Sinknode

C 125 Thermal resistanc€, R 125 Thermal capacitance,

401

57

SlSOsystem 8

Thermal systems 125

Sourcenode 57

Time response

Standard fomr of secpnd-order system 148

Time'response specifications 158

Standard test inputs 141

Transducer 2

State 505

Transfer function 11

State equations 505

Transmittance 60

model

State

1t[()

Transport lag 276

509

State space 506

Type 0 system 164

State traiectory fl)5

Type 1 system 164

State transition matrix (STM) 560

Type2system

State variable 505

Types of control systems 164

State vector 505

Types of feedback control system 7

Static acceleration error constant Static

error

4

158

155

$ 16 Static position error constant $ 76 Static velocity error constant (, 767 Static position error coefficimt

Steady-state

error

function

System

Undamped natural frequenry