Control Struct

Table of ContentsChapter· 1 Basics of Control System ~(1-1 ) to (1- 24)

1.1 Background 1 - 11.2 Definitions 1 - 11.3 Classification of Control Systems 1 - 31.4 Open Loop System ; 1 - 7

1.4.1 Advantages 1 - 81.4.2 Disadvantages 1- 81.4.3 Real Time Applications of an Open loop System 1 - 8

1.4.3.1 Sprinkler used to Water a lawn. . 1 -8

1.4.3.2 Stepper Motor Positioning System. 1-8

1.4.3.3 Automatic Toaster System. . . . 1-9

1.4.3.4 Traffic Ught Controller . . . . . 1-9

1.4.3.5 Automatic Door Opening and ClOSingSystem . 1- 91.5 Closed Loop System 1 - 9

1.5.1 Advantages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 - 111.5.2 Disadvantages 1 - 111.5.3 Real Time Applications of Closed loop System 1 - 11

1.5.3.1 Human Being . . . . . 1 - 11

1.5.3.2 Home Heating System. . . . 1 - 11

1.5.3.3 Ship Stabilization System. . . 1-121.5.3.4 Manual Speed Control System. 1 -12

1.5.3.5 D.C. Motor Speed Control . 1- 13

1.5.3.6 Temperature Control System 1 - 14

1.5.3.7 Missile launching System . 1 - 14

1.5.3.8 Voltage Stabilizer. . . . . 1 - 14

1.6 Comparison of Open Loop and Closed Loop Control System 1 - 161.7 Servomechanisms "" 1 - 161.8 Regulating Systems (Regulators) 1 - 17

y...... :":v~.~- < ~... :~.~

II \ nl

1.9 Feedback and Feed Forward System 1 - 191.9.1 Real Time Application of Feed Forward System 1 - 20

1.10 Multivariable Control Systems 1 - 21Review Questions ........ . ......... "" .." 1 - 23

- ~'9 ..... ~!' •

Chapter· 2 Basics of LaplaceTransform ,. h ,<

... .

< (2·1) to (2·20)2.1 Background 2 - 12.2 Definition of Laplace Transform 2 - 12.3 Properties of Laplace Transform 2 - 2

2.3.1 linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . .. 2 - 22.3.2 Scaling Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2 - 32.3.3 Real Differentiation (Differentiation in Time Domain) 2 - 32.3.4 Real Integration , 2 - 32.3.5 Differentiation by s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2 - 42.3.6 Complex Translation 2 - 42.3.7 Real Translation (Shifting Theorem) 2 - 42 3 8 InitialValueTheorem 2-4

2 3 9 FinalVallie Theorem 2 - 52.4 Inverse Laplace Transform 2 - 8

2.4.1 Simple and Real Roots. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2 - 92.4.2 Multiple Roots 2 -102.4.3 Complex Conjugate Roots 2 -12

2.5 Use of Laplace Transform in Control System 2 - 142.6 Special Case of Inverse Laplace Transform 2 - 15Examples with Solutions ..: 2 - 17

.. ' "'. ~'1W!(' .0J>' .Chapter..3' Transfer FUflction.andImpulseResponse.'.~.'.~ ',>.,~ . (3 ..1)to (3·26)3.1 Background 3 - 13.2 Concept of Transfer Function 3 - 13.3 Transfer Eunct;on " 3 - 2

3.3.1 Definition , , . .. . ! •••• ! ••••••••••••••••••••• 3 - 23.3.2 Advantages and Features of Transfer Function 3 - 43.3.3 Disadvantages 3- 53.3.4 Procedure to Determine the Transfer FuncUcn of a Control System 3 - 5

3.4 Impulse Response and Transfer Function 3 - 73.5 Some Important Terminologies Related to the T.F 3 - 10

II \ nl

3.5.1 Poles of a Transfer Function 3-103.5.2 Characteristic Equation of a Transfer Function 3 - 11

. I If., .•. • '.3.5.4 Pole-Zero Plot. . . . . . . . . . . . . . . . . . . . .. . . .. . 3 - 12. ,. ,. ... . I.3.5.6 D.C. Gain 3-12

3.6 Laplace Transform of Electrical Network 3 - 14Examples with Solutions 3 - 15Review Questions... .. " ." " 3 - 24

Chapter· 4. MathematicalModeling of Control Systems ~,?,~,ry., ~,\: (4 ..1) to (4 - 64)4.1 What is Mathematical Model? .4 - 14.2 AnalysiS of Mechanical Systems 4 - 2

4.2.1 Translational Motion , 4 - 24.2.2 Mass (M) 4 - 24.2.3 Unear Spring 4 - 34 2 4 Friction.. .. . "'''.'....... 4 - 4

4 3 RotationalMotion 4 - 5

4.4 Equivalent Mechanical System (Node Basis) .4 - 6: - II -. •• ~ •• _ IJ - ••• . -

4.6 Gear Trains 4 - 8.. .'. '. r. I ••• ·.... 1 • '.

. - .4.6.2 Belt or Chain Drives ..........•........... ' • . . . .. . 4 - 114 6 3 Levers .. .. . 4 - 11

4.7 Electrical Systems 4 - 114.8 Analogous Systems 4 - 13

4.8.1 Mechanical System 4 -134.8.2 Force Voltage Analogy (loop Analysis) .................................•... 4- 134.8.3 Force Current Analogy (Node Analysis) .................................••.. 4-15

4.9 Steps to Solve Problems on Analogous Systems 4 -16r

4.10 Servomotors 4 - 204.10.1 Requirements of Good Servomotor 4 - 20

4.11 Types of Servomotors 4 - 204.12 D.C. Servomotor 4 - 21

4.12.1 Field Controlled D.C. Servomotor 4 - 214,12.1.1 Features of Field Controlled D.C. Servomotor . . . . . . . . . . . . . . . . 4 - 21

r

J 1t ...4.12.2Armature Controlled D.C. Servomotor •.•..•...•.••.•.•...........•.•..•..•. 4 - 21: -u l:!' , ..III.·! ,: ,',I .", ,:.'. .:;l 0,11.' .' . - ~

. -4.12.4Applications of D.C. Servomotor .•••.....•.....•••......•..•...••....•.... 4 - 23-. . I '.' ., '. , -, • u' , . -

-. .I ' • I.'!. i.._ I I I ••• L~I I . -•- 'llllil '- ,

4.15.1 Construction ' ' , . , 4- 274.152 Rotor ..••.......................•••.•.........•..................... 4·274.15.3 Torque-speed Characteristics •.•.....••....••.••...••...•....••......•... 4 - 274.15.4 Features of A. C. Servomotor ........••...••....•..••....•.•.......••.... 4 - 284.15.5 Applications ••...••.••••..•.••.••••.•••••.•.•••.•••..•••••.•••••..•... 4 - 284.15.6 Transfer Function of A.C. Servomotor .••....••..•...•........•.•...••.••.• 4 - 28

4.16 Comparison of Servomotors .4 - 304.16.1 Comparison between A.C. and D.C. Servomotor •.••....••...•.•..••••.••.... 4- 304.16.2 Comparison between Armature Controfted and Field Controlled D.C. Servomotors 4 - 30

4.17 Models of Commonly used Electromechanical Systems .4 - 304.17.1 Generators ...••.•.•.•.•....•....•..••.•........•••.....•..•...•..•.. 4 • 314.17.2 Generator Driving ~tor ••.•..•••.••••..•.••......•••.•.••............. 4 - 324.17.3 Position Control System .•.........•.•.....•........•.......•........... 4 - 344.17.4 Position Control with Field Controlled Motor •...........•................... 4 - 354.17.5 Speed Control System •..•........•..•..•.•.•....•...•.................. 4 - 374.17.6 Speed Control using Generator Driving ~tor ............•................•. 4 - 394.17.7 A Typical Position Control System used in Industry •.•.....••..•....•••....... 4 - 41

4.18 D.C. Motor Position Control System .4 - 424.18.1 Transfer Function of D.C. Motor Position Control System ....•...•.....•...... 4·43

4.19 Models of Thermal Systems 4 - 494,19.1 Heat Transfer System ••.••.•............••......•......•...•...•.....• 4 - 494.19.2 Thermometer ..•...•..••..•...••...•••••........•.........•..•...... 4 - 51

4.20 Actuators 4 - 514.20.1 Hydraulic Actuator 4 - 524.20.2 Pneumatic Actuator ..........•.......................•......•........ 4 - 534.20.3 Comparison between Pneumatic and Hydraulic Systems . . . . . • . . . . . • . . . • . . . . .. 4 - 54

Examples with Solutions 4 - 54• ~:r.

Review Questions :..: 4 - 61

l I 'nl

.' ........

~ Ch.ter.;~J::'"~ Diagram·~entatlon~':('; :~:.- "~.Y~." ";., ": :: ~ ~., ,~.~.: .~. ' •• ~ (5..:1")t~:(.5-_la)5.1 Background 5 - 1

5.1.1 Illustrating Concept of Block Diagram Representation .......................•... 5 - 25.1.2 Advantages of Block Diagram 5 - 45.1.3 Disadvantages ....•...................................•................. 5 - 4

5.2 Simple or Canonical Form of Closed Loop System 5 - 45.2.1 Derivation of T.F. of Simple Closed LoopSystem .........................••.... 5- 5

5.3 Rules for Block Diagram Reduction 5 - 55 3 1CriticalRilles 5 - 13

5.3.2 Converting Nonunity Feedback to Unity Feedback .................•.•....•.... 5 - 155.3.3 Procedure to Solve Block Diagram Reduction P,roblems 5- 16

5.4 Analysis of Multiple Input Multiple Output Systems 5 - 18

5.5 Block Diagram from System Equations 5 - 21Examples with Solutions 5 - 24:.. ". .. . .• I

Chapter ·-6.. Signa('FIow Graph Representation :'~~~;:,1<,:~,,~::,':,~~~< ~ :: ~~:?"~.(6 ..1) to (6 • 66)6.1 Background 6 - 16.2 Properties of Signal Flow Graph 6 - 2.6.3 Terminology used in Signal Flow Graph 6 - 3E>.4Methods to Ot>tainSignal Flow Graph 6 - 5

6.4.1 From the System Equations .....................•......................... 6 - 56.4.2 From the Given Block Diagram ......•..•...•.........•........... ' 6 - 6

• M •• ' .~ I • III ~ .-

6.E>Comparison of Block Diagram and Signal Flow Graph Methods E>- 9E>.7Application of the General Gain Formula t>etweenOutput

Nodes and Non Input Nodes E>- 106.8 Application of Mason's Gain Formula to Electrical Network 6 - 12

6.9 Obtaining Block Diagram from Signal Flow Graph 6 - 14Examples with Solutions 6 - 17.- .

Chapter ..7 Time Response Analysis of Control Systems Y ."! (7·1) to (7.130)

7.1 Background 7 - 17.2 Definition and Classification of Time Response 7 - 1

.. ~..,I I nl

Control System EngineeringISBN 9788184314632

All rights reserved with Technical Publications. No port af this book should bereproduced in any form, Eledronic. Mechanical, Photocopy or any information slorage endretrieval systemwilhout prior permission in writing, from Technical Publications, Pune.

Published by :Technlcal Publications Pune8#1, Amit Residency, 412, ShaniwarPeth, Pune- 411 030, India.

Printer:AJertDTPrintasSr.no. 10/3,SinhagadRoed,Pune- 411041

:"..:, .., ".L -,', ",' :':" ,.,,:. ~.,...:. -.-, ;{ .t.. .\:' ,:,.~"" ~-~~. :.,' " ,,'_- ,,: "(if)." .,-.:.:.,'-:: :,.~.~..~;.~~~;:'h .~~:._. , ••~'.>":::,'~~:~.'~.....•.., : :.,~.:'...~: :.:~:..~.::';,t~·~'..~'.~~~.'A,," ,~.~. ~ .j,.;',.

_",' ,'",',. .::"':.:.....:.' ::::.::<~:,,':: ,," ......:: <.:' ', ' "" .':' ".:=-:,," -"t-: "" " .... .__ ~ .,., " " ..." .. '" ~\ :,'.

l I

7.3 Standard Test Inputs : 7 - 47.4 Steady State Analysis 7 - 67.5 Derivation of Steady State Error 7 - 67.6 Effect of Input (Type and Magnitude) on Steady State Error

(StatiCError Coefficient Method) 7 - 87.7 Effect of Change in G(s)H(s) on Steady State Error

(TYPE of a System) 7 - 107.8 Analysis of TYPE 0,1 and 2 Systems 7 -117.9 Disadvantages of Static Error Coefficient Method u_, 7 - 167.10 Generalised Error Coefficient Method (or Dynamic Error

Coefficients) 7 - 177.11 Transient Response Analysis 7 - 23

7.11.1 Method to Determine Total Output c(t) 7 - 237.12 Analysis of First Order System 7 - 24

7.12.1 Unit Step Response of First Order System 7 - 247.12.2 Closed loop Poles of First Order System 7 - 26

7.13 Analysis of Second Order System 7 - 287.14 Effect of l;on Second Order System Performance 7 - 29

7.15 Derivation of Unit Step Response of a Second Order System .(Underdamped Case) 7 - 34

7.16 Tra~sient Response Specifications 7 - 387.17 Derivations of Time Domain Specifications 7 - 39

7.17.1 Derivation of Peak Time TD 7 - 397.17.2 Derivation ofMp 7 -417.17.3 Derivation ofT, ............................•.......................... 7 -427.17.4 Derivation ofTs ......•...••.•......•..•.••..•....•.........•....•.•••. 7 -43

7.18 Feedback Characteristics of Control Systems 7 - 447.18.1 Effect of Parameter Variations in an Open loop Control System ••.•.•.•.•.....•. 7 - 457.18.2 Effect of Parameter Variations in a Closed loop System 7 - 457.18.3 Sensitivity of a Control System -........•...... 7 - 467.18.4 Effect of Feedback on Time Constant of a Control System ...............•.....• 7 - 487.18.5 Effect of Feedback on Overall Gain ..•.•................................... 7 - 501.18.6 Effect of Feedback on Stability 7 - 507.18.1 Effect of Feedback on Disturbance .............••.•••...•. ' 7 - 51

I I nl

7.19 Performance Criterion 7 - 577.19.1 Mean Square Error (EMS) 7 - 577.19.2 Integral Square Error Criterion (ISE) 7 - 587.19.3 l'ltegral of Time Multiplied Square Error Criterion (ITSE) 7 - 587.19.4 Integral of Squared Time Multiplied by Squared Error (ISTSE) 7 - 587.19.5 Integral of Absolute Value of Error (IAE) 7 - 597.19.6 Integral of Time Multiplied by Absolute Value of Error (ITAE) 7 - 59

Examples with Solutions 7 - 60Review Questions 7 - 124

':.. .-. • 'q~ :.):.1 <l- --~",,:: ~. ""1» '" :.r.-;: ~ • ',... • ... ~:'«I].C: ~ • ''':: ~":«. : ~

Chapter, 8 Stability Analysis '. . . .. . .. :; ,.,'. <. (8 -1) to (8 ·46)

8.1 Background 8 - 18.2 Concept of Stability 8 - 18.3 Stability of Control Systems 8 - 38.4 Zero Input and Asymptotic Stability 8 - 8

8.4.1 Remarks about Asymptotic Stability 8 - 88.5 Relative Stability 8 - 98 6 Routh-Hurwitz Criterion . . 8 - 9

8.6.1 Necessary Conditions 8 - 10862 Hurwitz's Criteoon : " 8-108.6.3 Disadvantages of Hurwitz's Method 8 - 11

8.7 Routh's Stability Criterion 8 - 1187.1 Routh'sCriterion , , 8- 12

8.8 Special Cases of Routh's Criterion 8 - 138.8.1 Special Case 1 8 - 138.8.2 Special Case 2 8 - 14

8.8.2.1 Procedure to Elirrinate this Difficulty . . . . . 8 -158.8.2.2Impcrtance of an Auxiliary Equation. . . . . . . . . . . . . . . . . . . . . 8 -158.8.2.3 Change in Criterion of Stability in Special Case 2 . . . . . . . . _ . . . . . . 8 - 16

8.9 Applications of Routh's Criterion 8 - 178.9.1 Relative Stability Analysis 8 - 178.9.2 Determining Range of Values of K 8 - 17

8.10 Advantages of Routh's Criterion 8 - 188 11 Limitations of Routh's Criterion 8 - 188.12 Marginal K and Frequency of Sustained Oscillations 8 - 20Examples with Solutions 8 - 21Review Questions _ _, 8 - 43

'.. .". c ~ :::_" . -". ........ '"x .

l I

hi rIt-w:···'· ··:9·'R· ott .AA.f.' '>."." y.", :~/_ c .. <. ~. ",0 ? ' z: , 119' 4\6.· (a 102)• UlIHI.I'.loer~:.'"~.. 0 ..~! ' ~ ~'... ~ ~, ~., .::":, '. :'\ ,a_ ".:~Q '.•. J<I~ ••

9.1 Background 9 - 19.2 Basic Concept of Root Locus 9 - 19.3 Angle and Magnitude Condition 9 - 3

9.3.1 Angle Condition .......•................................................. 9 - 39.3.2 Use of Angle Condition ........••.•.•..................................... 9 - 49.3.3 Magnitude Condition 9 - 59.3.4 Use of Magnitude Condition 9 - 5

9.~ <:;raJ)t1i~1Mettlod of [)eterrTlining 't(1 9 - E5• .. . ...:: .. . • - ... : - • •• I

. .•• • • •• • .- .9.7 Grapt1i~1 [)etermination of 'K' for Specified [)amping Ratio's' 9 - 209.8 <:;eneralSteps to Solve the Problem on Root locus 9 - 219.9 Effect of Addition of OJ)enLooJ)Poles and Zeros 9 - 28

9 9 1Additionof Pole 9 - 28•• !••..•I' .•t.t-

9.10 Advantages of Root locus Method 9 - 319.11 Obtaining G(s)H(s) from Ctlaracteristic Equation 9 - 319.12 Cancellation of Poles of <:;(s)with Zeros of H(s) 9 - 32

9.12·.1Gain Margin 9 - 339.13 Root Sensitivity 9 - 339.1~ Root Contour 9 - ~9.15 Inverse Root Locus "" " 9 -~9.1E5System with Positive Feedback (K is Positive) 9 - 37ExamJ)leswith Solutions 9 - 40Review Questions 9 - 100

Chapter ~:·;10·:··1lasica:of FtegueriCY:·:i)omain Analysis ' .'.-.' -;.' (10 ~1) to ( 10 '_28)10.1 Background 10 - 110.2 Advantages of Frequency DOrTlainApproach 10 - 110.3 Limitations of Frequency Response Mettlods 10 - 210.4 ConceJ)tual Approactl to Frequency Response 10 - 3

10.4.1Steady State Response to Sinusoidallnpul : (Frequency Response) 10- 310.5 Apparatus Required for Frequency Response 10 - 5

;~~-.-_. tt -: .0- ".... - '- • a,;, .~ _--; -: -, -_ ,f~_-.. ->

I nl

10.6 Relation between Transfer Function and Frequency Response 10 - 610.7 Transfer Function and Frequency Response of a

Series R-C Circuit.......................... .. .. 10 - 610.7.1 Generating Frequency Response Data 10 - 7

10.8 Frequency Domain Methods 10 - 710.9 Co-relation between Time Domain and Frequency Domain for

Second Order System 10 - 810.9.1 DerivationsofMr andro[ 10-810.9.2Comments on Co-relation between Time Domain and Frequency Domain 10 - 9

10.10 B.W. (Bandwidth) 10 -11Examples with Solutions 10 - 13Review Questions 10 - 27

~~-.~er:..~1.1 -;BodlfPlots -. ,-:~'--: ----~~:~~':~'--.-:~-,-_--.: -:-, :' -:~-~--..' --:'/_~"'~- (11".;.1lto '(11 •:90)I ,. ',., :,.-.,

11.1.1Magnitude Plot. 11-111.1.2The Phase Angle Plot. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 11- 1

11.2 Logarithmic Scales (Semilog Papers) 11 - 211.3 Standard Form of Open Loop T.F. G(jro)H(jro) 11 - 311.4 Bode Plots of Standard Factors of G(jro)H(jro) 11 - 4

11.4.1 Factor 1 : System Gain IK' . • . . . . . . . . . . • • . . • . . . • . . . . . . . . . . . . . . . . . . . . . . . . .. 11- 511.4.2 Factor 2 : Poles or Zeros at the Origin Oro)!p 11- 611.4.3 Factor 3 : Simple Poles or Zeros (First Order Factors) . . . . . . . . . . . . . . . . . . . . . . .. 11- 1311.4.4 Factor 4 : Quadratic Factors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 11-20

11.5 Steps to Sketch the Bode Plot 11 - 2511.6 Frequency Response Specifications 11 - 2511.7 Calculation of G.M. and P.M. from Bode PloL........... 11 - 2711.8 What should be Values of G.M. and P.M. of a Good System? 11 - 2911.9 How to Improve the G.M. and P.M. ? 11 - 30

11.10 Determination of rogcand P.M. for Standard Second OrderSystem , ,.... 11 - 30

11.11 Calculation of Transfer Function from Magnitude Plot 11 - 3511.12 Advantages of Bode Plots 11 - 3511.13 Determination of Kp, Kv and Ka from Bode Plot.......................... 11 - 37

J' _.oc" - .. ..»..,.~~:fJ" ... ';"'=. "j.' >~I·~:IJ.~· ~.......¥It): ..JS~ ~ • of ~.~.(- . ;N:~);'-1··•.... ~ "'=.r~" (··..::U) '0- oQCY"'{';' ." ~!_~ ..'.:' . --= ~~'- •• • • ~ •• I'."~ •• ~ J ."'. ,. :, ••• : ' •• ~ ,: " ~: •• "(:~ '. :... AJJJ "'.~'. •

I nl

11.14 Bode Plot of Systems with Transportation Lag 11 - 39Examples with Solutions 11 - 41Review Questions .... ..., ' , 11 - 88

Chapter -12 -- 'Polar'and Nyquist PlotS' '--12.1Background 12 - 112_2 Polar Plot 12 - 112.3 rogtand ropein Polar Plot 12 - 812.4Determination of G.M. and P.M. from Polar Plot.. 12 - 9

12.4.1 Determiningo}pc Mathematically 12 - 1012.5 Nyquist Plot Analysis :: 12 - 1212.6 Pole-Zero Configuration '" 12 - 1312.7Encirclement 12 - 14

12.7.1 Counting Number of Encirclements , , . , . . . . . . .. 12 - 1412.8Analytic Function and Singularities : 12 - 1612.9Mapping Theorem or Principle of Argument.................................. 12 - 1612.10Nyquist Stability Criterion , 12 - 1912.11Generalized Nyquist Path and its Mapping 12 - 2112.12Steps to Solve Problems by Nyquist Criterion............................. 12 - 2312.13 Behaviour of Right Half Pole 12 - 3512.14Advantages of Nyquist Plot.......................................................... 12 - 4012.15 Log Magnitude - Phase Plots....................................................... 12 - 41

12.15.1 Stability Analysis using Magnitude-Phase Plot. _. . . . . . . . . . . . . . . . . .. 12 - 42Examples with Solutions.......... 12 - 45Review Questions 12 - 85

Chapter ·-13 Closed Loop F~quency Response"'. (13~1) to ( 13~20)13.1Closed Loop Frequency Response 13 - 113.2M Circles [Constant Magnitude Loci] : 13 - 213.3 N Circles [Constant Phase Loci] 13 - 513.4 Use of M Circles 13 - 713.5 Use of N Circles 13 - 913.6 Nicholls Chart 13 - 10

(xiv) »,.. "t

II \ nl

13.7 Frequency Specifications from the Nichol's Chart 13 - 12Review Questions................................................................................. 13 - 20

Chapter: 14 . Compe-nsation ·ofControl Systems~ ~.! .. , ....• ~ ~ ::' (14 -1) to ( 147.48).14.1 Background 14 - 114.2 Types of Compensation 14 - 1

14.2.1 Series Compensation 14 - 214.2.2 Parallel Compensation 14 - 214.2.3 Series-Parallel Compensation 14 - 2

14.3 Compensating Networks 14 - 314.4 Lead Compensator 14 - 3

14.4.1 Maximum Lead Angle ¢m and 0: 14 - 414.4.2 Polar Plot of Lead Compensator 14 - 614.4.3 Bode Plot of Lead Compensator 14 - 714.4.4 Steps to Design Lead Compensator 14 - 814.4.5 Effects of Lead Compensation 14 - 914.4.6 Limitations of Lead Compensation 14 - 10

14.5 Lag Compensator 14 - 1414.5.1 Maximum Lag Angle and P 14 -1514.5.2 Polar Plot of Lag Compensator 14 -1614.5.3 Bode Plot of Lag Compensator 14 - 1714.5.4 Steps to Design lag Compensator 14 -1814.5.5 Effects and Limitations of Lag Compensator 14 - 19

14.6 Lag-Lead Compensator 14 - 2414.6.1 Polar Plot of Lag-Lead Compensator 14 - 2614.6.2 Bode Plot of Lag-Lead Compensator 14 - 2714.6.3 Effects of Lag-Lead Compensator 14 - 27

14.7 Compensation using Root Locus 14 - 3414.8 Designing Lead Compensator using Root Locus 14 - 3514.9 Designing Lag Compensator using Root Locus 14- 4014.10 Designing Lag-Lead Compensator usi~g Root Locus 14 - 44Review Questions : 14 - 46

Chapter..15 .. :St8te·Vari~~le;Analysi$.,-;.0<(. .. ;". :' h w ; h ~ ': •• d".",-~., (15.1) to ( 15.62)15.1 Background 15 - 1

15.1.1 Advantages of State Variable Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15 - 215.2 Concept of State 15 - 2

~-:.,!~.·~· ..t_ ... _·.·~.~::~~·~· ..~~ ...I0•• ~.:.~: .. - : ••• _< .... -'"~_~._~._: .. ;. (xv ..... "":~~ ~-; .c~ .~~: .. .......~~~.~»: '; .: "tJ..~ ....... :.:.":" .. : ..... ". e

I I nl

15.2.1 Important Definitions ...••............................................ " 15 - 415.3 State Model of Linear Systems 15 - 5

15.3.1 State Model of Single Input Single Output System••.....................•... , 15 - 915.4 State Variable Representation using Physical Variables 15 - 9

15.4.1 Advantages ..•...............•....••........ " •..................... 15 -1115.5 State Diagram Representation 15 - 11

15.5.1 State Diagram of Standard State Model ...•.............................. 15 -1215.6 Non Uniqueness of State Model 15 - 1315.7 State Space Representation using Phase Variables 15 - 14

15.7.1 State Model from Differential Equation. . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15 -1415.7.2 State Model from Transfer Function .........•............................ 15 -1815.7.3 Advantages. . • . . • . . • . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . . . . • . . . . . . . . .. 15 - 2815.7.4 limitations 15 - 28

15.8 State Space Representation using Canonical Variables 15 - 2915.8.1 Jordan's Canonical Fonn . . . • . . . • . . . . . . . . . . . . • . . • . . . . . . . . . . . . . . . . . . . . .. 15 - 3215.8.2 Advantages of Canonical Variables ...............•...................... 15 - 3615.8.3 Disadvantages of Canonical Variables .•.................................. 15 - 36

15.9 State Model by Cascade Programming 15 - 3815.10 Derivation of Transfer Function from State Model 15 - 40

15.10.1 Characteristic Equation ...•..•................... . . . . . . . . . . . . . . . . . . .. 15 - 4115.10.2 MIMO System .......•.............................................. 15 -42

15.11 Solution of State Equations 15- 4315.11.1 Homogeneous Equation 15 - 4415.11.2 Nonhomogeneous Equation .........................................•. 15 - 44

15.12 Review of Classical Method of Solution 15 - 4415.12.1 Zero Input Response ....•...•.•.••.•..••.......•..................... 15 - 46

15.13 Solution of Nonhomogeneous Equation 15 - 4615.14 Properties of State Transition Matrix 15- 4715.15 Solution of State Equation by laplace Transform Method 15 - 4815.16 Computation of State Transition Matrix 15 - 49

I .15.17 laplace Transform Method 15 - 50Examples with Solutions 15 - 51Review Questions 15 - 59

l I nl

Chapter· 16 Control Components and Controllers .~ (16 -1) to ( 16 • 64)16.1 Introduction to Stepper Motors 16 - 116.2 Permanent Magnet Stepper Motor 16 - 116.3 Variable Reluctance Stepper Motor 16 - 3

16.3.1 Reduction Gear Stepper Motor . . .. . . . .. . . . . . .. .. . . .. 16 - 416.3.2 Multistack Stepper Motor 16 - 4

16.4 Important Definitions Related to Stepper Motor 16 - 516.5 Stepper Motor Characteristics 16 - 6

16.5.1 Static Characteristics. .. .. . . . . .. .. 16 - 616.5.2 Dynamic Characteristics 16 - 7

16.6 Difference between Stepper Motor and D.C. Servomotor 16 - 716.7Applications of Stepper Motor 16 - 816.8 Synchros 16 - 8

16.8.1 Synchro Transmitter 16 - 816.8.2 Synchro Control Transformer 16 - 916.8.3 Error Detector using Synchros '.' , 16 - 9

16.9 Potentiometer 16 - 1116.9.1 Potentiometer as an Error Detector , 16 -1316.9.2 Types of Potentiometers 16 - 1316.9.3 Characteristics of Precision Potentiometer 16 -1416.9.4 Loading in Potentiometers 16 -15

16.10 Rotating Amplifiers H 16 - 1616.10.1 Single Stage Amplifier 16 -1616.10.2 Two State Rotating Amplifier (Amplidyne) 16 -1616.10.3 Transfer Function 16 -17

16.11Magnetic Amplifier 16 - 1816.12 Servoamplifiers 16 - 21• ••• ••• •• • - 16 - 24

16.14 Properties of Controller 16 - 2516 14 1Error . . . 16 - 2516.14.2 Variable Range 16 - 2616.14.3 Controller Output Range 16 - 2616.14.4 Control Lag 16 - 2716145 Dead Zone 16 - 27

16.15 Classification of Controllers 16 - 27

... }~,. ... ~.

I nl

16.16 Discontinuous Controller Modes 16 - 2816.16.1 Two Position Mode (ON-OFF Controller) r •••• 16 - 2816.16.2 Multiposition Mode 16 - 29

•.•• • • •• • - u •• - .-

16.18 Proportional Control Mode 16 - 3116.18.1 Characteristic of Proportional Mode 16 - 3216.18.2 Offset. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. . 16- 3316.18.3 Applications 16 - 33

16.19 Integral Control Mode 16 - 3416.19.1 Step Response of Integral Mode 16 - 3516.19.2 Characteristics of Integral Mode 16 - 3516.19.3 Applications 16 - 36

16.20 Derivative Control Mode " " ,,16 - 36• I I..··· I •• i .'. •I I 11.1I • .-

16.20.2 Applications 16 - 3816.21 Composite Control Modes " 16 - 3816.22 Proportional + Integral Mode (PI Control Mode) 16 - 38

16.22.1 Characteristics of PI Mode " 16 - 4016.22.2 Applications 16 - 40

16.23 Proportional + Derivative Mode (PO Control Mode) 16 - 4016.23.1 Characteristics of PO Mode.. " ,,'" 16 - 4116.23.2 Applications 16 - 42

16.24 Three Mode Controller (PID Control Mode) 16 - 4216.25 Response of Controllers to Various Inputs 16 - 4716.26 Effect of Composite Controllers on 2nd Order System 16 - 4716.27 PD Type of Controller 16 - 4816.28 PI Type of Controller 16 - 5016.29 PIO Type of Controller 16 - 5116.30 Rate Feedback Controller (Output Derivative Controller) 16 - 51Examples with Solutions 16 - 53Review Questions 16 - 62

.- .~(A -1) to (A.· 8)

l I nl

Basics of Control System

1.1 BackgroundIn recent years, concept of automatic control has achieved a very important position in

advancement of modem science. Automatic control systems have played an important rolein the advancement and improvement of engineering skills.

Practically, every activity in our day to day life is influenced by some sort of controlsystem. Concept of control systems also plays an important role in the working of spacevehicles, satellites, guided missiles etc. Such control systems are now integral part of themodem industrialization, industrial processes and home appliances. Control systems arefound in number of practical applications like computerised control systems, transportationsystems, power systems, temperature limiting systems, robotics etc.

Hence for an engineer it is absolutely necessary to get familiar with the analysis anddesigning methods of such control systems.

This chapter includes the concept of system and control system. Then it gives theclassification of control systems. It includes the discussion of various types of controlsystems supported with number of real time applications.

1.2 DefinitionsTo understand the meaning of the word control system, first we will define the word

system and then we will try to define the word control system.

System: A system is a combination or an arrangement of different physical components whichact together as an entire unit to achieve certa!n objective.

Every physical object is actually a system. A classroom is a good example of physicalsystem. A room along with the combination of benches, blackboard, fans, lightingarrangement etc. can be called a classroom which acts as an elementary system.

Another example of a system is a lamp. A lamp made up of glass, filament is aphysical system. Similarly a kite made up of paper and sticks is an example of a physicalsystem.

Similarly system can be of any type i.e, physical, ecological, biological etc.

(1 - 1)

I I nl

Control System Engineering 1 -10 Basics of Control System

In such system, output or part of the output is fedback to the input for comparisonwith the reference input applied to it.

Closed loop system can be represented as shown in the Fig. 1.12.

Error detector

r(t)

~ Forward path

met) Process to becontrolled

Reference .:!:.input

bet)

c{t) Controlledoutput

Commandinput

Referencetransducer ......-t--.c

Feedback"--------1 element t--_-...JFeedback signal

bet) ¢:=::I Feedback path

Fig. 1.12 Representation of closed loop control systemThe various signal s are,

ret) = Reference input e{t) = Error signal

c(t) = Controlled output m(t)::; Manipulated signal bet) = Feedback signal

It is not possible in all the systems that available signal can be applied as input to thesystem. Depending upon nature of controller and plant it is required to reduce it oramplify it or to change its nature i.e. making it discrete from continuous type of signal etc.This changed input as per requirement is called reference input which is to be generatedby using reference transducer. The main excitation to the system is called its commandinput which is then applied to the reference transducer to generate reference input.

Practically many electronic integrated circuits work on the d.c. voltage range of5 to 10V. The supply available is 230V a.c, Hence the reference input voltage in the rangeof 5 to 10 V d.c. is obtained from the command input 230V a.c. and proper rectifying unit.

The part of output, which is to be decided by feedback element is fed back to thereference input. The signal which is output of feedback element is called 'feedback signal'bet).

It is then compared with the reference input giving error signal e{t) = ret) ± bet)When feedback sign is positive, systems are caned positive feedback systems and if it

is negative systems are called negative feedback systems.

This error signa] is then modified by controller and decides the proportionalmanipulated signal for the process to be controlled.

This manipulation is such that error will approach zero. This signal then actuates theactual system and produces an output. As output is controlled one, hence called controlledoutput c(t),

I I nl


1.5.1AdvantagesThe advantages of closed loop system are,

1) Accuracy of such system is always very high because controller modifies andmanipulates the actuating signal such that error in the system will be zero.

2) Such system senses environmental changes, as well as internal disturbances andaccordingly modifies the error.

3) In such system, there is reduced effect of nonlinearities and distortions.4) Bandwidth of such system i.e. operating frequency zone for' such system is very

high.

1.5.2DisadvantagesThe disadvantages of closed loop system are,

1) Such systems are complicated and time consuming from design point of view andhence costlier.

2) Due to feedback, system tries to correct the error from time to time. Tendency toovercorrect the error may cause oscillations without bound in the system. Hencesystem has to be designed taking into consideration problems of instability due tofeedback. The stability problems are severe and must be taken care of whiledesigning the system.

1.5.3 Real Time Applications of Closed Loop System

1.5.3.1 Human BeingThe best example is human being. If a person wants to reach for a book on the table,

closed loop system can be represented as in the Fig. 1.13.Position of the book is given as the reference. Feedback signal from eyes, compares the

actual position of hands with reference position. Error signal is given to brain. Brainmanipulates this error and gives signal to the hands. This process continues till theposition of the hands get achieved appropriately.

Input eBrain Desired

Hands t---t~ .. positionof the hands

Reference -_--Ipositionof book

Eyes

Fig. 1.13 Human being1.5.3.2 Home Heating System

In this system, the heating system is operated by a valve. The actual temperature issensed by a thermal sensor and compared with the desired temperature. The differencebetween the two, actuates the valve mechanism to change the temperature as per therequirement.

l I nl


Desired

temperatureValve Heaijng House

t------1 system t--- ......- temperature

'--__ ----I Temperaturet-------'sensor

Fig. 1.14 Domestic heating system1.5.3.3Ship Stabilization System

In this system a roll sensor is used as a feedback element. The desired roll position is

Shipstabilisation system

\\,

Shlp-

\\\

Roll position

Fig. 1.15 Ship stabilization systemselected as 6r while actual roll position is Oc which is compared with Or to generatecontrolling signal. This activates fin actuator in proper way to stabilize the ship.

Desired roll9r Fin

actuator

Actual controlled rollt-----,.--+ 9cShip

Rollsensor

Fig. 1.161.5.3.4Manual Speed Control System

A locomotive operator driving a train is a good example of a manual speed controlsystem. The objective is to maintain the speed equal to the speed limits set. The entiresystem is shown in the block diagram in the Fig. 1.17.

II I nl


Speed

limit

Brain Hands

Actual._.....-Speed

Control 1----4Actuator t---1 Vehicleaction mechanism

Sensor

1.5.3.5D.C. Motor Speed Control

Potentiometer

Eyes

Fig. 1.17

Shaft

Tachometer:/couPling

Tachometer

Forspeed

feedback

Field currentconstant

Fig. 1.18 Speed control systemThe D.C. shunt motor is used where field current is kept constant and armature

voltage is changed to obtain the desired speed. The feedback is taken by speed tachometer.This generates voltage proportional to speed which is compared with voltage required tothe desired speed. This difference is used to change the input to controller whichcumulatively changes the speed of the motor as required.

ErrorJ----i Amplifier t----i Armature t----t Load

Actualspeed

.__---------I Tac iorneter1---------1

Fig. 1.19 Speed control system

nl


1.5.3.6 Temperature Control SystemThe aim is to maintain hot water temperature constant. Water is coming with constant

flow rate. Steam is coming from a valve. Pressure thermometer 'P' is used as a feedbackelement which sends a signal for comparison with the set point. This error actuates thevalve which controls the rate of flow of steam, eventually controlling the temperature ofthe water.

I. water intetSet

point --_.. Comparator t-f---.I

-------------Fig. 1.20 Temperature control system

Desiredtemperature Valve Steam

flowActual

1----1 Process t--_e_- temperature

Pressurethermometer

Fig. 1.211.5.3.7 Missile Launching System

This is sophisticated example of military applications of feedback control. The enemyplane is sighted by a radar which continuously tracks the path of the aeroplane. Thelaunch computer calculates the firing angle intenns of launch command, which whenamplified drives the launcher. The launcher angular position is the feedback to the launchcomputer and the missile is triggered when error between the command signal arid missilefiring angle becomes zero. The system is shown in the Fig. 1.22.

1.5.3.8 Voltage StabilizerSupply voltage required for various single phase appliances must be constant and high

- fluctuations are generally not permitted. Voltage stabilizer is a device which acceptsvariable voltage and outputs a fixed voltage.

1I \ nl

------ ------------------------- __---------

Control System Engineering 1-15 Basics of Control System

Flighlpalh.----7<--------------0'\Q)L ..

't...-: -,~L \

~~\ Lhead angle ,,'~, \ I'iii \-0 \ , Actual01 \ ..___~ I poSItion

\ I\ I

Trackmgcontrol

launchcomputer

launchcommand

Feedback

Poweramplifier

Fig. 1.22 Missile launching system,

Principle of such stabilizer is based on controlling number of secondary turns as perrequirement to increase or decrease the output yoltage. The actual output voltage lS sensedby a transformer and potential divider arrangement. The reference voltage is selectedproportional to the desired output level. The actual output is compared with this togenerate error which in tum is inputted to the controller. The controller takes the properdecision to increase or decrease the number of turns so as to adjust the output voltage.The scheme is shown in the Fig. 1.23.

N

Controlled signal

Controllerandrelays

Isolation transformer

Fig. 1.23 Voltage stabilizerI I nl


The other examples of closed loop system are machine tool position control,positioning of radio and optical telescopes, auto pilots for aircrafts, inertial guidancesystem, automatic electric iron, railway reservation status display, sunseeker solar system,water level controllers, temperature control system. So in closed loop feedback controlsystems cause and effect relationship between input and output exists.

1.6 Comparison of Open Loop and Closed Loop Control System

Sr. Open loop Closed LoopNo.

1. Any change in output has no effect on the Changes in output, affects the input which isinput I.e. feedback does not exists. possible by use of feedback.

2. Output measurement is not required for Output measurement is necessary.operation of system.

3. Feedback element is absent. Feedback element is present.

4. Error detector is absent. . Error detector is necessary .

5. It is inaccurate and unreliable. Highlyaccurate and reliable.

6. Highlysensitive to the disturbances. Less sensitive to the disturbances.

7. Highlysensitive to the environmental changes. Less sensitive to the environmental changes.

8. Bandwidth is small. Bandwidth is large.

9. Simple to construct and cheap. Complicated to design and hence costly.

10. Generally are stable in nature. Stability is the major consideration whiledesigning

11. Highlyaffected by nonlinearities. Reduced effect of nonlinearities.

1.7 ServomechanismsDefinition : It is a feedback control system in which the controlled variable or the output is a

mechanical position or its time derivatives such as velocity or acceleration.A simple example of servomechanism is a position control system. Consider a load

which requires a constant position in its application. The position is sensed and convertedto voltage using feedback potentiometer. It is compared with input potentiometer voltageto generate error signal. This is amplified and given to the controller. The controller inturn controls the voltage given to motor, due to which it changes its position.

l I

tControl SystemEngineering 1 -17 ~ Basics of Control System

The scheme is shown in the Fig. 1.24.

Amp.

Gears

Generator Motor

+

potentrorneter

Errorvoltage

Fig. 1.24Position control system

Few other examples of servomechanisms are,

1) Power steering apparatus for an automobile.

2) Machine tool position control.

3) Missile launchers.

4) Roll stabilization of ships.

1.8 Regulating Systems (Regulators)Definition : 1t is a feedback control system iff which for a preset oalue of tire reference input,

the output is kept constant at its desired value.

In such systems reference input remains constant for long periods. Most of the timesthe reference input or the desired output is either constant or slowly varying with time. In. a regulator, the desired value of the controlled outputs is more or less fixed. Similarly thereference input is also fixed and called set point. Thus the regulator maintains a constantoutput for a fixed reference input. The problems due to disturbances arc mainly rectifiedby the regulator. A simple example of such regulator system is servostabilizer. We have_ seen earlier that in voltage stabilizer position of tap on secondary is adjusted by usingrelay controls. But instead of fixed tap, the entire secondary can be smoothly tapped using

II I nl

Control System Engineering 1·18 Basics of Conb'Ol System

a servomotor drive. The servomotor drives the shaft and controls the position of tap onsecondary as per the controller signal. Due to the fluctuations in the main input if the loadvoltage changes, such effects are rectified by the regulator to keep load voltage constant.

The actual scheme is shown in the Fig. 1.25, while its block diagram representation isshown in the Fig. 1.26.

Feedback

Referencevoltage

Isolationtransformer

II

ControlledServo sinnal+-__ ",;;__:I_---.

motor

To loadL--.......-----O stabilised voltage

Referenceinput

Fig. 1.25 Regulating system

Error ----1----1 Controller t---I Servo

motorSecondary ..... _.. Actualtap load voltage

L...- -I Load voltage t-----~sensing

Fig. 1.26

.nl

Control System Engineering, .,! 'J Basics of Control System

Few other examples of regulating system arc,

1) Temperature regulators.

2) Frequency controllers.

3) Speed governors.

1.9 Feedback and Feed Forward SystemIn the control systems considered up till now, it is considered that the disturbance has

affected the output adversely. Such- an output is measured and compared with thereference input to generate an error. This error is fed to the controller which issuccessively operating the system to correct the output.

Thus such systems in which the effect of the disturbance must show up in the errorbefore the controller can take proper corrective action arc called feedback systems.

If the disturbance is measurable, then the signal can be added to the controller inputto modify the actuating signal. Thus, a corrective action is initiated without waiting for theeffect of the disturbance to show up in the output i.e, cumulatively in the error. Thus theundesirable effects of measurable disturbances by approximately compensating for thembefore they affect the output. This is much more advantageous as in normal feedbacksystem the corrective action starts only after the output has been affected. .

Key Point: Such systems in whiclz such corrective action is taken before distutbances affectthe output are called feed forward system,A block diagram with feed forward concept is shown in the Fig. 1.27.

r-----------i Feed forwardblock

Disturbance

Controller Actuator ProcessOutput

Fig. 1.27The two difficulties associated with feed forward system are,

i) In some systems, the disturba~c~ may not be measurable.

ii) The feed forward compensation is an open loop technique and if actuator transferfunction is not known accurately, then such compensation cannot be achieved.

r


Control system: To control means to regulate, to direct or to command. Hence a controlsystem is an arrangement of different physical elements connected in such a manner so as toregulate, direct or command itself or some other system.

For example, if in a classroom, professor is delivering his lecture, the combinationbecomes a control system as; he tries to regulate, direct or command the students in orderto achieve the objective which is to impart good knowledge to the students. Similarly iflamp is switched ON or OFF using a switch, the entire system can be called a controlsystem. The concept of physical system and a control system is shown in the Fig. 1.1 andFig. 1.2.

Lamp 0+----- Supply -----...0Fig. 1.2 Control systemFig. 1.1 Physical system

When a child plays with the kite, he tries to control it with the help of string andentire system can be considered as a control system.

In short, a control system is in the broadest sense, an interconnectionof the physicalcomponents to provide a desired function, involving some kind of controlling action in it.

Plant : Tire portion of a system whic11is to be controlled or regulated is called the plant orthe Process.

Controller: The element of the system itself or external to tire system uhid: controls the plantor the process is called controller.

For each system, there must be an excitation and system accepts ~i:as an input. And.for analyzing the behaviour of system for such input, it is necessary to define the outputof a system.

Input : It is an applied signal or an excitation signal applied to a control system from anexternal energy source in order to produce a specifiedoutput.

Output : It is the particular signal of interest or the actual response obtainedfrom a controlsystem wizen input is applied to it.

Disturbances : Disturbance is a signal which tends to adversely affect the value of theoutput of a system. If such a disturbance is generated within the system itself, it is calledan internal disturbance. The disturbance generated outside the system acting as an extrainput to the system in addition to its normal input. affecting the output adversely is calledan external disturbance.

Control systems may have more than one input or output. From the informationregarding the system, it is possible to wen define all the inputs and outputs of thesvstems,

r


1.9.1 Real Time Application of Feed Forward SystemIn a particular process control industry, it is necessary to maintain the temperature of

a molten metal constant before giving it to the next process. For this a general temperaturecontrol feedback system is used as shown in the Fig. 1.28. -

StorageI tank--

Feedback

"'---:--:i-- Constant/rrr,rrrrnL_..r--:::-='--:- temperature

required

Heatexchanger

Fig. 1.28 Feedback system for temperature controlPractically if the rate of flow of metal from tank to heat exchanger gets disturbed, due

to change in level then the temperature gets affected. But it takes a long time to see itseffect at the output to take the corrective action.

Key Point: In feedback system, the correctiveaction can not be taken unless and untill outputgets disturbed.

Due to the time lag, it is not possible to keep the output temperature constant withinlimits.

In such a case, feed forward system is used. The inlet flow rate is measured with theflowmeter. Immediately when there is a change in rate of flow, it is indicated to thecontroller through flowmeter before it is going to disturb the output. The controller takesthe corrective action in advance by adjusting the steam flow. Thus the output temperaturegets maintained constant within the limits. :

- .Key Point: The feed forward compensation, compensates the effect of disturbance before itactually disturbs the output.

The system is shown in the Fig. 1.29 (a) while its block diagram representation isshown in the Fig. 1.29 (b).

The feed forward minimizes the transient error due to measurable disturbances. Whilefeedback compensates for unmeasurable disturbances and other effects. Thus it is advisable,to incorporate both feedback and feed forward schemes in a system.

- I

nl

Control System Engineering I! Basics of Control System

Controller t-----,

Feedbacksignal

...- ..... ., Temperature'-...--' sensor

Feed forwardsignal

Storagetank I~ .. :I

Inlet"'---r-"""" Constant

temperature

(a) Feed forward system

Flow Disturbance

meter (change Inflow rare)

<. .I Feed forward

Reference -input -®-- Controller f-- Value - Heat

exchanger

Temperaturesensor

Constanttemperature

< )Feedback(b) Block diagram

Fig. 1.29 Practical example of feed forward system

1.10Multivariable Control SystemsThe control system in which there is only one output of the interest is called single

variable system. But in many practical applications more than one variables are mvolved.A control system with multiple inputs and multiple outputs in called a multivariablesystem.

The block diagram representation of a multivariable control system is shown in theFig. l.30. The part of the system which is required to be controlled is called plant. Thecontroller provides proper controlling action depending on the reference inputs. There arcn reference inputs rl, r2, rn-

There are n output variables Cl(t), C2(t),,__,.....cn(t}. The values of these variablesrepresent the performance of the plant. The control signals produced by the controller areapplied to the plant. With the help of feedback elements the closed loop control of theplant is also possible. Due to the feedback, the controller takes into account the actualoutput values to decide the control signals.

_' nl

Control System Engineering Basics of Control System

PlantController or

(Analog or digital) . System to becontrolled

...

-Feedback:elements:

Controlsignals

Fig. 1.30 Block diagram of multivariable control system

In case of multivariable systems, sometimes it is observed that a single inputconsiderably affects more than one outputs. The system is said to be having stronginteractions or coupling. nus coupling is nothing but the disturbances for the separatesystems. The interactions inherently present between inputs and outputs can be cancelledby designing a decoupling controller. Thus the resulting multivariable system isconsidered to have proper number of single input single output systems and the controlleris designed for each system. The another way is to design a controller which will take careof all the inherent interactions present in the multivariable system.

In multivariable linear control system, each input is independently considered. Onlyone input and one output is considered and the total effect on any output because of allthe inputs acting simultaneously is determined by addition of the outputs due to eachinput acting alone. Thus law of superposition is used to analyse multivariable linearcontrol systems.

In many practical control systems, control is achieved by more than one input and thesystem may have many outputs. In chemical processes simultaneous control of pressure,temperature and concentration is required by commanding various inputs. Air crafts andspace crafts are other examples where movement is controlled by various inputs. Powergenerators, atomic reactors and jet engines are some of other examples of multivariablesystems.

Consider the block diagram of multivariable autopilot system shown in the Fig. 1.31.

The system shown in the Pig. 1.31 keeps a track of rocket vehicle in response toreference inputs given to it. The pcsition, velocity and acceleration of the vehicle are fed tothe digital controller using motion st.~rs. The controller takes appropriate decision andsends a controlling signal which will drive the actuator, which will move the engine. Thusthere are three output variables which art. to be observed and controlled and there arecorresponding reference inputs hence the system is multivariable system.

\ I

Control System Engineering Basics of Control System1-23

ReferenceInputs

Controlsignal

IAngularposition

IL DigitalDigital r--- to - Actuator -~ Rocket f-----eoo Vehicle ~~

~ controller analogconverter

Positionfeedback

Analog Velocityto feedbackdigitalconverter

Accelerationfeedback

Fig. 1.31 Multivariable autopilot system

Review QuestionsI, Dtftne the folluwmg terms

(i) System (ii) Control system (iii) Input (iv) Output (v) Disturbance.2. Explain 'lOWtile control systems are classified3. Define linear and nonlinear control systems.4. What is lime variant system? Give suitable example. How if is different than time invariant

system?5. Dtftne open IDOp and closed loop system by giving suitable examples.6. Differentiatebetween open loop and closedloop systems giving suitable examples.7. With referenceto feedbackcontrol system define tirefollowing terms

i) Command input (ii) Referenceinput (iii) Forwardpat'l (iv) FeedbackpatllB. Explain thefollowiJlg terms giving sllitableexample

i) Seroomechanism (ii) Regulator9. Distinguish betweenfeedbackcontrol system and feed forward control system.

10. Differentiate between:1. Linear and Nonlinear systems 2. Continuous and Discrete data systems

11. Explain what is closed loop control system.12. Write a note on multioariablecontrol systems.

: I.

000

nl

(1 - 24)

nl

I ~ Control System Engineering 1-3 Basics of Control System

The input variable is generallyreferred as the Reference Inputand output is generally referred asthe Controlled Output.

Reference """'~ PLANTInput "l 10-----''''- Controlled

v' output

Cause and effect relationshipbetween input and output for a plant can be shown as in the Fig. 1.3.

Fig. 1.3

1.3 Classification of Control SystemsBroadly control systems can be classified as,

1) Natural Control Systems : The biological systems, systems inside human being areof natural type.

Example 1: The perspiration system inside the human being is a good example ofnatural control system. This system activates the secretion glands, secreating sweat andregulates the temperature of human body.

2) Manmade Control Systems : The various systems, we are using in our day to daylife are designed and manufactured by human beings. Such systems like vehicles,switches, various controllers etc. are called "manmade control systems.

Example 2 : An automobile system with gears, accelerator, braking systcm is a goodexample of manmade control system.

3) Combinational Control Systems: Combinational control system is one, havingcombination of natural and manmade together i.e. driver driving a vehicle. In suchsystem, for successful operation of the system, it is necessary that natural systemsof driver alongwith systems in vehicles which are manmade must be active.

But for the engineering analysis, control systems can be classified in many differentways. Some of the classifications are given below.

4) Time Varying and Time - Invariant Systems : Time varying control systems arethose in which parameters of the systems are varying with time. It is notdependent on whether input and output are functions of time or not. Forexample, space vehicle whose mass decreases with time, as it leaves earth. Themass is a parameter of space vehicle system. Similarly in case of a rocket,aerodynamic damping can change with time as the air density changes with thealtitude. As against this if even though the inputs and outputs are functions oftime but the parameters of system are independent of time, which are notvarying with time and are constants, then system is said to be time invariantsystem. Different electrical networks consisting of the elements as resistances,inductances and capacitances are time invariant systems as the values of theelements of such system are constant and not the functions of time. The complexityof the control system design increases considerably if the control system is of thetime varying type. This classification is shown in the Fig. 1.4.

\ I "nl

Control System Engineering 1-4 Basics of Control System.I , ... ~~ I ...... '

Parameters Outputof systemare

constantsand not f{t)functionsof time

f(t}

Parametersof system

are functionsof lime

Input Input Output

(a) Time invariant systemFig. 1.4

(b) Time variant system

5) Linear and Nonlinear Systems: A control system is said to be linear if it satisfiesfollowing properties.

a) The principle of superposition is applicable to the system. This means theresponse to several inputs can be obtained by considering one input at a timeand then algebraically adding the individual results.

Mathematically principle of superposition is expressed by two properties,

i) Additive property which says that for x and y belonging to the domain of thefunction f then we have,

f(x + y) = f(x) + fey)

ii) Homogeneous property which says that for any x belonging the domain of thefunction f and for any scalar constant exwe have,

f(a x} = exf(x}

b) The differential equation describing the system is linear having its coefficientsas constants.

c) Practically the output i.e. response varies linearly with the input i.e. forcingfunction for linear systems. L

Real time example: A resistive network shown in the Fig. 1.5 (a) is a linear system.The Fig. 1.5 (b) shows the linear relationship existing between input and output

R1

vln ? R2

~

vln = Input, I =Output

(8) linear system (b) Response of system

Fig. 1.5 Example of linear system

\ I nl

,'! ,~ I I. IfControl System Engineering 1·5 Basics of Control System

A control system is said to be nonlinear, If.a. It does not satisfy the principle of superposition.

b. The equations describing the system are nonlinear in nature.The function f(x) = x2 is nonlinear because

f(XI +X2) = (Xl +x2f ;e(Xl)2 +(X2)2

and f(a x) = (a x)2 ;eo. x2 where a = constant

The equations of nonlinear system involves such nonlinear functions.

c. The output does not vary linearly for nonlinear systems.

'The various nonlinearities practically present in the system are shown in theFig. 1.6 (a), (b) and (c).

Output

(a) Saturabon

Output

(b) Dead zone

Output

(c) Exponential orsquare law

Fig. 1.6 NonlinearitiesThe saturation means if input increases beyond certain limit, the output remains

constant i.e. it does not remain linear. The flux and current relation i.e. B-H curve showssaturation in practice. In some big valves, though force increases upto certain value, the.valve does not operate. So there is no response for certain time which is called dead zone.

The voltage-current equation of a diode is exponential and nonlinear thus diode circuitis an example of nonlinear system. This is shown in the Fig. 1.7.

R

(8) System

. Diode Exponentialrise

(b) Response

Fig. 1.7 Example of nonlinear system

I I ~I.

Control System Engineering 1·6. .

Basics of Control System ,~ I

It can be seen that as long as Vin increases upto certain value, current remains almost'Zero. This is a dead zone and thereafter voltage-current are exponentially related to eachother which is a nonlinear function.

Key Point: In practice it is difficult to find perfectly linear system. Most of thephysical systems are nonlinear to certain extent.But if the presence of certain nonlinearity is negligible and not affecting the system

response badly, keeping response within its linear limits then the nonlinearity can beneglected and for practical purpose the system can be treated to be linear.

Procedures for finding the solutions of nonlinear system problems are complicated andtime consuming. Because of this difficulty most of the nonlinear systems are treated aslinear systems for the limited range of operation with some assumptions andapproximations. The number of linear methods, then can be applied for analysis of suchlinear systems.

6) Continuous Time and Discrete Time Control Systems : In a continuous timecontrol system all system variables are the functions of a continuous timevariable 't'. The speed control of a d.c. motor using a tachogenerator feedback isan example of continuous data system. At any time 't' they are dependent on time.In discrete time systems one or more system variables are known only at certaindiscrete intervals of time. They are not continuously dependent on the time.Microprocessor or computer based systems use such discrete time signals. Thereasons for using such signals in digital controllers arc,

1) Such signals are less sensitive to noise.

2) Time sharing of one equipment with other channels is possible.

3) Advantageous from point of view of size, speed, memory, flexibility etc.

The systems using such digital controllers or sampled signals are called sampled datasystems.

Continuous time system uses the Signals as shown in the Fig. 1.B (a) which arecontinuous with time while discrete system uses the signals as shown in the Fig. 1.8 (b).

Signal

(a) Continuous signalFig. 1.8

Signal

t

(b) Discrete signal

r


7) Deterministic and Stochastic Control Systems : A control system is said to bedeterministic when its response to input as well as behaviour to externaldisturbances is predictable and repeatable. If such respon.c;e is unpredictable,system is said to be stochastic in nature.

8) Lumped Parameter and Distributed Parameter Control Systems : Control systemthat can be described by ordinary differential equations is called lumpedparameter control system. For example, electrical networks with differentparameters as resistance, inductance, etc. are lumped parameter systems. Controlsystems that can be described by partial differential equations are calleddistributed parameter control systems. For example, transmission line having itsparameters resistance and inductance totally distributed along it. Hence descriptionof transmission line characteristics is always by use of partial differential equations.The lumped parameters are physically separable and can be shown to be locatedat a particular point while representing the system. The distributed parameters cannot be physically separated and hence can not be represented at a particular place.

9) Single Input Single Output (SISO) and Multiple Input Multiple Output (MIMO)Systems : A system having only one input and one output is called single inputsingle output system. For example, a position control system has only one input(desired position) and one output (actual output position). Some systems may havemultiple type of inputs and multiple outputs, these are called multiple inputmultiple output systems.

10)Open Loop and Closed Loop Systems : This is another important classification.The features of both these types are discussed in detail in coming sections.

1.4 Open Loop SystemDefinition: A system in which output is dependent on input but controlling action or

input is totally independent of the output or changes in output of the system, is called anOpen Loop System.

In a broad manner it can be represented as in Fig. 1.9.

Reference Iinputr(t)--~" Controlleru.. Process Controlled1--- ....output c(t}

u = Actuating signal

Fig. 1.9 Open loop control systemReference input [ret)] is applied to the controller which generates the actuating signal

(u) required to control the process which is to be controlled. Process is giving out thenecessary desired controlled output c(t}.

_\ nl


1.4.1 AdvantagesThe advantages of open loop control system are,1) Such systems are simple in construction.2) Very much convenient when output is difficult to measure.3) Such systems are easy from maintenance point of view.4) Generally these are not troubled with the problems of stability.5) Such systems are simple to design and hence economical.

1.4.2 DisadvantagesThe disadvantages of open loop control system are,

1) Such systems are inaccurate and unreliable because accuracy of such systems aretotally dependent on the accurate precalibration of the controller.

2) Such systems give inaccurate results if there are variations in the externalenvironment i.e. such systems cannot sense environmental changes.

3) Similarly they cannot sense internal disturbances in the system, after the controllerstage.

4) To maintain the quality and accuracy, recalibration of the controller is necessaryfrom time to time.

To overcome all the above disadvantages, generally in practice closed loop systems areused.

The good example of an open loop system is an electric switch. This is 'open loopbecause output is light and switch is controller of lamp. Any change in light has no effecton the ON-OFF position of the switch, i.e. its controlling action.

Similarly automatic washing machine. Here output is degree of cleanliness of clothes.But any change in this output will not affect the controlling action or will not decide theoperation time or will not decide the amount of detergent which is to be used. Some otherexamples are traffic signal, automatic toaster system etc.

1.4.3 Real TIme Applications of an Open Loop SystemThe various illustrations of an open loop system are discussed below,

1.4.3.1Sprl~kler used to Water a LawnThe system is adjusted to water a given area by opening the water valve and

observing the resulting pattern. When the pattern is considered satisfactory, the system is"calibrated" and no further valve adjustment is made.

1.4.3.2Stepper Motor Positioning SystemThe actual position in such system is usually not monitored. The motor controller

commands a certain number of steps by the motor to drive the output to a previouslydetermined location. . I

\ I

Control System Engineering 1-9 Basics of Control· System

1.4.3.3 Automatic Toaster System po:l.

In this system, the quality of toast depends upon the time for which the toast isheated. Depending on the time setting, bread is simply heated in this system. The toastquality is to be judged by the user and has no effect on the inputs.

Power input

D1

esired Relay or Heating process

time Controller of breadActualtoast

Controller Process

Fig. 1.10

1.4.3.4Traffic Light ControllerA traffic flow control system used on roads is time dependent. The traffic on the road

becomes mobile or stationary depending on the duration and sequence of lamp glow. Thesequence and duration arc controlled by relays which arc predetermined and notdependent on the rush on the road.

Power

Des1

ired time Relays tocontrol sequence lights Actual

trafficcontrol

Fig. 1.11

1.4.3.5Automatic Door Opening and Closing SystemIn this system, photo sensitive devices are used. When a person interrupts a light,

photo device generates actuating signal which opens the door for specific time. Whenperson passes through the door, light becomes continuous closing the door. The openingand closing of the door is the output which has nothing to do with the inputs, hence anopen loop system. . ..

The room heater, fan regulator, automatic coffee server, electric lift, theatre lampdimmer, automatic dryer are examples of open loop system.

1.5 Closed Loop SystemDeflniHon : A system in which the controlling action or input is somehow dependent on the

output or changes ipt output is called closed loop system.To have dependence of input on the output, such system uses the feedback property.

Feedback : Feedback is a property of the system by which it permits the output to becompared with the reference input to generate the error signal based opt which tile appropriatecontrolling action captbe decided.

I I nl

Basics of Laplace Transform

2.1 BackgroundThe various methods used to solve the engineering problems are based on the

replacement of functions of time by certain frequency dependent variables. This makes thecomputation job very easy. The known example of such method is the use of Fourier seriesto solve certain electrical problems.

The transformation technique relating the time functions to frequency dependentfunctions of a complex variable is called the Laplace transformation technique. Suchtransformation is very useful in solving linear differential equations. The transfer functionof a system, which is heart of the control system analysis is based on the Laplacetransform. This chapter gives the definition of Laplace transform, some commonly usedfunctions and Laplace transform pairs and useful properties of Laplace and inverse Laplacetransform. Some examples are also included demonstrating the superiority of Laplaceapproach over the conventional approach.

2.2 Definition of LaplaceTransformThe Laplace transform is defined as below :

Let f(t) be a real function of a real variable t defined for t > 0, then

F(s) = L{f(t)}= 1 f(t).e-st dto

Where F(s) is called Laplace transform of f(t). And the variable's' which appears inF(s) is frequency dependent complex variable. It is given by,

I S = (Hjm IWhere 0' = Real part of complex variable '5'.

eo = Imaginary part of complex variable '5'.

The time function f(t) is obtained back from the Laplace transform by a process calledInverse Laplace transform and denoted as L-l. Thus,

. (2 - 1)

II \ nl

Control System Engineering 2 -10 Basics of Laplace Transform

".. Example 2.3 : Find the inverse Laplace transform of given F(s).

F(s) = s(s+ 3)(s+ 4)

Solution: The degree of N(s) is less than 0(5). Hence F(s) can be expressed as,

KI K2 KJ

F(s) = s + (s+3) + (8+4)

(s+2)

Where I (8+2) 2 1KJ = s. F(s) s= 0 = s . s(s+ 3)(s+4) = 3x4 ="6

5",,0

I (s+ 2) (-3+2) 1K2 -= (s+3) . F(s) __ = (s+ 3). ( 3)( 4) =

(-3)( -3+4) = 35- 3 S s+ s+s= -3

I (s+ 2) (-4+2) 1K3 = (s+4) . F(s) = (s+ 4). ( 3)( 4) = (-4)(-4+3) = 2"8=-4 S s+ s+

''''' 4

1/6 1/3 1/2F(s) = -+( ) ( )

5 s+3 s+4

Taking inverse Laplace transform,

1 1 1f(t) = 6 + 3" e-Jt -'2 e-4t

2.4.2 Multiple RootsThe given function is of the form,

N(s)F(s) -- (s-a) n O'(s)

Here there is multiple root of the order 'n' existing at s = a. The method of writing thepartial fraction expansion for such multiple roots is,

Ko KI K2 Kn_1 N'(s)F(s) = (s+a}" + (s-a)n-J + (s-a)n- 2 + ... + (s-a) + O'(s)

N'(s)Where "O'(s)represents remaining terms of the expansion of F(s).

Key Point: Thus a separate coeJfident is assumed for each power of repetaiioe Toot, startingfrom its higllfSt power n to 1.

I I nl"

Control System Engineering -" 2 -11 Basics of Laplace Transfonn

Here find the L.C.M. of the entire right hand side and express numerator interms ofKo ,K, ,K2, ••• The numerator N{s) on left hand side is known. Compare the coefficients ofall powers of s in the numerator of both sides which will give simultaneous equationsinterms of Ko ,K. ,K2, ••• Solving these equations we can obtain the coefficientsKo ,K1 ,K2,.··

For ease of solving simultaneous equations, we can find out the coefficient Ko by thesame method as discussed for simple roots.

Ko = (8- a) n • F(s)t = a

Similarly coefficients for simple roots present if any, can also be calculated by themethod discussed earlier, for ease of solving simultaneous equations.

While finding Laplace inverse transform. of expanded F(s) refer to standard transformpairs,

n!= --;;J andsn! n!

and L[e+ 011 tn 1 = (S-a)n+l( )

n+1s+a

,_ Example 2.4 : Obtain the inverse Laplace transform of given F(s).

F(s) =(5-2)5(5+ 1)3

Solution: The given F(s) can be expressed as,Ko KI Kz K)

F(s) :: (s+ 1)3 + (8+ If + (8+ I) +S

Finding L.C.M. of right hand side,

(s- 2) Ko (5)+ K. (8+ 1) s+ Kz (s+ 1)2 s+ K) (s+ 1)3s(s+ 1)3 - s(s+ 1)3

:. (s - 2) = Kos + K. S2 + Kj s + K;!s3 + 2K;!s2 + K2s + K3S3 + 3K3S2 + 3 KJs + K)Comparing coefficients of various powers of s on both sides,

For 53, K2 + KJ = 0 •.. (1)

For S2, K, + 2 K2 + 3 KJ =0 ... (2)

For s', Ko + K, + K2 + 3 KJ = 1 ... (3)

For s0, K,J = - 2 ... (4)

As K3 = -2

from (I), K2 = 2. from (2), Kl = 2..

I I nl

, Control System Engln .. rlng 2 -12 Basics of Laplace Transfonn

. from (3), Ko = 3

pes) 3 2 2 2= + +----(s+ 1)3 (s+ 1)2 (s+ 1) s

L[e-at t n] n!Now = (s+a)n+ I

[ 1 ]e-a' t n·. L-I = n!(S+8)n+ I

1 1 1 1·. F(s) = 3 . ( 1)3 + 2 . < 1)2 + 2 . (]) - 2 . Ss+ s+ s+

3 2. f(t) = L-I [F(s)]= 2!e-1 • t2 + 1! e-I • t + 2 e-t - 2

3·. f(t) = -t2e-I+2te-t +2e-t-22

2.4.3 Complex Conjugate RootsIf there exists a quadratic term in O(s) of F(s) whose roots are complex conjugates then

the P(s}is expressed with a first order polynomial in s in the numerator as,

As+B N'(s)F(s) = +-

(s2+as+J3) O'(s)

)N'(s)

Where (s2 +c:xs+i3 is the quadratic whose roots are complex conjugates while O'(s)

represents remaining terms of the expansion. The A and B are partial fraction coefficients.

The method of finding the coefficients in such a case is same as discussed earlier forthe multiple roots. Once A and B are known then use the following method for calculatinginverse Laplace transform.

Consider A and B are known

Now complete the square in the denominator by calculating last term as,(M.T, )2

L.T. = ( )4 F.T.

Where L.T = Last term

M.T = Middle term

F.T = First term

l I 'rrl

Control System Engineering r;.~ 2 -13 Basics of Laplace Transform

L.T.

Where[(i2

(I) = VJj-T

Now adjust the numerator As + B in such a way that it is of the form,

or L [e-al coscot] = ( )2 2s+a +(0

(s+ a)LIe-a: sin cot] = ( )2 25+ a +(1)

(l)

Key Point: Thus inverse Laplace transform of F(s) having complex conjugate roots of D(s),always contains sine, cosine or damped sine or damped cosine functions.

,_ Example 2.5: Find the inverse Laplace transform oj

S1 +3F(s) = (s1 +2s+5)(s+ 2)

Solution: Thegiven F(s)can be writteI1: ~,.. ~.... I I

As+B CF(s) - +-- s2+28+5 s+2

As 82 + 28+ 5 has complex conjugate ,roots. To find A, Band C find L.C.M. of righthand side,

F(s) = (s+ 2)(As+ B)+C(s 2 +2s+ 5)(s2 + 2s+ 5)(s+ 2)

=(52 + 28+ 5)(s+ 2)

AS2+2 As+Bs+2 B+Cs2 + 2sC+ 5Cr (s 2 + 2s+ 5)(s+ 2)... '-.

Comparing the coefficients of various powers of 5, of the numerators of both sides

· 52+3 = S2 (A+C)+s(2A+ B+2C)+(2B+SC)·.· A+C = 1 ... (1)·.

I: •

· 2A + B + 2C = 0 .•. (2)..·. 2B + SC = 3 ... (3)To solve the equations quickly, the coefficient C corresponding to the simple, real root

can be obtained as,

··. C = F s .(s+2)j = (s2+3)(s+2) = (4+3) = ~() 5=-2 (s2+28+5)(8+2)5=_2 (4-4+5) 5

l I 'rrl

Control System Engineering 2 ·14 ~: ..~ Basics of Laplace Transfonn

Substituting in (1) and (2),

2A = -5

and B = - 22 '7

-"'5 s-2 "5= +--s 2 + 28+ 5 (8+2)F(s)

Consider

Completing square in the denominator,

2 2-5 s-2 -5~-2Fl (s) = ., =?

s-+2s+1+5-1 (s+1)- +(2)2

= -i[(s+s~~:~)' ] split 4 as 2 x 2

... F(s)

7= _ ~{ (s+ 1) +2x 2 }+_l_

5 {s+ 1)2+(2)2 {s+ 1)2+(2)2 (s+ 2)

As [(s+a) ] [ ]L-I ., = e-al cosrot

(s+ a):' +ro2and

L-I[ ~ ] = [e-al sin rot](s+ a)- +(1)2

Hence taking inverse Laplace transform of F(s),

2 • 7f(t) = -5[e-t cos2t +2 e-t sm 2t]+S e-2t

2.5 Use of Laplace Transform in Control SystemThe control systems can be classified as electrical mechanical, hydraulic, thermal and

so on. All systems can be described by integrodifferential equations of various orders.While the output of such systems for any input can be obtained by solving suchintegrodifferential equations. Mathematically, it is very difficult to solve such equations in

l I nl

Control System Engineering 2 -15 Basics of Laplace Transfonn

time domain. The Laplace transform of such integrodifferential equations converts theminto simple algebraic equations. All the complicated computations then can be easilyperformed in s domain as the equations to be handled are algebraic in nature. Suchtransformed equations are known as equations in frequency domain.

Then by eliminating unwanted variable, the required variable in s domain can beobtained. Then by using technique of Laplace inverse, time domain function for therequired variable can be obtained. Hence making the computations easy by converting theintegrodifferential equations into algebraic is the main essence of the Laplace transform.

). Example 2.6: Obtain the expressionfor y(t) which is satisfying the differential equation

d2 y(t) dy(t)dt2 + 6---cit + 8 y(t) = 16e-'. Neglect initial conditions.

Solution: Taking Laplace transform of both sides of the given differential equation andI . . .. 1 di . . La 1 £ f d2y(t) d dy(t)neg ectmg initia con Ilion terms In p ace transform 0 dt2 an -at we get,

s2y(S) + 6sY(s) + BY(s) = 16s+ 1

(S2 + 6s+ 8) Y(s)16

= (5+1)

16yes) = (s+ 1) (s2 + 6s+ 8)

Yes)16

:=: (s+ 1) (s+ 2) (s+ 4)

Yes)at a2 a3

= --+--+--s+ 1 s+2 s+4

5.33 8 2.66Yes) :=: ---+--s+l 8+2 s+4

Taking inverse Laplace transform of Y(s),

y(t) = 5.33e-t - 8e-2t + 2.66e-4t

This is the required solution of differential equation.

2.6 Special Case of Inverse Laplace TransformLet us see now if the order of pes) and Q(s) of the function F(s) is same. In such case

pes) must be divided by Q(s), to obtain the seperation of F(s) as a constant term which isresult of the division and the remainder polynomial P'(s) having order less than Q(s).

pes)So F(s) = Q(s) ... order of pes) and Q(s) same

\1 I nl


P'(s)=: K+ Q(s) .... after dividing pes) by Q(s)

Now Laplace inverse of constant term is impulse function. Refer last pair in theTable 2.2

L-l IK} = KO(t) where 0 (t) = unit impulse. IWhile P' (s) / Q(s) can now be expressed to obtain partial fraction expansion" to get its

inverse very easily.

Note: The same method is to be applied F(s) with order of numerator polynomialpes) is greater than denominator polynomial Q(s) ........ s3+18s2+3s+5)..,... Example 2.7: Find the Laplace inverse of F(s) = :\ 8.2 17 10·s + ~ + s+Solution: Divide pes) by Q(s}.

s3+8s2+17s+10) 53+1852+ 35 + 5 (1 ~K53 + 852+ 175 + 10

...

10S2-145 - 5 --+ P'(s)]

F(s) 1+10s2-14s-5= s3 +8s2 +175+10

10s2 -14s-5 ABC= 1+(s+2) (5+1) (5+5) = 1+s+2+s+1 +5+5

A 10s2-14S-51= (s+1) (s+5) s =-2 = - 21

B 1052-14S-51= (5+2) (s+5) = 4.75. s --I

C 10s2-14S-51= (s+1) (s+2) = 26.25s ",-5

F(s)21 4.75 26.25= 1-5+2+s+1 + 5+5

f(t) = L-1 tF(s»)= 0(t)-21e- 2t + 4.75e-t +26.25e- 5t

L-l (U = oCt) = Unit impulse function.Where

nl

Control System Engineering 2 -17. Basics of Laplace Transfonn

Examples with Solutions

".. Example 2.8: In the circuit given, the values of R and Care 1MQ and 1 J.l1·respectively.Obtaitl the expressionfor the current flowing in the circuit if it is supplied with an input ojstep voltage of 1 V at t = O.

R

f"I'M Iei(;,)Step of

1var=o TFig. 2.1

Solution: Let us write down the equation for the circuit given; asswning supply voltageas vet).

R

f."NVv_ Iei(;,)vet}

! TFig. 2.1(a)

Applying Kirchhoff's voltage law we get,

vet} = i(t) R + ~ I i(t) dt

Where voltage across capacitor is ~ I i(t) dt

While vet} = step of 1 V at t = 0 as shown in the Fig. 2.1(b).

Taking Laplace transform of both sides of above equation,

Yes) = l(s) R + ~ [I~)]

v<ll,}t- _

oFig. 2.1(b)

Neglecting initial conditions hence neglecting the term of initial condition in Laplacetransform of I i(t) dt.

Now Yes) = .!. as vet) is step of 1 Vs ... Refer pair 1 in Table 2.2

1 1s = 1(5) R + sC I(s)

l I nl


1 I(S>[ R + s~]=5

1 1(5) [1+ sRC]- =s sC

I(s) C C= l+sRC =RC[S+ R~]

I(s) = i[s+l~JTaking Laplace inverse, referring Table 2.1

.() 1 -..!..tIt = -c", ...R

Substituting values of R and C,

i(t} = __ 1~ e-1)( 1(6)( 1)( 10-6

rx 10+6

i(t) = 1X 10-6 e-t A

Key Point: It can be observed in this example that solving the equation in time domain forthe current involves calculation of complementary function, then particular integral andarbitrary constants, independently. In Laplaceapproachwe get the answers of all in a singlestep. Hence Laplaceapproachproves to be superior over a normal approach.

). Example 2.9: A series circuit consisting of resistanceR and an inductance of L henry isconnected to a supply of v(t) volts. Find the expression of the current in s domain. Alsocalculate the value oj current at t = 0.5 msec with R = 1 X 103 n, L = 25 mH and supplyis a step voltage of 50 V. Neglect initial condition.

Solution: The circuit is shown in the Fig. 2.2.

Applying KVL we get,

vet) = i(t) R + L diet}dt

R

i(~ Lvet)

Taking Laplace and neglecting initialconditions of current we get, Fig. 2.2

V(s) = l(s)R + LsI(s)

V(s) = 1(5) [R + sL]

I I nl

Control System Engineering 2 -19 Basics of Laplace Transform

l{s) =yes)

R+sL

Now R = lxl03 n L = 25xlO-JH50

and yes) =s

l{s) =50

s[lx 103 + sx 25x 10-3]

50...Taking 25xlO-3 outsidel{s) = 25x 10-3 sx [s+ 4x 104]

I{s) = s [5+4x 104]

.. A = 0.05 and B = - 0.05

2000 A B= -+s [s+4xl04] ... Partial fractions

0.05 0.051(5) = -s-- s+ 4x 104

Taking inverse Laplace transform,

i(t) = 0.05 - 0.05 e-olx 1041 A

.·. i(t) = 0.05 (1_e-4xl04t) A

At t = 0.5 msec = O.5x 10-3 sec

. i(t) = 0.05 (l_e-4Xl04xo.5Xlo-3 ) A

i{t) = 0.05 A.

).. Example 2.10: Calculate Laplaceinverse of F(s) = (2 2~ 5)55 + s+

20Solution: pes) = S(52 + 2s+ 5)

Now quadratic s2+2s+ 5 has complex conjugate roots.

Therefore partial fraction expansion of F{s) is,

20 al a2 s+a 3F(s) = s(s2+2s+5) =5 + 52+18+5

·. 20 = al (s2+2s+5) s+ s(a2 s+a3)

· 20 = s2 (al +a2)+s(2al +a3)+5al·.· Sal = 20 equating constants·.

2al + a 3 = 0 equating coefficients of s

a1 +a2 = 0 equating coefficients of s2

II \ nl

Controt System Engineering 2·2 Basics of LaplaceTran.fonn

L-l [F(s)] = L-l (L(f(t»} = £(t)

The time function f(t) and its Laplace transfotm F(s) is called transform pair.

'. Example 2.1: Find the Laplace transform of rot and 1for t ~ O.

Solution: (i) f(t) = e-at

- -F(s) = L{f(t)} = I f(t) e-!lt dt = J e-at . e-st dt

0 0

- [1 ]-= I e-(s+a)tdt = _--. e-(s+a)to (s+a) 0

( -1 ) 1= o - s+a = s+a

L{e-at}1 and L-l{_l }· = = e-al.. s+a s+a

(ti) f(t) = 1

- - [e-st ]- 1· F(s) = J f(t) e- st dt = J e" st dt = _s _.. so 0 0

L{l}1

and L-l{~}=l· = -.. s

2.3 Properties of Laplace TransformNumber of important properties of the Laplace transform are discussed in this section.

The table of Laplace transform pairs is developed using these properties.

2.3.1 LinearityThe transform of a finite sum of time functions is the sum of the Laplace transforms of

the individual functions.

So if Ft(s), ~(s), , Fo(s) are the Laplace transforms of the time functions ft(t),f2(t), , fn(t) respectively then,

"L{fl(t) + h(t) + + fn(t)} = Ft(s) + F.2(s) + + Fn(s)

I The property can be further extended if the time functions are multiplied by theconstants i.e.

Where aI' a 2,.••••.•••.••, a n are constants.

II \ nl

Control System Engineering 2-20 Basics of Laplace Transform

equating coefficients of s2

al=4 a2=-4 a3=-8

F(s) = ! + -45-8 =! _4 [ s+ 2 ]S S2 + 2s+ 5 s S2 + 2s+ 5

Completeing square in the denominator

F{s) - 4 _ 4 [ s+ 2 ] _ 4 _ 4 [ s+ 2 ]- S S2 + 2s+ 1 + 5 - 1 - s (s+ 1)2 + (2)2

Now the denominator is in the form (s+ a)2 +W. Let us adjust "numerator to get theexpression in standard form.

F(s)=! 4 [(s:~::~)2]··.

4 4 [ s+ 2 1]= S - (5+ 1)2 + (2)2 + (s+ 1)2 + (2)2 w,

Now first term is in the form (s~ a) ro2 which is Laplace of e-at cos rot.(s+ a)- +

While the second term needs adjustment of constant to get in the form ~.,(s+ a)- + or

which is Laplace of e-al sin (l)t.

··. F( ) 4 [ (5+ 1) 1 2 ]s = S - 4 (s+ 1)2 + (2)2 +"2' (s+ 1)2 + (2)2

, Taking inverse Laplace transform we get.

f{t) = 4- 4[e-t cos 2t+ 0.5 e-t sin 2t]··. f(t) = 4- e - t (4cos 2t+ 25in 2t)

000

II \ nl

Control System Engineering 2-3 Basics of Laplace Transfonn

2.3.2 Scaling TheoremIf K is a constant then the Laplace transform of K f(t) is given as K times the Laplace

transform of f(t).

I L {K f(t») = K F(s) I ... K is constant

2.3.3 Real Differentiation (Differentiation In Time Domain)Let F(s) be the Laplace transform of f(t). Then,

L {dd~t}} = s F{s) - f(O-)

Where f (0-) indicates value of f(t) at t = 0- i.e. just before the instant t = o.The theorem can be extended for nth order derivative as,

{d n ttt)}L ~ = sn f(S)-S"-1 fto-)_sn-2 f'(O-)- ... - f(II-1) (0-)

Where f(ll- 1) (0-- ) is the value of (n - l)th derivative of f(t) at t = 0-.

. {d2f{t)}i.e for n = 2, L dE 2 = s2 F(s)-s f(O-)- f'(O-)

{d3ftt>}for n = 3, L dt3 = S 3F(s)-s 2 f (0- )- sf' (0- )- f* (0- ) and so on.

This property is most useful as it transfonns differential time domain equations tosimple algebraic equations, along with the initial conditions, if any.

2~3.4 Real IntegrationIf F(s) is the Laplace transform of f(t) then,

{I} F( )L !ftt)dt = ~

This property can be extended for multiple integrals as,

{I ll In } F(s)L J I ftt)dtJ' dt2, dtn =-n

o 0 0 s

l \ -nl

Control System Engineering 2-4 Basics of Laplace Transform

2.3.5 Differentiation by sIf F(s) is the Laplace transform of f(t) then the differentiation by s in the complex

frequency domain corresponds to the multiplication by t in the time domain.

L{tf(t)} =-dF(s)

ds

Thus, L{t} d d [1] 1 1!= L{ tx 1}= --d rL{!}] =--d - = -:;-= -1-1S S S s- s"

d d [ 1 ] 2 2!L{t2} = L{txt}=-ds [L{tlJ=-ds s2 =$"')= s2t-l

2.3.6 Complex TranslationIf F(s) is the Laplace transform of f(t) then by the complex translation property,

F(s - a) = L{e31 ttt)} and F(s + a) = L {e-3 I ttl)}

F(s =1= a) = F(S(= !I+a

Where F(s) is the Laplace transform of f(t).

2.3.7 Real Translation (Shifting Theorem)This theorem is useful to obtain the Laplace transform of the shifted or delayed

function of time.

If F(s) is the Laplace transform of f(t) then the Laplace transform of the functiondelayed by time Tis,

I~-L-{-f-(t---T)-~-=--e--T-S-F(-~~

2.3.8 Initial Value TheoremThe Laplace transform is very useful to find the initial value of the time function f(t).

Thus if F{s) is the Laplace transform of f(t) then,

f(O+) = Lim f(t) = Lim s F(s)1-.0+ S-O}""

The only restriction is that f(t) must be continuous or at the most, a step discontinuityat t = O.

r

Control System Engineering 2·5 Basics of Laplace Transfonn

2.3.9 Final Value TheoremSimilar to the initial value, the Laplace transform is also useful to find the final value

of the time function f(t). Thus if F(s) is the Laplace transform of f(t) then the final valuetheorem states that,

Lim f(t) = Lim s F(s)t ~"" s- ....o

The only restriction is that the roots of the denominator polynomial of F(s) i.e. poles ofF(s) have negative or zero real parts.

'. Example 2.2: Find the Laplace transform of sin (J)t.

Solution: The sin (Iltcan be expressed using Euler's equation as,

Now

ej(JlI(-e-i(lJtsin rot ::;

2j

L{ sin cot} {e~-e-~}= L 2j

{ eiM } _ {e - jM }= L ')' L 2'.... J J

Ueal}1

Ue-Jt} = 1= ands-a s+a

.,. using Linearity property

...

L{ sin rot} l[ 1 ] 1[ 1] 1[ 1 1]= 2j s-jro - 2j s+jro = 2j s-jro - s+jro

1 [ S+jro-S+jro] 1[ 2jre ]= 2j (s-jro)(s+j<o) = 2j s2_(jro)2

w= 52 +<02

L{ sin rot} eo= s2+ro2

I I nl


Table of Laplace Transforms:

f(t) F(s) Wavefonn1 1

1 t-s0

..Constant K K

Kt-s

0..

K f(t), K is constant K F(s)

t 1 1/tS2

0..

tn n!Sn+ 1

a-al 1

"-----s+a0

eat 1

~-s-a

0..

e-al tn n!(s + a)n+ 1

sin rot ros2+oo2 h /"'\. r:

0 '-/ '"cos rot S r ~ /s2+oo2 ..

0--0-0-

e-al sin rot (I)

(s+a)2+r02 "'--/Y-:n.-_.0 .y__y--'=-"

.....-e-81 cos rot (s+a)

(s+a)2+r02 Y~l\~-7'\-"_'o .')l...Y--V.-

sinh rot rosLr02

cosh rot ss2-r02

Table 2.1 Standard Laplace transfonn pair

r


Function f(t) Laplace WavefonnsTransform F(s)

Unit step = u(t) 11 t-s0 •

A u(t) AAt

-5

0 •Delayed unit step = u(t - T) e-Ts

t- 15

I •0 T

Au(t-T) Ae-Ts

+-1--s..

0 T

Unit ramp = ret) 'I: t u(t) 152

~lope=1o ..

At u(t) A tL::=A52

0

Delayed unit ramp :: r(t - T) = (t - T) u(t - T) e-Ts

52 t ~IOpe=10 T •

A (t - T) u(t - T) Ae-Ts52 ~A

o T

Unit impulse = & (t) 1+IlL_t=O

II \ nl


Delayed unit impulse = l)(t- T) e-Ts LL_I o t=T

Impulse of strength K i.e. K 5 (t)"' LO!:' K .KLt = 0

Table 2.2 Laplace transfonns of standard time functions

2.4 Inverse Laplace TransformAs mentioned earlier, inverse Laplace transform is calculated by partial fraction

method rather than complex integration evaluation. Let F(s} is the Laplace transform of f(t)then the inverse Laplace transform is denoted as,

F(t) = L-I [F(s)]

The F(s), in partial fraction method, is written in the form as,

pes)N (5) ..= D(s)

I· s ,

Where N(s) = Numerator polynomial in s

and D(s) = Denominator polynomial in s

Key Point: The given function F(s) can be expressed in partial fraction form only wizendegree of N(s) is less than D(s).

Hence if degree of N(s) is equal or higher than O(s) then mathematically divide N{s)by D(s) to express F(s) in quotient and remainder form as,

F(s) = Q + FI(s)

N'(s)= Q + O'(s)

Where Q = Quotient obtained by dividing N(s) by D(s)

and FI(s)N'(s)

= lY(s) = Remainder

l I


Now in the remainder, degree of N'(s) is less than D'(5) and hence FI (s) can beexpressed in the partial fraction form. Once F(s) is expanded intenns of partial fractions,inverse Laplace transform can be easily obtained by adjusting the terms and referring tothe table of standard Laplace transform pairs (Table 2.1).

The roots of denominator polynomial D(s) play an important role in expanding thegiven F(s) into partial fractions. There are three types of roots of D(s). The method offinding partial fractions for each type is different. Let us discuss these three cases of rootsof D(5).

2.4.1 Simple and Real RootsThe roots of 0(5) are simple and real. Hence the function F(s) can be expressed as,

N(s) N(s)F(s) = D(s) = (5-a)(s-b)(s-c) ...

Where a, b, c ... are the simple and real roots of D(s). The degree of N(s) should bealways less than O(s). This can be further expressed as,

F(s)N~) K, K2 KJ

= =--+--+--+(s-a)(s-b){s-c) '.. (s-a) (s-b) (s-c) ...

Where Kif K2, KJ, •.. are called partial fraction coefficients. The values of Kif K2,

KJ ••• can be obtained as,

K, = (S-d) F(S(; a

K2 = (s-b) .F(S(=b

K3 = (S-C) . F(S)t= c

In general, Kn = (s-sn) . F(s~ and so on.5='n

Where 511 = n III root of 0(5)

L{ e±al} 1= (s+ a)

is standard Laplace transform pair. Hence once F(s) is expressed in terms of partialfractions, with coefficients Kif K2 ... Kn, the inverse Laplace transform can be easilyobtained.

II \ nl

Transfer Function andImpulse Response

3.1 BackgroundThe mathematical indication of cause and effect relationship existing between input

and output means to decide the transfer function of the given system. It is commonly usedto characterize the input-output relationship of the system.

Transfer function explains mathematical function of the parameters of system,performing' on the applied input in order to produce the required output. Laplacetransform plays an important role in making mathematical analysis easy. Laplace transformand its use in control system analysis is thoroughly discussed in Chapter-2. In this chapterconcept of transfer function, transfer function models and impulse response models of thesystems are discussed.

3.2 Concept of Transfer FunctionIn any system, first the

system parameters aredesigned and their valuesare selected as per therequirement. The input isselected next, to see theperformance of thedesigned system. This isshown in the Fig. 3.1.

Now perfonnance of the system can be expressed in terms of its output as,Output = Effect of system parameters on the selected input

Selection ofsystem parameters

Selected• System parameters 1-----4.-- Output

input

Fig. 3.1

Output = Input x Effect of system parameters,

OutputEffect of system parameters = Input

This effect of system parameters, role of system parameters in the performance ofsystem can be expressed as ratio of output to input. Mathematically such a function

(3 - 1)

II \ nl

Control System Engineering 3-10 Transfer Function and Impulse Response

C(s) = 0.4 _ 0.4s 5+5

Taking Laplace inverse of this equation,

c(l) = 0.4 - 0.4 e-St ... Output

3.5 Some Important Terminologies Related to the T.F.As transfer function is a ratio of Laplace of output to Laplace of input and hence can

be expressed as a ratio of polynomials in 's',

P(5)T.F. = Q(s)

This can be further expressed as,

=aosm + alSm-1 + a2Sm-2 + + am

bosn + b15n-1 + b2Sn-2 + + bn

The numerator and denominator can be factorised to get the factorised form of thetransfer function as,~--------------------------~... T.F. =

K(s - sa)(s - 5b) (5 - Sm)(S- SI) (5- 52) (5- Sn)

where K is called system gain fador. Now if in the transfer function, values of '5' aresubstituted as 51 , 82 , 53 ...••. Sn in the denominator then value of T.F. will become infinity.

3.5.1 Poles of a Transfer Function

Definition : The values of '5',which make the T.F. infinite after 5ubstit"tion in thedenominator of a T.F. are called Ipoks' of that T.F.

So values 51 , 52 , 83 .••.••5n are called poles of the T.F.

These poles are nothing but the roots of the equation obtained by equatingdenominator of a T.P. to zero.

For example, let the transfer function of a system be,

T(s) :: 2(5+ 2)5(5+ 4)

The equation obtained by equating denominator to zero is,

s(s + 4) = 0... s=O and 5=-4

If these values are used in the denominator, the value of transfer function becomesinfinity. Hence poles of this transfer function are 5 = 0 and - 4.

l \ °nl

Control System Engineering 3 ·12 Transfer Function and Impulse Response

2(s+ 1}2 (5+ 2}(S2 + 2s+ 2) = 0

Simple zero at s = -2

Repeated zero at 5 = -1 (twice)

Complex conjugate zeros at s = -1 ± jl.

The zeros are indicated by small circle or zero '0' in the s-plane.

i.e.

3.5.4 Pole-Zero Plot

Definition : Plot obtained by locating all poles and zeros of a T.F. in s-plane iscalled pole-zero plot of a system.

3.5.5 Order of a Transfer Function

Definition : The highest power of 's' present in the characteristic equation i.e. inthe denominator polynomial of a closed loop transferfunction of a system is called'Order' of R system.

3.5.6 D.C. GainThe value of the transfer function obtained for s = 0 i.e. zero frequency is called the

d.c. gain of the system.rl-D-.C-.-G-a-in-=-T-(-S}-I-s"'-0'-',

Key Point: It is not possible to indicate the value of d.c. gain on pole-zero plot as it is aconstant value. It is required to be separately specified,alongwith the pole-zero plot.The d.c, gain decides the resultant constant of the system transfer function.

1For example, consider example 3.1 discussed earlier. The system T.F. is 1+ sRC

So 1 + sRC = 0 is its characteristic equation and system is first order system.Then s = - lIRe is a pole of that system

and T.F. has no zeros. ImaginaryThe corresponding pole-zero plot can be

shown as in the Fig. 3.9.Similarly for example 3.2, the T.F.

calculated is,

1-Fre

Is - Plane I

-~f-----+--'" Real

1 llLeT.F. = s2LC+ sRC+ 1 = R 1

52 + s L + LCFig. 3.9

_' nl

Control System Engineering 3 -13 Transfer Function and Impulse Response

The characteristic equation is,R 1·

52 + s L + LC = 0

So system is 2nd order and the two poles are, -~ ± (2~ r -(krT.F. has no zeros.

Pole2 Pole1

Now if values of R, Land C selectedare such that both poles are real, unequaland negative, the corresponding pole-zeroplot can be shown as in the Fig. 3.10.

Imaginary

Is-Planel

For a system having T.F. as,C(s) (s+ 2)

R(s) = S[S2 +2s+2] [52+75+12]

The characteristic equation is,Fig. 3.10

S (82 + 2s + 2}(s2 + 75 + 12) = 0 Le, s (S2 +2s+2) (s+3)(s+4) = 0

i.e. System is 5th order and there are 5 poles. Poles are 0, -1 ± j , - 3, - 4 while zero islocated at '-2'.

The corresponding pole-zero plot can be drawn as shown in the Fig. 3.11.After getting familiar with introductory remarks about control system, now it is

necessary to see how overall systems are represented and the methods to represent thegiven system, based on the transfer function approach.

Imaginary

¥---II

-*""~f-<J--I----i __---+ Real-1: 0"k--- -j

Is - PlaneI

Fig. 3.11

II \ nl

Control System Engineering 3 ·14 Transfer Function and Impulse Response

3"· Laplace Transfonn of Electrical NetworkIn the use of laplace in electrical systems, it is always easy to redraw the system by

finding Laplace transform of the given network. Electrical network mostly consists of theparameters R, L and C. The various expressions related to these parameters in timedomain and Laplace domain are given in the table below.

Element Time domain Laplace domain Laplace domainexpression for voltage expression for voltage behaviour

Resistance R I(l) x R l(s)R R

Inductance L d i(t) \ sUes) sLL ---at

Capacitance C ~ J i(t) dl 1 1~I(s) sC

Table 3.1From the table it can be seen that after taking Laplace transform of time domain

equations, neglecting the initial conditions, the resistance R behaves as R, the inductance1

behaves as sL, while the capacitance behaves as sc and all time domain functions get

converted to Laplace domain like i(t) to l(s), vet} to V(s) and so on.

By using these transformations, the parameters can be replaced by their Laplacetransform to get Laplace transform of the entire network. Once this is obtained, simplealgebraic equations relating Laplace of various voltages and currents can be directlyobtained. This eliminates the step of writing the integrodifferential equations and takingLaplace of them.

e.g. Consider a network shown below,

r rrto-) -"- __ C_~r

Fig. 3.12

The Laplace of the above network can be obtained by following replacements.

Rl ~ RJ L ~ sL1

R2 -+ R2 C -+ sC

nl

Control System Engineering 3 -15

The other variables then can beintroduced which will be directlyLaplace variables to obtain theLaplace domain equations directly.Such Laplace of network is shown inthe Fig. 3.13.

rrExamples with Solutions

Transfer Function and Impulse Response

)1(5)

Fig. 3.13

1sC

,,.. Example3.6: TI,e transfer function of a system is given by,K(5 + 6)

T(s) = 5 (5+ 2)(5 +5)(52 + 7s+12)Determine i) Poles ii) Zeros iii) Characteristicequation and iv) Pole-zeroplot in s-plane

Solution:i) Poles are the roots of the equation obtained by equating denominator to zero i.e.

roots of,

5 (s+ 2) (5+5) (s2 +7s+ 12) = 0

ie. S(5 + 2) (s + 5) (5 + 3) (s + 4) = 0

So there are 5 poles located at 5 = 0, - 2, -5, -3 and -4

ii) Zeros are the roots of the equationobtained by equating numerator tozero i.e. roots of K (5 + 6) = 0

Imaginary

i.e. s = - 6

There is only one zero.ill) Characteristic equation is one, whose

roots are the poles of the transfer -6 -5 -4 -3 -2function. So it is,

S (5 + 2) (5 + 5) (52 + 75 +12) = 0

Le, S (52 + 75 + 10)(52 + 7s+ 12) = 0

i.e s5 +1454+ 7153 + 154 52 + 1205 = 0

iv) Pole-zero plot

This is shown in the Fig. 3.14

o

Fig. 3.14

Is-Planel

l \ °nl


'. Example 3.7: The unit impulse responseof a system is given by T(t) = e: t (1- cos2tJDetermine its transfer function.

Solution: Laplace transform of the impulse response is the transfer function.

T(s) = L (T(t)}

= L {e-t (1 - cos 2t»)= L {c- t} - L {e- t cos 2tl1 (s+1) 1 (s +1)

= s+ 1 - (5+ 1)2 + (2)2 = S+ 1 - (S2 + 2s+5)

T(s) = __ ---=-4 _(s+ l}(s2 + 2s+5)

'. Example 3.8: Obtain the transferfunction of the lead network shown in the Fig. 3.15.

-R-1

C

Fig. 3.15

Solution: Take Laplace transform of the network, as shown in the Fig. 3.16.. R,

1sC

Fig. 3.16

The parallel combination of Rl and s~ gives impedance of,

1ZlZ2 = RI Xsc _ RlZI+Z2 R 1 -1+sR1C

1 + sCz =

l I nl


Once the parameters are replaced by corresponding Laplace domain conversions, allthe parameters R, 5L and 1/ sC behave as impedances. Hence their series and parallelcombinations can be easily obtained by the algebraic calculations.

~--~: z :r------r----~

Fig. 3.17

Applying KVL to the circuit,

E, (5) = Z 1(5) + I(s) R2 ... (1)

... (2)Eo(s) = l(s) R2

I(s)E.,(s). = R2.. from (2)

Substituting in (1) we get,

= I(s) [Z + R21 = Eo(s) IZ+ R21R2E,(s)

Eo(s)Ej(s) =

R...L.

Substituting Z, T. F. =R2 R2 (1+ s Rl C)

Rl = Rl + R2 (1 + s Rl C)1+ sRI c+ R2

where

s Rt R2 C+ R2 s+a= Rl + s R, R2 C+ R2 = 5+ 13

1 (Rl + R2)a = Rl C' P = Rl R2 C

This circuit is also called lead compensator.

I I nl


II" Example 3.9: Find the transfer function of the given circuit. (M.U.: Dec.-2003)

t Cl R2 tr

eoC2 !T0 0

Fig. 3.18

Solution: Draw the Laplace domain network of the given system.

Rl • ",.. ParallelSeries

t t1

E.(s) sCl t;,(s); 1!. "r sC2 I0

Fig. 3.19

Eo(s)

t tEt(s)

~

EJs)

I IFig. 3.20

1RI x-ZI = Rt II _1_ = sCII

SCI R1 +

SCI

= 1+sR1C1

Z2 = R') +_1_ = 1+sR2C2- SC2 SC2

E, (s) - Eo(s)I(s) = ... (1)

Eo(s) = 1(5) Z2 ... (2)

. E ( ) - Ej(s)-Eo(s) X Z2•• 0 S - Zt

r. Eo(s) [1+ ~:] = ~: Ej(s)

Eo(s) Z2r. Ei(s) = Zi +Z2

1+SR2C2

...

l I nl

Control Systems Englneerina. : I 3 -19 Transfer Function and Impulse Response

n.... Example 3.10 : Find the transfer function ~~:~ of a syst~ hIlving differential equationgiven below.2d1 c(t) + 2 d c(tJ + c(tJ = r(t) + 2r(t - 1)

d t2 dt

Solution: Taking Laplace transform of the given equation and assuming all initialconditions zero we get,

2 s2C(s) + 2s C(s) + C(s) = R(s) + 2 e- S R(s)

Laplace transform of delayed function is,

L {f(t - T)} = e-sT F(s)... L {ret - I)} = e-s.1R(s)

(ReferTable 2.2)

Combining terms of C(s) and R(s)we get,

(252 + 2s + 1) C(s) = R(s) (1+ 2 e-S)

C(s) 1+ 2e-s=R(s) 2 S2 + 25+1

'.. Example 3.11 : Find V,,(s) / Vj(s). Assume gain of buffer amplifier as 1.

1 MQ

0.5J.lF

Bufferamplifier

Fig. 3.21

. Solution: Taking Laplace transform of the network,

1-6

S)( 1)(10

61)( 10 {2

1-6

s)(0.5x10

61 x 10 n Vo(s)Buffer

amplifier

Fig. 3.22

_' nl


explaining the effect of system parameters on input to produce output is called transferfunction. Due to the own characteristics of the system parameters, the input getstransferred into output, once applied to the system. This is the concept of transfer function.The exact definition of the transfer function is given in the next section.

3.3 Transfer Function

3.3.1 DefinitionMathematically it is defined as the ratio of Laplacetransform of output (response) of the system

to the Laplace transform of input (excitation or driving function), under the assumption that allinitial conditions are zero.

Symbolically system can be represented as shown in the Fig. 3.2{a). While the transferfunction of system can be shown as in the Fig 3.2{b).

ret) --+I System 1--_ c(t) R(s)--+--I T(s) "'_-C(s)

Fig. 3.2(8) Fig. 3.2(b)

Transfer function of this system is ~~:~, where C(s) is Laplace of c(t) and R(s) is

Laplace of r(t).

H T(s) is the transfer fw.lction of the system then,

Laplace transform of output C(s)T(s} = Laplace transform of input = R(s)

'.. Example 3.1 : For a system shoum in the Fig. 3.3. calculate its transfer function whereV,l(t) is output and Vi(t) is input to the system.

R

fVWv Ie f)viet) vo(t)

1 i(t) T 1Fig. 3.3

Solution: We can write for this system, equations by applying KVL as,

Vi (t) = R· i(t) + ~J i(t) dt ... (1)

and vo(t) = ~I i{t)dt ... (2)

l \ °nl

Control System Engineering 3·20 Transfer Function and Impulse Response

Let us divide the network into two parts,

Part 1) :

1-6

S x 1 x 10

61 x 10 n

Fig. 3.23

Applying KVL,

Vi (S) = 1-s-x-l-x-l-0--"""6I(s) + 1 X 106 I(s) ... (l)

·..VI (S) =

I{s) =

1X io- l(s)Vt(s)

1 X 106

... (2)

__ [10s6 + 106]I(S) __[106 +ss 106] [V110(6S)]Substituting in (1) Vi (s)

·.. VI (s) s=

VieS) 5+ 1

Part 2) :

= s

61x10 U

.·.

Fig. 3.24

[2XI06 ]V2(S} = I(s) s + 1 X 106

Vo{s) = I(s) 1 X 106

1(5) Vo{s)= 106

... (1)

... (2)

r

Control System Engineering· - 3 -21 Transfer Function and Impulse Response

Substituting in (1)

= Vo(SHL2+SlJI06106 s

VO(s) sV2{s) = s+ 2

Now gain of buffer amplifier is 1 (unity)

• VI (s) = V2(s)

(s ) (s+ 2)

. s +1 Vi(s) = s Vo(S)

Vo{S) 52

Vi(s) = (s+ l)(s+ 2) ... T.F.

".. Example 3.12: Determine the transfer junction if the d.c. gain is equal to 10 for tilesystem whose pole-zero plot is shown below.

Imaginary

x---------IIII

o-o---'--~,__-+_-_Real-3 -2 -1

II

~--------- -j

Fig. 3.25

Solution: From pole-zero plot given, the transfer function has 3 poles at 5 = -1, -2+jand -2-;. And it has one zero at s = -3.

K~+~ K~+~T(s} = (s+1)(s+2+j)(s+2-j) = (s+1)[{s+2)2-G)2J

K(s+ 3)= (s + 1) [52 + 4s + 5)

Now d.c. gain is value of T(5} at s = 0 which is given as 10.

d.c. gain = T(s) lat 5=0

I I nl


10Kx3= 1xS

K = 5~ = 16.667

T(s)16.667 (s + 3)

=(5+ 1) (52 + 45+ 5)

This is the required transfer function.

).. Example 3.13: If the system transfer function isy(s) s+ 4x(s) :: S 2 + 25+5

Obtain the differential equation representing the system.

Solution:yes)Xes)

s+4=52 + 2s+ 5

(S2 + 2s +5) Yes) = {s + 4} Xes)

s2 yes) + 2s yes) +5Y(5) = 5X(s) + 4X(s)

Replacing variable 5 by :t and yes) by yet) and Xes} by x(t} we get,

d2 d d-., y(t) + 2 -d yet) +5y(t) = -d x(t) + 4x(t)dt- t t

dx- +4xdt ... Differential equation

J" Example 3.14: For the two port network shown in the Fig. 3.26, obtain the transferfunctions

.) V2(S) d ..) VI (s)1 VI (5) an II 11 (5)

I, 1H 1H

v, 1

J..4 2F

Fig. 3.26

\ I nl

Control System Engln~nng 3-23 Transfer Function and Impulse Response

Solution: Draw the Laplace transformof the given network as shown by replacing L by1

sL,C by sc and all time variablesby the Laplacevariables,

.".. "11(5)S s(s+2)

s+S+T = s+~ J_1_ + _1_

<- 25 2(5+2)V2{1(s} 1~ 5+1~:ooAMy = s(5+2) j1

1

[ SX1]Parallel s+1combination

+ Q-----t_-_...J

1

Fig. 3.26(a)

11(5}+0----...--'

1

1

Fig. 3.26(b)

Parallel combination___..__

+

v

[.lx.1.]4 2s

Parallel 1 1combination 4" +2s

1

+

s}

-------Parallel combination

Fig. 3.26(c)

r

Control System Engineering 3·24 Transfer Function and Impulse R.sponse•

+11(S)

1x s(s+2}s(s+2)(s+1)

~1 5(5+2) S2+35+1

11(s)+--(5+1)

(s)

~

5+1 V2(

(s2+35+1)

+

s)

Fig. 3.26(d)

11(5) = _::---:=:--V_I_(S_) = VI(S) = V1(5)s(s+ 2) s+ 1 [s2 + 3s+ 1]--:::---:---+ -:---_

s2 + 3s+ 1 s2 + 3s+ 1 52 + 35+ 1

VIeS)II(s) = 1

I ( ) (s+ 1) V ( ) (5+ 1)= I 5 X = I 5X----='---(s2 + 35+ 1) (s2 + 3s+ 1)

V2(s) s+ 1=

VI(S) 82 + 35+ 1

... V2(S)

'. Example 3.15: What will be the transferfunction oj a system whose unit step responseis a unit impulse function ? (M.U. : Dec.-2003)

Solution: The output c(t) = B(t) = unit impulse function

and the input ret) = 1 = unit step function

C(s) L{B(t)} 1T.F. = R(s) = L{l} = I = S

s

1) Define the transfer function of a system.2) ErplRin the signifiCll1lCeof II trllnsferfunction stating its advanltlges and features.3) What are the limitations of tTt!nsfer function approach?4) How transfer function is related to unit impulse response of a system?S) Define and erplRin the following terms related to the transfer function of a system.

(i) Poles (ii) Zeros (iii) Characteristic equation (iv) Pole-zero plot (v) Order.6) The unit impulse response of a system is e-7t• Find its transfer function. 1Ans.:--7s+

Review Questions

\ I nl

l. I I ;., r (' [ . I; . ;-1'" I

Control System Engineering 3 • 25 Transfer Function and Impulse Response

7) A certain system is described by a differentilll equation

d~t~{t) + 3d~~t)+ l1yW = 5x(t)

where y(t) is the output and the x(t) is the input. Obtain the transfer function of tile system.yes) 5

Ans.: Xes} = S2 + 3s+ 11

8) A certain system has its transfer function asCis) 25 + 1Rfs) = s2 + s + 1

Obtain its differential equation. A . d2c(t) ddt) (l)= 2~r(t) (t)ns.. dt2 + dt + C dt + r

9) if a system equation is given asddt)

3 lit + 2c{t) = r(t - T)

Where c(t) is output and rit) is input shifted by T seconds. Obtain its transfer function.e sT

Ans.: 3s+ 2

10) A system when excited by unit step type of input gives followirlg responsec(t) = 1- 2e-1 + 4e-31Obtain its transfer function C(s)!R(s) 3s2 + 2s+ 3

An5.: (s+ 1) (s+ 3)

11) Derive the transfer function of the system shown in Fig. 3.27. The amplifier gain is K.

v

1rot) C_1_=~=__ _;

AmplifierofgainK

Fig. 3.21

Eo(s) KAns.: Ei(5) = (1+ sRtC1) (1 + sR2C2)

12) The transfer function of a system is given byT(s) = 10(s + 8)

S(S + 4)(s2 + 6s+ 25}

Obtain its (i) Poles (ii) Zeros (iii) Order.Sketch its pole-zero plot. Ans. : Poles at 0, -41 -3 ± j~ Zero at -8, Order 4

r

Control System Engineering 3 -26 Transfer Functloi1 ~d Impulse Response

13) Obtain the transfer /unction of the network shown in the Fig. 3.28.O',

Fig. 3.28I:o(s) RIR2CIC2s1+ R2(CI + C2) s+ t

Ans.: Ej(s) = Rl R2CICIsl + (R1cI + RZCI + R2CI)S+ 1

14) Obtain the tra~er function of the network shown in the Fig. 3.29.

--..-

R1

1Lj,~cJ 1I

el(t) <- eo(t)~R2-?J

i{t)

i(t)

Fig. 3.29

Eo(s)AM.: Ej(s) = K(l + TIs) h

(1+ TIS) were

K= RzRI + R2

Rl RlCI + RIR2ClRt + R2

000

\ I °nl


We are interested in ~~~:~where V,,(S) is Laplace of vo{t) and Vi (s) is Laplace of Vi (I)

and initial conditions are to be neglected.

So taking Laplace of above two equations and assuming initial conditions zero we canwrite,

Vi(s) 1 .., (3)= R I(s)+ sC I(s)

Vo(s) 1 .,. (4)= sC I(s)

. I(s) = -cv, (s)..Substituting in equation (3),

Vi(s) = sCVo(s)[R+ s~]

Vi (5) = sCR Vo(s) +Vo(s) = Vo(s)[l + sCR}

Vo(s) 1VieS) - 1 + sCR

We can represent above system as in the Fig. 3.4, which is called transfer functionmodel of the system.

V,es} ~ Vo(s)~

Fig. 3.4

)I• ., Example 3.2: Find out the T.F. of the given network.

L__~___..T_C.J_ ~t) _ 1Fig. 3.5

Solution: Applying we get the equations as,

( ) 'R L di 1I ide, t = 1 + dt +cit

Input = Ci (t) ; Output = Co (t)

... (1)

\ I nl


I F(s)Laplace transform of F(t)dt = -5-'

df(t)and Laplace transform of (ft=sF(s)

Take Laplace transform,

..... neglecting initial conditions

... neglecting initial conditions

Now

Ej(s) r 11= I(s)l R+ sL+ scJI(s) 1Ej(s) = r 11

LR+ sL+ scJ

Co(t) = ~J idt1

Eo(s) = sC res)

1(5) = sCEo(s)

... (2)

...

... (3)

... (4)

Substituting value of 1(5) in equation (2),

1= r 11

LR+ sL+ scJ

1 1

So we can represent the system as in the Fig. 3.6.

Fig. 3.6 Transfer function model

3.3.2 Advantages and Features of Transfer FunctionThe various features of the transfer function are,i) It gives mathematical models of all system components and hence of the overall

system. Individual analysis of various components is also possible by the transferfunction approach.

il) As it uses a Laplace approach, it converts time domain equations to simplealgebraic equations.

I I nl


iii) It suggests operational method of expressing equations which relate output toinput.

iv) The transfer function is expressed only as a function of the complex variable's'. It is not a function of the real variable, time or any other variable that isused as the independent variable.

v) It is the property and characteristics of the system itself. Its value is dependenton the parameters of the system and independent of the values of inputs. Inthe example 3.1, if the output i.e, focus of interest is selected as voltage acrossresistance R rather than the voltage across capacitor C, the transfer functionwill be different. So transfer function is to be obtained for a pair of input andoutput and then it remains constant for any selection of input as long asoutput variable is same. It helps in calculating the output for any type of inputapplied to the system.

vi) Once transfer function is known, output response for any type of referenceinput can be calculated.

vii) It helps in determining the important information about the system i.e. poles,zeros, characteristic equation etc.

viii) It helps in the stability analysis of the system.•ix) The system differential equation can be easily obtained by replacing variable IS'

hy d/dt.

x) Finding inverse, the required variable can be easily expressed in the timedomain. This is much more easy than to analyse the entire system in the timedomain.

3.3.3 DisadvantagesThe few limitations of the transfer function approach are,

i) Only applicable to linear time invariant systems.

ii) It does not provide any information concerning the physical structure of thesystem. From transfer function, physical nature of the system whether it iselectrical, mechanical, thermal or hydraulic, cannot be judged.

ill) Effects arising due to initial conditions are totally neglected. Hence initialconditions loose their importance.

3.3.4 Procedure to Detennine the Transfer Function of a Control SystemThe procedure used inEx. 3.1 and Ex. 3.2 can be generalised as below :

1) Write -down the time domain equations for the system by introducing differentvariables in the system.

II \ nl


2) Take the Laplace transform of the system equations assuming all initial conditionsto be zero.

3) Identify system input and output variables.

4) Eliminating introduced variables, get the resultant equation in terms of input andoutput variables.

5) Take the ratio of Laplace transform of output variable to Laplace transform ofinput variable to get the transfer function model of the system.

). Example 3.3: Find out the T.F. of the given network.

L Cr roooooo' I1..r R r'

.. Fig. 3.7Solution: Applying KVL we can write,

ej(t) diet) 1 Ji(t) dt + i(t)R ... (1):: L(lt+ C

while eo(t) = i(t)R ... (2)

where ej (t) = input and

eo(t) = output

Taking Laplace of equations (1) and (2), neglecting the initial conditions.

1 1(5)Ej(s) = sUes) + Cs+ RI(s) ... (3)

... (4)

Substituting

Eo(s) = I{s)R

Ej(s) [ 1 1:: I{s) sL+ sC + RJ from (3)

l(s)Eo(s)

= "}{" from (4) in the above equation we get,

Ej(s)Eo(s) r 1 1

= --LsL+-+RJR sC

l \ °nl


This is the required transfer function.

Key Point: The network in Ex. 3.2 and Ex. 3.3 is same but as focus of interest i.e. output ischanged, the transfer junction is changed. For a fixed output, transfer junction is constant andindependent of any type of input applied to the system. If the output variable is changed, theT.F. also changes accordingly.

3.4 Impulse Response and Transfer FunctionThe impulse function is defined as,

f(t) = A for t = 0\ = 0 for t je 0

A unit impulse \ f~dion l)(t) can beconsidered a narrow pulse (of any shape)occuring at zero time such that area under thepulse is unity and the time for which the pulseoccurs tends to zero. In the limit t -+ 0, thepulse reduces to a unit impulse l)(t). Consider anarrow rectangular pulse of width A and heightYA units, so that the area under the pulse = 1, asshown in the Fig. 3.8(a).

... Ej(s) = Eo(s) x [S2LC+ 1+ SCR]R sC

Eo(s} sRCEj(s} = s2 LC+sRC+l

Now if we go on reducing width A andmaintain the area as unity then the height YAwill go on increasing. Ultimately when A -+ 0,YA -+ 00 and it results the pulse of infinitemagnitude. It may then be called an impulse ofmagnitude unity and it is denoted by l)(t). It isnot possible to draw an impulse function onpaper, hence it is represented by a verticalarrow at t = 0 as shown in the Fig. 3.8{b).

So mathematically unit impulse is definedas,

l)(t) = I, t = 0

= 0, t ~ 0

~

....1/A

...1----A ----.111'1 t-+

Fig. 3.8 (a)

1

oSymbol of ~t)

Fig. 3.8 (b)

t __.,

nl


oSymbol of K~t)

t-.

If in the above example the area underthe narrow pulse is maintained at K unitswhile the period of pulse is reduced, it iscalled to be an impulse of magnitude 'K' andis denoted by K B(t), as shown in theFig.3.8(c).

An important property of impulsefunction is that. if it is multiplied by anyfunction and integrated then the result is thevalue of the function at t = O.

K

Fig. 3.8 (c)

+- t 0+Thus J f(t) s(t) dt = I f(t) s(t) dt = J f(t) s(t) dt = f(t) It.o·

0- 0-

This is called Isampling' property of impulse. Hence if we define Laplace transform ofB(t) as,

-L[8(t)1 = I ~(t) e-S1dt ... by definition

o

= e-S1It..o

= e-O = 1

... by sampling property.

U~(t» = 1

Thus Laplace transform of impulse function S(t) = 1

Now T(s)C(s)

= R(s)

:: R(s)· T(s)... C(s)

So response C(s) can be determined for any input once T(s) is determined.

Key Point: The equation IC(s) I::: R(s) • T(s)] is applicable only in Laplace domain and canrwtbe used in time domain. The equation [c(t) = ret) . t(t)] is not et all VQUd in time domain.Now consider that input be unit impulse i.e.

ret) :: l)(t) = unit impulse input

R(s) = L {8(t»)= 1...Substituting in above,

C(s) = 1· T(s) = T(s)

Now c(t) = L-l lees)} = L-l (T(s)} = T(t)

r


Thus we can say that for impulse input, impulse response C(s) equals the transferfunction T(s). So impulse response is c(t) = T(t) as C(s) = T(s) hence we can conclude that,

Key Point: Laplace transform of impulse response of a linear time invariant system is itstransferfunction with all the initial conditions assumed to be zero.

'. Example 3.4 : The unit impulse response of a certain system is foundto be e-4t. Determine its transferfunction.

Solution: Laplace of unit impulse response is the transfer function.

= T(s)

T(s) =

). Example 3.5: The Laplace inverse of the transfer function in time domain of a certainsystem is e-51 while its input is rit) = 2. Determine its output cit).

Solution: Let T(s) be the transfer function.

But

L-l [T(s)] = T(t} = e-5t

ret} = 2

c(t) * ret) x T(t),

... given

It is mentioned earlier that :~:~= T(t) is not at all valid in time domain, so

c(t) ~ 2e-5t

Hence the equation valid according to the definition of transfer function must be used,C(s)R(s)

T(s) = L{T(t)}= L{e-St } = s! 5

T(s) =

so

R(s) 2 ... as r(l) "= 2= -s1 C(s).

s+5 =..(~)

C(s) 2 al a2= =-+ --.. s(s+ 5) s s+5

II \ nl

MathelTIatical Modeling of Systems

4.1 What is Mathematical Model ?To study and examine a control system, it is necessary to have some type of

equivalent representation of the syste!!"Such a representationcan be obtained from themathematicalequations,governing the behaviourof the system.Most of such mathematicalequations are differential equations whether the system may be electrical, mechanical,thermal,hydraulic etc.

Key Point: The set of mathematical equations, describing the dynamic characteristics of asystem is calledmathematical model of the system.Obtaining the mathematicalmodel is the first step in analysing a given system. In the

mathematical model, the various operations in the system are represented by themathematicalequations.

Most of the control systems containmechanicalor electricalor both types of elementsand components.To analyse such systems, it is necessary to convert such systems intomathematicalmodels based on transfer function approach. From mathematical angle ofview,models of mechanicaland electricalcomponentsare exactlyanalogous to each other.Not only this, but we-can show that for given mechanical system there is always ananalogous electricalnetwork exists and vice versa. The mathematicalequations describingboth the systemsare exactlysame in nature.

As we are well familiar with the behaviour of electrical networks and methods ofwriting equations for it, it will be better if we can draw equivalent electricalnetworks forgiven mechanical systems. This will help us in writing system equations in Simplifiedmanner and with more detailed understanding.

This chapter explains the concept of analogous networks, method of writingdifferential equations for various physical systems and derives the transfer functions ofvarious commonlyused control systems.

(4 - 1)

II \ nl

• -co Control System Engineering 4-10 Mathematical Modeling of Systems

Torque equation of side 1 is,

T-J d20t(t) B d01(t) 'E()- t dt2 + 1 dt + t t ... (I)


... (2)

Now TI Nt _ 02T2 = N2 -~

T2 N2 TJ= Nt

Substituting in equation (2)

·..

·.. 11 = Nl J., d202 + Nl B? d02 + Nt TLN2 - dt2 N2 - dt Nz

Substituting value of Tt in equation (1)

... (3)

Substituting O2 =

d201:. T = It dt2 + HI

:. T

JIe = Equivalent inertia referred to primary side

Jle = 11+(~~r12Ble = Equivalent friction referred to primary side

.·.

and

Ble = BI + (~~rB2

T d201 dOl (Nt)= lie (It2 + Bit,Cit + N2 n

l I nl

Control System Engineering, 4 -11 Mathematical Modeling of Systems

Similarly the equation can be written referred to load side also, where applied torque

gets transferred to load as (~: T).

(~: )T

( N?)2where 12.(.' = 12 + N; JI and

4.6.2. Belt or Chain DrivesBelt and chain drives perform same function as that of gear train. Assuming that there

is no slippage between belt and pulleys we can write,

T@-o ~---~8)r,

Fig. 4.16

for such drive

4.6.3 LeversThe lever system is shown in the

Fig. 4.17. This transmits translationalmotion and forces, similar to geartrains.

By law of moment,

I fl 11 = f2 hBy work done fl XI = £2 X2

Fig. 4.17

Hence

4.7 Electrical SystemsSimilar to the mechanical systems, very commonly used systems are of electrical type.

The behaviour of such systems is governed by Ohm's law. The dominant elements of anelectrical system are,

l I nl

Control System Engineering ~tt 4 -12 Mathematical Modeling of Systems

i) Resistor ii) Inductor iii) Capacitor

i) Resistor : Consider a resistance carryingcurrent 'I' as shown, then the voltage dropacross it can be written as,

Suppose it carries a current (11 - 12) thenfor the polarity of the voltage drop shown itsequation is, Ir--V-=-( -11---1-2)-R--'1

ii} Inductor : Consider an inductorcarrying current 'I' as shown, then the voltagedrop across it can be written as,

V = L dldt

I = ~J V dt

or

Io----~.-----~----------~o+ R -+------ V -----------+

_1112-o~----------~---------~o+ R -

------ V -----------+

Fig. 4.18

L'ODMooo' • oo

._------ V --"'"------- ...

Fig. 4.19

. If it carries a current (11 - 12) then for the polarity shown its voltage equation is ,

V = L d(11d~12)

or (11 -}z) = ~ I V dt

iii) Capacitor : Consider a capacitorcarrying current 'I' as shown, then the voltagedrop across it can be written as,

or

V = ~ Iidt

I = C dVdt

If it carries a current (11 -Iz) then for thepolarity shown its voltage equation is,

or

V = ~f (11 -12) dt

(II -12) = C dVdt

o

_11 12_

+ -'oOOObOo' ol

-------- V -------+Fig. 4.20

C 1~o----------~II~------~·----~o+---------- V -------+

Fig. 4.21

_11

12--co-----+~I~I----------~o------- V ------+

Fig. 4.22

l \ °nl

Control System Engineering 4-13 Mathematical Modeling of Systems

4.8 Analogous SystemsIn between electrical and mechanical systems there exists a fixed analogy and their

exists a similarity between - their equilibrium equations. Due to this, it is possible to drawan electrical system which will behave exactly similar to the given mechanical system, thisis called electrical analogous of given mechanical system and vice versa. It is alwaysadvantageous to obtain electrical analogous of the given mechanical system as we are wellfamiliar with the methods of analysing electrical network than mechanical systems.

There are two methods of obtaining electrical analogous networks, namely

1) Force - Voltage Analogy i.e. Direct Analogy.

2) Force - Current Analogy i.e. Inverse Analogy.

4.8.1 Mechanical System

~x(t)Consider simple mechanical system as

shown in the Fig. 4.23.

Due to the applied force, mass M willdisplace by an amount x(t) in the directionof the force f(t) as shown in the Fig. 4.23.

According to Newton's law of motion,applied force will cause displacement x(t)in spring, acceleration to mass M againstfrictional force having constant B

Fig. 4.23

. f(t) = Ma + Bv + K x(t)..Where, a = Acceleration, v = Velocity

f(t) M d2 x(t) d x(t) K.. = ? +B-d-+ x(t)dt- t

Taking Laplace, I F(s) = Ms2 Xes) + BsXes) + KX(s)

This is equilibrium equation for the given system.

Now we will try to derive analogous electrical network.

4.8.2 Force Voltage Analogy (Loop Analysis)

In this method, to the force in mechanicalsystem, voltage is assumed to be analogous one.Accordingly we will try to derive other analogousterms. Consider electric network as shown in theFig. 4.24.

Fig. 4.24

r

Control System Engineering 4·14 Mathematical Modeling of Systems

The equation according to Kirchhoffs law can be written as,

diet) 1J .vet) = let) R + Ldt+ C l(t)dt

Taking Laplace,

yes) = I(s)R+ Ls I(s) + I!~But we cannot compare F(s) and yes) unless we bring them into same form.

For this we will use current as rate of flow of charge.

. i(t) = dq.. (it

i.e. I(s) = sQ(s) or Q(s) = I(s)s

Replacing in above equation,

I V(s) = L s2Q(S) + Rs Q(s) + ~ Q(s) IComparing equations for F(s) and Yes) it is clear that,i) Inductance 'L' is analogous to mass M

ill Resistance 'R' is analogous to friction B.

iii) Reciprocal of capacitor i.e, llC is analogous to spring of constant K.

Translational Rotational Electrical

Force Torque T Voltage V

Mass M Inertia J Inductance L

Friction constant B Tortional friction Resistance Rconstant B

Spring constant K N/m Tortional spring constant Reciprocal ofK Nmlrad capacitor 1/C

Displacement 'K 8 Charge q

• dx e=:=co Current I • dqVelocity x = df dt

Table 4.2 Tabular form of force-voltage analogy

nl

Control System Engineering 4 -15 Mathematical Modeling of Systems

4.8.3 Force Current Analogy (Node Analysis)In this method, current is treated as

analogous quantity to force in the mechanicalsystem. So force shown is replaced by a currentsource in the system shown in the Fig. 4.25.

The equation according to Kirchhoff'scurrent law for above system is,

I = IL +IR +IeLet node voltage be V,

1 f V dV. I = - Vdt+--+C-L R dt

I t

Fig. 4.25

Taking Laplace,yes) yes)

I(s) = -+-+sCV(s)sL RBut to get this equation in the similar form as that of F(s) we will use,

d41v(t) = where 41 = flux

V(s) = s q,(s) "'(5) = V(s)i.e. "I' sSubstituting in equation for I(s)

1 1I(s) = CS2 ~(s) + R s ~(s) + L ~(s)

Comparing equations for F(s) and l(s) it is clear that,

i) Capacitor 'C' is analogous to mass M.

ii) Reciprocal of resistance ~ is analogous to frictional constant B.

iii) Reciprocal of inductance ~ is analogous to spring constant K.

Translational Rotationa. Electrical

F Force T Current I

MMall J C

B friction B 1/R

K Spring K 1/L

x displacement 6 ~

• . dx • dO Voltage 'e' = de)!x Velocity = - O=-=(J)dt dt dt

Table 4.3 Tabular form of force-current analogy

Ie

c

\ I nl


Key Point: The elements which are in series in F - Vanalogy, get connected in parallel inF - I analogous network and which are in parallel in F - Vanalogy, get connected in series inF- 1analogous network.

4.9 Steps to Solve Problems on Analogous SystemsStep 1: Identify all the displacements due to the applied force. The elements spring and

friction between two moving surfaces cause change in displacement.Step 2: Draw the equivalent mechanical system based on node basis. 11:.eelements under

same displacement will get connected in parallel under that node. Eachdisplacement is represented by separate node. Element causing change indisplacement (cither friction or spring) is always betwcen the two nodes.

Step 3: Write the equilibrium equations. At each node algebraic sum of all the forcesacting at the node is zero.

Step 4: In F-V analogy, use following replacements and rewrite equations,

F~V, M~L,•

B ~ R, K ~ l/C, x ~ q, x ~ i (current)

Step 5: Simulate the equations using loop method. Number of displacements equal to

number of loop currents.Step 6: In F-J analogy, use follOWingreplacements and rewrite equations,

F ~I,•

M ~ C, B ~ I/R, K ~ IlL, x ~ C), x=e(e.m.f.)

Step 7: Simulate the equations using node basis. Number of displacements eqUC11to

number of node voltages. Infact the system will be exactly same as equivalent

mechanical system obtained in step 2 with appropriate replacements.

,,.. Example 4.1: Draw the equivalent mechanical system of the given system. Hence writethe set of equilibrium equations for it and obtain electricalanalogous circuits using,

i) F-V Analogy and ii) F-J Analogy

B1

f(t)

I I nl


Solution: The displacement of Ml is Xl (t)and as Bl is between M1 and fixed supporthence it is also under the influence of Xl (t).While B2 changes the displacement from Xl(t)to X2(t) as it is between two moving points.And M2 and K are under the displacementX2(t) as K is between mass and fixed support.

:EF = 0

KF

Equivalent system

F = M} s2 Xl + Bl s Xl + B2 S(Xl - X2)

o = M2 52 X2+ KX2 + B2 S(X2- Xl)

At node 1,

At node 2,

... (1)

... (2)

Now (i) F - V Analogy M ~ L B ~ R K ~ l/C X ~ q

V(s) = Ll s2 ql + Rl S ql +R2 S(ql - q2) ... {3j

0 = L2 s2q2 + (l/q q2 + R2 S(q2 - qt) ... (4)

l(s) Q(s) i.e, 1(5) = 5 Q(s)=s

V(s) = Ll sI1(s) + RIll (5) + R2 [II(5) - h(s)] ~ Loop 1

Replacing

Number of loop currents equal to numberof displacements.

Hence,

C) R,

Loop 1 Loop 2

C(1/K)

(II) F • I AnalogyF -+ I M ~ C B -+ l/R K-+ 11L

I(s) 1 1= Cl S2 91 + Rl S • 1+ R2 S (~l - ~2)

0 :;:: ~2 S(.2-9l)+C2S2cj)2+~. ~2

Replacing 5 4' (5) = V(s)

•.. (1)

... (2)

r


I(s)Vl(S) 1= CISV1(S)+~+R2 [Vl(S)-V2(S)]

1 1o = R2 [V2(S)- Vl(s)]+ C2 s V2(S)+ sL V2(S)

...Node 1

...Node 2

Hence,

Number of node voltages equal to numberL(1/K) of displacements.

'.. Example 4.2: Draw the equivalent mechanical system and analogous systems based onF-V and F-J methods for the given system.

Equivalent system

Solution: Two displacements : No element under Xl (t) alone as force is directlyapplied to a spring Kl. So it will store energy and hence is the cause to change the forceapplied to M2. Hence displacement of M2 is X2 and as B2 and K2 are connected to fixedsupports both are under X2(t) only as shown in the equivalent system.

At node 1,

At node 2,

... (1)

..• (2)

M2 I 82 , K2 are under same displacement.

\ I °nl


(i) F-V analogy: M ~ L B -4 R

1V = C1 (ql - q2)

O 1 ) ., 1= C1 (q2 - qI + L2 s- q2 + C2 q2 + R2 sQ2

I~

C2(11K2)

~ Same displacement same current.

K~l/C

... (3) Loop (1)

... (4) Loop (2)

Equations in terms of II and 12 can be written by using i(t) = ~~ Le. l(s) :: s Q(s) as

explained earlier.

(ii) F - I analogy : M ~

l(s)

C, B ~ 1/R,

1= 1;(~l - q,2)

K~ llL

... (1) Node (1)

... (2) Node (1)

Equations in terms of VI (s) and V2(s) can be obtained by using the relation,

d41vet) = dt i.e. Yes) = s q,(s) as explained earlier.

v,

Same displacement-same voltage.

I I nl


4.2 Analysis of Mechanical SystemsInmechanical systems, motion can be of different types i.e. Translational, Rotational or

combination of both. The equations governing such motion in mechanical systems areoften directly or indirectly governed by Newton's laws of motion

4.2.1 Translational MotionConsider a mechanical system in which motion is taking place along a straight line.

Such systems are of translational type. These systems are characterised by displacement,linear velocity and linear acceleration.

Key Point: According to Newton's law of motion, sum of forces applied on rigid body orsystem must be equal to sum of forces consumed to produce displacement, velocity andaccelerationin various elements of the system.

The following elements are dominantly involved in the analysis of translationalmotion systems.

i) Mass ii) Spring iii) Friction

4.2.2 Mass (M)

1--- f(t) Force

This is the property of the system itselfx{t) Displacement which stores the kinetic energy of the

translational motion. Mass has no power tostore the potential energy. It is measured inkilograms (kg). The displacement of massalways takes place in the direction of theapplied force results in inertial force. This forceis always proportional to the accelerationproduced inmass (M) by the applied force.

Fig. 4.1

Consider a mass 'M' as shown in the Fig. 4.1 having zero friction with surface, shownby rollers.

The applied force f(t) produces displacement x(t) in the direction of the applied forcef(t). Force required for the same is proportional to acceleration.

... . d2x(t)f(t) = M x acceleration = M 2

dt

Taking Laplace and neglecting initial conditions we can write,

I f(s) = MS2 X(s) IAlso mass cannot store potential energy so there cannot be conswnption of force in the

mass e.g. if two masses are directly connected to each other as shown in the Fig. 4.2 and ifforce f(t) is applied to mass M, then mass M2 will also displace by same amount as Ml·

I


4.10 ServomotorsThe servosystem is one in which the output is some mechanical variable like position,

velocity or acceleration. Such systems are generally automatic control systems which. workon the error signals. The error signals are amplified to drive the motors used in such.systems. These motors used in servosystems are called servomotors. These motors arcusually coupled to the output shaft i.e, load through gear train for power matching.

These motors are used to convert electrical signal applied, into the angular velocity ormovement of shaft.

4.10.1 Requirements of Good ServomotorThe servomotors which are designed for use in feedback control systems must have

following requirements :

i) Linear relationship between electrical control signal and the rotor speed over awide range.

ii) Inertia of rotor should be as low as possible. A servomotor must stop runt).ing'without any time delay, if control signal to it is removed. For low inertia, it isdesigned with large length to diameter ratio, for rotors. Compared to its framesize, the rotor of a servomotor has very small diameter.

iii) Its response should be as fast as possible. For quickly changing error signals, itmust react with good response. This is achieved by keeping torque to weight ratiohigh.

iv) It should be easily reversible.

v) It should have linear torque - speed characteristics.vi) Its operation should be stable without any oscillations or overshoots.

4.11 Types of ServomotorsThe servomotors are basically classified depending upon the nature of the electric

supply to be used for its operation.

The types of servomotors are as shown in the following chart :

+A.C.Servomotors

Servomotors

t ~Special ServomotorsD.C.Servomotors

!Armature controlled Field controlled

nl

Control System Engineering 4-21 Mathematical Modeling of Systems '~

4.12 D.C. ServomotorBasically d.c. servomotor is more or less same as normal d.c.motor. There are some

minor differences between the two. All d.c. servomotors are essentially separately excitedtype. This ensures linear torque-speed characteristics.

The control of d.c, servomotor can be from field side or from armature side.Depending upon this, these are classified as field controlled d.c, servomotor and armaturecontrolled d.c. servomotor.

4.12.1 Field Controlled D.C. ServomotorIn this motor, the controlled signal obtained from the servoamplifier is applied to the

field winding. With the help of constant current source, the armature current is maintainedconstant. The arrangement is shown in the Fig. 4.26.

la (Constant)

Vf(t) fromservoamplifier

\Constantcurrent....__---------o source

r-'''--I

Fig. 4.26 Field controlled d.c. servomotor

TIlls type of motor has large Le IRe ratio where Lf is reactance and Rf is resistanceof field winding. Due to this the time constant of the motor is high. This means it can notgive rapid response to the quick changing control signals hence this is uncommon inpractice.

4.12.1.1 Features of FieJd Controlled D.C. ServomotorIt has following features :

i) Preferred for small rated motors.

ii) It has large time constantiii) It is open loop system. This means any change in output has no effect on theinput.

iv) Control circuit is simple to design.

4.12.2 Annature Controlled D.C. ServomotorIn this type of motor, the input voltage 'Va' is applied to the armature with a

resistance of Ra and inductance La. The field winding is supplied with constant currentIe. Thus armature input voltage controls the motor shaft output. The arrangement is shownin the Fig. 4.27.

r

Control System Engineering "'~I . 4-22 Mathe~atical Modeling of Systems

(Constant)

Constantrcurrentsource 0--------' __ ""'-t

Fig. 4.27 Armature controlled d.c. servomotor

The constant field can be supplied with the help of parmanent magnets. In such caseno field coils are necessary.

4.12.2.1 Features of Armature Controlled D.C. ServomotorIt has following features:

i) Suitable for large rated motors.

il) It has small time constant hence its response is fast to the control signal.

iii) It is closed loop system.

iv) The back e.m.f. provides internal damping which makes motor operation more,stable.

v) The efficiency and overall performance is better than field controlled motor.As the armature controlled d.c. servomotor is closed loop system, in comparison with

open loop field controlled system, generally armature controlled motors are used.

4.12.3Characteristics of D.C. ServomotorsThe characteristics of d.c. servomotors are mainly similar to the torque-speed

characteristics of a.c, servomotor. The characteristics are shown in the Fig. 4.28.

Torque(Nm)

Va = Amlature voltage

Ea4 > Ea3 > Ea2 > Ea1

Ea2-......_.

Va = Ea1-+-,",o Speed (r.p.m.)

Fig. 4.28 Torque-speed characteristics for an annature controlled d.c. servomotor

\ I 'nl

Control System Engineering ..• c , 4·23 Mathematical Modeling of Systems

4.12.4 Applications of D.C. ServomotorThese are widely used in air craft control systems, electromechanical actuators, process

controllers, robotics, machine tools etc.

4.13 Transfer Function of Field Controlled D.C. Motor

+

Assumptions :

+

Jm = InertiaBm = Friction

Fig. 4.29

(1) Constant armature current is fed into the motor.

(2) ~f ccII. Flux produced is proportional to field current.

I ~f = Kf If I(3) Torque is proportional to product of flux and armature current.

I Tm oc ~Ia ITm = K' ~Ia = K' Kf If r,Tm = Km Kill I ... (1)

Where Km = K'Ia = ConstantApply Kirchhoff's law to field circuit.

difLf dt + Rf If = ef ... (2)

\Now shaft torque Tm is used for driving load against the inertia and frictional torque.

Inertia force

T = , d 20m + B dOm'.1 m <tt2 m tit ... (3)

J d26m imil t d2x= m--2-S1 ar om d'"dt t-

_, nl

, Control System Engineering , 4-24 Mathematical Modeling of Systems ,.,

F ·' I ' B dO m • 'I B dxnctiona rorce = m ""(it smu ar to dt

Finding Laplace Transforms of equations (1), (2) and (3) we get,

Tm(s) = Km K, If(s)

Ef(s) = (sl., + Rf) If(s)

Tm(s) = J ms2am(S)+Bmsa m(s)

Eliminate I,(s) from equations (4) and (5)

Tm(s) = Km Kf Ef(s}(SLf + Rf)

Eliminate Tm(s) from equations (6) and (7),

(s2 J m + sBm)a m(S)Km Kf Ef(s)

= (sl., + R,)

Input = Ef(S)

Output = Rotational displacement Om(s)

:. Transfer function6m(s)

= E(s)

ames) Km KfEf(s) = amS2 + sBm} (Rf + sLf)

= Km KfsRf Bm [1+ s'tml [1+ S'tf1

Where "tm = ~: = Motor time constant

'tf = ~: = Field time constant

T.F. = Om(S) = x, Km 1Rrll + S'tf)

._Ef(s) Bm(l + s'tm) S

... (4)

... (5)

... (6)

... (7)

Block diagram for field controlled d.c. motor is as shown in Fig. 4.30.

Et(s)

~Kt I~S) Km (I)m(s) 1-

R,(1 + st,} Bm(1 + Stm} S...__

Fig. 4.30 Block diagram

, I

- .1 Control System Engineering .~......,~.~. 4 - 25 Mathematical Modeling of Systems,. .~.);

4.14 Transfer Function of Armature Controlled D.C. Motor

Assumptions :(i) Flux is directly proportional to current through field winding,

I Cj)m = Kr I r = Constant I(ii) Torque produced is proportional to product of flux and armature current.

T = K'mlJ)Ia

T = K'm x, If r,(iii) Back e.m.f. is directly proportional to shaft v~locity (Om, as flux ~ is constant.

d8(t)as rom = crt

Eb = Kb (l)m(s) = Kb sOm{s)

Rt Ra La

-+ la

Constant

+

Fig. 4.31

Apply Kirchhoff's law to annature circuit :

R L diaea = Eb + Ia ( a) + a dt


Now

Ea(s) = Eb(S) + Ia (s) [R, + S Lei]

l,(s) E, (s) - Ep(s)= R"\ + SLa

13 (S)E,(S) - Kb s8m(s)

= R, + S La

Tm = K'm Kf If In

Tm = K' K I {Ea - x, sames)}m f f R +sLa a

I I nl


Also ... from equation (3)

Equating equations of TIt\J

K'm Kf If Ea(s} K'm Kf If Kb sames) 0 2 B)a ()= + mS +s m mS(R, + S La) (R, + S La )

K'm K( If [K'ml<jl,Kt,s 2 ].. (Ra + S La) E.,(s} = (R, + 5 La) + Jms + S Bm ames)

Km. am (s) s R, Bm {I + S'tm}(1 + s 't'a) G(s)

= =..Ell (s) Km·sKb 1 +G(s)H(s)1+

s R.1 Bm (1+ s 't'm)(l + s 't'a)

where r., = 1m/Em and La't'a =-Ra

Km = K'm Kf

G(s) = KmS R., Bm (1 + S 't'm}{l + S 'ta)

H(s) = SKb

Therefore can be represented in its block diagram form as in Fig. 4.32.

1

fig. 4.32 Blockdiagram

Key Point: Field controlled d.c. motor is open loop while armature controlled is closed loopsystem. Hence armature controlled d.c. motors are prejnTed over field controlled type.

4.15 A.C. ServomotorMost of the servomotors used in Jow power servomechanisms are a.c, servomotors.

The a.c. servomotor is basically two phase induction motor. The output power of a,c,servomotor varies from fraction of watt to few hundred watts. The operating frequency is50 Hz to 400 Hz.

, I


The stator carries two windings,uniformly distributed and displaced by 90°, inspace. One winding is called main windingor fixed winding or reference winding. Thisis excited by a constant voltage a.c. supply.The other winding is called control winding.It is excited by variable control voltage, whichis obtained from a scrvoamplifier. This

voltage is 900 out of phase with respect to the voltage applied to the reference winding.This is necessary to obtain rotating magnetic field. The schematic stator is shown in theFig 4.33.

4.15.1 Construction

Controlwinding

Fig. 4.33 Stator of A.C. servomotor

4.15.2 Rotor

It is mainly divided into two partsnamely stator and rotor.

The rotor is generally of two types. The one is usual squirrel cage rotor. This has smalldiameter and large length. Aluminium conductors are used to keep weight small. Itsresistance is very high to keep torque-speed characteristics as linear as possible. Air gap iskept very small which reduces magnetising current. This cage type of rotor is shown withskewed bars in the Fig. 4.34 (a). The other type of rotor is drag cup type. There are twoair gaps in such construction. Such a construction reduces inertia considerably and hence

Aluminium bars

Shaft

(a) Squirrel cage rotor

L"tator

(b) Drag cup type rotor

Fig. 4.34such type of rotor is used in very low power applications. The aluminium is used for thecup construction. The construction is shown in the Fig. 4.34 {b}.

4.15.3 Torque-speed CharacteristicsThe torque-speed characteristics of a two phase induction motor, mainly depends on

the ratio of reactance to resistance. For small X to R ratio i.e. high resistance 10\\0 reactance

r


large ~ Torque(Nrn)

Control voltageE24> E23> E22> E2l

t XSmall R

- Speed Speed (r.p.m.)

Fig. 4.35 Fig. 4.36

motor, the characteristics is much more linear while it is nonlinear for large X to R ratio asshown in the Fig. 4.35.

In practice, design of the motor is so as to get almost linear torque-speedcharacteristics. The Fig. 4.36 shows the torque-speed characteristics for various controlvoltages. The torque varies almost linearly with speed. All the characteristics are equallyspaced for equal increments of control voltage. It is generally operated with low speeds.

4.15.4 Features of A. C. ServomotorThe a.c, servomotor has following features :

i) Light in weight. ii) Robust construction. iii) Reliable and stable operation.iv) Smooth and noise free operation. v) Large torque to weight ratio. vi) Large R to Xratio i.e, small X to R ratio. vii) No brushes or slip rings hence maintenance free. viii)Simple driving circuits.

4.15.5 ApplicationsDue to the above features it is widely used in instrument servomechanisms, remote

positioning devices, process control systems, self balancing recorders, computers, trackingand guidance systems, robotics, machine tools etc.

4.15.6 Transfer Function of A.C. ServomotorThe various approximations to derive transfer function are,

(i) A servomotor rarely operates at high speeds. Hence for a given value of controlvoltage, T oc: N characteristics are perfectly linear.

(ii) In order that T ee N characteristics are directly proportional to voltage applied toits control phase, we assume T oc N characteristics are straight lines and equallyspaced.

Torque at any speed 'N° is,

nl


T. K E demm = tm 2t + mdt"

where, d!; is speed of motor.

If load consists inertia 1m and friction Bm we can write,

Tm(s) = 1m s2 6m + Bm s6m

Now Laplace transform of equation (1) is

Tm{s) = KtmE2 (s) + m s6m(s)

Equating equations (2) and (3)

... (1)

.... (2)

... (3)

:. Ktm E2 (s) +m s6m (s) = Jm s2 ames)+ Bm sOmes)

. 6m(s) Ktm= =E2(S) s(s Jm - m + Bm)

[sJm]s (Bm - m) 1+ (Bm _ m)

am (s) Km=E2 (s) s (1+ 'tm s)

KmKtm

= Bm-m

'tm = 1mBm-m

where

and

2S Om S9m

E2~O------••---~~~---'_---~~~.--~09mKlJm 1/s 1/s

tm(m-Bm)

JmFig. 4.37 Signal flow graph of a.c.

servomotor

Key Point : As slope is negative, in theabove equation [Bm - m] shows that totalfriction increases due to m. As it adds morefriction, the damping improoes, improoingstability of'the motor. This is called InternalElectric Damping of 2 ph A.C. servomotor.

E2(s)Ktm 1 1 9m(8)-Jm s s

Bm-m----Jm

Fig. 4.38 Block diagram of a.c. servomotor

_, nl


Due to mass, there cannot be any change in force from one mass to other hence nochange in displacement.

Rigidconnection

(No friction and no elastic action)

Fig. 4.2

Key Point: The displacement of rigidly connected masses is always same.

4.2.3 Linear SpringIn actual mechanical system there may be an actual spring or indication of spring

action because of elastic cable or a belt. Now spring has a property to store the potentialenergy. The force required to cause the displacement is proportional to the netdisplacement in the spring. All springs are basically nonlinear in nature but for small

deformations their behaviours can beapproximated as linear one. Henceassuming linear spring constant 'K' for thespring, we can write equation for the

~-Jft1mrcrnif'-----f(t) spring in the system.Consider a spring having negligible

mass and connected to a rigid support. Itsspring constant be 'K' as shown in theFig. 4.3.

Fig. 4.3

ij+ x,(t)

. Force requireddisplacement x(t) in theproportional to displacement.

I f(t) =. K x(t)

Ito causespring is

Fig. 4.4

Now consider the spring connectedbetween the two moving elements havingmasses Ml and M2 as shown in the Fig. 4.4where force is applied to mass MI.

nl


Signal flow graph for A.C. servomotor is as shown in the Fig. 4.37. Hence blockdiagram of A.C. servomotor is as shown in Fig. 4.38

4.16 Comparison of Servomotors

4.16.1 Comparison between A.C. and D.C. Servomotor

Sr. A.C. Servomotor D.C. ServomotorNo.

1) Low power output of about Deliver high power output1'2 w to 100 W.

2) Efficiency is less about 5 to 20 %. HIgh efficiency.

3) Due to absence of commutator Frequent maintenance required due tomaintenance is less. commutator.

4) Stability problems are less, More problems of stability.

5) No radio frequency noise Brushes produce radio frequency noise.

6) Relatively stable and smooth operation. Noisy operation.

7) A.C. amplifiers used have no drift. Amplifiers used have a drift.

4.16.2 Comparison between Armature Controlled andField Controlled D.C. Servomotors

Sr. Field Controlled Armature ControlledNo.

1) Due to low power requirement amplifiers High power amplifiers are required toare simple to design. design.

2) Control voltage is applied to the field. Control voltage is applied to the armature

3) Time constant is large. Time constant is small.

4) This is open loop system. This is closed loop system.

5) Armature current is kept constant. Field current is kept constant.

6) Poor efficiency. Better efficiency.

7) Suitable for small rated motors Suitable for large rated motors.

8) Costly as field coils are must Permanent magnet can be used insteadof field coils which makes the motor lessexpensive.

4.17 Models of Commonly used Electromechanical SystemsIn this article the transfer functions of very commonly used systems are derived. This

will help the reader to find out the transfer functions of the different practical systems.

, I 'nl


4.11.1 GeneratorsConsider a separately excited generator

which is many times used in various practicalmechanical systems. Generators are required todrive the motors because vacuum diodes,transistor amplifiers are not suitable because oftheir low ratings. Consider a generator as shownin the figure.

f

I'....__--o

Fig. 4.39 Generator

Rf = Field resistance

Lf = Field inductance

ef = Applied voltage (input)

eg = Generated voltage (output)

Now for a generator,

where, = flux

Flux is directly proportional to current passing through the field winding say if.

Let K, be the generator constant in V/ A

... (1)

Applying Kirchhoff's Law to field circuit.

. R L difef = If f + f dt ... (2)

Taking Laplace of both the equations (l) and (2)

Eg(s) = x, If(s)

Er(s) = R, If{s) + Lr{s) II(s) neglecting Ir(O)

I,(s) =

Ka Er{s)Rf + S LrEg(s) =...

Eg(s) =Er{s) R, + s L,

This is the T.F. of separately excited generator.

r


4.17.2 Generator Driving Motor. It is very common to find generator driving motor in practical mechanical systems. So .let us discuss the T.F. of a system with generator driving motor.

+

Fig. 4.40 Generator driving motor

R, and Lf - Resistance and inductance of generator field

eg - Generated voltage

Kg - Generator constant in V/ A

R, and La - Resistance and inductance of motor armature

J - M.I. of Load and f is frictional force

Now.T. F. of generator is,

Eg(s) _ KgEf(s) - Rf + S Lf

Now consider armature controlled motor,Torque produced by motor is dependent on ia and let KT be torque constant.

T = KT ia

This torque is utilised to drive a load.

T = Jd2e + f dedt2 dt

Now Eg is voltage applied to armature.

... p_ = i R + L dia + ebo..g a a aCit

where eb is back e.mJ.

nl


Let Kb be the back e.m.f, constant,

K · J d20 + f dOT ta = dt2 dt '

Taking Laplace transform

KT I, (5) = J 52 e (s) + f s B (5)J:;". • R L di , K dO~ = I... a + ...dt + '" dt

... (3)


Eg(5) = la(s) n, +L.l sla(s)+KI:> sO(s) ... (4)

finding 101(5) from equation (3)

(J s2 + fs) O(s}I., (s) =

Kr


sO (5)U 5+ f)~g(5) = K (R, + S La) + Kb sO (5)

T •

Eg(S) = 0 (s) s [(f + ]"S) c:: + 5 La) + Kh ]

Eg (s) _ S Kb e (s) = 5 (Ra + 5 La) (s J + f) 0 (5)KT

0(5)Eg(s) - S Kh O(S) =

S (R, + 5 L" ) (5 J + f)

1sJ+ f f--04'__- 9(5)

Fig. 4.41 •...

Ir


Therefore using both the transfer functions of generator and armature controlled motorwe can develop the combined block diagram as below :

1sJ + f ~,__ ....8(s}

Fig. 4.42Reducing the block diagram, solving feedback loop of motor which is

KT

=

... 6(5)=Ef(s) (Rf + s Lt>{s [(Ra + 5 La ) (J s+ f) + Kb KT 1}

4.17.3 Position Control SystemAnother very common system used in practice is position control system. This is used

to control position of shaft, by use of potentiometer as error detector. The error is to beamplified by amplifier and then must be given to armature controlled motor whose shaftposition will get controlled as per the controlled signal. The motor shaft is coupled to theload through gearing arrangement with ratio Nt/N2.

Load has M.l. J and friction as f while 6 r is the reference position while 0 n is theactual position of shaft.

The circuit diagram can be drawn as below :La

e = error~I I,const. 1

---Gear

~N2

Fig. 4.43 Position control system_, nl


Let Kp be the potentiometer sensitivity, in V /radThe corresponding block diagram can be drawn as shown in the Fig. 4.44.

Ores)1

sJ+f

Fig. 4.44Reducing the block diagram as shown in thc Fig. 4.45.

fl,(s)

G(s) =

Fig. 4.45K Kp KT (Nl/N2)

K Kr KT (N1/N2) = Ks = System gain constant

H(s) = 1

r. Over all T. F. can be calculated as below,

O,(s) =OrCs)s [CRa + S La) Us+ f) + Kb Kr )

1+ Kss [(Ra + S La HJs + f) + Kb Kr ]

Oa (s) _ KsOrCs) - s [(Ra + S La) Us+ f) + Kb KT ]+ K,

4.17.4 Position Control with Field Controlled MotorIn the above case if instead of armature controlled motor, field controlled motor is

used then derive the T.P. of overall dosed loop system.Consider field controlled motor,

L · difc( = if Rf + f dt

Now T ex ~ ia and ia is constant

T oc q, oc if

r

Control System Engineering 4 ..36 Mathematical Mod.Ung of Systems

ia = constant

Fig. 4.46

Let Kf be constant in N-m/ A

T = K, if

This torque is utilised to drive a load of moment of inertia J and friction f.

··. d2S deT = J dt2 + f CIt

Kf if = J d2e + f dedt2 dt··.

Finding Laplace of all the equations,

Ef(s) = If{s) Rf + S Lf If(s)

and x, If(s) = J s2 O(s)+ s fO(s)

· If(s)SO(s) [s J+ f)·. = Kf

Substituting in Ef(s)

Ef(s)s O(s) [s J + f) [R, + s L,]

= Kf

· 6(s) Kf.-- =·. Ef(s) s [s J + fURi + s If I

Now using the same block diagram as derived in case (ill) replacing armaturecontrolled motor T. F. by the field controlled we can get the new block diagram.

s (sJ + f) (R, + sLf)

Fig. 4.47

nl


Let

G(s) Ks- s (s J + f) CRt + s Lf) H(s) = 1

KsG(s) S (5 J + f) (R, + sLd

--=--,--=--------1+ G(s)H(s) 1+ K,S(5 J + f)(Rf + 5 L()

Oa (5) =OreS) S (5 J + f) (R, + sLf ) + K,

4.17.5 Speed Control SystemIn s~me of the practical applications it is necessary to drive a load at a desired speed

(1) rad/sec. For this application an electromechanical system can be used which is' shown asbelow. .

/ Constant

Ra~\1\I\.r\,--. ....-_,

Tachometer

.Fig. 4.48 Speed control system

System uses armature controlled motor and tachometer feedback. Let KA be theamplifier gain. Let us derive the transfer function of this system. The objective of system isto move the load at a desired speed.

The d.c, tachometer output voltage is proportional to output speed (1).

e, is reference voltage and et is tachometer voltage.

error

.~.(5)

... (6)et = K, (1) where K, is constant.

Neglecting inductance of armature

l \ 'nl

Control System Engineering ;. 4-38 Mathematical Modeling of Systems

eb = Kb (1)

ca = ia Ra + Kb ro. Ka (e, - et) = ia Ra + Kb (1)

-. (ol

K, e, - K, K,(1) = ia R, + Kb (1)

substituting from equation (5),


Taking Laplace

K.. Er(s) :: (K01 K, + Kb ) roes)+ i.l (s) R,

Now torque produced by motor

'" (7)

T oc la

... (8)

Now this drives a load

... dOwhere ro= dt '" (9)

:. Taking Laplace

KT ia (s) = Js roes)+ f (1)(5)

.) U s+ f] f ()101(5 = KT + (J) S ... (10)


Us+f](1)(s)= (K, K, + Kb ) (1)(5)+ KT R,

(K01 K. + Kb ) 00(5) U s+ f] roeS) n,Er{s) = + -~------

K01 Ka KT

m(s)=Er(s)

'This is the required transfer function of the speed control system.

nl

Control System Engineering I 4-39 Mathematical Modeling of Systems

4.17.6Speed Control using Generator Driving MotorA position control system described in Fig. 4.49 in which the armature of motor is

applied with a control voltage through a generator. The field current of generator controlsthe voltage generated by the generator.

e1(l)R


Let us determine the transfer function of the system

Input is ej (t).

Nowd(O)

cc{t) == Cl (t) - K" dt ... (11)


Ec(s) = El (s) - Kb $6 (s)

Now E () K I die R':c t· 2 = ~f dt + I If ... (12)

applying Kirchhoff's law to the field circuit of generator.

Now C.1 = K" if where K, is constant.


Ea (s) = x, If (s)

Taking Laplace transform of (12)

Ec(s) K2 = If (s) [Rf + sLI ]

... (13)

Eliminating If (5)

Ec(S) K21,(5)=

(Re + s Lr)

E.) (s)ic, Ec(s) K2= (R, + s Lf)

E..(s) Ka K2=

Ec{s) (Rr + sLf)

r

Control System Engineering ". 4-4 MathematicalModeling of Systems

Now mass Ml will get displaced byXI(t) but mass M2 will get displaced byX2(t)as spring of constant K will storesome potential energy and will be thecause for change in displacement.Consider free body diagram of spring asshown in the Fig. 4.5.

Net displacement in the spring is Xt(t)- X2(t) and opposing force by the spring isproportional to the net displacement i.e. XI(t) - x2(l).

Fig. 4.5 Spring between two moving pointscauses change in displacement

Taking Laplace,

FsprinJ; = K[ XI(l) - X2(t)]

Fo;prinr. = K[XI (s) - X2(S)]

Key Point: The spring between the moving points causes a change ill displacementfrom onepoint to another.Spring behaves exactly same in rotational systems, only the linear spring constant

becomes torsional spring constant but denoted as 'K' only.

4.2.4 FrictionWhenever there is a motion, there exists a friction. Friction may be between moving

element and fixed support or between two moving surfaces. Friction is also nonlinear innature. It can be divided into three types,

i) Viscous friction ii) Static friction

Dash-potl..---~II1--n;"p2r'8'

Fig. 4.6

Fig. 4.7

iii) Coulomb friction

Viscous friction as dominant out of thethree is generally considered, neglecting othertwo types. Viscous friction is assumed to belinear, with frictional constant 'B'. This haslinear relationship with relative velocitybetween two moving surfaces.

The friction is generally shown by adash-pot or a damper as shown in the Fig. 4.6.

This is the symbolic representation of afriction.

Consider a mass M as shown in the Fig. 4.7having friction with a support with a constant '8'represented by a dash-pot.

Friction will oppose the motion of mass Mand opposing force is proportional to velocity ofmass M.

II \ nl

Control System Engineering ~~!1_- !i-. 4 - 40 Mathemabcal Modeling of Systems

Now consider field constant armature controlled motor. Armature is energized byoutput of the generator.

Applying Kirchhoff's law to armature circuit,.E" (s) =: Eb(S) + Ia (s) [Ra + s L" ]

Back e.m.f is proportional to em

Eb(S) = K dOmbdt

Taking Laplace Transform

Et>(s} = Kb s9m (s)

In armature controlled d.c, motor as field current is constant so torque produced isdirectly proportional to the armature current.

T = K I" (s) where K is motor constant.

.This torque is utilized to drive the load with moment of inertia J and friction Fm·

T = J d20m + F. d9m Taking Laplace transform fromdt2 m dt

KI,,(s) = Js29m(s)+sFm9m{s)

fa (5) = ~ Os + Fm)9m. (s)

Substituting in equation f?r E..(s)

Ea(s)

Omes) K=Ea{s) s KKb + seRa + s La)(Js+ Fm)

Km= s[l+Tms)... Neglecting L. of armature

KKm = [KKb + Ra Fm1 = Motor constant

I

'T' a, J T' f1m = (KK R F. ) =: ime constant 0 motor

s+ a m /'

Neglecting 'La' of armature motor,

Om(s) _ KmEa (s) -- s (1 + Tm s)

...

l I

f ~ Control System Engineering' nl,·Jl!ttC(I 4 -41 Mathematical Modeling of Systems

Ea{s) K" K2=Ec(s) [Re + s Ld

0 E...(s)x, K2 Ec(s) ames)s (1 + Tm s)

.0 = :::(R, + S Lf) Km

and Ec(s) = El (s) - Kb sam

[Ko x, K2 x; s]

= am (5) s (1+ Tm s) + (Rr + S Lf)

= e (5) [Rr S (1+ Tm s) (1+ Tr s) + Kb K" K2 Km s]m [1+ Tf{s)]

ames)=Ej(S)

Kwhere ~; = R; generator constant.

4.17.7 A Typical Position Control System used in IndustryA position control feedback system has a potentiometer bridge with a sensitivity of

Kp V/radian as error detector. It feeds a doc. amplifier with an open circuit gain K. ThisSupplies to a field of generator which has resistance Rf and inductance Lfo Generatorconstant is KgV/ Ao.This is connected to an armature controlled doc. motor with motortorque constant KT and back e.m.f constant Kb V/rad/sec. It drives a load of inertia J andfriction f.

Draw the Simplified block diagram and hence calculate its transfer function.

9r

I~Consti

R. ~m

e

Fig. 4.50 Typical position control system

r


e r = Reference position , e a = Actual position

T. F of generatorEg (s) _ KgEc(s) - Rf + S Lf

Block diagram can be drawn from the analysis of the different cases discussed.

Fig. 4.51

Neglecting the inductance of armature winding of motor.

Fig. 4.52

G(s) =x, K ~ KT

H(s) = 1s (R, + sLf) [Ra Us + f) + Kb KT 1

Let

T.F =

4.18 D.C. Motor Position Control SystemIn industry to conrol the position of the shaft, a d.c. motor position control system is

commonly used.

I I nl

Control System Engineering 4·43 MathematicalModeling of Systems

4.18.1 Transfer Function of D.C. Motor Position Control SystemConsider the D.C. position control system which is controlling position of the shaft.

Assume that the input and output of the system are the input shaft position and outputshaft position respectively.

Assume following system constants,

r = Angular displacement of the reference input shaft

c = Angular displacement of the output shaft

e = Angular displacement of the d.c. motor shaft used

K} = Gain of potentiometric error detector

Kp == Amplifier gain

eol = Applied armature voltage

eb == Back e.m.f.

Rol = Armature winding resistance

La = Armature winding inductance

ia = Armature winding current

Kb = Back e.m.f, constant

K = Motor torque constant

Jm = Moment of inertia of motor.

bm = Viscous friction coefficient of motor

Il = Moment of inertia of load

bL = Viscous friction coefficient of load

n = Gear ratio Nt/N2

~constant

Fig. 4.53 D.C. Motor position control system

r


Equations describing above system can be written as follows :Output shaft position is to be controlled so as to keep that at required position. Output

is sensed by angular displacement 'c' and compared with input which is r. Error isamplified by amplifier with gain 'Kp' and given as input to the d.c, motor which in tumcontrols the angular position of the shaft of the motor 'e' which in tum controls outputposition of shaft 'c' so as to modify the error.

For potentiometric error detector we can write.

E(s) = Kl [R(s) - C(s)]

For amplifier

E, (s) = Kp E(s)

For armature controlled d.c. motor

11 = constant so flux q, is constantT = xt, whcre K = motor torque constantCb ee e

. Cb = K de.. bili

For armature circuit

L di, . RCb + ,1 dt + la a

Taking Laplace transforms

Eb(S) = Kb 56 (s)

s,(5) = s, (s) + Ia (s)[Ra + SLa 1T = KIa (s)

Now torque is utilised to drive load + shaft of motor.

Jeq = J rn + n 2 J L = Equivalent moment of inertia

beq = bm + n 2 bj, = Equivalent frictional coefficient

T d2S d6= KIa (s) = Jeq -2 + beqddt t

= [Jc.tq S2 + beqS16(s) Laplace transform

r,(s) = ~Ueq S2 + bt.'qs16(s)

Ea{s)6(5)

= Kb s6 (5)+K UeqS2 + bcqs][Ra + S La )

nl


6 (s) K=Ea (s) s [JeqS + bcq ](Ra + sola 1+ KKb S

K::;:

'La' is generally small hence neglected.

K=

= s (1+ Tm s)

KKm = (bl.'qRa + KKb) = Motor constant

R, J..-q .Tm ::;: 0 R KK) ::;:Motor time constant

cq a + b

Now C(s) ::;: n 6 (s)

C(s) 6(s)Ea(s) = n--

Ea(s)

C(s) nKm=a, (s) s(l + Tm s)

C(s} nKmwhere Ea (s) = Kp E(s)=KpE(s} s(l + Tm s)

C{s) =n Km x, E(s)

s(1 + Tm 5)

C{s) = n Kill Kp KI (R(s) - C(s)]s(1+ Tnt s)

n Km Kl' K, n Km Kp K, R(s):. C(s) + (1 T. .)- C(s) = --. (1 T )s + mS S + mS

[S(1 + Tm 5)+ n Km Kp Kl] n Km Kp Kl R(s)

:.C(S) 5(1+ Tm S) = s(1 + Tm S)

C(s)=R(s)

nKmKpKls(l + Tms)nKmKpKt

1 +~~=---:-5(1+ Tms}

G(s)=--~--1+C(s)H(s)

_, nl


Block diagram :

R(s)

Fig. 4.54

'-. Example 4.3: A d.c. motor drives a pointer which is spring loaded, to return to thereferenceposition. If Kb = Back e.m.f. constant, KT = Torque constant J<.~ = Spring constantand J = Moment of inertia. Find the transferJunction.

1fo----___.I

(M.U. : Nov.-93, Nov.-94)

Solution: Writing the system equations.

Va(t)at,

= Ia R, + La crt +s,

Taking Laplace,

Va(s) = Eb(S) + Iii(s) [Ra + SLa] ... (1)

. r,(5) Va (S) - Eb(S) .... (2)=.. (R, + sLa) •

Now Tm = KT r.. Tm ::: Motor torque ... (3)

and Eb(S) Kb sees) as s, d6 ... (4)= oc -dt

r,(s) Va(s)- 6(s).. = (R, + S La)

and Tm{s) = KT [V,(S) - KbS6(S)] .. (5).(Ra + S La)

, I


This torque is used to drive a pointer with inertia J and spring load of constant Ks.d20{t)

Tm :: J dt2 + K:;O(t)

Taking Laplace

Tm{s) = J 52 O(s) + x, 6(s)

Equating (5) and (6)

...(6)

:. J s2 O{s)+ x, 6(5) = KT [Va(S}-KbS9{S)]R, + sl.,

? KT Kb s9(s) KTJ 5- 9 (5)+ K~9(s)+ (R L ) = (R L ) v,(s)_,+S a .,+s .1

• 6 ) [Js2 (Ra + S La) + K~(Ra + sLa) + KT KbS] _ Kr .• • (s (R L ) - (R L ) V" (s)a+5 a .,+s a

So transfer function is,0(5) Kl

... 9(s)

)_ Example4.4: A d.c. generator supplies its output to a separately excited d.c. motor. Thefield current of the motor is constant. The voltage applied to the field circuit of d.c. generatoris e/ (f). Write the differential equations of this Ward-Leonardsystem relating the input ef (I)

to the output 6 III (f). (Refer to figure shoum below) Hence ootain expressionfor 6 III (5).(M.U.-May-96[PTDC] and April-96(Electrical])

KA Generator

Lau

Motor if Constant

C1

r emL-.. ...-.I Ward-Leonard system

Solution: Torque produced by motor,

T = KT Ia(t) ... (1)

Voltage applied to the motorar,

Eg(t) = r, Ra + L dt+ Eb(t) ... (2)

r


Back emf is proportional to the angular velocity

:. Eb{t) = x, dad~(t) ... (3)

Torque produced T is used to drive a load of inertia J and friction BT = J d2am + B dem.. dt2 dt ... (4)


... (5)

Substituting in equation (2) we can get resultant differential equation 'relating Eg(t) andam(t).

Now Ef(t) '" (6)

and Eg(t) ... (7)

. Substituting If(t) = ~t) in equation (6) we can get, equation relating Ef(t) and

Eg(t).

Substituting this in equation (2) we can obtain the final differential equation relatingEt(t) and emet). It is very difficult to obtain in time domain so let us obtain it in Laplacedomain.

Taking Laplace transform of all the equations

T(s) = KT I,(5)

Eg(S) = 101 (s)[Ra + 5 La]+ Eb(S)

Eb(S) = Kb Sem(s)

T (s) = emes)[Js2 + BS]·. KT r,(5) = ameS)[Js+ B] 5

. Ia (s) ;:::ames}s[Js+ B]·. KT

Eg(s)em(s) s[Js + B][Ra + S La]·. = KT + Kb sames)

and Ef(s) = If(s) [R, + s Lt]

... (1)

... (2)

... (3)

... (4)

... (5)

... (6)

... (7)

Control System Englneenng 4 ..49 Mathematical Modeling of Systems

So

Eg(s) = Kg If(s)

IE(S) = Eg(s)

~

Ef(s) = EgCs)[Rf + sLt]•.• (8)Kg

Substituting Eg(s) in equation (8) we get

Ef(s) Kg = ames) [sas + B)~~a + S La + KbS][Rf + s Ltl

...(R, + S Lf) {s (Ra + S La) Us+B)+SKbK T}

Kg KT Ef(s)

4.19 Models of Thermal Systems

4.19.1 Heat Transfer System

,-------IIIOutletw;;- :r---"""

o90 C

I_I

Surrounding temp. 8 .

_ Thermal insulation

Ir.:=::J - Inlet :,ater01 C

Water

Heater

Fig. 4.55

A thermal system used for heating flow of water is shown below.Electric heating element is provided in the tank to heat the water. The tank is

msulated to reduce heat to the surroundings.The necessary simplifying assumptions are :1) There is no heat storage in the insulation.2) All the water in the tank is perfectly mixed and hence at a uniform temperature.

r


... d x(t)FfrictJonal = B~

Taking Laplace and neglecting initial conditions,

I Ffrictiollill (s) = Bs X(s) ISimilarly if friction is between two moving surfaces, it is shown in the Fig. 4.8.

Fig. 4.8 Friction between two moving points causes change in displacement

In such a case, opposing force is given by,

~ . . = 5 [d Xt(t) _ ~ X2(t)]Irlctional dt dt

Taking Laplace,

I F/rictional (s) = Bs [X1(s) - X2(S)]

Thus if the force applied to mass M2 is f(t) then due to friction between the masses Mland M2, the force getting transmitted to Ml is always less than f(t). Hence thedisplacement of mass Ml is different than the displacement of mass M2•

Key Point: The friction between two moving points, causes a change in displacement fromone point to other.Frictional force also behaves exactly in same manner, in rotational systems, only linear

frictional constant becomes torsional frictional constant but denoted by same symbol '5' only.

4.3 Rotational MotionThis is the motion about a fixed axis. In such systems, the force gets replaced by a

moment about the fixed axis i.e. (force x distance from fixed axis) which is called Torque.

So extension of Newton's law states that the sum of the torques applied to a rigidbody or a system must be equal to sum of the torques consumed by the different elementsof the system in order to produce angular displacement (e), angular velocity (ro) andangular acceleration (n) in them. As previously stated, spring and friction behaves in samemanner in rotational systems. The property of system which stores kinetic energy in

_, nl


6 i = Inlet water temperature in "C,

00 = Outlet water temperature in "C.

o = Surrounding temperature.

q = Rate of heat flow from heating clement in JIsec.q i = Rate of heat flow to the water.

q t = Rate of heat flow through tank insulation.

C = Thermal capacity in Jl°e.R = Resistance of thermal insulation.

So rate of heat flow for the water in tank is,

d60ql = Cdt ... (1)

The rate of heat flow from the water to the surrounding atmosphere throughinsulation is,

... (2)

As per the heat transfer principles,

q=qj+q,

Substituting equation (1) and (2)

C d6" 00-6q = ---crt + R

... (3)

... (4)

Neglecting the term aIR from the equation (4) this is because the variation of watertemperature a0 is over and above ambient temperature 0 (I).

. q = C dOo + 00dt R

Taking Laplace transform,

60 (s)Q(s) = Cs60(s)+-,r-

:. Transfer function is,

60(s) R=Q(s) 1+ sCR

The time constant of the system is Re.

nl

Control System Engineering 4 ·51 Mathematical Modeling of Systems

4.19.2 ThermometerConsider a thermometer placed in a water

bath having temperature e J I as shown.eo is the temperature indicated by the

thermometer. The rate of heat flow into thethermometer through its wall is,

dq =citWhere R = Thermal

resistance of the thermometer wall.

The indicated temperature, rises at a rate of

dOo _ 1 dq(it - Cdt

where C is thermal capacity of the thermometer.

deo = ~. [Oi -00]dt C R

Taking Laplace of the equation,

1560(s) = RC [6.(5)-8(1(5)]

1= 1+ s RC

The time constant is RC.

4.20 Actuators

Bath

Fig. 4.56

The actuator is a device which receives input signal from the controller and itproduces the input signal to the plant according to control signal so that the output willapproach the reference input signal to reduce the error to zero. Thus an actuator isgenerally after the controller and before the plant in the control system. An actuator can beof two types :

i) Hydraulic actuator 2) Pneumatic actuator

_, nl

Control System Engin •• nng 4·52 Mathematical ModeUngof Systems

4.20.1 Hydraulic ActuatorThe structure of an hydraulic actuator is shown in the Fig. 4.57.

Input--+-----+-+------+-+------+-+-!f-displacement

Main piston

To sump From highpressure source

TosumpValve piston

Fig. 4.57

They are piston devices in which motion of the spool regulates the oil flow to eitherside of the power cylinder. When the spool moves to the right, the high pressure oil entersthe power cylinder to the left of the piston.

The differential pressure produced causes the power piston to move to the right,pushing the oil infront of it into the sump.

The load coupled rigidly to the piston moves a distance y from its reference positioncorresponding to the displacement x of the valve piston from its neutral position. The oil ispressurised by a pump and is recalculated in the system.

Equation of motion and transfer functionThe rate of flow of oil into the piston is proportionaJ to the rate of the movement of

the piston.

Q = A dy A = area of pistondt'

... (l)

If P is the differential pressure across the piston then the force on the piston is AP.This moves the load of mass M against friction B.

d2 y dyA x P = M dt2 + B dt ... (2)

For small values of the displacement X, if P is the differential pressure on the pistonand Q is the oil flow then,

..• (3)

dyKl x-Adt



K.,M K2 BA 52 Y (s) +--p;- S Y(s) = KJ XeS)- A yes) s

Yes) [s2 K2 M+ K2 B5 + A 2 5] = KJ A Xes)

yes) _ KIAXes) - s2 K2 M + s K2 B + A 2 5

yes)Xes)

(~:) .'=

This transfer function is similar to the electric motors.


A pes) = Ms2 yes) + as Yes)+ KY(s)

yes) A=pes) Ms2 + Bs+K

4.20.2 Pneumatic ActuatorPneumatic acting valve is used to

obtain linear displacement of a plungerwith pressurised air as input.

The air at pressure P is injectedthrough inlet. Pressurised air pushes thediaphragm and plunger. The plunger hasa mass M and friction on B with springconstant K.

Let A be the area of diaphragm thentransfer function can be obtained asbelow.

Force exerted on the plunger is A x P.This force is opposed by mass, frictionand spring.

d2y dyAxP = M -+B-+Kydt2 dt

pressure

Fig. 4.58

The advantages of pneumatic systems are fire proof, explosion proof, simplicity andeasy to maintain.

, I nl

Control System Engineering .. 4-54 Mathematical Modeling of Systems

4.20.3 Comparison between Pneumatic and Hydraulic Systems

Sr.No. Pneumatic systems Hydraulic systems

1. The fluid used is air. The fluid used is oil.

2. Air is compressible. Oil is incompressible.

3. Air does not have lubricating property. Oil acts as a lubricator.

4. The output power is much Jesscompared to The output power is much higher thanhydraulic. pneumatic.

S. At low velocities, the accuracy is poor. At all velocities, the accuracy is satisfactory.

6. No return pipes are required when air is used. The return pipes are must.

7. Can be operated for the temperature range of It is sensitive to the temperature changes and00 C to 200OC.It is insensitive to temperature the range is 20OCto 70OC.changes.

8. These are fire proof and explosion proof. These are not fire and explosion proof.

9. Easy from maintenance point of view. Difficult from maintenance point of view.

Note: The state space method of modeling the systems is separately covered in thechapter 15.


Example 4.5: The voltage generated by a d.c. generator is filtered by a low pass filterconsisting R:z and L2 as shown below. This filtered voltage is controlled by the voltageapplied to the field of generator.If Rf=40!l, Lf=60H, Ro=O.SQ, Lo=lH, [.2=lH,R2=2Q andgenerator constant Kg = 120 V / field Amp. Determine the transfer function£7(S)E;(s) of the system.

(M.U.: Dec.-97)

r---------

1e,(t) ) L,

1o---~_____,_________ J

low pass filter

Constant speed


Solution: For field circuit, d itCf(t) = ij(t) Rf + Lf --

dt... (1)

For armature circuit, ( ) . () R . ( ) R ( d i2 L d i., (2)£',1 I = 12 t cl' 12 t '2 ~ ',. -cit' + : It ...

For gencrator, Ca (I) = Kg if (1) '" (3)

... (4)

Taking Laplace transform of all the equations,

Ef{s) = It(s) lR, + s Lt]

E" (s) = 12(s) [(R.. + R2) + s (La + L2)]

a, (5) = ~ If{s)

... (5)

... (6)

... (7)

... (8)E2(S) = l::!(s) R2

Equation (6) and (7),

Kg It(s) = h(s) l(Rd + R2) + s(L" + L:d]

Substituting I a (s) from equation (8) and If (s) from equation (5),

Substituting 12(s) from equation (8) and Ids) from (5),

Substituting the values,~--------------------------------------~120x2 2.4

(40 + 60s) [(2.5+ 2s)) = (1 + 1.5s) (1 + 0.8s)

,,... Example 4.6: A higJrgain speed control system uses a tachogeneratorfor speed.sensing.The iachogenerator produces 5 V per 100 r.p.m. This voltage is compared with referencevoltage to produce error signal. If the referencevoltage is set to 10.8 volt. What is the talueof expected speed? (M.U. : Nov.-94)

Solution: The tachogenerator produces 5 V per 100 r.p.m.,

Its constant is 5100 = 0.05 VIr.p.rn.

r

Control System Engineering "1. 4-56 Mathematical Modeling of Systems .1

Now for the expected speed, error should be zero,

i.e VI - tachogenerator output = error

Tachogenerator output = Vr

Let N be expected speed in r.p.m.... N x 0.05 = 10.8

N = 216 r.p.m. ... Expected speed

". Example 4.7: An armature controlled d.c. motor has an armature resistance of 0.37 n.The moment of inertia is 2.5 X 10-6 kg - 1112. A back emf of 2.09 V is generated per 100r.p.m. of the motor speed. The torque constant of the motor is 0.2 N-m /A. Determine tiretransferfunction of the motor relating the motor shaft shift and the input voltage.

(M.U. : Dec.-96)

Solution : The motor can be shown as,

G:n~nl1Vaet) )1 i.(I)

Now back emf is 2.09 V per 100 r.p.m. , N = 100 r.p.m. is,

21t N(I) = --w- = 10.4719 rad/sec

:. Back emf constant Kb = W~4~9= 0.1995 V /rad/secKT = 0.2 N-m/A

For armature circuit Va(t) = Ia R, + Eto

Taking Laplace Va (s) = 13 (S) R, + El>(s)

Eb(t) = K d9bdt

Taking Laplace Eb(S) = KbS9 (s)

Tm = KrIa{t)

... (1)

... (2)

... (3)

nl


Taking Laplace l".n(s) = Kr 1.1 (s)

This torque drives a load of inertia J.dze

= J dt2

Taking Laplace Tm(s) = J 526 (s)

· Kl Ia (s) = J s2 B (s)·.

fa (s)2.5 X 10-6

52 e (s) = 1.25 X 10-5 52 e (5)· =.. 0.2

· V.I(s) = 1.25 X S2 10-5 e (S) X 0.37 + Kb S6 (5)·.· Va(s) = 4.625 X 10-tl S2 6(s) + 0.1995 S6(5)·.. 6(5) 1 216216.22..

Vol(s) = =4.625xl0-fi 52 + 0.1995 s S(5+ 43135.135)

.......E I 48 I I bId . E~(s)n...,... xamp e . : For t te system 5 toum e ow etermine E~( ).~f s

Separately excited d.c. generator

Solution:

For field circuit,

... (4)

(M.U. : Dec.-97)

For annature circuit,

For generator, ea (l) = Kg if{t)

e2(t) = i2(t) R2For output,

, I nl


Taking Laplace of all the equations we get,

Ef (s) = If (S)[Rf + s Lrl

Ea(s) = 12 (5) [(Ra + R2) + s(La + Lz)] = Kg Ids)

E2(S) = h (s) R2

Hence equating Ea (s) equations,

Kg Ids) = 12(s) [Iz (5) (Ra + R2) + 5 (La + L2)]

and hence using values of If(s) and 12(s) from remaining equations we get,

".. Example 4.9: A 50 Hz, 2 phase a.c. servomotor has the following parameters :

Starting torque = 0.186 NmRotor inertia = 1xl0-S kg=m?

Supply voltage = 120 VNo load angular velocity = 304 rad/sAssuming straight line torque-speed characteristicsof the motor and zero friction, obtain itstransfer function. (Gate)

Solution: The starting torque is nothing but locked rotor torque.Locked rotor torque _ 0.186

Ktm = ---Rated voltage 120...

= 1.55xlO-3

Let m be the slope of linearised torque-speed characteristics.

Locked rotor torque 0.186No load angular speed = 304m =

= - 6.118xlO-4

The torque at any angular speed rois given by,. de

Tm = Ktm E2l +m dt ... (1)

r


where [2t = rotor voltage

Taking Laplace, Tm(s) = Ktm E2t (s)+ m sO(s)

This torque is used to drive load of inertia J m- The friction is given zero.

... (2)

t, d2e, Tm., I = 1m dt2

· Tm(s) = s2JmO(s)...,. (3)

... (4)

Equating (2) and (4),

KtmE2t (s) +m sO (5) = s2J m8 (5)

··. Kim E2t(s) = sO(s)(sJm-m]

Hence the transfer function of the motor is,

8(s)E2t(s) =

1.55X 10-3= 5[1xlO-5 s-(-6.118xl0-4)]

E2t(s) s(s+61.18)O(s) 155=

This is the required transfer function.1_ Example 4.10: A two phase a.c. servomotor haoing a torque constant of D.lUS Nm/Vcontrols a position load tI,rough a gear ratio of 10:1. The effective moment of inertia andcoefficient of viscous friction referred to load side are 0.25 kg-m2 and 1.0 N-m/(rad/sec). Thesynchro error detector produces an error signal of 0.1V per degree error in misalignment.Develop the block diagram representation of the control system and there from obtain thetransferfunction. (M.U. : May-99)

Solution: For the error detector, KJ = 0.1 V/degree error

··. Kl = 0.1x 1807t

= 5.7295 V/ radian

Ktm = Torque constant

= 0.045 Nm/V

NJ 1N2 = 10

Jl = 0.25 kg - m2 on load side

Jcq = Motor side = (~~rx Jt = 1~0xO.2S··.

r


rotational system is called Inertia and denoted by T i.e, moment of inertia. Opposingtorque due to inertia T is proportional to the angular acceleration (a) of that inertia.

= J d28(t}dt2Tdue to inertia

Taking Laplace,~--------------------~Tduc to inertia (S) = J 52 O(S)

Sr. No. Translational Motion Rotational Motion

1 Mass (M) Inertia (J)

2 Friction (8) Friction (B)

3 Spring (K) Spring (K)

4 Force (F) Torquem5 Displacement (x) Angular displacement (0)

6 Velocity v = (~f) Angular velocity ((J) = ~n7

Acceleration (~t2x) Angular acceleration (ex = d2O)dl2

Table 4.1 Analogous Elements

4.4 Equivalent Mechanical System (Node Basis)While drawing analogous networks, it is always better to draw the equivalent

mechanical system from the given mechanical system. To draw such system use followingsteps:

Step 1 : Due to applied force, identify the displacements in the mechanical system.

Step 2 : Identify the elements which are under the influence of differentdisplacements.

Step 3 : Represent each displacement by a separate node, using Nodal Analysis,Step 4: Show all the elements in parallel under the respective nodes which are

under the influence of respective displacements.

Step 5: Elements causing same change in displacement will get connected in parallelin between the respective nodes.

l \ ·nl


= 2.5 X 10-3 kg-m2

BL = 1 N-m/(rad/sec)

(N r 1·. Beq = Motor side = N~ x B. = 100 x l

= 0.01 N-m/ (rad/sec)

The block diagram of the system is,

NtG(s) = K. x motorT.F.x N2

A.C.Motor

For motor, Tm = Kim E2. Tm(s) = Ktm E2 (5)·.and Tm{t)

d28 de= Jeq dt2 + B(_'q dt

. Tm(s) = (S2 Jcq + SBcq)8(s)·.

... (1)

... (2)

Equating equations (1) and (2),

Ktm E2 (s) = (S2 Jeq + 5 Bl'q) 0(5)

0(5) Ktm= = =

E2(S) S(SJeq + Beq) 5(SX 2.5x 10-3 + 0.01) S(5+ 4)0.045 18

G(s) = 5.7295x 18 x_!_= 10.3131s(s+4) 10 5(s+4)

ac(s)=SiCS)

10.31315(S+ 4)

1+ 10.3131s(s+ 4)

10.3131=

S2 + 4s+ 10.3131

nl


Review Questions1. Explain the derivation of analogous networks using

i) Force-voltage ii) Force-current analogy2. Write a shorf note on direct and inverse analogous networks.3. Draw the analogous electrical networks based on

a) F - V analogy b) F - I tl11alogy of the following mechanical systems

B \

I-_f(t)

K81

\ I nl


4. Distinguish between AC. servomotor and D.C. servomotor.5. Derive transfer function of a.c. servomotor stating the assumption made.6. State the applications of a.c. servomotor.7. Derive the transfer function of field controll~d d.c. seroomotor.8. Answer the following giving reasons :

aJ A.c. servomotor has a smaller diameter and more length.bJ Field controlled D.C. servomotor is pre/med than armature controlled.

9. Develop block diagram for armature controlled D.C. seroomoior and find its transfer function.10. An armature controlled d.c. motor is supplied in series with a resistance from a

24 V d.c. supply. The motor takes current of SA on smiling and the stalling torque being 0.915Nsm. The motor runs at 1000 r.p.m. taking a current of 1 A. The ualue of J = 4 X 10-3 Kg - ",2and friction constant as 1.5 x 10- 3 Nm/ (rad/sec)Determine the transfer function of tile motor. o (s) 2.9

(Ans. : v(s) = s(1 + 0.3s»

11. for the closed loop systt.'17I shoton heloto, draw the Mock diagram and determine

transfer function O["(s)/Or(S). Tile gioen ualues are Error detector gain K" = 8 V/rad, Amplifiergain KA = 10 VI\. Rf == 5 Q, I--f == 0.25 H, Kf == 0.05 Nm/A, J",oIor = 0.02 Kg - m2,It == 3 Kg - m2• flJUlttlf == 0.03 Nm/(rad /sec).ft, = 5.5 Nm/ (rad/sec)

la Constant+

o CIHOr

Am.: Oc(s) = 32.12 )( e,(s) (sl + 21.5s2 + 30.35 + 32.12)

12. The moment of inertia Jm and the coeffiCientof viscous friction fm for a field controll~d d.c. motorare motor respectively 5 x1O-" kg - m2 and 12.5 xlO-4Nm (md/sec). The motor torqru constantIC, being 2.5 Nm/A. Determine the transfer function relating the angular speed of the shllft andth fi Id t (l)(s) 2000e e curren. (Ans.: I,(5) = 1+0.4s)

nl


13. Figure given below shows a block diagram of speed regulator system. The accelerating torque is thedifference between the torque developed by the controller and load TL. The controller gaitl is0.00102 Nm/rad and the taC#logenerator constant is 0.191 V/rad/.<;ec.Tile load speed is adjltstl'd to

1000 r.p.m. The moment of inertia is J and friction is negligible.Calculate a) The reference voltage. b) TIll' speed if a constant load torque of 0.001 Nm issuddenly applied. (Ans.: a) 20 V b) 951 r.p.m.)

V(s) +1--+-- w{s)

14. An instrument servo consisting of motor, spring loaded shaft, etc is shown below

R=1rad +

Where V = Voltage in voltsR = Motor resistance 1Q

L = Motor inductance 0.1 HK/\ :;; Ampl~fier gain = 10 V/VK = Spring C01lStIZUt= 0.001 Nmfrcd

KI.I = Back. e.m]. constant == 0.01 V (rodtsec)

KT = Torque content = 0.01 N-m/A] = Moment of inertia = 0.005 N-m-s2

if tire input is 1 radians, what is the steady state error]

l I


15. The positions seroomedumism is shown in figure below. 8r is rtforence position and 9c actual

angular displaament oj sJuift in radiam. Obtain its closed loop tramJer junction ~~:~

+v-=-

~constantrlGiven e = angular displacement of the motor shllft

Kf = gain f error detector = 7.64 V/ rad

n, = 212, L" = negligibleKb = 5.5 X 10-2 V/ rad / sec

KT = 6xlO-S Nm/A

1m = 1 xlo-5 Kg - m2

fm = negligible1L = 4.4 X 10-3 Kg - m2

fr. = 4 xlQ-2 Nm/ (rad/sec)

Nt = 1N2 = 10

(AN.' Sc(s) ::I 42.3 )• 6r(s) S2 +-7.7 S + 42.3

16. Derive the transfer junction of a typical d.c. position control system.17. Derive the transfer fundion oj a typical d.c. speed control system.18. Obtain the mathematical model of heat transfer system.19. Obtain the mllfhematical model of thermometer.20. W1uIt is IlCtuator ? E:rpwn hydmulic and pneumatic actuator systems.21. Comptlre hydraulic and pneu'I1UIticactuators.

DOD

_l nl


4.5 Remarks on Nodal Methoda} The terms for an element

connected to a node 'X' andstationary surface (reference) is,

d2xFor mass ~ M dt2

For friction .; B [~tl - d;t2 ]

r--X(l) st@=>r(l} ::::;

77777777177

@=>f(t) fFnction "S" ......../777//7l777

I it---x(t)

~~Il :::::"

(a)

. B

F fricti B dxor tenon -4 dt

For spring ~ Kx

b) The term for an elementconnected between the two nodes'Xl • and ' X2' i.e. between twomoving surfaces is,

..... X2 r-x1I I

~f(t)::::;

/7~two nodes as due to mass there Fnction 'S'rX2 t-', X1cannot be change in force as masscannot store potential energy. ~ f(t) :::;

L2J Spv~~~'K' ~c) All elements which are /77777777777777777777777

under the influence of samedisplacement get connected inparallel under that node indicatingthe corresponding displacement.

e.g. consider the part of the system, shown in the Fig. 4.10(a).

No mass can be between the

For spring -4 K [ Xl - X2]

(b)

Fig. 4.9

All elementsunder samedisplacement -- B

M

x(t)

(a) (b)

Fig. 4.10

nl

Control System Engineering 4-8 MathematicalModeling of Systems

Here M, Band K all are under theinfluence of x(t). Hence in equivalent systemall of them will get connected in parallelunder the node 'X'. Consider anotherexample of the system shown in the Fig. 4.11.In this system M, , Bl and KI all are underthe influence of displacement Xl' This isbecause all are connected to rigid support.

While there is change from Xl to X2 dueto simultaneous effect of B2 and K2. So B2and K2 are under the influence of (Xl - X2)'But mass M2 is under the influence of X2alone. Mass cannot be under the influence

of difference between displacements. So in equivalent system the elements B1 , KJ andMI, all in parallel under Xl while B2 and K2 in parallel between Xl and X2 and clementM2 is under node X2 as shown in the Fig. 4.12 .

Fig. 4.11

4.6 Gear Trains

Fig. 4.13 Gear system

Fig. 4.12

A gear train is a mechanical device thattransmits energy from one part of a system toanother in such a way that force, torque, speedand displacement may be altered. The inertia andfriction of the gears are neglected in the ideal case.Consider a gear system as shown in the Fig. 4.13.

The number of teeth on the surface of thegears is proportional to the radii TI and f2 of thegears.

nl

Control System Engineering ~\'4 - 9 II" Mathematical Modeling of Systems 1~;

The distance travelled along the surface of each gear is same.

i.e. I 01 rl = 02r2 IThe work done by one gear is same as the other.

i.e,

:. we can say = N2

Remarks:1) The numbers of teeth N are

proportional to the radius rof a gear.

2) The distance travelled oneach gear is same.

3) Work done = 'ill by eachgear is same.

\Distance = re

1r-----~

Fig. 4.14

4.6.1.Gear Train with Inertia and FrictionIn practice, gears do have inertia and friction which can not be neglected. Consider

such practical gear arrangement connected to the load, shown below.

Primary side

load torque TL

load side

Fig. 4.15

T = Applied torque 01 ,02 = Angular displacements

T. I T2 = Torque transmitted to gears J 1 , J 2 = Inertia of gears.

N1 , N 2 = Number of teeth Bl , B2 = Friction coefficients

I I nl

Control System Engineering I,·4 - 9 .~ Mathematical Modeling of Systems i~:

The distance travelled along the surface of each gear is same.

i.e, I 81rl = 82r2 IThe work done by one gear is same as the other.

:. we can say

i.e,

Remarks:1) The numbers of teeth N are

proportional to the radius rof a gear.

2) The distance travelled oneach gear is same.

3) Work done = 1'8 by eachgear is same.

\Distance = rO

1r-----~

Fig. 4.14

4.6.1. Gear Train with Inertia and FrictionIn practice, gears do have inertia and friction which can not be neglected. Consider

such practical gear arrangement connected to the load, shown below.

81 N1

.) I) ~ IJ1

~InputtorqueT T1.B1

Primaryside T2·92~

~N2 82 load side

Fig. 4.15

T = Applied torque 9) ,92 = Angular displacements1i ,T2 = Torque transmitted to gears Jl , Jz = Inertia of gears.N1 , N2 = Number of teeth B) , B2= Friction coefficients

_\1 , ,:r I,

··c Control System Engineering 4-10 Mathematical Modeling of Systems


T - J d201(t) B dOI(t) "Ii ( )- 1 dt2 + 1 -----err- + t t ... (1)


To J d202(1} B d02(t) r ()2 = 2 dt2 + 2 ---cit + L t ... (2)

Now 'Ii Nl _ O2T2 = N2 - Ot"

T2 = N2 T)00 NI


... (3)

Substituting value of Tl in equation (1)

T J d2GI B dOl NI J d262 Nt B d02 N1 'Ii= I dt2 + l(it +N2 2 """"dt2"+N2 2dt" + N2 I.

00 T =

0 T =

..0

and

.

JIe = Equivalent inertia referred to primary side

B1e = Equivalent friction referred to primary side

_\1 , ,:r I,


Similarly the equation can be written referred to load side also, where applied torque

gets transferred to load as (~: T).

where J2l> = h+(~:rJ, and B2e = B2+(~:rBI

4.6.2. Belt or Chain DrivesBelt and chain drives perform same function as that of gear train. Assuming that there

is no slippage between belt and pulleys we can write,

T(®-O --=------r:C0))r,

Fig. 4.16

for such drive

4.6.3 LeversThe lever system is shown in the

Fig. 4.17. This transmits translationalmotion and forces, similar to geartrains.

By law of moment,

I f. I} = f2 12

By work done fl XI = f2 X2Fig. 4.17

Hence

4.7 Elecbical SystemsSimilar to the mechanical systems, very commonly used systems are of electrical type.

The behaviour of such systems is governed by Ohm's law. The dominant elements of anelectrical system are,

_'I , ,:1" ,

Control System Engineering ~tt4-12 .. MathematicalModeling of Syetems

i) Resistor ii) Inductor iii) Capacitori) Resistor: Consider a resistance carrying

current 'I' as shown, then the voltage dropacross it can be written as,

Suppose it carries a current (I, -12) thenfor the polarity of the voltage drop shown itsequation is,

,,..--v-=-( -1,---1-2)-R---'1

ii) Inductor : Consider an inductorcarrying current 'I' as shown, then the voltagedrop across it can be written as,

V = L dIdt

I = ~JVdt

or

,~O--~t-----~ __------~o

+ R -

-------v-------'I'2-~o---------~~------~o+ R -

------V------Fig. 4.18

l 1o~----~rOmDmOO~DO~O~'--~.----o

------v---'-----Fig. 4.19

. If it carries a current (I, -12) then for the polarity shown its voltage equation is ,

V = L d(lld~12)

or

iii) Capacitor : Consider a capacitorcarrying current 'I' as shown, then the voltagedrop across it can be written as,

or

V = ~ J I dt

I = C dVdt

If it carries a current (It - h) then for thepolarity shown its voltage equation is,

or

V = ~J (I, -12) dt

(I,-12) = CdVdt

o

_Ii

'2-+ (ooDObOo' - o

l

------v------Fig. 4.20

c~o--------~II~--~-~o------v--------

Fig. 4.21

co~--------+~I~I----------~o------v------

Fig. 4.22

_\1 , ,:r I,


4.8 Analogous SystemsIn between electrical and mechanical systems there exists a fixed analogy and their

exists a similarity between their equilibrium equations. Due to this, it is possible to drawan electrical system which will behave exactly similar to the given mechanical system, thisis called electrical analogous of given mechanical system and vice versa. It is alwaysadvantageous to obtain electrical analogous of the given mechanical system as we are wellfamiliar with the methods of analysing electrical network than mechanical systems.

There are two methods of obtaining electrical analogous networks, namely1) Force - Voltage Analogy i.e, Direct Analogy.2) Force - Current Analogy i.e. Inverse Analogy.

4.8.1 Mechanical System

Fig. 4.23

Consider simple mechanical system asshown in the Fig. 4.23.

Due to the applied force, mass M willdisplace by an amount x{t) in the directionof the force f{O as shown in the Fig. 4.23.

According to Newton's law of motion,applied force will cause displacement x(t)in spring, acceleration to mass M againstfrictional force having constant B

.. f(t) = Ma + Bv + K x(t)

Where, a = Acceleration, v = Velocity

f{t) d2 x(t) d x(t).. = M dt2 + B(:It+ Kx(t)

Taking Laplace, I F(s) = Ms2 Xes)+ BsXes) + KX(s)

This is equilibrium equation for the given system.Now we will try to derive analogous electrical network.

4.8.2 Force Voltage Analogy (Loop Analysis)

L

i~

Fig. 4.24

In this method, to the force in mechanicalsystem, voltage is assumed to be analogous one.Accordingly we will try to derive other analogousterms. Consider electric network as shown in theFig. 4.24.

_\1 , ,:r I,

Control System Engineering Mathematical Modeling of Systems

The equation according to Kirchhoffs law can be written as,

vet) '( ) R L diet) 1 I '()d= It + (it +cit t

Taking Laplace,

yes) = 1(5)R + Ls 1(5)+ I!~

But we cannot compare F(s) and Yes) unless we bring them into same form,For this we will use current as rate of flow of charge,

i(t) = dq.. dt

i.e. I(s) = sQ(s) or Q(s} = I(s)s

Replacing in above equation,I VIs) = L s'Q(s) +Rs Q(s) + ~ Q(s) IComparing equations for F(s) and yes) it is dear that,i) Inductance 'L' is analogous to mass Mii) Resistance 'R' is analogous to friction B.iii) Reciprocal of capacitor i.e. lie is analogous to spring of constant K.

Translational

Force

RotatIonal Electrical

Torque T Voltage V

Mass M Inertia J Inductance L

Friction constant B Tomonal friction ReSistanceRconstant B

Spring constant I< Nlm Tortional spnng constant Reciprocal ofK Nrnfrad capacitor 1/C

Dispfacement '~ o Charge q

Velocity; =~ 8 III de :I Q) CUrrent I • dqdf cit

Table 4.2 Tabular form of force-voltage analogy

_\1 , ,:r I,


4.8.3 Force Current Analogy (Node Analysis)In thls method, current is treated as

analogous quantity to force in the mechanicalsystem. So force shown is replaced by a currentsource in the system shown in the Fig. 4.25.

The equation according to Kirchhoff'scurrent law for above system is,

I = IL +IR +IeLet node voltage be V,

.. I = i I V dt + ~ +C ~~

Taking Laplace,

I t

Fig. 4.25

V(s) V(s)I(s) = T+T+sCV(s)

But to get this equation in the similar form as that of F(s) we will use.d41

vet) = dt where • = flux

yes) = s ~(s) . "'(s) = Yes)i.e, ... sSubstituting in equation for I(s)

1 1I(s) = Cs2 ~(s)+ R s ~(s)+I ,(s)

Comparing equations for F(s) and 1(5) it is clear that.i) Capacitor'C' is analogous to mass M.

ii) Reciprocal of resistance ~ is analogous to frictional constant B.

iii) Reciprocal of inductance iis analogous to spring constant K.Translational Rotational ElectrIcal

F Forca T Current I

MM ... J C

B friction B 1/R

K Spring K 1/L

x displacement 9 ,; Velocity = dx • dO Voltage 'e' = d~0""-=(1)dt dt dt

Table 4.3 Tabular form of force-current analogy

Ie

c

_\1 , ,:r I,

Control System Engineering 4-16 MathematicalModelingof Systems

Key Point: The elements which are in series in F - V tmIllogy,get connected in parallel inF - I tmIllogousnetwork and which are in parallel in F - Vanalogy, get connected in series inF - J analogous network.

4.9 Steps to Solve Problems on Analogous SystemsStep 1: Identify all the displacements due to the applied force. The elements spring and

friction between two moving surfaces cause change in displacement.Step 2: Draw the equivalent mechanical system based on node basis. The elements under

same displacement will get connected in parallel under that node. Eachdisplacement is represented by separate node. Element causing change indisplacement (either friction or spring) is always between the two nodes.

Step 3: Write the equilibrium equations. At each node algebraic sum of all the forcesacting at the node is zero.

Step 4: InF-V analogy, use follOwingreplacements and rewrite equations,

F -+ V, M -+ L,•

B -+ R, K -+ lIe, x -+ q, x -+ j (current)

Step 5: Simulate the equations using loop method. Number of displacements equal tonumber of loop currents.

Step 6: In F-J analogy, use following replacements and rewrite equations,

I F -+ I, •M -+ C, B -+ l/R, K -+ l/L, x -+ 41, x=e(e.m.f.)--------------------------------~------~---~~Step 7: Simulate the equations using node basis. Number of displacements equal to

number of node voltages. Infact the system will be exactly same as equivalentmechanical system obtained in step 2 with appropriate replacements.

,...,. Example 4.1: Draw the equivalent mechanicalsystem of the given system. Hence writethe set of equilibrium equationsfor it and obtain electricalanalogous circuits using,

i) F-V Analogy and ii) F-JAnalogy

-rx,(t}

f(t)

_\1 , ,:r I,


Solution: The displacement of Ml is Xl(t)and as 81 is between MI and fixed supporthence it is also under the influence of Xl[t).While 82 changes the displacement from Xt(t)to X2(t) as it is between two moving points.And M2 and K are under the displacementX2( t) as K is between mass and fixed support.

KF

Equivalent system

:EF = 0

At node 1, F = M} s2 Xt + 81 s Xl +B2 S(XI -X2)

o = M2 52 X2 + KX2 + B2 S(X2- XI)

••. (1)

... (2)At node 2,

Now (i) F - V Analogy M -+ L B -+ R K -+ 11C x -+ q

Replacing

Yes) = LI s2 ql + R) 5 qt +R2 S(ql - q2) ..• (3j

0 = L2 s2q2 + (1fQ qz +R2 S(q2 - ql) •.• (4)

I(s)Q(5) i.e. 1(5) = s Q(s)- =s

yes) = LI sl)(s) + R) 11(S) + R2 [ll(S) - h(s)] -+ Loop 1

0 1= L2SI2(S) + sC h(s) + R2 [h(s)-Il(S)] -+ Loop2

Hence,

Number of loop currents equal to numberof displacements.

Loop 1 Loop 2

C(1/K)

(II) F • I AnalogyB -+ l/R I< -+ IlL x -+ •

1 1= CJ S29) + Rl S 91 + R2 S (.1 - .2) •.• (1)I(s)

..• (2)

Replacing s • (s) = Yes)

_\1 , ,:r I,


I(s) ...Node 1

...Node 2

Hence,

Number of node voltages equal to numberL(1/K) of displacements.

,..,. Example 4.2: Draw the equiVlllent mechanical system and analogous systems based onF-V and F-I methods for the given system.

1X1(t)

X1

Jt)K2

82 K2

Equivalent system

Solution: Two displacements : No element under Xl (t) alone as force is directlyapplied to a spring Kl. So it will store energy and hence is the cause to change the forceapplied to M2. Hence displacement of M2 is X2 and as B2 and K2 are connected to fixedsupports both are under X2(t) only as shown in the equivalent system.

At node 1,

At node 2,

••• (1)

... (2)

M2 , B2 , K2 are under same displacement.

_\1 , ,:r I,


(i) F·V analogy: M _. L B _. R

1V= C1(QI-Q2)

o = ~1 (q2-ql)+L2S2q2+~2 Q2+R2sQ2

K41/C

... (3) Loop (1)

... (4) Loop (2)

Same displacement same current.

Equations in tenns of 11 and 12 can be written by using i(t) = ~~ i.e. I(s) = s Q(s) as

explained earlier.

(ii) F - I analogy: M -t C, B -t l/R, K _. IlL

11(5) = Ii" (~I - ~2) ... (1) Node (1)

... (2) Node (1)

Equations in tenns of VI (s) and Vz(s) can be obtained by using the relation,

d41v(t) = dt i.e. V(s) = s cjI(s) as explained earlier.

Samedisplacement-samevoltage.

_\1 , ,:r I,

Control System Engineering 4-20 Mathematigl Modeling of SY8tems

4.10 ServomotorsThe servosystem is one in which the output is some mechanical variable like position.

velocity or acceleration. Such systems are generally automatic control systems which workon the error signals. The error signals are amplified to drive the motors used in suchsystems. These motors used in servosystems arc called servomotors. These motors arcusually coupled to the output shaft i.e. load through gear train for power matching.

These motors are used to convert electrical signal applied, into the angular velocity ormovement of shaft.

4.10.1 Requirements of Good ServomotorThe servomotors which are designed for use in feedback control systems must have

following requirements :i) Linear relationship between electrical control signal and the rotor speed over a

wide range.ii) Inertia of rotor should be as low as possible. A servomotor must stop running

without any time delay, if control signal to it is removed. For low inertia, it isdesigned with large length to diameter ratio, for rotors. Compared to its framesize, the rotor of a servomotor has very small diameter.

iii) Its response should be as fast as possible. For quickly changing error signals, itmust react with good response. This is achieved by keeping torque to weight ratiohigh.

iv) It should be easily reversible.v) It should have linear torque - speed characteristics.vi) Its operation should be stable without any oscillations or overshoot s.

4.11 Types of ServomotorsThe servomotors are basically classified depending upon the nature of the electric

supply to be used for its operation.The types of servomotors are as shown in the following chart :

..AC.Servomotors

Servomotors

i +Speclal ServomotorsD.C.Servomotors~

Armature controlled Field controlled

_\1 , ,:r I,

Control System Engineering 4·21 Mathematical Modeling of Systems '..4.12 D.C. Servomotor

Basically d.c, servomotor is more or less same as normal d.c.motor. There are someminor differences between the two. All d.c. servomotors are essentially separately excitedtype. This ensures linear torque-speed characteristics.The control of d,c, servomotor can be from field side or from armature side.

Depending upon this, these are classifiedas field controlled d.c. servomotor and armaturecontrolled d.c. servomotor.

4.12.1 Field Controlled D.C. ServomotorIn this motor, the controlled signal obtained from the scrvoamplifier is applied to the

field winding. With the help of constant current source, the armature current is maintainedconstant. The arrangement is shown in the Fig.4.26.

Vf(t) fromservoamplifier

la (Constant)

\ConstantcurrentL..----------o source

r-""""'"

Fig. 4.26 Field controlled d.c. servomotor

This type of motor has large Lf /Rf ratio where Lf is reactance and Rf is resistanceof field winding. Due to this the time constant of the motor is high. This means it can notgive rapid response to the quick changing control signals hence this is uncommon inpractice.

4.12.1.1Features of Field Controlled D.C.ServomotorIt hass following features :i) Preferred for small rated motors.ii) It has large time constantiii) It is open loop system. This means any change in output has no effect on theinput.

iv) Control circuit is simple to design.

4.12.2 Annature Controlled D.C. ServomotorIn this type of motor, the input voltage 'Va I is applied to the armature with a

resistance of Ra and inductance La. The field winding is supplied with constant currentIf. Thus armature input voltage controls the motor shaft output. The arrangement is shownin the Fig.4.27.

_\1 , ,:r I,

Control System Enginearing tll~1 .4-22 MathematicalModeling of Systems

I,

(Constant)

Constantrcurrentsource0-------' ..........._.

Fig. 4.27 Armature controlled d.c. servomotor

The constant field can be supplied with the help of parmanent magnets. In such caseno field coils are necessary.

4.12.2.1 Features of Annature Controlled D.C. ServomotorIt has following features :i) Suitable for large rated motors.ii) It has small time constant hence its response is fast to the control signal.iii) It is closed loop system.iv) The back e.m.f. provides internal damping which makes motor operation more

stable.v) The efficiency and overall performance is better than field controlled motor.As the armature controlled d.c. servomotor is closed loop system, in comparison with

open loop field controlled system. generally armature controlled motors are used.

4.12.3 Characteristics of D.C. ServomotorsThe characteristics of d.c. servomotors arc mainly similar to the torque-speed

characteristics of a.c, servomotor. The characteristics are shown in the Fig. 4.28.

Torque(Nm)

Va =Armature voltageEa4 > Ea3 > Ea2 > Ea,

Speed (r.p.m.)

Fig. 4.28 Torque-speed characteristics for an amature controlled d.c. servomotor

_\1 , ,:r I,

Control System Engineering .,. 4·23 Mathematical Modeling of Systems

4.12.4 Applications of D.C. ServomotorThese are widely used in air craft control systems, electromechanical actuators, process

controllers, robotics, machine tools etc.

4.13 Transfer Function of Field Controlled D.C.Motor

+

Assumptions :

+

Jm= InertiaBm= Friction

Fig. 4.29

(1) Constant armature current is fed into the motor.(2) cIIf acIf. Flux produced is proportional to field current.

I .f = Kf If I(3) Torque is proportional to product of flux and armature current.

I Tm oc cilIa ITm = K' 'Ia = K' Kf If r,Tm = Km Kt If I •.. (1)

Where Km = K'la = ConstantApply Kirchhoff's law to field circuit.

difLf dt +Rf If = ef ... (2)

\Now shaft torque Tm is used for driving load against the inertia and frictional torque.

Inertia force

T = J d29m B dOmI r., m dt2 + m dt ..• (3)

_\1 , ,:r I,

Control System Engineering , 4-24 Mathematical Modeling of Systems

F ·· 1 " B dOm imil B dxnctiona rorce = m '"(it" sarto dt

Finding Laplace Transforms of equations (1), (2) and (3) we get,

Tm(s) = Km Kf If(S)

Ef(s) = (SLf + Rf) If(s)

Tm(s) = Jms20m(s)+ BmsOm(s)

Eliminate If(S) from equations (4) and (5)

Tm{s) = Km Kf Ef(s)(sLf + Rf)

Eliminate Tm{s)from equations (6) and (7),

(S2 Jm + sBm)0 m(s)Km x, Ef(s)= (sLr + Rf)

Input = Ef(s)

Output = Rotational displacement Om(s)

:. Transfer function ames)= Ef(S)

Om(s) Km KfEf(S) = UmS2 + sBm) (Rt + sLr)

= Km KfsRr Bm [1+ s'tm] [1+ s'tr]

Where 'tm = Jm = Motor time constantBm

'tf = i: = Field time constant

T.F. = Um(S) = Kr Km 1.-Ef(s) Rr[l + S1'r) Bm(1+ s'tm) S

... (4)

..• (5)

..• (6)

... (7)

Block diagram for field controlled d.c. motor is as shown in Fig. 4.30.

Et(s)

~K, It{s) K.n wm(s) 1- --R,{1 + s11} Bm{1 + Slm) S

,___


_\1 , ,:r I,

·, :, Control System Engineering ·~",,·r"mr. 4 -25 MathematicalModeling of Systems,. ,~,~,

4.14 Transfer Function of Armature Controlled D.C. Motor

Assumptions :(i) Flux is directly proportional to current through field winding,

I cIIm = Kr Ir = Constant I(li) Torque produced is proportional to product of flux and armature current.

T = K'm K, If r,T = K'm CIlIa

(iii) Back e.m.f. is directly proportional to shaft velocity rom, as flux 41is constant.

romd9(t)

as = crt

Eb = Kb (J}m(s)= Kb sOmes)

R, Ra La+

Constant

Fig. 4.31

Apply Kirchhoff's law to armature circuit :

L di;ea = Eb +Ia (Ra)+ a dt


Now

. Ea(s) = Eb(S) + la (s) lR.a + S La]

i,(s) E..(s) - Eb(S)= R .. +s La

la (s)Ea (s) - Kb sOmes)= Ra + S La

Tm = K'm Kr Ir r,

Tm = K' K I {Ea - Kb SOmes)}m f' Ra + S La

_\1 , ,:r I,


Also ... from equation (3)

Equating equations of TIt\I

K'm Kr If Ea(s) K'm x, r, Kb sOm(S) 0 2 B)O )(Ra + S La) = (Ra + S La) + m S + S m m(S

K'm Kr If.• (R,+SLa)Ea(s)

em(s) =Ell (s)G(s)sRa Bm(1 + stmHl + S 'fa)

1+ Km·sKa,SRa Bm (1+ S 'tm)(l + S 'fa)

= 1+G(s)H(s)

where r., = Im/Bm and La'ta =

Ra

G(s)S R, Bm(1 + S 'tm) (1+ s 'tot)

H(s) =Therefore can be represented in its block diagram form as in Fig. 4.32.


Key Point: Field controlled d.c. motor is open loop while armature controlled is closed loopsystem. Hence armature controlled d.c. motors are preferred over {reId controlled type.

4.15 A.C. ServomotorMost of the servomotors used in low power servomechanisms are a,c, servomotors..

TIle a.c. servomotor is basically two phase induction motor. The output power of a.c.servomotor varies from fraction of watt to few hundred watts. The operating frequency is50 Hz to 400 Hz.

_'I , ,:r' .


4.15.1 Construction

A.C.SupplyC9n~ol T ~ ReferenceW31ndlng~ winding

Control 9(?-: RotorVOI:~ti~ __~servo V-_·

amplifier :

It is mainly divided into two partsnamely stator and rotor.

The stator carries two windings,uniformly distributed and displaced by 90°, inspace. One winding is called main windingor fixed winding or reference winding. Thisis excited by a constant voltage a.c, supply.The other winding is called control winding.It is excited by variable control voltage, whichis obtained from a scrvoamplifier. This

voltage is 90" out of phase with respect to the voltage applied to the reference winding.This is necessary to obtain rotating magnetic field. The schematic stator is shown in theFig 4.33.

Fig. 4.33 Stator of A.C. servomotor

4.15.2 RotorThe rotor is generally of two types. The one is usual squirrel cage rotor. This has small

diameter and large length. Aluminium conductors are used to keep weight small. Itsresistance is very high to keep torque-speed characteristics as linear as possible. Air gap iskept very small which reduces magnetising current. This cage type of rotor is shown withskewed bars in the Fig. 4.34 (a). The other type of rotor is drag cup type. There are twoair gaps in such construction. Such a construction reduces inertia considerably and hence

Aluminiumbars

Shaft

(a) Squirrel cage rotor

Drag

(b) Drag cup type rotor

Fig. 4.34such type of rotor is used in very low power applications. The aluminium is used for thecup construction The construction is shown in the Fig. 4.34 (b).

4.15.3 Torque-speed CharacteristicsThe torque-speed characteristics of a two phase induction motor, mainly depends on

the ratio of reactance to resistance. For small X to R ratio i.e. high resistance 10\\ reactance

_\1 , ,:r I,


Large ~ TorqueeNm)

Control voltageE24 > E23 > E22 > E21

t XSmall"R

Speed (r.p.m.)- Speed

Fig. 4.35 Fig. 4.36

motor, the characteristicsis much more linear while it is nonlinear for large X to R ratio asshown in the Fig. 4.35.

In practice, design of the motor is 50 as to get almost linear torque-speedcharacteristics. The Fig. 4.36 shows the torque-speed characteristics for various controlvoltages. The torque varies almost linearly with speed. All the characteristics are equallyspaced for equal increments of control voltage. It is generally operated with low speeds.

4.15.4 Features of A. C. ServomotorThe a.c, servomotor has following features :i) Light in weight. ii) Robust construction. iii) Reliable and stable operation.

iv) Smooth and noise free operation. v) Large torque to weight ratio. vi) Large R to Xratio i.e, small X to R ratio. vii) No brushes or slip rings hence maintenance free. viii)Simple driving circuits.

4.15.5 ApplicationsDue to the above features it is widely used in instrument servomechanisms, remote

positioning devices, process control systems, self balancing recorders, computers, trackingand guidance systems, robotics, machine tools etc.

4.15.6 Transfer Function of A.C. ServomotorThe various approximations to derive transfer function are,(i) A servomotor rarely operates at high speeds. Hence for a given value of control

voltage, T oc N characteristicsare perfectly linear.(ii) In order that T oc N characteristicsare directly proportional to voltage applied to

its control phase, we assume T c>c N characteristics are straight lines and equallyspaced.

Torque at any speed 'N° is,

_\1 , ,:r I,


T. K E demm = tm 2t + m"""'dt

where, d:; is speed of motor.

If load consists inertia 1m and friction Bm we can write,

Tm(s) = 1m 52 Om+ Bm sOm

Now Laplace transform of equation (1) is

Tm(s) = KtmE2 (s) +m sem(s)Equating equations (2) and (3)

... (1)

.... (2)

... (3)

:. Kim E2 (s) +m sOm (s) = 1m S2 Om(S)+ BmsOmes)Om(S) Ktm= =E2(S) s(slm - m + Bm)

Om (5) KmE2 (S) =.. s(l + 'tm s)

where Km = KtmBm-m

and 'tm = 1mBm-m

2S em sem

~ 00-- ....--\1iSJo=-_...--o-:;.;._---t ..----ooemKit ~ 1/s

(m-Bm)Jm

Fig. 4.37 Signal flow graph of a.c.servomotor

[sJm]s (Bm - m) 1+ (Bm_ m)

Kay Point: As slope is negative, in theabove equation [Bm - m J shows that totalfriction increasesdue to m. As it adds morefriction, the damping improves, improvingstability of tire motor. This is allied InterJUllElectric Damping of 2 ph A.C. servomotor.

Fig. 4.38 Block diagram of a.c. servomotor

_\1 , ,:r I,


Signal flow graph for A.c. servomotor is as shown in the Fig. 4.37. Hence blockdiagram of A.C. servomotor is as shown in Fig. 4.38

4.16 Comparison of Servomotors

4.16.1 Comparison between A.C. and D.C. Servomotor

Sr. A.C. Servomotor D.C. ServomotorNo.

1} Low power output of about Deliver high power output12 W to 100 W.

2} Efficiency is less about 5 to 20 %. High efficiency.

3) Due to absence of commutator Frequentmaintenancerequired due tomaintenance is less. commutator.

4) Stability problems are less. More problems of stability.

5) No radio frequency noise. Brushes produce radio frequency noise.

6} Relatively stable and smooth operation. Noisy operation.

7) A.C. amplifiers used have no drift. Amplifiers used have a drift.

4.16.2 Comparison between Armature Controlled andField Controlled D.C. Servomotors

Sr. Field Controlled Annature ControlledNo.

1) Due to low power requirementamplifiers High power amplifiers are required toare simple to design. design.

2) Control voltage is applied to the field. Control voltage is applied to the armature

3) Time constant is large. Time constant is small.

4) This is open loop system. This is closed loop system.

5) Armature current is kept constant. Field current is kept constant.

6) Poor effiCiency. Better efficiency.

7) Suitable for small rated motors Suitable for large rated motors.

8) CosUyas field coils are must Permanentmagnet can be used insteadof field coils which makes the motor lessexpensive.

4.17 Models of Commonly used Electromechanical SystemsIn this article the transfer functions of very commonly used systems arc derived. This

will help the reader to find out the transfer functions of the different practical systems.

_\1 , ,:r I,

Control System Engineering 4-31

4.11.1 GeneratorsConsider a separately excited generator

which is many times used in various practicalmechanical systems. Generators are required todrive the motors because vacuum diodes,transistor amplifiers are not suitable because oftheir low ratings. Consider a generator as shownin the figure.

Rf = Field resistance

Lf = Field inductance

ef = Applied voltage (input)

eg = Generated voltage (output)

Now for a generator,

where. = fluxFlux is directly proportional to current passing through the field winding say if.

Let K, be the generator constant inV/ A

Applying Kirchhoff's Law to field circuit.

. R L diref = If r + r dt

Taking Laplace of both the equations (1) and (2)

Mathematical Modeling of Systems

IrrFig. 4.39 Generator

Eg(s) = Ka I!(s)

E,(s) = Rf If(s) + Lr(s) It(s) neglecting Ir(O)

If(5) Ef(s).. :::Rr + s L,

. Eg(s) =x, Ef{s).. Rr + s Lf

.. Eg(s)= Ka

E,(s) Rf + S Lf

This is the T.F. of separately excited generator.

... (1)

... (2)

_\1 , ,:r I,

- Control System Engineering 4-32 Mathematical Modeling of Systems

4.17.2 Generator Driving Motor. It is very common to find generator driving motor in practical mechanical systems. So .let us discuss the T.F. of a system with generator driving motor.

Fig. 4.40 Generator driving motor

Rf and L, - Resistance and inductance of generator field

eg - Generated voltage

Kg - Generator constant in V/ A

R, and La - Resistance and inductance of motor armature

J - M.I. of Load and f is frictional force

Now. T. F. of generator is,

Eg(s) KgEf(s) = Rf + 5 Lf

Now consider armature controlled motor,Torque produced by motor is dependent on ia and let I<T be torque constant.

T = KT ia

This torque is utilised to drive a load.

T = Jd20 + f dO.. dt2 dt

Now Eg is voltage applied to armature.

p_ -_ I' R + L dia + eb---g a a a crt

where eb is back e.mJ.dO

es ee c.o ee Cit

_\1 , ,:r I,


Let Kbbe the back e.m.f. constant,

eb = Kb d6dtd20 dO

KT ia = J dt2 + f (ft.


KT la(s) = Js26(s)+fs6(s)

~. . R L diil v. dO""g = Iii a + a Cit + ''I> dt

... (3)


Eg(s) = la(s) Ra + Lol s la(s) + Kto s O(s) ... (4)

finding 1.,(s) from equation (3)

(J s2 + f s) O(s)1.,(5) =

Substituting in equation (4)s6(s)(Js+f) ,

~g(s) = K (R, + S La) + K" sO (s)T .

Eg(s) = O(S)s[(f+fS)C:;+SLa)+Kh]

Eg (s) _ S Kb 6 (s) = s (Ra + S LoI) (s J + f) 6 (s)Kr

6(s) = s (Ra + SLa )(5 J + f)

1sJ + f I---'IP-" 9(s)

Fig. 4.41 •...

_\1' ir I


Tnerefore using both the transfer functions of generator and armature controlled motorwe can develop the combined block diagram as below :

~R,+ sL,

1 t--4,__... 9(s)sJ+ fE,(s)

Fig. 4.42Reducing the block diagram, solving feedback loop of motor which is

KT

= s (Ra + S La) 0 s + f)1+ sKt, KT

s(R, + S La) 0 S + f)

KT= -s"';;(R'a-+-s ";""""L'a)"O";""""s-=--+-f)+ s Kb KT

=

6(s) KT KgEf (8) = -:::(R=-f-+-s-L=-,-:'")-;:{s~(:-(=-R-a -+-s-=-L-a:;_)-::O:-S-+~f)=--+ -=-K=-b-=-K-=-T--:::;"l}

4.17.3 Position Control SystemAnother very common system used in practice is position control system. This is used

to control position of shaft, by use of potentiometer as error detector. The error is to beamplified by amplifier and then must be given to armature controlled motor whose shaftposition will get controlled as per the controlled signal The motor shaft is coupled to theload through gearing arrangement with ratio Nt/N2.

Load has M.I. J and friction as f while 6r is the reference position while all is theactual position of shaft.

The circuit diagram can be drawn as below :La

--Gear

e =error~l"const.l

Fig. 4.43 Position control system_\1 , ,:r I,


Let Kp be the potentiometer sensitivity, in V /radThe corresponding block diagram can be drawn as shown in the Fig. 4.44.

1sJ+f

Fig. 4.44Reducing the block diagram as shown in the Fig. 4.45.

9r(S)

G(s) =Fig. 4.45

KKp KT (Nl/N2)

K Kr KT (Nl/N2) = K, = System gain constant

H(s) = 1

r. Over all T. F. can be calculated as below,

Kss[(Ra +sLa)Us+f)+KbKT]

1+ I<ss [(Ra + S La) Us+ f) + Kb KT 1

6a(s) =Ores) s[(Ra + s La}Us+ f)+ Kb KT J+ K,

4.17.4 Position Control with Field Controlled MotorIn the above case if instead of armature controlled motor, field controlled motor is

used then derive the T.F. of overall closed loop system.Consider field controlled motor,

Now T De 41 ia and i" is constant

_\1 , ,:r I,

Control System Engineering 4-36 Mathematical ModeUng of Systems

ia= constant

Fig. 4.46

Let Kf be constant in N-m/ A

This torque is utilised to drive a load of moment of inertia 1and friction f.

• d20 deT = 1 dt2 + f dt

. K, if = 1 d26 + f d6dt2 dt

Finding Laplace of all the equations,

Ef(S) = I,(s) Rf + s Lf If(s)

and K, If(s} = J 526(5) + 5 f6(s)

If(s)s6(s) [51+ f).. = Kf

Substituting in Ef(s)

E,(s) s 6(s) [5J + f] [Rf + s L,]= Kf

6(8) Kf.. ' E,(s) = s [s J "'"fHRf + s Lf INow using the same block diagram as derived in case (iii) replacing armature

controlled motor T. F. by the field controlled we can get the new block diagram.

Fig. 4.47

_\1 , ,:r I,

. Control System Engineering 4-37 Mathematical Modeling of Systems

Let K!' = K Kp Kf (~;) .

G(S) K,. H(s) = 1- sCsI+ f)(Rf + sLf)

G(s) s(s] + f) (Rf + sLf~-=---:::::::-:~=-:-=-....;_;__....:.....:~-____:_;___1+ G(s)H(s) 1+ K~

s (s } + f) (R, + s Lr)

0'1(5)=OrCS) s (s J+ f) (R, + S Lr ) + K,

4.17.5 Speed Control SystemIn s~me of the practical applications it is necessary to drive a load at a desired speed

(l)rad/sec. For this application an electromechanical system can be used which is' shown asbelow. .

I Constant

RaV

Tachometer

-Fig. 4.48 Speed control system

System uses armature controlled motor and tachometer feedback. Let I<A be theamplifier gain. Let us derive the transfer function of this·system. The objective of system isto move the load at a. desired speed.

The d.c, tachometer output voltage is proportional to output speed (l).er is reference voltage and et is tachometer voltage.

error

('a = K" X e = KOI (e, - et )

e, = Kt (I) where K, is constant.

.~.(5)

... (6)

Neglecting inductance of armature

_\1 , ,:r I,

Control System Engineering :. 4-38 Mathematical Modeling of Systems


substituting from equation (6),ia Ra + Kb m~ f;J

K, e, - K, K, m = ia Ril + Kb10Taking Laplace

K, Er(s) = (K" K, + Kb) m(s)+ iii (s) R..

Now torque produced by motor

--- (7)

T DC ia

•._(8)

Now this drives a loaddBwhere m=Cft ___(9)

r, Taking Laplace

KT iii (s) = Jsm(s) + f m(s)

U s+ f)""---=;----" + f m( s)KT


ia (s) = ___(10)

U s+ f]m(s)= (K, K, + Kb )m(s) + KT Ra

(K...Kt + Kt. ) m(s) Us + f]m(s) RdEr{s) = + --=--=c:;---x, x, KT

m(s)=Er(s)

nus is the required transfer function of the speed control system.

_\1 , ,:r I,

Control System Engineering! 4-39 Mathematical Modeling of Systems

4.17.6 Speed Control using Generator Driving MotorA position control system described in Fig. 4.49 in which the armature of motor is

applied with a control voltage through a generator. The field current of generator controlsthe voltage generated by the generator.

R

L Load. -\--o::JJ1Fm

Om


Let us determine the transfer function of the systemInput is e, (t).

Now d(O)ce(t) = l'l(t) - Kbdt ... (11)


Ee(s) = EI (5) - Kb I>0 (5)

Now E () K L di, R·ct· 2 = ((ff+ I If ... (12)

applying Kirchhoff's law to the field circuit of generator.

Now e.1 = K, il where K" is constant.


E" (s) = K" If (s)

Taking Laplace transform of (12)

Ee(s) K2 = If (s) [Rf + sLf ]

Eliminating If (s)

... (13)

Ec(s) K2(Rf + sL,)

E.1(5)K" Ec(5) K2= (R, + 5L()

E,,(s) Ka K2Ec(s) = (Rt + sL()

_\1 , ,:r I,

Control System Engineering ..\: '.. !i-.. 4 - 40 Mathematical Modeling of Systems.j

Now consider fieJd constant armature controlled motor. Armature is energized byoutput of the generator.

Applying Kirchhoff's Jaw to armature circuit,,

Eil(s) = Eb(s)+Ia(s)[Ra+sL",]

Back e.m.f. is proportional to em

Eb{S) = K dOmb(ft

Taking Laplace Transform

Eb(S} = Kb sOm(s)

In armature controlled d.c. motor as field current is constant so torque produced isdirectly proportional to t1iearmature current.

T = K I a(s) where K is motor constant .

.This torque is utilized to drive the load with moment of inertia J and friction Fro.

T = J d20m + Po dOm Taking Laplace transform fromdt2 m dt

K1a(s) = Js20m(s)+sFmOm(s)

t, (s) = ~ (ls + Fm)Om,(s). .

Substituting in equation f~r E, (s)

EOl(s)

9m(s) K..Ea(s) = sKKb+s(Ra +sLa)Us+ Fm}

= Km ... Neglecting L. of armature5[1 + Tros]

Km K= [KKb+ R, ,Fm1= Motor constant

Tm Raj T f t= IKK Po 1= une constant 0 mo orb+R", m /

Neglecting 'La' of armature motor,

ames} _ KmEa(s)·- s(l+Tms)

_\1 , ,:r I,

r~"- Control System Engineering ~n"iJ~ttw 4 - 41 Mathematical Modeling of Systems

Ea(s) x, K2Ec(s) = JRr + s Lr]

E"\(s)K" K2 Ec(s) Om(s}s (1 + Tm s).. = =(R, + s Lf) Km

and Ec(s) = El (s) - Kb sOm

K" K2 lEI (s) - Kb sam (5)] Om (5) 5(1 +r;5).. -----.~~~~--~ =[R, + S L,)

x, K2 Km [ x, x, K., Km S][R, + s Ltl EI (5) = Om (S) s(l + Tm S) + (Rf + 5-Lf)

x, K2 x, a (s) [Rr s (1+r,s) (1 + Tr s) + Kb Kit K2 Km s].. [1 + sTd EI (5) = m [1+ Tf(s)]

Om(S) Ka K2 KmEj(s) = S [K, Ka K2 Km + Rf (1 + Tm s) (1 + Tf s)]

Om(S) =Ej(s)

Kwhere Kg = R; generator constant.

4.17.7 A Typical Position Control System used in IndustryA position control feedback system has a potentiometer bridge with a sensitivity of

KpV/radian as error detector. It feeds a d.c. amplifier with an open circuit gain K. ThisSupplies to a field of generator which has resistance Rf and inductance Lf. Generatorconstant is KgV/ A. .This is connected to an armature controlled d.c, motor with motortorque constant KT and back e.m.f constant Kb V/rad/sec. It drives a load of inertia J andfriction f.

Draw the simplified block diagram and hence calculate its transfer function.

Fig. 4.50 Typical position control system

_\1 , ,:r I,

Control System Englnearing 4-42 MathematicalModeling of Systems

e r = Reference position , ea = Actual position

T. F of generatorEg (5) KgEf(s) :::::Rf + S Lf

Block diagram can be drawn from the analysis of the different cases discussed.

1sJ + f ..........,_.. 98(s)

Fig. 4.51

Neglecting the inductance of armature winding of motor.

Fig. 4.52

G(s) = x, K ~ KTH{s) = 1SeRf + SLf) [Ra (Js+ f) + Kb KTJ

Let

T.F =

9a(s) =6r(s)

4.18 D.C. Motor Position Control SystemIn industry to conrol the position of the shaft, a d.c, motor position control system is

commonly used.

_\1 \ ,:r'


4.18.1 Transfer Function of D.C. Motor Position Control SystemConsider the D.C. position control system which is controlling position of the shaft.

Assume that the input and output of the system are the input shaft position and outputshaft position respectively.

Assume following system constants,

r Angular displacement of the reference input shaft

c = Angular displacement of the output shaft

e = Angular displacement of the d.c. motor shaft used

K) = Gain of potentiometric error detector

Kp = Amplifier gain

e", = Applied armature voltage

eb = Back e.m.f.

Ra = Armature winding resistance

La = Armature winding inductance

ia = Armature winding current

~ = Back e.m.f, constant

K = Motor torque constant

Jm = Moment of inertia of motor.

bm = Viscous friction coefficient of motor

}L = Moment of inertia of load

bL = Viscous friction coefficient of load

n = Gear ratio Nt/N2

~constant

Fig. 4.53 D.C. Motor position control system_\1 , ,:r I,


Equations describing above system can be written as follows :Output shaft position is to be controlled so as to keep that at required position. Output

is sensed by angular displacement 'c' and compared with input which is r. Error isamplified by amplifier with gain 'Kp' and given as input to the d.c. motor which in tumcontrols thc angular position of the shaft of the motor '0' which in turn controls outputposition of shaft' c' so as to modify the error.

For potentiometric error detector we can write.

E(s) = KJ [R(s) - C(s)]

For amplifier

E., (s) = Kr E(s)

For armature controlled d,c, motor

if = constant SO flux 41 is constant

T = Kia where K = motor torquc constant

eb ee adO.. Cb = Kb""dt

For armature circuit

ea = L di, . RCb + 1Tt+ la a

Taking Laplace transforms

Eb(S) = l<j, s8 (s)

Ea (s) = Eb(s) + Ia (s)[Ra + s La]

T = KIa (s)

Now torque is utilised to drive load + shaft of motor.

Jeq = Jm+ n 2 JL = Equivalent moment of inertia

beq = bm + n 2 bL = Equivalent frictional coefficient

T d20 dO= Kia (s) = Jeq dt2 + beqdt

= lJ..'q s2 + beqs}O(s) Laplace transform

la (s) 1= KUeq S2 + beqs]O(s)

Ea(s) 8(s)= Kbs8 (5)+ """'j(Ueqs2+ bcqslIRa + SLa]

_\1 , ,:r I,


o(S) K=Ea(s) s Ueqs + bcq](Ra + sola J+ KKb s

K=

'La' is generally small hence neglected.

K=S [KKb + s n, J l'<j + bc.-qn, J

Km= s(l + Tm s)

KKm = (b..-qRa + KKb) = Motor constant

Ra J,"'I .Tm = Ol'<jRa + KKb) = Motor time constant

Now C(s) = nO (5)

C(s) 9(s).. Ea(s) = n--Ea(s)

C(s) nKm=..Ea(S) s(l + Tm s)

C(s) nKmwhere Ea(s) = Kp E(s)=K"E(s) s(l + Tm s)

C(s} = n Km Kp E(s)s (1+ 1""m s)

n Km Kp Kl IR(s) - C(s))C(s) = s(l + Tm S)

[5(1 + Tm 5)+ n Km Kr Kl] n Km Kp Kl R(s)

:. C(s) s (1+ Tm s) = s (1+Tm s)

C(s) =R(s)

nKmKpKt5(1 + Tms)

nKmKpKt1+--:-=~;;;-"7s(1 +Tms)

G(s)=-....,...-;."..;-_

l+G(s)H(s)

_\1 , ,:r I,


Block diagram :

R(s)

Fig. 4.54

'. Example 4.3: A d.c. motor drives Q pointer which is spring loaded, to retum to thereference position. If Kb = Back e.m.f. constant, Kr = Torque constant K~ = Spring constantand J =Moment of inertia. Find the transferfunction.

1

fo-----.(M.U. : Nov.-93, Nov.-94)

Solution: Writing the system equations.

Va(t)ar,= I a RII+ Ladt+ s,

Taking Laplace,

v,(5) = Eb(S)+.1" (5) [RO!+ SLa]

r,(5) v,(5) - Eb(S).. = (Ra + sL.,) •

Now Tm = KTIa, Tm = Motor torque

and Eb(S) Kb s8(s) as Eb de= oc -dt

Ia (s)Va(s)- 8(5).. = (R, + S La)

and Tm(s) = KT [Va (s) - KbSe(S)].(Ra + 5La)

..• (1)

•... (2)

.•• (3)

•.. (4)

.. (5)

_\1 , ,:r I,


This torque is used to drive a pointer with inertia J and spring load of constant K",.d20(t)

Tm = Jdt2 +K 0 (l)

Taking Laplace

Tm(s) = J 520(5) + K, O(s)


...(6)

KT [Va(s) _ Kb sees)]:. Js2 O(s) + K:. O(s) =

RII + sl.,

2 KTKbS9(S)_ KT .J s 6 (5) + x, 0 (5) + (R L ) _ (R L ) V.,(S)_,+5 a _,+S .1

So transfer function is,0(5)

52 J (Ril +sLa )+K, (R, + sLa) + Kr K...sKT

=Va(S)

9(s)=Va (5)

,_ Example 4A: A d.c. generator supplies its output to a separately excited d.c. motor. Thefield current of the motor is constant. The voltage applied to the field circuit of d.c. gene!atoris eI (t). Write the differential equations of this Ward-LeotUlrdsystem relating the inpuief (I)to the output 9m (t). (Refer to figure shown below) Hence obtain expression for 0 III (s).

(M.U.-May-96(PTDC) and April-96(ElectricalJ)

Motor if Constant

Cem

L- --l Ward-Leonard system

Solution: Torque produced by motor,

T = KT Ia(t)

Voltage applied to the motord I,

Eg(t) = r, Ra +L Tt+ Eb(t)

... (1)

... (2)

_\1 , ,:r I,


Back emf is proportional to the angular velocity

E (t) = K dam(t)b b dt

Torque produced T is used to drive a load of inertia J and friction B

T = J dZSm + B dOm.. dt2 dt

... ... (3)

.•. (4)


KTla(t) = J d20m +B dOmdt2 dt

J d1am B damla(t) = KT '(ft2+ KT"dt ... (5)

Substituting in equation (2) we can get resultant differential equation 'relating Eg(t) andOm(t).

Nowar,

Ef{t) = If R, + L, dt ... (6)

... (7)and Eg(t) = .Kg I,(t)

.. Substituting If{l) = ~t) in equation (6) we can gel, equation relating Er(l) and

Eg(t).

Substituting this in equation (2) we can obtain the final differential equation relatingE,(t) and 0met).It is very difficult to obtain in time domain so let us obtain it in Laplacedomain.

Taking Laplace transform of all the equations

T{s) = KTr,(s)Eg(s) = la (5) [Ra + S La]+ Eb(S)

Eb(s) = KbSOm(s)

T (s) = em(s)[Js2 + BS]

.. KTIa(s) = Om(s) [Js +B] S

Ia (5) = Om(s)s[Js+ B].. KT

Eg(s)ames)s[Js + B][Ra + S La].. = KT + KbsOm(s)

and E(s) = 1,(5) [Rr + s Lt]

... (1)

... (2)

..• (3)

... (4)

... (5)

... (6)

... (7)

_\1 , ,:r I,


Eg(s) = Kg If(s)

So If(s) = Eg(s)~

Ef(s) = Eg(s) [Rf + sLr]••• (8).. Kg

Substituting Eg(s) in equation (8) we get

Ef(s) Kg = Om(s) [sOS + B)~~a + SLa + KbS][Rf + SL,l

(Rf + s Lf) {seRa + s La)Os+ B)+SKbKT}

Kg KT Ef(s)Om(S) =

4.19 Models of Thermal Systems

4.19.1 Heat Transfer System

II fsurrounding temp. 9 -

Thermal insulationWater

Outletwa~C: ==~II i '- In!etwater

90°c ~_-====::tFl======::t::==::::J-1 ' .. "cHeater

Fig. 4.55

A thermal system used for heating flow of water is shown below.Electric heating element is provided in the tank to heat the water. The tank is

insulated to reduce heat to the surroundings.The necessary simplifying assumptions are :1) There is no heat storage in the insulation.2) All the water in the tank is perfectly mixed and hence at a uniform temperature.

_\1 , ,:r I,


01 = Inlet water temperature in "C.

00 = Outlet water temperature in"C,

o Surrounding temperature.

q = Rate of heat flow from heating clement in JI sec.qi = Rate of heat flow to the water.

ql = Rate of heat flow through tank insulation.

C = Thermal capacity in JI"C.R = Resistance of thermal insulation.

So rate of heat flow for the water in tank is,ae,

q i = C"""(ii'"" •.. (1)

The rate of heat flow from the water to the surrounding atmosphere throughinsulation is,

... (2)

As per the heat transfer principles,

q = qi + ql

Substituting equation (1) and (2)

CdOo 00-0

q = (ft+-R-

... (3)

... (4)

Neglecting the term O/R from the equation (4) this is because the variation of watertemperature 00 is over and above ambient temperature 0 (1)'

q =CdOo+Oo.. dt R


Q(s}= CsOo(s)+ Oo~s)

:. Transfer function is,~--------------~Oo(s) R=Q(s} l+sCR

The time constant of the system is RC.

_\1 , ,:r I,


4.19.2 ThermometerConsider a thermometer placed in a water

bath having temperature eJ' as shown.00 is the temperature indicated by the

thermometer. The rate of heat flow into thethermometer through its wall is,

dqdi=

Where R = Thermalresistanceof the thermometer wall.

The indicated temperature, rises at a rate ofau, 1 dqcrt = Cdt

where C is thermal capacity of the thermometer.

d:to = b. [Oi ;00 ]Taking Laplace of the equation,

1sO" (5) = RC [Oi(S)-0,,(5)]

0o(s) 1=9j(s) l+sRC

The time constant is RC.

4.20 Actuators

Bath

Fig. 4.56

The actuator is a device which receives input signal hom the controller and itproduces the input signal to the plant according to control signal so that the output willapproach the reference input signal to reduce the error to zero. Thus an actuator isgenerally after the controller and before the plant in the control system. An actuator can beof two types :

i) Hydraulic actuator 2) Pneumatic actuator

_\1 , ,:r I,

Control SystemEngineering 4·52 Mathematical Modeling of Sy.....

4.20.1 Hydraulic ActuatorThe structure of an hydraulic actuator is shown in the Fig. 4.57.

Input--il----+-il------t-t------i--t--t-displacement

Mainpiston

Tosump From highpressure source

To sump Valvepiston

Fig. 4.57

They are piston devices in which motion of the spool regulates the oil flow to eitherside of the power cylinder. When the spool moves to the right, the high pressure oil entersthe power cylinder to the left of the piston.

The differential pressure produced causes the power piston to move to the right,pushing the oil infront of it into the sump.The load coupled rigidly to the piston moves a distance y from its reference position

corresponding to the displacement x of the valve piston from its neutral position. The oil ispressurised by a pump and is recalculated in the system.

Equation of motion and transfer functionThe rate of flow of oil into the piston is proportional to the rate of the movement of

the piston.dy

Q = A dt' A = area of piston .•. (1)

If P is the differential pressure across the piston then the force on the piston is AP.This moves the load of mass M against friction B.

d2 Y dyA x P = M dt2 +B Cit ... (2)

For small values of the displacement X, if P is the differential pressure on the pistonand Q is the oil flow then,

..• (3)

dyKlx-Adt

_\1 , ,:r I,


Taking Laplace transform,K M K.,B+S2 Y(S)+ --j:-S YeS) = K) X(s) -/\ Y(s) s

yes) [s2 K2 M+ K2 Bs-i- A2 s1 = K) A Xes)

yes) KIAXes} = S2 K2M+sK2 B+A2 S

yes) =Xes)

(~2A)

This transfer function is similar to the electricmotors.

4.20.2 Pneumatic ActuatorPneumatic acting valve is used to

obtain linear displacement of a plungerwith pressurised air as input.

The air at pressure P is injectedthrough inlet. Pressurised air pushes thediaphragm and plunger. The plunger hasa mass M and friction on B with springconstant K.

Let A be the area of diaphragm thentransfer function can be obtained asbelow.

Force exerted on the plunger is A X P.This force is opposed by mass, frictionand spring.

Fig. 4.58


A P(s) = MS2 Yes)+ Bs Y(s) + Kyes)

Y(s) A=pes) Ms2 + Bs+ K

The advantages of pneumatic systems are fire proof, explosion proof, simplicity andeasy to maintain.

_\1 , ,:r I,


4.20.3 Comparison between Pneumatic and Hydraulic Systems

Sr.No. Pneumatic systems Hydraulic systems

1. The fluid used Is air. The fluid used is oil.

2. Air is compressible. Oil is incompressible.

3. Air does not have lubricating property. Oil acts as a lubricator.

4. The output power is much less compared to The output power is much higher thanhydraulic. pneumatic.

5. At low velocities, the accuracy is poor. At all velocities, the accuracy is satisfactory.

6. No return pipes are required when air is used. The return pipes are must.

7. Can be operated for the temperature range of It is sensitive to the temperature changes and00 Cto 200 OC. It is insensitive to temperature the range is 20 OC to 70 OC.changes.

8. These are fire proof and explOSionproof. These are not fire and explosion proof.

9. Easy from maintenancepoint of view. Difficult from maintenancepoint of View.

Note : The state space method of modeling the systems is separately covered in thechapter 15.


Example 4.5: The voltage generated by a d.c. generator is filtered by a luw pass filterconsisting Rz and Lz as shan," below. This filtered voltage is controlled by the voltageapplied to the field of genenuor.If Rf = 40 n, Lf = 60H, Ro = 0.5 n, Lo = lH, la = IH, Rz = 2n andgmerator constant Kg = 120 V Ifield Amp. Determine the transfer functionE2(s)EI(s) of the system.

(M.U.: Dec.-97)

r---------R, Ra La

t ~) ~) Te,{t) Lr e2(t)

1 1.._--------Low pass filter

Constant speed

_\1 , ,:r I,


Solution: For field circuit, ... (1)

For armature circuit,

For generator, Ca (l) = Kg ir(l) ... (3)

... (4)For output, ez(t) = iz{t) R2

Taking Laplace transform of all the equations,

Er{s) = I,(s) lR, + s I., ]

Ea (5) = 12(s) [(Ra + R2) + s (La + 1-2)]

s, (5) =: Kg 1,(5)

... (5)

... (6)

... (7)

... (8)E2(S} = 12(S}R2

Equation (6) and (7),

Kglr(s} = h(s} l(RII+R2)+s(L,,+L2)1Substituting la (5) from equation (8) and 1,(5) from equation (5),

, [ Ef{s) ] E2(S)~ R L = -R- l(R., + R2)+ s(La + L2)]'I + S f 2

Substituting 12(S} from equation (8) and If(s) from (5),

E2(S} _ Kg R2Ec(s) - (R, + S Lr) [(K, + R2)+ s(La + L2)]

Substituting the values,r-------------------------------------~E2(s) =Ef(s)

120x2=(40 + 60s) [(2.5+ 2s)J

2.4(1+ 1.55) (1+ U.8;;)

,_ Example 4.6: A high gain speed control system uses a tadtogeneraior for speed.sensing.The tachogenerator produces 5 V per 100 r.p.m. This voltage is compared witll referencevoltJlge to produce error signal. If the reference voltage is set to 10.8 volt. What is the valueof expected speed? (M.U. : Nov.-94)

Solution: The tachogenerator produces 5 V per 100 r.p.m.,

Its constant is 5100 = 0.05 VIr.p.m.

_\1 , ,:r I,

Control System Engineering' I. 4-56 .JMathematical Modeling of Systems .1

Now for the expected speed, error should be zero,

i.e Vr - tachogenerator output = error

Tachogenerator output = Vr

Let N be expected speed in r.p.m.

N x 0.05 = 10.8

N = 216 r.p.m. ... Expected speed

I,... Example 4.7: An armature controlled d.c. motor has Rtf armature resistance of 0.37 n.The moment of inertia is 2,5 x 10-6 kg - m2• A back emf of 2.09 V is generated per 100r,p,m. of the motor speed. Tire torque constant of the motor is 0.2 N-m IA. Determine thetransfer function of the motor relating the motor shaft shift and the input voltage.

(M.U. : Dec.-96)

Solution : The motor can be shown as,

Ra G:-1 ;.(1')Va(t) load

1 J

Now back emf is 2.09 V per 100 r.p.m. , N = 100 r.p.m. is,21tN

(1) = ""60 = 10.4719rad/sec

:. Back emf constant x, = lO~40~9= 0.1995V/rad/sec

KT = 0.2 N-ml A

For armature circuit Va(l) = Ia Ra + Eb

Vats) = Ia(s) Ra +Eb(S)

... (1)

Taking Laplace

... (2)

Taking Laplace Eb(S) = KbS6 (s)

Tm = KT loa(t) ... (3)

_\1 , ,:r I,


Taking Laplace 1".n(5) = Kr I., (s)

This torque drives a load of inertia J.

Taking Laplace Trn(s) J s2 9 (5)

.. Klla(s) = J S29 (s)

fa (s)2.5 X 10-6

52 9 (s) = 1.25 X 10-5 S2 9 (s)=.. 0.2

.. V.,(s) = 1.25 X S2 10-59 (s) X 0.37 + Kb 59 (s)

.. Va(S) = 4.625 X 10"" S29(5) + 0.1995 s9(s)

.. 9(s) 1 216216.22= =Va(s) 4.625 X 10-6 S2 + 0.1995 s s(s + 43135.135)

..... E I h h bid . £,(5)II...,... xamp e 4.8: For t e system s own e ow eiermine E- ( )._f 5

ett)

Separately excited d.c. generator

Solution:

For field circuit,di(

ef (t) = if (t) Rf + L, dt

... (4)

(M.U. : Dec.-97)

For armature circuit,

For generator,

For output,

Ca (t) = Kg if(t)

C2(l) = i2(t}R2

_\1 , ,:r I,


Taking Laplace of all the equations we get,

Et(s) = If(S) [Rf + s Lr]

E,,(s) = I;2(5) [(Ra + R21+ s(La + L2)] = Kg Ir(s)

E2(S) = 12 (5) R2

Hence equating Ea (s) equations,

Kg If(s) = 12(s) [12 (s)(Ra + R2. + s(La + L2>]

and hence using values of If(s) and 12(s}from remaining equations we get,

'.. Example 4.9 : A 50 Hz, 2 phase a.c. servomotor has tile following parameters :

Starting torque = 0.186 NmRotor inertia = 1xlO-5 kg-m2

Supply voltage = 120 V

No load angular velocity = 304 rad/sAssuming straight line torque-speed characteristics of the motor and zero friction, obtain itstransfer junction. (Gate)

Solution: The starting torque is nothing but locked rotor torque.Lockedrotor torque 0.186

Ktrn = Ratedvoltage = 120

= 1.55xl0-3

Let m be the slope of lincarised torque-speed characteristics.

Lockedrotor torque 0.186No load angular speed = 304m =

= - 6.118xlO-4

The torque at any angular speed (J) is given by,. dB

Tm = Kim E21 +md't ... (1)

_\1 , ,:r I,


where £21 = rotor voltage

Taking Laplace, Tm(s) = Kim E21(s) + m sO (s)

This torque is used to drive load of inertia 1m- The friction is given zero.

... (2)

" d20• Tm.. j == 1m dt2

.. Tm(s) = s2JmO(s}

Equating (2) and (4),

KlmE21(S)+ msO(s) = s2JmO(s)

.. Kim E2I(s) = sO(s) IsJm -m]

... (3)

... (4)

Hence the transfer function of the motor is,

1.55xlO-3= s[lxlO-5 s-(-6.118xl0-"»)

O(s} =E2t(s)155

s(s+61.18)

This is the required transfer function.J. Example 4.10: A two phase a.c. servomotor having a torque constant of 0.045 Nm/Vcontrols a position load through a gear ratio of 10:1. The effective moment of inertia andcoefficient of viscous friction referred to load side are 0.25 kg-m2 and 1.0 N-m/(rad/sec). Thesynchro error detector produces an error signal of O.lV per degree error in misalignment.Develop the block diagram representation of the control system and there from obtain thetransfer function. (M.U. : May-99)

Solution: For the error detector, KJ = 0.1 V/ degree error

.. Kl = 0.1x 1801t

= 5.7295V /radian

Kim = Torque constant

= 0.045Nm/V

NJ 1N2 = 10

JL = 0.25 kg ..m2 on load side

(N r 1.. Jeq = Motor side = N: x It, = 100x 0.25

_\1 , ,:r I,


= 2.5x 10-3 kg-m2

BL = 1 N-m/(rad/sec)

Beq Motor side =(~~rx B,1.. = = 100xl

= 0.01 N-m/ (rad/sec)

The block diagram of the system is,

NtG(5) = Kt x motor T.F.x N2

For motor, Tm = KIm E2

.. Tm(s) = Ktm E2 (5)

and Tm(t) = d2e deJeq dt2 + Beq dt

.. Tm(s) = (s2 Jcq + s Bcq)e(s)

... (1)

... (2)

Equating equations (1) and (2),

Kim E2 (s) = (s2 Jcq + s Bcq)e(s)

O(s) = Kim _ 0.045 _ 18E2(5) s(sJeq + Beq) - s(sx 2.5x 10-3 + 0.01) - 5(5+ 4)

G(s) = S.7295x 18 x..!...= 10.3131s(s+ 4) 10 S(5+ 4)

10.31315(S+ 4)

1+ 10.3131S(5+ 4)

10.3131=......"....----S2+ 4s+ 10.3131

_\1 , ,:r I,. ,


Review Questions1. Explain the derivation of tm410gous networks using

i) Force-voltage ii) Fora-current analogy2. Write a short note 071 direct and inverse analogous networks.3. Draw the tmIllogous electrical nehvorks based on

a) F- V analogy b) F - I analogy of the following mechaniclll systems

B \

I-_f(t)

K81

_\1 , ,:r I,


4. Distinguish between A.C. servomotor and D.C. servomotor.5. Derive transfer /unction of a.c. servomotor stating the tlSSumption made.6. State the applications of a.c. SmJOmotor.7. Derive the transfer /unction of field controlled d.c. seroomotor.8. Ansrver the following gimng reasons :

a) AC. seroomotor has a smaller diameter and more length.b) Field controlled D.C. strVOmotor is prefemd than armature controlled.

9. Develop block dillgram for armature controlled D.C. servomotor and find its transfer function.10. An armature controlled d.c. motor is supplied in series with a resistance from II

24 V d.c. supply. Themotor takes current of 5A on stalling and the stall;,'g torque bei"g 0.915N-m. The motor runs at 1000 r.p.m. taking a current of 1 A. The value of I = 4 X10-3 Kg • ",1and friction constant as 1.5 x 10- 3 Nm/ (rDd/sec)

o (s) 2.9(Ans.: V(s) = $(1+ 0.3s)'

11. for the closed loop system shoo'll below, draw the l'lock diagram and determinetransfer fundion O~(s)/O,(s). The give" values are Error defector gain K.. = 8 V/rad, Amplifiergain KA = 10 V/\ R, = 5 n, Lt = 0.25 H, Kr = 0.05 Nm/A, Jm.,'or = 0.02 Kg - m2,

Determine the transfer /unction of the motor.

h, = 3 Kg - m2• fOlotor = 0.03 Nm/(rad /sec).fL = 5.5 Nm/ (rad/sec)

la Constant+

0(1 ....(AN.: Oc(s) = 32.12 )

9,,(5) (sl + 21.551 + 30.35 + 32.12)

12. The moment of inertia Jm and the coefficient of viscous friction fn. for IIfield controlled d.c. motorare motor respectively 5 x1O-' kg -m2 and 12.5xlO-'Nm (tad/sec). The motor torque constantI{, being 2.5 Nm/A. Determine the trdnsferfunction rtlating th4 angular speed of the shaft and

the field current. (Ans.: ~~ '" 1!~~J

_'I , ,:r' ,

Control System Engineering Mathematical Modeling of Systems4·63

13. Figure given below shows a block diagram oj speed regulator system. The accelertlting t01'que is thedifforence between ti,e torque developed by the controller and load TL. The controller gain is0.00102 Nm/rad lind the tachogenerator constant is 0.191 V/rad/sec. TIll! load speed is adjllSted to1000 r.p.m. The moment of inerti« is J and frictiim is negligi/JIe.Calcullzte a) TIll! rtference voltage. 1,) Tile speed if a constant load torque oj 0.001 Nm issuddenly applied. (Ans.: a) 20V b}951r.p.m.l

+V,.(s)

14. An instrument servo conSisting of motor, spring loaded shaft, etc is SJr(¥w1I below

R=1rad +

Where V = Voltage in volt"R = Motor resisumce 1 Q

L = Motor inductance 0.1 HK" = Al1Ipr~fiergaitl = 10 V/VK = SprirrK COflsttlflt = 0.001 Nm/radK" = Rack em]. rOflstard == 0.01 V (rtld/sec)

KT = Torque content = 0.01 N-m/AJ = Moment of inertia = 0.005 N-m-s2

If the input is 1 TildfullS, wlUlt lS the steady state error? 1 1(Ans. : e...= -[ K;1(:.-]=1(1)

1+(Kx It)

_\1 , ,:r I,


15. The positions serrxnneclurnism is shown in figure below. 6r is refmnce position and 9c: actualtmgular displtzttmmt 0/ shaft in radUms. Obtmn its closed loop trtJns/tr function ~~:~

+V-=-

I.~

If constant!lGive" a = angular displacement 0/ the motor shaft

K, = gain / error detector = 7.64V/radRa = 212, lu = mgligibleKb = 5.5 x1Q-2V/rad/secKT = 6x10-S Nm/AJm 1 xlO-~ Kg - rn2

fm = mgligibleJL = 4.4 X 10-3 Kg - m2

ft. = 4 X 10-2 Nm/ (rad/sec)

Nt = 1N2 = 10

(Ans.. ac(s) <= 42.3 )• EMs) 52 +,7.7 s + 42.3

16. Doive the transfer function of a typical d.c. position control system.17. Derive the trlmS/tr function of a typical d.c. speed control system.18. Obtain the PIUlfhenurticalmodel oJ heRt transfor system.19. Obtllin the rrurtlrmrllficlllmodel 0/ thermometer.20. W1ud is actuAtor ? Explain hydrtlUlit and pneumatic actuator systems.21. Compttre hydrllulic and pnerlmlltic actuators.

(J(JO

r "

Block Diagram Representation

5.1 BackgroundIf a given system is complicated, it is very difficult to analyse it as a whole. With the

help of transfer function approach, we can find transfer function of each and everyelement of the complicated system. And by showing connection between the elements,complete system can be splitted into different blocks and can be analysed conveniently.nus is the basic concept of block diagram representation.

Basically block diagram is a pictorial representation of the given system. It is verySimple way of representing the given complicated practical system. In block diagram, theinterconnection of system components to form a system can be conveniently shown by theblocks arranged in proper sequence. It explains the cause and effect relationship existingbetween input and output of the system, through the blocks.

To draw the block diagram of a practical system, each clement of practical system isrepresented by a block. The block is called functional block, It means, block explainsmathematical operation on the input by the element to produce the corresponding output.The actual mathematical function is indicated by inserting corresponding transfer functionof the element inside the block.

Key Point: For a closed loop systems, the function of comparing the different signals isindicated by the summing point while a point from which signal is taken Jor the feedbackpurposeis indicated by takeoff point in blockdiagrams.All these summing points, blocks and takeoff points then must be connected exactly as

per actual elements connected in the practical system. The connection between the blocksis shown by lines called branches of the block diagram. An arrow is associated with eachand every branch which indicates the direction of flow of signal along the branch.

Kay Point: The signal can travel along the direction of an arrow only.It cannot pass against the direction of an arrow, Hence block diagram is an unilateral

property of the system.

(5 - 1)

r "

Control System Engineering 5·2 Block Diagram Representation

In short any block diagram has following five basic elements associated with it :1) Blocks 2) Transfer functions of elements shown inside the blocks.3) Summing points 4) Takeoff points 5) Arrows.

5.1.1 Illustrating Concept of Block Diagram Representation

Pneumaticvalve

(!) Controller

Fig. 5.1 Liquid level control systemFor example : Consider the liquid level system as shown in the Fig. 5.1. So to

represent this by block diagram, identify the elements which are,i) Controller (ti) Pneumatic valve (iii) Tank (iv) Float.

In this system, the level of water is sensed by the float. Hence the float position acts asthe feedback. According to the float position, with respect to desired level of water, thecontroller operates the pneumatic valve controlling the flow of water in the tank. When therequired level is reached, controller operates the pneumatic valve in such a way that theflow of water in the tank, stops. U the output from the tank is taken i.e. the water fromthe tank is drained then the float position changes from the desired position andaccordingly the controller operates the pneumatic valve to start the flow of water in thetank.

Hence indicating all the elements by blocks, the block diagram of the system can bedeveloped as shown in the Fig. 5.2.

FOlWard pathg_. %.... up .x-__ :>

Actuallevel

Summingpoint

< Feedback path-i- - I

Fig. 5.2 Block diagram of liquid level control

r "

Control System Engineering 5-3 Block DiagramRepresentation

Consider another example of bottle filling mechanism. When bottle gets filled by thecontents upto the required level it should get replaced by an empty bottle. This system canbe made closed loop and hence can be as shown in the Fig. 5.3.

.---- .....----, Valvecontrollingsignal

Tank

sensor beltFig. 5.3 Automatic bottle filling mechanism

In the system shown, conveyor belt is driven by the controller as well as valveposition is also controlled by the controller.

When empty bottle comes at the specific position, weight sensor senses the weight andgives signal to controller. Controller stops conveyor movement and opens the valve sobottle starts getting filled. When required level is achieved, again weight sensor sensingthe proper Weight sends a signal to controller which sends signals to start movement ofbelt and also dosing the valve position with proper time delay till next empty bottlecomes at the proper position.

This system can be represented by a block diagram as shown in the Fig. 5.4.

RaJWesign

Comparator Positionof valve

/ atthe·tank

erence~ FilledboController r.-

Ight81 -

Belt drivingmechanism F

Weightsensor

ttle

eedback


r "

Control System Engineering 5.4 Block Diagram Representation

5.1.2 Advantages of Block DiagramThe various advantages of block diagram representation are,1) Very simple to construct the block diagram for complicated systems.2) The function of individual element can be visualised from block diagram.3) Individual as well as overall performance of the system can be studied by using

transfer functions shown in the block diagram.4) Overall closed loop T.F. can be easily calculated by using block diagram reduction

rules.

5.1.3 DisadvantagesThe various disadvantages of block diagram representation are,1) Block diagram does not include any information about the physical construction of

the system.2) Source of energy is generally not shown in the block diagram. So number of

different block diagrams can be drawn depending upon the point of view ofanalysis. So block diagram for given system is not unique.

5.2 Simple or Canonical Form of Closed Loop SystemKey Point: A block diagram in whic/l,forward path contains only one block,feedbackpathcontains only one block, one summing point and one takeoff point represents simple orcanonical form of a closed loopsystem.~ can be aclUeved by umng

block diagram reduction rules R(s)without disturbing output of thesystem. nus form is very useful as itsclosed loop transfer function can beeasily calculated by using standardresult. This result is derived in this

E(s)G(s) 11---.--+ C{s)

+8(8)

H(s)

section.The simple form can be as shown

in the Fig. 5.5.Fig. 5.5

where , R(s) -+ Laplace of reference input r(t)

C(s) -+ Laplace of controlled output c(t)

E(s) -+ Laplace of error signal e(t)

B(s) -+ Laplace of feedback signal b(t)

G(s). -+ Equivalent forward path transfer function

H(s) -+ Equivalent feedback path transfer function.

r "

Control System Engineering 5·5 Block Diagram Repl'8Sentation

Key Point: G(s) and H(s) can be obtained by reducing complicated block diagram by usingblock diagram reduction rules.

5.2.1 Derivation of T.F. of Simple Closed Loop SystemReferring to the Fig. 5.5, we can write following equations as,

E(s) = R(s) + B(s) .•. (1)

B(s) = C(s)H(s) ... (2)

C(s) = E(s)G(s) .•• (3)

B(s) = C(s)H(s) and substituting in equation (1)

E(s) = R(s) =1= C(s)H(s)

E(s)C(s)

= G(s)

C(s)R(s) =1= C(s)H(s)G(s) =

C(s) = R(s)G(s) =t: C(s)G(s)H(s)

:. C(s) (1±G(s)H(s») = R(s) G(s)

C(s) G(s)R(s) = 1±C(s)H(s)

Use + sign for negative feedback and use - sign for positive feedback.nus can be represented as shown in the Fig. 5.6.

R(s) --- ..., H::~S)Closed Loop T.F.

I---"'~C(s)

Fig. 5.6Key Point: This can be used as a standard result to eliminllte such simple loops in acomplicated system reduction procedure.

5.3 Rules for Block Diagram ReductionAny complicated system if brought into its simple form as shown in the Fig. 5.5, its

T.F. can be calculated by using the result derived eariler. To bring it into simple form it isnecessary to reduce the block diagram but using proper logic such that output of thatsystem and the value of any feedback signal should not get disturbed. This can beachieved by using following mathematical rules while block diagram reduction.

r "


Rule 1: Associative law : Consider two summing points as shown in the Fig. 5.7.

Fig. 5.7

Now change the position of two summing points. Output remains same.So associative law holds good for +

summing points which are directly R1 ~+ R1: R3 ~_connected to each other (i.e. there is nointennediate block between two summingpoints or there is no takeoff point in

R3 R2between the summing points).Fig. 5.8

Consider summing points with a block in between as shown in the Fig. 5.9.

Key Point: So associative lawis applicable to summing pointswhich are directly connected toeach other.Rule 2 : For blocks in series :The transfer functions of the blocks which are connected in series get multiplied with

each other,

Now interchange twosumming points, R1--__..

Now the output docs notremain same.

+

Fig. 5.9

+

Fig. 5.10

r "

-- - - --- -----


Consider system as shown in the Fig. 5.11.

R(s)

Fig. 5.11

C(s) = R(s) [Gl G2 G3]

So instead of three different blocks, only one block with T.F. [G1 q2 G 3] can be usedas shown in the Fig. 5.12.

R(S)II

C{s}•

Fig. 5.12

Output in both cases is same.Key Point: If there is a taJceoff or a summing point in beiueen the blocks the blockscannotbe said to be in series.Consider the combination of the blocks as shown in the Fig. 5.13.

Fig. 5.13

Key Point: In this combitUltionGIG2 are in series Imd can be combined as G1G2 but G3is now not in series with G}G2 as there is takeoff point in between.

To call G3 to be in series with G}G2 it is necessary to shift the takeoff point beforeG1G2 or after G3. The rules for such shifting are discussed later.

Rule 3 : Fat blocks in parallel:The transfer functions of the blocks which are connected in parallel get added

algebraically (considering the sign).Consider system as shown in the Fig. 5.14.

r "

Control System Engineering 5-8 Block Diagram Representation

R(s) R(s) G2C(s)

R(s)G3

Fig. 5.14

C(s) = - R(s) G1 + R(s) G2 + R(s) G 3

= R(s) [G2+G3-Gl]

Now replace three blocks with only oneblock with T.F. G2 + G 3 - G1 (Fig. 5.15)

C(s);::;:R(s) [G2 +G3-GI]

R_(_S)_._-IIIG, +G3_ ~ III----q+~ s)

Fig. 5.15

Output is same. So blocks which are in parallel get added algebraically.Identify the blocks in parallel correctly. The confusing cases are discussed here. Let

there exists a takeoff point as shown in the Fig. 5.16 along with blocks G1, G2 whichappear to be in parallel.

+

Fig. 5.16

Key Point: But unless Imd until this takeoff point is Shifted before the block, blocks cannot besaid to be in parallel.Shifting of takeoff point is discussed next. Secondly shifting a takeoff point after a

summing point needs some adjustments to keep output same. In above case the takeoffpoint cannot be shown after summing point without any alteration. This type of shifting isdiscussed as critical rules later as such shifting makes the block diagram complicated andshould be avoided as far as possible.

r "

IIII Control System Engineering" ,

+

(a) Avoid such shiftingas far as possible

5·9 Block Diagram Representation

Not valid

Fig. 5.17(b) Without any alteration

such shifting is invalid

Similarly consider a configuraion as shown in the Fig. S.lS.This combination is not the parallel

combination of GI and HI.Key Point: For a parallel combinationthe directionof signals through the blocksin parallelmust be same.

In this case direction of signal throughGI and HI is opposite. Such acombination is called minor feedbackloop and reduction rule for this isdiscussed later.

Fig. 5.18

R II G

" , ' C{s); RG+y•

Fig. 5.19Rule 4 : Shifting a summing point behind the block:

C(s) = RG + Y

Now we have to shift summing point behind the block.Key Point: The output must remainsame,while shifting a summing point.

(R + x)G '" C(s)

Fig. 5.20

r "


RG+xG = RG+y

xC = y

:. x = ~ so signa] y must be multiplied with ~ to keep output same.

R

C(s)=RG+y

Fig. 5.21

Thus while shifting a summing point behind the block i.e. before the block, add ablock having T.P. as reciprocal of the T.P. of the block before which summing point is tobe shifted, in series with all the signals at that summing point.

Rule 5 : Shifting a summing point beyond the block :Consider the combination shown in the Fig. 5.22.

C=(R +y)G

Fig. 5.22

Now to shift summing point after the block keeping output same, consider the shiftedsumming point without any change.

R ..C=RG+yG

Fig. 5.23

•. RG + x = RC + yG

x = yG

i.e. signal y must get multiplied with T.F.of the block beyond which sumnting point isto be shifted.

r "


R

C=RG +yG=(R + y)G

Fig. 5.24

Thus while shifting a summing point after a block, add a block having T.F. same asthat of block after which summing point is to be shifted, in series with all the signals atthat summing point.

Rule 6 : Shifting a takeoff point behind the block:Consider the combination shown in the Fig. 5.25.

R II II RG• G L .C

Fig. 5.25

C = RG

Y = RG

Key Point: To shift takeoff point behind the block villue of signlll takingoff must remainsame.Though shifting of takeoff point without any change does not affect output directly,

the value of feedback signal which is changed affects the output indirectly which must bekept same. But without any change it is just R as shown in the Fig. 5.26.

_----_#,,- ' ....,

_R __• -'--111 G II"~J • C=RG

y=R

Fig. 5.26

But it must be equal to RG. So a block with T.F. G must be introduced in series withsignal takingoff after the block.

Thus while shifting a takeoff point behind the block, add a block having T.F. same asthat of the block behind which takeoff point is to be shifted, in series with aU the signalstakingoff from that takeoff point.

r "


Rt--- ..C::RG

Fig. 5.27

Rule 7 : Shifting a takeoff point beyond the block:Consider the combination shown in the Fig. 5.28.

Rto

y=RG ~ =R

Fig. 5.28 Fig. 5.29

To shift takeoff point beyond the block, value of 'y' must remain same. To keep valueof 'y' constant it must be multiplied by 'lIG'.

While shifting a takeoff point beyond the block, add a block in series with all thesignals which are takingoff from that point, having T.F. as reciprocal of the T.F. of theblock beyond which takeoff point is to be shifted.

Rule 8 : Removing minor feedback loop :This includes the removal of internal simple forms of the loops by using standard

result derived earlier in section 5.2.Key Point: After eliminating such a minor loop if slimming point carries only one signalinput and one sigmzJ .output, it should be removed from the block diagram to avoid furtherconfusion.

C'(s) G'(S) C'(s)

1:!:G'(s) H'(s):!:

Fig. 5.30

r "

Control System Engineering 5 -13 Block DiagramRepresentation

Rule 9 : For multiple input system use superposition theorem :

R,~91 IIR2

System • C

Rn

Fig. 5.31Consider only one input at a time treating all other as zero.

Consider R1, R2 = R3 = R" = 0 and find output Ct.

R, = R~ = Rn = 0 and find output C2.Then consider R2,

At the end when all inputs are covered, take algebraic sum of all the outputs.

Total output C = Cl +G2 + Cn

Same logic can be extended to find the outputs if system is multiple input multipleoutput type. Separate ratio of each output with each input is to be calculated, assuming allother inputs and outputs zero. Then such components of outputs can be added to getresultant outputs of the system. In very few cases, it is not possible to reduce the blockdiagram to its simple form by use of above discussed nine rules. In such case there is arequirement to shift a summing point before or after a takeoff point to solve the problem.These rules are discussed below but reader should avoid to use these rules U.,IC5S anduntil it is the requirement of the problem. Use of these rules in simple problems mavcomplicate the block diagram.

5.3.1 Critical RulesRule 10 : Shifting takeoff point after a summing point. Consider a situation as shov

in Fig. 5.32.

• ~~J ·c

y z

Fig. 5.32 Fig. 5.33

Now after shifting the takeoff point, let signal takingoff be 'z' as shown in theFig. 5.33.

Now z = Rl ±y

r "

Control System Engineering 5·14 Block Diagram Repntlentatlon

But we want feedback signal as x = Rl only.So signal 'y' must be inverted and added to CI to keep feedback signal value same.

And to add the signal, summing point must be introduced in series with takeoff signal. Somodified configuration becomes as shown in the Fig. 5.34.

:!: +

:!:

y

Fig. 5.34

Rule 11 : Shifting takeoff point before a summing point:Consider a situation as shown in the Fig. 5.35.

.. c= R1.:!:y

x= R1.:!:y

Fig. 5.35

Now after shifting the takeoff point, let signal takingoff be IZ' as shown in theFig. 5.36.

Now z = Rl only because nothing is changed.

R1 ,Iz=R1

Fig. 5.36

But we want feedback signal x which is Rl ± y. Hence to z, signal 'y' must be addedwith same sign as it is present at summing point, which can be achieved by usingsumming point in series with takeoff signal as shown in the Fig. 5.37.

r "


+

J---- C=R1±y:!:

v

Fig. 5.37

Thus it can be noticed that shifting of takeoff point before or after a summing pointadds an additional summing point in the block diagram and this complicates the blockdiagram. No doubt, in some rare cases, it is not possible to reduce the block diagramwithout such shifting of takeoff point before or after a summing point. Apart from suchcases, do not use such shifting, which will complicate the Simple block diagrams.

Key Point: Do not use these rules unless and until problem really needs them.

5.3.2 Converting Nonunlty Feedback to Unity FeedbackConsider a nonunity feedback system,

shown in the Fig. 5.38.It is minor feedback loop and hence,

C GR = l+GH ... (1)

It is to be converted to unity feedbacksystem. But the denominator of dosed looptransfer function must remain same asbefore.

R •

Fig. 5.38 Nonunily feedback system

C1;~, with unity feedback i.e, H = 1.. R =

. 1 + G' = l+GH i.e. G' = GH

Thus,C GH ... (2)R = l+GH

1But transfer function must remain same hence it is necessary to introduce block of H

in series with R Thus converted unity feedback system is as shown in the Fig. 5.39.

r "

Control System Engin"ring 5·18 Block Diagram Representation

Unityfeedback

Fig. 5.39 Equivalent unity feedback system

C 1 GH GR = H x 1+GH = 1+GH ... same as before

5.3.3 Procedure to Solve Block Diagram Reduction Problems

Step 1 : Reduce the blocks connected in series.

Step 2 : Reduce the blocks connected in parallel.

Step 3 : Reduce the minor internal feedback loops.

Step 4 : As far as possible try to shift takeoff points towards right and summingpoints to the left. Unless and until it is the requirement of problem do notuse rule 10 and 11.

Step 5 : Repeat steps 1 to 4 till simple form is obtained.

Step 6 : Using standard T.F. of simple closed loop system, obtain the closed loop T.F.C(s)R(s) of the overall system.1. Example 5.1 : Determine the transferfunction C(s) / R(s) of the system shown in the

Fig. 5.40.

c

~--------------~H2~----_'----~

Fig. 5.40

r "

Control System Engineering 5 -17 Block Diagram Representation

Solution: The blocks G2 and G3 are in parallel so combining them as (G2+G3) weget,

Blocks in series

R-----1+

Minor feedback loop

R-----1

Blocks in series

R------i

G, G4 (G2 + G3)

1-G, G4H, ~---~--------c

Feedback loop

GI"G4 (G2+G3)C(s) I-GI Col HIRts) = GI G4 (G2 +G3) H2

1+ I-GI Col HI

"

Control System Engineering 5 ·18 Block Diagram Representation

C(s)R(s) = I-GI G4 HI +GI G4 (G2 +G3) H:z

... Ans.

5.4 Analysis of Multiple Input Multiple Output SystemsIn these problems, the law of superposition is to be used, considering each input

separately. While assuming the other inputs as zero, most of the times if only input isapplied to the summing point, summing point is to be removed if not necessary.

Key Point: While removing summing point if signof the signal present, at that summing point which isto be removed is negative then it must be carriedforward in the further analysis.

Yes)

R(s) +

Fig. 5.41

nus can be achieved by introducing a blockof transfer function -1 in series with thatsignal. This is the important step to beremembered while solving problems onmultiple input multiple output systems.

L....- )(

e.g. consider a part of system showing twoinputs R(s) and Yes).

Other details are not shown for simplicity.When R(s) is considered alone, Yes) must be

assumed zero and summing point at Yes)can beremoved as with Yes) = 0 there remains only asingle signal present at that point so systemgets modified as shown in the Fig. 5.42.

R(s}• •

Fig. 5.42

. Now sign of signal from block Gl is positive at the summing point which isremoved, hence there is no need of adding any other block.

Now when R(s) = 0 with Yes) active, the summing point at R(s) also can be removed.But now sign of the signal 'x' at that summing point is negative which must beconsidered and carried forward for further analysiS. This is possible by adding a block of-1 in series with x without altering any other sign. This avoids the confusion and problemcan be solved without any error.

Blockof '-1' Yes)

Fig. 5.43

r "


)... Example 5.2: Using block diagram reduction technique find the transfer junction fromeach Input to the output C for the system shown in the Fig. 5.44.

X(s)-----t

Fig. 5.44

Solution: With X(s) = 0, block diagram reduces as,

Minor loop Minor loop

c

-~ r:' t;d..___. C(s)

R(s}

Senes blocks

R(s) '~'___1_~_~_2 ........~ ...__, +_:_:_H_r._ _,1 I • C(s)

(1 +G2)(1 +GsHsl

G2G3 o,1+-~--=-~

(1 + G2)(1 • Gs Hs).... ---MlnorJoop Senes blocks

Fig. 5.44(8)

C{s)

"


C(s} GlG2G3GSR(s) = 1+G2+GsHs+G2GsHs+GzG3Gs

With R(s) = 0, G1 vanishes but minus sign at summing point must be considered byintroducing block of - I, as shown.

X(s)

Minor loopNote this block

~~

Minor loop

r(3l2

I(3l3 • I(3l5 C(s)

iii

Fig. 5.44(b)

Series blocks

rt--1~-C(s)

Fig. 5.44(c)

Rearranging the input - output we get,

X(s)-

Minor loop

Fig. 5.44(d)


C(s) G4Gs(1+G2)Xes) = 1+G5Hs+G2 +G2GsIi5+GzG)G5

5.5 Block Diagram from System EquationsThe block diagram can be constructed from the set of equations representing the

system. The addition and subtraction of the various terms is represented by the summingpoints while multiplication factors are represented by the blocks of the respectivetransfer functions.

Consider an equation,

V2 ;: 4Vl - 5V3

The corresponding block diagram has blocksof transfer functions 4 and 5 along with asumming point to represent the subtraction. It isas shown in the Fig. 5.45.

Fig. 5.45

The signals Vt and V3 arc taken from their generating points and available V2 can beused further for generating other variables and output.

Combining the block diagrams of all the system equations, the complete block diagramrepresentations of the system can be obtained.

Key Point: The block diagrams of electrical systems can be easily obtained by this method.As methods of writing equations for electrical system may vary, the block diagram is notunique for the electrical systems.

'. Example 5.3: Obtain the block diagram for the given electrical network.

R, R3r rvl(t) R4 vo(t)

L 1_

Fig. 5.46

Solution: Convert the given network into Laplace domain and assume the currents asshown in the Fig. 5.46(a).

r "


Fig. 5.46(8)

The KVLequations for the two loops are,

-laRa -I, Rz+lzR2 +V; = 0 i.e 11= Vi(Ra !Rz )+h(Rl~Rz ) ••.(1)

-12R3 -12R4-lzRz +11Rz 0 i.e [ Rz' 1 ...(2)= h= 11 Rz+R3+R4J

And v, = Iz a,The block diagrams for the three equations are,

Equation (1) Equation (2) Equation (3)

Fig. 5.46(b)

Thus the overall block diagram is as shown in the Fig. S.46(c).

VI(S) 1 ~ 12Vo(S)R,+R2 R2+R3+R4

~R,+~

Fig. 5.46(c)

,I r "


If the equations are written with differentmethod, the different block diagram can beobtained but the transfer function of the system remains same. For the above network wecan write equations as,

12

v,= h R.

The overallblock diagram is shown in the Fig.5.46(d).

Fig. 5.46(d)

r "



J_ Example 5.4: Reduce the block diagram and obtam its closed loop T.F. C(s)!R(s).

C(s)

Solution: No blocks are connected in series or parallel. Blocks having transfer functionsG2 and H2 form minor feedback loop so eliminating that loop we get,

'<, 1+G,H, /'

t----' .-.,-.------

C(s)

Key Point: Always try to shift takeoff point towards right i.e. output side and summingpoint towardsleft ie. input side.So shift takeoff point after GI to the right. While doing so, it is necessary to add a

block having T.P. equal to reciprocal of the T.F. of the block after which takeoff point is tobe shifted, in series with signal at that takeoff point. So in series with HI we get a block of

(G2 ) 1+G2H2

1/ 1+G2H2 i.e G2 after shifting takeoff point.

C(s)

r "


Minor feedback loop

C(s)

Minor feedbackloop

R(s) C(s)

Simplifying, C(s)R(s)

'.... Example 5.5: Reduce tire block diagram to its simple form and lienee obtain C(s)/R(s).

C(s)

r "


Solution: The blocks G4 and HI form minor feedback loop,

Shifting takeoff point as shown I

G4 C(s)

1+G"H,

The signal from takeoff point A reaching to summing point B is without any block Le.gain of that branch joining A to B is one. So introduce block of T.F. '1' in between A and Bto avoid further confusion as shown below.

Blocks in parallel

Minor feedback loop

"


R(s) (1 + G, G2 H2) (1 + G4 H,)

G, G4 (G2 + G3)1 + -:-:---:::'--:::'-:-:--:-:'~-:::--:-:-:"

(1 +G1 G2 H2) (1 + G4 H1)

After simplification

G2+G3

C(s) G2t- ......-

C(s)

Minorfeedback

leap

C(s)-C{s) G) G4 (G2 +G 3)

R(s) = 1+ GI G2 H2 +G4 HI +GI C2 G4 HI H2 + GI G4 (G2 +G 3)

".. Example 5.6: Reduce the block diagram using reduction rules and obtain C(s)!R(s).

R(s} C(s)

Solution: Block with T.F. G 3 and unity gain block are in ·parallel so combining themweget,

Minor feedback loopR(s) C(s)

"


R(s)

Shifting takeoff point,Blocks in series

~----~~~------~( )

R(s)

Blocks in parallel

R{s) C(s)

C(s)

qs)

___ ......H2+ H2G3+ 1

1 + G3

R(s)

H, (H2 + H2 G3+ 1)

(1 +GJ)

R(s)

Minorfeedback

loop

C(s)

Control System Engineering 5 - 29 Block Diagram Representation

R(s) G2 (1 + G3) C{s)

1 + G2 + H1 (H2 + H2 G3+ 1) (G2)

C(s)R(s) =

II" Example 5.7: Obtain C(s)!R(s) using block diagram reduction rules.

C(s)

Solution: Separating out the feedbacksat different summing points we can rearrangethe aboveblockdiagram as below.

R(s) ""•

Blocks in series

R(s)

C(s}

Minorfeedback

loop

C(s)

C(s)

Minorfee Iback

loop

"

Control System Engineering ,..- 5 ·30 Block DiagramRepresentation

• •

C(s)R(s)

C(s)R(s)

Minorfeedbackloop

R(

G1G2 G3

s) 1 + G3H1 H2 + G2G3H, C(S

G1G2 G3H1 H2 H31+

1 +G3H, H2 + G2G3H1

C(s)R(s) =

'. Example 5.8: Use of Rule No. 10, critical rule illustration. Reduce the blockdiagram. C(s)

and obtaan R(s)'

."

C(S)

"

Control System Engineering 5 -31 Block DiagramRepresentation

Solution: Shifting takeoff point before Gz.

..

C(s)

Shifing takeoff point before summing point using critical rule No. 10,

Separating the paths in the feedback path shown dotted,

Minorfeedback loop

C(s)

C(s)

Key Point: Remember that though the paths through summing point are separated, signs atthe summing points to those paths must be carried as it is.Hence after H2, carry the negative sign and then HI Gz to get - HI H2 G2.

AI '" r "

Control System Engineering 5·32 Block Diagram Representation ,I

Shifting summing point as shown and then interchanging the two summing pointsusing Associative law we get,

Blocks in parallel

Minor feedback loopd C(s)

C{s)

G1 (G2 + G3) C(s)(1+G2H2)(1 +H1G1G2)

C(s)R(s)

(1+ G2H2)(1 + HIG IGZ)c Gt(Gz +G3)(-HtH2G2)

1+ (1+ G2H2)(1 + HIGIG 2)

C(s) Gl (G2 +G 3)R(s) = 1+G2H2+HIGtG2-GIG2G3HIH2

r "


'. Example 5.9: Obtain the expression for CJ and C2 for the given multiple input multipleoutput system.

t-------c2

Solution: In this case there are two inputs and two outputs. Consider one input at atime assuming other zero and one output at a time. Consider Rl acting, R2 = 0 and C2

not considered R2 = 0 and C2 is suppressed (not considered). C2 suppressed does notmean that C2 = O.Only it is not the focus of interest while C1 is considered. As R2= 0,summing point at R2 can be removed but block of '-1' must be introduced in series withthe signal which is shown negative at that summing point.

I,.•.• ,..', .» Forward path

1--_-...c1(

Minorloop

0=

Feedbackpath

Gl Gll+[GI] [-G2 G3 G4] = l-Gt G2 G3 G4

C2For Rl I assume C1 suppressed.

r "


Forwordpath

'",;",-.1' ;- "y

Minor loop

C2 -GlG3G2 -GlG2GJ

Rt = I+l-Gt G3 G2][G41= I-Gl G 2 G3 G4

ClFor R2' Rl = 0 and C2 is suppressed.

Forwordftpath .~

Minor loop

~ -GlG2G4 -GtG2G4R2 =1+f-GlG2G41[G3J =1-GlG2G3G4

C2For R2' Rl = 0 and Ct is suppressed.

Feedback =....... '" path

"'''C-'' ','

;s~:: G2 ~.;)

~?.~:-,.-~~.;J;.~'~'-'J

Minor loop

C2 G2 G2R2 = 1+[G21 [-Gl G3 G4] = 1-Gl G2 G3 G4

r


,_ Example 5.10: Determine the transfer function C(s)!R(s) of the system shoum below byblock diagram reduction method. (M.U. : Nov.-94,Dec.-98)

Solution: Shifting summing point before G. and takeoff point after G4 we get,

R

1

Exchanging summing points and takeoff points using associative law and combiningseries blocks we get,

cR

• •'I'

Control8yatem Engineering 5-36 Block Diagram R.p ..... ntation II

Eliminating minor feedback loops we get,

R

C(s) GIG2G3G4R(s) = 1+G1G2H1+ G3G4H2+GtG2G3G4HIH2+G2G3H3

Series blocks

C(s)R(s)

r GIG2 u G3G4 1ll+GIG2HI Jll+G3G4H2J= 1.r GIG2 lr G"G4 lr H3 ]

"11+GIG2Hl Jll+G3G4H2J lGJG4

G1G2G"G4

'... Example 5.11 : Obtain the closed loop transfer function.

R(s) C(s)

Solution: No blocks are in series or parallel so shifting summing point towards left Le,before the block having transfer function Gl as shown in Figure.

R(s) C(s)

r "

Control System Engineering '5 -37 Block Diagram Representation

C(s}R(s)

Using Associativelaw for two summing points we get,

Minorfeedbackloop

R(s} C(s)

R(s)

.,,-_.-......G1G2 Aftersimplification

Parallelblocks

The two blocks with transfer function 1 and with transfer function ~: are in parallel.

So they will add to each other so we have,

R(sl

.--, ~--.r_---A-ft-e-rS-i-m-p-lif-ica-t-io-n--~

/ C(s) = _G_;.,_G.,;;;2_+_G...;3_;.G...;2;;__R(s) 1 + G2 H2 + G1G2H1C(s)

f--

Control System Engineering 5· 38 Block Diagram Representation

'. Example 5.12: Obtain the closed loop transfer function C(s)!R(s)

Res) C(s)

(M.U. : Dec.-97, D«.-2OO5, Dec.-2007)

Solution: Separating two feedback from second takeoff point which is after blockhaving transfer function G2 as shown, we get,

R(s) C(s)

R(s) C(s)

r "


Shifting summing point behind the block having transfer function 'eI' as shown weget,

R(s) C(s)

get,Use Associative law for the two summing points and interchange their positions, we

R(s)

R(s}

C(s)

Minor loop


r "


Mmorloop

..L...- ~ -J

1 + G,GzCi:3 Hz1+G,tI,-G,G2H; G, +

C(s)

C(s) =R(s)G4 +G4 G2 HI - G4 Gl G2 HI + G2 G 3 G4 H2 + GI G2 G 3

1+G2 HI -GI G2 HI + G2 C3 H2

'.... Example 5.13: Obtain the closed loop transfer function C(s)!R(s).

R(s)

"

Control System Engineering - 5 ·41 Block Diagram Representation

Solution: The blocks G1 and Gs are in parallel, so add them.

R(s)

Shiftingsumming point behind the block 'G2', towards left as shown we get,

R(s)

Using Associative law for the two summing points in between and interchanging theirposition we get,

R(s)

Minor loop

r "


R(s}

R(s}

R(s)


Minor loop

+

(G1 + G5)G2G3G4

1 + G2G3H2 + G3G4H3

+

The block G6 and equivalent block obtained are in parallel,

C(s) =R(s)G(>+~ ~ G(>H2 +G3 G4 Gil H3 +(~ +C;) G.! G~ G.. Gil H, +(~ +C;) G.! ~ G..

1+G2 G3H2 +C3C..H3 +(~ + C;) G.! C~G.. HI

"


II" Example 5.14: Obtam the closed loop transfer function C(s)!R(s).

C(s)

(M.U. : May-97)

Solution: The blocks G2 and H2 form minor feedback loop.

qs)

After SlmplJlicaliOO

G.+G2G.H2-oG,G21+G~2

Minor loop

r "

Control System Engineering Block Diagram Representation

C(s) G3 [C4 +G2 Col H2 +Cl C:d=

R(s) 1+G2 H2 +G3 HI [G-I +G2 G4 H2 + Gl G2]

'. Example 5.15: Obtain tire resultant output C(s) inierms of the inputs R(s) and Y(s).

yes)

C(s)

Solution: As there are two inputs, consider each input separately. Consider R(s),assumingyes) = o.

Minor loop

C(s)t

C(s) =R(s)

Now considerYes) actingwith R(s) = o.Now sign of signal obtained from H1 is negative which must be carried forward,

though summing point at R(s) is removed,as R(s)= 0, so we get,

Yes)

C(s)

r


Combining the blocks G1 HI and -1 as in series,

C(s)

Geq = G2

Hcq = -GI HIFeedback sign positive at input summing

point.

Yes)

CCs) Geq:. yes) = 1-GeqHl"q

So part of C(s) due to Yes) alone is,

C(s) = Y(S)[1+G~~2HJ

Hence the net output C(s) is given by algebraically adding its two components,

C(s) = G1G2R(s)+G2Y(s)1+G1G2Hl

,_ Example 5.16: Tile system block diagram is given below.

Find i) C(s) if N(s) = 0 ii) C(s) if N(s) = 0 iii) NC«(~)if R(s) = 0E(s) Ris) "

(M.U. : May-l'J95)

Solution: i) With N(s) = 0 block diagram becomes

R(s)

rMinor,ooP

10 C(s}

r "


Minor feedback loop =10

5 (5+ 1)--~-- = = -::-~10 52+ 8+58 52+ 651+ 5(5+ 1) 0.5s

10 10

R(s) C(s)102

s +6s

R{s) 10 C(s)2

s +6s

Assume output of second summing points as Xes).

Hence E(s) = R(s) - C(5)

C(s) = Xes)10(s+ 4)S2 + 65

Xes) = 3E(s)+ -4 R{s)s+

... (i)

•.. (ii)

.•. (iii)

Substituting value of Xes)and R(s) from (i) and (ii) in (iii) we get,

52+ 6s 3 310(5+ 4) C(s) = E(s) + s + 4 E(s)+ 5+ 4 C(s)

[ 82 + 65 3] C = (1 + s! 4) E(5)10(s+ 4) - (s + 4) (s)

-fST+ 6s - 30) C( ) = (s+7) E(s)10(5+4) s (s+4) .

. C(s) 10(s+7) when N(s) = 0..E(5) = s2 + 6s- 30

ii) To find ~~:~, w,: have to reduce block diagram solving minor feedback loop and

shifting summing point to the left as shown earlier in (i).

r "


10(S+4)2 t---...--

s +6s

So referring to block diagram after these two steps i.e,

R(s) 102

s +6s

Exchanging two .summing points using associative law,

Res)

3Parallel of '1' and -s+4

C(s)

C(s)

-10 (s + 4). G s2+6sMinor loop -- = -~--':...;;_-

1 + GH 1 + 10(s + 4)s2+ 6s

3=1+-s+4

.. Block diagram becomes,

R(s) 3 10(S+4) C(s1+-- r--- 25+4 s +165+40

C(s) (5 +7) (10 (5+ 4») 10(5+7)R(s) = s + 4 X 52 + 165+ 40 = 52 +165+ 40

iii) With R(s) = 0 bldt'k diagram becomes,

Ces)

"


The 'block of '3' will not exist as R(s) = O. Similarly first summing point will alsovanish but student should note that negative sign of feedback must be considered as it is,though summing point gets deleted.

-Parallel

Two blocks are in parallel, adding them with signs.

C(s)10s(s + 1)

C(s)

In general whiledeleting summing point, itis necessary to considerthe signs of the differentSignals at that summingpoints and should not bedisturbed. So introducingblock of '-1' to considernegative sign.

C(s)

L...---i - (s + 4) - 0.5 S ~ __ ...J--____.'......---After Simplification(-1.55 - 4)

Removing summing point, as sign is positive no need of adding a block.

C(s)10S(S+1)

C(s)

L...-_-I_(1.5 s + 4)-10(1.55 + 4)

'---" s(s + 1)

\Minor loop With G = 1

r


C(s) 1 1= =N(s) 1-[ -10(1.55+ 4)] 1+15s+ 40

s(s+ 1) s(s + 1)

C(s) s(s+ I)N(s) = 52+ 165+ 40

I'" Example 5.17; Use block diagram reduction technique and obtain the transfer functions.) C(S) "J C(S) .... J a»l R(s)' II xi»: nn yes)

Also find total output of the system. (M.U. : Nov.-95)

C(s)

Y(s)

Solution: In such multiple input systems, it is necessary to use superposition principleAnd it must be noted that while removing summing point, it is necessary to considersigns of various signals at that summing point.

i) Consider R(s) acting alone, X{s)= yes) = 0

So blocks G.. and H;\ will vanish along with the summing points and signs of allsignals are positive.

Minor loop

C(s)

C(s) =R(s} ... (i)

COntrol System Engineering 5 -50 Block Diagram Representation

ii) Consider Xes)acting alonge, R(s) = Yes)= 0When R(s) = 0 , Summing point at R(s) will vanish, but sign of feedback signal at that

summing point is negative. So it is necessary to carry on that sign by adding a block of'-1' in series with that signal.

All alein series

Senes~ ,..-------,

X(s)

Minor loop

C(s) ~.1--......- ~--

C(s)

C(s) G3 G.. ..:. Xes)= 1+ GI C2 G 3 HI H2 ... (ii)C(s)

iii) Consider yes) acting alone, R(s) = Xes)= 0

1-- --- ---- -- ---- - ----- ---- - - ..I

Allare :in series:

IIIIIIII1 ---------

Minolioop

I---....-C(_S) ~(~ 0-C(s)

•iii

C(s)• CCs) =

yes)-Gl Gz G3 Hz H3 (""")1+Gl Gz G3 HI Hz .... W

From (i), (ii) and (iii) we can write the total output as

G1Gz G3 R(S)+G3 G4 X(s)- Gl G2 G3 H2 H3Y(s)C(s) -. 1+G1 G2 G3 HI H2

"


'.... Example 5.18: Reduce the block diagram and obtain the transfer function C(sj!R(s).(M.U. : May-96),.

C(s)

3'------t s ...8 t--------'

s-4

Solution: No series, parallel combination and no minor feedback loop exists. So shifting

takeoff point before the block of (_S_) .~+ 10)

Minor loopwithG=1

and H=_1_s+10

Parallelblocks

C(s)

3'-------I s + 8 t--I-----'

S-4

_5 +_1_o x :--5-,:(2..",s:-:+_7..:..) =_ t-_..-C_(S..)5 + 11 (5 + 10) (5 + 31


~~

r


C(s) =R(s)

S(25+ 7)(5 + 11)(5 - 3) S(25+7)(5 - 4)

= (s+ 11)(s - 3)(s - 4) + s(2s+7) (53 + 8)5(25+ 7) (53 + 8)1+ . .....,----".,...

(s+11)(s-3) (s-4)

C(s) s(2s2 - S - 28)=R(s) 2s5 + 754 + 53 + 2052 - 9s+ 132

,_ Example 5.19: Determine C!R ratio for the system shown below. (M.U. : May-95)

c

Solution: Shifting takeoff point which is before G2 to, after Gl.

Parallel


c

r "

Control System Engineenng 5 -53 Block Diagram Representation

Shifting summing point before the block 'GI' and interchanging using associative law,

Minor loop

Series

~ MInor loop.e e

G1(G2G3+ G4)

1+G1G2H1

cR =

CR

G. (Gz G:l + Col)

"


II" Example 5.20: For tire block diagram slrown, obtain C(s)!R(s) by using reduction rules.(M.U. : Jan.-92)

Solution: Shifl the takeoff point, to the right of summing point. This is the "Criticalrule" discussed earlier as rule 10 and 11. In this problem it is necessary to use this rule,which is generally not used to solve simple problems.

New summingpoint due to -t-------+___.critical rule

C(s)

New summing point gets added due to use of critical rule . This summing point canbe eliminated by separating the two paths which are linked by that summing point. Thepaths are shown dotted. So block diagram reduces as,

_\1 , ,:r I,

Control System Engineering 5 - 55 Block Diagram Representation

WhiJe removing summing point, as sign of one of the signal is negative, the block oftransfer function '-1' is connected series with that signal.

So now there exists a minor feedback loop

Shifting take off to the right of G 2

Parallel

C(s)

Combining blocks in parallel and shifting summing point to the left of 'et'.

C(s)

r "


Interchanging the summing points using associative law.

Solving minor feedback loop. The block of 'I' and 'Ht' are in parallel so loop becomes,

+

}

The feedbackpaths H2and 1

G1are in parallel'-- ....1 but with

opposite signs

Combining the feedback blocks which are in parallel.

C(s)

+

_\1 , ,:r I,


C(s)R(s)

C(s) ~ Gz G3 +~ G"R(s) = I-G) Gz + G) Gz HI-G) c, G4 Hz +G) Gz G4H)Hz +G1 Gz G3+G) G4 -G2 G3Hz

,,... Example 5.21 : In the given block diagram,obtain the transferfunction of the systemCt/R1•

(M.U. : Nov.-96)

Solution: Solving the minor feedback loop of 'Col' and '1'.

c,

As C2 is not the focus of interest hence G6 becomes meaningless in the block diagram,thus it is removed.

r "


Minor loop

Solving minor feedback loop and shifting takeoff point.

SeriesG,G2 ~

C,

G4GsH,G,G2G3R, C,

.-oJ(1 + G4)(1 +G,G2)- +

H2G2G3

R,G4GS1+04

Gl G2 G3 G4 GS HI(1+Gl Gz) (1+G4)

1- Gl G2 G3 G4 G5 HJ. H2(l+G} G2) (l+G4) G2 G3

_\1 , ,:r I,

Control System Engineering 5·59 Block Diagram Representationt.. Example 5.22: Find C using block diagram reduction techniques(M.U. - May-98, Dec.-D6)

c

+

Solution: Consider R) alone R21 R3 I R4 are zero.Note : Whenever Rl is zero, as the sign of feedback at R) is negative, while

removing summing point at RII do not forget to insert a block of '-1' to consider effect ofnegative sign.

MinorJo0p

cc

G)G2C 1+G2H2 G1G2.. = = I+G2H2 +G1G2HlRl 1+ GIG2Hl

1+G2H2

C GIG2Rl ... due to Rl.. = I+G2H2 +GIG2Hl


Consider R2 alone, with R3 = RJ = R" = 0Notethis

c +

G2C 1+G2H2R2 =

1-(1 +~~H2}-GIHl)

C = G2R21+G2H2 +G1G2Hl

Consider R3 alone, Rl = R2 = R" = 0

Notethis

Combining two summing points we get,

Minor loop

R3 -- - C~

G2+ -

C III-G,H, I

+

c

c

nl

Control System Engineering 5 ·61 Block Diagram Rep.... entation

=-R3C

-R3G2C = 1+G2H2 +G1G2Hl

Consider Rl alone, with Rl =R2=R3 = O.

C=

1-(-G1H1l(1 +g~HJ

c = - GIG2HIR41+G2H2 +G1G2Hl

=R.t

c

c•

Combining all the values of C, we get

C = GIG2Rl +G2(R2 -R3)-G1G2H1R•1+G2H2 +G1G2Hl

a»I'", Exampel 5.23 : Determine the transfer function R(s) U6inf block diagram reductiontechnique for the block diagram shown - (M.U.: May·2003)

C(s)r-.--

r "


Solution: Shifting takeoff point of G3 to the right of G2 we get,

Parallel

F.

'_....--C(s}

1+~1-....--1 G2 '_......"j,__---C(s}" "'.... ,,'...._------

Shifting takeoff point towards C(s) and interchanging takeoff points,

..'~i'§-~.l:'" :",':;'~' ... ,.,,,..~ Minorloop

.:.. :. y ~ G2 + G3 <1.i<G2 c;;- ....-·~r:-:--.......----- C(s)

~.~: ,. i. After solving gives.

G2+G31+G2+ G3

r "


=

'. Example 5.24: Determine the closed loop transfer junction for the system shown.(M.U. : Df!c.-2003)

R(s)

Fig. 5.47

C{s)t--+--..

Solution: Interchange the positions of two internal summing points, using associativelaw.

I---+-__ C(S)

Minor feedback loop

R(S) I---+-__ C(s)

Shift tothe right

r "

Control System Engineering 5 ·64....,

Block Diagram Representation

1

- M:~~feedbaCk

t--?-- C(s)

"-Se" ->nes

I--_- ..C(s)

qs)

C(s) _ Gl G2G3R(s) - 1+G2HZ +G2G3H3 +GIG2Hl

.~~--------------~'... Example 5.25: Obtain the transfer function ~~:~. (M.U. : May-2004)

Fig. 5.48

Control System Engineering 5-65 Block DiagramRepresentation ~

Solution: Interchange the positions of two summing points using associative law.

~Parallelofr--1 '1' and G,

G, fMinor loop

R(s) +

+~qs)

G2 res)r--l H,+H2

H,

~G2

1 + G2(H,+H2)

Parallel

C(s) _ (1+Gt) GzR(s) - 1+ Gz (HI +H2)

'. Example 5.26: Find the overall transfer function. (M.U. : May-2005, May-20(1)

C(s)

Fig,5.49

Solution :. Separating the feedback paths,

"

" Control System Engineering 5·66 Block Diagram Representation

Separating the dotted paths from the summing point '5'.

C(s)

52 T2 C(s)

••I~e I,__ ...

IIIII:G>~.__--_.,...._-- ....----

Minor feedbackloop

r "

Control System Engineering . 5-67 Block Diagram Representation

C(s) r------,Shift thesumming point

C(s) --------.Interchangethesumming points

C(S)

Parallel Minor loop

G2G3

_1_--:::G:-=3,,::-H..;..'H....::2=--__ t-----1,........~C(S)

1+ ~G3 x H H1-G3H,H2 -, 2

, Control Syst.m Engineering 5·68 BlockDiagram Rep..... ntatJon'

• C(s)=R(s)

[G1G2 - HI (1+G2)]G;l1- G3HIH2 - G2G3H1H2[GIG2 - H1(1-!- G2)]G3

1+~~~~--~~~=-1-G3HIH;z - G2G3HIH2

1--_...-_CCs)

C(s) =R(s)

'.. Example 5.27: For the system represented by the block diagram shown, evaluate theclosed loop transferfunctions when the input R is i) at station 1and in ii) at station II.

(M.U. : May-20(6)

I--<t--_C(S)

....- ........,,' ..." ,, \

,/ 'CCS) r-----.

Fig. 5.50

Solution: i) Input R at station I

r "


iiiiii

Minor loop

R(s)

=

ii) Input R at station IINote that at station I, though input is zero while removing the summing point, keep

the negative sign of HJ as it is. Thus add a blockof '_ l' as shown in the system.

r


,I-_.....-_C(S)

.. ,,... _-----

Shift the summing point to the left of G2 and interchange the two summing points.

R(s)

R(s)

CIS)

."" BIoclIsin parallcl

+ C(S)t--t--

C(S)

r "


C(s)C(s) G1..R(s) = G;\ (- G1G;!H1 - H2)

1-1+ G2H3

C(s) G3 (1+G2H;\)..R(s) = 1+ G2H3 + G3H2 +GtG2G;\H1

-G1G2H1- H2

1 + G2H3

Review Questions1. What is block diagram representation ? Explain with suitable example.2. State advantages and disadvantages of tile block diagram reduction technique.3. Explain the block diagram reduction rules.4. Determine C!R ratio for the system shown below.

c

5. Determine the tran.~erfunction of the system using block diagram reduction of the systan shown.

C(s)

6. for the block diagram shown, obtain C(s)!R(s) by using reduction rules.

r "

~.,ControlSystem Engineering 5-72 Block Diagram Representabon

C(s)

7. Use block dillgrllm reduction ndes fo obtain the tratlSferfunction of the block diagram shown below.

C{s)

8. Reduce the block diagram of the mllitilcop system slwwn in figure beloW.

C(s)

(Ans . C(s) _ G 1 Gl G3 G4 )

•• R(s) -1+G3 c, IfJ +Gz G3 H2 +Gt c, Gl G4 III

9. Determine the uverall trmlSfa function relating C and R for the system whose block dillgram isshown in figure.

c

Ans. ~ = GI G1+GI G3 --.• R 1+G2 H1+GI G2111 -G I G2 G~ HI liz

r or

Control System Engineering Block Diagram Representation5-73

c cR'D system zuhoseoutput for thelind the totalthe10. Determine mtio

block diJIgramis shown in follawing figure.

c

(Ans .~ _ G1 G3 G4 +Gz G3 G4 _

•• R - 1 + G3 H2 + G 1 GJ G4 111 + Gz G3 G4 HI

C Goa(1 +GJ H:!)O=l+GJ H2 +G1 G3 G4 H, +Gz G3 G4 H,

~ tal I t G1 G3 G4 + G 2 G3 G4 Ro ou pu ". 1 +G3 Hz +G1 G3 G4 til +G:z GJ G4 II,

G4 (l +G3 H2) D)+ 1 +GJ H2 +G1 G3 G4 HI +Gz G3 G4 111

11. Derive 1111 expression for the total output for the system represented by tire block diagram infolIawing figure.

c

(A C GIGZR1+GZRZ)

ns.: = 1+G1 G21111h+G21l2

72. Use block diJIgramreduction methods to obtain the equivalent tran.1erfunction from R to C.

_\1 \ I :r I I


c

(Ans • ~ _ Gs G4 (G 2 + GJ) (GI) )

•• R - (1+G" "2) (1+Gl HI) +GS G4 (G2 +G3) GI

13. Use block diagram reduction methods to obtain the equivalent transfer fundion from R to C •

c

(Ans .~ _ Gt GJ (Gl +G4) )

•• R - (1 +G, G2 liz) (1+G3 HJ ) +G1G3 (Gl +G4,

14. Findthe equivalent transfer function for the figure shown below.

c

r "


15. Using block diagtam reduction, find the transferfunction from ead: input to the output C.

c

An (A C G1 Gl G3 C G4 G3s.: ns.: Ii = 1 + G1 Gl G3 HI Hz' X = 1+ G3 GI Gz HI H2'

C -GI Gl G] Hz H3 )

Y = 1+G1 GIG] HI HI

DOD

r "

(5 - 76)

r If

Signal flow Graph Representation

6.1 BackgroundThere is one more way of representating systems particularly when set of equations

describing the system is available. This representation which is obtained from theequations, which shows how signal flows in the system is called signal flow graphrepresentation. As it uses the equations of the system which consist of various variables ofthe system, the variables of the system plays a base role in signal flow graph. Thus we candefine signal flow graph as -

The graphical represenliltion of the variables of a set of linear algebraic equations reprn>entingthe system is called signal flow graph representation.

Let us see which are the important elements constituting the signal flow graph.As variables are important elements of the set of equations for the system, these are

represented first in signal flow graph by small circles called nodes of signal flow graph.Each node represents a separate variable of the system.

All the dependent and independent variables are represented by the nodes. Therelationships between various nodes are represented by joining the nodes as per theequations. The lines joining the nodes are called branches. The branch is associated withthe transfer function and an arrow. The transfer function represents mathematicaloperation on one variable to produce the other variable. The arrow indicates the flow ofsignal and signal can travel only along an arrow.

e.g. Consider a simple equation,

V = IR

Where V = Voltage V

I = Current Fig. 6.1

R = Resistance which is parameter of the system.

This is nothing but simple Ohm's law. Now while representing this equation by signalflow graph, first the variables voltage V and current I, are represented by nodes and theyare connected by the branch as shown in the Fig. 6.1.

(6 - 1)

r "

Control System Engineering 6·2 Signal Flow Graph Representation

1his represents that voltage V depends on value ofcurrent I.And the relationship between the two is throughresistance R. So signal I gets multiplied by R to generatevariable V. So R becomes branch transfer function orbranch gain joining I and V. The direction of arrow is fromI to V. This is shown in the Fig. 6.2.

So all the branches represent the cause and effect relationship existing between thevarious variables. The branch transfer function is also called branch gain or branchtransmiHance in signal flow graph terminology.

Ro-------~~----__ov

Fig. 6.2

6.2 Properties of Signal Flow Graph1) The signal flow graph is applicable only to linear time invariant systems.2) The signal in the system. flows along the branches and along the arrowheads

associated with the branches.3) The signal gets multiplied by the branch gain or branch transmittance when it

travels along it. .e.g. Consider signal flow graph shown in the x, 4 x2Fig. 6.3.The signal from XI gets multiplied by 4 when it Fig. 6.3

travels along the branch joining XI to X2. SO we can say value of X2 is 4 times thevalue of XI. It shows dependence of X2 on Xl.

4) The value of variable represented by any node is an algebraic sum of all thesignals entering at the nodee.g. Consider the variable X2. At that node,3 signals are entering from XI , X3 and X4· x,So value of X2 depends on the variables Xl, o:----__--~X3 and X4. The branch gains indicate theexact contribution of each variable ingenerating X2. So value of X2 is algebraicsum of all such signals entering. So we can x3write,

X2 = 4Xl - 2X3 + 3x 45) The value of the variable represented by any

node is available to all the branches leavingthat node. The number of branches leavinga node does not affect the value of variable XI Xs~----~--~~----~----~.represented by that node.e.g. Consider signal flow graph representedin Fig. 6.5. The value of X2 can be obtainedfrom signals entering at X2.

Flg.6A

i.e. Fig. 6.5

r "

Control System Engineering 6-3 Signal Flow Graph Representation

Now there are three branches leaving X2 , joining to X4, Xs and X6 that means x",

Xs and X6 variables depend on X2.

So we can write x" = X2 I X5 = 2X2 , X6 = 3X2

i.e. for all branches leaving from X2, the value of X2 available is same and numberof such branches do not affect the value of X2.

Key Point: The value of a variable represented by node depends only on the signnL-;entering and this value is as it is available to all the branches leaving from that node.

6) For a given system signal flow graph is not unique. Many other graphs can bedrawn by writing system equations in different manner.

Key Point: The signal flow graph is not the unique property of tire system.

6.3 Terminology used in Signal Flow GraphConsider a signal flow graph shown in the

Fig. 6.6i)

Xo 1

Source Node : The node having onlyoutgoing branches is known as sourceor input node. e.g. Xo is source node.

ii) Sink Node : The node havingonly incoming branches is knownas sink or output node. e.g. Xs issink node. Fig. 6.6

iii) Chain Node: A node having incoming and outgoing branches is known as chainnode. e.g. Xl , X2 , X3 and X4 •

iv) Forward Path : A path from the input to output node is defined as forward path.

xe - Xl - X3 - xs

First forward pathSecond forward path.Third forward path.

e.g. Xo - Xl - X2 - X3 - X.J - Xs

Xo - Xl - X3 - X4 - Xs

Xo - Xl - "'2 - X 3 - X;; Fourth forward path.v) Feedback Loop : A path which originates from a particular node and terminatingat the same node, travelling through atleast one other node, without tracing anynode twi~e is called feedback loop. For example, X2 -X3 -X4 -X2·

vi) Self Loop : A feedback loop consisting of only one node is called self loop. i.e, t 3.'at X3 is self loop. A self loop can not appear while defining a forward path orfeedback loop as node containing it gets traced twice which is not allowed.

vii) Path Gain : The product of branch gains while going through a forward path isknown as path gain. i.e, path gain for path Xo - Xl - X2- Xl - X4 - Xs is ,1X t12X t23X t 34 X t 45. This can be also called forward path gain.

r "


Kay Point: While tracing a forward path or a feedback loop, no node is to be trtlCed twice.viii) Dummy Node : H there exists incoming and outgoing branches both at first and

last node representing input and output variables, then as per definition these cannot be called source and sink nodes. In such a case a separate input and outputnodes can be created by adding branches with 'gain 1. Such nodes are calleddummy nodes.

e.g.

(a)

1.... 0 •

Source

1• 0

Sink

(b)

Fig. 6.7

In the signal flow graph Xl and X2 are input and output variables but as per definitionnot input and output nodes. Such independent nodes can be generated by adding branchesof gain 1 as shown in the Fig. 6.7 (b).

Note: Such creation of dummy nodes is not necessary. Without this also, signal flowgraph can be analysed to get the overall transfer function.

Key Point: AdditiolI of branches of gllin 1 is possible only before starting node and after thelast node. In between the chain nodes slIch branches of gain 1 cannot be added,ix) Non touching loops : If there is no node common in between the two or more

loops, such loops are said to be non touching loops.The Fig. 6.8 (a)&(b) show a combination of non touching loops of two and three loops.

(a) Two non touching loops (b) Three non touching loopsFig. 6.8

Similarly if there is no node common in between a forward path and a feedback loop,a loop is said to be non touching to that forward path. .

I

The Fig. 6.9 (a) and (b) shows such a loop which is non touching to a forward path.

.:«;~4 Xs

Non touching loopto forward path shown

(a)Fig. 6.9

(b)r "


x) Loop Gain : The product of all the gains of the branches forming a loop is calledloop gain. For a self loop, gain indicated along it is its gain. Generally such loopgains are denoted by 'L' e.g. L. I L2 etc.

-5Gl G2 G3 3

0 •X~X3 •

0

xl X4 Xl x2 2 x5

-Hl -2(a)

Fig. 6.10(b)

In the Fig. 6.10 (a), there is one loop with gain L. = G2 x.:...Hl = - G2H.In the Fig. 6.10 (b), there are two loops with gains.

LJ = 4 x -2 = - 8 and other self loop with L2 = - 5

6.4 Methods to Obtain Signal Flow Graph

6.4.1 From the System Equations

Steps:1) Represent each variable by a separate node.

2) Use the property that value of the variable represented by a node is an algt·braicsum of all the signals entering at that node, to simulate the equations.

3) Coefficicnts of the variables in the eqJlations are to be represented as the branchgains, joining the nodes in signal flow graph.

4) Show the input and output variables separately to complete signal flow graph.

Example: Consider the system equations as say,

V. = 2Vi + 3V2 •••(1)

V2 = 4Vl +SV3 + 2V2 •••(2)

V3 = SV2 + Vo

Vo = 6V3

...(3)

...(4)

Let output be Vo and input be Vi where Vl ,V2 , V3 are the system variables.

Equation (1) shows that V. depends on VIand V2. So there are two branches entering atnode Vt i.e. from Vi and V2. But branch from V2to VI has direction from output to input hence tobe shown as a feedback path as in the Fig.6.11(a). Similarly all equations are to besimulated and joined tu get complete signal flowgraph.Fig. 6.11(a)

r " 0""'


Signal from output side towards input i.e.from, V2 to VI, V3 to V2 and so on are to beindicated as feedback paths.In equation (2) there is component of V2

itself, contributing to generate the variable V2.This results in a self loop in a signal flowgraph. The complete signal flow graph isshown in the Fig. 6.11(b).

6.4.2 From the Given Block Diagram

~~v~ 02 3+3 +5 +1

Fig. 6.11 (b)

Steps:1) Name all the slimming points and take off points in the block diagram.2) Represent each summing and take off point by a separate node in signal floW graph3) Connect them by the branches instead of blocks, indicating block transfer functions a..s the

gains of the corresponding branches.4) Show the input and output nodes separately if required, to complete signal flow graph.

Example:

Fig. 6.12Naming summing and take off points

as shown in the Fig. 6.13.

Key Point: Make sure that if summing o--~1"_+_":""-6-_=--+--+;:"_~_--oand take off points are near each other in i/pa given block diagram, they are. to berepresented by separate nodes in thecorresponding signal flOW graph.

o/p

oIp

Fig. 6.13

r "

Control System EnSJineering 6-7 Signal Flow Graph Representation

6.5 Mason's Gain FormulaIt is seen earlier that in block diagram representation, we have to apply reduction

rules, one after the other to obtain simple form of the system and hence overall transferfunction. We have to draw the reduced block diagram after every step. This is timeconsuming. In signal flow graph approach, once S.F.G is obtained, direct use of one

formula leads to the overall system transfer function ~~:~.·Thisformula is stated by Mason

and hence referred as Mason's gain formula. The formula can be stated as :

Overall T.F. = n~AK Iwhere K = Number of forward paths

TK = Gain of Klh forward path

l!.. = System determinant to be calculated as :

l!.. = 1-(l; all individual feedback loop gains [including selfloops]] + [ I:gain x gain product of all possible

combinations of two non touching loops]

- [I: gain x gain x gain product ofcombinations of three non touching loops] +....

l!.." = Value of above l!.. by eliminating all loop gains and

associated products which are touching to the Klhforward path.

Explanation : H we have identified in signal flow graph following information that,Number of forward paths K = 3.The gains i.e. product of branch gains involved in defining various forward paths are

denoted as TI,T2 and T3'

Now number of loops including self loops are say 3 and their gains are say LII L21and L3•

Out of these three, Ll L2 and Ll L3 are the combinations of two non touching loops.There is no combination of three non touching loops.

l!.. = 1- [I:all individual and self loop gains)

+ [I:gain x gain product of all combinations of

two non touching loops]

l!.. = 1- [Ll + L2 + L3]+ [Ll L2 + Ll L3J

r "


Now for ilK i.e, ill, il2 and il3 consider each forward path separately.

For 11, say all loops LI, L2 and L3 are touching to 11 hence all loop gains arc to beeliminated from A to get AI.

For T2, say L2 is non touching and Lt, L3 are touching. So L1, L3 and associativeproducts i.e. LI L2 and Ll L3 are to be eliminated to get A2, L2 will exist.

For T3, say Ll and L2 both are non touching and L3 is touching. So L3 and associatedproduct LI L3 must be eliminated LI, L2 and product Ll L2 as both are non touching toT3 will exist in il3

Hence T.F. can be obtained by substituting these values in

TF = TIAI+T242+T3A3. • A

.... Example 6.1 : Find tire overall T.F. by using Mason's gain formula for the signal Jluwgraph given in the Fig. 6.14

~Fig. 6.14

Solution: Two forward paths, K = 2,11 = G1G3G-1GSG6

T2 = G1 Gl G6

Loops are, Ll = - G 4 HI

L2 = - G3 G4 GsH2

L3 = - G2 H2-H2

Fig. 6.14{a) Non touching loops

Out of these, L1and L3 is combination of 2 non touching loops

A = 1- [ Ll + L2+ L3] + [Ll L3]

r "


Al = Eliminate LI , L2 , L3 as all are touching to Tl from A

Al = 1

A2 = Eliminate L 2 and L3 , as they are touching to T2, from

A . But L. is non touching hence keep it as it is in A.

G4

W-11,

Fig. 6.14(b) L1 Non touching to T2

Substitute in Mason's gain formula,

T.F. =

T.F. =

6.6 Comparison of Block Diagram and Signal Flow Graph MethodsThe comparison of block diagram representation and signal flow graph is given in a

tabular form as :Sr. Block Diagram Signal Flow GraphNo.

1. Basic importance given is to the Basic importancegiven is to the variables of the systems.elements and their transfer

. functions •2. Each element III represented by a Each variable is representedby a separate node.

block.3. Transfer function of the element is The transfer function Is shown along the branches

shown inside the corresponding coMeCting the nodes.block.

4. Summing points and takeoff points Summing and takeoff points are absent Any node canare separate. I have any number of incoming and outgoing branches.

5. Feedback path Is present from Instead of feedback path. various feedback loops areoutput to Input. considered for the analysis.

r "


6. For a minor feedback loop present, Gains of various forward paths and feedback loops arethe formula aGGH can be used. just the product of associative branch gains. No such

formula 1:t GGA is necessary.

7. Block diagram reduction rules can The Mason's Gain Formula is available which can bebe used to obtain the resultant used directly to get resuHant transfer function withouttransfer function. reduction of signal flow graph.

8. Method is slightly complicated and No need to draw the signal flow graph again and again.time consuming as block diagram Once drawn, use of Mason's Gain Formula gives theis required to be drawn time to resultanttransfer functson.time after each step of reduction.

9. Concept of self loop is not existing Self loops can exist in signal flow graph approach.in block diagram approach.

10. Applicable only to linear time Applicableto linear til)1einvariant systems.Invariant systems.

6.7 Application of the General Gain Formula between OutputNodes and Non Input Nodes

It was derived earlier that Mason's gain formula is used to get a relation betweenoutput node and input node called transfer function.

But often, it is required to calculate the relation between output node variable and anon input node variable.

Example :

g6

Now it is required to calculate eoul i.e. dependence of t'oul on C2 and C2 is not inpute2

variable.

So eOUIcan be expressed ase2

~I~from ein toez

E!"~~from €?into COlit

r "

Control System Engineering 6 ·11 Signal Flow Graph Representation

Since 6 is independent of inputs and outputs,

xr 6 IK J< from '-'ill to CouL

zr A IK K from ejn 10C2

,,.. Example 6.2 : Calculate Yl of the system, w/w..<;esignal florv graph is given below.Yl.

-H3Solution: Forward paths for Yt to Y7 arc two

Tt = GtGZG3G4, ~ = G1GZG5

Individual feedback loops are

G1 G3 G1 G2 G3

\7 \7 C • ::J ~'0-H1 -H2 Self

• loop-H3

L1=-G1 H1 L2=-G3H2 L3=-G1G2G3H3 L4=-H4

Combinations of two non touching loops

Ll Lz = +G1 G 3 Ht lIz,

L2 L.. = + G 3 H Z"H4 ,

L1 L4 = +G1.H I H4

La L.. = + G1 G z G a H3 H4One combination of three non touching

Ll L2 L4 = - G t G 3 HI Hz H4

A = 1 - [l. + lz + L3+ ",,]+[L) L2 + Lt .... + l2 La + L3 Ld-[lt L214]

At = 1 aU loops are touching

A2 = 1 all loops are touchingL!... = Tl 61 +T262 = G1 G2 G3 G4 + Gl G2 GsYt A 6

r "


Now to find the ratio Y2YI

Forward paths for YI to Y2 is one. T. = 1 Now A is same as above.

and ..• L2 and L4 nontouching to 1j = 1

= 1 + G 3 H2 + H, + G3 H 2 H,

GI G2 G3 G4 + G. G2 Gs (1+G3 H2)t + G 3 H2 + H 4 + G 3 H2 H4

Y7Yl= =

6.8 Application of Mason's Gain Formula to Electrical NetworkTo apply Mason's gain formula, it is necessary to draw the signal flow graph of the

network. This section explains simple method of writing network equations from whichsignal flow graph can be easily obtained. Applying Mason's gain formula to the signalflow graph, transfer function of the network is obtained.

General procedure to solve such network is as follows,

i) Find out the lAplace transform oj tIre gioen fretwork and redraw the tzetwork ins-domain.

ii) Write down tire equations for the different branch currents and "ode voltages.iii) Simulate each equation by drawing corresponding signal flow graph.io) Combine all signal flow grapkt; to get total signal flow graph for the gtoen network.v) Use Mason's gain Formula to derioe tire transfer function oj the given network.

). Example 6.3: Find tire T.F. for tire given network.

1 tVI L Vo!_ __,____.__!

r "

Control System Engineering 6 -13 Signal Flow Graph Representation '~

Solution: Laplace Transform of the given network is as shown in following figure.

r rVI(s) sL Vo(s)

!0------1---1--01Key Point: Assume the network variables alternately as the loop current and node voltageand then write the equations by analysing the horizontal and vertical branches alternately.

11(s) (Vi - VI)... (I)= RI

VI1 ... (Il)= (11 -12) -sC

h = (VI - Yo)•••(111)Rz

Vo = 12 sL ...(IV)

S.F.G. for equation (I) :

+.1. v1VI

R1

1-R,

S.F.G. for equation (II) :

111 sC V,

-1sC

S.F.G. for equation (III) :

-1R2

S.F.G. for equation (IV) :

r "


Total Signal Flow Graph for the network is as follows.

1 1 +1R, 11 sC V1 R2 12 sL

VjOOO-....~CJCA:7~-~--+---9VO-1 -1 -1R, sC R2

Use Mason's gain formula to find ~~.

v, 1; TK aK N b f £ d thsVi A; urn er 0 rorwar pa = 1

Vo 'Ii At..Vi = A

TJL

= R. R2C

Individual feedback loops are,1 1

L. = - R C' L2 = - -R C'SIS 2

L, and L3 are non touching.

a = 1 - fL. + L2 + L3] + ILl L3)

1 1 sl, LA = 1 + --+--- +- +=-=--=sRI C sR2 C R2 R. R2 C

As all loops are touching to Tl , al = 1,L

= 1 1 sL L1+ -- + -- + - + -=-=---=sR. C SR2 C R2 R. R2 C

v, sL=

Vi sRt R2 C + R2 + R. + s2 L RI C + sL

Vo(s) sLVieS) = s2LRIC+s[L+RJR2CJ+(R. +(2)

6.9 Obtaining Block Diagram from Signal Flow GraphTo obtain the block diagram from given Signal flow graph, it is necessary to write the

set of system equations representing the given signal flow graph. Ass~c suitable nodevariables, write the equation for every node. While writing the equation remember that the

r "

Control System Engineering 6 -15 Signal Flow Graph Representation

value of the variable represented by a node is an algebraic sum of all the signalsentering at that node. The number of outgoing branches have no effect on the value ofthe node variable. For example consider the part of signal flow graph shown in theFig.6.15.

Z:=iv:\Y x w

(a) (b)

Fig. 6.15

In both the casex = ay + bz

And the number of outgoing brancheshave no effecton the value of node variablex.This set of equations can be represented by block diagram, simulating each equation

separately.Key Point: For any + or - sign in equation, there exists a Slimming point uihile for eachbranch gain of signal flow graph, there exists a block of slime transfer function as brandt gain,in the block diagram.For example,consider the equation.

VI = 2V2-4V:t

Here 2 and 4 are the gains; correspondingly there exists the blocks of transferfunctions2 and 4 and a summing point for a minus sign. The block diagram simulationofabove equation is shown in the Fig.6.16.

V, = 2V, -4V31----+

Fig. 6.16

Simulatingall the equations in the same manner and joining all of them, the requiredblockdiagram can be obtained.

r "


'. Example 6.4: Obtain tile block diagram for the signal J1uw grapll, shoum in the figurebekno.

5

-5

Solution: Write the equations for various node variables y, z, wand u. The mput nodeis x as there are only outgoing branches. •

y = (2s)x - (4)z - (S)u

z = (3)y - (3)w

... (1)

... (2)

w = (2)z

u = (l)w + (s)z

The block diagram simulations of various equations are,Equation (1)

.•• (3)

... (4)

Fromz

Fromu

Equation (2),

y

Fromw

Equation (3),

r "


Equation (4),

u•

Combining all the simulations and repositioning branches for convenience we get thecomplete block diagram as,

u


........ . C(s)I,...,... Example6.5: Fmd R(s) for S.F.G. shown in following figure.

Solution : Number of forward paths = K = 22LTK.6K

T.F. = K..I using Mason's gain Formula.. .6

-I = GIG2G3G4

T2 = GI Gs

r "

Control System Engineering ~~; 6-18 Signal Flow Graph Representation

Individual feedback loops,G1 G2

~-H3

L1 =-G1G2H3

G2 G3

~-H2

L2 =-G2G3H2

Gs

Loops Ll and La are non touching loops.

A = [LJ +L2+L3+L4] +[Ll L3]

= 1-[-GIG2H3 -G2G31-h -G4Hl +G5H•H21+[GIG2G4H)H31

Consider Tl , all loops are touching .. Al = 1Consider T2, all loops are touching .. A2 = 1:. T.F.= 1j.6I~T2Az = GIGzG3G .. ·1+GIGS·1

t + G1Gz H3 + Gz G:.1Hz + G-IHI - Gs HI Hz + GI Gz G. HI H3

C(s) _ Gl G2 G 3 G4 +Gl GsR(s) - 1 +Gl G2 H 3 + G2 G3 H2 + G4 HI - Gs HI Hz +GI G2 G4 HI H3

,..........Ex I 6 6 F' d C(s) by . M ' . fi I.".." amp e • : In R(s) ustng ason5 gam ormua.

R(s)o

1 C(s)• 0

Solution: Number of forward paths K = 2

Mason's gain formula,

T.F.

r "


li = Gl G2 G 3 G 4, T2 = GS G4

Individual feedback loops are,

Ll and L3 are two non touching loops.

:. A. = 1-[Ll+L2+L3J+[LlL3J

= 1-[-G2 HI -Gl G2 G3 G4 H2 -Gs G4 H2J+[G2 HI Gs G4 H2J

= 1 + G2 HI + GI G2 G3 G4 H2 + Gs G4 H2 + G2 Gs G4 HI H2

For T. all loops are touching:. £\1 = 1 eliminating all loop gains and prod,:!cts from ll.

Consider T2,

L1 Is non touching to T2'

C(s)=R(s)

C(s) GI C2 G3 G4 + G4 Gs (1 +-G2 HI)R(s) = 1+ G2 HI + Gl G2 G3G4 H 2+ Gs G4 H2 + G2 Cs G4 HI H2

r "

_ Control System Engineering. 6-20 Signal Flow Graph Representation

I'" Example 6.7: Find 0:;)R(:;)

R(s) 1o ..

1 C{s), 0

Solution: Number of forward paths = K = 1T.F. = 11 A]

A.... Mason's gain formula

T] = G1G2G3G"G5

Individual feedback loops,

-H2

L3 = -G1G2G3H2

Combinations of two non touching loops,i) L. and L2 ii) LI and Ls iii) LI and L4

iv) L2 and Ls v) L3 and LsCombination of three non touching loops, is L1, L2 and L,.

A = l-[LI + L2 + L3 +L4 +Lsl+[L]L2 + L1LS+ LIL., +L2LS + L3Lsl-[L]L2LsJ

11 = 1+ G I HI + G 3 H3 + G. G2 G 3 H2 + G 4 H4 + Gs Hs

+G] G3 HI H3+G. G4 H] H4+G. Gs HI H5+G3 Gs H3 Hs

+ G. G2 G3 c, H2 H,+Gl G3 c, HI H3 HsNow considering T. =Gt G2 G3 G4 Gs

All loops are touching to this forward path hence,

111 = 1

r or

Control System Engineering '.. :1 6·21 ,+ Signal Flow Graph Representation

C(s) =R(s)Tl.61 Gl G2 G 3 G4 Gs ·1-/},.- = /},.

C(s) C1C2G3G4GsR(s) = -=-1-+"""G=-I-H=-1-+--=G:-3-:H=-=-3-+-G=-1-=C:-2-G=-3-=H-=-2-+--=G=-4-:H=-4-+-G=-S-=H-=-;;

+GIG3HIH3+GIG4HIH4+GIGsHIHs+ G3 Gs H3 Hs + Gl G2 G3 G 5 H2 Hs+GI G3 Gs HI H3 Hs

,_ Example 6.8: Construct the signal flow graph for the followi'lg set of system equation».

Y2 = c. YI +G3 Y3

Y3 = G4 Yl +G2 Y2 +Gs Y3

... (1)

... (2)

... (3)Y4 = G6 Y2 +G7 Y3

where Y4 is output. Find transfer function ~; .

Solution: System node variables are Y1 I Y2 I Y3 I Y4

Consider equation 1 : nus indicates Y2 depends on YI and Y3

S.F.G. for equation (1)

Consider equation 2 : This indicates Y3 depends on YI , Y2 and Y3


Consider equation 3: This indicates Y4 depends on, Y3 and Y2


r "


Combining all three we get, complete S.F.G.as shown,G4

No. of forward paths = K = 4

~ TKAK = Tl fl.] +T2 A2 +T3 fl.3+T4.tQT.F. = kK-l fI. fI.

Tl = G1 C2 C7, T2 = C" C7, T3= C1 C", T4 = C"C ::IGt;Individual loops are,

SELF LOOP

No non touching loop combinations.

Consider T1, both loops are touching

T2, both loops are touching

T3, for this 'Cs' self loop is non touching,

TI, both loops are touching

L2 non touching to T3

... Mason's gain formula

.. At = 1

•• l!.2 = 1

:. A3 = l-C 5

:. 114 = 1

r "


Gl G2 G7 + G-a G7 +G1 G6 (1- Gs)+ G4G3Gt;l-G2G~-Gs

'. Example 6.9 : Find c(s)R(s)

R(s) 1 1 C(s)

Solution: Number of forward paths = K = 1

~ Tf( £\K

T.F. = Mason's gain formula

Tl = G) G2 G 3 G 4 Gs

Individual feedback loops :

Combinations of two non touching loops.

i) Ll and L3 ii) L2 and L3

L\ = l-[LJ +L2+L3+L4] +[Ll L3+L2 L3]

Consider All loops are touching :. £\1 = 1C(s) =R(s)

r If

Control System Engineering ~!.' 6·24 Signal Flow Graph Representation

C(s) =R(s) 1+ G 2 H. + G. G2 H2+G4 H3 + G2 G 3 G4 H4+G2G4HIH3+GIG2G4H~H3

'. Example 6.10 : Find C{~)R(s)

R{s) 1

-H1

Solution : Number of forward paths = K = 1

T.F. = ... By Mason's gain formula

Individual feedback loops

G4

Q-H3

L3=-G4 H3

G2 G3 G4

L4=-G2 G3 G4 H2Combinations of two non touching loopsi) Ll and L2 ii) Ll and L3 iii) L. and LsCombination of three non touching loops.i) Ll , L2 and Ls

t,=Gs

iv) L2 and Ls

A = I-[L. + L2+ L3+ 4 +4] + [Ll L2+ L. L3+ 4 ~ + L2Ls] -[LI L2~]

Ii = l+GIH. +G3H .. +G .. H3+\.2G3G4 H2-GS+G. G3 H) Hoi

+G) G.. HI H3-G. HI GS-G3 H .. Gs-G. G3 Gs HI H4

r "

Control Systems Engineering 6 ·25 Signal Flow Graph Repre&entation

~l = 1 ... All loops are touching to the forward path.C(S) =R(s)

C(s) _ GI G:! G3 G-&R(s) - 1+ G) H I + G:\ H.. + G-&H3 + G:! G 3 G4 H:! + Gs + G1 G;\ HI H..

+G I G-&HI 113-G, HI GS-G3 G5 H4-G, G3 G5 HI HI

n.. Example 6.11 : Fill,t C(s)R(s)

Solution : Number of forward paths = K = 2

T.F. = ... By Mason's gain formula

TI =G. G:! G3 G4 and T:!=GI G7 G4

Individual loops

SELF LOOP

LI = +G3 G1 H,

No combination of non touching loops

.. II = l-[LI + L2] -= I-G3 G., HI - H2

Consider 11 ~ both loops are touching .. ll. =T2 ~ both loops are touching .. 112 = 1

C(s) T, 111 +T2 ~2 GIG2G3G4·1+GIG7G .. ·1..R(s) = =~ II

r "

Control System Engineering :'~'\tI. 6·26 Signal Flow Graph Representation

C(s) G1 G2 G3 G4 + G. G1 G4R(s) = 1- G 3 G4 HI - H2

'. Example 6.12 F '.,1 a»"m R(s)

R(s) 1 10 1 C(s)

-1 -2

Solution : Number of forward paths K = 2

C(s)T.F. = R(s) = ... Mason's gain formula

.TJ = 5·4·2·10 = 400 and T2 = 1·10·2·10 = 200

Individual loops arc

4 2

V V-1 -2

Ll =-4 L2 = -4

Q-'SELF LOOP

L1=-1

Combinations of two non-touching loops are

i) L} and L3 and ii) L2 and L3

No combination of three non-touching loops :

A = I-[Ll + L2 + L3] + [LJ L3 + L2 L31

= 1 - [- 4 - 4 -11 + [4 + 41 = 1 + 9 + 8 = 18

Consider T. , L3 loop is non touching

Al = 1 - L3 = 1 - [ -11 = 2

1o a 05 T1 4 2 10• 0 • 0 a 0 • 0

r "


Consider T2, Ll is nontouching

1 10

-1 L1 (Nontouching)

t\2 = 1 - Ll = 1 - 1- 4] = 5C(s)

":

R(s)TJ t\l +T2li2 4OOx2+200x5 HOO+l000

Il = 18 = 18 = 100

))~ Ex I 6 13 FO d C(s) by M ' ° Jo l-r ampe. : tn R(s) ason s gam 'mit Qo

R(s) G1 Ga C(s)

Solution: Number of forward paths = K = 221: TK ilK

K =1C(s) =R(s)... Mason's gain fonnula

Individual loops are,

-HeL16Ge

Ll = - G6 H6

Both Ll and Lz are non touching to each other.

r "

~»:I;·~:~ntrol System Engineering , .." r 6 ..28 'fit :,. Signal Flow Graph RePl:ftsntation I it••:!II.,.U

Consider 11 I L2 is non touching

T G6

o ,~:;y ·~.,~p 0

G3

l2v:;;J-H3

Consider T2 I L. is non touching-H6

~G6

~.~'G3

.6.2 = 1 - L. = 1+ G6 H6

C{s)=R(s)

C(s) G. Gs G6 G7 Gs (1+G3 H3)+G1 G2 G,3 G4 Gs (1+G6 H6)

R(s) = 1 + G6 H6 +G.3 H3 + G 3 G6 H3 H6

,....... Exam Ie 6 14 F· d C(s)_", p.: In R(s)

R(s) 1 1 C(s)

Solution: Number of forward paths = K = 3

r "


C(s) =R(s)K '"' 1

11 06.1 + Tz flz + T3 o6.'Jfl

TI = CI G;? G3 G" GS G6

T::! = GIG z C 7 G"

T3 = CI GZ G3 G4 GilIndividual feedback loops are :

... By Mason's gain formula

o-H3

L7 = - GIG 2 G 3 G-I C II H:\Combinations of two non touching loops are :

i) Ll and L4 ii) L-I and Ls ill) LI and L6No combination of three non touching loops

.. fl = 1 - [Ll + L2+ L) + L., + ~ + L(, + L, + Lft] + II.I J-l + 1.4 LII+ L, L61

+GIG2C3C4G~G61l3 +CIG:!G3C4G8H~ +GIG2C7G6H~ +GIlHl

+ C-I H~ G2 C,;, H2+ Gz G, Ih Gil HI + G~ B-1G, Cl G7 Cit 113

r "

: Control System Engineering 6 ·30 Signal Flow Graph Representation

Consider TIl all loops are touching :. £\1 = 1For T2 I only LI is non touching.

For T3, aU loops are touching :..6. 3 ;;;;1

C(s) "Ii .6.1+ T2 .6.2 + T3 .6.3:. R(s) = a

C(s) Gl G2 G3 GolGs Go + Cl C2 G7 G" (1+ Gol 1-Iol)+ G1 G2 G3 ColGsR(s} = 1+ G4 H4 + Gs Go HI + G2 G 3 G" Gs H2+ C2 G7 H2

+ G1 G2 G3 C4 Gs Go Ih + G1 C2 C3 ColCs H3 + C. C2 C1 Go H3+ GR HI +G2 G" G7 H2 Hoi+ G2 G7 c, HI H2 + CI G2 C7 G4 c, H3 H4

'. Example 6.15: Using Mason's gain formula, find the gain of the following system infigure below.

I C(s)

-H2

Solution: The number of forward paths are K = 6The forward path gains are,

Tt = GIG2G), T2 = G4GSG6

T3 = GIG7G6, T4 = G4GgG3Ts = G4Gs~-H2)G7G61 T6 = G.G1(-H.)GgG3

The feedback loop gains are,Ll _ - Gs HI , L2 = - ~ H2 , L3 = + G7 HI Gs H2

r "


The two non touching loops are Lt ~

4 = I-fl, +l] +L31+[l, L2]

= 1+ GsH, +G2H2 -G7GgH,H2 +G2GsH,H2

For TI, LI is non touching.

4 1 = 1 - L, = 1+ Cs HI

For T2 ' ~ is non touching.

For T3 to T6 all loops are touching to all forwardpaths.

.. fl.3 = fl.. =fl.s =fl.6= 1

Cain = LTKfl.)i. = T)J11 + r242 + T343 + T-Ifl.1+ T5ll.5 + I'"ll..,..Il fl.

.. G,G~G3 {I+GsHd+G,GSG6 (I + G2H:d+

Cain G,G,G6 + G"GgG)-G4GKG7G6H2 - G,G3G,GslI,... Ans= I +G5H, + G]1-l2- G;G)lH,112+ G~G;H,I~

)I_ Example 6.16: Obtain the overall transfer function of the system shoum by blockdiagram, using signal JlO'w graph technique.

R(s)--.c+ I----t-- C(s)

Solution: Name the various summing and take off points to draw the signal flow graphas shown,

r "


R(s)-~~Input

t----+- ..C(s)..... _.... Output+

The corresponding signal flow graph is,

Forward path gains are,

T) = GI G2 T2 = G 3G2The various loops are,

Ll =-GI G2 Hl L2 = -G2 G3 H2No combinations of non touching loops.

:.Ii = 1-[L1+L2+L3)= 1+GIG2H2+GZG3Hz+G2Hl

For TI, All loops are touching, :. iiI = 1For Tz, All loops are touching, :. liz = 1According to Mason's gain formula,

C(s) T) Al + T2 1i2R(s) = Ii

C(s) G1 G2 +G2 G3

R(s) = 1 +G) G2 Hz +G2 G3 H2 +G2 HI

r "

Control System Engineering 6-33 r'I I I I. • t

Signal Flow Graph ~epresentation .

'.. Example 6.11: Determine the transfer junction for the block diagram shown below byMason's gain formula.

Solution: Name the various summing and takeoff points to draw the signal flow graph.

1 C(s)R(s) 1

-1

The fo~ard path gains arc,

The various loop gains are,

and

LI = - Gl ~ HI 14 = - ~ GJ H2

La = - GI~G3 L. = - G,

Ls = 1 x G" x 1 X (- H2)X ~ X (- HI) = ~ G, HI H2

r "

Contiol8ystem Engineering 6·34 Signal Flow Graph Representation

No combinations of non touching loops,

A. = 1 - [Li + L2 + L.1 + Lt + Ls] = 1 + Gt ~ Hi

+ ~ ~ 1-12+ GI G G3 + G. - G G. HI H2

All the loops arc touching 19Tl to T2 hence,

Using Mason's gain formula,

C(s) TIA.1 +TzA.2R(s) = A.

C(s) GIGZG3 +G4R(s) = 1+G1GzHl +GzG3Hz +G1GZG3 +G-l-GzG4HIH2

I'. Example 6.18: For the signal flow graph of figure below, determine the transfer junctionC(s]/ R( s) using Mason's gain formula.

, t •

: I

Solution: Number of forward paths K= 4

Tl = l·GI·GZ,}=GtGz

Tz = 1·G3·G .. -t=G3G-lT3 = I-GI-G6-G4-1=G1 Gt>G ..T.. = t-G3-Gs-Gz-l=Gz GJ G5

Individual loops areLl = -Gz HI,L.. = -G4 HI G6,

Lz = - G3 H2, L3 = Gs G6Ls =-Gl G6 H2

Combinations of two non touching loops arei) II and Lzii) No combination of three non touching loops

A. = 1-(ll+Lz +L3 +L-I +Ls)+(LI Lz)

= 1-[(- G2 HI -G3 Hz +G5 G6 -G4 G6 HI -GI G6 Hz)]+{(- Gz HI )(- G3 Hz)]

_\1 , ,:r I,


.. ll. = 1+G2 HI +G3 H2 -Gs Gli +G4 G6 HI +GI G6 H2 +G2 G"I HI H2

For all the forward paths an the loops are touching hencell.1=ll.2=ll.3=l!.4=1

Hence C(s)R(s)

C(s)=R(s) ... (2)

". Example 6.19 : Find ~~:~ (M.U. : M.ay-97)

R(s) 1

1

1 C(s)

Solution : Number of forward paths = K = 2C(s) =R(s)

... Mason's gain formula


G1 G2 G3 -H4

Q Q Q U-H1 -H2 -H3 Gs

Ll = - GI HI L2 = - G2 l'h L3 = - G3 H3 L., = - Gs H4

Combinations of two non-touching loopsi) LI and L3 ti) LI and L4 iii) L2 and L4 iv) L3 and L4

Combinations of three non-touching loops ~ LJ ,L) and L4•

r "

Contr0I~Y"l"" Engineering 6-36 Signal Flow Graph Representation

Consider TI, For this 'L ..' is non-touching

-H4

WL4Gs

T, 1 G, G2 G3 10 • 0 • 0 .. 0 .. 0 .. 0

For T2, L2 and L3 are non-touching

C(s) =R(s)

C(s)R(s)

G1 G2 G J (1 + Gs H ..) + G4 Gs G6 (1 + G2 J-h + G3 }-I3)= ( 1+GI HI +G2 Ih + G3 H3 + Gs I1-l + GI G 3 HI 113 +GI Gs HI H-I+G2 G5 1-[2H" +G3 G5 H3 H-I +GI G3 c, HI H3 HI]

II. Example 6.20: Find tile oatue of C(s). (M.lJ. : May-07, Dec.-06)

U(s)

R(s) 1 1 C(s)

W(s)

Solution: As system has three inputs, considering each input separately,Assuming U(s) = W{s) = 0, S.F.G. becomes,

TI = GI G2

and LI = GI G2 HI H2

R(s) 1 G, G2 1 C(s)0 ..¥ II 0

_\1 , ,:r I,

6 -37~ t •.• •


Ii = 1- [LJ] = I-GIG2HIH2

IiI = 1 as loop is touching to TJTJ iiI G} G2

= ~ = 1- G} G2 HI H2. C(s)R(s)

C(s)

Assume U(s) = R(s) = 0

. . S.F.G. becomes as shown in following figure

TI = HI GI G2

LI = I-h HI GJ G2

Ii = 1- LI = 1- GIG2 HJ H2

£\1 = 1

C(s) =W(s)TJ £\} HI GI G2-ll- = I-G} G2 HI H2

C(s)

Assume R(s) = W(s) = 0

LI = G2' HI H2 G}

£\ = I-Ll = I-HI H2GI G2

£\1 = 1C(s)U(s)

TI iii= -ll- =

Signal Flow Graph Representation

•... (1)

W(s)

... (2)

U(s)

1 C(s)

... (3)

• Total output is combination of all three individual output.

C(s) = GI G2 R(s)+H} GI G2 W(S)+G2 U(s)1- c, G2 III H2

r "

Control System Engineering 6 -38 Signal Flow Graph RaprGS!"tattDn I J

•• J ,;.'. Example 6.21 : Construct the signal flow graph for the following set of simultaneous

equations.

X2 = AZI XI + A2.1 XJX3 = A.u XJ +A32 Xl +A1.1 X3

X4 = A42 Xz + A4_~ X.1 (M.U. : Dec.-96)

Solution: The value of the variable is the algebraic sum of all the signals entering at thenode, representing that variable. The variables are XI , X2 I X;I I X" while the gains areA21, A 2;1, A 31, A 32, A 33, A..2 and A..3·

So selecting nodes representing variables and simulating differential equations.

,!

... Equation (I)

... Equation (II)

... Equation (III)

Iherefore resultant signal flow graph is

'.... Example 6.22: Find transfer functions for signal flow graphs given below

i)

_ G1 O-~2_o__.~__~~__G~1~~~G~2~__.~~o

-H2

(M.U. : Nov.-95)

_\1 , ,:r I,

ContrUl S~ Engineering 6 ·39 Signal Flow Graph Representation

~1W:ion: i) Forward paths K = 1Forward path gain Tt = GIG 2


n-H,

V Selfloop

Lt = -G1 G2 H2

There is no non touching loop combinations..!J. = 1-[Ll+L2] = 1-[-GtG2H2-Ht] = 1+G.G2H2+Hl

Both the loops are touching to the forward path

!J.l = 1Using Mason's Gain Formula

C(s)R(s)

ii)

Ro1• 1 C

" 0

Solution: Forward paths K :;;:1

"Ii = c, G2 G3 G4

Individual feedback loops

_\1 , ,:r I,

6 • 40 Signal Flow Graph Representation ~t 1\ II

~' .. ~

Control System Engineering

-H,

L4 =-Gl G2G)G4 III

All loops are touching each other, no combination of non touching loops

= 1+ G.j HI + G" G4 H3 + G2 G:\ G.. 112+ Gl G2 G3 G4 HI

All the loops are touching to the forward path hence AJ = 1

C(s)=R(s)

iii)

Ro

1•

Solution: Forward path K = 111 = G,G2G3G4


~G, G:!

Ll = -GJ G2 H2

1 C.. 0

r or


The loops L2 and L4 forms combination of two non touching loops. Similarly Ll andL.. also forms combination of two non touching loops.

L2 L4 = G2 G4 HI H~

L. L4 = G. C2 Col H) H::!

11 = 1-[Ll+L2+L3+L4]+[L2L4+Ll L4]

= 1 + Gl G2 lh + C2 II. + G2 G3 G..+ H4 G4 H3 + Gz Col HI II:. + G. G2 C., Hz 1-13AUloops arc touching to forward path hence 111 = 1

C(s) Tll11=R(s) =r:C(s) G. G2 G:. G.,

.. R(s) "" 1+ G1 G2 112+ c. li-;-+ G2l~., Col II ..+ G4 H" + l.2"G7iI.I-13 + G, G2 G. Ib 113"

,,.. Example 6.23 : Use Mason's Cain Formula to obtain ~~~ of the system snoum below.

C(s)

(M.U. : Jan-92)

Solution : Name all the summing and take off points and assuming each as a separatenode draw the signal flow graph.

r "


1C(s)R(s)

-1

Number of forward paths K = 2

Forward path gains


~-1.~/•-1

-1

L.a = -G1 G..

L7 = G1 G2 G4 I-}z

There is no combination of nontouching loops.

•• A = 1- [Ll + L2 + L3 + L.. + Ls + L6 + L7]

All loops are touching to both the forward paths

:. Eliminating all loop gains from A we get,

AI = A2 = 1

r "

Control Systei.1 Engineering :..:: 6 ·43 Signal Flow Graph Representation

...Using Mason's Gain formula,

C(s)R(s)

--

C(s) _ C. G2 G~ t (~tG..r. I{(s) - 1- G1 (';2 - G2 G) H2 + G1 C2 G3 +G1 G4 + ~ G2 G4 III III + G1 G2 I-IJ - G. G2 Col1-12

'. Example 6.24: Draw the signal flow graph and hence obtain the transfer funaion of thesystel1J shoum below. (M.U. : July-91)

R(s) C(s)

Solution: Name all the summing and take off points as shown below and representingeach separately as a node draw the signal flow graph .

...

R(s)

H,

R(s)

-H2-H,..

-1

Number of forward paths are K = 2

Forward path gains areTl = G)G2G3

T2 = G4

1C(s)

• I • I_\1 \ "r ,

Control System Engineering 6-44 •Signal Flow Graph Representation I ~,


-H2

1..,,2= -G2 G3 H2 L3 = +G2 G. I-II 1'12.

-1

L4 = -G1 G2 G)

There are no combination of non touching loops

-1

Ls = -C ..

•••

All the loops are touching to both the forward paths

••• ~1 = 112 = 1

... Using Mason's gain formula

C(s) Tt A) + T2 A2--R(s) 11

•••C(s)R(s)

'. Example8.25: The small signal equitalent circuit of Q common emitter transistoralnplifier ;s shoum below. The transistor amplifier includes a feedback resistor RI: Obtain asignal flow grapll model of the feedback amplifier and determine the input-output ratio(Va / Vinje (M.U. : Dec.-96)

+

Vee- -

-

• I • I_\1 \ "r ,

Control System Engi ng 6-45 Signal Flow Graph Representation

Solution: Let input current be iin and load current be it .

• •

" If

Rs hieR, •

A C F IL H

B •

'in I,Vee- --

• E o J I

At node B, • • •lin + If = Ib ... (1)

Apply KVL to loop ABCDEA ,

... (2)

Substituting • ••lin = Jb -If from (1) in (2) we get

Yin = (ib - if) Rs + Ilie ib + hre Vee

Vin = ib (R, + hie) - if R, + hre Vee ... (3)

The output equation is

Vee = iL Rl ... (4)

At node G,

... (5)

••• it = - (ic + if) substitute in (4) we get

Vee = -(ic + if) RL ... (6)

Now

and as 110e is admittance

•• • ... (7)

Substituting (7) in (6) we get

Vee = - [11ceVo+~ib+if]Rl

Applying KVL to the loop GBCDIG

•

... (8)

•I.e. ... (9)

• I • I_\1 \ "r ,


Now the different variables are

Yin = input•Vee = output

and if lib are intermediate currents.

To sketch the signal flow graph let us obtain the equations for ib , if and Vee. Vi" isinput hence no equation for Vin is expected.

From (3) we can write equation for ib as

•Ib =

Vin + if Rf -Ilrc Vee(R" + hie)

•I.e. •Ib = 1 VI." + .Rf · h re V

I ) .' R I ) 1f - (R I.) ce(R~+ 'ic ( s + 1ie s + 'Itt.•. (10)

This can be simulated by signal flow graph as

1

From (8) we can write equation for if as,

. - hoe l{l Vee -Ji RL ib - VeeIf = Rl

•I.e. ... (11)

This can be simulated as

• -J) • V~Ib If0 • "- /• (1 + hoeRL)-

R._

•

Equation (9) is directly the output equation

•I.e.

Vee = if Rf + ib hie + hre Vee

Rf• hie.If + Ib1- h re 1- Ilre

... (12)

•

• I • I_\1 \ "r ,


This can be simulated as,

•

Combining all three signal flow graphs, the complete signal flow graph can beobtained as,

1•If Vee-J~

-h'eR + h.s ~


•

-~ -~

IL3 = + Jl Rf hre _

(Rs + Il ic) (1 - 11rc)

• I. • I_\1 \ "r ,


L -hie hre4 = (R, + 11 ic) (1- 11re)

Ls = -Rf hie(l+h(lt>RI)RL (R, + 11i~)(1 - 11re)

•• • A = 1 - [Lt + L2 + L3 + L.a+ Ls ]

As there are no combinations of two nontouching loops.

AI) the loops are touching to both the forward paths r. ~1 = ~2 = 1

...According to Mason's gain formula

Vee T1 A1 + Tl ~2--- -Vin A

Simplifying A ,

I!. = 1+ PRf +Rf(l+hoeRt>(Rs + h ic.-) RL (1 - h r~)

•

J3 Rf hre hie hre [{f Ilip (1+ hoe RL)- + I ------...,._--~-(R, + 11ie) (1-11 re) (R,+ 11ic) (1- 11r~) RL (1{~+ 11ie) (1- 11re)

[RL (Rs + hie) (1- hre) +P Rf Rl (1 - llre) + n, (1+ hoc Rt) (Rs + hie)

:. I!.= -~ Rr hre Rt + hie hre RL + Rf hie (1 + hoe Rl»(Rl (Rs + 11ie) (1-11re)]

:. I!. = Rf (1+ hoe RL> (Rs + 2 hie) + Rl IRs (1 - h re)+ h ieJ+ Ii Rf Rl (1 - 2 h re)RL (Rs+ llie) (1- hre)

Vc~ (Rs + hie) (1- hre). --------•• Vin - ~

hie RL -13 R, RL=-----------------------------------------------------Rf (1 + hoe RL)(Rs + 2 hie) +Rl (Rs (1-11re) + hie]+ ~ Rf R. (1- 2 h,e)

Inpractice for analysis of such network the values of h oe and 11re are neglected.

:. Simplified ratio ~: neglecting hoe and h re is

• I. • I_\1 \ "r ,


n, (Rs + 2h ie)+ RL (Rs + hie) + ~ Rf Rl

III. Example 6.26: Find T.F. for the given network, using Mason's gain fonnula

(M.V. : May-97)

c

Solution: Laplace transform of the given network is,

_ Parallel R,1 + sR,C

---r Z

I~R2 V.(s) R2 Vo(s)

1Rl x- RIz., = sC =1 1+ Rl CsR1+-sC

I(s) (Vi - Vol C+ sRI C)= = (VI - Vo) . RlZ

V(I(s) = I(s) R2

For equation (1) : S.F.G. is,

... (1)

... (2)

For equation (2) : S.F.G. is,

I(s)o

_\1 , ,:r I,

Control System Engineering 8-SO Signal Flow Graph Representation

Combined S.F.G.

""'= '!>:7v,•_(1 +:~,c)

Use Mason's gain formula. Number of forward path = K = 11i = R2 (1+ SRI C)I Rl

_R2( 1+RSRIIC»)Individual loop Ll =

VI 0

1 + sR,CR,..

R2 (1+ SRIC)11 = 1 - [Ll] = 1+ Rt = Rl + R2 (1+ sRI C)

Rt

R:r (1+ sRI C)Rl + R2 (1+ sRt C)

,,_. Example 6.27: Draw the signal flow graph for the following network and find the,I'. ft . Vo(s)transfer tnction Vi (S)'

Solution: Convert the given network in its laplace form and assume different loopcurrents and node voltages as shown.

1 1sC, SC2

1 -1;.) R1

I~VI(S) R2 Vo(s)

1 "

_\1 , ,:r I,

.' Control System Engineering 6-51 Signal Flow Graph Representation

Writing down equations for 11 , Vl ,h,Vo we get,

(Vi - VI)I I = 1 = SCI (Vi - V0)

sCt

VI = (lI-lz)Rl

h = (VI ~ Vol = sCz (V1 - Vol

sCz

v, = 12 R2

Simulating above equations by signal tlow graph.

Vi 0sC, 1, v,..

"'= '7 For equation (I)

"-sC,"

R, V,

For equation (II)

-R,

V, sC2

For equation (III)

sCz

12 R2 Vo For equation (IV)0 II 0

Combining we get signal flow graph for given network.

To find T.F. apply Mason's gain formula

T.F. :ETK ~l Number of forward path = K = 1= ~

T.F. 'Ii ~I.. = =x:Tl = sCt Rl sCz Rz = sZ Rt Rz Ct Cz

••• (1)

...(ll)

...(Ill)

... (IV)

_\1 , ,:r I,

Control Systems Engineering 6-52»>

Signal Flow Graph Representation ~,

Individual loops :

Lt = - Rt Ci 5, L2 = - sRt C2Out of three, Ll and L3 are nontouching

A = 1 - [Lt + L2+ L3]+[Lt L3]= 1-[ -sRt Ct -sRt C2 -SR2 C2]+[S2 Rt Ct R2 C2l= 1+ S [Rl C1 + RJ C2+ R2C2] + 82 [Rl Cl R2 Cd

V('I(s) 11 AtVieS) = ~

All loops arc touching to "Ii, :. At = 1

,,... Example 6.28: Determine the transferJunctions CJRt and C!R2 from the block diagramshoum by drawing its signal flow graph and using Mason's gain formula. (M.U. : May-99)

R, c

Solution: Let us draw the signal flow graph to use the Mason's gain formula.The signal flow graph is,

~G2For qRt, let R2 = 0

R, 1C.. TI = GIG3

and T2 = G2G3

LJ = -G3 HI -H,

A = l-[LI] = 1-[-G3 HI] = 1+G3 HI

Al = 1 and A2 :;;:;1

C "Ii At +T2A2 = Gl G3+G2 GJ G3(GI+G2).. R1:;;:; =Ii 1+G3 HI 1+G3 HI

For CI R2, let Rl = 0

_\1 , ,:r I,

·,.Control System Engineering 6-53 Signal FlowGraph Representation

R2Hence signal flow graph reduces to,TI = G3Ll = - G3 Ht

A = 1- LI = 1+G3 HI

At = t

=R2 =c

,_... Example 6.29: Pind the T.F. of tiregiven network. (M.U. : May-2003)

1 1VI(S) R2 R4 Vo(s)

10---1---'--01Solution: LaplaceTransform of the given network is,

R, v, R3

Equations for different currents and voltages are

1 1S.F.G. (I)Vi

R, I, V, II = (Vi - VI) x - ... (I)RI

-1

I, if,S.F.G. (II) VI = (lJ -12) R2 ... (II)

1-R2

V, R3 . 1S.F.G. (ill) 12 = (VI - Vo) X R3 ... (III)-1R3

S.F.G. (IV)12 R" Va

Vo :: 12 R4 •.. (IV)0 .. D

r "

Control System Engineering • 6-55 Signal Flow Graph Representation

Solution: For the given current distributions,•

For branch rl, ... (1)

For branch r3, ... (2)

_ Now current through f2 is (i2 - a it). Hence according to Ohm's law,

. . V2 - V312 - a 11 =

••• ...(3)

And . .. (4)

The simulations of all the equations are,

• a1

- 1 - 1

Equation (1) Equation (2) Equation (3) Equation (4)

Thus the overall signal flow graph is as shown,

a

1 1-- --

There are two forward paths,

,Tl =

• I • I_\1 \ "r ,


•The various loops and loop gains are,

a

1 - 1 1-- --r,

One combination of two non touching loops is Ll L3•

•• •

All loop are touching to all the forward paths,

•• • ~1 = !J,2 =1

•• • --r3 f4 + a r..rl r2 rl---------------~~--~- l-[LI +L2 +L3 +L ..l+[Ll L3]

-- rs r3 It a r., r~ f..11+ + + .. - ...+-....._...r) f2 f2 fl r1 1-2

•• •V3 r3r4+ar4r2= ------------------------------VI ft r2 + fl r2 + r3 rl + rl r4 - a r2 r) + r=, r4

J. Example 6.31 : Obtain the tramp junction ~:~:; for the network shoum ill the figurebelow. (M.U. : May-2003)

R

c cR

~----------~------------o

• I. • I_\1 \ "r ,

-Control System Englne.ring Signal Flow Graph

Solution: The Laplace domain representation of the given network is shown below.

The various branch currents are shown,

R

V1(s)

11(5) 1

sC 1(11+12)

R

•• • 11(S) = = sC Vt (5) - sC Vr (s) ... (1).a

Then, 1 1= RVI (s) - RV,,(s) ... (2)

... (3)

and also, Vo(s)- V1(S)1 = sCVo(s)-sCV.(s)

sC

From this obtain the equation for Vo(s) as 12(5) equation is already obtained.Note: Write the separate equation for separate branch and each element must be

considered at least once.

••• ... (4)

Hence the signal flow graph is,

+1

The forward path gains are,1

11 = sCR T2 = sCR

" • I • I_\1 \ "r ,

Control System Englne.rlng 6-58 Signal Flow Graph Representation

The various loop gains are,1Ll = - sCR L2 = -_sCR

The loops Ll and L2 are non touching

1LJ= --xRxl = - 1R

•• ••

Hence system determinant is,

t,. = l-[Ll + L2 +L3]+[L1L2J

= 1 CI' 1 1 1 = 3sCR+ s2C2R~ +1+5 '+sCR+ + sCR

...

A1 = 1 all loops touching to .lj

112 = 1 - Ll as L) is non touching to T2= (1 + sCR)

all loops touching to T:\

According to Mason's gain formula,

Vo(s) Tll1J +T2~2 +T363--Vi (s) l!.

3sCR+S2C2R2 +1sCR

S2C2R2 +1 +sCR+sCRsCR=---------3sCR+s2C2R2 +1sCR

1sCR 1+sCR)+1

--

••• -- S2C2R2 +2sCR+ls2C2R2 + 3sCR+ 1

'. Example 6.32: Obtain ooerall transfer function.

1 1R(s) C(s)

(M.V. : May - 2005)

• I. • I_\1 \ "r ,

Control System Engineering 6-59 Signal Flow Graph Representation 'tr

... Only one forward path

-H3

L3 = -G,G2G3G4H3

No combination of non touching loops.

... All loops touching to Tl

C(s) =R(s)1i~1 G]G2G3G4----x-- = 1+G2Hl +G3H2 +G1G2G3G4H3

'. Example 6.33: Draw the signal flow graph and derive the transfer function usingM1lson's gain formula.

Solution: Name the summing and take off points.

C(s}

(M.U. : May - 2006)

r "

--


The signal flow graph is,

R(s) 1 1 C(s)

... K = 2 forward paths

All loops are touching to each other, so no combination of nontouching loops.

•••

... All loops touching to both forward paths

•• • C(s)R(s)

G1G2+G1G3-- 1+GIG2Hl +G2H2 -GtG2G3HtH2

_\1 \ I :r I I

• I


Review Questions

1. Define signal fluw graph.2 Defirrefollowitlg terms related to signal fluw graph

(i) Source node (ii) Sink node (iii) Chain node (iv) Forward path and its gain(v) Feedback loop and its gain (vi) Self loop (vii) Nontouching loops.

3. Explain the various properties of signal flow graph representation.4. Explain how to construct signal fluw grapIJfrom

(i) Set of equations (ii) Block diagram. Give silitavie example.5. State and explain Mason's Grin Pormula.6. Compare block diagram representation witll signal fluw grapl! representation.7. Construct the signal fluw graph for the folluwing !et of simultaneous equations.

Xa =AI2 X2+ An X38. Find transfer /unctions for signal fluw graphs given beluw

1,o , o

-H2

(A C(s) Gt G1 )ns. : R(s) = 1+GI Gz Hz+Ht

9.1, 1 C

, 0Ro

10.

r "

Control System Engineering .. 6-62 Signal Flow Graph Representation

11. Draw signalflow graphfor the system shown below.Find overall transferfunction using Mason's Gain Formula.

C(s)R(s)

(A C(s) 10 (s +7) )

ns, : R(s) :: 52+165+40

12. For system shown, obtain the closedloop transferfunction by Mason's Gain Formula

(A C(s) GI G'1.G3 )

ns. : R(s) = I-G. G'1.HI +Gz G3l1z +G1 Gz G3

13. UseMason's Gain Formula tofind ;~.

1

(Ans..' Xz .. G.GZG,G.G7G. +GIG CG•G7G• +G,GzG,GsG ... GIG.G,G. )

. XI 1- GZG3G,C7G,1ft - GcG,G,C.H J - GzG,GsG.1ft - GeGsG alit - G,G6~ 7Hz- C,GsHz

r "


14. Use Mason's gain formula to calculateC/R oj tile system shown below.

c

(Ans . C _ G1 G2 G3 +GI G4 )

.. R -l+Gt G4 +Gt c , G3 +G1 Gz HI +Gz G3 Hz +G4 H2

15. Draw the signal flow graph and obtain tire transfer function of the system shawn below.

C(s)

16. UseMason's Gain Formula to obtain C(s)/R(s) oj the system shown below.

C(s)

r "

Control System Engineering 6-64•

Signal Flow Graph Representation

17. For the signal flow graph shown in following figure determine "If ratio ~/Xl. Use Mason's gainformula for signal flow graphs.

X1

A Xs G 1 G1G3 G4 +G1 G4 G, +G1 G7 (l-Gs)ns. : Xt = t - G1 H1 + G2 Gs HI - Gs

18. Using MIlSon's gain formula dete,nzine tile system output for input R and disturbancesDt and ~ for the syste,,, described in the block diQgrDm•

•

°l(s)

15+1

1 ~__-..j10--.Rs + 2 C(s}

205

Ans. : C(s) =~ (R(s) + (s + 1)01(s)+ Dz(s)8](s3(s +1) (s+2)- 20]

19. Draw the signal flow graph lind obtain 1M transfer function.

A

R{sJ 1 C(s)

..8

A+B+2ABAns.: t-A- 8-3AB

• I. • I_\1 \ "r ,


20. Drau: tilt signal flow graph and obtain the transfer function.

ODD

• •

• I • I_\1 \ "r ,

(6·66)

_\1 , ,:r I,

Time Response Analysis ofControl Systems

7.1 BackgroundMost of the control systems, use time as its independent variable, so it is important to

analyze the response given by the system for the applied excitation which is function oftime. Analysis of response means to see the variation of output with respect to time. Theevaluation of system is based on the analysis of such response. This output behaviour withrespect to time should be within specified limits to have satisfactory performance of thesystem. The complete base of stability analysis lies in the time response analysis. Thesystem stability, system accuracy and complete evaluation is always based on the timeresponse analysis and corresponding results.

This chapter explains the concept of time response, steady state analysis, transientresponse and derives the various transient response specifications.

7.2 Definition and Classification of Time ResponseTime response of a control system means, how output behaves with respect to time. So

it can be defined as below.Definition : Time Response : The responsegiven by the system which is function of the

time, to the appliedexcitation is called time responseof a control system.In any practical system, output of the system takes some finite time to reach to its final

value. This time varies from system to system and io; dependent on different factors.Similarly final value achieved by the output also depends on the different factors likefriction,mass or inertia of moving elements, some nonlinearities present etc.

For example consider a simple ammeter as a system. It is connected in a system so asto measure current of magnitude 5 A. Ammeter pointer hence must deflect to show us 5 Areading on it. So 5 A is its ideal value that it must show. Now pointer will take somefinite time to stabilise to indicate some reading and after stabilising also, it depends onvarious factors like friction, pointer inertia etc. whether it will show us accurate 5 A ornot.

Basedon this example, we can classify the total output response into two parts. First isthe part of output during the time, it takes to reach to its final value. And second is the

(7 - 1)

r

Control System Engineering 7·2 Time ResponseAnalysis ofCentrol Systems

final value attained by the output which will be near to its desired value if system isstable and accurate.This can be further explained by considering another practical example. Suppose we

want to travel from city A to city B.So our final desired position is city B. But it will takesome finite time to reach to city B.Now this time depends on whether we travel by a busor a train or a plane. Similarly whether we wilt reach to city B or not depends on numberof factors like vehicle condition, road condition, weather condition etc. So in short we canclassify the output as,i) Where to reach ?il) How to reach ?Successfulness and accuracy of system depends on the final value reached by the

system output which should be very close to what is desired from that system. Whilereaching to its final value, in the mean time, output should behave smoothly.Thus final state achieved by the output is called steady state while output variations

within the time it takes to achieve the steady state is called transient response of thesystem.

Definition : Transient Response

Tireoutput variation during the time, it takes to achiel'eits final value is calledas transientrespoJlse. Tile time required to achieve the final value is called transient period.

This can also be defined as that part of the time response which decays to zero aftersome time as system output reaches to its final value.

Key Point : The transient response may be exponential or oscillatory in nature. Symbolicallyit is denoted as cr (t).

To get the desired output, system must pass satisfactorily through transient period.Transient response must vanish after some time to get the final value closer to the desiredvalue. Such systems in which transient response dies out after some time are called StableSystems.

Mathematically for stable operating systems,

Limct(t) = 0

t ---+ 00

From transient response we can get following information about the system,i) When the system has started showing its response to the applied excitation?ii) What is the rate of rise of output? From this, parameters of system can bedesigned which can withstand such rate of rise. It also gives indication aboutspeed of the system.

iii) Whether output is increasing exponentiallyor it is oscillating.

r "

Control System Engineering 7-3 Time Response Analysis ofControl Systems

iv) If output is oscillating, whether it is over shootting its final value.v) When it is settling down to its final value ?

All this information matters much at the time of designing the systems.

Definition : Steady State Response

. It is that part of the time response which remains after complete transient response vanishesfrom the system output.

This also can be defined as response of the system as time approaches infinity fromthe time at which transient response completely dies out. The steady state response isgenerally the final value achieved by the system output. Its significance is that it tells ushow far away the actual output is from its desired value.

Key Point: The steady state response indicates the accuracy of tire system. The symbol forsteady state output is C/I5'

From steady state response we can get following information about the system :i) How much away the system output is from its desired value which indicateserror?

ii) Whether this error is constant or varying with time? So the entire informationabout system performance can be obtained from transient and steady stateresponse.

Hence total time response c(t) we can write as,

The difference between the desired output and the actual oulput of tire system is called steadystau error which is denoted as e"•. This error indicates the accuracy and plays an importantrole in designing the system.

The above definitions can be shown in the waveform as in the Fig. 7.1 (a), (b) whereinput applied to the system is step type of input.

c t) c(t)

'1(l)css(t)-

Step Step

css(t)-

St d lime 0 limeea y_Transient

, state of Transienttime system time

(a) C1(t)'s exponential (b) ct (t) 's oscillatoryFig. 7.1

'" r "

Control System Engineering 7-4 Time ResponseAnalysis ofControl Systems

7.3 Standard Test InputsIn practice, many signals arc available which are the functions of time and can be used

as reference inputs for the various control systems. These signals are step, ramp, sawtoothtype, square wave, triangular etc. But while analysing the systems it is highly impossibleto consider each one of it as an input and study the response. Hence from the analysispoint of view, those signals which are most commonly used as reference inputs aredefined as Standard Test Inputs. The evaluation of the system can be done on the basis ofthe response given by the system to the standard test inputs. Once system behavessatisfactorily to a test input, its time response to actual input is assumed to be upto themark.

These standard test signals are,

i) Step Input (Position function) :

It is the sudden application of the input at a specified time as shown in the Fig. 7.2.Mathematically it can be described as,

r(t) = A for t ~ 0

=0 fort<O

If A = 1, then it is called unit step functionand denoted by u(t).

Laplace transform of such input is A.s

Ii) Ramp Input (Velocity function) :

It is constant rate of change in input i.e. I)gradual application of input as shown in theFig. 7.3.

Magnitude of Ramp input is nothing but itsslope. Mathematically it is defined as,

ret) = At for t z D

= 0 for t c O

If A = 1" it is called Unit Ramp input. It isdenoted as r(t). Its Laplace transform is ~.

s

iii) Parabolic Input (Accelerationfunction) :

This is the input which is one degree faster thana ramp type of input as shown in the Fig. 7.4.

r t}

A1--------

oFig. 7.2 Step

Fig. 7.3 Ramp

t}

-Slope=AI

oFig. 7.4 Parabolic

r "

Control System Engineering 7·5 Time ResponseAnalysis ofControl Systems

Mathematically this function is described as,

A t2,2r{t) = for t ~o

= 0 I for t<O

where A is called magnitude of the parabolic inputKey Point: Parabolicfunction is expressed as ~ t2 so that in Laplace trans/onns 0/ differentstandard inputs, similarity will get maintained.

IfA = 1, i.e. r{t) = ~ it is called unit parabolic input. Its Laplace transform is ~.

Iv) Impulse Input:

It is the input applied instantaneously (for shortduration of time) of very high amplitude as shownin the Fig. 7.5

It is the pulse whose magnitude is infinite whileits width lends to zero i.e, t -+ 0, appliedmomentarily .

Area of the impulse is nothing but itsmagnitude. If its area is unity it is called UnitImpulse Input, denoted as B(t).

Mathematically it can be expressed as,

r{t) = A, for t = 0

= 0, for t ~ 0

ret)

1A

Fig. 7.5 Impulse

The Laplace transform of unit impulse input is always 1. (Refer Chapter-2). The unitimpulse is denoted as l)(t).

r(t) Symbol R(s,

Unit step u(t} 11s

Unit ramp ret) 1/52

Unit parabolic - 1/53

Unit impulse ~t) 1

Table 7.1

r "

Control System Engineering 7-6 Time ResponseAnalysISofControl Systems

7.4 Steady State AnalysisAs discussed earlier, steady state is that part of the output which remains after

transients completely vanish from the output.Mainly the steady state response has following two specifications,i) How much time system takes to reach its steady state which is called seHling time

which is discussed later in connection with transient response. It is related totransient response also because same time will be required by the transients to dieout completely from the system output.

ii) How far away actual output is reached from its desired value which is calledsteady i!!tateerror (ess)'

Out of the two specifications, the steady state error is the most importantspecification which is related only to the steady state. So let us see on whichfactors it depends, how to calculate it and how to reduce it.

Definition : Steady State Error : It is the difference between the actual output and thedesired output.

Now reference input tells us the level of desired output and actual output is fed backthrough feedback element to compare it with the reference input. Hence to be precise itcan be defined as the difference between reference input and the feedback signal (actualoutput).

and L ( e(t) }= E(s) = R(s) - C(s), for unity feedback systems.

Mathematically it is defined in Laplace domain as,

L ( e(t) }= E(s}= R(s)-C(s)H(s), for non unity feedback systems

7.5 Derivation of Steady State ErrorConsidera simple closed loop system using negative feedback as shown in the Fig. 7.6

C(s)

B{s)

Now,

Fig. 7.6

E(s) = Error signal, and B(s) = Feedback signal

E(s) = R(s) - B(5)

B(5) = C(s)H(s)

where

But

r "


" E(s) = R(s) - C(s)H(s)

and C(s) = E(s)G(s)

.. E(s) = R(s) - E(s) G(s)H(s)

.., E(s) + E(s)G(s)H(s) = R(s)


R(s)E(s) = 1+G(s) for unity feedback

R(s)E(s) = 1+ G(s)H(s) for nonunity feedback

This E(s) is the error in Laplace domain and is expression in '5', We want to calculatethe error value. In time domain. corresponding error will be e(t). Now steady state of thesystem is that state which remains as t -t 00 •

LimSteady state error, ess = e(t)

t-too

Now we can relate this in Laplace domain by using final value theorem which statestha~

Lim LimF(t) = 0 sF(s)t 400 s 4 where F(s) = u F(t) t

Therefore,Lim Lim

ess = e{t) = 0 sEes) where E(s) is LI e(t) I,t 4- S4

Substituting E(s) from the expression derived, we can write

Lim sR(s)e =liS S -to 1+ G(s)H(s)

For negative feedback systems use positive sign in denominator while use negativesign in denominator if system uses positive feedback.

From the above expression it can be concluded that st_y state error depends on,i) R(s) i.e. reference input, its type and magnitude.il) G(s)H(s) i.e. open loop transfer function.iii)Dominant nonlinearities present if any.Now we will study the effect of change in input and product G(s)H(s) on the value of

steady state error. As transfer function approach is applicable to only linear systems, theeffed of nonlinearities is not discussed.

r "

Control System Engineering,. 7·8 Time RasponseAnalysis ofControl Systems

7.6 Effect of Input (Type and Magnitude)on Steady State Error(Static Error CoefficientMethod)

Consider a system having open loop T.F. G(s)H(s) and excited by,

a) Reference Input Is step of magnitude A :A t) A

R(s) = - ess= 'i"+'i<s pA )lim s R(s)

Cst =.. s-+O 1+ G(s)H(s) Desired

Limoutput

SoA/s= s -+ 0 1+ G(s)H(s)0 Time

Lim A= 1+ G(s)H(s) Fig. 7.7s-+O

Aess =.. Lim1+ 0G(s)H(s)s-+

LimFor a system selected, 0 G(s)H(s) is constant and called Positional Errors-+

Coefficient of the system denoted as ~.

LimKp = s -+ 0 G(s)H(s) = Positional error coefficient

And corresponding error is,

I e.=~ ISo whenever step input is selected as a reference input, positional error coefficient ~

will control the error in the system along with the magnitude of the input applied.

bJReferem:e Input ,. tamp of magnitude "A' :

Lim s.A/s2= 5-+ 0 1+G(s)H(s)

Actualoutput

R(s) = A/52Lim sR(s)s -+ 0 1+G{s)H(s)

Fig. 7.8r "


Lim A= s [1+ G(s)H(s)]5--+0

Lim A= s + s C(S)H(s)5-)0

Ae~~ =.,

Limo sC(s)H(s)s --)

LimFor a selects..ed system s G(s)H(s) i" constant and called Velocity Error Coefficient

5-)0

as K ...

LimK,. = ) s G(s)H(s) = Velocityerror coefficient

s --+ \


~

~

So whenever ramp input is selected as a reference input, velocity error coefficient K,will control the error in the system alongwith the magnitude of input applied.

c) Reference input is parabolic of magnitude ·A· :

Le, R(t) = ~t2

R(s) A= s3 c{l)

Lim sR(s)ess = s~O 1+G(s)H(s)

Lim soA/s3= s -) 0 1 + G(s)H(s)

Lim A= 5-)0 S2 [1 + G(s)H(s)]

Lim A= s-+O S2 + S2 G(s)H(s)

Lim AellS = s-)O S2 G(s)H{s)

Actualoutput

e - A55 - K;

Fig. 7.9

r "

Control System Engineering 7 -10 Time Response Analysis ofControl Systems

Lim .,So for a selected system 0 s- G(s)H(s) is constant and called Acceleration Error

s~

Coefficient as K I •~---------------------------------------.... K =a

LimOS2 G(s)H(s) = Acceleration error coefficient

s-t


So whenever parabolic input is selected as a reference input, acceleration errorcoefficient K" will control the error in the system along with magnitude of input applied.So static error coefficients are given in Table 7.2.

Static Error Coefficient Corresponding S.S. error 8••

Kp s lim G(s)H{s) As ....0 1+ Kp

K" = Urn sG(s)H(s) A• __0K"

Ka = Lim 52 G(5)H(5) A..... 0 Ka

Note : 'A' denotes the magnitude of the input appliedTable 7.2

7.7 Effect of Change in G(s)H(s) on Steady State Error(TYPEof a System)This ~ be studied by focusing on to the dominant elements of G(s)H(s) from error

point of view. Such elements are constant of system 'K' and poles of G(s)H(s) at origin ifG(s)H(s) is expressed in a particular form called time constant form. This is as shownbelow,

G(s)H(s) = K(1+Tls) (1+T2s)....si (1+TasH1+Tbs)•••••••

K = Resultant system gain and j = TYPE of the systemwhere

"'TYPE' of the system means number of poles at origin of open loop T.F. G(s)H(s) ofthe system.

r "

Control System Engineering 7 -11 Time Response Analysis ofControl Systems••

So j = 0, 1'l'PEzero system

j = 1 I TYPE one system

j = 2 I TYPE two system

••

••

j = n , ITPE "n' system

Key Point: Th!IS 'TYPE' is the propertv (if opel' loop "r.F. G(sj/-l(s) tt'/,i/e "Order' is the. l;(s) ·

property of closed loop T.F. i± C(s)H(s)"~--------------------------------------------------------------,----~

This is because, as defined earlier, order is the highest power of s present in thedenominator polynomial of closed loop T.F. of the system.

7.8 Analysis of TYPE 0, 1 and 2 SystemsNote: A popular method to assess stead)' state performance of servomechanisms or

unity feedback systems is to find their error co-efficients K~"K\. and K3•

K.., = Position error constant,

K\ = Velocity error constant and

K, = Acceleration error constant.

Obviously in order to find these error constants the system must be stable, because foran unstable system there is no steady state and Kpl K, and K, are undefined.

Hence before we proceed to find Kp, K\. and K \ we must ensure (either by polelocation or by Routh table of the closed loop system) that it is stable.

Thus the concept of Kp, K, and K, is applicable only if,

i) System is represented in its simple form.

ii) Only if the system is stable.

Consider the input selected as step of magnitude 'A'.

i) Let us assume that the system is of TYPE "0'.

. K(l + ~ s) (1+ T2s) .I e G(s)H(s) =· · (1+ Tas) (1+ Tbs) ..•....

LimOG(s)H(s) = K

54For step input ... using above G(s)H(s)

A A••• e~ = --=--l+Kp 1+1<

• I. • I_\1 \ "r ,


i.e. TYPE "0' systems follow the step type of input with finite error 1~ K which can be

reduced by change in 'A' or 'KI or both as per requirement.

Now 'K' can be increased by introducing a variable gain amplifier in the forward pathand error can be reduced. But there is limitation on the increase in value of 'K' formstability point of view which is discussed later. But increase in 'K' is one way to reducethe error. Corresponding response will be as shown in the Fig. 7.10 (a) or (b).

•

A+-----------~---•

Aess= 1+K

(finite)(finite)

o Time o Time

(a)Fig. 7.10

(b)

ii) If for the same input now TYPE is increasedto one' by adding pole at origin. In G(s)H(s).

TYPE 1: G(s)H(s)K(l + Iis) (1+ T2s) .-- s(1+T, s) (1+ Tbs) .

As input is step, Kp =Linl

OG(s)H(s) = 00

S~

•• •

iii) Similarly If now TYPE is further increased to 'two' i.e. G(s)H(s) with 2 polesat origin,

TYPE2: G(s)H(s)

As input is step, Kp = LimOG(s)H(s) = 00

S~

eS$ = L\ = ~ = 01+ K 00p

• I • I_\1 \ "r ,

Control System Engineering'[J

7 -13 Time Response Analysis ofControl Systems

In general, for any TYPE of system mOTC than zero, Kp will be infinite (00) and errorwill be zero. Though mathematically answer for error is zero, practically small error willbe present but it will be negligibly small. Such type of responses rna} take one of theforms shown in the Fig. 7.11 (a) and (b).

c(t) CCl)

Time

(a)Fig. 7.11

(b)

Thus TYPE 1 and above systems follow a step type of reference input of anymagnitude, successfully, with negligibly small error.

Let us now change the selected input from step to ramp of magnitude .t\' so K,. willcontrol the error.

iv) Let the system be of TYPE O.

K(1+ 1;s)(l + T2s) ....G(s)H(s) =

(1 t T.. s] (1+ T['>s) .

LimK,. = s ~ 0 s G(s)H(s) = 0

i.e. TYPE 0 systems will not follow ramp input of any magnitude and will give largeerror in the output which may damage the parameters of system or may cause thesaturation in parameters. Hence ramp input should not be applied to TYPE '0' systems.The output may take the form as shown in the Fig. 7.12 (a) and (b).

c(t) c(t)

Time Time

(a)Fig. 7.12

(b)

.\1 , , :r ' , ale'~i?111/".


v) If TYPE 1 System is subjected to Ramp input then,

G( ) K(I+1js)(I+T2s) .s)H(s = s (1+ Ta 5) (1+ TbS) .....

Limo s G(s)lI(s) = l(

S~l( =v :. e~~ A A fini=- = - teKy K

i.e. TYPE 1 systems follow the ramp type of input of magnitude 'A' with finite errorA/K which can be reduced as discussed earlier. The output may take the form as shownin the Fig. 7.13 (a) and (b).

CCt) c t)

Time Time

(a)Fig. 7.13

(b)

vi) If TYPE2 system Is excited by Ramp Input then,

i.e.

K =vLim

o s G(s)H(s) = 00

s~A A

e$S = ~ = -;; = 0

This is true for any system of TYPEmore than one. Hence all systems of TYPE 2 andmore than two follow ramp type of input with negligible small error and may take theform as shown in the Fig. 7.14 (a) and (b).

c t) c(t)

Time

(a)Fig. 7.14

(b)

r "


Let us now change the selected input from ramp to parabolic input of magnitude Ahence coefficient K.l will control the error.

vii) Consider TYPE 0 system :

G )K(l+ rls) (1+T2s) .

(s H(s) =(1+ T..s) (1+ TbS) .

Lim .,K, = s" G(s}H(s} == 0s~o

A A.. c =_=_=00s, Ka 0

viii) Consider TYPE 1 system :

K (1+ TIS) (1+ T2s) .G(s}H(s) =

s(l + Ta s) (1+Tbs) .

K~ =Lim .,o "- G{s)H(s) = 0S~

A:. Css =K = 00

OJ

For both TYPE '0' and '1' systems, error will be very large and uncontrollable jfparabolic input is used. Hence parabolic input should not be used as a reference to exciteTYPE '0' and TYPE '1' systems. The output may take the form as shown in theFig. 7.15 (a) and (b) if excited by such input.

c(t) c(t)

Time Time

(a)Fig. 7.15

(b)

ix) If TYPE 2 system is used i.e.

G(s)H(s) = K (1+ Tls) (1+T2s) .s2(1+ T"s) (1+Tbs) .

LimK, = s2 G(s)H(s) = K

s~OThen A A fini.. c~,== K = K te

a

Hence TYPE 2 systems will follow Parabolic input with finite error A/K which can becontrolled by change in A or K or both and output may take form as shown in theFig.7.16 (a) and (b).

r "


c(t) c(t)

e - ASS- K(finite)

e - ASS- K(finite)

Time Time

(a)Fig. 7.16

(b)

And for any system of TYPE 3 or more if parabolic input is used, error will benegligiblysmall.The results can be summarised as shown in the Table 7.3.

Typeof Error Coefficients Erroresa forSystem

Kp K~ Ka Step input Ramp input Parabolic input

0 K 0 0 A DO DO

1+ K

1 DO K 0 0 A -K2 00 00 K 0 0 A

K

Table 7.3

7.9 Disadvantages of Static Error Coefficient MethodThe disadvantages of Static error Coefficient Method are :

1) Method cannot give the error if inputs are other than the three standard testinputs.

2) Most of the times, method gives mathematical answer of the error as '0' or'infinite' and hence does not provide precisevalue of the error.

3) Method does not provide variation of error with respect to time, which will beotherwise very useful from design point of view.

R(s)---+{

+ C{s)

Fig. 7.17 r "

· Control System Engineering 7 -17 Time Response Analysis ofControl Systems

4) Error constants are defined for the loop transfer function G(s)H(s), strictly, hence themethod is applicable to only the system configuration shown in the Fig. 7.17.

5) As final value theorem is used to calculate steady state error SO before applying itis necessary to check if sEes)has any poles on the jro axis or in the right half of thes-plane. This means before applying this method, the system must be checked forits stability. The method can not be applied to unstable systems.

6) When the system configuration is different than as shown above then it isnecessary to establish the expression for the error signal and apply the final valuetheorem directly, without the use of error coefficients.

7) The method is applicable only to stable systems.Because of these disadvantages Dynamic error coefficient method (Error series method)

is developed which eliminates above said disadvantages.

7.10 Generalised Error Coefficient Method (or Dynamic ErrorCoefficients)

As explained earlier,

R(s)E(s) = 1+G(s)H(s)

Let us assume that this is the product of two polynomials of's'.

E(s) = F,(s)·F2(s)1

:;--~-=-:-:--.-, F2 [s].= R(s)1+C(s)H(s)Where FI(s) =

Now if, F(s) = FI(s)·F2(s) then using convolution integral,t

L-IIF(sH = F(t) = IFI(T) l~(t- T) dr(I

SimilarlyI I

e(t) = IFI(T} F2(t-t} dT= Jfj(t) R(t-T) dr(J 0

R (t - or)can be expanded by using Taylor series form as,2 3

R(t - or) = R(t) - orR'(t)+ ~! R"(t) - ;!RIII(t) + .....

Substituting I [ 2 3 ]eCt) = !11 (T) R(t)-TR'(t)+ ~! R"(t)- ;! R"'(t)+ ....... Jt

r "


I I= fR(t)F1(T)d'C - J TR'{t) F({T)d'C+ .

o 0

Nmv ess = Lime(t)

t-+co

.. 00 GO .,

= R(t) JFt(T) dT-R'(t) fTl)(T) dT+R"(t)J ~~ F('C) dr .000

Where-

Kl = - JT Fj (T) dt;o

...Kz = J'C2 F.(T) de .....

o

Substituting these values we have,

es~ = Ko R(t)+ Kl R'(l)+ ~~ R"(t) + .

where Ko ,K1 ' K2 are caned dynamic error coefficients.

To calculate these coefficients use the following method:

According to definition of Laplace transform,..

F(5) = f Fj (T) e-"''t dto

Now

Multiplying by c - S't to both sides,eo

Ko c-n = JF1(T) dr e-~"'f= F(s)o

Taking limit as s -+ 0 of both sides,

Lim Ko e - S't = Lim F((5)s -to 8 -to

Ko = Lim Fj (s)5_0

where 1Fl (5) = """"1+--=C-:-(s::-:"")H=(s"'""'")

Taking derivative of Ko e - 5t w.r.t. 'Sf we get,

r "

C;:ontrolSystem Engineering 7 -19 Time Response Analysis ofControl Systems

Substituting

-- 't J Fi (t) dr e - s'(

o

i.e,

Taking limit as s -+ 0 of both sides,

~ = lim d F1_(s)~-to ds

In general,

This method eliminates the disadvantages of static error coefficient method.Advantages of this method is,i) It gives variation of error as a function of time andii) For any input, other than standard input error can be determined.

. ~~+VJ_ Example 7.1 : A unrty feedback system has G(s) = (_ 1) (- 4)'s:;+ :;+

Determine (i) Type of the system, (ii) All error coefficients and(iii) Error for ramp input with magnitude 4.

Solution: To determine type of system arrange G(s)H(s) in time constant form.

40(s + 2) 40(2) (1+ 0.55)G{s)H(s) = s(s + 1) (s + 4) = s(1 + s) (4) (1 + 0.25s)

20(1 +O.5s)= s(1 + s) (1 + 0.25 5)

Comparing this with standard form,

G(K(l + lis) (1+ Tzs) .s)H(s) =si (1+ T..s) (1+ Tbs) ••••..

where = Type of system

= 1 so given system is type 1 system.

r "

Control System Engineering Time ResponseAnalysis ofControl Systems

7-23f·

es.(t) = KoR(t)+ K. R'(t) + ~~ R"(t)

R(t) = 6 + 5t + 6;2 I R'(t) = 5 + 6t I R"(t) = 6

7.346x 10-3es,(t) = 0.1428[6+5t+312)+0.0857{5+6t1+ '" x6.....

= 0.8568+ 0.714t + 0.4284 t2 + 0.4285 ....0.5142 t + 0.02203

e",(t) = 0.4284 t2 + 1.2282I + 1.3073

7.11 Transient Response AnalysisTransient response is the part of the total output which can not be separated out and

hence to analyse it, it is necessary to calculate total output c(t) for the applied input. Thiscontains steady state value along with transient response. From the expression of cCl) only,the transient response Ct(l) can be analysed.

7.11.1 Method to Detennine Total Output c(t)

. . C(s)1. Detemune the closed loop transfer function of the system R(s)'

C(s)R(s)

C(s)= 1± C(s)H(s)

2. Find the expression for R(s) from the information of reference input ret) to beapplied to the system.

3. Substitute R(s) in the closed loop T.F. to obtain expression for C(s).

4. Take Laplace inverse of C(s) by using partial fraction method to obtain totalc(t) of the system to the applied input ret).

AsThe transient output can be studied from the above expression. Transient response

may be exponential or oscillatory in nature.Lim

Key Point: TRking of c(t), the final steady state value Css of the output also can bet -+00obtained.

r "

Control System Engineering ,"J !..•.

7-24 Time Response Analysis ofControl Systems

7.12 Analysis of First Order SystemOrder: Order of system is the highest power of 's' in the denominator of a closed looptransfer Junction.

Consider a simple system shown in theFig. 7.18 (a).

v j (t) = 1, t 2: 0

= 0, t < 0

Find v oCt) i.e. rcsponc;e if it is excited byunit step input.

Vj (s) = 1/5 Fig. 7.18 (a)

Now first calculate system T.F. The Laplacenetwork is shown in the Fig. 7.18 (b). R

Vies) 1= 1(5) R + sCres)

... (1)

Vo(s)1= scI(s)

... (2)

Vo(s) 1 1Vj(S) = 1+ sRC = 1+ Ts..

Fig. 7.18 (b)

where T = RC

7.12.1Unit Step Response of First Order SystemLet input applied Vi (t) is unit step voltage.

Substituting Vj(S) = 1/5 in the transfer function

1 A' 5'= = - + ..,...---=-=.s (1+sRC) S 1+ sRC' A' = 1 and B' = - RC

1 1= ;- 5+(1 / RC)

r "

'., . . '''-Control System Engineering 7-25 Time ResponseAnalysis of~;(

Control Systems

Taking Laplace inverse,

v oCt) = 1 - e -lIRe ~ Css + ci (t) form

c, = 1and c1(t) = e-t/RC ISo

Now transient term is totally dependent on the values of R and C and its rate ofexponential decay will get controlled by '-1IRC' which is pole of the system.

Key Point: The values of R and C will not affect the steady state part.

The response will be as shown in the Fig. 7.18 (c).

t vJt)

0 0

RC 0.632

2RC 0.860

3RC 0.950

4RC 0.982- 1 Fig. 7.18 (c) Unit step response of firstorder system

The response is purely exponential.Now suppose input is changed to step of 'A' units.

Then Vies)A= 5

Vo(s) A A' 8'.. = = -+5 (1+sRC) s 1+ sRC

.. A = A '(1 +sRC)+s8'

.. A'RC+ B' = 0 and A'= A

B.

-ARC.. =V.,(s)

A ARC= _-.. 1+sRCs

A A= s+{l I RC)s

.. vo(t) = A [l_e-t/RC]

r "

Control System Engineering:~~'Io.f·.,.? 41'';~ • '•.=

7·26 Time ResponseAnalysis ofControl Systems

So rate of decay is not changed but thesteady state value has changed. Thecorresponding response can be shown as in theFig. 7.19.

But if values of Rand C are changed i.e.location of pole s = - l/RC is changed, thetransient output will behave in a diffurent way asrate of decay will get affected without change inits steady state.

The transient tenn is vanishing as it contains exponential term of negative indexwhich is only because pole of the system is having real negative part.

Fig. 7.19

Key Point: In general transient responsedepends on,

i) Poles of the closed loop T.F.ii) Location of the poles of the closed loop T.F. and it is independent of the magnitude of tIreinput applied. Any change in the magnitude oJ the selected input will not have any effecton the transient responseof the system.

7•.12.2 Closed Loop Poles of First Order SystemThe closed loop transfer function of a system is given by,

C(s) (;(s)R(s) = 1±G(s)H(s)

The equation which gives poles of system is called characteristic equation which is,

I 1+ G(s)H(s) = 0 IFor first order system this equation is also first order in general of the form,

I 1 + Ts = 0 IAs closed loop poles are the roots of characteristic equation, so for first order system

there is only one dosed loop pole i.e.I s = - i Imag.""y

The pole-zero plot is as shown in the Fig. 7.20.-1fT

Real

Fig. 7.20

r "


Key Point: The value of closed loop pole appears as an exponential index in the transientoutput.

e.g, in the system considered above the closed loop pole is s = - it and in the output

c(t) = 1 - e-I/I{{', the exponential index is - de which is nothing but the dosed loop pole.

Consider two systems, one with closed loop pole at s, = -2 while other having closedloop pole at S2 = - 4. In the output of two systems we will get the terms (,-2\ in firstsystem and e- 41 in second system. Now as t -+ 00,_ <,-II wiu vanish earlier than e-21. Hencesecond system transients will vanish quickly and will stabilise to its final value earlier thanthe first system.

From this we can conclude that locations of closed loop poles affect the transientbehaviour.

Key Point: As closed loop pole moves in tire left half anoayfrom the imaginary axis m tiles-plane, transients die out more quickly making system more stable.

). Example 7.4: FaT a first order system, find out the output of the system wizen the inputapplied to lire system is unit ramp input. Sketclr ti,e rtt) and dt) and show the steady stateerror.

Solution: Let the first order system has transfer functionC(s) I=R(s) (s+T)

r(t) =Input

R(s) =52

unit ramp input = t

1

1\ B C-:--- = '+-+--Ts2(s+1) s- s s+

. . A (5 + T) + Bs (s + T) + C 52 = 1

•• (B + C) s2 + (A + BT) s + AT = 1

AT = 1

C(s) =

1A=T

A+BT = 0 B=

B +C = 0 1C=T2

C(s)(VT) VT2 VT2= -----+--52 S S+ r

r "



.. c(t) = (~}- T~+T~e-rt

The sketch of c(t) and ret) is shown below. Take T = 1for plotting c(t).

c(t), r(t)

CCl) forT = 1

7.13 Analysis of Second Order SystemEvery practical system takes finite time to reach to its steady state and during this

period, it oscillates or increases exponentially. The behaviour of system gets decided bytype of closed loop poles and locations of closed loop poles in s-plane, The closed looppoles arc dependent on selection of the parameters of the system. Every system has atendency to oppose the oscillatory behaviour of the system which is called damping. Nowthis tendency controls the type of dosed loop poles and hence the nature of the response.

This damping is measured by a factor or a ratio called damping ratio of the system.This factor explains us, how much dominant the opposition from the system is to theoscillations in the output. In some systems it will be low in which case system willoscillate but slowly i.e. with damped frequency. In some systems it may be so high thatsystem output will not oscillate at all and not only that it will be exponential, so slow thatit will take very long time to reach the steady state. This damping ratio, explaining suchbehaviour is denoted by a greek symbol (Zeta) l;. Now as this measures the opposition bythe system to the oscillatory behaviour, if it ismade zero, (t; = 0) system will oscillate withmaximum frequency. As there is no opposition from system, system naturally and freelyoscillates under such condition. Hence this frequency of oscillations under ~ = 0 conditionis called natural frequency of oscillations of the system and denoted by the symbol(l)n rad / sec .

For a second order system the denominator of closed loop T.F. is quadratic and thecoefficients of this equation are directly related to ~ and (l)n as explained below.

r "

Control System Engineering 7 -29 Time Response Analysip.Control Sy9l..

The CL.T.F. (closed loop transfer function) for a standard second order system ~.Ikcsthe form as,

C{s) =R(s)

(fJ2n ... standard 2nd order system

Where characteristic equation is, s2 +2; (1)n s + (1)~ = 0The standard second order system is that where in C.L.T.F. numerator is (1)~.

Key Point: In practice it is 1I0t necessary tha! numerator must be alllJay.t;(J)~. It may beother constant or polynomial of '5' but denominator middle term coefficient and last ten"coefficient always reflect '2; (1)n ' and '(J)~' of the system respectively.

Hence always denominator of a I.F. must be compared with the standard formS2 + 2<;(1)n s + (fJ~ = 0 to decide the values of; and (l)n of the system. The numeratorshould not be used for comparison to obtain the values of t; and (1)n'

e.g. :

10 C(s)(5+1)(5+2)

Fig. 7.2110

C{5} G(s} (s+ l)(s + 2) 10.. R(s) = = =1...C(s)H(s) 1+ 10 52 + 3s + 12(s+ 1) (s 4- 2)

This CL.T.F. is not standard as numerator term is not (I)~ but denominator alwaysreflects ; and (J)n' The values can be decided by comparing the denominator with thestandard characteristic equation S2 + 2; (l)n s + (I)~ = o.

.. (1)2 = 12 i.e. (I)n = Jf2 rad / S(.'Cn

While 2~ (1)n = 3 .. ~ = Ju = 0.4332 12

7.14 Effect of S on Second Order System PerformanceConsider input applied to the standard second order system is unit step.

R(s) = lis

WhileC(s) =R(s)

r "


(IlC(s) = nS(S2 +2~(On s+ro~)

Finding the TOOts of the equation S2 + 2; ron s + (O~= 0

- 2~ ron ± .J4l;2 ro! - 4ro~i.e, 2

i.e,

We can write, C(s) = (02n

Now nature of these roots is dependent on damping ratio l; . Consider the followingcases,

I case1:1<li<-1

The roots are,

i.e. real, unequal and negative, say - Kl and - K22

C(s) = (On = A +_B_ + _s _S (5 + K1) (s + K2) .. S of KJ S + K~

Taking Laplace inverse, cCl)will take the following form,

cCl) = Cs,+Be-KJt+Ce-K2I,

where C'$ = Steady state output = AThe output is purely exponential. This means damping is SO high that there are no

oscillations in the output and is purely exponential. Hence such systems are called"Overdamped'.

Hence nature of response will be as shown in the Fig. 7.22.C(I) Overdamped

Critically damped

oFig. 7.22 l;~ 1

r "

Control Sysmm Engineering 7-35 Time Response AnalYSIs ofControl Systems

The partial fraction can be calculated for the Laplace inverse as below,

C(s) = a1 a2s+a3-+--=---=--_..:;:__-:-S S2 + 2; (J)ns + (J)~

a1(s2 + 2;(J)n !\+(J)~) + s(a2 s+ a 3)=5(52 + 2; (J)nS + ro~)

ro2• n•• S(S2 + 2; ron s + (J)~)

equating numerators on both sides,

ro~ = s2(a1 +32)+s{a1 2;ron +a3)+alro~

equating constant

equating coefficients of 52

equating coefficients of s

.• a 1 = 1, a 2 = - 1, a 3 = - 2l;ronAs assumed earlier for ease of computations.

at = 1, a 2 = - 1, a 3 = - 2a1 -s-2a

C(s) = - + -::-----::-S S2 + 2a..c; + (1)~

C(s) = ~-{s2 +s~~rod

Now consider s 2 + 2as+ (1)2n'

.... (Taking negative sign outside.)

For completing square,

(middle term):! 4a 2 ~Last term = = 'x 1 = a-4 x first term "t

So adjusting denominator as s2 + 2a.s + a 2 + ro~ - a 2

i.e. denominator = [s + a)2 + (J)~- ex:!

but

Substituting in above we get, (s +a)2 + ro~_;2 ro;

i.c. denominator = (s + a) 2 + (J)~ (1 _;2)

Now (J)d = (J)nJt _1;2

Substituting this in the expression of C(s) we get,

C(s) = ~-{(s+:~;:rod

r "

~~!'trol~.tem Engineering 7-37 Time ResponseAnalysis ofControl Systems

{H)e = tan-I ~ radians

Key Point : Substituting l; = 0 in this expression,we can find out expressionfor the outputfor undamped systems.

Important Remarks

1) The result derived is applicable for standard second order systems which isunderdamped and excited by unit step input.

2) If in the problem, the input is given as step of A units rather than unit step theneach term of the final expression of c(t) gets multiplied by A as,

o c(t) = A [1 ~,"n("'dt+e)] ... For step of A units

3) If the system is not in the standard fonn i.e. numerator of closed loop transferfunction is not (J)~but some other constant as,

<:(s) }(

R(s) = S2 + 2~(J}ns+(J)! K = constant

Then the above result can be applied by expressing qs)/R(s) as,

And in such case, the standard expression of c(t) gets multiplied by the constantK/ (I)!.

c(t) = ~[1- e~Sin«(J}dt+8)](J}n ,,1-~-

The values of ~ and (l)n must be obtained by comparing denominator of C(s)/R(s)witll the standard form.

-4) If numerator of C(s) / R(s) has some polynomial in s as,

C(s) P(s). C(s) 2s+4-- = I.e., -- = -::----R(s) s2+2~(I)s+(I)2 R(s) s2+105+64':t n n

Then the above result cannot be applied to get c(t). In such case, though l; and (l)n

values are to be obtained by comparing denominator with the standard fonn, theexpression for c(t) must be obtained by actually calculating partial fractioncoefficients,after substituting the proper value of R(s).

r "


5) For any other input, other than step the derivation is not applicable but the stepsof the derivation can be used as a guide line to obtain expression for c(t) in anyother condition.

7.16 Transient Response SpecificationsThe actual output behaviour according to the expression derived can be shown as in

the Fig. 7.26.

c t)Peakovershoot Mp

Toleranceband .t 2%~ Insteady stale, output

100%-li----f-i!-+-_-_-_+_*_\-_-_-_-_-=. =_ .... remainswithin!.2% error~~~ --i.---t band

Fig. 7.26 Transient response specifications

Let us define the various time response specificationsreferring to the Fig. 7.26.1) Delay Time Ttl : It is the time required for the response to reach 50% of the /tnal

value in the first attempt. It is given by,

I Td = 1+~~ I2) Rise Time T, : It is the time required for the response to rise from 100" to 90% ofthe final VIdue for overdamped systems and 0 to 100% of the final valm forunderdamped systems. The rise time is reci. ""Dcalof the slope of the response atthe instlmt, the response is equal to 50% of the final value. It is given by,

r "

Control System Engineering 7 -39 ~ Time ResponseAnalysis ofControl Systems

1t -0 sec where 0 must be in radians.

3) Peak Time Tp : It is the time required lor the response to reach its peak value. Itis also defined as the time at which response undergoes the first overshoot whichis always peak overshoot.

T =p1t 1t

-ro-d = ron ~1_ ~2 sec

4) Peak Oversl,oot M" : It is tI,e largest error between reference input and outputduring tI,e transient period.It also can be defined as the amount by which output overshoots its referencesteady state value during the first overshoot.

Mp = { c(t) It c 1p} - 1 1 is for unit step input

S) Settling Time Ts : This is defined as the time requir...d lor the response to decreaseand stay within specified percentage of its final value (within tolerance band).

Time constant of system

Ts = 4 x Time constant

Practically the setting time is assumed to be 4 times, the time constant of the system.

..... for a tolerance band of ± 2% of steady state

Key Point: 1 Time constant 7' is the time required by tire system output to reach63.2%of its final value during the first attempt.

7.17 Derivations of Time Domain Specifications

7.17.1 Derivation of Peak TimeTpTransient response of second order underdamped system is given by,

-~w Ie nc(t) = 1- r.;--;:r sin (rod t +0),,1_;2

r "

:Control System Engineering 7·40 Time Response ~nalysisofControl Systems

Where~e = tan"! "l/l-'l.

~

As at t = Tp' c(t) will achieve its maxima. According to maxima theorem,

So differentiating c(t) w.r.t. "t' we can write,- ~(I) t

• C n (-~ron)sin(rodt+e)t.e, - r:--;;:;,,1 _ ~2

.. ; sin (rodt + e) - ~1- ;2 cos (rodt + e) = 0

tan (rodt + 8) =~1_~2.. ;

Now e = t -1 ~1- ;2an ~

~1_;2= tane.. ;,

.. tan (rodt + e) = lane

From trignometric formula,

tan (n 7t +e) = tan ewhere n = I, 2, 3

But Tp , time required for first peak overshoot. •. n = 1rod Tp = 7t

r "


7.17.2 Derivation of MpFrom the Fig. 7.27, Mp = qTp) - 1

tl

Fig. 7.27

{ -to.T,sin (ol. Tp + Ol}-1Mp = l_c

~1_~2

-~CilnTp

Mpe sin (rod Tp + 6)=..~1_~2

But Tp = ~ , substituting above we get,(J)d

-~lIIn Tp

Mp-e

sin (n +6).. =~1_~2

Now, sin (rt +0) = - sin (0) J1- f,2-~CllnTp

Mpe sinO.. ;;;;

~1_~2 f,

Jl-~2. h . the F 728fig. 7.28

We know, tan 0 = ~ ass ownm 19•••

sinO oppositeside ~g 2 2.. = h but hypotenuse =( 1-~ ) +~ = 1ypotenuse

.. sinO = Jl-~2-tllln Tp ~1_~2

Mpe=..~1_~2 1

= -~(l)n Tpe

'" r "

,. : j Control System engineering Time ResponseAnalysis ofControl Systems

Substitute T = ~P rod

-n~

% M = 100 e Jl-~2p

7.17.3 Derivation of T,Time required by output to achieve 100% of its final value, starting from zero during

the first attempt is the rise time.

c(t)

Fig. 7.29

i.e. {c(t~l ..T,I = 1 for unit step input-I;,Clln r,

1 = 1- e sin(wd Tr+9}J1-~2

Equation will get satisfied·only if,

sin (wd Tr + 9) = 0

Trigonometrically this is true only if,

wd Tr+8 = n 7t where n = I, 2 .._.

r "


defining the term sensitivity for the system. Let us study the effect of parameter variationsin an open loop control system.

7.18.1 Effect of Parameter Variations in an Open Loop Control SystemConsider an open loop control system shown in the Fig. 7.30. The overall transfer

function of the system is given by,C(s) R(s) .~ C(s)

G(S) •= R(s)Fig. 7.30.. C(s) = G(s)·R(s)

Let !! G(s) be the change in G(s) due to the parameter variations. The correspondingchange in the output be !! C(s).

C(s) + A. C(s} = [G(s) + A. G(s)] R(s)

C(s) + A. C(s) = G(s)·R(s) + A. G(s)·R(s)

C(s) + A. C(s) = C(s) + AG(s)· R(s)

I A C(s) = A G(s)·R(s) '.• (1)

The equation (I), gives the effect of change in transfer functi0:r:"due to the parametervariations, on the system output, in an open loop control system. Let us discuss now, theeffect of such parameter variations in a closed loop control system.

7.18.2 Effect of Parameter Variations in a Closed Loop SystemConsider a closed loop system as shown in

Fig.7.31.The signal E(s) is the Laplace transform ofthe error signal e(t}. The overall transfer function ofthe system is given by,

= C(s) _ G(s)R(s) - 1+ G(s) lies)

C(s)

T(s)

Fig. 7.31

Let AG(s) be the change in G(s) which is due to the parameter variations in thesystem. The corresponding change in the output be AC(s).

[G(s) + !! G(s)]C(s) + A. C(s) = .R(s)

1+[G(s)+AG(s)] lies)

[G(s) +AG(s)]= 1+G(s)i-i(s)+!! G(s)H(s}.R(s)C(s)+A C(s)

r "


The term AG(s)H(s) is negligibly small as compared to G(s)H(s), as the change AG(s) isvery small compared to G(s). Neglecting the term A G(s)H(s) from the denominator, we get,

G(s}+ A G(s) .R(s)C(s) + A C(s) = 1+ G(s)H(s)

C(s) + A C(s) = G(s) R( ) + A G(s) R( )1+ G(s)H(s)· s 1+ G(s)H(s)· s

AG(s)C(s)+ A Os) = C(s) + 1 + G(s)H(s) . R(s)

AG(s)A C(s) = 1+ G(s)H(s) . R(s) ... (2)

The equation (2), gives the change in the output due to the parameter variations inG(s), in a closed loop system.

In practice, the magnitude of 1+G(s)H(s) is very much greater than unity.

IG(s)H(s)1» 1

Hence it can be observed from the equations (1) and (2), that in a closed loop system,due to the feedback, the change in the output, due to the parameter variations in G(s), isreduced by the factor [1+G(s)H(S)]. In an open loop system, such a reduction does notexist, due to the absence of the feedback.

7.18.3Sensitivity of a Control SystemThe parameters of any control system cannot be constant throughout its entire life.

There are always changes in the parameters due to environmetal changes and otherdisturbances. These changes are called parameter variations. Such parameter variationsaffect the system performance adversely. Such an effect, in the system performance due toparameter variations can be studied mathematically defining the term sensitivity of acontrol system. The change in a particuJar variable due to the parameter variations can beexpressed in terms of sensitivity as below :

Let the variable in a system which is varying be 'T, due to the variations in theparameter 'K' of the system. The sensitivity of the system parameter T to the parameter Kis expressed as,

S = "10 change in T0/0 change in K ... (3)

Mathematically, it can be expressed as

ST _ dln(T)J( - din (K)

r "


(a T/1)(aK/K) ... (4)

The symbolic representation s~ represents the sensitivity of a variable T with respectto the variations in the parameter K. In practice, the variable T may be an output variablewhile the parameter K may be the gain, the feedback factor etc. The representation si isalso called the sensitivity function of a system. For a good control system, the value ofsuch a sensitivity function should be as minimum as possible,

Let T(s) be the overall transfer function of a control system. The forward path transferfunction G(s) is varying. Then the sensitivity of overall transfer function with respect to thevariations in G(s) is defined as,

a T(s) / T(s)a G(s) / G(s)

ST = G(s) a T(s)G T(s) . a G(s)

For the open loop system,T(s) = G(s)ST = G(s) a G(s)

G G(s) . a G(s)= 1 .•. (5)

Thus the sensitivity of T(s) with respect to G(s) for an open loop system is unity.Let us find out the sensitivity function for a dosed loop system. For a closed loop

system,

T(s) C(s) G(s)= --=R(s) 1+ G(s)H(s)

aT(s) [1+ G(s)H(s}I[1]-[ G(s)][H(s)] = 1aGes) =

[1+ G(s)H(s)t [1+ G(s)H(s)rST G(s) aT(s)G = T(s) . a G(s)

G(s) 1=[ G(s) ] .[1+ G(s)H(s)_t1+ G(s)H(s) .

'" r "


ST _ 1G - 1+G(s)H(s) ••. (6)

Comparing the two equations (5) and (6), it can be observed that due to the feedbackthe sensitivity function gets reduced by the factor 1/ [1+G(s)H(s)] compared to an openloop system. And less the value of sensitivity function, Jess sensitive is the system to thevariations in the forward path transfer function G(s).

Sensitivity of T(s) with respect to H(s)

Let us calculate the sensitivity function which indicates the sensitivity of the overalltransfer function T(s) with respect to the feedback path transfer function H(s). Such afunction can be expressed as,

ST _ H(s) aT(s)II - T(s)' a H(s)

For a closed loop system,

T(s) = G(s)1+G(s)H(s)

aT(s)a H(s}

[1+G(s)H(s)][01-[G(s)][G(s)]= [1+G(s)H(s)t

H(s) -[G(S)ts~ = ---.--~--~~T(s) [1+G(s)H(S)]2

[1+G(s)ll(s)t

H(s) -[ G(s)t -G(s)H(s)= [ G(s) ]. [1+G(s}H(s)r = 1+G(s)H(s)

1+ G(s)H(s)

... (7)

It can be observed from the equations (6) and (7) that the closed loop system is moresensitive to variations in the feedback path parameters than the variations in the forwardpath parameters. Thus, the specifications of the feedback elements must be observedstrictly as compared to the specifications of the forward path elements.

7.18.4 Effect of Feedback on Time Constant of a Control SystemConsider an open loop system with overall transfer function as,

KG(s) = 1+ sT

r "


When this system is subjected to unit step input, its response can be obtained as,C(s) K ..... as open loop systemR(s) = 1+sT

R(s)1 unit step input= s

C(s) K..s(l+ s1)

c(t) = L-'[C(s)]= c,[~-Kl]= K[l-e 'IT]..(s+T)

So T is the time constant of the open loop system.Now the feedback is introduced in the

system with feedback transfer function asH(s)= h. This is shown in the Fig. 7.32.

Let us obtain the response of this closedloop system for unit step input. The overalltransfer function of the system is,

KC(s) f+STR(s) = K1+--·hl+sT

C(s}

Fig. 7.32

C(s) _ K = K = K/ TR(s) - 1+sT+Kh t+Kh+sT [s+e+~)]

1R(s) = -s

(K!1+Kh>j(1+Kh)s+ -T-

K K t (I+ Kh) K [ t]= -----.c--T- = __ 1-e (T/1+.Kh)1+Kh l+Kh 1+ Kh .

r "


Thus it can be observed that the new time constant due to the feedback is (T / 1+ Kh).Thus for positive value of hand K>1, the time constant (T / 1+ Kh) is less than T. Thus itcan be concluded that the time constant of a closed loop system is less than the open loopsystem.

Key Point : Less the time constant faster is the response. I fence the feedback improves thetime response oj a system.

7.18.5 Effect of Feedback on Overall GainConsider an open loop system with overall transfer function as Cfs), The overall gain

of such system is nothing but G(s).If the feedback with transfer function H(s) is introduced in such a system, then its

overall gain becomes [G(s) / 1± G(s)H(s)] . The positive or negative sign in thedenominator gets decided by the sign of the feedback.

For a negative feedback, the gain C(s) reduces by a factor [1 / (1+G(s)H(s»] .Due to the negative feedbackoverall gain of the system reduces.

7.18.6 Effect of Feedback on StabilityIt is discussed earlier that the feedback reduces the time constant and makes the

system response more fast. Hence the transient response decays more quickly.Consider an open loop system with overall transfer function as,

KC(s) = s+T

The open loop pole is located at s= -T.Now let a unity negative feedback is introduced in the system. The overall transfer

function of a closed loop system becomes,K

C(s)=R(s)

'S+T _ K1+~ - s+(K+ T)

s+T

Thus the closed loop pole is now located at s= -(K+ 1). This is shown in theFig. 7.33(a) and (b).

Imj Imj

---~,__-+---Real-(K +T)-T-~I-----+------ Real

(a) Open loop system (b) Closed loop systemFig. 7.33

r "


Thus the feedback controls the time respono;ei.c, dynamics of the system by adjustinglocationof its poles. The stability of a system depends on the location of poles in s-plane.Thus it can be concluded that the feedbackaffects the stability of the system. The feedbackmay improve the stability and also may be harmful to the system from stability point ofview. The closed loop systemmay be unstable though the open loop system is stable.

Key Point: Thus the stability of the system can be controlled by proper design andapplication of the feedback.

7.18.7 Effect of Feedback on DisturbanceEverycontrol system has some nonlinearitiespresent in it. The dominant nonlinearitics

like friction, dead zone, saturation etc. affect the output of the system adversely. Someexternal disturbance signals also make the system output inaccurate. Ihe examples of suchexternal disturbances are high frequencynoise in electronicapplications, thermal noise inamplifier tubes, wind gusts on antenna of radar systems etc. The disturbance may be inthe forward path, feedbackpath or output of a system.

(a) Disturbance in the forward path

Let us assume that there is a disturbance in the forward path of a control systemproduced due to varying properties of forward path elements or due to effect ofsurrounding conditions. Fig. 7.34 shows the disturbance signal Tel(s) produced in theforward path.

C(s)

Fig. 7.34

Assuming R(s) to be zero, let us obtain the ratio C(s)/ Td(s) to study the effect ofdisturbance on output. With R(s);;;;; 0, systembecomes.

C(s)

Fig. 7.35r "


The resultant elements are,

Gv;) = G2(S)

H'(s) = - G1 (s)H(s)

Positive feedbackNegative input

C(s) G2(s).. -= Td(S) = 1-[ G2(s)( - G1 (s)H(s»]C(s) - G2(s)..Td(s) = 1+G1 G2H(s)

C(s) = - Td(s)G2..1+G1 G2H

In the denominator assume that 1<<G1 G2 H hence we get,- Td(s)

C(s) = G. H(s)

Thus to make the effect of disturbance on the output as small as possible, the G I(s)must be selected as large as possible.(b) Disturbance In the feedback path

These are produced due to the nonlinear behaviour of the feedback path elements. TheFif!' 7.36 shows the disturbance signal Td(s)produced in the feedback path.

C(s)

Td(s)

Fig. 7.36With R(s)= 0, the effect of Td (s) on output can be obtained.'The system becomes,

Fig. 7.36(8)

r "

Control System Engineering 7·53 nme ResponseAnalpls ofControl Systems

C(s)Td(s)

For large values of GI' G2' HI ' H2, in the denominator 1 can be neglected.C(s) 1Td(s} = - H1(s)

Thus designing proper feedbackelement HI (s), the effect of disturbancein feedback path on output can bereduced.

C(s)

(c) Disturbance at the output

Consider that there is disturbanceTd(s) affecting the output directly asshown in the Fig. 7.37.

with R(s}= 0, we get

Fig. 7.37

C(s}

Fig. 7.38

C(s) =Td(s)1 1-:------=;;~=-o-=~

1-[ - G(s)H(s)] 1+G(s)H(s)

For large values of G (s) H (s), 1 in denominator can be neglected.

Td(s)•• C(s) = G(s)H(s)

Thus if disturbance is affecting the output directly then by changing the values ofG(s), H(s) or both the effect of disturbance can be minimised.

The feedback minimizes the effect of disturbance signals occurring in the controlsystem.

r "

-Control System E;ngineering..... "1 •••

7-54 Time ResponseAnalysis ofControl Systems

'-.. Example 7.5: The negative feedback control system has the forward path trallsjerfunction as,

10G(s) = s (s+1)

While the feedback path transfer function H(s) is 5. Determine the sensitivity of theclosed loop transfer function with respect to G and H at (1) = 1 radtsec. (A.M.LE.-97)

Solution: For the given closedloop system,10

G(s) = s(s+ 1)' H(s}=5

10= G(s) _ s (s+1) = 10

1+ G(s)H{s)- 1+ 10 .5 S2+ s+ 50s(s+1)

.The sensitivity of T(s) with respect to G(s) is given by,ST _ 1 = 1 :; s (s+ 1)G - 1+ G(s)H(s) 10 s2 + s+ 50

1+ s(s+ 1}·5

T(s)

To calculate S~ at (1)=I, substitute s = j (I)to convert the time domain function to thefrequency domain.

s~ = j (1) (1+ jro) =_ (1)2+ j (I)..

(jro)2+ j (1)+50 (50_ (1)2)+ j (I)

For the value of (1)= I,

ST =-1+ j1

G 49+ j 1

IS~I .Jf+T.. =~492 +1

= 0.02885The sensitivity of T(s), with respect to H(s) is given by,

10 .5S~ = -G(s)H(s) = s(s+1)

1+G(s)H(s) 1+ 10 .5S(5+ 1)

-so=

Replacing s by j (1), to convert the function to the frequency domain we get,-50 -50

SJI = = -:----=-:--\(jro)2+j(l)+50 (5O-(I)2)+j(l)

r "


At 00=1, S~I-so= 49+jl

ISill I-SOl.. =~492 + (1)2

...

= 1.0202

It can be observed that SJI is more than S~ i.e. the system is more sensitive to thevariations in H(s) rather than G(s).

'. Example 1.6: A position control system is shown in tile figure.

~KI s (s + aJC(s)•I

Evaluate the sensitivities s~I S!.K = 20 and a = 4

Solution : For a given system,

G(s) = and H{s)= 1S(5+ a)K

Now

K

T(s)C(s) 5 (5+a)

= R{s)=I+ K5 (s+a)

K= 52 + as+ K

a T(c;)

aK(52 + as+ K) (l)-K (1)_ 52+ as

= 2 - ?

(52 + as+ K) (52 + as+ Kf

5(5+a)

51+as+K

5 (s+ 4)(52 + 45+ 20)

r "

Control System Engineeringu: 7 ·62 TimeResponseAnalysis ofControl Systems

0.005 0.1=K

For any value of K greater than 20, e'5 will be less than 0.005. Hence the range ofvalue of K for c"s s 0.005 is,

20 s K<oo

'. Example 7.11: For a unity feedback system G(s) = /008) and ret) = 2t determines s +steady state error. If it is desired to reduce this existing error by 5% find new value of gainof the system.

Solution : The input is 2t i.e, ramp of magnitude 2, soK, will control the error.

Lim Lim 200Ky = 0 sG(s)H(s) = 0 s- ( 8)1 = 25s-+ s4 ss+

A 2C~S = K = 25 = 0.08

v

This error is to be reduced by 5% of existing value, with new gain of G(s) as K2instead of 200.

( 5) 5xO.08e!o_-<;l = cs" - 100 xe,,~ = 0.08 - 100 = 0.076

New error is 0.076.

KNew G(s) = 2 and H(s) = 1 with same inputs(s+8)

Lim s·K2 K.... Ky = sG(s)H(s) = -- = -~S40 5(s+8) 8

A 2 16.. e...",1 =-=--=-Ky (~1) K2

160.076 =

K2 = 210.52

So new gain is 210.52.

r "

Control System Engineering 7 - 63~.'I .t.ILli~1

Time Response Analysis oft Control Systems

). Example 7.12: The control SlJstem is shown belota. If ti,e input to the system isi) Ilni! step and ii) Unit ramp, find e.,.". (M.U. : Nov. - 96)

C(5)1

0.1 200t-----t + - I-------t

5 (5+1)(5+2).___--,

Solution: G(s) = (1 0.1) ( 200 ) =+S (5+ 1) (s+ 2)

(s + 0.1)200s(s+ 1) (s+ 2)

H(s) 0.2

G(s)H(s) 200(s + 0.1) 2= --- x 00 =S(5+ 1) (5+ 2) .

0.2(1+ 105)s(l + s) (1+O.5s)

200x 0.02x 0.1 (1 + lCs)-~----:~- x - -1x2 s(1+s) (1+0.5s)

i) For unit step input

LimKp = s ---+0 G(s)H(s) =

_1_=01+ Kp

Lim 0.2(1 + lOs) = 00s ---+0 s(l + s) (1 + 0.55)

e =S!'

ii) For unit ramp inputLim

K, = 0 sG(s)H(s) =s---+

Lim 0.2( I + lOs) = 0.2s -} 0 s(I + s){1+ 0.55)

e =~1 1-=- = 5x, 0.2

'. Example 7.13: Find the steady state error E, if T is unit step input and R = O.

c

r "

Control System Engineering. ,. 7-64 Time Response Analysis ofControl Systems

Solution : When system is not in the simple closed loop form then we can not apply theerror coefficients.

In such case, we have to use final value theoremLim Lim

i.e, e = e(t) = sEes}"'~ t~oo s~OI -

In the system given E(s) = - C(s) when R(s) = 0i....-.

Now let us find out i~:;,so that for unit step disturbance we can calculate C(s) and.hence E(s). When R = 0, summing point at R(s) can be removed and block of '-1' is to beadded to consider sign of the signal at that summing point.

Shifting summing point to right.

1+ 0.1 s10

C(s)

C(s)

0.1sx10'----t 1+ 0.18 1-----1

Combining the two summing points and redrawing the diagram.

-100(1 +O.1s)

1001-----1 (1 + O.lS) t----.J

CIs)

r "

Control System Engineering 7·65 nme Response Analysis ofControl Systems

Negative sign of (1 ~l~s) can be taken out to change sign of the signal at the

summing point from positive to negative.Now the two blocks are in parallel, in the feedback path.

C(s)

100 + s(1+0.1s)

100C(s) 7.. T(s) = 100 (100+ s)

1+-·S2 (1+0.1s)

. C(s) 100 (1 + O.ls)T(s) =

(0.ls3 + S2+ 100s+ 10000)

1 C(s) 1 100 (1 + O.ls)For T(s) = -, = -x

s S (0.153 + s2 + l00s+ 10000)

Now E(s) - C(s) =- 100 (1 + O.ls)=

s(0.ls3 + S2+ l00s+ 10000)

Lim.c"" = OS E(s)s-+

Lim { ,-100 (1+ O.ls) } 100e';'1 = s-+Osx s(0.ls3+s2+100s+10ooo) =-10000

Steady state error = - 0.01

' .. Example 7.14: Determine Kp, KI1alld KaJor a system with

Solution:

Gf ) 100 , H(s) = 1ts = 5(5+ 0.5) (4- 5) (5+ 1000)

100(M.U. : Nov. - 95)

G(s)H(s) =sx O.SX(I+O~S)X4X(I-i)X lOOX(l+l~)

G(s)H(s) 0.5= -s(=1-+-.::2,.....,s),.-(~1--'-0.25s) (1+O.Q1s)

r "

Control System Engineering:' I~t , ' 7 - 66 Time Response Analysis ofControl Systems

Lim Lim 0.500 K = o G(s)H(s) = = 00p s~ s ~ 0 s(l + 2s) (1- 0.25s) (1+ O.Ols)

Lim Lim sO.5 = 0.5K" = sC(s)H(s) =s~o s-+O s(l + 25) (1- O.25s) (1+ O.01s)

Lim .. LIm ~2 x 0.5K = s~G(s)H(s) = = 0

oJ S -40 S ~ 0 s(1 + 2..) (1- O.25s) (1+O.oIs)

'. Example 7.15: Find the steady state error for various types of standard test inputs for almity feedback system witll

G(s) = K·s(s+ 5) (s+ 10)

(a) K = 10 (b) K = 200 (M.U. ~May-1(95)

Solution: KG(s)H(s) = s(s + 5) (s+ 10) = K

=(5~)

s{1+ 0.2s) (1+ O.ls)

LimO

G(s)H(s) =S-4

(5~),.---=-:-_ = 00

s -+ 0 s(1:;' 0.2s) (1+ O.ls)Lim

LimO

sC(s)H{s)s-+

Lim Lim s2(5~)K" = 5 -40 s2C(s)H(s) = s -+ 0 s(l + 0.25) (1+O.ls) = o.

:. For step input of magnitude I,I 1e = --=-- = 0

S$ l+K 1+-p•••for any value of K.

For ramp input of magnitude I,1 50

eo;s = K" = K

a) For K = 10, e~~ = 5

b) For K = 200, c!!>,= 0.25

r "


For parabolic input of magnitude 1,

1 1e.... = K = '0 = 00.. ... for any value of K.

II'" Example 7.16: 111 the system gioen, the command input is R = 10 and disturbance signalis li. = 4, what is tire steady state error? (M.U. : Nov. - 94)

51+0.1s

100 C(s)t-_- ...

(s + 1)(1 .. 02s)

Solution: Using superposition principle, consider inputs separately.

a) R acting,

51 .. 0.1s

100 C(s)

(s + 1)(1 .. 0.2s)

G(s)H{s) 500.. = (1+ 0.15) (5+ 1) (J + 0.2s)

For step input

LimK., = G(s)H(s) = SOO

s-+O

A . 10 10.. e"",l 1+ Kp where A = magrutude of step = -- = --1+500 501

b) TL acting, R = 0

5 100 C{s)--1 + 0.1s (s + 1){1 + 0.2s)

TL

E(s) = - C(s)

r "


As system is not in standard fonn, error coefficientmethod cannot be used.

5 100 C(s)......._--1 + 0.1s (5 + 1)(1 + 0.2s)

100 C(s)

-51 + 0.15

+(s + 1) (1 + 0.2s)

100(1+ 5) (1+ 0.25)

1- 100 x( -5 )(1+ s) (1+ 0.2s) 1+ O.ls

100(1 +O.ls)C(s) =T(s)

Now

(1+ s) (1+ O.2s)x (1+ O.ls)+ 500

-4s -ve sign as TI. applied with -ve sign.T(s)

C(s)-400(1 +O.ls)= 5[(1+ 5) (1+ 0.25)X (1+ 0.15)+ 5001

= _ C(s) = +400(1 + 0.15)5Hl + s) (1+ 0.2s) x (1 + O.ls)+ 5001but E(s)

LimOsE(s) =s~

Lim 400(1 + 0.15)s ~ 0 (1 + s) (1+ 0.25)X (1+O.ls)+ 5001

:. Total error

= [~] = +4001+500 501

10 400 410ess = cuI + css2 = 501 +50r = 501 = 0.8183

,........En I 717 A d rd t . -, by C(s) 25~ F' d i . ....". mpe. : secon 0 er sys em IS glvm R( ) = 2 ~• In ,ts rise time,s s +6s+25

ptJlk time, peak overshoot and settling time if subjected to unit step input. Also calculateexpression for its output response.

Solution : Comparing the denominator of T.F. w'th the standard form S2 + 2; (l)n 5 + (I)~.,

25 and 2; (l)n 6(1)- = =n. , 1 . i 'I· '1

(l)n = 5 .. l; = 0.6

'" r "

Control System EngJneeri{'l9 t 1 7-69 Time Response Analysis ofControl Systems

e = tan " [JN2]~ = 0.9272 radians

Tr7t-9 7t-0.9272= --= = D.5535 secrod 4

Tp7t 7t=- = - = 0.785 sec

rod 4

%M = e -"f./Jl-f,2 x 100 = 9.48 %p

T:I4= -- = 1.33 sec

l;,ron

and c(t)e-f,"'nt e-3t= 1 - ~ sin (rod t+9) = 1 - sin(4t+ 0.9272),,1-l;,2 ~1-(0.6)2

= 1 - 1.5625 e-31 sin (4t +0.9272)

'... Example 7.18: For the system shown in thefigure obtain the closedloop T.F., dampingratio, natural frequency and expressionfor the output response if subjected to unit stepin~t. .

20 ,R(s)(5+1)(5+4)

20

Solution:1+ 20 s2 + 5s + 24

(s+ 1)(s+4)

Key Point : Now tlwugh T.F. is not in standardform, denominator always reflect 2~ ronand ro! from middle term and the last term respectively.

C(s) =R(s)(s+ 1)(s+ 4) 20=

.. ro2 = 24 .. (I) = 4.8989 l'adl!'ecn nr I

2~ron = 5 .. l;, = 0.51031

rod = ron Jl-~2 = 4.2129 radlsec

Now for c(t) we can use standard expressi~n for ~~:~ in standard form. So writing

r "


CCs) 20 { 24 }-R(s) = 24' s2+5s+24

For the bracket term use standard expression,multiplying this expression by constant ;~.

20 [ e -~fIIn l.. c(t) = -2 1- ~

4 ,,1_~2

and then c{t) can be obtained by

e = tan-1 ~ radians = 1.03radians

c(t) = ~~ [1- 1.1628e - 2.51 sin (4.2129 t + 1.03)]

'.. Example 7.19: For a control system shown in figure, find the values of KI andK2 sothat Mp = 25% and Tp "" 4 sec. Assume unit step input.

C{s)

Solution: G(s) = ~ , H(s) = 1 + K2 sS2

T.F. =

:. Comparing with standard fonn,

Now Mp is function of ~ alone,

.. % Mp = e-",VJI-f.2 xl00

25 = e-"'VJl-f,,2 x 100

0.25

r "

Control System Engineering 7·74 Time Response Analysis ofControl Systems

Now the time for 1 cycle is known and if it is known to us that what is the timerequired by the system to achieve steady state, we can find how many cycles output willperform before reaching steady stale.

4T = --=2secII ~ ron

So 1.6792 sec for one cycle, how many cycles output will perform in 2 sec.2

:. Total no. of cycles = 1.6792 = 1.191

Output will perform 1.191 cycles before reaching the steady state.

This can be shown as in Fig. 7.39.

c(t)

time

Fig. 7.39

The frequency in Hz if required can be obtained as,

1fd T = 0.5955 Hz

fd = Frequency of damped oscillations in Hz.

'. Example 7.22 : A second order system is representedby the transferjunction,

Q(s) 1I(s) = }S2 + fs+ K

A step input of 10 Nm is applied to the system and the test results are,a) Maximum overshoot = 6 %b) Time at peak overshoot = 1secc) The stetldy state value of the output is 0.5 radiansDetermine the values of 1,f and K.

_'I , ,:r' ,

•'l.. Control System Engineering 7-75 Time Response Analysis ofControl Systems

Solution : Arrange the given transfer function as,

Q(s) =l(s)

ro2 K i.e. ron = If= Tn

and 2~(I)n f i.e. f= j ~ = 2JKf

Now Mp = 6% i.e., 0.06

0.06 = -1tVJl-~2.. e

. In (0.06) -1t~=Jl-~2

... (1)

.•. (2)

Solving for ~ , ~ = 0.667 ... (3)

1tron = = 4.2165 rad/sec~1-(0.667)2

... (4)

The Laplace transfonn of output is Q(s).Now input is step of 10 Nm hence corresponding Laplace transform is,

I(s) 10= -sQ(s) 1

e~)= Js7+fs+K

Q(s) = 10S(JS2+ fs+ K)

The steady state of output can be obtained by final value theorem.Steady state output = Lim sQ(s)

....0

0.5 = Lim s·10 = 10.... os(Js7+fs+K) K

_\1 , ,:r I,

Control System Engineering 7·76 Time Respons4f Analysis ofControl Systems

.. K = 20

Equating (1) and (4),4.2165 = ~

.•• (5)

.. 4.2165 = F1. J = 1.1249

From equation (2), 0.667 = M2 KJ

0.667 f.. =2.J20x1.1249

.. f = 6.3274

•,.. Example 7.23 : The open loop T.F. of unity feedback system is G(s) = s (1~ Ts) . For thisl:>ystem overshoot reduces from 0.6 to 0.2 due to change in 'K' only.

TK -1Show that ' = 43.33 where Kl and K, are values of K for 0.6 and 0.2TK2-1 ~

overshoot respectively.• G(s)

Solution: Closed loop T.F. = 1+ G(s} , H(s) = 1

K Ks(l + Ts) K T= K = = 2 1 K1+ s2T+s+Ks(l + Ts) S +Ts+ T

fiiComparing with standard fonn 2 n 2

S +2; (J)n S+ (l}n

002K

(J) =~= Tn n T

21;(l)n 1 1;= 1 1= =T2~'T

2.JKT

Now for Mp = 0.6 Let l; = l;J

.. 0.6 = e -ft~tNl-~

i.e, -0.51 = 7t l;t

~1-l;~...

_\1 , ,:r I,

Control System Engineering 7.78 Time ResponseAnalpls ·ofControl Systems

.,

As l; > 1, system is overdamped, hence the output will not contain any oscillations.Hence standard expression for c(t) cannot be used.

Now input is unit step, so Xes) = l/s Substituting in T.F.12yes) =

s{s2+ 7s+ 12)Use partial fraction method

= 12 ABC--:----::=-:--= = - + -- + --s{s+ 3) (s+ 4) s s + 3 s + 4

where, A = 1, B= - 4, C = 3143

yes) = s- s + 3 + s + 4


yet) = 1- 4e- 31 +3e- ct

... Example7.25: For a control system shown in figure, find the values of K and K, 50 thatthe damping ratio of system is 0.6 and settling time is 0.1 sec. Use T. - 3.2 Assume- !;(I),,'unit step input.

C(s)

Solution : Using block diagram reduction rule, reduction of inner loop is,

1001+ 0.2s 100100

1+ 1+ 0.2s .K,= 1+ 0,2s+ 100x,

Overall G(s) = K. 100 ._1_= 5K1+0.2s+ 100Kt 20s s[1+100Kt +O.2s]

H{s} = 15K

T.F.G(s) s (1+ 100K, + 0.2s)= =1+ G(s)H(s) 1+ 5K

s{l + 100KI + 0.28)

r "

Control System Engineering 7 -79 TIme Response Analysis ofControl Systems

5K=0.252+ 5(1+100K, ) + 5K

But coefficient of s2 in the denominator must be 'unity' to compare it with thestandard form. So dividing it by 0.2.

Now

Using

i.e.

and

C(s) 25KR(s) 52 + 5s (1 + 100 Kt) + 25K

ol = 25 K, ron = 5.JKn

2 ~ron = 5 (1 + 100 I<t)

l; = 5(1 + 100Kt)

lOJK

~ = 0.6 and Ts = 0.1 sec

r; 3.2~ron

0.1 3.2= 0.6 ron

ron = 53.33 rad/sec

ron = s.JKK = 113.78

~ = 5(1 + 100 1(1) = 0.610.Ji(

Kt = 0.118

'. Example 7.26: For the system shown in figure show that system is always overdamped,independent of the selection of Rand C.

Solution : To prove that system is overdamped means to prove ~ > 1 and not dependenton the values of Rand C.

So first to find its C.L.T.F. ~~~:~use signal flow graph method.

r "

Control System Engineering-,


I}(s) = Vj -VI.. R

Vt(s)1= (It -12). sC

12(s) = VI -VoR

Vo(s) 1= 12.- sC

r, Signal flow graph is shown in the following figure.

VIo

11R•

-l/R -1 -11RsC

1

L1 and L.) is combination of 2 nontouching loops.


At = 1 as all loops are touching to T,1

C(s) =R(s)

1

=

Comparing with standard form,

1C2 R2

1•• wn= CR rad/sec

r "


2 J: "'n = 3':t'" CR"

3 1 3l; = CR .2: .CR= 2" = 1.S

As l; > 1 system is overdampcd and l; is independent of R and C values.For this, two resistances must be equal, and two capacitor values must be equal.

,,.. Example 7.27: A position control system driVesa load through a 50 : 1 gear ratio. Theinertia of motor shaft is 20 x 10-5 k.~·nl and friction is 60 x 10-5 N-m sec. Tne torqueconstant is 0.04 N-m per degreeof error, if output speed is 20 rpm. CalculateiJ l; ii) ro" •

Solution: Let T = Torque produced proportional to '8; -80'

where, 9i = Reference input position

80 = Output position

Used to drive load of M.I. T and friction 'B'

T = J d2 eo + B deodt2 dt

K [0 •.-90] = J d260 + B dOodt2 dt

...

Taking Laplace we have,

K e; (s) KO o(s}+ J s2 0o(s} + BsO 0(5)

KT~------ = --~~-=

}S2 + Bs+ K 2 B Ks +TS+T

K

K = 0.04 N-m/deg = 0.04 x 57.3 N-m/radian

Gear ratio = 1/50

:. K at motor shaft = 0.04 x 57.3/50

r "

Control System Engineering Tim. Response Analysis ofControl Systems

0.04 x 57.3I-----=-5 = 15.13 radlsec5Ox20x10-

B~ = -- = 0.0992JfKI)" ExMIple 7.28: A system has the following transfer function

C(s) 20=R(s) s + 10

Determine its unit impulse, step and ramp response with zero initial conditions. Sketch theresponses. (M.U. : May-%)

Solution : a) Unit impulse input

120

s+ 10

1-,.s, 20s(s+ 10)

A ==2,

R(s) =C(s)as R(s) =

C(s) =

.. c(t) =

.. c(t) =b) Unit step input

.. R(s) =

C(s) =

where2C{s) ==

20s+ 10

L-1 {~}. s+ 10

e-tDt

A B+ -8+10--s

2

cCl) = L-1 {~- S;lO}c(t) = 2 - 2e -lOt

c) Unit ramp input1R(s) =S2

C(s) = 20 ABC-=---==- +- +--52(5+10) 52 S 5+10

r "

Control S,.eem engineering 7-83 Time R_pon .. Analysis ofControl Systems..... -,

A(5+ 10) + Bs(5+ 10) + Cs2 = 20

.. B+C = 0, A + lOB = 0, lOA = 20

.. A= 2, B = - 0.2, e =0.2

. C(s) 2 0.2 02= ---+--52 5 5+ 10

.. c(t) = L-1 {.3.. _ 0.2 + ~}52 5 5+10

.. dt) = 2t - 0.2+O.2e-lot

The responses are,

c(t)

20

Unit impulse

o

c(t)

2t---------~~~==~Unitslep

c(t) _--_ ...-..........................,,"Unit ramp

r "


'... Example 7.29: Derive an expressionfor the time responseof a second order systemsubjectedto a unit impulse input for ~ < 1, ~ >1, where~ is a damping ratio.

Solution : The second order system transfer function is

C(s) =R(s)oln

The characteristic equation is,

s2+2~(I)ns+ro~ = 0

where ~ = Damping ratio

The roots of this equation are,

- 2~(I)n± ~ 4~2(I)~- 4(1)!51•2 = 2 =-l;(I)n±(I)n~l;2-1

= - ~ (l)n± (l)nJl -I;2(l)d = (l)n~1-l;2 .•where

Case i): ~<1, under damped.

The roots are complex conjugates. The input is unit impulse,

R(s) = 1

(1)2C(s) = n

52 + 21; (1)n5+ (1)~

(JlC(s) = n

S2 + 2 ~ ro 5+ ~2(1)2+ (1)2_ ~2(1)2n n n n

(1)2 .,(1)- rodC(s) = n = ..2!..{S+~ron)2 +(I)!(1_~2) rod {s+ ~ (l)n)2 + (I)~

L-1{ co } = e - at sin (l)t..(s+a)2+(1)2

[ 2 ]L-1[ees» = L-l~. CIld..(J)d (s» lJ.CI\t)2 + ~

2c(t) = m;; 'e-tcon lsin (J) t..

CIl"Jl- ~2 d

'" r "

C~~ISysam engineering~ .T_' ,. _'I •

7-85 Tim. Response·Analysis ofControl Systems

c(t)

Case ii) ~ = 1, critically damped

R(s) = 1as unit impulse input

C(s)ro2 ~;;;;; n =

52+2ro s+ro2 (s+ ron)2n n

Now L-1{e-altl 1= (s+a)2

L-1[C(s)] = L-'[ 2 1 ]ron· (S+ron)2

c(t) = ro2. t e-lIIntn

~ > 1, overdamped

R(s) = 1 as unit impulse input

Case iii)

The roots are real and negative

Finding partial fractions we get,

.. L-1[C(s)j = c(t)

c(t) = ~ [J-e .. - .. N-' ]0J~..,,_J~2_,].]..2CJ)n~~2-1

cCt) = ro,. [ -e-..".(=i..«~-R=i)]..2~~2 -1

r "


The responses are shown in the following figure.r(t)

Unit impulse

c(t)

'... Example 7.30: The open loop transfer function of a unity FBCS (Feed Back Control

System) is given by G(s)= '(: 1)'s s +

i) By what factor the amplifier gain K should be multiplied so that damping ratio isincrea:sed from 0.2 to O.S.

ii) By what factor the time constant T should be multiplied so that the damping ratio isreducedfrom 0.6 to 0.3. (M.U. : Jan.-92,. May-97)

Solution: Given G(s) = s (S~+1) and H(s) = 1

KC(s) G(s) s(sT+ 1)..R(s) = =1+ G(s)H(s) 1+ K

s(sT+ 1)

K KIT= =Ts2 +s+K 2 1 KS +-s+-T T

Comparing denominator with S2 +2;(I)n s+ro;,

2 K(I) = - andn T

and

r "

Control System Engineering-. :--'!'- ,-


Case i) : 2;1 = 0.2, K = K}while ;2 = 0.8, K = K2, T = constant1

0.2 = 2/1<.,=1-

1

and 0.8 = 2.J K2 T1

0.20.8 -

1 K2 ... squaring both sides.. 16 ::;

~

K2 = 1.. 16K1

So K must be multiplied by Vl6 to increase; from 0.2 to 0.8.Case ii) : ;1 = 0.6, T = T} while ;2 = 0.3, T = T2,K = constant.. 0.6 = 1 and 0.3 = 12.J KTI 2.J KTz

!f;0.3 = ..rr;0.6 . 2 rr;

I.C. = V"T;

i.e.

So T must be multiplied by 4 to reduce; from 0.6 to 0.3.

'. Example 7.31: For the system shown in the given figure, calculate the sensitivity of theclosed loop system with respect to the function G(s). Does the sensitivity depend on U(s) orM(s) ?

M(s)

-1

Solution': From given signal flow graph,

Tl = U(s)Q(s}G(s}, T2 = M(s)G(s}

Ll = - Q(s)G(s),

T(s)= Y(s) _ U(s) Q(s)G(s)+M(s) G(s)

R(s) - 1+Q(s) G(s)

G(s}[U(s)Q(s) +M(s}]::;

l+Q(s)G(s)T(s)

r "

tK. Control System Engineering 7 -91 l!Time Response Analysis ofControl Systems

u.. Example 7.34: A second order system lias unity feedback and open loop Irmrsfi'" [unction.

G(sJ= 500s(s+ 15)

a) Draw the block diagram for closed loop system.

b) What is characteristic equation?c) What is damping ratio and natural frequency values?d) Calculate Tp (peak time), M" (peak overslzoot) and T; (settling time) for the system outputresponse tohen excited by unit step input.e) Sketch the transient response for unit step input./) If input is ramp of 0.5 rad/sec, calculate steady state error. (M.U. : Nov. - 95)

Solution : a) Block diagram is,

C(s)•I

b) Characteristic equation is

1 + C(s)H(s) = 0 i.e. 1 + s{:2~5)= 0 i.e, s:!+ 15s+5110= 0

c) Comparing above with standard equation

S2 + 21;(I)ns + (I)~ = 0.,

(I)~ = 500 i.e, (l)n = 22.36068rad/sec

i.e.

System is underdamped.

d) (l)d (l)n ~1-l;2 = 22.36 x F- (0.335-1) '2 = 21.0648 rad/sec.

Tp1t 0.1491 sec.= - =(l)d

%Mp = e-rV./Fl;2 x 100 = 32.677%

r; 4 4:: £;(I)n' :: O.335fxi2.36 :: 0.5333sec.

r "

Control Sysl!.m Engineering 7-92 Time Response Analysis of r..Control Systems

e}

c(t)

._.__. ./. ± 2%

1t=t=~t==t==~~~====--r-------·---I-fI---...;..._----__._,-----_ time

0.533sec

0.149sec

f) ret) = 0.5t

R(s) 05= 70.5

Lim sR(s) Lim s·_?

eS.~ = = s-s-+O 1+ G(s)H(s) s-+O 1+ 500

s(s+ 15)

Lim 0.5 0.5 xIS= sxSOO = 500s-+O s+ s(s + 15)

Css = 0.015

,_ Example 7.35 : For a unity feedback system G(s) = s(s :'~7i)'Determine characteristic

equation and hence calculatedamping ratio, peak time, settling time, peak overshootandnumber of cycles completed beforeoutput settlesJor a unit step input. (M.U. : May - 93)

Solution : Characteristic equation is 1 + G(s)H(s) = 0

. 1+ 36 0s(s+ 0.72) =

i.e. s2 + 0.125 + 36 = 0

Comparing with? ?s - + 2l;wns + w~ = 0

? 36, con = 6 rad/sec.. w- ::: ..n

and 2l;con = 0.72, .. l; = 0.06

COd = {J)n ~l- ;2

.. rod = 5.9891 red/sec.

'" r "

:. eq;wirol System Engineering,.~ I '.

7-93.,t_ Time ResponseAnalysis of

Control Systems

Peak time Tp 1t 1t.. = rod = 5.9891 = 0.5245 sec.

Settling time Ts 4 4= --= 0.06 x 6 = 11.11 sec.l;ron

Peak overshoot Mp = -rtVJI-T._i = 0.8279e

i.e, %Mp = 82.79%

rod = 5.9891 rad/sec,

= 2ft fd

roEd = -2~ = 0.9531 cycles/sec.

Td1

1.0491 sedcycle ... Time for one cycle.= 0.9531 -

Now

i.e.

Settling time is 11.11 sec therefore output will perform,

11.111.0491 = 10.59 ... Cycles before it settles.

'. Example 7.36: Followitl8 expression denotes the lime response of a servomechanism.

c(t) = 1 + D.2e-60t - 1.2 e-lOt

iJ Obtain tire expression for tire closed loop transferjunction of tire system.ii) Determine tile undamped natural frequency and damping ratio. Assume unit step input

(M.U. : Nov. - 96)

Solution:

i} c(t) = 1 + 0.2C-6l' - 1.2c-10l


C( ) 1 0.2 1.2S = S + s+ 60 - s+10

= (s + 10)(s + (0) + 0.2S(8 + 10)- 1.25(S + (0)s{s+ 60) (s+ 10)

= S2 + 70s +600 +0.2s2 +25- 1.252 - 725s(s+ 60) (s+ 10)

r "

Control System Engineering ,;. 1-94 Time ResponseAnalysis ofControl Systems

1R(s) =5

600

Closed loop transfer functionC(s) s(s+ 60}(s+ 10)

= = = ----=-----R(s) 1 52 + 70s+ 600

s

ii) Characteristicequation is

52 + 70s +- 600 = 0

Comparing with

S2 + 2~ rons+ (I)~ = 0.,

600, :LOn = 24.4948 rad/sec.(I)~ =n

and 2~(I)n = 70, ~ = 1.4288

600

As l; > 1, system is overdamped hence output is purely exponential withoutoscillations.

I'.... Example 1.31: A position servomechanism has as error detector gerlerating 1V/rad error.It is followed by an amplifier of gain 'K'. The amplifier feeds the armature of a motor whichhas armature resistance oj 1nand negligible inductance. TILemotor generates O.IV/(rod/sec)back e.m.f and lias 1= 0.1 kg _",2 and viscous damping is negligible. The motor is coupledto the load through a reduction gear 'luitl, ratio 5. Tire load has1= 1kg-m2 and viscous damping constant B = 0.5 Nm!(rad/sec). Determine the closed looptransfer function of the system. Also find the damping factor. Take K = 100. What is theeffect of increase in K on damping factor. (M.U. : Nov. - 95)

Solution : The circuit diagram is

Error detector

Reduction gear

N, f

J=1B =0.5

Jeq Jmolor + hOOd (~: r cr 0.14kg-m2= = 0.1+ 1 x 5" =

Bt.") Bml)IOr + BlQad (~: r cr 0.1 Nm/(rad/sec)= = 0+ 0.5x 5 =

'" r "


The block diagram is

Ke Error

detector

• Kb = 0.1V!(rad!sec), • K 100

• K\! = ] V! rad I

• Assume torque constant KT of the motor as 1.

15(0.145+0.1)

1 + S (0.145+ 0.1) x 0.1s

15 (0.145+ 0.1)

120 x --:-::--::---,--:,- 1---....-....S (0.145 + 0.2)

20eC<5) 5(0.1-15+0.2)ares)

=1+ 20

5(0.145+ 0.2)

6c(s) 20 142.857= =6r(s) 0.14s2 + 0.25+ 20 S2 + 1.42855+ 142.857

The characteristic equation is

52 + 1.42855+ 142.857 = 0

Comparing with

52 +2e ron 5+ ro~ = 0

r "

Control. System EngineeringI· ~. l"1li

7 -96 Time Response Analysis of t:.').Control Systems •

.,(J)~ == 142.857, ron ==11.952rad/sec

and 2~m" == 1.4285, :. ~ ==0.0597

Damping factor is 0.0597.

If gain K is increased, then (O~ value will increase but '21; (l)n' will remain constant.Therefore increase in K will cause decrease in damping factor.

n_ Example 7.38 : A system has 30% ooersnoo: and srttlirtg time of 5 seconds,Jor a unitstep input. Determine the transfer function. Calculate peak time and output response.Assume <; as 2%. (M.U. : Nov. - 95)

Solution:

30 == e ,uJ.[t:E.2 x 100

== e -rrE,!Jt - 1.20.3

. In(O.3) ==~1 - Z;:

l;2 == 0.128, i.e, l; == 0.3578.JT. == --==5!;(J)n

(J)n == _~ = 2.2358 rad/sec:l."

C(s)T.F. ==R(S)

.,(1)- ".9988= n_~__ = .__ _

52 + 2):(1). ..:+ I,;!. 52 + 1.599s+ 4.99RR., .,

Where

a = tan-1 [~] radians = 1.2048 rad

c(t)C-O·81

= 1 - sin(2.OS77t + 1.2048)~l - (0.3578)2

c(t) = 1 - 1.07 e -O.8t sin(2.0877t + 1.2048)

r "

- l-

.i Control SystemEngineering~:t:

7 -97 Time Respon.. Analysis OfControl System.

'... Example 7.39: A closed loop system has two complex conjugate poles at 51' ~2= -2 ± j 1.Determine the form of transfer function and values of 0>" ' 'fp' T, ' ~< and M" assumingstandard second order system. (M.U. : Nov. - 95)

Solution:C(s) 0>2nR(s) =

s2 + 2~O>ns+ (I.)~ .while -2 ±jl '.

Sl ' 52 =• Denominator of T.F = (5 + 2 + jl) (s + 2 - jl) = (5+ 2)2 - (jI)2

= s2 + 4s+5

<:(s) 5=R(s) S2 + 4s+ 5

Comparing with standard form

.. (1.)2 = 5, :. (l.)n= 2.236n

and 2~(I.)n 4, 4= .. ~ = -- = 0.89442x O>n

(l.)d = (l,)n~I- ~2 = 0.9999 red/sec ""1 rad/sec

G [R'] radians= tan -1 ~ rad = 0.4636

Tp7t

= - = 3.1415 secrod

Ts4= -=2sec

~ron

r; 7t-G= -- = 2.6779 sec

rod

% Mp = c-fr1jJI-F.2 X 100 = 0.1867%

.-

'. Example 7.40: find the impulse responseof electricalcircuit given below.(M.U. : May-95)

1 F 1U

1° I 1el 2U 2F eo

1 1 '" ''"I . r "

Control System Engineering 7 -98 Time Response Analysis of . 'FControl Systems

Solution: Taking Laplace of the network

1/5 V1 eor 1I-------r---'\MI\r--r------o 1fo-s_> __ I_J__ 2__.___ ')__ ---,_;_5_-01"'5)

II = Cj - "1 = s{c, -VI>1s

vI = 2(1]-12)

(2 = \~I - eo1

121

eo = X -2s

•••.. (1)

.... (2)

..... (3)

..... (4)

-s -2

[I = 1

L] -2s L2 = - 2 L3 = 1= - 25

:. Signal flow graph is

L) L3 = Non touching = ~

= 1+2S+2+21 +1s

1=4+2s+2s

L\1 = 1 as all loops are touching to T]

eo(s) = T) L\1 Using Mason's gain formulaej (s) =r:

1 25= =1 4s2 + 8<:+ 14+ 25+'2 s

eo(s) 0.55=C, (s) S2 + 25+ 0.25

r "

Control System Engineering 7-99 , Time RespOnse Analysis ofControl Systems

For impulse input ells) = 1

e.,(s)O.5s 0.5s=

S2 + 2s+ 0.25= (s+ 0.1339) (s+ 1.866)

A B= - +(s"t" 0.1339) (s + 1.866)

A = - 0.0386, B = 0.5386

(~,,(s) -0.0386 0.5386= +..(s+ 0.1339) (s + 1.866)

.. cCl(t} :::; L-1{c,,(s}}

eo(t) = 0.5386 e -1.6661 - 0.0386 e~.13391

eo(t)

,_ Example 7.41 : A system has unit step response of c(t) = 1 - e-(lIl. Determine itsimpulse response and ramp response. Assume zero initial conditions. (M.U. : June - 94)

Solution : For unit step input

R(s) = !sTaking Laplace of c(t),

C(s) :::; ~ - s+10.1 ". S(S~~.l)

0.1TF C(s) :::;. R(s)

s(s + 0.1) 0.11 ". 5+0.1s

For impulse response, R(s) = 1

C(s} = 0.1s+ 0.1

c(t} = L-1 {C(s)} = 0.1 e-()·lI

r "

Control System Engineering 7 -100 Time Response Analysis of ~J;""Control Systems

For ramp, 1R(s) =

ABC=-+-+--52 S s+0.1

A(s + 0.1) + Bs(s + 0.1) + Cs2 = 0.1

B+C = 0, A + O.IB = 0, O.lA = 0.1.. A = 1, B = -10, C = 10

C(s) 1 to 10.. = - - - +_ .._--s2 5 S-4-0.1

.. c(t) = L-J (C(s)J

= t - 10 + 10e-l).u

n_ Example 7.42: Theopen loop transfer function of lmity feedback system is

ao = --:=,.....K---=::(f!'+ 1)

By what factor the gairz K should be multiplied so that damping ratio is increased from 0.3to 0.8. By U'/ratfactor time constant slzould be multiplied so that damping ratio is reduced

from 0.6 to 0.4. (M.U. : Jan.-92, May-97)

Solmon : The characteristic equation is 1 + G(s)H(s) = 0

1 + K = 0•• 5(T5+ 1)

TS2 + s + K = 0

Dividing by T, to compare it with 52 + 2~(1)n5+ (1)~= 02 1 K5 +- s+- = 0T T

(J)2 K:. (J)n = ~.. = rn

and 2~(1)n1 1= 1" .. ~ = 2m

a) ~ is increased from 0.3 to 0.8

lit = 0.3, letK=Kt and ~2 = 0.8, K = K2

l;J1 1= ~2 = 2..JK,.T..

2.JK;T

r "

C9ntrol System Engineering 7 -101 T"me Response Analysis ofControl Systems

~J =~

0.3=~..li2

.. O.B KJ

.. K2 = 0.1406 KJ

So gain should be multiplied by 0.1406.

b) ~ is reduced from 0.6 to 0.4

~l = 0.6, letT=Tt and ~2 = 0.4, T = T2

li1 1 1.. =2.JKTJ ~l = 2JKT2

lil =~

O.b=~..li2

..0.4 Tl

.. T2 = 2.25 TJ

So time constant should be multiplied by 2.25.

II. Example 7.43: For tile centro system shown

Showi) Trajectory on which parabolic error constant K" is 40 ~ec-2ii) Trajectory on wlziclz roll = 40 rad/sec.on the parameter plane K, - Kn' Take KIJ on x axis and Kp on yaxis. (M.U. : Nov.-96)

Solution:

G(S)H(5) ;;::

10(Kp + Kns)52

Lim52 G(s)H(s}i) Ka ;;::

5--70

Lim s2.lO(Kl, + K[)s)

= S-70 ~s~

40 ;;:: 10 Kp

Kp = 4 is line on which Ka = 40 sec-2

r "

Control System· Engineering 7 -102 Time Response Analysis ofControl Systems! .

ii} Characteristic equation is 1 + G(s}H(s) = 0

lO(Kp + Kosj1 + 2 = 0

S

52 + 10KDs+ 10Kp = 0

Compare with

s2 + 2~rons+ ro! = 0

ro~ = 10Kp

40 = 3.162.JK-;:ron= 3.162 .JK;

.JK; = 12.649i

.. Kp = 160 is line on which ron= 40 rad/sec.

KpKp = 160 for (110= 40 rad/sec

Kp= 4 forKa= 4~-----------------------------+Ko

,,... Example 7.44: A seroomechanism is used to control the angular posuion 60 of a mass,using 0 i as tire referettce input signal. The moment of inertia of the moving parts referred toload is 200 kg _",2 and the motor torque at the load is 2 x 10" Nnr/rad of error. Thedamping torque coefficient referred to load is 3 x 103 Nm/(radlsec). Find:

i) The step response of tire system to a step input of one radian.ii) The natural freqUe1lCY of ascillaiions, damped frequency of ascillations, peak time andovershoot.iii) Tire steady state error which exists when a step torque of 1000 Nm is applied to toadshaft·iv) The steady state error if the reference input i~ a constant angular velocity of 1 r.p.m.

Solution : The open loop transfer function of the system with,torque constant KT = 2 X 104 Nm/rad of errormoment of inertia J I. = 200kg - m 2

damping torque constant Bl ""3 x 103 Nm/(rad/sec) is

KG(s) = T

s(s Tl + Bl)

r "


Assume lies) = 1

The system can be shown as below :

KT 0o(s)

s (s JL + BL)

K,0..(5)0; (5)

( KT)It=S2 +s(~)+(K1)

J I. J J

Substituting the values of KT, J I and BJ

°n(S) 100=..

8;(5) S2 + 155+ 100.,

100 i.e, 10 rad/sec., (1)- = (l)n =n

and 2l; (l)n 15 i.e. l; 15= = 2xlO = 0.75

.. (l)d = (l)n ~1 -l; 2 =10~1 - (0:75)2 6.6143 rad/sec

i) The step response for input of one radian is

-~Wn t8,,(t) = 1g sin({l)Jt+O)

l-l;2

where 0 _,[~Hl'_,[~l-{O75)'1 use radian mode= tan l; - an 075

= 0.722 radians

60(t) = 1- 1.511e- 7.5 t sin (6.6143t+ 0.722)radians

Tp 1t= -=0.47sec

Wd

r "

-s.Control' System Engineering 7-104 Time Response Analysis ofControl Systems

and

= 2.83%

iii) Load torque is 1000 Nm i.e, T1. = 1000Nm

But K... = fad of error

. di TL 1000 0 OS d· . 2.860error m ra tans = KT ="2 x 10" =. ra lans i.e,

iv) The input now is ramp input of magnitude 1 r.p.m.

i.e, 7t(I) = 1 r.p.m. = 30 red/sec

As 1 revolution i.e. 27t radians in 60 sec.

:. Magnitude of ramp input = A = ;0K, = lim s G(s)H(s) = lim s- KT

s ...o .....0 s(sJ. +BI.)Now

= KT _ 2xlO"' =1\- 3xlO:\ 6.67

e =ssA (:0)K..-= 6.67 = 0.0157 rad == 0.9"

'. Example 7AS: A servomechanism is designed to keep a radar antenna pointed at a flyingaeroplane. If the aeroplane is flying with a velocity of 600 Km/hr, at a range of 2 Km andttu maximum tracking error is to be within 0.1°, determine tile required velocity errorcoefficient. (Gate)

Solution : The linear velocity of aeroplane is,

V = 600 km/hr

The range of the aeroplane is,

r = 2 km

Hence the angular velocity of the aeroplane is,.V 600

(l) = -=- = 300 rad/hr =r 2300 1 dl3600 == Ii ra sec

This is the ramp type of input to the system.

r "

· Control System.,Engineering 7 -105 Time ResponseAnalysis of t,JrControl Systems

ret} I= wt = 12 t

R(s) =12s:!

The velocity error coefficient for system is to be calculated, for the ramp input. Hencefor finite existence of velocity error coefficient, system must be of TYPE 1.

KG(s)H(s) = s(l+sf)

with H(sJ = 1, the system can be shown as,

C(s)•INow K.. = lim s G(s)H(s} = lim s. __ K___ = K

s .....CJ < ..... (I s (1+ sT)

Ac"" K where A = magnitude of ramp input

v

e"..(1/12)

K

and C<is == 0.10 allowed

10.1 = 12K

1K = 1.2 / degree

As x, = K, the require velocity error coefficient is 1~2I degree.

'... Example7.46: The open loop transfer function of a unity feedback system is given by,G(s) = I!- 2,. Sketch the output of the feedback system for a unit step input. Assume that thesystem is initially relaxed. (Gate)

Solution : As system has unity feedback,

Hts) = 1

C(s) =R(s)G(5) G(s)

1+ G(s)J:-I(s)= 1+G(5)

r "

"J Control System Engineering 7-106 Time Response Analysis ofControl Systems

=

Dividing both numerator and denominator by e-2s,

C(s) . 1R(s) = 1+ e2s

The system is excited by unit step input

ret) = 1 for t~O1

R(s) = -s

Substituting in C(s)

C(s) 1 1.. = 5'1+ e2s

. C(S)[l+ e2s] 1= -s

C(s)+ C(s) e"ls1.. = s

Now L {c (t+ T)} = C(s) eTs

.. L-t { C(s) e2s} = c(t+ 2)

Hence taking Laplace inverse of equation (1),

c(t)+ c(t+ 2) = u(t)

where u(t) = unit step function

.. c(t+ 2) = u(t) - c(t)

... (1)

... (2)

To obtain the output waveform let us obtain the values of c{t) for various values of t.Now at t = 0, c(t) has its initial value. Using initial value theorem,

. 1c(O)= Lim cCt) = lim s C(s) = lim s- = 0

t-.o s-.- ~-... S (1+ e2~)c(O) = 0

Substituting t = 0 in equation (2),

c(2} = u(t)-c(O)=1-0

= 1 as u(t)= 1 at t = 0

r "



c(4) = u(t)-c(2)=1-1=O


c(6) = u(t)-c(4)=1-0=1


c(8) = u(t)- c(6)= 1-1 = 0 and so on.

Hence the sketch of the output response of the system for unit step input is as shownin the figure.

c(t)Input---1.._ __;~I

o 2 4 6 8 10

u_.. Example 7.47: For the control system shown ill figure.

R(s) +

1) Determinei) /vip for a unit step inputii) e!OSfor unit ramp input.

11) Calculate tire steQdy state value of the output when the input slUlft is held fixed and asudden torque T,. = 1 Nm is applied. (M.U. : Dec. - 98)

r "

Control System Engineering 7·108 Time Response Analysis ofControl Systems

Solution : Assume TL as zero for [I]i) To calculate Mp,

G(s) = _--::-6 __ H(s) = 10.15 s2 + 0.9s

6C(s) G(s) 5 (0.15 s+ 0.9).. R(s) = =1+ G(s)H(s) 1+ 6

s (0.155+0.9)

6 40= =0.1552+ 0.95+6 52+ 6s+ 40

Comparing the characteristic equation with,_

S2 +2l;ron5+(,l)n 2 = 0(I)~ = 40 Le, (l)n = 6.3245

~ = 62(1)n

i.e. ~ = 0.4743

Now,

-nxa4743

eJt-W474312 xlOO 18.4"10

ii) For unit ramp input,

K, = Lim 5G(s)H(5) = Lim s. (0 1 6 09).1 = 6.67s_.o ,_.o 5 .55+.

tess = - = 0.15 radKv

m For TL is t N-m, assume R(s) as zero. With R(s) zero, the system gets modified as

1 C(s)1---..__-

From the figure we can write.

1G(s) =0.1552+0.95

r "

Control System Engineering 7 -110 Time ResponseAnalysis of. Control Systems

K K -atc(t) = _-_ ea a

From graph, ~~ c(t) = 22 = K_ K e- = K

a a a

From graph,

c(t) = 2 - 2 e- at

c(t) = 0.8 at t = 2

0.8 ::;;;;2 - 2 e- 2a

0.2554 and K = 0.5108Solving we get, a =C(s}

=R(s)0.510985+0.2554

'. Example 7.49: For what mlues of K is tire time constant of the closed loop system lesstitan 0.2 sec. (M.U. : May-2003)

~_ ,,!S) r:-L_j 3 IL ~2S+1: 1C(s)•

Solution : The closed loop transfer function is,

C(s) =R(s)

3K [ 3K ]2s+1 3K 1+3K3K = 2s+1+3K= :::"[-=--2--=~]

1+ 25+ 1 1+1+3K5

2Comparing denominator with l+Ts, the time constant is 1+ 31( which must be

maximum 0.2.

S 0.2

K 2: 3II" Example 7.50: A pair of complex conjllg'ltepoles in the s-plane is required to meet thevarious specifications.For each specificatiOttsketch tire region ill tire s-plane ill which polesshould be located

i) ; 2:: 0.707, ro" 2: 2 md/sec (positive damping)

ii) 0 s; s 0.707, ro" s 2 rad/sec (positive damping)iii) ; S 0.5, 1 S roll S 5 rad/sec (positive damping)io) 0.5 S; S 0.707, roll s 5 Tad/sec(positive and negative damping) (M.U.: May-2003)

_\1 , ,:r I,

Control System Engineering 7 -111 Time Response ~alysis ofControl Systems

Solution:

----•.•.--- .•--. j 1.4144

-1.4144-------. - j 1.4144

j2

-1.4144

-j2

-2.5

jO.866

------------- j 4.3301

-0.5

-------------- - j 4.3301

i) ~ ""0.707 con = 2

Oli ::; (ll, ~1-l;2

"" 1.4144 rad/sec

::; - 1.4144 ± j 1.4144

As ~ increases and becomes unityboth the poles converge at - 2 andthereafter become real.

ii) ; = 0.707 and <q, = 2

.'. Poles = - 1.4144 ± j 1.4144But for 0 s l; s 0.707 the poles

remain complex and at ; = 0 becomespure imaginary. For ron S; 2, theimaginary part of the poles must beless than 2. So region in s-plane is asshown in adjacent figure.

iii) ~ :s 0.5 and 1 S ron S 5

Let l; = 0.5 and ron ::;1

then rod = 0.866

r.Poles are - 0.5 ± j 0.866

For ~ = 0.5 and (J)n ::; 5

then (J)d = 4.3301.

:. Poles are - 2.5 ± j 4.3301So region is as shown in the

adjacent figure.

_\I , ,:r', oIj


iv) 0.5 S ~ S 0.707 and Oh S 5

For ~ = 0.5 and ron = 5 the poles are - 2.5 ± j 4.3301.For ~ = 0.707 and ron = 5 the poles are - 3.535 ± j 3.536.

For positive and negative damping the region is as shown in the figure.

11.+ Example 7.51: Consider the ullity feedback control system whose open loop transferfunction is C(s) = s (1:~.1s) . Determine tile steady state error and its variation with time

when the input is r(t) = 1+ t + t2•

Solution: For steady state error e5-~'C(s)H(s) = s (1:~.1s)"(M.U.: Dec.-2003)

Kp = Lim C(s)H(s) = -5-.0

Lim sC(s)H(s)'= Lim 50Kv = s---+0 s---+O

sXs(l+O.ls) = 50

~a = LimO s2C(s)H(s) = 0s---+

In the input ret) = rl(t)+r2(t) + r3(t)

r1(t) = 1 i.e. AI = 1

r2(t) = t i.e. A2 = 1

r3(t) ? 1 ?i.e. A3 = 2= t- = -x2t-

2

... step

... ramp

... parabolic

r "

Control System Engineering 7 -113 Time Respons~A~alysisofControlSystems

A.. 1e=--<2 = ~ = 50'

For obtaining variation of e"i!'with time use general error series method.

1 1F.(s) = 1+ G(s)H(s) = 1+ 50

5 (1+ 0.1 s)

es~(t) = Ko r(t) + K. r'(t)+ ~~ r"(t)+ ....

0.1 s2 + s= ----=----

0.152 + s+ 50

0.1 S2+ sK - Lim F.(s) = Lim

o - 5--+0 s--+O 0.152 + s+ 50 = 0

L. d FI (5) L' {(0.ls2 + s+ 50) (0.2s+ 1)- (0.152 + 5) (0.2s+ I)}lin --= 1m

Kl = s--+O ds 5--+0 (0.ls2+5+50)2

Lim { 50 (0.25+ 1) } = .2Q_ =_!_ = 0.02= 5--+0 (0.152 + s+ 50)2 (50)2 50

. {<0.152 + 5+ 50)2 (10)- (10 s+ SO)(2) (0.152 + s+ SO)(0.2s+ I)}_ Lim ? 4- 5--+0 (0.1S-+5+SO)

K2 = (50)2 (10)- (5O)(2)(SO) (1) = 0.0032(50)4

ret) = 1+ t + t2, r'(t) = 1 + 2t , r"(t) = 20.0032

.• e:;s(t) = 0.02 ( 1+ 2t)+-2- (2) = O.04t+ 0.0232

r "

Control Syst~";'Engineering.!


'.. Example 7.52: In the block dillgram given below, determine fire output rate factor thatyields a response fa a step input command having a maximum overshoot of 10 %.

(M.U. : Dec.-20(3)

Solution : Solve the internal minor feedback loop,

21+--xsKs(s + 2)

25 x """"2___;;;"'_-s +2s+2sK

2s(s+2) C(s) C(s)

It's a unity feedback system with G(s) = , 10s- + 2s+ 2sK

]0C(s) =R(s)

S2 +2s+2sK1+ ]0

S2 + 2s+ 2sK

10=-::-------S2 + S (2+ 2K)+ 10

Comparing denominator with 52 + 2l;wns+w~,

w; = ]0 i.e, wn = .J1O = 3.16227 rad/sec

J::2+2K=1+K2l;wn = 2 + 2 K i.e. ~ 2x..JTO .JTO

.•. (1)

... (2)

Given, Mp = 100/0

.. 10 = 100 e-K~/b-f.2

Solving for l;, ~= 0.5911

Using (2), 0.5911 l+K= JTO.. K = 0.86922 ... Output rate factor

r "

Contt:ol System Engineering 7-115 Time Response Analysis ofControl Systems

'. Example 7.53 : A feedback control system is represented by the differential equation,

d2c de-2 + 6.4-d = 160 e where e = r - 0.4 c.dt t

The variable c denotes output. Find the value of tire damping ratio and what informationdoes this COlrvey about the trans~t behaviour of the system.

(M.U. : Dec-2003, Dec.-2006)

Solution : Substituting value of e,

d2e + 6.4de = 160 (r - 0.4 e)dt2 dt

d2c de:.-2 + 6.4 dt + 64c = 160 rdt .

Taking Laplace transform of both the sides and neglecting initial conditions,

52 CCs)+6.4 s C(s)+ 64C(s) = 160 R(s)

C(5) .160R(s) = 52 + 6.4 s+ 64

... Transfer function

Comparing denominator with S2 + 2~ron s+ro~,

ro2 = 64 and 2~ron= 6.4n

i.e. Oln = 8 rad/sec. and ~ = 26~48= 0.4.

c(t) As the damping ratio l; is less than unity,the system is underdamped and will producethe time response with damped oscillatorytransients for the step input as shown in thefigure.

'.. Example 7.54: For a second order undamped system, show that c(t) = 1 - cos ro,l.(M.U. : May-20(4)

Solution : For an underdamped second order system, the output is given by,-~mnt

e(t) = 1 h- sin (rodt+B) ... (1)1-l;2

r "

c(t) = 1- cos (J)n t ... Proved

Control System Engineering 7 ·116 Time Response Analysis of~Control System~

For an undamped system, ~ = O.

(J)d = (J)n and 0 = tan - 1 .,.; = ; rad = 90".

Using in (1),

'... Example 7.55: For the block diagram shown below first obtain steady state output andthen obtain steady state error. (M.U. : May-20M)

20.000 C(s)(1 + 8)(1 + 0.18)(1 + 0.2s)

S luti F the ei t G( ) 20000 H(s)= 1o on: rom e gIven sys em, s = (T+s) (1+ 0.1 s) (1+ 0.2 s)'

20000C(s) =R(s)

G(s) (1+ s) (1+ 0.1 5) (1+ 0.2 s)=1+ G(s) 1+ 20000(1+ 5)(1+ 0.1 s) (1+ 0.2 s)

C(s)=R(s)

20000 d R() 1000(1+ 8)(1 + 0.1 8)(1 + 0.2 s)+ 20000 an 8 = -s-

• C(s) - [ 20000 ] x 1000 = 20000000•. - (1 + s)(1 + O.Is) (1+ 0.2s) + 20000 s s [(1+ 8) (1+ 0.1 s) (1+ 0.25) + 20000J

L· C() L' {. 20000000 }im s S = im sx --=-~--:-.",......-=--:::--:--=---:--=-:--==Cs... = s~ 0 S~ 0 s [(1+ s) (1+ 0.1 s)(1 + 0.2 s)+ 20000J.C""

= 20000000 = 999 951+ 20000 • .... Steady state output

e5S = Reference input - C" = 1000 -; 999.95 = 0.05

r "

-Control System Engineering

"7 -117 Time Response Analysis of

Control Systems•

- -).. Example 7.56: The open loop transfer junction of a 1lnity feedback control system is

G(s) = s (:~010)and input applied ;s r(t) = a + bt + c~2.Obtain generalised error series.

(M.U. : May-2004)

Solution : For generalised error series,

F,(5) =.,

1 1 s" + 10s-~-- = ---- = -----1+ G(s)H(s) 1+ 100 52 + 10 5+ 100

S (5+ 10)

--.,

Lim Ft(s) = Lim 5- + 105 = 0s-+O 5-+0 s2 + 105+ 100

-- L· dFl (5) L·rm ds = rms~o s-e O

(S2 + 10s+ 100) (2s+ 10)- (S2 + 105)(25+ 10)(S2 + 10s+ 100)2

--Lim 100 (25+ 10) _ = Lim5-+ 0 (52 + 105+ 100)2 s-+ 0

200 s+ 1000

--. d2fj(s) . dLIm ds2 = LIm dss~O s~O

2008+ 1000(52 + 105+ 100)2

-- Lim

5-)0

(52 + 105+ 100)2(200)- (200s+ 1000) (2) (52 + 10 s+ 100) (2s+ 10)= 0

(52 + 105+ 100)4

••• -- KKo r(t)+ Kl r'(t)+ 2~r" (t) = 0+ 0.1 [b+ c t]+ 0

es, = 0.1 (b + et] •... error scnes•• •

'. Example 7.57: A control design has to meet following specifications: damping ratio of0.5, natural frequency of JTO rad/sec and steady state error of 10 %. For Q unity feedback

K (s+a)control syste71' G(s) = 2 ' find values oj K, a and tJ.

(5+a) (M.U. : May-20M)

Solution : The closed loop transfer function is,

K(s+a)C(s) G(s) (S+~)2 K (s+a)- - ----------- - -R(s) 1+ G(s)H(s) 1+K(s+a) 52 + s (2P+ K)+ (ka +p2)

(S+p)2• • ~. • ." • (":l. .....I I I I _ I I I I ._ I I c • " .

I

-Control System Engineering 7 ·118 Time Response Analysis of. t , ! ,

Control Systems '

Comparing denominator with S2 + 2~(l)n s+(I)!,

fJ)2 Ka+p2 •(l)n = ~ Ka +132 .•. (1)- I.e.-n

I

I2~(1)n = .2{3+K • 2P+K ... (2)I.e. -- 2xJ Ka +132

II For unit step input, A = 1.

Lim G(s)H(s) = Lim K (s+a) = K~Kp = (5+13)2 ~-

5-+0 s-+o

•• •A 1 p2

Cs:; = ~ = Ka = Ko~2

•.. (3)

But elL~is 100/0 hence,

~20.1 = Ka i.e. Ka = 10~2 ... (4)

From (1), JW ~10fJ2 +132 · 132 10 • ~ = 0.9534- I.e. = 11 I.e.-

From (2), 0.52x (0.9534)+ K

i.e. K = 1.2552-- 2xJ lOx (0.9534)2 + (0.9534)2

From (4), 1.2552 x a = lOx (0.9534)2 i.e. ex= 7.2416

. . d2 yll) 8dl/(t)Example 7.58 : A control system IS descnbed by - ,+ d + 25ylt) = 50xtt).dt- t

Eval~te the output reSpOJ1Sey(t) and its peak.value for Q input xlt) = 2.5 JI(t).(M.U. : May-2004, Dec.-2006)

Solution : Taking Laplace transform of both sides,~

5" Y(s) + 8 s Y(s) + 25 Y(s) = 50 X(s)

yes) 50--Xes)•• •

50 25----=52+85+25 2S 52+85+25

Comparing denominator with S2 + 2~(J)n s+ro~,

Le, (1)n = 5 rad/ sec.

i.e, l; = 0.8

•• •,

• • ~. • •• • (":l. ",.I I I I _ I I I I ._ I I c • • •

I

Control System EngineeringJ

7 -119..


Hence the output response iSI

•• •

= 2{1-1.6b7e-41 sin (3t+ 3b.869°»)

v(t) = 2- 3.3334 e- 4 t sin (3t + 36. 8690)~

... Output response

If the input is 2.5 u(t) Le, step of 2.5 instead of unity then the response changes as,

y(t) = 2.5x (2-3.33.Ie-4t sin (3t+36.869°») = 5-8.3325e-.J' sin(3t +36.869°)

The peak value is attained at t = Tp = 1t =..!!.3 secrod

•• •

ny(t)It'ca~ = 5 - 8.3325 e- ..)('3 sin ( 3xi + 0.6434 rad)

= 5.0758 ... Usc radian mode to calculate sin

". Example 7.59: For the signal flow graplt shoum beloto obtain the impulse response.(M.U. : ~{ay-2004}

R(s) 11-s

1

3 1 C(s)1-s

-12

Solution : The Laplace transform of impulse respon.c;eis the transfer function of the givensystem.

Using Mason's gain formula for the given signal flow graph,

113Tl = lx-x-x3xl= 2S S S

1 1T., = 1x - x 1x 1 = -• s s1-s 1-s 1-5 Only two individual

loops without anynontouching cornbina tion.

-7

L - 71---S

• • ~. • •• • (":l. ",.I I I I _ I I I I ._ I I c • • •

•-


and

As ~< 1, the system is underdamped in nature,

I'" Example 7.61 : Find value of Kl and K2 so as to obtain peak time = 2 sec, and settlingtime = 5 sic.

50s(s + 2)

C(s)R(s) +

(M.V. : May-200S, May-2007)

Solution: Solving inner minor feedback loop,

50s(s + 2)

0( ~.;;.

1 50)( sK+ s(s + 2) 2

C(s)50R(s)

•••C(s)R(s)

-- 5(5+2)+ 50 K2s 50Kl-------------=--------~-------1 50K1 52 + s (2 + 50 K2) + 50 K1+ 5(8+ 2)+ 50K25

Comparing denominator with 52 + 2~(1)ns+(J)!,

ro! = 50K. i.e. ron =.J50K.2l;ron = 2 + 50K2 Le. ~ = 2 +SOK2

2.JSOKJ

...(1)

...(2)

Given: Tp - 2sec and T,= 5 sec-

Now, Tp7t =2 and Ts= ~

4= 5 i.e. ron 0.8- -- -

(t)d ~ron

•• •

•

•i.e.

• • 1 :11.1 1 I' 1 (":l. ",II I I I _ I I I I ._ I I c •

• 1

Control System Engineering 7 -122 Time Response Analysis of _.•Control Systems

••• 3.8553 i.e, 4.8553 ~2 = 1

~2 0.20595 • ~ = 0.4538 and ron = 1.76277 radl sec• - I.e.II • -.J5OKIUsing (1), 1.7627 • x, =0.06214- i.e,-

Using (2), 0.4538 - 2+50K2i.e. K2 = - 0.008-

2xJSOx 0.0621

)'.. Example 7.62: A feedback system wllich uses a rate feedback controller is shown in thefigure.

Motor

1 C(s)Amplifier

KA 1- ... s (s + 2)R(s)

- -

a) For KA = 10, in absence of derivative feedback (Ko = 0) determine the damping ratio lindnatural frequency of oscillations. Also find the s.s error for unit ramp input.b) Determine the constant Ko if the damping factor required is 0.6, witll K/\ = 1.0. Will. thisoalue of Ko' determine ss errorfor unit ramp input. (M.V. : May-2006)

Solution : i) With Ko = 0, the system is,

KAC(s) s(s+ 2)--R(s) 1+ KA

s(s+2)

R(s) 1KA .,__,.s(s + 2)

Comparing denominator withs2 + 2~(l)ns+(I)!,

i.e. ron =..jK; = JfO = 3.16 radl sec

i.e. ~ = ~ = 0.31622x 10

G(s)H(s) = KA =~---s(s+ 2) S(5+ 2)10

•••Lim 10

K, = 5-+0 sG(s)H(s) = 2 = 5

•••A 1-=-=0.2Kv 5

...For unit ramp A = 1

• 1 ::lI.1 1 I' 1 (":l. ",II I I I _ I I I I ._ I I c •

• 1

•

II


..

Time Response Analysis of· Control Systems

ii) With Ko, the system becomes,•

Minor loop

R(s) C(s) R(s)----1-s(s + 2)

10

•• •C(s)R(s)

S2 + s(2+Ko>1+ 10

s2 + S (2+ Ko>

=------52 + S (2+ Ko)+ 10

10--..

•• • i.e. ron = JfO rad/ sec

and

But given J: - 06 h 0.6 = 2+ I<o~ -. ence, 2M

•• • = 1.7947

•• • G(s)H(s) = 10 10------ =----52 + 5(2+ 1.7949) 5(S+ 3.7949)

•• •Lim 10

= s~ 0 sG(s)H(s) = 3.7949 = 2.6351

•• •A 1

ess = ~ = 2.6351 = 0.3795

•

1

15(5 + 2)SKq

1 + s(s + 2)

• I

C(s)

•

•"

•


Review Questions1. What is tile d~'Jerence between steady state response mId transient response of a control system?2. Define steady stafe response and steady state error.3. How steadystate error of a control system is defennined? Ho» it can be reduced ?4. State how type of a control system is determined? How it affects the steady state error of a

system?5" Derive the expressions for static error coefficients? How these coefficients are lise!ul in determining

steady state error? State the limitations of static error coefficient method ..

6. How damping ratio affects the time response of a second order system?7. Show the locus 0/ closed loop poles of a second order system as ~ is varied from 0 to DO.

B. Define tI,e fo11ou,ing systems sketching their output waveform for Q unit step input :i) Ilnderdamped system ii) UndAmped systemiii) Ooerdamped system io) Critically dIl',rped system.

9. With a neat sketch explain all the time domain specifications.10. Derive the relafionslzip between maximum overslJoof and damping ratio of a second order systeln.

Draw the graph or Dvershoot versus damping ratio11. Derive the expressions for peak time and rise time interms of ; a11d (t)n for a second order control

system"12. Discuss the adva,rtllges of feedback in a control system.13. An "nity feedback system has the loop trans!tr function

Ka» = ( )5 s+ a

If its time response is to have (i) an overshoot of less than 5% and (ii) seuling time Ilot exceeding4 seconds I determine silitable values for K and Q. lAnse : K = 2; a ~ 2]

14. A position control system is stabilised by means of acceleration feedback. If the system has amoment of inertia of 10-5 Kg - m 2; viscous frictional '''''1w~of 10-1Nm/rad - sec-I and the motor

torque is gioen by

T = 4 X 10-3 Kd2 00 N - mn' e+ dt2

where

K d2 80 = Acceleration fi-eedbackdt2

i) Draw its block diagram.jj) Write the system D.E.iii) Determine K fOT critical damping andiv) Determine the steady state error e» when the input is a constant velocity of 20 rpm.

[Ans. : K = 2.34 x 10- 3 seC I 3°]

• 1 :11.1I I

1 I'. (":lo ,.I I _ I I I I ._ I I c • • •

•


15) Determine the step, ramp, and paral'olic error const'lnts of the !oIIOlvi,rg Ilnily - feedl1ack contrdsystems. The open loop transfer funct ions are gillen

a) G(s) 1000-- (1 + 0.15) (1+ lOs)

b) G(s) 100--S(52 + 105+ 100)

c) G(s) 100-- 5(1+ 0.15) (1+ 0.55)

d) G(s) 100--52(52 + 105+ 100)

e) G(s) 1000-- 5(5+ 10) (5+ 100)

fJ G(s) K(l + 25) (1 + 4s)--S2(S2 + s+ 1) IAns. : a) Kp = 1000 K,..= 0 K., = 0

b) K., = 00 K",= 1 Ka == 0c) K.,= 00 K, = K Ka = 0d) K.,= 00 ~"= 00 K.1 = 1e) K., = 00 K,= ao Kd =0

f) K., = 00 ,,= 00 K.1= 1 KJ16. A two phase ac. servometer haoing (l torque constant of 0.045 N,n!V controls It position load

througll a gear ratio of 10:1. The effectittt moment of inertia tz",1 co-efficient of oiscous frictionreferred to load side are 0.25 Kg - m2 and 1.0 Ntn(rad/sec). TIlt syneJITo (n,tgslipJ t"9T detectorproduces an error signal of 0.1 V per degree l..'TrOT in misalignment. Develop tI,e block diagramrepresentation of the control systenl and there from obtain the transfer function.

oj Calculate the natura! freqJlnrcy of oscillations and damping ratio.b) If damping is to be made critical using a deril1ative error control, determine the time constant of

the phase advance circuit. (Ans.: (l)n ; 3.21 radlsec , ~= 0.623 I Td = 0.234 sec]

17. For the control system shoum in figure.

R(s) +--

f -

1 C(s)..6- ..... 20.15s + 0.95

J ... -.~-

Il) Determine (i) "'\, for a uni: step input (ii) es.~foT unit ramp inpu!b) Calculate the steady state value of the output iLthen the input shaft ;s held fixed and suddentorque Tl = 1Nm is applied fAns.: Mp -18.390/0, ess = 0.15 rad, Css = - 0.166 rad]

. . ~.I I

..". (":lo ,.I I _ I I I I ._ I I c • " .

II


J

I"

~

rI

18. The open loop transfer function of II unity feedback control system is given byK

G(s) = s{l + sT)

The input to the system is 1 R.P.M. lind the steady slate error being 0.25.. Calculate the naturalfrequency of oscilllltions if the system is critically damped. (Ans. : 48 od/s)

19. The block diagram of a position control system witl, velocity feedback is shown in figure. Determinethe value of a so that the step response has maximum overshoot of 10 percent. WIUlt is the steadystate error?

R(s) C{s)10s (s + 2)

1+cxs

[Ans. : a = 0.1739,ess = 0]

20. The closed loop transfer function of a second order system u'itll proportional plu« error-rate

feedback is given by ~~s~= ~(s +z) where parameters K and z are adjustables s + 45+ 8

a) If rit) = t; determine the values of K and Z sud: thai the steady state error is zero.b) For these txdue« of K and z, determine the steady state error to a unit parabolic input; i.e.

r(t) = - 1/l.t2•lAnse : (a) K =4, z = 2 (b) 0.251

21. For the system shown in jigrlre determine tlte steady state error for

R(s) + E(s)

8{s)

0.4 C(s)(0.2s + 1)(0.5s + 1) 2

S + 3s + 4

Motor-

Amplifier Amplifier

•i) A unit step input

ii) A unit step 'Velocity inputiii) A unit step acceleration input IAns. : (i) 0.0475 (ii) 00 (iii) oa ]

22. The block diagram of a fire control system t(Jith unity feedback is described i" figJ1re. Usinggeneralised error series determin« the steady slate error of the system loJren the systenr input is

Preamplifierand Demodulator Network Preamplifier Motor and load

+s) C(.. 10 s+1 5 200 0.015.. ... .. .. ..- .. ... - - - ...s+5 (5 + 2) (s + 1)-~•

Amplidyne

R( s)

• • ::lI..I I..". (":lo ,.

I I _ I I I I ._ I I c • " .,

Control System Engineering 7 -127 Time~~spons•.~"alysisofControl Systems

•

3t2r(t) = 5 + -It + ?- lAnse : 0.5 + 0.373t + 0.09315t2)

23. The block diagram of a simple servo system is shoum in !ollou1ing figure. Determine tIrecnaracteristic equation of tIle system. Hence calculate the undamped frequency of oscillations,,illnlpillg ratio, damping factor, maximum overshoot first undershoot, time intervals after whichmaximum a'Jd minimum OCCJlTS, settling time and the number of cycles completed before the outputis settled toiihin 2% of the final talue. The input to the system is a unit step.

R(E( )

)+ ss) 1.2 C(s

20- - .. .- -.. - ...5 (s + 1) (0.25 + 1) -

- .•5-6

[Ans. : 51 + 65 + 25 = 0; (1)n = 5 rad/sec;

~= 0.6; rod = 4; Mp = 9.5%; 0.94"0;0.785 sec; 1.57 sec i 1.33 see 0.85 cycles)

24. The open loop transfer function of a unity feedback control system IS given byK

G(s) = s(sT + 1)

i) By what factor theamplifier gain K should be multiplIed so that the danrping ratio is increasedfrom 0.2 to O.B.

;i) By wlUlt factor fl,e time constant T sllould be multiplied so that the thI"'ping ratio is reduced from0.6 to 0.3.

iii) For the system overshoot of tire unit step response 10 reduce from 6()Ok to 20%. Slww thatTK -1TK: _ 1 = 43.22 where Kl and Kz are the values of K for 60% lind 200.4

1IAns. : 16' 4]

25. Calculate static error coefficients for a unity feedback system with G(s) = S(S~ 6)" If input given is

r(t) ~ .. + 3t determine steady stale error. For the above system of ess is to be reduced to 10% of

existing value, what u'ould be the percentage chilnge in gain?IAns. : I<p = 00, K,. =2, ess= 1.5,9000/0 change is need)

26. Given:What· is the steady stJJteerror? (Ans. : 0.1721

R=O + E c

-TL = 0.172

. . ~.I I

• ••• (":l. ,.I I _ I I I I ._ I I c • • •

Control System Engineering 7 -128 Time Response Analysis ofControl Systems.·i ,\.

27. Given:

E

2

R 100 c(0.05 s + 1) (0.05 s + 1)

•

s1 +0.25-

The input is a ramp, R = 1.St . W},af ;s the steady-state error?

28. Gioen :[Ans. : 0.0151

T(s)

R c

If R = 0, uJJuztis the sttady - state error?

29. Find error coeDicil'llts for tile gioen llnily feedback system haoing

a» = 4(52 + l?s + 100)S (5 + 3) (s' + 2s + 10)

lAnse :-11

(Ans.co, ~, 0)

30. Determine error coefficients for the system haoing(s +2)

G(s)H(s) = 5(1 + 0.55) (1 + 0.2s) (Ans. 00, 2, 0)

31. Find error coefficients for a system IlaVing G(s) = 10 and steady state52 (1+ s)

error i/input to the system;s ao+ a. t+ az t2• (Ans.:oo,oo, 10, az/lO)

32. A systeln has 40% overslzoot and requires a settling time of 4 sees when given (I step input. Thesteady stale error 2%. Determine fire transfer function of the second order system. Also find risetime and peak time. (M.U.: Nov.-94)

(Ans.: 0.5409 sec, 0.9163 sec)

33. A standard second order system has 50% overshoot and settling time of 3 seconds- Determinew", l;, 1"" and t; (M.U.: Jan.-94) I

(Ans. : 6.19, 0.2154, 0.53.97,1.29.

• 1 ::lI.1I I

1 1'1 (":l. ,II I _ I I I I ._ I I c •

• 1


34. The open loop transfer function of a Ullity feedback control system is given byK

G(s) = ~(5T+ 1)

iJ By fuha' factor the amplifier gain K SI'Ollld be multiplied so that the damping ratio is increasedfrom 0.2 to 0.8.

ii) By what factor the time constant T should be multiplied so tllnt damping ratio is reduced from 0.6to 0.3.

(M.U. : May-97)(Ans. : V16, 4)

35. TI,e block diagram of a speed control system is shown in the figure. The disturbance is present inits jonoard path.

--

V1(s) : ~"'__-I 1. ~ G1(s)= s+ 2

«(s)...:

--~--------------~+Kb~-------------~

Determine the sensitivity of the systemi) ~d ii) ~d

where Md (5) is the ratio of (1)(5) and Td(s)./ ,

\ A. 52 + 35+ 2 K Kbns. 2 ' 2 ( )

", 5 + 35+ (K Kb + 2) 5 + 35+ K Kb 'J

36. The block diagram of a position control system is shown in the figure. Determine the sensitivity ofclosed loop transfer function T(s) with respect to G(s) and 1-1(5) fOT (l) = 1 rad / sec.

(Ans. : 0.029, 1.02)

R(s) - -C(s)...10G(s) = s (s + 1)

'____--t H(s) = 5

37. A flywheel is driven by an electric motor. It is controlled Ilutomatically by a mcroement ofhandtoheel and follows its mDl1ement. The effective moment of inertial of the flywheel is150 kg- m 2• The torque developed by the motor is 2400 Nm/rad of misalignment between flywheel

and handwheel. Viscous friction of the system is 600 Nm rad-1 sec. If the Iu2ndwheel is suddenly

moved through j rad from rest, determine fhe expression for the subsequent angular position of the

flywheel. " (Ans. : 1.05[1- e- 2 t cos3.5 t+ 0.6 sin 3.5 t) ),

• 1 :11.1, ,1 I" I I" ,.:lo .::""I I _ I I c •

• I

Control System Engin_rlng 7-130 Time Response Analysis ofControl Systems

38. A servo mecMnism shown in the figure has moment of inertia of the moving paris referred to loadslulft as 150 kg - m2• The motor load torque is 4 x 104 Nm/rad of e"OT. The dnmping torquecoefficient referred to load slulft is 4 x tel Nm/rlld/sec.

e~s)• -- 2

Js +Bs

Find :a) The step response if input is step type 0/ one radian.. Also determine rise time, peak time and peak

uoershoot.b) The S.S. error if input is constant angular velocity of 1r.p.m .

•

c) The S.S. error which exists when II tDrque of 1200Nm is applied to the load shaft.(Ans, :1- 1.75e-13.4t sin (9.3St+ 0.61),0.60, 1.7r)

000

•

•

• 1 ::lI.1 1 I' 1 (":l. ",II I I I _ I I I I ._ I I c •

•

• 1

Stability Analysis

8.1 BackgroundAs we have seen earlier that every system, for small amount of time has to pass

through a transient period. Now whether system will reach to its intended steady stateafter passing through transients or not? The 311..c;WCrto this question means to definewhether system is stable or unstable. This is stability analysis.

For example, a meter is connected in a system to measure a particular parameter.Before showing the final reading, the pointer of meter will pass through the transients. Thefinal reading is the steady state of the pointer. But during transients, it is possible that thepointer may become stationary due to certain problems in the moving system of thatmeter .. So to achieve steady state, the system must pass through the transient periodsuccessfully.

Key Point: Tile analysis of tohether the gioen systell1 call reach steady state; IJQssingthrough tile transients sZlccessftllly is called Stability A"alysis of the systet".

8.2 Concept of StabilityConsider a system i.e. a deep container with an object placed inside it as shown in the

Fig. 8.1.

ForceF

(a)

Fig. 8.1

(8 - 1)

..... lIP··.r >; ~ ...:. ti' :.. J....,.. ....,..

• •4 •• •..--'. ...-......• • • •

~••••I •••••~.: I, ,• •./....... r" .....

• • • •· , . ......~.... ',.e"-~ 4".. - ...... ..-

(b)

. . ~.I I

• ••• (":lo "I I _ I I I I ._ I I c • • •

Control System Engineering 8·2 I Stability Analysis- -_. •

Now if we apply a force to take out the object, as the depth of the ,c~ntainer _Ie; more, itwill oscillate and will settle down again at its original position.

If we assume that the force requiredto take out the object tends to infinityi.e. always object will oscillate whenforce is applied and will settle down butwill not come out, such a system iscalled absolutely stable system. Nochange in parameters, disturbances,changes the output. As against this,consider a container which is pointedone, on which we tty to keep a circularobjects shown in the Fig. 8.2.

In this case object will fall down without any external application of force. So if we tryto keep the circular object, we will always fail to do so. Such system is called unstablesystem.

••••r•••••'f

•••t•i••

(a) (b)

Fig. 8.2

While in certain cases the container is shallow then there exists a critical value of forcefor which object will come out of container.

I

oF F F

F < Fcritical

Fig. 8.3

As long as F < Fcriticat, object regains Its original position but if F > FCTltiCl.' object willcorne out. Stability depends on certain conditions of the system hence system is calledconditionally stable system. I

There are few systems e.g. : Pendulum where system keeps on oscillating whencertain force is applied. Such systems arc neither stable nor unstable and hence calledcritically stable or ",arginal'y stable systems.

Now let us see on which factors exactly the stability depends in a control system.

. . ~.I I

• ••• (":lo ,.I I _ I I I I ._ I I c •

•

• •

.. • . ~.. ...Control System Engineering 8·3 Stability Analysis

... -8.3 Stability of Control Systems

The stability of a linear closed loop system can be deter mined from the locations ofclose~~loop poles in the s-plane.

For example: If system has closed loop T.F.

C(s) 10--R(s) (s+2) (5+4)

Let us find out, output response for unit step input .•+. R(s) = l/s

••• C(s) = 10 ABC-----= + +--5(5+2)(5+4) 5 s+2 5+4

• C(s) 10 1/8 1/4 1/8- - +• • -S 5+2 s+4

C(s) 1.25 25 1.25• - - +• • - s s+ 2 s+ 4

... Finding the partial fractions

c(t) = 1.25 - 2.5 e-2t +1.25 e-4t = C$S + Ct (t)Steadystate Transient

As closed loop poles are located in left half of s-plane, in output response there areexponential terms with negative indices i,e, e-2t and e- 4t •

Now as t -+ 00 both exponential terms will approach to zero and output will be steadystate output .

•

i.e. as t ~ co, c, (t) = 0

Transient output = 0

Such systems are called absol"tely stable systems.Now transient terms are exponential terms with negative index because closed loop

poles are located in left half of s-plane. For the above system under consideration, theclosed loop poles are s = - 2 and s = - 4 and the negative indices of exponential terms arealso - 2 and - 4.

Key Point: TI'!lS if closed loop poles are located in left half, exponential indices itl outputare negative. And if indices are negative, exponential transient terms will vanish when t ~ 00.

Now let us have a system with one closed loop pole located in right half of s-plane.C(s) 10--R(s) (s-2) (s+4)

Find out unit step response of above system.

10 ABCC(s) = s(5-2)(5~4),~ s, + 5-2+ 5+4

.

• 1 :11.1 1 I' 1 (":l. ",II I I I _ I I I I ._ I I c •

• 1

Control System Engineering 8-4 Stability Analysis.

C(s) 1.25 0.833 0.416= - + +--s s-2 s+4

•• • c{t) = - 1.25 + 0-833 e+2t + 0.416 e-4t

Now due to pole located in right half, there is one exponential term with positiveindex in transient output.

while C~c;(t) = - 1..25

,.t c(t)

0 0

1 + 4.91

2 + 44.23

4 + 2481.88

00 00

As it is clear from the table that instead of approaching to steady state value as t ~ co,

due to exponential term with positive index, transients go on increasing in amplitude. Sosuch system is said to be unstable,

Key Point: So it is clear that if any of the closed loop poles lie in righ! half of s-plane, thenit gives the exponential term of positive index and due to that, transient response of increasingamplitude, making systenz unstable.

In such systems output is uncontrollable and unbounded one. Output response ofsuch systems is as shown in the Fig. 8.4.

c(t) 00

.................................................... OR Steady state output

c(t) 00

Steady state output

_,..--------~ t

(8) Increasing exponentially

~--------------------~t(b) Increasing amplitude oscillations

Fig. 8.4 Uncontrollable responseFor such unstable systems, if input is removed output may not return to zero. And as

soon as input power is turned on, output tends to 00. If no saturation takes place in systemand no mechanical stop is provided then system may get damaged and failed.

Remember that the stability depends on locations of closed loop poles. And the closedloop poles are the roots of the characteristic equation of the systenl.

So, Closed loop poles = Roots of the characteristic equation

• • ~. • •• • (":l. ",.I I I I _ I I I I ._ I I c • • •

Control System Engineering 11·76 Bode Plots

1 1 1v) --s-' simple pole, T2 = 20' c.oC2= T2 = 20.

1+20

The straight line of slope - 20 dB/dec for c.o~20.

vi) (1 + :0> , simple zero, T3 = io' c.oC3= i3 = 30.

The straight line of slope + 20 dB/dec for O)~ 30.

vii) 1 s ,simple pole , T4 = ;0 ' c.oC4=; = 90(1+ 90) 4

The st:raight line of slope - 20 dB/dec for c.o~90.Resultant slope table :

Range of (I) Resultant slope

o < (I) < 1 (roc,) - 20 dB/dec

1 < CO< 20 (ron) - 20 -20 = - 40 dB/dec

20 < (I) < 30 (We3) - 40 - 20 = - 60 dB/dec

30 < (I) < 90 (Wc4) - 60 + 20 = - 40 dB/dec

9O<ro<00 - 40 - lO = - 60 dB/dec

Step 3 : Phase angle table

GGo:t =K e-O.1jro (1+ j~)

30

jro (1+ j~ (1+j ;) (1+ j ~)

ro 1 -O.1rorad -tan-1 ro - tan-1~ -1 (I) -tan-1 ~ 4IRJCj) 20 + tan 30 90

0.5 -goo - 0.05 rad - 26 .5ao - 1.430 + 0.88° - 0.31° - 120.28°= - 2.86°

10 -900 -1 rad - 84.28° - 26.sao + 18.43° - 6.34° - 246.05°= - 57.3°

5 -90" - 0.5 rad - 78.6go - 14.03° +9.46° - 3.1ao - 211.31°= - 28.6°

2 -900 - 0.2rad = - 63.43° - 5.71° + 3.81° - 1.2]0 - 1680- 11.40

4 -9()0 - 0.4 rad -75.96° - 11.3° + 7.590 - 2.54° -194.75°= - 22_SO

• I ~" r:::ll. ,,'

Control System Engineering 11· 81 Bode Plots

At (I) = 1, the magnitude is 36.123dB as obtained by equation (1).

:. 36.123= - 60 Log (I) + C2 i.e. C2 = 36.123.:. Magnitude in dB = - 60 Log (I)+ 36.123. ... (2)

At (I)= 5, the slope changes by + 20 dB/dec hence there is simple zero at (l)C2 = 5.

Thus T2 = _1_ = 0.2. The factor is (1+ 0.25).(l)C2

The magnitude at (I)= 5 can be ontained by equation (2).

Magnitude in dB (at (I) = 5) = - 60 Log 5 + 36.123 = - 5.815 dBThe equation of the third line is - 40 Log (I) + C3'r. - 5.815 = - 40 Log 5 + C3 i.e. C3 = + 22.143 dBr, Magnitude in dB = - 40 Log (l)+ 22.143. ... (3)

At (I)= 10, the slope changes by - 40 dB/dec. As correction is zero (; = 0), it is aquadratic factor without middle term. The comer frequency is roo = 10. Hence

ro. = roo = 10.Thus the factor is [11 (1+~2 )] i.e. [u(1+ ;:o)}Hence the overall transfer function is,

ii) With correction (~ = 0.316)Let the quadratic factor is s 2 + 2; (l)n S + (l)n 2•

As comer frequency is 10, 00n = 10 rad/ sec.•• 2;(I)n = 2x 0.316x 10 = 6.32

Factor _ [ 1 ]- s2 + 6.32 s+ 100

In time constant form = 1 2s(1 + 0.06325+100)

Other analysis remains same. Hence the overall transfer function is,

G(s)H(s) = 64 (1+ 0.2 s)s2 ( 1+5) (1+ 0.0632 s+ 0.01 52)

•• I ~" " _" •• r:::lio ,,'

Control System Engineering 11·82 Bode Plots

'.. Example11.22: Derive the transfer function from the given Bode plot.(M.U. : Dec.-2003)

-12-21

Fig. 11.43Solution:

The inital slope is - 20 dB/dec hence there is one pole at the origin (Vs). The equationof the initial line is y = - 20 Log (I) + C1where y = mag in dB.

y = -20Log(l)+C1 •.. (1)

Consider second line whose equation is y = - 40 Log (I) + C2 . And on this liney = 0 dB at (I)= 4.

o = - 40 Log 4 + C2 Le, C2 = 24.0823

Y = - 40 Log (I) + 24.0823 ... (2)

At It)= cot, Y= 36 dB which is common to both the lines.

36 = - 40 Log IDt + 24.0823 i.e. (1)1 = 0.5035.

Using (1)1 in (1), 36 = - 20 Log 0.5035+ C1 i.e. C1 = 30.04

y = -20 Log (I) + 30.04 ... (la)

At CI.) =. 1, the equation (1a) gives y = 30.04 dB which is the contribution by K Le,20 Log K. ,I

... 20 Log K = 30.04 i.e. K = 31.768•

The slope changes by - 20 dBIdec at (I) = (1)1 = 0.5035.Thus there is simple pole with

(l)Cl = 0.5035. Thus Tl = CiJ~ = 1.986.The factor is [(1 +1~986s)]

• I ~" " _" •• r:::lio ,,'

Control System Engineering 11-83 Bode Plots

Now on the second line, at (I)= ~, the magnitude y = - 12 dB.Using in (2),

- 12 = - 40 Log roz + 24.0823 Le, roz = 1.981...8The slope changes by +20 dB/dec at (1)2;::; 8. Thus there is simple zero with

000 = 8. Thus T2 = _1_ = 0.125The factor is (1+0.1255).(l)C2

The equation of third line is y = -20 Log 00+ C3. At (1)2 = 8, Y= - 12 dB hence,

- 12 = - 20 Log 8 + C3 i.e, C3 = 6.0618

y = - 20 Log (I)+ 6.0618 ... (3)

On this line, at 00= (1)3the magnitude is - 21 dB.

- 21 ;::; - 20 Log (1)3 + 6.0618 i.e, (1)3 = 22.541

The slope changes by - 20 dB/dec at 003. Thus there is simple pole with roo = 22.547.

Thus T3 = _1_ = 0.0443. The factor is [ 1 -Jroo 1 + 0.04435 .

Thus the overall transfer function is,

G( )H( ) 31.768 (1+0.1255) =s s = s(1 + 1.9865)(1+0.0443s)

45.135 (5+8)s(s +0.5035)(s +22.547)

II.... Example 11.23: Obtain the Bode plot for,G(s)H(s) = 10(1- s)

5(5+2) (52 +2s+25)

Hence find G.M. and P.M.

Solution : Step 1 : Time constant form of G(s)H(s).10 (l-s)

( M.U.: May - 2C06)

G(s)H(s) = 0.2(1-s)5 2 52

s X 2 X (1+2)X 25 (1+ 25 s+ 25)

Step 2 : Analysis of factorsi) K = 0.2, 20 Log K :: - 13.98dB, straight line parallel to the Log roaxis.

u) ~ , one pole at origin, straight line of slope -20 dB/dec passIng through the

=5(1+0.55) (1 + 0.085+ 0.0452)

intersection of (I) == 1 and 0 dB.1

"iii) (1- s), simple zero, Tl = I, roCl = Tl = 1

The straight line of slope + 20 dB/des for ro~1.

• I ~" " _" •• r:::ll. ,,'

Control System Engineering Bode Plots11·88

Corresponding to (J)pc1=20,AB = + 64 dB

.. 20LogK = 64

.. K = 1585

Corresponding to (J)pc2= 400, CD = 100 dB.. 20 Log K = 100

.. K = 100000

So range of values of K is,

I 1585< K < 100000

Review Questions1. WIult are Bode plots?2. State the advantages of Bode plots.3. Explain tIre nature of Bode plots for

i) Poles at origin ii) Simple pole iii} Simple zero4. Explain the concept of gain margin tmd phase margin. Explain how these lJtliues help in studying

relative stability.5. Write a note 011 frequency domain specifications.6. Draw lhe Bode diagram for

G(s) = lOO(O.02s + 1)(s+ 1)(0.15+ 1)(0.015+ 1)2

Mark the following on the Bode diilgram, recording the numerialllJtllues.1) Gain cross-ooer frequency2) Phase margin3) Ph4se c:ross-overfrequerzcy4) CIZin margin (Ans. : lI\c = 32; (l)pe = 60; G.M.". 10; P.M. - 15° stable)

7. Given

0) Draw the Bode diagram.b) Is the system stable? (Ans. : Qlse = 45; Qlpc- = 00; G.M. = 00; P.M. = 18" stable)

200 (0.15 + 1) CS (0.25+ 1)(0.055 + 1)

Fig. 11.46

• I ~" r:::lio ,,'

Control System Engineering 14·42 Compensation of Control Systems

IKI = 11-0.4+jO.693IIO.6+jO.693113.6+jO.6931

K =10.8x 0.9167 x 3.667

K = 2.688

The transfer function of uncompensated system is,

2.688G(s) = s(s+l)(s+4)

= Lun s G s) _ Um s.2.688Ky s~o (- 5...0 S(5+1)(s+4)

= 0.672

It is desired to have Ky of 5 sec-). The factor by which static error constant is to beincreased is,

Factor = K ..desired 5= 0.672K v of uncompensated system

= 7.44

Let us choose this factor be 10.

~ = 10

Place the zero and pole of the lag compensator very close to the origin.Let zero of compensator at s = - 0.1

:. Pole of compensator at 5 = - 0.01

So transfer function of the lag compensator is,

Gc{s) = i< s+0.1c s+O.OI

Hence the transfer function of the compensated system becomes,

G ( )G( ) K 8+0.1. 2.688c s s = c s+O.01 s(s+1)(s+4)

.'0

K(s+0.1)S(5+1)(s+ 4)(s+0.01)

Where, K = 2..688 Kc

• I ~" " _" •• r:::lio ,,' .c

Control System Engineering 14·44 Compensation of Control Systems

The new dominant pole is - 0.43 + j 0.67.Apply magnitude condition at this new location of dominant closed loop pole.

IGc(s)H(s)1 f;=-O.43+jO.67 ;.: 1

·. IKII-O .33+ j O. 671 = 11-0.43+; 0.671I0.57+j 0.67113.57+j 0.67I1-O·42+j 0.671

KxO.7468 = 1.. 0.7961 x 0.8786 x 3.6323 x 0.7907

·. K = 2.69

But K = 2.688

·. i<c = 1.0019

Hence the compensated system has the transfer function,

G G 2.69(5+0.1)c(s) (s) = s(5+0.01) (s+l) (s+4)

This gives static error constant of 6.725which is greater than 5.

14.10 Designing Lag-Lead Compensator using Root LocusIt is known that the lead compensator increases the speed of the response and

improves stability. The lag compensator improves the steady state response. Ifimprovement in both transient as well as the steady state response is desired then thelag-lead compensator is used. Lag-lead compensator combines the advantages of both leadand lag compensators.

Assume the transfer function of the lag-lead compensator as,

f} > 1, Y > 1

Let us consider two cases to design such compensator.Casel:y~pIn this design, determine the locations of dominant closed loop pol~ from the given

specifications.From uncompensated G(s), calculate the angle deficiency 4» which must be contributed

by lead portion of the compensator.Choose T2 sufficiently large so that magnitude of lag portion is unity .

• I ~" " _" •• r:::lio ,,'

Control System Engineering 14-45 Compensation of Control Systems

= 1

Where s = sl is one of the dominant closed loop pole. From the angle deficiency ,determine values of Tl and y.

I5. +T,"L • = •Y5. +T,"

•The determine value of I<c from the magnitude condition.

Is. +T.K __ I G(s) = tc Y 15. +T.

I

If the static velocity error constant K, is given then.

K, = Lim s Ge(s) G(s)s_o

= Lim..... 0

( I) ( I)S+- 5+-sI<c T.. T2• G(s)

(5+ i.) (5+P~J= Lim s Kc . ~ G(s)

" ...0 y

As Kc and y are known, f3 can be obtained.Using this value of 13,choose T2 such that the magnitude of the lag portion is unity.Case2:y=13Determine the locations of dominant closed loop poles according to given

specifications.As y = 13.the transfer function of compensator becomes,

P>l

From the requirement of Ky, Kc can be determined.• I ~" " _" •• r:::lio ,,'


To have the dominant closed loop poles at desired location, calculate the angledeficiency~

Choose T2 very large at the end so that

(51 +*)(SJ +P~J = 1

Where, s = st is one of the dominant closed loop pole.

Now = 1

and = 41>

These two equations give the values of Tl and (3.

Once P is known, choose large T2 so that the magnitude of the lag portion isapproximately unity and

151+y;1

sJ +PT2

PT2should be high but should be physically realizable.

-so < L

1. What is compensation ? What is compmslltor ? Which are the various compensation schemes rLwdin practice ?

2. Which are the "important electrical networks used practiC4lly for the compensation of the controlsystems ?

3. Derive the transfor junctions of,II. Ltad ~twork b. Lag network c. Lag-lead network

4. Draw and expltlin the polar plot of,tl. Lead network b. Lng network c. Lng-lead network

Review Questions

• I ~" r:::lio ,,'


5. Draw Imd upltlin the Bode plot of,a. Lead network b. lAg network c. lAg-lead network

6. Derive fhe reltltion between .m and a for the lead compensator.7.Compare the characteristics of three types of compensators.8. Explain the design in frequency domain of,

a. Lead compensator b. Lag compensator c. lAg-/ead compensator9. Explain the design of three compensators using Root locus metlwd.10. Explain the effects and limitations of the three types of electrical compensators.11. Design a suitable lead compensator for system with,

G(s) =-( 4 2) to meet the specifications as,ss+-1-1 bQ. K = 20 sec • P.M.=+ 5(J' c. C.M.';! + 10dB

12. Design Q suitable lag compensator for a system with,

G(s} = lO(s+1) (~+ 0.55) to meet the following specifications:

Q. K ';!5 sec-1 b. P.M. :<!: + 4(J' c. C.M. :<!: + 10dB13. Design a lead compensator using root locus for the system with,

G(s) = (42) to meet the following specifications:ss+

a. Damping ratio = 0.5 b. Setting time = 2 sec14. Design a suitable lag compensator root locus for the system with,

G(s) = s(s+ ;f(s+ 2) to meet tirefolWwing specifications:

-1a. Damping ratio = 0.5 b. K C!: 5 sec c. Undamped natural frequency = 0.7 ,ad/sec

000

•• I ~I

Control System Engineering P-47 Polar and Nyquist Plots

Q.iS Obtain Nyquist plot -

(i) G(s)H(s) = S(;~6) (ii) G(s)H(s) = :g=~~(May-2ooS, 12 Marks)

. ./

Hence comment on stability and number of poles on R.11.5. of jro axis.Ans.: Refer example 12.30.

Q.i6 Explain: ,,"(i) Principle of arguments(ii) Selection of Nyquist path is s-plane(iii) Mapping of Nyquist path in F-plane(iv) Nyquist stability criteria.

Ans.: Refer sections 12.9, 12.10and 12.11.

Q.i7 How is gain Imargin and phase margin found from m.agnitude-phase plot?

(May-2ooS, 10 Marks)

(May-2006, S Marks)

Ans.: Refer section 12.15.

a.iS

,

Draw the approximate polar plots of the following transfer ftmctions :

( 'J 1 (..) 1t 1 . T. II . (1 . T.)+Jro I Jro +Jro 1

(iv) 1(jro)2 (1+jro1j)(1 +jroT2)

\.-

(iii) (1+jroT1)(1+j~T2)(1 "!-jroT3)

1(v) .(joo) 3 (1 +jroT1) (May-2006, 10 Marks)

Ans. : Refer example 12.2, 12.3, 12.4 and 12.5 for procedure.

, Q.i9 Discuss the stability of the system using Nyquist plot jor G(s)H(s) :;::;K(s - ~(5+1)

(May-2006, 10 Marks)Ans. : Refer example 12.33.

Q.20 Draw Nyquist plot for G(s) R(s) = ( ~)~ ~)and hence comment on stabilih/.5 5+ 5- (Dec.-2006, Dec.-2001, 10Marks)

Refer example 12.36.

If G(s)R(s) = K(s +1) using polar plot determine the range of K for stability.52 (s +2)(5 +4) .

Verify yOUT results by Routh's criterion. (Dec.-2006,10 Marks)

, Ana. : Refer example 1234.

Ans. :

Q.21

/• I ~I " _" •• r:::ll. .__":,

Control System Engineering P-48 Polar and Nyquist Plots

Q.22 If G(s)H(s) = s (1+2S)~1+0.1s) , using polar plot determine the range of K for stability.

Verify your results by Routh criterion. (May-2007, 10 Marks)Ans.: Referexample 12.35.

IIIIII

"IIIIIIII,I

Q.23 Draui lire Nyquist plot for ;G(s)H(s) = (4s+1)

52(5+1)(25+1)

(May-2007, 10 Marks)

Hence, comment on stability.Ans.: Referexample 12.8 for the procedure and verify N = + 2, unstable.

Q.24 Draw the polar plot for a system given by,

G(s)H(s) = oS (5+2)(~0~4)(5 +S)" Find whether tlte system is stable and if so find G.M. and

P.M. 1/ 100 is replaced by K find critical value of K by Routh criteria and also verify G.M.and P.M. (Dec.-2007, 12 Marks)

Ans. : Refer example 12.32for the procedure.QQQ

.••ttc,l.

• I ~"" _" II r:::lio ,,'

Technical Publications Pune~

Contents• Control System Analysis

Examples of control systems, Open loop conrrol system ... Closed loop control systems Transferfunction Types of feedback and feedback control ~stem characteristics Noise ",'ection. Gain.Sensitivity. Stabtlity.

• Mathematical Modeling of SystemsImportance of a mathematical model. Block diagrams, Signal flow qraphs, Ma~an's gmn formulaand its applicallon 10block diagrdm reducoon State space method. Solving time-invanamsystem.Transferrnamx

• Transient and Steady State • Response AnalysisImpulse response function. Firsl order 5!,.lorem.Second order "Ystem. Time domain SpeClllLdllons

of :oYl'tems.Analysis of transient-response ~,"g second order modelClasslflcllion of control systems according 10 'Type of S!. '{.mc•Sllo!ddy tar errors, Static rrorconstants Steady . 51 te analy is of different Iypb of ~~Iems usIng step. ramp and parabolicInput signal'

• Stability AnalysisConcept of ~tabillty, ~wblhly anai~l!> uSing Routh's stability entenon Ah.'IOiulec;tablhty.R" <111..,stability. Root·L()cus plo~ ~ummary of genua' rull 5 f( c"':lnrnac..in Rvo, Roo -Locuanalysis of control !>ystemcCompensauon t chruque -Log, lead. Ioq-I d

• Frequency-Response Analysisl roquenr- domam specifk.chlons. R, IdllC ' "",..It and pe"lk n, onCiung lI ..u C' , Relauonsl IPb..tw en time "md fnlf"uencv domain speCIfication of i terns Bode plo., Paldr plote,. Logmagnltud .. Vs phase plo~. Nyqul~ stabIlity cit Ion. "'Iabihr.ydnc:li~ R otll L bihtuflam marglO.Phase margm. Stab..lty <mal!! "u. -n W>lng Bode 1'10 Ci d-loopfrequt'nLV rcsponse-Constent gain and phe 10. Nichol'!>ch. rl and then U!.. In •." ..Ity Iud.OfSJ,lstLIn

• Control Components and ControllerD.C and A C s -rvomosors.Servoamplifler. P"lt ..nnorneter, C nchro tran:.mlttl:P.J Synchroreceivers. Synchro control transformer, otcpper motors DI .onnnuous controller modeCuntinuous colltroll 'I' mode •Composue 1..0n,,011•

First Edition: 2008Rs.560/-

ISB~ 978-81-8431-463-2

# 1, Amit Residency. 412 Shaniwar Peth, Pune· 411030. M.S., India.Telefax: +91 (020) 24495496/97, Email: [email protected]

Visit us at: www.vtubooks.com

Control Struct

Documents

Transcript of Control Struct