Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:


 Beverley Cunningham
 1 years ago
 Views:
Transcription
1 Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Ph: CLASS TEST 089 ELECTRONICS ENGINEERING Subject : Control System Date of test : 30/07/08 Answer Key. (c) 7. (b) 3. (b) 9. (a) 5. (b). (d) 8. (c) 4. (b) 0. (b) 6. (c) 3. (b) 9. (a) 5. (d). (b) 7. (c) 4. (c) 0. (b) 6. (b). (b) 8. (b) 5. (c). (d) 7. (a) 3. (a) 9. (b) 6. (d). (d) 8. (b) 4. (b) 30. (b)
2 CT08 EC Control System 7 Detailed Explanations. (c) t r π φ ω d where φ tan ξ ξ and ω d ωn ξ. (d) characteristic equation (si A) s 0 (si A) ( s )( s ) 0 s comparing with second order characteristic equation ω n ξω n 3 s 3s ξ.5.06 thus the system is over damped. 3. (b) Number of infinite semicircle thus number of pole at origin 4. (c) For the circuit shown G(s) α s R RC R R R s R R RC R R 3Ω Ω 5.50 R Ω 5. (c) Slope at highest frequency ( number of poles number of zeros) 0 db/decade 40 db/decade
3 8 Electronics Engineering 6. (d) Cs () Rs () or C(s) s 5 s s 5 Taing inverse Laplace transform. we get, c(t) 0 t e 5 or τ sec 7. (b) Comparing with standard transfer function of a compensator Here, τ 0.04 ατ 0. or α φ α sin 30 α 8. (c) characteristic equation G(s)H(s) 0 s 4 3s 3 (4s ) 0s 9 0 using Routh s tabular form S 4 S 3 S S S Number of sign change Hence, two poles lies on the RHS. 9. (a) The intersection point centroid Σp Σz ( p z ) magnitude (b) Characteristic equation is given by s 6s 6 0 s ξω n s ω n 0
4 CT08 EC Control System 9 ω n 4 rad/s ξ 3 ξ 0.75 underdamped system For underdamped system T sec ξωn 3 settling time of the system for % tolerance band 4T 4.33 sec 3. (d) The poles of the three systems will have same real part and two poles for each system indicate a second order system. The envelope of the second order system with unit step input is governed by e ξω n t. Thus the second order system will have same envelope for the following systems. ct () 3 t. (d) Since the phase margin is the phase required for the system to be unstable, or the phase required to mae the total phase of the system equal to 80 at gain crossover frequency. 3. (b) G(s)H(s) s ( st ) Phase crossover frequency for GH(jω) 90 tan ω pc T 80 jω ( jωt) is given by tan ω pc T 90 ω pc Since the phase crossover frequency is not finite thus gain margin of the system does not exist. We can also see this from the nyquist and root locus plot which are shown in the figure below. 4. (b) A system is always stable if its root locus lies on the left side of (jω) axis.
5 0 Electronics Engineering 5. (d) Characteristic matrix I G(s)H(s) 0 s s s 0 s s 0 s s s s ( s 3) ( s ) T(s) Gs () I Gs () Hs () G(s) [I G(s)H(s)] Here, T(s) s 3 s s s s s s s s 5s ss ( ) 3s 9s 4 ( s ) ( s ) s 5s ( ss ) 3s 6. (b) From the characteristic equation the location of open loop poles and zeros are P j P j P 3 3 θ P Z thus angle of departure 80 φ θ d φ θz θp φ tan () 90 tan (/) φ 45 tan (/) θ d tan (/) 08.4 θ z P P 3 θ P jω j j σ 7. (a) comparing it with the standard equation T(s) T(s) ωn ξω n ω Cs () Rs () s s n 0 s ( 0 ) s 0 D Rs () 0 Cs () ss ( ( 0 )) D
6 CT08 EC Control System ω n 0 0 D ξω n ξ 0 D D (b) Put s (p ) then the system becomes: (p ) 3 3( ) (p ) (7 5) (p ) p 3 3p p( ) 4 0 The Rooth s array is 3 p ( ) p 3 4 p p 4 For stability: 3 > 0 and > 0 > > 0 > 0.53 and <.5 > (a) G(s) ( s)0 ss ( ) 0 0s T(s) s s( 0 ) 0 ω n rad/sec ξω n 0 ξωn 0 ξπ ξ now, M p % 9.49% e ξ 0.6 thus 0.8 v lim sg( s) 5 s 0 e ss 0. v 0. (b) By calculating the characteristic equation, we get G(s) H(s) 0 K s ( ) 3 ( 4) s 0
7 Electronics Engineering (s ) 3 (s 4) K 0 s 4 7s 3 5s 3s (K 4) 0 s 4 5 (4 K) s s s K 3.4 (4 K) 0 0 s 0 4 K Range of K 4 < K < 0.4 a 4 b 0.4 b a 5.. (b) Transfer function C[sI A] B T (s) [ ] s 0 0 s 3 s 3 0 ( s )( s 3) 0 s [ ] ( s )( s 3) s [ ] T(s) ( s ) For unit step input U(s) s Y(s) s ( s ) e ss lim sy( s) s 0. (b) Before closing the switch s Cs () Rs () ζ and e ss s as a sr() s s s lim se( s) lim 0 0 GH ( s )
8 CT08 EC Control System 3 e ss s 0 ( ) s s as lim s s as 3. (a) 4. (b) e ss a After closing the switch s Cs () Rs () s ( a ) s a ζ and e ss lim se( s) s 0 e ss a Therefore e ss and ζ both will increase. controllability matrix Q c [ B: AB ] for controllable Q c 0 αα αα α 0 α α α α α α 0 or α 0 observability matrix Q 0 C T : A T C T β β β β β for observability Q 0 0 or β (β β ) β β 0 β ββ ββ 0 or β 0 G(s) and H(s) ss ( 4) GH s 4s 0 ss ( ) SH T GH ss 4 ( ) at T H s j S ω ( ) 0 jω 4jω 0 ω 0.5 rad/sec 0 ω 4jω 0
9 4 Electronics Engineering 0 0 ( ) j j 4 T S H (b) The signal flow graph of the above system is U /s /s X X X X Y Ẋ U X Ẋ X X U The state model Chec for controllability: Y X X X U X 0 X X y [ 0] X Q c [B : AB] Q c Q c 0 and ρ(q c ) ρ(a) noncontrollable Chec for observability Q 0 [C T : A T C T ] Observable Q 0 0 Q 0 0 and ρ(q 0 ) ρ(a) 6. (c) Brea away point can be calculated as dk 0 ds The characteristic equation for unity feedbac system is G(s) 0 4 K s s ( s ) 0
10 CT08 EC Control System 5 or s ( s ) K s 4 or K dk ds 0 s ( s ) 4 s 4 (3 4 ) ( ) s s s s s 4 s 4 or 3 ss ( 8) s s ( s ) 0 or s 0, 4, 4 The brea point 4 Equal roots are at s 4 Hence K(s 4) (c) For second order system M PO 0.6 / e πξ ξ / e πξ ξ or ξ 0.5 Because of maximum overshoot occur at pea time t p or ωn ξ 5 3 π π π ω d ω ξ ω n (0.5) ω n 0 rad/sec for sinusoidal input, the frequency response is the steady state response ω r ωn ξ n rad/sec 8. (b) To unstablize the system, the critical point must be located between 0 to 0.5. Additional gain required 0.5 K K or 0 log K 6 db
11 6 Electronics Engineering 9. (b) Here, ω r 4 rad/sec ωn ξ ; ξ () m r.5...() ξ ξ from equation () 4ξ ( ξ ) 0.64 ξ ξ ξ ξ 0 ξ 0.8 and 0. ξ 0.894, for ξ ω n 5.6 rad/sec π θ π cos (0.447) τ r sec ωd (b) R ( GG G4) G3 C H 3 G H H R ( GG G4) G3 C G H H3 H H R ( GG G4) G3 C H H G HH 3 (G G G )G 4 3 C G (GG G 4)G3 R GH GH(H 3 GGH) 3 GGHH 3 4 If G G 3 5 G 9 G 4 H H 4 and H 3 6 C R ( 9 )5 ( 4)( )
Homework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationControl System (ECE411) Lectures 13 & 14
Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 SteadyState Error Analysis Remark: For a unity feedback system
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial :. PT_EE_A+C_Control Sytem_798 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubanewar olkata Patna Web: Email: info@madeeay.in Ph: 4546 CLASS TEST 89 ELECTRICAL ENGINEERING Subject
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
More informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More informationController Design using Root Locus
Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers
More informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
More informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More informationCHAPTER # 9 ROOT LOCUS ANALYSES
F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closedloop system is closely related to the location of the closedloop poles. If the system
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : CH_EE_B_Network Theory_098 Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubaneswar Kolkata Patna Web: Email: info@madeeasy.in Ph: 056 CLASS TEST 089 ELECTCAL ENGNEENG Subject : Network
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real
More informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
More informationRoot Locus Design Example #3
Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll
More informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
More information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More information10ES43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A
10ES43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationControl Systems, Lecture04
Control Systems, Lecture04 İbrahim Beklan Küçükdemiral Yıldız Teknik Üniversitesi 2015 1 / 53 Transfer Functions The output response of a system is the sum of two responses: the forced response and the
More informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
More informationLecture 1 Root Locus
Root Locus ELEC304Alper Erdogan 1 1 Lecture 1 Root Locus What is RootLocus? : A graphical representation of closed loop poles as a system parameter varied. Based on RootLocus graph we can choose the
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
More informationGATE : , Copyright reserved. Web:www.thegateacademy.com
GATE2016 Index 1. Question Paper Analysis 2. Question Paper & Answer keys : 080617 66 222, info@thegateacademy.com Copyright reserved. Web:www.thegateacademy.com ANALYSIS OF GATE 2016 Electrical Engineering
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationIntroduction to Root Locus. What is root locus?
Introduction to Root Locus What is root locus? A graphical representation of the closed loop poles as a system parameter (Gain K) is varied Method of analysis and design for stability and transient response
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : ND_EE_NW_Analog Electronics_05088 Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubaneswar Kolkata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTCAL ENGNEENG Subject
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
More informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More informationEngraving Machine Example
Engraving Machine Example MCE44  Fall 8 Dr. Richter November 24, 28 Basic Design The Xaxis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationRoot Locus Design Example #4
Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
More informationControl Systems. EC / EE / IN. For
Control Systems For EC / EE / IN By www.thegateacademy.com Syllabus Syllabus for Control Systems Basic Control System Components; Block Diagrammatic Description, Reduction of Block Diagrams. Open Loop
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
More informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More informationChapter 7 : Root Locus Technique
Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci
More informationFATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY
FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY Senkottai Village, Madurai Sivagangai Main Road, Madurai  625 020. An ISO 9001:2008 Certified Institution DEPARTMENT OF ELECTRONICS AND COMMUNICATION
More informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
More informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
More informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More information12.7 Steady State Error
Lecture Notes on Control Systems/D. Ghose/01 106 1.7 Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More informationEE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation
EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginallystable
More informationNPTEL Online Course: Control Engineering
NPTEL Online Course: Control Engineering Ramkrishna Pasumarthy Assignment11 : s 1. Consider a system described by state space model [ ] [ 0 1 1 x + u 5 1 2] y = [ 1 2 ] x What is the transfer function
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationChapter 12. Feedback Control Characteristics of Feedback Systems
Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an openloop system (a system without feedbac) and a closedloop
More informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
More informationResponse to a pure sinusoid
Harvard University Division of Engineering and Applied Sciences ES 145/215  INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles
More informationCHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System
CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationProblem Value Score Total 100/105
RULES This is a closed book, closed notes test. You are, however, allowed one piece of paper (front side only) for notes and definitions, but no sample problems. The top half is the same as from the first
More informationControl System. Contents
Contents Chapter Topic Page Chapter Chapter Chapter3 Chapter4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of
More information