CHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS


 Felix Nichols
 1 years ago
 Views:
Transcription
1 CHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS Objectives Students should be able to: Reduce a block diagram of multiple subsystems to a single block representing the transfer function from input to output Analyze and design for transient response, a system consisting of multiple subsystems 5.1 INTRODUCTION We have been working with individual subsystems represented by a block with its input and output. More complicated systems, are represented by the interconnection of many subsystems. Since the response of a single transfer function can be calculated, we want to represent multiple subsystems as a single transfer function. Then, we can apply the analytical techniques of the previous chapters and obtain transient response information about the entire system. 5.2 BLOCK DIAGRAMS Many systems are composed of multiple subsystems. When multiple subsystems are interconnected, a few more elements such as summing junction and pickoff points must be added to the block. All component parts of a block diagram for a linear, timeinvariant system are shown in Figure
2 Figure 5.1 We will now examine some common topologies for interconnecting subsystems and derive the single transfer function representation for each of them. These common topologies will form the basis for reducing more complicated systems to a single block. Cascade Form Figure 5.2(a) shows an example of cascaded subsystems and Figure 5.2(b) shows the equivalent single transfer function. Figure 5.2 2
3 Parallel Form Figure 5.3(a) shows an example of parallel subsystems. The equivalent transfer function, Ge(s) appears in Figure 5.3(b). Figure 5.3 Feedback Form The third topology is the feedback form as shown in Figure 5.4(a) and Figure 5.4(b) shows a simplified model. We obtain the equivalent, or closedloop, transfer function shown in Figure 5.4(c). 3
4 Figure 5.4 Moving Blocks to Create Familiar Forms Figure 5.5 shows equivalent block diagrams formed when transfer functions are moved left or right past a summing junction. Figure 5.6 shows equivalent block diagrams formed when transfer functions are moved left or right past a pickoff point. 4
5 Figure 5.5: Block diagram algebra for summing junctions equivalent forms for moving a block a. to the left past a summing junction; b. to the right past a summing junction Figure 5.6 Block diagram algebra for pickoff points equivalent forms for moving a block a. to the left past a pickoff point; b. to the right past a pickoff point 5
6 Ex 5.1 Reduce the block diagram shown in Figure 5.7 to a single transfer function. Answer: Figure 5.7 6
7 Ex 5.2 Reduce the block diagram shown in Figure 5.8 to a single transfer function. Figure 5.8 7
8 Answer: 8
9 Exercise Find the equivalent transfer function, T(s)=C(s)/R(s) for the system shown in Figure 5.9. Figure 5.9 Answer: s3 1 T (s) 4 2s s 2 2s 9
10 5.3 ANALYSIS AND DESIGN OF FEEDBACK SYSTEMS Consider the system shown in Figure 5.10, which can model a control system such as the antenna azimuth position control system. Figure 5.10 : Secondorder feedback control system K The transfer function (open loop), s( s a ) in Figure 5.10 can be model the amplifiers, motor, load and gears. The closed loop transfer function T (s ) for this system is T (s) K / s( s a) 1 K / s ( s a) K K s( s a ) s( s a ) T (s) K s(s a) K 1 s ( s a) s( s a ) T (s) K s 2 as K where K models the amplifier gain, that is, the ratio of the output voltage to the input voltage. 10
11 As K varies, the poles move through three ranges of operation of a second order system : overdamped, critically damped and underdamped. For example: For 0 < K < a2/4, the poles of the system are real and located at: s1, 2 a a 2 4K 2 2 It is overdamped secondorder system. For K = a2/4, the poles of the system are real and equal, located at: s1, 2 a 2 The system is critically damped. For K > a2/4, the poles are complex and located at: s1, 2 a 4K a 2 j 2 2 The system is underdamped. If K increases, the real part remains constant and the imaginary part increases. Thus, The peak time (Tp) decreases The percent overshoot (%OS) increases The settling time (Ts) remains constant 11
12 Ex 5.3 (Finding transient response) For the system shown in Figure 5.11, find the peak time, percent overshoot and settling time. Figure 5.11 Answer: Tp wn 1 %OS e / Ts second % second wn
13 Ex 5.4 (Gain design for transient response) Design the value of gain, K for the feedback control system of Figure 5.12 so that the system will respond with a 10% overshoot. Figure 5.12 Answer: K = SIGNAL FLOW GRAPHS Signal flow graphs are an alternative to block diagrams. Unlike block diagrams, which consist of blocks, signals, summing junction, and pickoff points, a signal flow graph consists only of branches, which represent systems and nodes, which represent signal. These elements are shown in Figure 5.13(a) and (b). Figure 5.13(c) shows the interconnection of the systems and the signals. Each signal is the sum of signals flowing into it. For example, The signal V(s) V ( s ) R1 ( s )G1 ( s) R2G2 ( s ) R3G3 ( s ) The signal C1(s) C 1 ( s ) V ( s )G 4 ( s ) R1 ( s )G 1 ( s )G 4 ( s ) R 2 G 2 ( s )G 4 ( s ) R 3G 3 ( s )G 4 ( s ) 13
14 The signal C2(s) C 2 ( s ) V ( s )G5 ( s) R1 ( s )G1 ( s)g5 ( s ) R2G2 ( s )G5 ( s) R3G3 ( s )G5 ( s) The signal C 3 ( s ) V ( s )G 6 ( s ) R1 ( s )G 1 ( s )G 6 ( s ) R 2G 2 ( s )G 6 ( s ) R 3G 3 ( s )G 6 ( s ) Figure 5.13 Signalflow graph components: a. system; b. signal; c. interconnection of systems and signals Ex 5.5 (Converting common block diagrams to signalflow graphs) Convert the cascaded, parallel and feedback forms of the block diagrams shown in Figures 5.2(a), 5.3(a) and 5.4(b), respectively into a signal flow graphs. Solution: In each case we start by drawing the signal nodes for the system. Next, we interconnect the signal nodes with system branches. 14
15 Figure 5.14 Building signalflow graphs: a. cascaded system nodes (from Figure 5.3(a)); b. cascaded system signalflow graph; c. parallel system nodes (from Figure 5.5(a)); d. parallel system signalflow graph; e. feedback system nodes (from Figure 5.6(b)); f. feedback system signalflow graph 15
16 Ex 5.6 (Converting block diagram to a signalflow graph) Convert the block diagram of Figure 5.8 to a signalflow graph. Answer: Figure 5.15: Signalflow graph development: a. signal nodes; b. signalflow graph; c. simplified signalflow graph 16
17 If desired, simplify the signal flow graph by eliminating signals that have a single flow in and a single flow out. 5.5 MASON S RULE 17
18 Mason s rule is a technique by using one formula that was derived by S.J.Mason for reducing signalflow graphs to a single transfer function. Mason s formula has several components that must be evaluated and these definitions of the components must be well understood. Definitions 1. Loop gain the product of branch gains found by traversing a path that starts at a node and ends at the same node without passing through any other node more that once and following the direction of the signal flow. For examples of loop gains (see Figure 5.16). There are four loop gains: G2 ( s ) H 1 ( s ) G4 ( s ) H 2 ( s ) G4 ( s )G5 ( s) H 3 ( s) G4 ( s )G6 ( s) H 3 ( s ) Figure 5.16 : Signalflow graph for demonstrating Mason s rule 18
19 2. Forwardpath gain the product of gains found by traversing a path from the input node to the output node of the signal flow graph in the direction of signal flow. Examples of forwardpath are also shown in Figure There are two forwardpath gains: G1 ( s )G2 ( s )G3 ( s )G4 ( s)g5 ( s )G7 ( s ) G1 ( s )G2 ( s )G3 ( s )G4 ( s )G6 ( s )G7 ( s ) 3. Nontouching loops loops that do not have any nodes in common. In Figure 5.16, loop G2 ( s ) H 1 ( s ) does not touch loops G4 ( s ) H 2 ( s ), G4 ( s )G5 ( s) H 3 ( s) and G4 ( s )G6 ( s ) H 3 ( s ) 4. Nontouchingloop gain the product of loop gains from nontouching loops taken two. Three, four, etc., at a time. In Figure 5.16, the product of loop gain G2 ( s ) H 1 ( s ) and loop gain G4 ( s ) H 2 ( s ) is a nontouchingloop gain taken two at a time are a time. In summary, all three of the nontouchingloop gains taken two at a time are: [G2 ( s ) H1 ( s )][G4 ( s ) H 2 ( s )] [G2 ( s ) H1 ( s )][G4 ( s )G5 ( s ) H 3 ( s )] [G2 ( s ) H1 ( s )][G4 ( s )G6 ( s ) H 3 ( s)] Now, we ready to state Mason s rule. 19
20 Mason s Rule The transfer function, C(s)/R(s). of a system represented by a signalflow graph is C ( s) G (s) R( s) T k k k Where k = number of forward path Tk = the kth forwardpath gain = 1 loop gains + nontouchingloop gains taken two at time nontouchingloop gains taken three at time nontouchingloop gains taken four at time. k = loop gain terms in that touch the kth forward path. In other words, k is formed by eliminating from those loop gains that touch the kth forward path. 20
21 Ex 5.7 (Transfer function via Mason s rule) Find the transfer function, C(s)/R(s), for the signalflow graph in Figure Figure 5.17 Solution: First, identify the forwardpath gains. There is only one: T1 G1 ( s )G2 ( s )G3 ( s )G4 ( s )G5 ( s ) Second, identify the loop gains. There are four, as follows: 1. G2 ( s ) H 1 ( s)  loop 1 2. G4 ( s ) H 2 ( s)  loop 2 3. G7 ( s ) H 4 ( s)  loop 3 4. G2 ( s )G3 ( s )G4 ( s )G5 ( s )G6 ( s )G7 ( s )G8 ( s )  loop 4 Third, identify the nontouching loops taken two at a time. Refer the loops defined in second step and Figure
22 We can see that: Loop 1 does not touch Loop 2, Loop 1 does not touch Loop 3, and Loop 2 does not touch Loop 3 Notice that loops 1, 2 and 3 all touch loop 4. Thus, the combinations of nontouching loops taken two at time are as follows: 1. G2 ( s ) H 1 ( s )G4 ( s ) H 2 ( s ) Loop 1 and Loop 2 2. G2 ( s) H1 ( s)g7 ( s) H 4 ( s ) Loop 1 and Loop 3 3. G4 ( s) H 2 ( s )G7 ( s ) H 4 ( s ) Loop 2 and Loop 3 Finally, the nontouching loops taken three at a time are as follows: G2 ( s) H1 ( s )G4 ( s ) H 2 ( s)g7 ( s ) H 4 ( s) Loops 1, 2 and 3 Therefore; [G2 ( s ) H1 ( s) G4 ( s) H 2 ( s) G7 ( s ) H 4 ( s) 1 G ( s)g ( s )G ( s )G ( s )G ( s )G ( s )G ( s )] [G2 ( s ) H1 ( s)g4 ( s) H 2 ( s ) G2 ( s) H1 ( s )G7 ( s ) H 4 ( s) G4 ( s) H 2 ( s )G7 ( s ) H 4 ( s )] [G2 ( s) H 1 ( s )G4 ( s ) H 2 ( s )G7 ( s ) H 4 ( s )] We form k by eliminating from the loop gains that touch the kth forward path: 1 1 G7 ( s ) H 4 ( s ) The transfer function yields: G( s) 22 T1 1 [G1 ( s )G2 ( s)g3 ( s)g4 ( s )G5 ( s )][1 G7 ( s ) H 4 ( s )]
23 Ex 5.8 (Transfer function via Mason s rule) Use Mason s rule to find the transfer function of the signalflow diagram shown in Figure 5.19(c). Notice that this is the same system used in Example 5.2 to find the transfer function via block diagram reduction. Answer: T (s) 23 G1 ( s )G3 ( s )[1 G2 ( s )] [1 G2 ( s ) H 2 ( s ) G1 ( s)g2 ( s ) H 1 ( s)][1 G3 ( s ) H 3 ( s )]
Control Systems, Lecture 05
Control Systems, Lecture 05 İbrahim Beklan Küçükdemiral Yıldız Teknik Üniversitesi 2015 1 / 33 Laplace Transform Solution of State Equations In previous sections, systems were modeled in state space, where
More informationSECTION 2: BLOCK DIAGRAMS & SIGNAL FLOW GRAPHS
SECTION 2: BLOCK DIAGRAMS & SIGNAL FLOW GRAPHS MAE 4421 Control of Aerospace & Mechanical Systems 2 Block Diagram Manipulation Block Diagrams 3 In the introductory section we saw examples of block diagrams
More informationGoals for today 2.004
Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
More informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
More informationVideo 5.1 Vijay Kumar and Ani Hsieh
Video 5.1 Vijay Kumar and Ani Hsieh Robo3x1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior
More informationControl Systems (CS)
Control Systems (CS) Lecture0 Signal Flow raphs Dr. Imtiaz ussain Associate Professor Mehran University of Engineering & Technology Jamshoro, Pakistan email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/
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 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 informationSFG and Mason s Rule : A revision
SFG and Mason s Rule : A revision Andersen Ang 2016Nov29 SFG and Mason s Rule Vu Pham Review SFG: SignalFlow Graph SFG is a directed graph SFG is used to model signal flow in a system SFG can be used
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 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 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 informationLab # 4 Time Response Analysis
Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an
More informationJ א א J א א א F א א א א
J CHAPTER # 4 SIGNAL FLOW GRAPH (SFG) 1. Introduction For complex control systems, the block diagram reduction technique is cumbersome. An alternative method for determining the relationship between system
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 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 informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
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 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 informationReview: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Plan of the Lecture Review: transient and steadystate
More informationLecture 7:Time Response PoleZero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 FirstOrder Specs: Step : Pole Real inputs contain
More informationFeedback Control part 2
Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open and closedloop control Everything before chapter 7 are openloop systems Transient response Design criteria Translate criteria
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 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 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 informationStudy Material. CONTROL SYSTEM ENGINEERING (As per SCTE&VT,Odisha new syllabus) 4th Semester Electronics & Telecom Engineering
Study Material CONTROL SYSTEM ENGINEERING (As per SCTE&VT,Odisha new syllabus) 4th Semester Electronics & Telecom Engineering By Sri Asit Kumar Acharya, Lecturer ETC, Govt. Polytechnic Dhenkanal & Sri
More informationSRV02Series Rotary Experiment # 1. Position Control. Student Handout
SRV02Series Rotary Experiment # 1 Position Control Student Handout SRV02Series Rotary Experiment # 1 Position Control Student Handout 1. Objectives The objective in this experiment is to introduce the
More informationCONTROL SYSTEMS LECTURE NOTES B.TECH (II YEAR II SEM) ( ) Prepared by: Mrs.P.ANITHA, Associate Professor Mr.V.KIRAN KUMAR, Assistant Professor
LECTURE NOTES B.TECH (II YEAR II SEM) (201718) Prepared by: Mrs.P.ANITHA, Associate Professor Mr.V.KIRAN KUMAR, Assistant Professor Department of Electronics and Communication Engineering MALLA REDDY
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 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 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 informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad  500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III
More informationReduction of Multiple Subsystems
F I V E Reduction of Multiple Subytem SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Deigning a CloedLoop Repone a. Drawing the block diagram of the ytem: u i +  Pot 0 Π Pre amp K Power amp 50
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
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 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 informationINSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad ELECTRICAL AND ELECTRONICS ENGINEERING TUTORIAL QUESTION BANK
Course Name Course Code Class Branch INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad 500 043 ELECTRICAL AND ELECTRONICS ENGINEERING TUTORIAL QUESTION BAN : CONTROL SYSTEMS : A50 : III B. Tech
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 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 information7.1 Introduction. Apago PDF Enhancer. Definition and Test Inputs. 340 Chapter 7 SteadyState Errors
340 Chapter 7 SteadyState Errors 7. Introduction In Chapter, we saw that control systems analysis and design focus on three specifications: () transient response, (2) stability, and (3) steadystate errors,
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 ENGINEERING Sixth Edition International Student Version
CONTROL SYSTEMS ENGINEERING Sixth Edition International Student Version Norman S. Nise California State Polytechnic University, Pomona John Wiley fir Sons, Inc. Contents PREFACE, vii 1. INTRODUCTION, 1
More informationSchool of Engineering Faculty of Built Environment, Engineering, Technology & Design
Module Name and Code : ENG60803 Real Time Instrumentation Semester and Year : Semester 5/6, Year 3 Lecture Number/ Week : Lecture 3, Week 3 Learning Outcome (s) : LO5 Module Coordinator/Tutor : Dr. Phang
More informationDesign via Root Locus
Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steadystate error (Sections
More informationRoot Locus Techniques
Root Locus Techniques 8 Chapter Learning Outcomes After completing this chapter the student will be able to: Define a root locus (Sections 8.1 8.2) State the properties of a root locus (Section 8.3) Sketch
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
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 informationFeedback Control Systems
ME Homework #0 Feedback Control Systems Last Updated November 06 Text problem 67 (Revised Chapter 6 Homework Problems attached) 65 Chapter 6 Homework Problems 65 Transient Response of a Second Order Model
More informationEC Control Systems Question bank
MODULE I Topic Question mark Automatic control & modeling, Transfer function Write the merits and demerits of open loop and closed loop Month &Year May 12 Regula tion Compare open loop system with closed
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 informationProblems XO («) splane. splane *~8 X 5. id) X splane. splane. * Xtg) FIGURE P8.1. jplane. JO) k JO)
Problems 1. For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4] *~8 XO
More informationEEE 184 Project: Option 1
EEE 184 Project: Option 1 Date: November 16th 2012 Due: December 3rd 2012 Work Alone, show your work, and comment your results. Comments, clarity, and organization are important. Same wrong result or same
More informationDesign via Root Locus
Design via Root Locus 9 Chapter Learning Outcomes After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steadystate error (Sections
More informationIMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION. K. M. Yanev A. Obok Opok
IMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION K. M. Yanev A. Obok Opok Considering marginal control systems, a useful technique, contributing to the method of multistage compensation is suggested. A
More informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationSecond Order and Higher Order Systems
Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical secondorder control system to a step input. In terms of damping ratio and natural
More informationC(s) R(s) 1 C(s) C(s) C(s) = s  T. Ts + 1 = 1 s  1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain
analyses of the step response, ramp response, and impulse response of the secondorder systems are presented. Section 5 4 discusses the transientresponse analysis of higherorder systems. Section 5 5 gives
More informationVALLIAMMAI ENGINEERING COLLEGE
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK V SEMESTER IC650 CONTROL SYSTEMS Regulation 203 Academic Year 207 8 Prepared
More informationQuestion paper solution. 1. Compare linear and nonlinear control system. ( 4 marks, Dec 2012)
Question paper solution UNIT. Compare linear and nonlinear control system. ( 4 marks, Dec 0) Linearcontrol system: obey super position theorem, stability depends only on root location, do not exhibit
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
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 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 informationBangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory
Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system
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 informationTest 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010
Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the
More informationEEE 480 LAB EXPERIMENTS. K. Tsakalis. November 25, 2002
EEE 480 LAB EXPERIMENTS K. Tsakalis November 25, 2002 1. Introduction The following set of experiments aims to supplement the EEE 480 classroom instruction by providing a more detailed and handson experience
More informationPoles, Zeros and System Response
Time Response After the engineer obtains a mathematical representation of a subsystem, the subsystem is analyzed for its transient and steady state responses to see if these characteristics yield the desired
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 informationCONTROL * ~ SYSTEMS ENGINEERING
CONTROL * ~ SYSTEMS ENGINEERING H Fourth Edition NormanS. Nise California State Polytechnic University, Pomona JOHN WILEY& SONS, INC. Contents 1. Introduction 1 1.1 Introduction, 2 1.2 A History of Control
More informationHomework 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 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 informationECE382/ME482 Spring 2005 Homework 1 Solution February 10,
ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 1 Solution to HW1 P2.33 For the system shown in Figure P2.33 on p. 119 of the text, find T(s) = Y 2 (s)/r 1 (s). Determine a relationship that
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 informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 09Dec13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More information(a) Torsional springmass system. (b) Spring element.
m v s T s v a (a) T a (b) T a FIGURE 2.1 (a) Torsional springmass system. (b) Spring element. by ky Wall friction, b Mass M k y M y r(t) Force r(t) (a) (b) FIGURE 2.2 (a) Springmassdamper system. (b)
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 informationTransient Response of a SecondOrder System
Transient Response of a SecondOrder System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a wellbehaved closedloop
More informationEC6405  CONTROL SYSTEM ENGINEERING Questions and Answers Unit  I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
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 informationYTÜ Mechanical Engineering Department
YTÜ Mechanical Engineering Department Lecture of Special Laboratory of Machine Theory, System Dynamics and Control Division Coupled Tank 1 Level Control with using Feedforward PI Controller Lab Date: Lab
More informationMATHEMATICAL MODELING OF CONTROL SYSTEMS
1 MATHEMATICAL MODELING OF CONTROL SYSTEMS Sep14 Dr. Mohammed Morsy Outline Introduction Transfer function and impulse response function Laplace Transform Review Automatic control systems Signal Flow
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 informationManufacturing Equipment Control
QUESTION 1 An electric drive spindle has the following parameters: J m = 2 1 3 kg m 2, R a = 8 Ω, K t =.5 N m/a, K v =.5 V/(rad/s), K a = 2, J s = 4 1 2 kg m 2, and K s =.3. Ignore electrical dynamics
More information06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance.
Chapter 06 Feedback 06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Lesson of the Course Fondamenti di Controlli Automatici of
More informationCO Statement. Book No [Page No] C C C C
IC6501 CONTROL SYSTEMS L T P C 3 1 0 4 OBJECTIVES: To understand the use of transfer function models for analysis physical systems and introduce the control system components. To provide adequate knowledge
More informationComputer Aided Control Design
Computer Aided Control Design ProjectLab 3 Automatic Control Basic Course, EL1000/EL1100/EL1120 Revised August 18, 2008 Modified version of laboration developed by Håkan Fortell and Svante Gunnarsson
More informationPID Control. Objectives
PID Control Objectives The objective of this lab is to study basic design issues for proportionalintegralderivative control laws. Emphasis is placed on transient responses and steadystate errors. The
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 informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
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 informationProblem Set 2: Solution Due on Wed. 25th Sept. Fall 2013
EE 561: Digital Control Systems Problem Set 2: Solution Due on Wed 25th Sept Fall 2013 Problem 1 Check the following for (internal) stability [Hint: Analyze the characteristic equation] (a) u k = 05u k
More informationHomework Assignment 3
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full
More informationENGR 2405 Chapter 8. Second Order Circuits
ENGR 2405 Chapter 8 Second Order Circuits Overview The previous chapter introduced the concept of first order circuits. This chapter will expand on that with second order circuits: those that need a second
More informationReviewer: prof. Ing. Miroslav Olehla, CSc. Osvald Modrlák, Lukáš Hubka Technical University of Liberec, 2014 ISBN
Bibliographic reference to this document: MODRLÁK, O. a L. HUBKA. Automatic Control in Mechatronics. 1st edition. Liberec: Technical University of Liberec, Faculty of Mechatronics, 2014. ISBN 978807494175
More informationSchool of Mechanical Engineering Purdue University. ME375 Feedback Control  1
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 informationQUIZ 1 SOLUTION. One way of labeling voltages and currents is shown below.
F 14 1250 QUIZ 1 SOLUTION EX: Find the numerical value of v 2 in the circuit below. Show all work. SOL'N: One method of solution is to use Kirchhoff's and Ohm's laws. The first step in this approach is
More informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
More information