Time Response Analysis (Part II)


 Frederica Morrison
 1 years ago
 Views:
Transcription
1 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 axis (d) A pair of complex conjugate poles [GATE 988: Marks] Soln. For critically damped continuous time second order system roots of denominator are: s + ξω n s + ω n = 0 when ξ =, s + ω n s + ω n = 0 (s + ω n ) = 0 Root are s = ω n, ω n Double pole on real axis Option (b) 5. A secondorder system has transfer function given by G(s) =. If S +8s+5 the system, initially at rest, is subjected to a unit step input at t = 0, the second peak in the response will occur at (a) π sec (c) π 3 sec (b) π 3 sec (d) π sec [GATE 99: Marks] Soln. G(s) = 5 S +8s+5 ω n = 5 ω n = 5 ξω n = 8
2 ξ = 8 0 = 0. 8 ω d = ω n ξ for nd peak n = 3 = 5 (0. 8) = 3 t p = nπ ω d = 3π 3 = π sec Option (a) 3. The poles of a continuous time oscillator are.. [GATE 994: Mark] Soln. The poles of a continuous time oscillator are imaginary 4. The response of an LCR circuit to a step input is (a) Over damped (b) Critically damped (c) Oscillatory If the transfer function has (i) poles on the negative real axis (ii) poles on the imaginary axis (iii) multiple poles on the positive real axis (iv) poles on the positive real axis (v) multiple poles on the negative real axis [GATE: 994: Marks] Soln. The response of an LCR circuit to a step input (a) over damped case ξ > () poles on the negative real axis C(s) R(s) = ω n s + ξ ω n s + ω n
3 Roots of denominator are s = ξ ω n ± ω n ξ (b) critically damped ξ = (5) C(s) R(s) = ω n s + ω n s + ω n = ω n (s+ω n ), s = ω n, ω n It has multiple poles on the negative real axis (c) oscillatory () poles on the imaginary axis ξ = 0 s = ± jω n Option a, b 5, c 5. Match the following codes with ListI with ListII: List I (a) Very low response at very high frequencies (b) Over shoot (c) Synchrocontrol transformer output List II (i) Low pass systems (iv) Phasesensitive (ii) Velocity damping modulation (iii) Natural frequency (v) Damping ratio [GATE 944: Marks] Soln. (a) (b) very low response at very high frequencies low pass system over sheet Damping ratio (c) synchro control transformer output phase sensitive modulation Option: a, b 5, c 4
4 6. For a second order system, damping ratio (ξ), is 0 < ξ <, then the roots of the characteristic polynomial are (a) real but not equal (c) complex conjugates (b) real and equal (d) imaginary [GATE 995: Mark] Soln. ξ < underdamped system, roots of denominator s + ξ ω n s + ω n = 0 Are s = ξω n ± jω n ξ Roots are complex conjugate Option (c) 7. If L[f(t)] = (S+) then S +S+5 f(0+ ) and f( ) are given by (a) 0, respectively (c) 0, respectively (b). 0 respectively (d) /5, 0 respectively [Note: L stands for Laplace Transform of] [GATE 995: Mark] Soln. F(s) = (s+) s +s+5 f(0 + ) = lim s SF(s) = s +s s +s+5 + s = lim = +0 = s + s + 5 s +0 f( ) = lim s 0 SF(s) = s +s s +s+5 = 0 Option (b)
5 8. The final value theorem is used to find the (a) steady state value of the system output (b) initial value of the system output (c) transient behavior of the system output (d) none of these [GATE 995: Mark] Soln. The final value theorem is used to find the steady state value of the system. Option (a) 9. If F(s) = ω, then the value of lim f(t), S +ω t {where F(s)is the L[f(t)]} (a) cannot be determined (b) is zero Soln. F(s) = F(s) = ω s +ω ω s +ω (c) is unity (d) is infinite [GATE 998: Mark] has poles s ± jω (pure imaginary) it is oscillatory function hence final value lim t f(t) can not be determined 0. Consider a feedback control system with loop transfer function K( + 0.5s) G(s)H(s) = s( + s)( + s) The type of the closed loop system is (a) zero (c) two (b) one (d) three [GATE 998: Mark] Soln. G(s) H(s) = k(+0.5s) s(+s)(+s)
6 The number of poles at the origin of open loop transfer function given the type of the system. It is a type one system Option (b). The unit impulse response of a linear time invariant system is the unit step function u(t). For t > 0, the response of the system to an excitation e at u(t), a > 0 will be (a) ae at (b) (/a)( e at ) (c) a( e at ) (d) e at [GATE 998: Marks] Soln. H(s) = s System excitation r(t) = e at u(t) R(s) = s+a Response of the system C(s) = R(s) H(s) = s(s+a) C(s) = a [ s (s+a) ] C(t) = a [ e at ] Option (b). If the characteristic equation of a closedloop system is s + s + = 0, then the system is (a) Overdamped (c) Under damped (b) Critically damped (d) undamped
7 Soln. Characteristic equation of closed loop system is s + s + = 0 Comparing with nd order equation s + ξω n + ω n = 0 ξω n = ξ = ω n ω n =, ω n =, ξ = which is less than It is case of under damped Option (c) 3. Consider a system with the transfer function (s) = s+6 ratio will be 0.5 when the value of K is (a) /6 (b) 3 (c) /6 (d) 6 Ks +s+6. Its damping [GATE 00: Mark] Soln. Transfer function G(s) = s+6 ks +s+6 Damping ratio ξ = 0. 5 Comparing with nd order equation s + ξω n s + ω n G(s) = s+6 k[s + s k +6 k ] ω n = 6 k, ω n = 6 k ξω n = k
8 ξ 6 k = k k = k or 6 k = k or k = 6 Option (c) 4. A casual system having the transfer function H(s) = s+ is excited with 0u(t). The time at which the output reaches 99% of its steady state value is (a).7 sec (c).3 sec (b).5 sec (d). sec [GATE 004: Marks] Soln. H(s) = s+ r(t) = 0μ(t) R(s) = 0 s C(s) = H(s) R(s) = 0 s(s+) C(s) = 5 s 5 s+ C(t) = 5[ e t ] Steady state value = 5 99% of the steady value reaches at
9 5[ e t ] = ( e t ) = e t = Or e t = 0. t = l n (0. ) t =. 3 sec Option (c) 5. The transfer function of a plant is (s) = 5 (s+5)(s +s+) approximation of T(s) using dominant pole concept is (a) (b) (s+5)(s+) 5 (s+5)(s+) Soln. Transfer function = 5 (s+5)(s +s+) (c) (d) 5 (s +s+). The second order (s +s+) [GATE 007: Marks] = 5 5( s 5 +)(s +s+) The poles nearer to imaginary axis dominates nature of time response and are called dominant poles. The factor that has to be eliminated should be in time constant form so, T(s) = Option (d) s +s+
10 6. Group I lists a set of four transfer functions. Group II gives a list of possible step responses y(t). Match the step responses with the corresponding transfer functions. Group I P = 5 s + 5 R = 36 s + s + 36 Q = 36 s + 0s + 36 S = 36 s + 7s + 49 y(t) y(t).. t t y(t) y(t) (a) P 3, Q, R 4, S (b) P 3, Q, R 4, S (c) P, Q, R 4, S 3 (d) P 3, Q 4, R, S t [GATE 008: Marks] t
11 Soln. The figure is a case of critically damped, Figure underdamped, Figure 3 undamped, figure 4 is overdamped The given transfer function P = Q = 5 s s + 0s + 36 R = S = 36 s + s s + 7s + 49 Comparing the given transfer function with ω n s + ξ ω n s+ω n, P = ω n = 5, ξ = 0 5 s +5 P is undamped, matched to figure 3 so P 3 Q = 36 s +0s+36, ω n = 6 ξ ω n = 0 ξ = 0 = Q is overdamped matched to fig. 4 Q 4 R = 36 s +s+36, ω n = 6 ξ ω n = ξ = 6 = R is critically damped matched to fig, R
12 S = 49 s +7s+49, ω n = 7 ξ ω n = 7 ξ = = 7 7 = 0. 5 S is underdamped matched to fig Option (d) P 3, Q 4, R, S 7. The unit step response of an underdamped second order system has steady state value of. Which one of the following transfer functions has these properties? (a).4 s +.59s+. (c).4 s.59s+. (b) 3.8 s +.9s+.9 (d) 3.8 s.9s+.9 [GATE 009: Marks] Soln. Steady state value =  The nd order T F = ω n s + ξ ω n s+ω n For underdamped system ξ < Option (c) and (d) are wrong because the system is unstable as ξ is negative In option (a) ω n =. =. 06 ξ ω n =. 59 ξ = =.
13 ξ > hence it is wrong In option (b) T. F = 3.8 s +.9s+.9 ω n =. 9 =. 38 ξ ω n =. 9 ξ =.9.38 = ξ < so system is underdamped Option (b) 8. The differential equation 00 d y dy 0 + y = x(t) describes a system dt dt with an input x(t) and an output y(t). The system, which is initially relaxed, is exited, by a unit step input. The output y(t) can be represented by the waveform. (a) (b) (c) (d) [GATE 0: Mark]
14 Soln. 00 d y dy dt 0 + y = x(t) dt Taking Laplace transform of both sides 00 s y(s) 0 s y (s) + y(s) = X(s) = s (00s 0s + )y(s) = s y(s) = Poles are at s = 0,, s(0s ) 0 0 Poles are on the right hand side of s plane so given system is unstable Option (a) represents unstable system. 9. The openloop transfer function of a dc motor is given as ω(s) V a (a) = 0 + 0s when connected in feedback as shown below, the approximate value of K a that will reduce the time constant of the closed loop system by one hundred times as compared to that of the openloop system is (a) (b) 5 (c) 0 (d) 00 [GATE 03: Marks] Soln. The open loop transfer function of a dc motor is given as ω(s) V a (s) = When connected in feedback is shown below. 0 +0s R(s) +  k a 0 + 0s ω(s)
15 Open loop transfer function = 0 k a +0 s Loop time constant τ 0l = 0 G(s) = k a s+ 0 g(t) = e t 0 Closed loop transfer function = H(s) = G(s) +G(s) H(s) = 0k a +0s = +(0k a ) +0s H(s) = 0k a +0s+0k a = h(t) = k a e ( 0k a +0s+0k a k a s+(k a + 0 ) ka+0. )t Time constant of closed loop system = (k a + 0 ) τ cl = τ ol 00 = = 0 00 = k a +0. k a +0. or k a = 0 0. = Option (c)
16 0. For the second order closedloop system shown in the figure, the natural frequency (in rad/s) is U(s) s(s+4) Y(s) (a) 6 (b) 4 (c) (d) [GATE 04: Mark] Soln. Y(s) U(s) = 4 s(s+4) +4 s(s+4) = = 4 s(s+4)+4 4 s +4s+4 Comparing with nd order system transfer Function ω n s + ξ ω n s+ω n ω n = 4, or ω n = Option (c)
LTI Systems (Continuous & Discrete)  Basics
LTI Systems (Continuous & Discrete)  Basics 1. A system with an input x(t) and output y(t) is described by the relation: y(t) = t. x(t). This system is (a) linear and timeinvariant (b) linear and timevarying
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 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 informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
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 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 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 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 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 informationGATE EE Topic wise Questions SIGNALS & SYSTEMS
www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)
More informationTime Response of Systems
Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) splane Time response p =0 s p =0,p 2 =0 s 2 t p =
More informationOutline. Classical Control. Lecture 2
Outline Outline Outline Review of Material from Lecture 2 New Stuff  Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New
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 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 informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationagree w/input bond => + sign disagree w/input bond =>  sign
1 ME 344 REVIEW FOR FINAL EXAM LOCATION: CPE 2.204 M. D. BRYANT DATE: Wednesday, May 7, 2008 9noon Finals week office hours: May 6, 47 pm Permitted at final exam: 1 sheet of formulas & calculator I.
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 informationIntroduction & Laplace Transforms Lectures 1 & 2
Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Control System Definition of a Control System Group of components that collectively
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 informationAnalysis and Design of Control Systems in the Time Domain
Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.
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 informationTransform Solutions to LTI Systems Part 3
Transform Solutions to LTI Systems Part 3 Example of second order system solution: Same example with increased damping: k=5 N/m, b=6 Ns/m, F=2 N, m=1 Kg Given x(0) = 0, x (0) = 0, find x(t). The revised
More informationDr Ian R. Manchester
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 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 informationEE Experiment 11 The Laplace Transform and Control System Characteristics
EE216:11 1 EE 216  Experiment 11 The Laplace Transform and Control System Characteristics Objectives: To illustrate computer usage in determining inverse Laplace transforms. Also to determine useful signal
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 informationControls Problems for Qualifying Exam  Spring 2014
Controls Problems for Qualifying Exam  Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
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 informationME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II
ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
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 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 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 informationUnit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace
Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, 010 1 Pair Table Unit, Part : Unit,
More informationFrequency Response of Linear Time Invariant Systems
ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z
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 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 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 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 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 informationAn Introduction to Control Systems
An Introduction to Control Systems Signals and Systems: 3C1 Control Systems Handout 1 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie November 21, 2012 Recall the concept of a
More informationLecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.
ISS0031 Modeling and Identification Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design. Aleksei Tepljakov, Ph.D. September 30, 2015 Linear Dynamic Systems Definition
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 informationRoot Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus  1
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position
More informationModule 4. Related web links and videos. 1. FT and ZT
Module 4 Laplace transforms, ROC, rational systems, Z transform, properties of LT and ZT, rational functions, system properties from ROC, inverse transforms Related web links and videos Sl no Web link
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informatione st f (t) dt = e st tf(t) dt = L {t f(t)} s
Additional operational properties How to find the Laplace transform of a function f (t) that is multiplied by a monomial t n, the transform of a special type of integral, and the transform of a periodic
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 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 informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
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 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 informationStability. X(s) Y(s) = (s + 2) 2 (s 2) System has 2 poles: points where Y(s) > at s = +2 and s = 2. Y(s) 8X(s) G 1 G 2
Stability 8X(s) X(s) Y(s) = (s 2) 2 (s 2) System has 2 poles: points where Y(s) > at s = 2 and s = 2 If all poles are in region where s < 0, system is stable in Fourier language s = jω G 0  x3 x7 Y(s)
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 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 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 informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
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 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 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 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 informationLaplace Transforms Chapter 3
Laplace Transforms Important analytical method for solving linear ordinary differential equations.  Application to nonlinear ODEs? Must linearize first. Laplace transforms play a key role in important
More information26 Feedback Example: The Inverted Pendulum
6 Feedback Example: The Inverted Pendulum Solutions to Recommended Problems S6. Ld 0(t) (a) Ldz6(t) = g0(t) a(t) + Lx(t), Ld (t) dt  ga(t) = Lx(t) Taking the Laplace transform of both sides yields szlo(s)
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
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 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 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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #4 Monday, January 13, 2003 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Impulse and Step Responses of ContinuousTime
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 informationMA 266 Review Topics  Exam # 2 (updated)
MA 66 Reiew Topics  Exam # updated Spring First Order Differential Equations Separable, st Order Linear, Homogeneous, Exact Second Order Linear Homogeneous with Equations Constant Coefficients The differential
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 informationSchool of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See
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 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 informationFirst and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015
First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage
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 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 information( ) ( = ) = ( ) ( ) ( )
( ) Vρ C st s T t 0 wc Ti s T s Q s (8) K T ( s) Q ( s) + Ti ( s) (0) τs+ τs+ V ρ K and τ wc w T (s)g (s)q (s) + G (s)t(s) i G and G are transfer functions and independent of the inputs, Q and T i. Note
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS
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 informationOutline. Control systems. Lecture4 Stability. V. Sankaranarayanan. V. Sankaranarayanan Control system
Outline Control systems Lecture4 Stability V. Sankaranarayanan Outline Outline 1 Outline Outline 1 2 Concept of Stability Zero State Response: The zerostate response is due to the input only; all the
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 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 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 informationBasic Procedures for Common Problems
Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available
More informationEE/ME/AE324: Dynamical Systems. Chapter 7: Transform Solutions of Linear Models
EE/ME/AE324: Dynamical Systems Chapter 7: Transform Solutions of Linear Models The Laplace Transform Converts systems or signals from the real time domain, e.g., functions of the real variable t, to the
More informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
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 informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212  Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
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 informationProblem Weight Score Total 100
EE 350 EXAM IV 15 December 2010 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total
More informationChapter 6: The Laplace Transform. ChihWei Liu
Chapter 6: The Laplace Transform ChihWei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace
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 information8 sin 3 V. For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0.
For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0. Spring 2015, Exam #5, Problem #1 4t Answer: e tut 8 sin 3 V 1 For the circuit
More informationSome of the different forms of a signal, obtained by transformations, are shown in the figure. jwt e z. jwt z e
Transform methods Some of the different forms of a signal, obtained by transformations, are shown in the figure. X(s) X(t) L  L F  F jw s s jw X(jw) X*(t) F  F X*(jw) jwt e z jwt z e X(nT) Z  Z X(z)
More informationChapter three. Mathematical Modeling of mechanical end electrical systems. Laith Batarseh
Chapter three Mathematical Modeling of mechanical end electrical systems Laith Batarseh 1 Next Previous Mathematical Modeling of mechanical end electrical systems Dynamic system modeling Definition of
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 information