EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
|
|
- Andrea McGee
- 3 years ago
- Views:
Transcription
1 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 to drive the step response error to zero. (a) Draw the root locus for the uncompensated system. Answer: (pt) 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (b) (c) Determine the gain K and location of the poles at the operating point. Answer: (pt) K = 242, poles are.49,.4926 ± j Using sisotool, we have a damping ratio of. when K 242. The poles are.49,.4926± j G = tf(,poly([ 3 8])); 2 K = 242; 3 poles = pole(feedback(k*g,)) Add an ideal integral compensator with a pole at the origin and a zero at.. Draw the compensated root locus. Answer: (2pt) G c = K(s+.) s(s+)(s+3)(s+8) of 6
2 EE C28 / ME C34 Fall 24 HW 6.2 Solutions 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (d) (e) Again, determine the gain K, and the locations of the poles at the operating point. Answer: (pt) K 234, poles:.9387,.4848 ± j4.828,.98 Use MATLAB to simulate the step responses. Did you meet the design specifications? (Remember that the system is in negative feedback!) Answer: (pt) Step Response.6.4 Uncompensated compensated.2 Amplitude Time (seconds) Indeed the compensated system has zero steady-state error. The response reaches. at t 85 s. 2. Lag Compensator For the system G = K (s+)(s+3)(s+8) in negative feedback, we are going to design a lag compensator reduce the steady-state error by a factor of, operating with a damping ratio of. (a) Using the gain K from Problem 2(b), determine K p and the steady-state error for a step input for both the uncompensated system and the desired lag-compensated system. 2 of 6
3 EE C28 / ME C34 Fall 24 HW 6.2 Solutions Answer: (pt) K p =.8, e step ( ) =.9, K des p = 9.83, e des step( ) = K p = lim s (s + )(s + 3)(s + 8) =.8, e step ( ) = =.9, + K p e des step( ) = e step( ) K des p = =.9, e des = step ( ) (b) (c) Using these values and a pole placed at., determine the location of the required zero. Answer: (pt) s =. K des p = lim s s + z c s + p c G(s) = z c p c K p. If we place a pole at s =., then p c =., so z c = Kdes p =.. Thus the zero must be placed at s =.. p c K p Draw the compensated root locus. Answer: (2pt) 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (d) (e) Again, determine the new gain K, and the new locations of the poles at the operating point Answer: (pt) K 234, poles:.9388,.485 ± j4.83,.9 Use MATLAB to simulate the step responses. Did you meet the design specifications? (Remember that the system is in negative feedback!) Answer: (pt) 3 of 6
4 EE C28 / ME C34 Fall 24 HW 6.2 Solutions Step Response.6.4 Uncompensated compensated.2 Amplitude Time (seconds) The steady-state error of the uncompensated and compensated systems are.92 and.93, respectively, a reduction by a factor of 9.7 ( ). 3. Lead Compensator For the system G = (s+3)(s+4) in negative feedback, we are going to design a lead compensator to achieve a settling time of less than.7 s and a percent overshoot of less than.5%. (a) Draw the uncompensated root locus. Answer: (pt) Imaginary Axis (seconds ) Real Axis (seconds ) (b) A lead compensator moves the closed-loop poles to a new desired location. Where should the closed-loop poles be located to satisfy the design specifications? Answer: (2pt) s = 5.7 ± j4.27 for T s =.7 s and %OS =.5% Using a second-order approximation, the poles are located at s = ζω n ± jω n ζ 2. The settling time can be approximated by T s = 4 ζω n = 4 Re(s), 4 of 6
5 EE C28 / ME C34 Fall 24 HW 6.2 Solutions so we need Re(s) > 4.7 = 5.7. For at most.5% overshoot, we need damping ratio ζ > ln(%os/) = π 2 + ln 2 (%OS/) ln(.5) π 2 + ln 2 (.5) =.8 Using the representation of s = ζω n ±jω n ζ 2 in the complex plane,to ( satisfy ) the overshoot condition, the poles must lie below the line which makes angle θ = cos ζωn = cos (ζ) = 36.8 with the negative real axis. The poles should be in the region bounded by the solid lines in the graph below, with the poles at 5.7 ± j4.27 corresponding to the lower bounds. ω n Im(s) Re(s) (c) (d) In what general region should the compensator pole and zero be placed to shift the root locus in that direction? Why? Answer: (pt) The compensator zero needs to be put to the left of the current poles to pull the root locus to the left. The compensator pole should be placed even further off to the left to give the zero ample time to bring the root locus to the left before its effect is canceled by the compensator pole. For this example, we choose to place a zero at 25 and a pole at 6. Find the value of K that gives us an overshoot of.4% Answer: (pt) K = 9.2 For an overshoot of.4% we need ζ = ln(.4) =.85. π 2 + ln 2 (.4) Using sisotool, ζ =.85 when K = 9.2. (e) Draw the compensated root locus. Answer: (2pt) 5 of 6
6 EE C28 / ME C34 Fall 24 HW 6.2 Solutions 2 5 Imaginary Axis (seconds ) Real Axis (seconds ) (f) Use MATLAB to simulate the compensated step response and find the overshoot and settling time. Did you meet the design specifications? (Remember that the system is in negative feedback!) Answer: (pt) T s =.82s, %OS =.44 Step Response Amplitude Time (seconds) The overshoot is.44% and setting time is.82s. We did not meet the design specifications, though we were close. We used a second-order approximation for the design step. Even though the pole at 6 is far away from the dominant poles of the system, in this case the approximation was not close enough. 6 of 6
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
EE 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
EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following open-loop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the state-variable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
Transient 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
Proportional 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
MASSACHUSETTS 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:
Due Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
Software 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
Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system
ME45: Control Systems Lecture Time response of nd-order systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
Discrete 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},
Outline. 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?
Compensator 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
Meeting Design Specs using Root Locus
Meeting Design Specs using Root Locus So far, we have Lead compensators which cancel a pole and move it left, speeding up the root locus. PID compensators which add a zero at s=0 and add zero, one, or
Root 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
Design of a Lead Compensator
Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc - Funded by MHRD The Lecture Contains Standard Forms of
CHAPTER 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.
ME 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
MAS107 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
Root 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
6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.
6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
AMME3500: 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
Control Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
Unit 8: Part 2: PD, PID, and Feedback Compensation
Ideal Derivative Compensation (PD) Lead Compensation PID Controller Design Feedback Compensation Physical Realization of Compensation Unit 8: Part 2: PD, PID, and Feedback Compensation Engineering 5821:
7.4 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)
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
Chapter 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, time-invariant system. Let s see, I
Systems 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
Dynamic 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
Root 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
Control 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
Homework 11 Solution - AME 30315, Spring 2015
1 Homework 11 Solution - AME 30315, Spring 2015 Problem 1 [10/10 pts] R + - K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closed-loop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
If 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
PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada
PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop
Example on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44 - Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
(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.
100 (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
EEE 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
Outline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
EE 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 Phase-lead and Phase-lag compensators using
Root 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
1 (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
EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO
EE 380 EXAM II 3 November 2011 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
Systems 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
Controller 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
The 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
CYBER 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)
Chapter 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 open-loop system (a system without feedbac) and a closed-loop
Systems 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
Module 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus
Introduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System
VALLIAMMAI 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
Exercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31-3-17 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
Design 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 steady-state error (Sections
Chapter 2 SDOF Vibration Control 2.1 Transfer Function
Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 02-760-4253 Fax:02-760-4435 Introduction In this lesson, you will learn the following : The
Systems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12: Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain
Systems 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
IMPROVED TECHNIQUE OF MULTI-STAGE COMPENSATION. K. M. Yanev A. Obok Opok
IMPROVED TECHNIQUE OF MULTI-STAGE COMPENSATION K. M. Yanev A. Obok Opok Considering marginal control systems, a useful technique, contributing to the method of multi-stage compensation is suggested. A
Systems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 15: Root Locus Part 4 Overview In this Lecture, you will learn: Which Poles go to Zeroes? Arrival Angles Picking Points? Calculating
Bangladesh 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
Course 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: Steady-State Error Unit 7: Root Locus Techniques
Control 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
ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques [] For the following system, Design a compensator such
Digital Control: Summary # 7
Digital Control: Summary # 7 Proportional, integral and derivative control where K i is controller parameter (gain). It defines the ratio of the control change to the control error. Note that e(k) 0 u(k)
MAE 143B - Homework 8 Solutions
MAE 43B - Homework 8 Solutions P6.4 b) With this system, the root locus simply starts at the pole and ends at the zero. Sketches by hand and matlab are in Figure. In matlab, use zpk to build the system
Transient Response of a Second-Order System
Transient Response of a Second-Order System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a well-behaved closed-loop
Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequency-response design method Goal: wrap up lead and
Frequency 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
IC6501 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
Pitch Rate CAS Design Project
Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability
CHAPTER # 9 ROOT LOCUS ANALYSES
F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closed-loop system is closely related to the location of the closed-loop poles. If the system
PM diagram of the Transfer Function and its use in the Design of Controllers
PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude
Lecture 7:Time Response Pole-Zero 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 First-Order Specs: Step : Pole Real inputs contain
CHAPTER 7 STEADY-STATE RESPONSE ANALYSES
CHAPTER 7 STEADY-STATE 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
Robust 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
Lecture 1 Root Locus
Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the
Professor Fearing EE C128 / ME C134 Problem Set 4 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley. control input. error Controller D(s)
Professor Fearing EE C18 / ME C13 Problem Set Solution Fall 1 Jansen Sheng and Wenjie Chen, UC Berkeley reference input r(t) + Σ error e(t) Controller D(s) grid 8 pixels control input u(t) plant G(s) output
R10 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
KINGS 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
ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop 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 closed-loop system when the gain K changes from 0 to 1+ K G ( s)
R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage 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, Force-current and Force-Voltage analogies..
Root 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
INTRODUCTION 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 linear-time-invariant
EE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
INSTITUTE 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
Homework 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
UNIVERSITY 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
AN INTRODUCTION TO THE CONTROL THEORY
Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter
a. Closed-loop 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
Essence of the Root Locus Technique
Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general set-up, namely for the case when the closed-loop
Solutions to Skill-Assessment Exercises
Solutions to Skill-Assessment 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
6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson
Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closed-loop behavior what we want it to be. To review: - G c (s) G(s) H(s) you are here! plant For
Chapter 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 Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral
EE3CL4: Introduction to Linear Control Systems
1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We
2.004 Dynamics and Control II Spring 2008
MT OpenCourseWare http://ocw.mit.edu.004 Dynamics and Control Spring 008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts nstitute of Technology
Frequency (rad/s)
. The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer
ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
ECE-320: Linear Control Systems Homework 8. 1) For one of the rectilinear systems in lab, I found the following state variable representations:
ECE-30: Linear Control Systems Homework 8 Due: Thursday May 6, 00 at the beginning of class ) For one of the rectilinear systems in lab, I found the following state variable representations: 0 0 q q+ 74.805.6469
Chapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steady-state error, and transient response for computer-controlled systems. Transfer functions,
Engraving Machine Example
Engraving Machine Example MCE44 - Fall 8 Dr. Richter November 24, 28 Basic Design The X-axis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional