Homework 7 - Solutions

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Homework 7 - Solutions"

Transcription

1 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 gain, K, for the uncompensated system to operate with a 30% overshoot. Find the peak time and K v for the uncompensated system. Design a lag-lead compenator to decrease the peak time by a factor of 2, decrease the percent overshoot by a factor of 2, and improve the steady-state error by a factor of 30. Specify all poles, zeros and gains. Solution: Searching the 30% overshoot line (ζ = 0.358, arcsin (0.358) = ) for angle sum of 180 yields pole at, j3.818 and gain, K = T p = π = π = seconds. ω d K v = lim s 0 sg(s) = = (c) Required specifications: %OS = 15% T p = seconds K v = = Lead Design: Using the required specifications for %OS and T p, we can calculate the damping ratio and the natural frequency: ln(%os/100) ζ = π 2 + ln 2 (%OS/100) = ω d = π T p = ω n 1 ζ 2 = Hence ω n = Thus, the desired pole is located at ζω n + jω n 1 ζ 2 = j7.634 Assume a lead compensator zero at 5. Summing the angles of the uncompensated system s poles as well as the compensator zero at 5 yields Therefore, the compensator pole Rev. 1.0, 04/14/ of 13

2 must contribute ( ) = 8.8. Using the geometry below, tan(8.8 ) = p c Hence, p c = The compensated open-loop transfer function is now: K s(s + 11)(s ) Evaluating the gain for this function at the desired pole, we get K = Lag Design: The lead compensated K v is given by, K v = lim s 0 sg(s) = = Thus, we need an improvement over the lead compensated system of /7.469 = Choosing p c = 0.001, we get z c = Thus, the compensator is given by G lag (s) = s s The final open-loop transfer function is 2. Compensators (9 points) The unity feedback system shown in Problem 1, with G(s) = to meet the following specifications: 4430(s ) s(s + 11)(s )(s ) K (s 2 is to be designed + 4s + 8)(s + 10) Overshoot: Less than 25% Settling Time: Less than 1 second K p = 10 Do the following: (c) Evaluate the performance of the uncompensated system operating at 10% overshoot. Design a passive compensator to meet the desired specifications. Use MATLAB to simulate the compensated system. Compare the response with the desired specifications. Solution: Searching along the 10% overshoot line (ζ = 0.591) the operating point is found to be j2.53 with K = A third pole is at Thus, the estimated performance before Rev. 1.0, 04/14/ of 13

3 EE C128 / ME C134 Spring 2014 HW7 - Solutions compensation is: 10% overshoot, Ts = 4/1.85 = 2.16 seconds, and Kp = 21.27/(8 10) = Lead Design: Since the overshoot of the uncompensated system is already lower than the desired specification, we can keep the same damping ratio (ζ = 0.591). Place compensator zero arbitrarily at 3. Now, with the required specifications, we can find the natural frequency: 4 4 = = ωn = ζts p Thus the desired pole is located at ζωn + jωn 1 ζ 2 = 4 + j5.46. The angular contribution of the system poles and compensator zero at the design point is Thus, the compensator pole must contribute = Using the geometry below, tan(13.04 ) = Hence, pc = pc 4 K(s + 3) + 4s + 8)(s + 10)(s ) Evaluating this function at 4+j5.46 yields K = 1092 with higher order poles at and Lag Design: For the lead compensated system, Kp = Thus, we need an improvement of a factor of 10/1, 485 = Choosing the compensator pole at 0.01, we get Glag (s) = s Finally, the equivalent forward path transfer function is: s Thus, the compensated open-loop transfer function is Gc (s) = Rev. 1.0, 04/14/2014 (s2 (s2 1092(s + 3)(s ) + 4s + 8)(s + 10)(s )(s ) 3 of 13

4 EE C128 / ME C134 Spring 2014 HW7 - Solutions (c) System Response: Uncompensated system Compensated system Rev. 1.0, 04/14/ of 13

5 3. Bode Form (3 points) Express the following transfer functions in their Bode forms. 1 G(s) = s(s + 2)(s + 4) G(s) = (c) G(s) = Solution: Thus, (s + 5) (s + 2)(s + 4) (s + 3)(s + 5) s(s + 2)(s + 4) G(s) = G(jω) = M(ω) = 1 s(s + 2)(s + 4) 1 6ω 2 + j(8ω ω 3 ) 1 (8ω ω 3 ) 2 + (6ω 2 ) 2 Thus, φ(ω) = π arctan 8 ω2 6ω G(s) = G(jω) = (s + 5) (s + 2)(s + 4) jω + 5 ω 2 + 6jω ω 2 M(ω) = (8 ω 2 ) ω 2 (c) Thus, φ(ω) = arctan ω 5 G(s) = G(jω) = (s + 3)(s + 5) s(s + 2)(s + 4) (15 ω2 ) + 8jω 6ω6 2 + j(8ω ω 3 (15 ω M(ω) = 2 ) ω 2 36ω 4 + (8ω ω 3 ) 2 φ(ω) = arctan 8ω 8 ω2 arctan 15 ω2 6ω Rev. 1.0, 04/14/ of 13

6 EE C128 / ME C134 Spring 2014 HW7 - Solutions 4. Sketching Bode Plots (9 points) For each of the functions given in Problem 3, sketch the Bode asymptotic magnitude and asymptotic phase plots. Solution: Rev. 1.0, 04/14/ of 13

7 EE C128 / ME C134 Spring 2014 HW7 - Solutions 5. Nyquist Plots (6 points) Sketch the Nyquist diagram for each of the systems given below, and using the Nyquist Criterion, determine whether or not each system is stable. Solution: P = 0, N = 0 Using Nyquist Criterion, Z = P N = 0. Thus, the system is stable. Rev. 1.0, 04/14/ of 13

8 EE C128 / ME C134 Spring 2014 HW7 - Solutions P = 0, N = 2 Using Nyquist Criterion, Z = P N = 2. Thus, the system is unstable. 6. Bode Plot Analysis (6 points) For the Bode plots shown below, determine the transfer functions by hand, and confirm your answers via MATLAB. Rev. 1.0, 04/14/ of 13

9 Solution: Step 1: Initial slope (ω 0) is 20 db/decade. Thus, we can say that there exists a pure integration, 1/s. G(s) = K s Step 2: Final slope (ω ) is 40 db/decade. Looking for the intersection of the asymptotes K yields that the pole is at 5.3 rad/s. G(s) = s(s + 5.3) Step 3: 1/s at ω = 0.01 rad/s should contribute +40 db/decade. From the magnitude plot, we see that G(j0.01) = 66.5 db/decade. Thus, writing the function in bode form: G(s) = K 1 we see that K/5.3 contributes 26.5 db/decade ), s( s 5.3 Therefore, the transfer function is given by: 20 log K 5.3 = 26.5 K = G(s) = s(s + 5.3) Rev. 1.0, 04/14/ of 13

10 Step 4: MATLAB Simulation Step 1: By inspection, ω n = 10 rad/s and there exists a second order term of the form Kω 2 n s 2 + 2ζω n + ωn 2. And there is a zero at 1 rad/s. Therefore, there is a (s + 1) term in the numerator. 100K(s + 1) Step 2: Let ζ = 0.3, ω n = 10rad/s, we get: s 2 + 6s To compensate for the slope near the natural frequency, add a pole at 8.5 db. Therefore, a (s + 8.5) term exists in the denominator. Step 3: To compensate for the steep slope after the natural frequency, we add a zero at 23. Thus, a (s + 23) term exists in the numerator. Since these terms do not affect the gain of the bode plot, we need to make the DC gain of these terms equal to 1. Thus, we need a 8.5/23 850(s + 1)(s + 23) term. Therefore, G(s) = K 23(s 2 + 6s + 100)(s + 8.5) Rev. 1.0, 04/14/ of 13

11 Step 4: Check the Bode Plot to add appropriate gain: At ω 0, G(jω) = 10. Therefore, 20 log K = 10 or K = Therefore, 850(s + 1)(s + 23) G(s) = (s 2 + 6s + 100)(s + 8.5) = (s + 1)(s + 23) (s 2 + 6s + 100)(s + 8.5) Step 5: MATLAB Simulation Rev. 1.0, 04/14/ of 13

12 7. Bode Plot Analysis (6 points) The Bode plots for a plant, G(s), used in a unity feedback system are shown below. Do the following: Find the gain margin, phase margin, zero db frequency, 180 frequency, and the closed-loop bandwidth. Use your results in Part to estimate the damping ratio, percent overshoot, settling time, and peak time. Solutions: From the Bode Plots: Gain Margin = db; phase margin = ; zero db frequency = rad/s; 180 frequency = rad/s; bandwidth 7 db point) = 3.8 rad/s. Using the following relation (10.73 in Nise): φ M = tan 1 2ζ 2ζ ζ 4 we get, ζ = Using the damping ratio, we find the %OS = 17.93%. Using the following relation (10.55 in Nise): ω BW = 4 (1 2ζ T s ζ 2 ) + 4ζ 4 4ζ Rev. 1.0, 04/14/ of 13

13 we get, T s = 2.84 s. Using the following relation (10.56 in Nise): we get T p = 1.22s. π ω BW = (1 2ζ 2 ) + 4ζ 4 4ζ T p 1 ζ 2 Rev. 1.0, 04/14/ of 13

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

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

More information

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

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

More information

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

EE 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 information

Systems Analysis and Control

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

More information

8.1.6 Quadratic pole response: resonance

8.1.6 Quadratic pole response: resonance 8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Second-order denominator, of the form 1+a 1 s + a s v 1 (s) + C R Two-pole low-pass filter example v (s) with

More information

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

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.

More information

1 (20 pts) Nyquist Exercise

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

More information

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 K(s +1)(s +2) G(s) =. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:

More information

MAS107 Control Theory Exam Solutions 2008

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

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007

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

More information

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)

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)

More information

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques

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

More information

EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =

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

More information

Frequency Response Techniques

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

More information

Systems Analysis and Control

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

More information

EE3CL4: Introduction to Linear Control Systems

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

More information

Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system

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

More information

Solutions to Skill-Assessment Exercises

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

More information

1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I

1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant

More information

If you need more room, use the backs of the pages and indicate that you have done so.

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

More information

Exercises for lectures 13 Design using frequency methods

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)

More information

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.

More information

ECE 486 Control Systems

ECE 486 Control Systems ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following

More information

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. 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

More information

ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]

ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1] ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative

More information

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:

(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 information

r + - FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic

r + - FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of hand-written notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers

More information

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

Delhi 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: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONICS ENGINEERING

More information

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

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial : 5LS_EC_S_Control Systems_08 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONICS ENGINEERING Subject

More information

Frequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability

Frequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods

More information

EE 4343/ Control System Design Project LECTURE 10

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

More information

Outline. Classical Control. Lecture 1

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

More information

Software Engineering 3DX3. Slides 8: Root Locus Techniques

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

More information

Lecture 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 Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This

More information

AMME3500: System Dynamics & Control

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

More information

Compensator Design to Improve Transient Performance Using Root Locus

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

More information

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =

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

More information

Chapter 2 SDOF Vibration Control 2.1 Transfer Function

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:

More information

Lecture 11. Frequency Response in Discrete Time Control Systems

Lecture 11. Frequency Response in Discrete Time Control Systems EE42 - Discrete Time Systems Spring 28 Lecturer: Asst. Prof. M. Mert Ankarali Lecture.. Frequency Response in Discrete Time Control Systems Let s assume u[k], y[k], and G(z) represents the input, output,

More information

Design of a Lead Compensator

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

More information

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

Radar 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 information

INTRODUCTION TO DIGITAL CONTROL

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

More information

ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =

ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) = ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.

More information

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10

More information

MAE 143B - Homework 9

MAE 143B - Homework 9 MAE 143B - Homework 9 7.1 a) We have stable first-order poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straight-line

More information

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

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

More information

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type- servomechanism:

More information

R10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1

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

More information

Controls Problems for Qualifying Exam - Spring 2014

Controls 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 information

Dynamic Compensation using root locus method

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

More information

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.

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..

More information

ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION

ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned

More information

Module 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions

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

More information

Due Wednesday, February 6th EE/MFS 599 HW #5

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)]

More information

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD 206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)

More information

ECE382/ME482 Spring 2005 Homework 8 Solution December 11,

ECE382/ME482 Spring 2005 Homework 8 Solution December 11, ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 1 Solution to HW8 P1.21 We are given a system with open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1) and unity negative feedback. We are

More information

Root Locus Techniques

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

More information

Root Locus Methods. The root locus procedure

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

More information

Control Systems I Lecture 10: System Specifications

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

More information

The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain

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

More information

EECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.

EECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators. Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 5-8 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3

More information

Asymptotic Bode Plot & Lead-Lag Compensator

Asymptotic Bode Plot & Lead-Lag Compensator Asymptotic Bode Plot & Lead-Lag Compensator. Introduction Consider a general transfer function Ang Man Shun 202-2-5 G(s = n k=0 a ks k m k=0 b ks k = A n k=0 (s z k m k=0 (s p k m > n When s =, transfer

More information

SKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot

SKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot SKEE 3143 CONTROL SYSTEM DESIGN CHAPTER 3 Compenator Deign Uing the Bode Plot 1 Chapter Outline 3.1 Introduc4on Re- viit to Frequency Repone, ploang frequency repone, bode plot tability analyi. 3.2 Gain

More information

Root Locus Design Example #4

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

More information

Homework 11 Solution - AME 30315, Spring 2015

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

More information

Frequency (rad/s)

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

More information

Proportional plus Integral (PI) Controller

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

More information

Analysis and Design of Analog Integrated Circuits Lecture 12. Feedback

Analysis and Design of Analog Integrated Circuits Lecture 12. Feedback Analysis and Design of Analog Integrated Circuits Lecture 12 Feedback Michael H. Perrott March 11, 2012 Copyright 2012 by Michael H. Perrott All rights reserved. Open Loop Versus Closed Loop Amplifier

More information

(a) Find the transfer function of the amplifier. Ans.: G(s) =

(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 closed-loop system

More information

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.

6.1 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)

More information

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur

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

More information

Digital Control Systems

Digital Control Systems Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist

More information

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 information

Solution: K m = R 1 = 10. From the original circuit, Z L1 = jωl 1 = j10 Ω. For the scaled circuit, L 1 = jk m ωl 1 = j10 10 = j100 Ω, Z L

Solution: K m = R 1 = 10. From the original circuit, Z L1 = jωl 1 = j10 Ω. For the scaled circuit, L 1 = jk m ωl 1 = j10 10 = j100 Ω, Z L Problem 9.9 Circuit (b) in Fig. P9.9 is a scaled version of circuit (a). The scaling process may have involved magnitude or frequency scaling, or both simultaneously. If R = kω gets scaled to R = kω, supply

More information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

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 STAFF NAME: Mr. P.NARASIMMAN BRANCH : ECE Mr.K.R.VENKATESAN YEAR : II SEMESTER

More information

Automatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year

Automatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback

More information

Control Systems I. Lecture 9: The Nyquist condition

Control Systems I. Lecture 9: The Nyquist condition Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute

More information

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control

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

More information

Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10

Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:

More information

Dynamic circuits: Frequency domain analysis

Dynamic circuits: Frequency domain analysis Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution

More information

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 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

More information

Introduction to Feedback Control

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

More information

Chapter 7. Digital Control Systems

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,

More information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

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

More information

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design

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

More information

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method .. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to

More information

PID controllers. Laith Batarseh. PID controllers

PID controllers. Laith Batarseh. PID controllers Next Previous 24-Jan-15 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 information

The Frequency-Response

The Frequency-Response 6 The Frequency-Response Design Method A Perspective on the Frequency-Response Design Method The design of feedback control systems in industry is probably accomplished using frequency-response methods

More information

Stability of CL System

Stability of CL System Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed

More information

AN INTRODUCTION TO THE CONTROL THEORY

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

More information

Homework Assignment 3

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

More information

CDS 101/110 Homework #7 Solution

CDS 101/110 Homework #7 Solution Amplitude Amplitude CDS / Homework #7 Solution Problem (CDS, CDS ): (5 points) From (.), k i = a = a( a)2 P (a) Note that the above equation is unbounded, so it does not make sense to talk about maximum

More information

Classify a transfer function to see which order or ramp it can follow and with which expected error.

Classify a transfer function to see which order or ramp it can follow and with which expected error. Dr. J. Tani, Prof. Dr. E. Frazzoli 5-059-00 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,

More information

H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )

H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at ) .7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a

More information

Outline. Classical Control. Lecture 5

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?

More information

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

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

More information

Example on Root Locus Sketching and Control Design

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

More information

Chapter 9: Controller design

Chapter 9: Controller design Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer function from disturbances to the output 9.2.2. Feedback

More information

APPLICATIONS FOR ROBOTICS

APPLICATIONS 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 information