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

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

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

Transcription

1 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: Stabilizing Network Servoamplifier + servomotor i s (+0.5s) o Part A: Draw the Bode plots of the uncompensated open-loop transfer function and conclude that the closed-loop system is unstable. Can this system be stabilized by a network consisting of a preamplifier with gain K? Part B: Design a stabilizing network so that:. The compensated closed-loop system is stable with a phase margin (PM) greater than 5 o and a gain margin (GM) greater than 0 db.. The low-frequency gain of the compensated open-loop system is equal to that of the uncompensated system. 3. The bandwidth of the compensated open-loop system is twice that of the uncompensated system. Solution: Part A: Stability Analysis of the Uncompensated System: The magnitude and phase of the uncompensated open-loop transfer function are given by: G(jω) = = o (jω) = ( + jω/) ω 80 o + (ω/) tan ω/ 0.5 ω + (ω/) 80 o tan ω = G(ω) φ(ω) The Bode plots (magnitude and phase) of the uncompensated open-loop transfer function are shown in solid line on the next page. The closed-loop system is unstable because the open-loop transfer function has negative phase margin at the crossover frequency ω c = / rad/s. Because of this negative phase margin, the system can never be stabilized by a preamplifier with a real gain K. The Bode magnitude plot would be shifted upwards and the croossover frequency would increase. Clearly, the phase will always be less than -80 o resulting in a negative phase margin.

2 6 db/oct 0 /8 / 6 db/oct / ω (rad/s) MAGNITUDE (db) 36 db/oct 8 db/oct deg/oct 5 deg/oct /8 / / ω (rad/s) PHASE (deg) 90 5 deg/oct 5 deg/oct deg/oct 70 5 deg/oct

3 Part B: Stabilizing-Network Design Stabilizing Network Servoamplifier + servomotor i + K +st α +s α T 0.5 s (+0.5s) o Preamplifier Lead Network We need to add some phase lead to the uncompensated open-loop transfer function in order to obtain positive phase margin. To realize the stabilizing network, we select a preamplifier followed by a lead network. The overall compensated open-loop transfer function has the following form: G(s) = K α + st + sαt 0.5 s ( + s/) = K α + s/ω s/ω s ( + s/) The low-frequency approximation of the compensated open-loop transfer function should equal that of the uncompensated system. Therefore, G(s) = K α + st + sαt 0.5 s ( + s/) Therefore, the preamplifier gain K should be selected such that = Kα 0.5 s = 0.5 s = Kα = K = α From the previous Bode plots, the bandwidth of the uncompensated open-loop transfer function is equal to the zero-db crossover frequency, that is ω B = / rad/s. The new open-loop bandwidth should be twice that of the uncompensated bandwith, that is, ω B = = rad/s As a rule of thumb, we want the compensated open-loop transfer function to cross the zero-db line with a slope of -6 db/oct (equivalently, -0 db/dec). Using the previous Bode plots, the phase margin of the uncompensated open-loop transfer function at ω c = rad/s is equal to -5 o (that is, it is 5 o below the 80 o -line). Since we want a minimum phase margin of 5 o, the phase lead to be added at the crossover frequency ω c = rad/s should be at least 5 o + 5 o = 60 o. Recalling the relationship between the maximum phase of a lead network and α, we should have tan α α 60o α α tan 60o = 3 α 3 α + α α α α α + 0 = α 7 8 = or α = 3.9 Since α <, we choose α = 6 = = K = 6 3

4 The maximum phase occurs at ω c = T α = rad/s = T α = T 6 = = T = ω = = T rad/s ω = = = rad/s αt /6 Finally, the open-loop transfer function of the compensated system can be expressed as G(s) = 6 + s 6 + s/ 0.5 s ( + s/) = 0.5( + s) s ( + s/) The magnitude and phase of the compensated open-loop transfer function are given by: G(jω) = 0.5( + jω) (jω) ( + jω/) = (ω) tan ω ω 80 o [ + (ω/) ] tan ω/ = ω ω ( + ω /6 tan ω 80 o tan ω = G(ω) φ(ω) The Bode plots (magnitude and phase) of the compensated open-loop transfer function are shown in dashed line next to the uncompensated Bode plots. Clearly, the system is now stable with a positive phase margin at the crossover frequency ω c = rad/s. We can either use the Bode straightline approximations to compute the gain and phase margins of the compensated system, or we can do exact calculations using a calculator. We will do both below. Phase Margin: Assuming that the crossover frequency is exactly ω c = rad/s, we read an approximate phase margin of PM = 5 o on the graph. An exact computation yields φ(ω c ) = tan ω c 80 o tan ω c = tan 80 o tan PM = 80 o + φ(ω c ) = 80 o 3. o = 7.9 o = 3. o The exact crossover frequency is such that G(ω) = ω ω ( + ω /6) = This expression can be reduced to the following polynomial equation: ω 8 + 3ω ω 56ω 6 = 0 which has ω c = rad/s as a solution. The exact phase margin at this frequency is φ(ω c ) = tan ω c 80 o tan ω c φ(ω c ) = 3.8 o PM = 80 o + φ(ω c ) = 80 o 3.8 o = 8. o = tan o tan

5 Gain Margin: From the Bode plots, the compensated open-loop transfer function has a phase of 80 o at ω o = rad/s. The actual phase at ω o = rad/s is φ(ω o ) = tan ω o 80 o tan ω o The approximate gain margin at this frequency is GM (db) = 0 log G(ω o ) = tan 80 o tan = 0 log G(ω o ) = ( ) = db = 83. o Using the fact that there is a maximum error of 6 db at the double pole ω o = rad/s, we can conclude that the gain margin is truly GM = 6 = 8 db An exact computation yields G(ω o ) = ωo ωo( + ωo/6) = ( + = 0.5 /6) GM = 0 log G(ω o ) = 0 log 0.5 = 8.0 db To be more accurate, we need to compute the exact frequency at which the phase is equal to 80 o. Using a calculator, and, by trial and error: By interpolation, ω o ω o (rad/s) φ(ω o ) (deg) 68.5 o 76 o 78 o 79.5 o 80.8 o = 3.7 rad/s. We check that φ(ω o ) = tan ω o 80 o tan ω o φ(ω o ) = o = 80 o = tan o 3.7 tan G(ω o ) = ωo ωo( + ωo/6) = ( = 0.3 /6) GM = 0 log G(ω o ) = 0 log 0.3 = 6.9 db Therefore, all of the criteria have been met and this design is satisfactory. Exercise #: Redo the design using decade straight-line approximations. Exercise #: Redo the design using a PD controller of the form G c (s) = K p + K d s = K p ( + sk d /K p ) = ( + sk). (Notice that K p must be equal to using Criterion #.) Design for an approximate gain K that will meet the design objectives. There is no need for a preamplifier in this case. 5

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

Homework 7 - Solutions

Homework 7 - Solutions Homework 7 - Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the

More 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

Active Control? Contact : Website : Teaching

Active Control? Contact : Website :   Teaching Active Control? Contact : bmokrani@ulb.ac.be Website : http://scmero.ulb.ac.be Teaching Active Control? Disturbances System Measurement Control Controler. Regulator.,,, Aims of an Active Control Disturbances

More information

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

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

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

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

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

Module 5: Design of Sampled Data Control Systems Lecture Note 8

Module 5: Design of Sampled Data Control Systems Lecture Note 8 Module 5: Design of Sampled Data Control Systems Lecture Note 8 Lag-lead Compensator When a single lead or lag compensator cannot guarantee the specified design criteria, a laglead compensator is used.

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

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

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

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

Frequency Response Analysis

Frequency Response Analysis Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions

More information

Exercise 1 (A Non-minimum Phase System)

Exercise 1 (A Non-minimum Phase System) Prof. Dr. E. Frazzoli 5-59- Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Non-minimum Phase System) To increase the rise time of the system, we

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

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

Exercise 1 (A Non-minimum Phase System)

Exercise 1 (A Non-minimum Phase System) Prof. Dr. E. Frazzoli 5-59- Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Non-minimum Phase System) To decrease the rise time of the system,

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

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

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

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

FREQUENCY-RESPONSE DESIGN

FREQUENCY-RESPONSE DESIGN ECE45/55: Feedback Control Systems. 9 FREQUENCY-RESPONSE DESIGN 9.: PD and lead compensation networks The frequency-response methods we have seen so far largely tell us about stability and stability margins

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

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

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

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

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

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

ECE 388 Automatic Control

ECE 388 Automatic Control Lead Compensator and PID Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage: http://ece388.cankaya.edu.tr

More information

LINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad

LINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad LINEAR CONTROL SYSTEMS Ali Karimpour Associate Professor Ferdowsi University of Mashhad Controller design in the frequency domain Topics to be covered include: Lag controller design 2 Dr. Ali Karimpour

More information

Engraving Machine Example

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

More information

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 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

More information

Nyquist Criterion For Stability of Closed Loop System

Nyquist Criterion For Stability of Closed Loop System Nyquist Criterion For Stability of Closed Loop System Prof. N. Puri ECE Department, Rutgers University Nyquist Theorem Given a closed loop system: r(t) + KG(s) = K N(s) c(t) H(s) = KG(s) +KG(s) = KN(s)

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

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

K(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s

K(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s 321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify

More information

2nd-order filters. EE 230 second-order filters 1

2nd-order filters. EE 230 second-order filters 1 nd-order filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth-, first-, or second-order polyinomials in the numerator. Use two reactive

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

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

ECEN 326 Electronic Circuits

ECEN 326 Electronic Circuits ECEN 326 Electronic Circuits Stability Dr. Aydın İlker Karşılayan Texas A&M University Department of Electrical and Computer Engineering Ideal Configuration V i Σ V ε a(s) V o V fb f a(s) = V o V ε (s)

More information

Professor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

Professor 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

16.30/31, Fall 2010 Recitation # 2

16.30/31, Fall 2010 Recitation # 2 16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R + - E G c (s) G(s) C Figure 1: The standard block diagram

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

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

ESE319 Introduction to Microelectronics. Feedback Basics

ESE319 Introduction to Microelectronics. Feedback Basics Feedback Basics Feedback concept Feedback in emitter follower Stability One-pole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability

More information

ESE319 Introduction to Microelectronics. Feedback Basics

ESE319 Introduction to Microelectronics. Feedback Basics Feedback Basics Stability Feedback concept Feedback in emitter follower One-pole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability

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

Theory of Machines and Automatic Control Winter 2018/2019

Theory of Machines and Automatic Control Winter 2018/2019 Theory of Machines and Automatic Control Winter 2018/2019 Lecturer: Sebastian Korczak, PhD, Eng. Institute of Machine Design Fundamentals - Department of Mechanics http://www.ipbm.simr.pw.edu.pl/ Lecture

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

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

Frequency domain analysis

Frequency domain analysis Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 2010-2011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 2010-2011

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

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

The Frequency-response Design Method

The Frequency-response Design Method Chapter 6 The Frequency-response Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)

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

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

Control Systems. Control Systems Design Lead-Lag Compensator.

Control Systems. Control Systems Design Lead-Lag Compensator. Design Lead-Lag Compensator hibum@seoulteh.a.kr Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design. Frequeny Domain Design Frequeny response

More information

FREQUENCY RESPONSE ANALYSIS Closed Loop Frequency Response

FREQUENCY RESPONSE ANALYSIS Closed Loop Frequency Response Closed Loop Frequency Response The Bode plot is generally constructed for an open loop transfer function of a system. In order to draw the Bode plot for a closed loop system, the transfer function has

More information

An Internal Stability Example

An Internal Stability Example An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several pole-zero cancellations between the controller and

More information

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson

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

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

MAE 143B - Homework 9

MAE 143B - Homework 9 MAE 43B - Homework 9 7.2 2 2 3.8.6.4.2.2 9 8 2 2 3 a) G(s) = (s+)(s+).4.6.8.2.2.4.6.8. Polar plot; red for negative ; no encirclements of, a.s. under unit feedback... 2 2 3. 4 9 2 2 3 h) G(s) = s+ s(s+)..2.4.6.8.2.4

More information

EKT 119 ELECTRIC CIRCUIT II. Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016 Dr. Mohd Rashidi Che Beson

EKT 119 ELECTRIC CIRCUIT II. Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016 Dr. Mohd Rashidi Che Beson EKT 9 ELECTRIC CIRCUIT II Chapter 3: Frequency Response of AC Circuit Sem 05/06 Dr. Mohd Rashidi Che Beson TRANSFER FUNCTION (TF Frequency response can be obtained by using transfer function. DEFINITION:

More information

Robust fixed-order H Controller Design for Spectral Models by Convex Optimization

Robust fixed-order H Controller Design for Spectral Models by Convex Optimization Robust fixed-order H Controller Design for Spectral Models by Convex Optimization Alireza Karimi, Gorka Galdos and Roland Longchamp Abstract A new approach for robust fixed-order H controller design by

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

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

Robust Control 3 The Closed Loop

Robust Control 3 The Closed Loop Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time

More information

Poles and Zeros and Transfer Functions

Poles and Zeros and Transfer Functions Poles and Zeros and Transfer Functions Transfer Function: Considerations: Factorization: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the

More information

ECEN 607 (ESS) Op-Amps Stability and Frequency Compensation Techniques. Analog & Mixed-Signal Center Texas A&M University

ECEN 607 (ESS) Op-Amps Stability and Frequency Compensation Techniques. Analog & Mixed-Signal Center Texas A&M University ECEN 67 (ESS) Op-Amps Stability and Frequency Compensation Techniques Analog & Mixed-Signal Center Texas A&M University Stability of Linear Systems Harold S. Black, 97 Negative feedback concept Negative

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

STABILITY ANALYSIS TECHNIQUES

STABILITY ANALYSIS TECHNIQUES ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to control-system design: 1 Stability, 2 Steady-state response, 3 Transient response

More information

FEEDBACK AND STABILITY

FEEDBACK AND STABILITY FEEDBCK ND STBILITY THE NEGTIVE-FEEDBCK LOOP x IN X OUT x S + x IN x OUT Σ Signal source _ β Open loop Closed loop x F Feedback network Output x S input signal x OUT x IN x F feedback signal x IN x S x

More information

MAE 142 Homework #2 (Design Project) SOLUTIONS. (a) The main body s kinematic relationship is: φ θ ψ. + C 3 (ψ) 0 + C 3 (ψ)c 1 (θ)

MAE 142 Homework #2 (Design Project) SOLUTIONS. (a) The main body s kinematic relationship is: φ θ ψ. + C 3 (ψ) 0 + C 3 (ψ)c 1 (θ) MAE 42 Homework #2 (Design Project) SOLUTIONS. Top Dynamics (a) The main body s kinematic relationship is: ω b/a ω b/a + ω a /a + ω a /a ψâ 3 + θâ + â 3 ψˆb 3 + θâ + â 3 ψˆb 3 + C 3 (ψ) θâ ψ + C 3 (ψ)c

More information

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III

More information

Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...

Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature... Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number..........................

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

CYBER EXPLORATION LABORATORY EXPERIMENTS

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)

More information

= rad/sec. We can find the last parameter, T, from ωcg new

= rad/sec. We can find the last parameter, T, from ωcg new EE572 Solution to HW#22. Keep working on your project!! 1. Consider the following system: W(s) + T s =1 msec G lead (z) G zoh (z) 8 ( s+ 4) - a) Design a lead compensator, G lead (z), which meets the following

More information

The loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)

The loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2) Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)

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

Loop shaping exercise

Loop shaping exercise Loop shaping exercise Excerpt 1 from Controlli Automatici - Esercizi di Sintesi, L. Lanari, G. Oriolo, EUROMA - La Goliardica, 1997. It s a generic book with some typical problems in control, not a collection

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

DIGITAL CONTROLLER DESIGN

DIGITAL CONTROLLER DESIGN ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steady-state accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some

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

Second-order filters. EE 230 second-order filters 1

Second-order filters. EE 230 second-order filters 1 Second-order filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth-, first-, or second-order polynomials in the numerator. Use two

More information

IC6501 CONTROL SYSTEMS

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

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

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 2: Drawing Bode Plots, Part 2 Overview In this Lecture, you will learn: Simple Plots Real Zeros Real Poles Complex

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

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using

More 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

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

Generalized Fractional Order Reset Element (GFrORE)

Generalized Fractional Order Reset Element (GFrORE) ENOC 217, June 25-3, 217, Budapest, Hungary Generalized Fractional Order Reset Element (GFrORE) Niranjan Saikumar and Hassan HosseinNia Precision and Microsystems Engineering, 3ME, TU Delft, The Netherlands

More information

VALLIAMMAI ENGINEERING COLLEGE

VALLIAMMAI ENGINEERING COLLEGE VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK V SEMESTER IC650 CONTROL SYSTEMS Regulation 203 Academic Year 207 8 Prepared

More 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