Control Systems. Control Systems Design Lead-Lag Compensator.

Size: px
Start display at page:

Download "Control Systems. Control Systems Design Lead-Lag Compensator."

Transcription

1 Design Lead-Lag Compensator

2 Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design.

3 Frequeny Domain Design Frequeny response approah an impose the transient response performane indiretly Bode diagrams: useful for the frequeny domain speifiation Nyquist plot: useful for system (relative) stability analysis.

4 Lead Compensator Appropriate improvement in transient response and a small hange in steady-state auray G s Ts ( s) K T K (0 ) Ts s T Polar plot (K C = ) jt G ( j) K jt α determines the maximum phase angle φ m whih ours at ω m Zero:- T Pole: - T sin m 2 2

5 Lead Compensator Bode diagram (K C =, α=0.) G ( j) m jt K jt T G ( j) 0. jt j0. T Lead ompensator: High-pass filter Goal of the lead ompensator: to provide suffiient phase-lead angle to offset the exessive phase lag in the unompensated system

6 Lead Compensator The maximum phase lead angle φ m ours at ω m To inrease phase margin, magnitude of the ompensated system to ross 0 db at this frequeny ω m but need to are for magnitude distortion added by lead ompensator 50 Bode Diagram Gm = 9.93 db (at 4.0 rad/s), Pm = 04 deg (at rad/s) Magnitude (db) 0-50 Phase (deg) Phase Margin PM 80 G( j ) H ( j ) ω : Gain rossover frequeny Frequeny (rad/s)

7 Lead Compensator The magnitude of the lead ontroller at: m T G ( j) jt jt m j j T This is the amount that the lead ompensator will shift the magnitude plot. To have the gain rossover point (0 db) at the right point, we have to make sure that the geometri frequeny mean falls at the point where the open-loop unompensated system is 20 log 0 ( α) below 0 db. If is absorbed into the ontrol gain, then the previous maximum magnitude would be

8 Lead Compensator Design Step 0: Assume following lead ompensator Open-loop Plant: Gs () Compensator: G Ts Ts ( s) K K ( K K) Ts Ts Loop Gain of Compensated System: Ts Ts GC ( s) G( s) K G( s) G ( s) Ts Ts with G( s) KG( s)

9 Lead Compensator Design Steps Step : Determine K to satisfy stati error onstants (K p or K v ) Step 2: Using this K, draw a Bode diagram of G (s), and evaluate the phase margin Step 3: Determine the neessary phase angle needed to meet design spes.

10 Lead Compensator Design Steps Step 4: Determine by using ( φ m =required phase lead angle + 5~2deg) Find the frequeny where: sin Selet this frequeny as the new gain rossover frequeny, where the maximum phase shift ours Step 5: Determine the lead ompensator m T m G ( s) 20log0 - T - T

11 Lead Compensator Design Steps Step 6: Solve for K K Step 7: Chek the gain margin. If it is not satisfatory, one may have to iterate.

12 Example Ex. Gs () 4 s s 2 Design a lead ompensator to give K v = 20 s -, PM 50 deg, and GM 0 db Step 0: Find low-frequeny gain: Using G () s K C C Ts Ts 4K G ( s) KG( s) K K s s 2 C

13 Example Step : Ts 4K Kv lim sgc ( s) Gs lim s 2K s0 s0 Ts s s 2 Gain requirements: 2K 20 K 0 Step 2: draw a Bode diagram of G (s), and evaluate the phase margin G ( j) j( 40 j 2)

14 Example Step 3: Determine the neessary phase angle needed to meet design spes. φ m = 33 (neessary)+5 (additional) deg (extra 5 deg for ompensation of the shift in gain rossover frequeny.) Step 4: Determine sin sin 38 m sinm 0.24 sin sin 38 m

15 Example Step 5: Determine the orner frequenies First we find ω by noting that G ( s) 20log0 The graph gives at db G ( j) 6. 2 db m 9 rad/se The zero at /T T T rad / se The pole at /T T rad / se

16 Example Step 6: Solve for K C = K/ KC G ( s) K 0/ Ts K K Ts s T s T 4.7 s s Step 7: Verify the design PM 50 deg and GM infinite

17 Example Nyquist

18 Leture Outline Lag ompensator design in frequeny domain Lag ompensator design steps. Example on lag ompensator design.

19 Lag Compensator An appreiable improvement in steady-state auray at the expense of inreasing the transient-response time G s Ts ( s) K T K ( ) Ts s T Polar plot (K C = ) Zero:- T Pole: - T G ( j) K jt jt

20 Lag Compensator Bode diagram (K C =, β=0) G ( j) 0 jt j0t G ( j) K jt jt Lead ompensator: Low-pass filter Goal of the lead ompensator: to provide attenuation in high frequeny range to give a system suffiient phase margin

21 Lag Compensator Design Steps Step 0: Assume the following lag ompensator Compensator: G Ts Ts ( s) K K ( K K) Ts Ts Loop Gain of Compensated System: G Ts Ts ( s) G( s) K G( s) ( ) Ts Ts G s with G( s) KG( s) Step : Determine K to satisfy stati error onstants.

22 Lag Compensator Design Steps Step 2: Find C for G (s) Chek PM and GM to see if they meet spes If not, find the frequeny where G( jw) = -80 deg + required PM Required PM = speified PM + 5~2 deg. for safety margin This frequeny is the new gain rossover frequeny C Step 3: Choose the orner frequeny =/T of the zero We want to hange the magnitude plot without hanging the phase plot at the new rossover frequeny Therefore, hoose the zero at /T to be around deade below the new orner frequeny C

23 Lag Compensator Design Steps Step 4: Determine and the pole loation... We now examine than 0 db. G ( j ) to find out how muh it is greater 0dB G ( j log ) 20 0 Choose and then the pole is at /T Step 5: Form the lag ompensator... The atual ompensator gain K K

24 Example Ex. G( s) s( s )(0.5s ) Design a lag ompensator K v = 5 se -, PM > 40 degrees and GM > 0 db Step 0: G s ( s) K T,( ) s T G ( s) KG( s) s( s K )(0.5s )

25 Example Step : Gain K v lim s0 lim s0 K 5 sg sg ( s) ( s) G( s) lim s0 lim s0 s sk s( s )(0.5s Ts G Ts K ) ( s) Step 2: draw a Bode diagram of G (s) G ( s) s( s 5 )(0.5s )

26 Example Step 3: The phase margin is 20 degrees whih means the unity feedbak of the gain adjusted unompensated system G (s) is unstable. If we want to have a phase margin of 40 deg, we should add another 5~2 deg to be safe. Choose orner frequeny =/T=0. rad/se (zero of the lag ompensator) The phase of G (s) is 28 deg around 0.5 rad/s Therefore, the new rossover frequeny should be about 0.5 rad/s.

27 Example Step 4: Based on the new ross over frequeny, the zero should be plaed at = /T = 0.05 rad/s about a deade below. The gain of G (s) at = 0.5 rad/s is about 20 db so that s how muh the lag ompensator must attenuate. 20log The pole of the lag ompensator 0.0 T rad/se

28 Example Step 5: G ( s) K s 0 s 00 K K G ( s) G( s) 5(0s ) s(00s )( s )(0.5s )

29 Example Lag ontroller auses slower transients

30 Outline Lead-Lag ompensator Qualitative system responses.

31 Lead-Lag Compensators Whether to use the Lead or Lag ontroller depends on the nature of your plant Lead for improved transient performane Lag for improved steady-state performane when you need improved transient performane and steady-state traking lead-lag ompensator G ( s) K s s s T T2 (, ) s T T 2

32 Lead Lag Compensators Phase lead adds phase at the unompensated gain rossover frequeny thereby inreasing the phase margin Phase lag provides attenuation allowing an inrease in gain at low frequeny to improve steady-state performane K =, γ = β = 0, T 2 = 0T. K =, γ = β.

33 Lead Lag Compensators Qualitative system responses y (t) y (t) y (t) y(t) y (t) y (t) y (t) y(t) unompensated lead lag lead-lag

CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS. Professor Dae Ryook Yang

CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS. Professor Dae Ryook Yang CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 208 Dept. of Chemial and Biologial Engineering 0- Road Map of the Leture X Stability of losed-loop ontrol system

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

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

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

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

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

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

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

ECE-320 Linear Control Systems. Winter 2013, Exam 1. No calculators or computers allowed, you may leave your answers as fractions.

ECE-320 Linear Control Systems. Winter 2013, Exam 1. No calculators or computers allowed, you may leave your answers as fractions. ECE-320 Linear Control Systems Winter 2013, Exam 1 No alulators or omputers allowed, you may leave your answers as frations. All problems are worth 3 points unless noted otherwise. Total /100 1 Problems

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

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

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

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

(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

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

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

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

ECSE First order low pass filter First order high pass filter. As ω : Z 0; v = 0. Let's look at this in the s-domain. V (s) Find H(s) = + AC + C

ECSE First order low pass filter First order high pass filter. As ω : Z 0; v = 0. Let's look at this in the s-domain. V (s) Find H(s) = + AC + C ESE- First order low pass filter First order high pass filter Leture 4 ontinued sawyes@rpi.edu www.rpi.edu/~sawyes A v in v A v in v As (D): Z ; v = vin As : Z ; v = Let's look at this in the s-domain

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

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

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

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

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

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

Today we re going to talk about time delays, affectionately referred to as e -st in the Laplace domain. Why e -st? Recall: time.

Today we re going to talk about time delays, affectionately referred to as e -st in the Laplace domain. Why e -st? Recall: time. 6.302 Feedba Systems Reitation 23: ime Delays oday we re going to tal about time delays, affetionately referred to as e -st in the Laplae domain. Why e -st? Reall: time h(t) delay h(t-) Laplae transform:

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

Controller Design Based on Transient Response Criteria. Chapter 12 1

Controller Design Based on Transient Response Criteria. Chapter 12 1 Controller Design Based on Transient Response Criteria Chapter 12 1 Desirable Controller Features 0. Stable 1. Quik responding 2. Adequate disturbane rejetion 3. Insensitive to model, measurement errors

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

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

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

Step Response Analysis. Frequency Response, Relation Between Model Descriptions

Step Response Analysis. Frequency Response, Relation Between Model Descriptions Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step

More 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

AC : A GRAPHICAL USER INTERFACE (GUI) FOR A UNIFIED APPROACH FOR CONTINUOUS-TIME COMPENSATOR DESIGN

AC : A GRAPHICAL USER INTERFACE (GUI) FOR A UNIFIED APPROACH FOR CONTINUOUS-TIME COMPENSATOR DESIGN AC 28-1986: A GRAPHICAL USER INTERFACE (GUI) FOR A UNIFIED APPROACH FOR CONTINUOUS-TIME COMPENSATOR DESIGN Minh Cao, Wihita State University Minh Cao ompleted his Bahelor s of Siene degree at Wihita State

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

Control System Design

Control System Design ELEC ENG 4CL4: Control System Design Notes for Lecture #11 Wednesday, January 28, 2004 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Relative Stability: Stability

More information

EE 321 Project Spring 2018

EE 321 Project Spring 2018 EE 21 Projet Spring 2018 This ourse projet is intended to be an individual effort projet. The student is required to omplete the work individually, without help from anyone else. (The student may, however,

More information

Topic # Feedback Control

Topic # Feedback Control Topic #4 16.31 Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests. Fall 2007 16.31 4 2 Frequency Stability

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

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 (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

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

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

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

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

UTC. Engineering 329. Proportional Controller Design. Speed System. John Beverly. Green Team. John Beverly Keith Skiles John Barker.

UTC. Engineering 329. Proportional Controller Design. Speed System. John Beverly. Green Team. John Beverly Keith Skiles John Barker. UTC Engineering 329 Proportional Controller Design for Speed System By John Beverly Green Team John Beverly Keith Skiles John Barker 24 Mar 2006 Introdution This experiment is intended test the variable

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

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

Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback

Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Topic #17 16.31 Feedback Control State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall

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

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

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

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

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

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

= 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

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

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

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

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

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

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

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

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

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

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

Desired Bode plot shape

Desired Bode plot shape Desired Bode plot shape 0dB Want high gain Use PI or lag control Low freq ess, type High low freq gain for steady state tracking Low high freq gain for noise attenuation Sufficient PM near ω gc for stability

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

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

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

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

ME 475/591 Control Systems Final Exam Fall '99

ME 475/591 Control Systems Final Exam Fall '99 ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does

More information

MEM 355 Performance Enhancement of Dynamical Systems

MEM 355 Performance Enhancement of Dynamical Systems MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions

More information

ECE317 : Feedback and Control

ECE317 : Feedback and Control ECE317 : Feedback and Control Lecture : Steady-state error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace

More information

Prüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Prüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid

More information

Time Response Analysis (Part II)

Time Response Analysis (Part II) Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary

More 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

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

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

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour

More information

Simple FIR Digital Filters. Simple FIR Digital Filters. Simple Digital Filters. Simple FIR Digital Filters. Simple FIR Digital Filters

Simple FIR Digital Filters. Simple FIR Digital Filters. Simple Digital Filters. Simple FIR Digital Filters. Simple FIR Digital Filters Simple Digital Filters Later in the ourse we shall review various methods of designing frequeny-seletive filters satisfying presribed speifiations We now desribe several low-order FIR and IIR digital filters

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

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

D(s) G(s) A control system design definition

D(s) G(s) A control system design definition R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure

More 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

Chapter 6 Control Systems Design by Root-Locus Method. Lag-Lead Compensation. Lag lead Compensation Techniques Based on the Root-Locus Approach.

Chapter 6 Control Systems Design by Root-Locus Method. Lag-Lead Compensation. Lag lead Compensation Techniques Based on the Root-Locus Approach. hapter 6 ontrol Sytem Deign by Root-Lou Method Lag-Lead ompenation Lag lead ompenation ehnique Baed on the Root-Lou Approah. γ β K, ( γ >, β > ) In deigning lag lead ompenator, we onider two ae where γ

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

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

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

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

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

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

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

Visual Servoing Based on Multirate Sampling Control

Visual Servoing Based on Multirate Sampling Control (), () Visual ervoing Based on Multirate ampling Control tability Improvement by Intersample Observer and Periodi Disturbane Rejetion Hiroshi Fujimoto (Nagaoka University of Tehnology), Yoihi Hori (The

More information

EE3CL4: Introduction to Linear Control Systems

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

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

MEM 355 Performance Enhancement of Dynamical Systems

MEM 355 Performance Enhancement of Dynamical Systems MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Intro Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /5/27 Outline Closed Loop Transfer

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

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

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