12.7 Steady State Error

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "12.7 Steady State Error"

Transcription

1 Lecture Notes on Control Systems/D. Ghose/ Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there was a steady state error between the output and the reference input. In fact, the same steady state error occurs even in the case of the second order system. Why does this steady state error occur? Note that the plant (which has the transfer function G(s)) is driven by u e, where e is the error signal. If the error goes to zero then the output will also be driven to zero. So there should be a non-zero input to the plant to eep it close to the reference input. But our objective is to mae sure that the steady state output matches the reference input. How can we ensure this? There are basically three possible methods by which we can achieve this objective by driving the steady state error to zero or to very small values. Two of these are almost obvious but are fraught with practical difficulties. The third one is the best and gives rise to the idea behind integral control. Method 1: Use large control gain. Since, for both first and second order systems, the steady state output of the closed-loop system is given by,wehave, 1as So, by using large values of, we can ensure that the steady state output is as close to 1 as we want. In fact, using large also improves system response as we have seen earlier. But there is a serious problem. Loo at the input to the system u(t) e(t) [r(t) y(t)] Initially, y(0) 0 and r(0) 1. So u(0) r(0), that is, the input to the plant is initially times the reference step input. This signal later drops to low values. Now, thin of any common electrical device that operates at the line voltage of 30 V. If 10, then initially the voltage applied to the device will be 300 V, enough to damage it completely! Same problem for electronic components that operate with voltages ranging between 5 V to 1 V. None of them are designed to withstand voltages that are 10 times the specified voltage. Method : Command shaping: Scale the commanded or reference input by. 1 Y (s) τs+1 +1 R(s) τs+1 1 R(s) For which, when R(s) 1/s, and y(t) 1 e t/ τ 1ast

2 Lecture Notes on Control Systems/D. Ghose/ Figure 1.9: Command shaping control This appears to offer the solution we are looing for. However, this method will wor only when we now the open-loop gain exactly, which is unliely since in a practical system, gains are measured inaccurately and they may vary over operating ranges and differing conditions. Suppose, the actual open-loop gain is not but is off by an amount δ. So,theopenloop gain is + δ. Then, the closed-loop gain is, ( ) + δ +1 + δ +1 + δ +1 + δ +1 + δ δ +1 + ( +1)δ ( + δ +1) 1 δ 1+ δ + ( ) 1+ δ ( 1 δ ) δ ( 1 δ ) +1 + (Expanding in Taylor series) δ δ ( 1 δ ) (Ignoring nd order terms in δ) +1 δ δ (Again ignoring nd order terms in δ) δ 1+ ( +1) Thus, a non-zero steady state error still remains unless is nown exactly and does not change during operation. Method 3: Integral control. For y 0 in steady state, G(s) has to be driven by u 0. But, if we ill the steady state error completely then this condition is no longer satisfied. So, instead of u e let us try u e. Then e r y can go to zero while u e 0.

3 Lecture Notes on Control Systems/D. Ghose/ Figure 1.10: Integral control The corresponding bloc diagram is as shown below. The plots alongside show that even when e 0, the area under the e curve which represents e is non-zero. Consider a first-order system. The open loop response is given by, The closed loop response is given by, G(s) 1 τs+1 Y ol (s) G(s)R(s) 1 τs+1 R(s) Y c (s) G(s)K(s) 1+G(s)K(s) τs + s + R(s) 1 s τs s τs+1 τ s + s τ + τ R(s) which is a second order system. Since, ζω n 1 τ is a constant, the settling time ( 4/(ζω n )), is a constant too, irrespective of the value of. What happens when you use the exact expression? Steady state response to a step input is given by, Y c (s) τ s + s τ + τ 1 s

4 Lecture Notes on Control Systems/D. Ghose/ The system is stable for all positive values of. This can be shown by a simple application of the Routh-Hurwitz criterion. s 1 τ s 1 1 τ 0 s 0 τ 0 The final value theorem can be applied to show that So there is no steady state error. Finally, we can see that y c ( ) lim s 0 sy c (s) 1 ω n which implies that the system is τ ζ 1 τ Underdamped if ζ<1 > 1 4τ Overdamped if ζ>1 < 1 4τ So, for a first order system, integral control produces zero steady state error for step inputs but runs the ris of oscillations in the output if the gain value is large. Let us consider a general second order system and examine how integral control affects its performance. Let the open loop system be given by, G ol (s) ω n s +ζω n s + ω n Then, the closed loop system with integral control would be, G c (s) ωn s +ζω n s+ωn s ωn s +ζω ns+ω n s 1+ ωn s(s +ζω n s + ωn)+ω n ωn s 3 +ζω n s + ωns + ωn

5 Lecture Notes on Control Systems/D. Ghose/ This is a third order system which has 3 real poles or one real pole and a pair of complex conjugate poles. Let us chec the stability of the system. s 3 1 ω n s ζω n ω n s 1 ζω n ω n ζ 0 s ω n 0 From which we can see that the system will be stable if 0 <<ζω n. So, a second order system with integral control is stable only when this condition on is met. Let us apply a step input and chec the steady state response by applying the final value theorem, assuming that the value of is so chosen as to maintain stability. Then, So, Y (s) which ills the steady state error. ωn s 3 +ζω n s + ωns + ωn 1 s y( ) lim s 0 sy (s) Summary of Results In this section we summarize the results obtained in the previous sections. consider, We will Class of systems: First order and Second order Class of control: Proportional and Integral Class of inputs: Step

6 Lecture Notes on Control Systems/D. Ghose/ Step input P-Control K(s) Integral Control K(s) s First Order System Open loop: Closed loop: Closed loop: G(s) 1 G τs+1 c (s) 1 τ G s+1 c (s) /τ s +(1/τ)s+(/τ) (Second order system) Gain 1 Gain Gain 1 Time constant τ Time constant τ T r.τ Tr.τ T r π ω n π τ T s 3.9τ Ts 3.9τ T s 4 ζω n 8τ Steady state error 0 Steady state error 1 Steady state error 0 Second Order System Open loop: Closed loop: Closed loop: ωn G(s) s +ζω ns+ωn ω G c (s) n s +ζω ns+ωn() G c (s) ωn() s +ζω ns+ωn() ω n s 3 +ζω ns +ω ns+ω n y( ) 1 y( ) y( ) 1 Stable Stable <ζω n (stable) ζω n (unstable) Problem Set 6 1. Let G(s) 1 be a first order system. Assume some value for τ and plot the sτ+1 open loop response to a step and a ramp input.. For the same system consider a feedbac with just P-control. For three different values of gain K plot the time response for a unit step and a unit ramp input. 3. Now consider a integral control. Select two values of K, one maing the system

7 Lecture Notes on Control Systems/D. Ghose/01 11 underdamped and the other overdamped and plot the time responses to step and ramp inputs. 4. Consider a second order system G(s) s +ζω ns+ω and assume reasonable values n for ζ and ω n. Do the same as above. Only in PI-control, select K such that one system is stable and the other unstable. 5. Compare the rise time, settling time, steady state error, etc. of the above time responses and comment on them. ω n Note: Use MATLAB. Select parameter values that are different from other students! 1.9 Disturbance Signals One of the properties of a feedbac system is that of disturbance rejection. Disturbance signals can appear at the system input, the system output, or at the sensor. Figure 1.11: A feedbac system with many input signals In order to find the effect of these disturbance signals we need to find the closed-loop transfer functions from each noise term to the output Y. In the following, we will drop the argument s (the laplace variable) from the equations for clarity. Y D o + Y D o + G(U + D i ) D o + G(KE + D i ) D o + G[K(R R )+D i ]

8 Lecture Notes on Control Systems/D. Ghose/ D o + G[K{R H(Y + N)} + D i ] D o + G[KR KHY KHN + D i ] D o + GKR GKHY GKHN + GD i So, where, Y 1 D KG o + R G Do D o + G R R + G N N + G Di D i G Do G R G N G Di 1 KG KGH G KGH N + G D i are the closed-loop transfer functions from the corresponding disturbance inputs to the output, when all other inputs are considered to be zero. So, a design objective would be to select K(s) andh(s) so that KG Y R R should have a good response Y Do,Y N,Y Di should all be small However, there has to be a trade-off since these four transfer functions are not independent and there are only two parameters K(s) andh(s). For example, note that, G Do + G N 1 + KGH 1 So, if the effect of the sensor noise N on the output is small, then the effect of the output disturbance D o has to be large, provided that the level of both disturbance signals are the same.

Chapter 12. Feedback Control Characteristics of Feedback Systems

Chapter 12. Feedback Control Characteristics of Feedback Systems Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an open-loop system (a system without feedbac) and a closed-loop

More information

Time Response of Systems

Time Response of Systems Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) s-plane Time response p =0 s p =0,p 2 =0 s 2 t p =

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

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

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

EEE 184: Introduction to feedback systems

EEE 184: Introduction to feedback systems EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)

More information

Lecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion

Lecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 First-Order Specs: Step : Pole Real inputs contain

More information

ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II

ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and

More 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

FEEDBACK CONTROL SYSTEMS

FEEDBACK CONTROL SYSTEMS FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles

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

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

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

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

Analysis and Design of Control Systems in the Time Domain

Analysis and Design of Control Systems in the Time Domain Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.

More 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

06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance.

06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Chapter 06 Feedback 06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Lesson of the Course Fondamenti di Controlli Automatici of

More information

Dynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.

Dynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology. Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control

More information

School of Mechanical Engineering Purdue University. ME375 Feedback Control - 1

School of Mechanical Engineering Purdue University. ME375 Feedback Control - 1 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

Laplace Transform Analysis of Signals and Systems

Laplace Transform Analysis of Signals and Systems Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.

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

Root Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus - 1

Root Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus - 1 Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics

More information

Proportional, Integral & Derivative Control Design. Raktim Bhattacharya

Proportional, Integral & Derivative Control Design. Raktim Bhattacharya AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University

More information

Root Locus Design Example #3

Root Locus Design Example #3 Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll

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

Notes for ECE-320. Winter by R. Throne

Notes for ECE-320. Winter by R. Throne Notes for ECE-3 Winter 4-5 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................

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

Alireza Mousavi Brunel University

Alireza Mousavi Brunel University Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched

More information

Lab # 4 Time Response Analysis

Lab # 4 Time Response Analysis Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an

More information

Chapter 7 : Root Locus Technique

Chapter 7 : Root Locus Technique Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci

More information

Chapter 5 HW Solution

Chapter 5 HW Solution Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, time-invariant system. Let s see, I

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

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

Transient Response of a Second-Order System

Transient Response of a Second-Order System Transient Response of a Second-Order System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a well-behaved closed-loop

More information

ME 375 System Modeling and Analysis. Homework 11 Solution. Out: 18 November 2011 Due: 30 November 2011 = + +

ME 375 System Modeling and Analysis. Homework 11 Solution. Out: 18 November 2011 Due: 30 November 2011 = + + Out: 8 November Due: 3 November Problem : You are given the following system: Gs () =. s + s+ a) Using Lalace and Inverse Lalace, calculate the unit ste resonse of this system (assume zero initial conditions).

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

Lab-Report Control Engineering. Proportional Control of a Liquid Level System

Lab-Report Control Engineering. Proportional Control of a Liquid Level System Lab-Report Control Engineering Proportional Control of a Liquid Level System Name: Dirk Becker Course: BEng 2 Group: A Student No.: 9801351 Date: 10/April/1999 1. Contents 1. CONTENTS... 2 2. INTRODUCTION...

More information

Performance of Feedback Control Systems

Performance of Feedback Control Systems Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steady-state Error and Type 0, Type

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

CHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System

CHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages

More information

An Introduction to Control Systems

An Introduction to Control Systems An Introduction to Control Systems Signals and Systems: 3C1 Control Systems Handout 1 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie November 21, 2012 Recall the concept of a

More information

Laboratory handouts, ME 340

Laboratory handouts, ME 340 Laboratory handouts, ME 340 This document contains summary theory, solved exercises, prelab assignments, lab instructions, and report assignments for Lab 4. 2014-2016 Harry Dankowicz, unless otherwise

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

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

Pole placement control: state space and polynomial approaches Lecture 2

Pole placement control: state space and polynomial approaches Lecture 2 : state space and polynomial approaches Lecture 2 : a state O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.fr www.gipsa-lab.fr/ o.sename -based November 21, 2017 Outline : a state

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

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

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -

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

a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a

a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a Root Locus Simple definition Locus of points on the s- plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation

More information

Solutions for Tutorial 4 Modelling of Non-Linear Systems

Solutions for Tutorial 4 Modelling of Non-Linear Systems Solutions for Tutorial 4 Modelling of Non-Linear Systems 4.1 Isothermal CSTR: The chemical reactor shown in textbook igure 3.1 and repeated in the following is considered in this question. The reaction

More information

Feedback Control of Linear SISO systems. Process Dynamics and Control

Feedback Control of Linear SISO systems. Process Dynamics and Control Feedback Control of Linear SISO systems Process Dynamics and Control 1 Open-Loop Process The study of dynamics was limited to open-loop systems Observe process behavior as a result of specific input signals

More information

Richiami di Controlli Automatici

Richiami di Controlli Automatici Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici

More information

16.31 Homework 2 Solution

16.31 Homework 2 Solution 16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p

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

Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur. Module 3 Lecture 8

Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur. Module 3 Lecture 8 Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur Module 3 Lecture 8 Welcome to lecture number 8 of module 3. In the previous

More information

Analysis of SISO Control Loops

Analysis of SISO Control Loops Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities

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

Dynamic System Response. Dynamic System Response K. Craig 1

Dynamic System Response. Dynamic System Response K. Craig 1 Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. Non-LTI Behavior Solution of Linear, Constant-Coefficient, Ordinary Differential Equations Classical

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

Linear Control Systems Solution to Assignment #1

Linear Control Systems Solution to Assignment #1 Linear Control Systems Solution to Assignment # Instructor: H. Karimi Issued: Mehr 0, 389 Due: Mehr 8, 389 Solution to Exercise. a) Using the superposition property of linear systems we can compute the

More information

Chapter 3. 1 st Order Sine Function Input. General Solution. Ce t. Measurement System Behavior Part 2

Chapter 3. 1 st Order Sine Function Input. General Solution. Ce t. Measurement System Behavior Part 2 Chapter 3 Measurement System Behavior Part 2 1 st Order Sine Function Input Examples of Periodic: vibrating structure, vehicle suspension, reciprocating pumps, environmental conditions The frequency of

More information

Singular Value Decomposition Analysis

Singular Value Decomposition Analysis Singular Value Decomposition Analysis Singular Value Decomposition Analysis Introduction Introduce a linear algebra tool: singular values of a matrix Motivation Why do we need singular values in MIMO control

More information

Transfer func+ons, block diagram algebra, and Bode plots. by Ania- Ariadna Bae+ca CDS Caltech 11/05/15

Transfer func+ons, block diagram algebra, and Bode plots. by Ania- Ariadna Bae+ca CDS Caltech 11/05/15 Transfer func+ons, block diagram algebra, and Bode plots by Ania- Ariadna Bae+ca CDS Caltech 11/05/15 Going back and forth between the +me and the frequency domain (1) Transfer func+ons exist only for

More information

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control DC Motor Control Trainer (DCMCT) Student Manual Table of Contents 1 Laboratory Objectives1 2 References1 3 DCMCT Plant

More information

Control System. Contents

Control System. Contents Contents Chapter Topic Page Chapter- Chapter- Chapter-3 Chapter-4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of

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

School of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:

School of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by: Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See

More 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

PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada

PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop

More information

Pitch Rate CAS Design Project

Pitch Rate CAS Design Project Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability

More information

Introduction & Laplace Transforms Lectures 1 & 2

Introduction & Laplace Transforms Lectures 1 & 2 Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Control System Definition of a Control System Group of components that collectively

More information

Lecture 1 Root Locus

Lecture 1 Root Locus Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the

More information

RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS. Ryszard Gessing

RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS. Ryszard Gessing RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS Ryszard Gessing Politechnika Śl aska Instytut Automatyki, ul. Akademicka 16, 44-101 Gliwice, Poland, fax: +4832 372127, email: gessing@ia.gliwice.edu.pl

More information

Chapter 6 Steady-State Analysis of Continuous-Time Systems

Chapter 6 Steady-State Analysis of Continuous-Time Systems Chapter 6 Steady-State Analysis of Continuous-Time Systems 6.1 INTRODUCTION One of the objectives of a control systems engineer is to minimize the steady-state error of the closed-loop system response

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

2.010 Fall 2000 Solution of Homework Assignment 7

2.010 Fall 2000 Solution of Homework Assignment 7 . Fall Solution of Homework Assignment 7. Control of Hydraulic Servomechanism. We return to the Hydraulic Servomechanism of Problem in Homework Assignment 6 with additional data which permits quantitative

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

CBE507 LECTURE III Controller Design Using State-space Methods. Professor Dae Ryook Yang

CBE507 LECTURE III Controller Design Using State-space Methods. Professor Dae Ryook Yang CBE507 LECTURE III Controller Design Using State-space Methods Professor Dae Ryook Yang Fall 2013 Dept. of Chemical and Biological Engineering Korea University Korea University III -1 Overview States What

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

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

Chapter 7 - Solved Problems

Chapter 7 - Solved Problems Chapter 7 - Solved Problems Solved Problem 7.1. A continuous time system has transfer function G o (s) given by G o (s) = B o(s) A o (s) = 2 (s 1)(s + 2) = 2 s 2 + s 2 (1) Find a controller of minimal

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

1 (30 pts) Dominant Pole

1 (30 pts) Dominant Pole EECS C8/ME C34 Fall Problem Set 9 Solutions (3 pts) Dominant Pole For the following transfer function: Y (s) U(s) = (s + )(s + ) a) Give state space description of the system in parallel form (ẋ = Ax +

More information

Second Order and Higher Order Systems

Second Order and Higher Order Systems Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical second-order control system to a step input. In terms of damping ratio and natural

More information

Laboratory 11 Control Systems Laboratory ECE3557. State Feedback Controller for Position Control of a Flexible Joint

Laboratory 11 Control Systems Laboratory ECE3557. State Feedback Controller for Position Control of a Flexible Joint Laboratory 11 State Feedback Controller for Position Control of a Flexible Joint 11.1 Objective The objective of this laboratory is to design a full state feedback controller for endpoint position control

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

Lecture 9. Welcome back! Coming week labs: Today: Lab 16 System Identification (2 sessions)

Lecture 9. Welcome back! Coming week labs: Today: Lab 16 System Identification (2 sessions) 232 Welcome back! Coming week labs: Lecture 9 Lab 16 System Identification (2 sessions) Today: Review of Lab 15 System identification (ala ME4232) Time domain Frequency domain 1 Future Labs To develop

More information

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)

More information

Goals for today 2.004

Goals for today 2.004 Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation

More information

K c < K u K c = K u K c > K u step 4 Calculate and implement PID parameters using the the Ziegler-Nichols tuning tables: 30

K c < K u K c = K u K c > K u step 4 Calculate and implement PID parameters using the the Ziegler-Nichols tuning tables: 30 1.5 QUANTITIVE PID TUNING METHODS Tuning PID parameters is not a trivial task in general. Various tuning methods have been proposed for dierent model descriptions and performance criteria. 1.5.1 CONTINUOUS

More information

The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location.

The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location. Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location. Mapping. If we take a complex number

More information

ECE 380: Control Systems

ECE 380: Control Systems ECE 380: Control Systems Course Notes: Winter 2014 Prof. Shreyas Sundaram Department of Electrical and Computer Engineering University of Waterloo ii Acknowledgments Parts of these course notes are loosely

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

Power System Control

Power System Control Power System Control Basic Control Engineering Prof. Wonhee Kim School of Energy Systems Engineering, Chung-Ang University 2 Contents Why feedback? System Modeling in Frequency Domain System Modeling in

More information

Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc.

Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc. Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc. Electrical Engineering Department University of Indonesia 2 Steady State Error How well can

More information

10 Transfer Matrix Models

10 Transfer Matrix Models MIT EECS 6.241 (FALL 26) LECTURE NOTES BY A. MEGRETSKI 1 Transfer Matrix Models So far, transfer matrices were introduced for finite order state space LTI models, in which case they serve as an important

More information

Tradeoffs and Limits of Performance

Tradeoffs and Limits of Performance Chapter 9 Tradeoffs and Limits of Performance 9. Introduction Fundamental limits of feedback systems will be investigated in this chapter. We begin in Section 9.2 by discussing the basic feedback loop

More information