6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0."

Transcription

1 6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i) To sketch the z-domain root locus by hand, we apply the root locus rules. The plots can more easily be obtained using the MATLAB command rlocus. For Gz K () z 4., the closed-loop characteristic equation is K + z.4 At z, we obtain the critical gain, K cr.4 (see also the root locus plot) Root Locus.8.6 Imaginary Axis System: g Gain:.4 Pole: Damping:.755 Overshoot (%): 99.8 Frequency (rad/sec): (ii) For Gz Real Axis Figure 6- Root locus for Problem 6.(i) K () ( z+ 9. )( z 9. ), the closed-loop characteristic equation isk + z.8 On the unit circle, z and we obtain K cr.8

2 Root Locus.8 System: g Gain:.8 Pole: -8.3e-7 + i Damping: -.79 Overshoot (%): Frequency (rad/sec): 57 Imaginary Axis Real Axis Figure 6- Root locus for Problem 6.(ii) Kz (iii) For G() z, the closed-loop characteristic equation is ( z. )( z ) K z + z. z +.. On the unit circle, z, K and the critical gain is K cr.4 Root Locus Imaginary Axis System: g Gain:.4 Pole: - Damping: -.94 Overshoot (%): Frequency (rad/sec): Real Axis Figure 6-3 Root locus for Problem 6.(iii) (iv) For G() z Kz ( + 9. ) ( z. )( z 8. ), the closed-loop characteristic equation is

3 z z K z +.9 K. On the unit circle, z.9 K +.6 and we obtain K cr Root Locus Imaginary Axis.5 System: g Gain:.934 Pole: i Damping:.39 Overshoot (%): 99 Frequency (rad/sec): Real Axis Figure 6-4 Root locus for Problem 6.(iv)

4 6.5 Design proportional controllers for the systems of Problem 6. to meet the following specifications where possible. If the design specification cannot be met, explain why and suggest a more appropriate controller. (a) A damping ratio of.7. (b) A steady-state error of % due to a unit step. (c) A steady-state error of % due to a unit ramp. K (i) Gz () z 4. (a) A damping ratio of.7 This is a first order system and does not have an oscillatory response. Hence, the design specification cannot be met with proportional control.

5 A PI controller would give a system with the desired damping ratio but it is probably preferable to use proportional control to obtain a good time response with the appropriate time constant unless the steady-state error is required to be zero. (b) A steady-state error of % due to a unit step The position error constant is K p K G( ) 9.4 e( )% Hence, the gain for % error due to step is K 5.4. Unfortunately, the critical gain for the system is.4 and a gain of 5.4 would make the system unstable and the design specifications cannot be met with proportional control. Using a PD controller would allow us to increase the gain without causing instability. Alternatively, we could use a PI controller to reduce the error due to a step to zero, which would meet the desired specifications. (c) A steady-state error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steady-state error due to ramp. (ii) K G ( z) ( z +.9)( z.9) (a) A damping ratio of.7 The closed-loop characteristic polynomial is ζωnt z.8+ K z cos ω T e z + e ζωnt ( ) d For a gain K>.8, we have imaginary poles so that the angle of the pole is ω T ω T ζ ω T.5 π / d n The square magnitude of the poles is ζω K.8 e n T e.4599 The required gain is K n Note that we only need the value of the product ω n T to solve the problem rather than each of the two values. Hence, the result can be verified with MATLAB using a unity sampling period and the command rlocus.

6 Root Locus Imaginary Axis System: g Gain:.856 Pole: -.54e-7 +.4i Damping:.7 Overshoot (%): 4.59 Frequency.7 (rad/sec): Real Axis Figure 6.5 Root locus for Problem 6.5 (ii) and the gain for a damping ratio of.7. (b) A steady-state error of % due to a unit step. The steady-state error is given by e ( )% + K p The error constant for the system is K p 9 K G() ( +.9)(.9) We solve for the gain K The specification can be met because the gain is less than the critical gain. (c) A steady-state error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steady-state error due to ramp. (iii) K z G ( z) ( z.)( z ) (a) A damping ratio of.7 The closed-loop characteristic polynomial is

7 z + ζωnt ζωnt ( K.) z +. z cos( ω T ) e z + e Equating coefficients, we have e ζωn T ω T n cos..447 ln(. ) ( ω T ) cos ω T ζ cos(.496.5). 685 d K. cos n ( ω T ) d e ζω n T d K We obtain approximately the same answer using the MALAB command rlocus. Root Locus Imaginary Axis System: g Gain:.595 Pole: i Damping:.698 Overshoot (%): 4.66 Frequency (rad/sec): Real Axis Figure 6.6 Root locus for Problem 6.5 (iii) and the gain for a damping ratio of.7. (b) A steady-state error of % due to a unit step. The steady-state error is zero since the system is type I and the specification can be met for any stable gain. The gain can be selected as in Part (a) to obtain a satisfactory transient response. (c) A steady-state error of % due to a unit ramp. The system is type and can track a ramp input. The velocity error constant is K v ( z ) G( z) T K 8T z T K (.) The sampling period is typically sufficiently small to result in a small gain that is less than the critical value and the specifications can be met.

8 Kz ( + 9. ) (iv) Gz () ( z. )( z 8. ) (a) A damping ratio of.7 cos The closed-loop characteristic polynomial is z + K z K z cos ω dt Equating coefficients, we have e ζω ω T ln n n T ζωnt ζωnt ( ) ( ) e z + e.6 +.9K (.6 +.9K ). 4 ( ω T ) cos ω nt ζ cos ln(.6.9k ) d + (.5.4) ( ln.6 +.9K.5.4) (.6. K ) ζω T ( ω T ) e cos ( ) K cos + d n 9 This is a nonlinear equation in K and must be solved numerically. By trial and error, we obtain gain value K.434. Approximately, the same value can be obtained using the MATLAB command rlocus..5 Root Locus.5 System: g Gain:.4 Pole: i Damping:.699 Overshoot (%): 4.64 Frequency (rad/sec): Imaginary Axis Real Axis Figure 6.7 Root locus for Problem 6.5 (iv) and the gain for a damping ratio of.7.

9 (b) A steady-state error of % due to a unit step. The steady-state error is given by e ( )% + K p The error constant for the system is K p 9 K( +.9) G() (.)(.8) We solve for the gain K The specification can be met because the gain is less than the critical gain. (c) A steady-state error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steady-state error due to ramp.

10 6.8 Design a digital PID controller (with T.) for the plant G( s) e s + 5s by applying the Ziegler-Nichols tuning rules of Table 5.. The Ziegler-Nichols tuning rules shown in Table 5. can be applied by considering K, τ, L5. We obtain K p.4, T i, T d.5. Thus, by applying (6.3) we have. z + z.4z C z) z. z + ( 4z z

11 6. Use direct control design for the system of Problem 5.7 (with T.), namely, design a controller for the transfer function G ( s) ( s + )( s + 5) to obtain (i) zero steady-state error due to step, (ii) a settling time of less than s, and (iii) an undamped natural frequency of 5 rad/s. Obtain the discretized and the analog output. Then, apply the designed controller to the system G ( s) ( s + )( s + 5)(.s + ) and obtain the discretized and the analog output in order to verify the robustness of the control system. First, we find the discretized process transfer function: ( z ) G( s) G ZAS ( z) Z s.47( z +.889) ( z.948)( z.665) Then, the desired analog characteristic polynomial is s ζω + ωn ns T s 4 ζω n where, according to the specifications, ω n 5 and, which implies that ζ.4. Thus, by taking into account that a zero steady-state error is required, the desired closed-loop transfer function is

12 G cl ( s) s 5 + 4s + 5 Then, the desired closed-loop transfer function is obtained by using z e st, namely, G cl ( z).9 z By applying (6.43) we have z +.469z C ( z).576( z.948)( z.665)( z + ) ( z )( z +.889)( z.5694) The obtained discretized and analog closed-loop system output are shown in Figure 6-6 and Figure 6-7 while the corresponding control variable is plotted in Figure 6-8. If the same controller is applied to the system G ( s) ( s + )( s + 5)(.s + ) the process output obtained is that in Figure 6-9. It can be seen that the additional lag causes a more significant overshoot and an increment of the settling time..4. process output time [s] Figure 6-6 Discretized process output for Problem 6.

13 6.3 Design a deadbeat controller for the system of Problem 5.7 to obtain perfect tracking of a unit step in minimum finite time. Obtain the analog output for the system and compare your design to that obtained in Problem 5.7. Apply then the controller to the process G ( s) ( s + )( s + 5)(.s + ) in order to verify the robustness of the control system. We consider the same discretized process transfer function of Problem 6., but in this case we set G cl ( z) z. Thus, by applying (6.43) we have C ( z) ( z.948)( z.665) ( z )( z +.889) The corresponding discretized and analog output are shown in Figure 6-3, the control variable in Figure 6-3. We observe that the deadbeat controller causes wide intersampling oscillations and requires a much higher control effort. It results in an increased overshoot without significantly decreasing the settling time. When the system G ( s) ( s + )( s + 5)(.s + ) is considered, the process output obtained is plotted in Error! Reference source not found...5 process output time [s] Figure 6-3 Discretized and analog output for Problem 6.3

14 3 control variable time [s] Figure 6-3 Control variable for Problem 6.3. process output time [s] Figure 6-3 Analog output for Problem 6.3 with uncertainty

15 6.8 To examine the effect of the sampling period on the relative stability and transient response of a digital control system, consider the system G ( s) ( s + )( s + 5) a) Obtain the transfer function of the system, the root locus and the critical gain for T.,.5,. s. Using the MATLAB commands zpk and cd, we obtain:

16 >> gzpk([],[-,-5],) Zero/pole/gain: (s+) (s+5) >> gdcd(g,.) Zero/pole/gain: 4.93e-5 (z+.98) (z-.99) (z-.95) Sampling time:. >> gdcd(g,.5) Zero/pole/gain:.37 (z+.949) (z-.95) (z-.7788) Sampling time:.5 >> gd3cd(g,.) Zero/pole/gain:.47 (z+.889) (z-.948) (z-.665) Sampling time:. Using the command >> [Gm,PM]margin(gd) We obtain the gain margin, which is the critical gain value. The values obtain are summarized in the following table. Sampling Period Zero Pole Pole K cr e b) Obtain the step response for each system at a gain of 5.

17 .4 Step Response. Amplitude Time (sec) Figure 6-4 Step response for sampling period T. (blue), T.5 (green), and T. s (red). c) Discuss the effect of the sampling period on the transient response and relative stability of the system based on your results from (a) and (b). The transfer functions for the different sampling period with generally faster poles for slower sampling. Faster poles are poles that are closer to the origin. The systems with slower sampling does not monitor the process as closely as ones with faster sampling. As a result, the step response of the system is more oscillatory.-

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

(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

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

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

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

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

Introduction to Feedback Control

Introduction to Feedback Control Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System

More information

Chapter 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

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

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

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

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

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

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

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

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

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

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

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

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

2.010 Fall 2000 Solution of Homework Assignment 8

2.010 Fall 2000 Solution of Homework Assignment 8 2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)

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

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

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

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

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

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

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

9/9/2011 Classical Control 1

9/9/2011 Classical Control 1 MM11 Root Locus Design Method Reading material: FC pp.270-328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),

More information

Department of Mechanical Engineering

Department of Mechanical Engineering Department of Mechanical Engineering 2.14 ANALYSIS AND DESIGN OF FEEDBACK CONTROL SYSTEMS Fall Term 23 Problem Set 5: Solutions Problem 1: Nise, Ch. 6, Problem 2. Notice that there are sign changes in

More information

Homework 11 Solution - AME 30315, Spring 2015

Homework 11 Solution - AME 30315, Spring 2015 1 Homework 11 Solution - AME 30315, Spring 2015 Problem 1 [10/10 pts] R + - K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closed-loop pole locations as the parameter k is varied. Θpsq Ipsq k ωn

More information

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

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

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

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

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

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

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

Open Loop Tuning Rules

Open Loop Tuning Rules Open Loop Tuning Rules Based on approximate process models Process Reaction Curve: The process reaction curve is an approximate model of the process, assuming the process behaves as a first order plus

More information

Additional Closed-Loop Frequency Response Material (Second edition, Chapter 14)

Additional Closed-Loop Frequency Response Material (Second edition, Chapter 14) Appendix J Additional Closed-Loop Frequency Response Material (Second edition, Chapter 4) APPENDIX CONTENTS J. Closed-Loop Behavior J.2 Bode Stability Criterion J.3 Nyquist Stability Criterion J.4 Gain

More information

1 Chapter 9: Design via Root Locus

1 Chapter 9: Design via Root Locus 1 Figure 9.1 a. Sample root locus, showing possible design point via gain adjustment (A) and desired design point that cannot be met via simple gain adjustment (B); b. responses from poles at A and B 2

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

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

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

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

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

Mechanical Systems Part A: State-Space Systems Lecture AL12

Mechanical Systems Part A: State-Space Systems Lecture AL12 AL: 436-433 Mechanical Systems Part A: State-Space Systems Lecture AL Case study Case study AL: Design of a satellite attitude control system see Franklin, Powell & Emami-Naeini, Ch. 9. Requirements: accurate

More information

Vibrations: Second Order Systems with One Degree of Freedom, Free Response

Vibrations: Second Order Systems with One Degree of Freedom, Free Response Single Degree of Freedom System 1.003J/1.053J Dynamics and Control I, Spring 007 Professor Thomas Peacock 5//007 Lecture 0 Vibrations: Second Order Systems with One Degree of Freedom, Free Response Single

More information

Design via Root Locus

Design via Root Locus Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steady-state error (Sections

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

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequency-domain analysis and control design (15 pt) Given is a

More information

Tuning Method of PI Controller with Desired Damping Coefficient for a First-order Lag Plus Deadtime System

Tuning Method of PI Controller with Desired Damping Coefficient for a First-order Lag Plus Deadtime System PID' Brescia (Italy), March 8-0, 0 FrA. Tuning Method of PI Controller with Desired Damping Coefficient for a First-order Lag Plus Deadtime System Yuji Yamakawa*. Yohei Okada** Takanori Yamazaki***. Shigeru

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

12.7 Steady State Error

12.7 Steady State Error Lecture Notes on Control Systems/D. Ghose/01 106 1.7 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

More information

Chemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University

Chemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this

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

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

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators. Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8- am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total

More information

Robust Performance Example #1

Robust Performance Example #1 Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants

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

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

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

University of Science and Technology, Sudan Department of Chemical Engineering.

University of Science and Technology, Sudan Department of Chemical Engineering. ISO 91:28 Certified Volume 3, Issue 6, November 214 Design and Decoupling of Control System for a Continuous Stirred Tank Reactor (CSTR) Georgeous, N.B *1 and Gasmalseed, G.A, Abdalla, B.K (1-2) University

More information

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2

More information

Control System Design

Control System Design ELEC4410 Control System Design Lecture 19: Feedback from Estimated States and Discrete-Time Control Design Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science

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

PID Control. Objectives

PID Control. Objectives PID Control Objectives The objective of this lab is to study basic design issues for proportional-integral-derivative control laws. Emphasis is placed on transient responses and steady-state errors. The

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 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 Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering

More information

Problem 1 (Analysis of a Feedback System - Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function 1.

Problem 1 (Analysis of a Feedback System - Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function 1. 1 EEE480 Final Exam, Spring 2016 A.A. Rodriguez Rules: Calculators permitted, One 8.5 11 sheet, closed notes/books, open minds GWC 352, 965-3712 Problem 1 (Analysis of a Feedback System - Bode, Root Locus,

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

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

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

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

Feedback Control part 2

Feedback Control part 2 Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open- and closed-loop control Everything before chapter 7 are open-loop systems Transient response Design criteria Translate criteria

More information

Chapter 13 Digital Control

Chapter 13 Digital Control Chapter 13 Digital Control Chapter 12 was concerned with building models for systems acting under digital control. We next turn to the question of control itself. Topics to be covered include: why one

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

DIGITAL CONTROL OF POWER CONVERTERS. 3 Digital controller design

DIGITAL CONTROL OF POWER CONVERTERS. 3 Digital controller design DIGITAL CONTROL OF POWER CONVERTERS 3 Digital controller design Frequency response of discrete systems H(z) Properties: z e j T s 1 DC Gain z=1 H(1)=DC 2 Periodic nature j Ts z e jt e s cos( jt ) j sin(

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

Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis

Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis J. Alexandre dit Sandretto, A. Chapoutot and O. Mullier U2IS, ENSTA ParisTech SYNCOP April 11, 2015 Closed-loop

More information

Design of a Heading Autopilot for Mariner Class Ship with Wave Filtering Based on Passive Observer

Design of a Heading Autopilot for Mariner Class Ship with Wave Filtering Based on Passive Observer Design of a Heading Autopilot for Mariner Class Ship with Wave Filtering Based on Passive Observer 1 Mridul Pande, K K Mangrulkar 1, Aerospace Engg Dept DIAT (DU), Pune Email: 1 mridul_pande000@yahoo.com

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

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

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

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

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

Simulation Study on Pressure Control using Nonlinear Input/Output Linearization Method and Classical PID Approach

Simulation Study on Pressure Control using Nonlinear Input/Output Linearization Method and Classical PID Approach Simulation Study on Pressure Control using Nonlinear Input/Output Linearization Method and Classical PID Approach Ufuk Bakirdogen*, Matthias Liermann** *Institute for Fluid Power Drives and Controls (IFAS),

More information

PID Tuning of Plants With Time Delay Using Root Locus

PID Tuning of Plants With Time Delay Using Root Locus San Jose State University SJSU ScholarWorks Master's Theses Master's Theses and Graduate Research Summer 2011 PID Tuning of Plants With Time Delay Using Root Locus Greg Baker San Jose State University

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

The output voltage is given by,

The output voltage is given by, 71 The output voltage is given by, = (3.1) The inductor and capacitor values of the Boost converter are derived by having the same assumption as that of the Buck converter. Now the critical value of the

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. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid

More information

Poles, Zeros and System Response

Poles, Zeros and System Response Time Response After the engineer obtains a mathematical representation of a subsystem, the subsystem is analyzed for its transient and steady state responses to see if these characteristics yield the desired

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

EEL2216 Control Theory CT1: PID Controller Design

EEL2216 Control Theory CT1: PID Controller Design EEL6 Control Theory CT: PID Controller Design. Objectives (i) To design proportional-integral-derivative (PID) controller for closed loop control. (ii) To evaluate the performance of different controllers

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

Topic # Feedback Control Systems

Topic # Feedback Control Systems Topic #14 16.31 Feedback Control Systems State-Space Systems Full-state Feedback Control How do we change the poles of the state-space system? Or, even if we can change the pole locations. Where do we

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