# Systems Analysis and Control

Size: px
Start display at page:

Transcription

1 Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop

2 Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect on Stability Stability Effects Gain Margin Phase Margin Bandwidth Estimating Closed-Loop Performance using Open-Loop Data Damping Ratio Settling Time Rise Time M. Peet Lecture 21: Control Systems 2 / 31

3 Review Recall: Frequency Response Input: u(t) = M sin(ωt + φ) Output: Magnitude and Phase Shift Linear Simulation Results y(t) = G(ıω) M sin(ωt + φ + G(ıω)) Amplitude Time (sec) Frequency Response to sin ωt is given by G(ıω) M. Peet Lecture 21: Control Systems 3 / 31

4 Review Recall: Bode Plot Definition 1. The Bode Plot is a pair of log-log and semi-log plots: 1. Magnitude Plot: 20 log 10 G(ıω) vs. log 10 ω 2. Phase Plot: G(ıω) vs. log 10 ω Bite-Size Chucks: G(ıω) = i G i (ıω) M. Peet Lecture 21: Control Systems 4 / 31

5 Complex Poles and Zeros We left off with Complex Poles: 1 G(s) = ( ( ) ) 2 s ω n + 2ζ s ω n + 1 M. Peet Lecture 21: Control Systems 5 / 31

6 Closing The Loop Now we examine the effect of closing the loop on the Frequency Response. Use simple Unity Feedback (K = 1). Closed-Loop Transfer Function: G cl (ıω) = G(ıω) 1 + G(ıω) We are most concerned with magnitude: G cl (ıω) = G(ıω) 1 + G(ıω) - u(s) + k G(s) Figure : Unity Feedback y(s) M. Peet Lecture 21: Control Systems 6 / 31

7 Closing The Loop On the Bode Plot 20 log G cl (ıω) = 20 log G(ıω) 20 log 1 + G(ıω) Which is the combination of The original bode plot The new factor log 1 + G(ıω) Bode Diagram We are most concerned with the effect of the new term 20 log 1 + G(ıω) Specifically, as 1 + G(ıω) 0 lim 20 log 1 + G(ıω) = 1+G(ıω) 0 An unstable mode! Magnitude (db) Phase (deg) Frequency (rad/sec) Figure : Open Loop: Blue, CL: Green M. Peet Lecture 21: Control Systems 7 / 31

8 Closing The Loop Stability Margin Instability occurs when 1 + G(ıω) = 0 For this to happen, we need: G(ıω) = 1 G(ıω) = 180 Stability Margins measure how far we are from the point ( G = 1, G = 180 ). Definition 2. The Gain Crossover Frequency, ω gc is the frequency at which G(ıω c ) = 1. This is the danger point: If G(ıω c ) = 180, we are unstable M. Peet Lecture 21: Control Systems 8 / 31

9 Closing The Loop Phase Margin Definition 3. The Phase Margin, Φ M is the phase relative to 180 when G = 1. Φ M = 180 G(iω gc ) ω gc is also known as the phase-margin frequency, ω ΦM M. Peet Lecture 21: Control Systems 9 / 31

10 Closing The Loop Gain Margin Definition 4. The Phase Crossover Frequency, ω pc is the frequency (frequencies) at which G(ıω pc ) = 180. Definition 5. The Gain Margin, G M is the gain relative to 0dB when G = 180. G M = 20 log G(ıω pc ) G M is the gain (in db) which will destabilize the system in closed loop. ω pc is also known as the gain-margin frequency, ω GM M. Peet Lecture 21: Control Systems 10 / 31

11 Closing The Loop Stability Margins Gain and Phase Margin tell how stable the system would be in Closed Loop. These quantities can be read from the Open-Loop Data. M. Peet Lecture 21: Control Systems 11 / 31

12 Closing The Loop Stability Margins: Suspension System Bode Diagram Gm = Inf db (at Inf rad/sec), Pm = 42.1 deg (at rad/sec) Magnitude (db) Phase (deg) Frequency (rad/sec) M. Peet Lecture 21: Control Systems 12 / 31

13 Closing The Loop Stability Margins Φ M = 35 G M = 10dB M. Peet Lecture 21: Control Systems 13 / 31

14 Closing The Loop Stability Margins Note that sometimes the margins are undefined When there is no crossover at 0dB When there is no crossover at 180 M. Peet Lecture 21: Control Systems 14 / 31

15 Transient Response Closing the Loop Question: What happens when we Close the Loop? We want Performance Specs! We only have open-loop data. Φ M and G M can help us. Unity Feedback: Step Response u(s) + k G(s) y(s) We want: Damping Ratio Settling Time 0 Peak Time Amplitude Time (sec) M. Peet Lecture 21: Control Systems 15 / 31

16 Transient Response Quadratic Approximation Assume the closed loop system is the quadratic Then the open-loop system must be ω 2 n G cl = s 2 + 2ζω n s + ωn 2 G ol = Assume that our open-loop system is G ol Use Φ M to solve for ζ ω 2 n s 2 + 2ζω n s Set 20 log G cl = 3dB to solve for ω n M. Peet Lecture 21: Control Systems 16 / 31

17 Transient Response Damping Ratio The Quadratic Approximation gives the Closed-Loop Damping Ratio as Φ M = tan 1 2ζ 2ζ 2 + 4ζ Φ M is from the Open-Loop Data! A Handy approximation is ζ = Φ M 100 Only valid out to ζ =.7. Given Φ M, we find closed-loop ζ. M. Peet Lecture 21: Control Systems 17 / 31

18 Transient Response Bandwidth and Natural Frequency We find closed-loop ζ from Phase margin. We can find closed-loop Natural Frequency ω n from the closed-loop Bandwidth. Definition 6. The Bandwidth, ω BW is the frequency at gain 20 log G(ıω BW ) = 20 log G(0) 3dB. Closely related to crossover frequency. The Bandwidth measures the range of frequencies in the output. For 2nd order, Bandwidth is related to natural frequency by ω BW = ω n (1 2ζ 2 ) + 4ζ 4 4ζ M. Peet Lecture 21: Control Systems 18 / 31

19 Transient Response Finding Closed-Loop Bandwidth from Open-Loop Data Question: How to find closed-loop bandwidth? Finding the closed-loop bandwidth from open-loop data is tricky. Have to find the frequency when the Bode plot intersects this curve. Heuristic: Check the frequency at 6dB and see if phase is = 180. M. Peet Lecture 21: Control Systems 19 / 31

20 Finding Closed-Loop Bandwidth from Open-Loop Data Example Bode Diagram Magnitude (db) Phase (deg) Frequency (rad/sec) At phase 135, 5dB, we get closed loop ω BW = 1. M. Peet Lecture 21: Control Systems 20 / 31

21 Transient Response Bandwidth and Settling Time We can use the expression T s = 4 ζω n to get ω BW = 4 (1 2ζ T s ζ 2 ) + 4ζ 4 4ζ Given closed-loop ζ and ω BW, we can find T s. M. Peet Lecture 21: Control Systems 21 / 31

22 Transient Response Bandwidth and Peak Time We can use the expression T p = π ω n 1 ζ 2 to get π ω BW = (1 2ζ 2 ) + 4ζ 4 4ζ T p 1 ζ 2 Given closed-loop ζ and ω BW, we can find T p. M. Peet Lecture 21: Control Systems 22 / 31

23 Transient Response Bandwidth and Rise Time Using an expression for T r, we get a relationship between ω BW T r and ζ. Given closed-loop ζ and ω BW, we can find T r. M. Peet Lecture 21: Control Systems 23 / 31

24 Transient Response Example Question: Using Frequency Response Data, find T r, T s, T p after unity feedback. First Step: Find the phase Margin. Frequency at 0dB is ω gc = 2 G(2) = 145 Φ M = = 35 M. Peet Lecture 21: Control Systems 24 / 31

25 Transient Response Example Step 2: Closed-Loop Damping Ratio ζ = Φ M 100 =.35 Step 3: Closed-Loop Bandwidth Intersect at = ( G = 6dB, G = 170 ) Frequency at intersection is ω BW = 3.7 M. Peet Lecture 21: Control Systems 25 / 31

26 Transient Response Example Step 4: Settling Time and Peak Time ω BW = 3.7, ζ =.35 ω BW T s = 20 implies T s = 5.4s ω BW T p = 4.9 implies T p = 1.32 M. Peet Lecture 21: Control Systems 26 / 31

27 Transient Response Example Step 5: Rise Time ω BW = 3.7, ζ =.35 ω BW T r = 1.98 implies T r =.535 M. Peet Lecture 21: Control Systems 27 / 31

28 Transient Response Example Step 6: Experimental Validation. Use the plant G(s) = 50 s(s + 3)(s + 6) We find T p = 1.6s predicted 1.32 T r =.7s predicted.535 T s = 4s Predicted Figure : Step Response M. Peet Lecture 21: Control Systems 28 / 31

29 Steady-State Error Finally, we want steady-state error. Steady-State Step response is lim G(s) = lim G(ıω) s 0 ω 0 Steady-state response is the Low-Frequency Gain, G(0). Close The Loop to get steady-state error e ss = G(0) M. Peet Lecture 21: Control Systems 29 / 31

30 Steady-State Error Example A Lag Compensator lim 20 log G(ıω) = 20dB ω 0 So lim ω 0 G(ıω) = 10. Steady-State Error: e ss = G(0) = 1 11 =.091 Phase (deg) Magnitude (db) Bode Diagram Frequency (rad/sec) M. Peet Lecture 21: Control Systems 30 / 31

31 Summary What have we learned today? Closing the Loop Effect on Bode Plot Effect on Stability Stability Effects Gain Margin Phase Margin Bandwidth Estimating Closed-Loop Performance using Open-Loop Data Damping Ratio Settling Time Rise Time Next Lecture: Compensation in the Frequency Domain M. Peet Lecture 21: Control Systems 31 / 31

### Systems Analysis and Control

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

### 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

### 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

### 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

### 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

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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.

### 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

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance

### 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

### 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

### 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

### 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

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 13: Root Locus Continued Overview In this Lecture, you will learn: Review Definition of Root Locus Points on the Real Axis

### 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

### 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:

### Digital Control Systems

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

### 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

### 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)

### CDS 101/110a: Lecture 8-1 Frequency Domain Design

CDS 11/11a: Lecture 8-1 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve

### 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

### 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

### Dynamic Compensation using root locus method

CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the

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

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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,

### 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

### 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

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 6: Generalized and Controller Design Overview In this Lecture, you will learn: Generalized? What about changing OTHER parameters

### 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)

### Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types

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

### 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

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

6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)

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

Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus

### (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.

### Discrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture

Discrete Systems Mark Cannon Hilary Term 22 - Lecture 4 Step response and pole locations 4 - Review Definition of -transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},

### 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

### 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

### Review: transient and steady-state response; DC gain and the FVT Today s topic: system-modeling diagrams; prototype 2nd-order system

Plan of the Lecture Review: transient and steady-state response; DC gain and the FVT Today s topic: system-modeling diagrams; prototype 2nd-order system Plan of the Lecture Review: transient and steady-state

### 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)

### = 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

### 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

### 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

### 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,

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

EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller

. 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

### 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

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

126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closed-loop system

### Return Difference Function and Closed-Loop Roots Single-Input/Single-Output Control Systems

Spectral Properties of Linear- Quadratic Regulators Robert Stengel Optimal Control and Estimation MAE 546 Princeton University, 2018! Stability margins of single-input/singleoutput (SISO) systems! Characterizations

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12: Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain

### MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions

MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noise-cancelling headphone system. 1a. Based on the low-pass filter given, design a high-pass filter,

### Response to a pure sinusoid

Harvard University Division of Engineering and Applied Sciences ES 145/215 - INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid

### 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

### 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

### 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

### 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

### 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.

### 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

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

Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]

### Procedure for sketching bode plots (mentioned on Oct 5 th notes, Pg. 20)

Procedure for sketching bode plots (mentioned on Oct 5 th notes, Pg. 20) 1. Rewrite the transfer function in proper p form. 2. Separate the transfer function into its constituent parts. 3. Draw the Bode

### 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

### 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

### 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

### 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..

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 15: Root Locus Part 4 Overview In this Lecture, you will learn: Which Poles go to Zeroes? Arrival Angles Picking Points? Calculating

### 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

### Automatic Control (TSRT15): Lecture 7

Automatic Control (TSRT15): Lecture 7 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13-282226 Office: B-house extrance 25-27 Outline 2 Feedforward

### 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

### 9. Two-Degrees-of-Freedom Design

9. Two-Degrees-of-Freedom Design In some feedback schemes we have additional degrees-offreedom outside the feedback path. For example, feed forwarding known disturbance signals or reference signals. In

### Compensator Design to Improve Transient Performance Using Root Locus

1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning

### 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

### 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.

### 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.

### 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

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

Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral

### 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

### 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

### 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

### Control of Manufacturing Processes

Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection

### Control Systems Design

ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:

### ECE317 : Feedback and Control

ECE317 : Feedback and Control Lecture : Routh-Hurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling

### 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