Chapter 12. Feedback Control Characteristics of Feedback Systems

Similar documents
12.7 Steady State Error

Controller Design using Root Locus

Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system

Time Response of Systems

Systems Analysis and Control

AN INTRODUCTION TO THE CONTROL THEORY

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

Systems Analysis and Control

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.

Homework 7 - Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013

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

ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II

Systems Analysis and Control

Time Response Analysis (Part II)

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design

EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =

Digital Control Systems State Feedback Control

Lecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =

Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 5. An Introduction to Feedback Control Systems

Control of Manufacturing Processes

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

Outline. Classical Control. Lecture 1

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

Some special cases

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

Control of Manufacturing Processes

Dr Ian R. Manchester

Systems Analysis and Control

Analysis and Design of Control Systems in the Time Domain

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

EE3CL4: Introduction to Linear Control Systems

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

Homework Assignment 3

Dynamic Behavior. Chapter 5

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

Control Systems Design

Control Systems I Lecture 10: System Specifications

EE402 - Discrete Time Systems Spring Lecture 10

Example on Root Locus Sketching and Control Design

PID Control. Objectives

MAS107 Control Theory Exam Solutions 2008

Meeting Design Specs using Root Locus

STABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse

Problem Value Score Total 100/105

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

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Lecture 25: Tue Nov 27, 2018

Topic # Feedback Control Systems

Unit 8: Part 2: PD, PID, and Feedback Compensation

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:

Automatic Control (TSRT15): Lecture 7

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

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

Automatic Control (TSRT15): Lecture 4

INTRODUCTION TO DIGITAL CONTROL

R10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1

Outline. Classical Control. Lecture 5

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.

DO NOT DO HOMEWORK UNTIL IT IS ASSIGNED. THE ASSIGNMENTS MAY CHANGE UNTIL ANNOUNCED.

ECE317 : Feedback and Control

Introduction to Feedback Control

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.

AMME3500: System Dynamics & Control

Software Engineering 3DX3. Slides 8: Root Locus Techniques

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27

APPLICATIONS FOR ROBOTICS

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus

Chapter 5 HW Solution

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)

Step input, ramp input, parabolic input and impulse input signals. 2. What is the initial slope of a step response of a first order system?

DC-motor PID control

CHBE320 LECTURE XI CONTROLLER DESIGN AND PID CONTOLLER TUNING. Professor Dae Ryook Yang

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

LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593

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

Design of a Lead Compensator

BASIC PROPERTIES OF FEEDBACK

Controls Problems for Qualifying Exam - Spring 2014

Introduction to Process Control

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.

FEEDBACK CONTROL SYSTEMS

Lec 6: State Feedback, Controllability, Integral Action

If you need more room, use the backs of the pages and indicate that you have done so.

PID controllers. Laith Batarseh. PID controllers

Performance of Feedback Control Systems

Systems Analysis and Control

C(s) R(s) 1 C(s) C(s) C(s) = s - T. Ts + 1 = 1 s - 1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BSC (HONS) MECHATRONICS TOP-UP SEMESTER 1 EXAMINATION 2017/2018 ADVANCED MECHATRONIC SYSTEMS

Dynamic Compensation using root locus method

Robust Performance Example #1

Proportional plus Integral (PI) Controller

EEE 184: Introduction to feedback systems

Automatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year


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

Chapter 7 Control. Part Classical Control. Mobile Robotics - Prof Alonzo Kelly, CMU RI

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

Transcription:

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 system (a system with feedbac). Figure 1.1: Open loop and closed-loop system The H bloc is the transfer function that represents the dynamics of the feedbac loop. 1.1 Characteristics of Feedbac Systems Advantages 1. Faster time response.. Better stability. 97

Lecture Notes on Control Systems/D. Ghose/01 98 3. Less sensitive to noise. 4. Less sensitive to system parameter variations. Disadvantages 1. May have a tendency to oscillate.. Cost becomes higher. 1. Objectives of Feedbac Control 1. Speed (Rise time). Accuracy (settling time and steady-state error) 3. Stability (Overshoot) 4. Robustness (will be treated in frequency response methods) 1.3 Types of Feedbac Control 1. Proportional control (P-control). Proportional-Integral or Integral control (PI-control) 3. Proportional-Integral-Derivative control (PID-control) 1.4 What Motivates Feedbac Control? Let us reason this out through an example. Consider the first order transfer function of the linearized model of a missile autopilot. G(s) 1+s where is the autopilot time constant. Now, let us apply a unit step input R(s) 1/s.

Lecture Notes on Control Systems/D. Ghose/01 99 Figure 1.: A first order system Then the output is, Y 1 (s) G(s)R(s) 1+s 1 s 1 s +1/ 1 s ( ) 1 s 1 s +1/ y 1 (t) ( 1 e t/ ) The response is fairly sluggish. What can we do to mae the response faster without actually changing the autopilot? Figure 1.3: What happens when we double the input? Let us double the input, that is, R(s) /s. Then, Y (s) G(s)R(s) 1+s s

Lecture Notes on Control Systems/D. Ghose/01 100 1 s +1/ 1 s ( ) 1 s 1 s +1/ y (t) ( 1 e t/ ) There is actually no difference between the two time responses. Both have the same sluggish response as they have the same type of damping. But, loo at the figure above. You can see that y (t) attains the level much earlier than y 1 (t) does. This gives rise to the following idea: Why don t we apply a high input initially so that the system responds quicly and then decrease the input later? In other words, instead of driving G(s) with a step input, drive it with an input that is high initially and then drops down gradually. One way to do this would be to use the difference between the input step signal and the output and then drive the plant G(s) with a high gain on this difference. This difference is usually nown as the error signal. Note that initially this error is high (actually, it is the same as the reference input) and then it gradually decreases with time as the output attains values close to the reference input. The following bloc diagram achieves this and is called the feedbac configuration. It is also called the closed-loop configuration. The closed-loop transfer function is, Figure 1.4: A closed loop configuration G c (s) K(s)G(s) 1+K(s)G(s) Let us examine the performance of this configuration with our example. let, K(s) 1, which is a pure DC gain.

Lecture Notes on Control Systems/D. Ghose/01 101 Since, we have G(s) 1+s G c (s) 1 1+s 1+ 1 1+s Now, apply a unit step input R(s) 1/s. 1 1+s+ 1 1 ( s + 1+1 ) where, Y (s) 1 s 1 ( ) s + 1+1 y(t) 1 1 +1 1 1 +1 1 1 +1 1 1 +1 c 1 +1 [ 1 s 1 s + 1+1 [ 1 e ( ] 1 +1 )t [ ] 1 e t /( 1 +1) [ ] 1 e t c ] is the time constant of the closed-loop system. One can observe that by selecting the value of 1 we can reduce the time constant of the system. But there is a problem here. The steady state value of the output is 1 1 +1 which is less than the reference input 1 to the system. Since we want the system to follow the reference input, this steady state error is a matter of concern and there could be different ways to tae care of this problem. We will address this problem a little later. In the above example, the feedbac control is just a simple gain, but it served our purpose of decreasing the time constant quite effectively. This configuration is called proportional control or P-control.

Lecture Notes on Control Systems/D. Ghose/01 10 Figure 1.5: An unwanted steady state error 1.5 P-Control of First Order Systems Let, G(s) 1 1+s Figure 1.6: P-control configuration Then, G c (s) +1 +1 1 1 s +1 1+ s

Lecture Notes on Control Systems/D. Ghose/01 103 Open Loop Closed Loop Gain 1 +1 (decreases slightly) Time Constant +1 Rise Time (T r ).. +1 Settling Time (T s ) 3.9 3.9 +1 there is considerable improvement in terms of rise time and settling time even though the DC gain reduces slightly giving rise to a non-zero steady state error. 1.6 P-Control of Second Order Systems Figure 1.7: P-control configuration for second order system Let us begin with an example shown in the above figure. The figure represents a missile autopilot with the missile lateral acceleration as its output. This lateral acceleration is integrated to obtain the missile angular velocity. The open loop transfer function is, The open-loop poles are: G ol (s) s(s+1) p 1 0; p 1

Lecture Notes on Control Systems/D. Ghose/01 104 Figure 1.8: Pole positions and responses The impulse response and the unit step response of the open-loop system are given by, y impulse ( 1 e t/ ) andareshownintheabovefigure. y step ( + t + e t/ ) Note that the open loop response is not oscillatory as there are no complex conjugate poles. The closed-loop transfer function is, The closed-loop poles are, G c (s) s(s+1) 1+ s(s+1) p 1, 1 ± 1 4 1 s + s + [ 1 ± 1 4 ] The closed-loop system is still second order but, depending on the value of, the response may oscillate. The poles are complex when, 1 4 < 0 > 1 4 This shows that the choice of is very crucial to the ind of time response we are looing for. Now, consider the effect of P-control of a general second order system. The open-loop system is,

Lecture Notes on Control Systems/D. Ghose/01 105 G ol (s) ω n s +ζω n s + ω n The closed-loop system with P-control gain of is, where, G c (s) G ol (s) 1+G ol (s) ωn s +ζω n s +( +1)ωn ( +1)ωn +1 s + ζ +1 +1ωn s +( +1)ωn ω n s + ζ ω n s + ω n +1 ζ ζ +1 ω n +1ω n Open Loop Closed Loop Gain 1 +1 (decreases slightly) Natural Frequency ω n ω n ω n + 1 (increases) Damping Ratio ζ ζ ζ +1 (decreases) Rise Time (T r ) π ω n π +1 ω n (decreases) Overshoot (M p ) 1 ζ 0.6 1 ζ 0.6 +1 (increases) Settling Time (T s ) 4 ζω n 4 ζω n (no change) However, a word of caution here is necessary. We have used only the approximate relationships, and in some cases they may not be exactly valid. You may need to chec with the exact expressions.