Dr Ian R. Manchester

Similar documents
Course Outline. Time vs. Freq. Domain Analysis. Frequency Response. Amme 3500 : System Dynamics & Control. Design via Frequency Response

Dr Ian R. Manchester Dr Ian R. Manchester Amme 3500 : Root Locus Design

CDS 101/110: Lecture 9.1 Frequency DomainLoop Shaping

LECTURE 2: PD, PID, and Feedback Compensation. ( ) = + We consider various settings for Zc when compensating the system with the following RL:

Bode Plot for Controller Design

CDS 101/110: Lecture 8.2 PID Control

ECE317 : Feedback and Control

JUNE 2014 Solved Question Paper

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

ANNA UNIVERSITY :: CHENNAI MODEL QUESTION PAPER(V-SEMESTER) B.E. ELECTRONICS AND COMMUNICATION ENGINEERING EC334 - CONTROL SYSTEMS

SECTION 6: ROOT LOCUS DESIGN

Frequency Response Analysis and Design Tutorial

Classical Control Design Guidelines & Tools (L10.2) Transfer Functions

Lecture 7:Examples using compensators

EC CONTROL SYSTEMS ENGINEERING

Position Control of DC Motor by Compensating Strategies

ME451: Control Systems. Course roadmap

Andrea Zanchettin Automatic Control 1 AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Winter Semester, Linear control systems design Part 1

ME 375 System Modeling and Analysis

Andrea Zanchettin Automatic Control 1 AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Spring Semester, Linear control systems design

BSNL TTA Question Paper Control Systems Specialization 2007


1.What is frequency response? A frequency responses the steady state response of a system when the input to the system is a sinusoidal signal.

EES42042 Fundamental of Control Systems Bode Plots

EEL2216 Control Theory CT2: Frequency Response Analysis

CHAPTER 9 FEEDBACK. NTUEE Electronics L.H. Lu 9-1

Microelectronic Circuits II. Ch 9 : Feedback

Välkomna till TSRT15 Reglerteknik Föreläsning 8

CDS 101/110a: Lecture 8-1 Frequency Domain Design. Frequency Domain Performance Specifications

(1) Identify individual entries in a Control Loop Diagram. (2) Sketch Bode Plots by hand (when we could have used a computer

Bode and Log Magnitude Plots

Chapter 5 Frequency-domain design

Loop Design. Chapter Introduction

Readings: FC: p : lead compensation. 9/9/2011 Classical Control 1

Rotary Motion Servo Plant: SRV02. Rotary Experiment #03: Speed Control. SRV02 Speed Control using QuaRC. Student Manual

Module 08 Controller Designs: Compensators and PIDs

Positive Feedback and Oscillators

TUTORIAL 9 OPEN AND CLOSED LOOP LINKS. On completion of this tutorial, you should be able to do the following.

EE 435. Lecture 16. Compensation Systematic Two-Stage Op Amp Design

CHASSIS DYNAMOMETER TORQUE CONTROL SYSTEM DESIGN BY DIRECT INVERSE COMPENSATION. C.Matthews, P.Dickinson, A.T.Shenton

EECE488: Analog CMOS Integrated Circuit Design Set 7 Opamp Design

Implementation and Simulation of Digital Control Compensators from Continuous Compensators Using MATLAB Software

ECE317 Homework 7. where

Pole, zero and Bode plot

JNTUWORLD. 6 The unity feedback system whose open loop transfer function is given by G(s)=K/s(s 2 +6s+10) Determine: (i) Angles of asymptotes *****

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

ME 5281 Fall Homework 8 Due: Wed. Nov. 4th; start of class.

Reduction of Multiple Subsystems

Magnetic Levitation System

P Shrikant Rao and Indraneel Sen

Chapter 10 Feedback ECE 3120 Microelectronics II Dr. Suketu Naik

and using the step routine on the closed loop system shows the step response to be less than the maximum allowed 20%.

Designing PID controllers with Matlab using frequency response methodology

Electrical Engineering. Control Systems. Comprehensive Theory with Solved Examples and Practice Questions. Publications

A Comparison And Evaluation of common Pid Tuning Methods

Homework Assignment 10

This article presents a review of PLL transfer functions with attention to the conditions required for steady-state stability

Outline. Digital Control. Lecture 3

DESIGN AND ANALYSIS OF FEEDBACK CONTROLLERS FOR A DC BUCK-BOOST CONVERTER

EE 482 : CONTROL SYSTEMS Lab Manual

Consider the control loop shown in figure 1 with the PI(D) controller C(s) and the plant described by a stable transfer function P(s).

Homework Assignment 13

ADJUSTING SERVO DRIVE COMPENSATION George W. Younkin, P.E. Life Fellow IEEE Industrial Controls Research, Inc. Fond du Lac, Wisconsin

Homework Assignment 13

Lecture 18 Stability of Feedback Control Systems

Lecture 8 ECEN 4517/5517

Bode plot, named after Hendrik Wade Bode, is usually a combination of a Bode magnitude plot and Bode phase plot:

CDS 101/110: Lecture 10-2 Loop Shaping Design Example. Richard M. Murray 2 December 2015

Microelectronic Circuits - Fifth Edition Sedra/Smith Copyright 2004 by Oxford University Press, Inc.

EE 370/L Feedback and Control Systems Lab Section Post-Lab Report. EE 370L Feedback and Control Systems Lab

Bode Plots. Hamid Roozbahani

DEGREE: Biomedical Engineering YEAR: TERM: 1

Laboratory Assignment 5 Digital Velocity and Position control of a D.C. motor

GE420 Laboratory Assignment 8 Positioning Control of a Motor Using PD, PID, and Hybrid Control

CONTROLLER DESIGN FOR POWER CONVERSION SYSTEMS

Feedback. Operational amplifiers invariably are incorporated within a circuit with negative feedback. Consider the inverting amplifier configuration :

Analog Integrated Circuits Fundamental Building Blocks

Another Compensator Design Example

This manuscript was the basis for the article A Refresher Course in Control Theory printed in Machine Design, September 9, 1999.

CHAPTER 6 INTRODUCTION TO SYSTEM IDENTIFICATION

ELECTRICAL CIRCUITS 6. OPERATIONAL AMPLIFIERS PART III DYNAMIC RESPONSE

Experiment 9. PID Controller

SYLLABUS. osmania university CHAPTER - 1 : CONTROL SYSTEMS CLASSIFICATION

Lecture 10. Lab next week: Agenda: Control design fundamentals. Proportional Control Proportional-Integral Control

Implementation of Proportional and Derivative Controller in a Ball and Beam System

Today s topic: frequency response. Chapter 4

Lecture 5 Introduction to control

Phys Lecture 5. Motors

Testing Power Sources for Stability

ITT Technical Institute ET245 Devices II Unit 5 Chapter

ECEN 474/704 Lab 5: Frequency Response of Inverting Amplifiers

LINEAR MODELING OF A SELF-OSCILLATING PWM CONTROL LOOP

Stabilizing and Robust FOPI Controller Synthesis for First Order Plus Time Delay Systems

DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING BANGLADESH UNIVERSITY OF ENGINEERING & TECHNOLOGY EEE 402 : CONTROL SYSTEMS SESSIONAL

ECEN 474/704 Lab 8: Two-Stage Miller Operational Amplifier

1. Consider the closed loop system shown in the figure below. Select the appropriate option to implement the system shown in dotted lines using

SAR (successive-approximation-register) ADCs

Optimal Control System Design

LECTURE FOUR Time Domain Analysis Transient and Steady-State Response Analysis

[ á{tå TÄàt. Chapter Four. Time Domain Analysis of control system

Transcription:

Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign 2 Due 8 Bode Plots 9 Bode Plots 2 10 State Space Modeling Assign 3 Due 11 State Space Design Techniques 12 Advanced Control Topics 13 Review Assign 4 Due Dr. Ian R. Manchester Amme 3500 : Introduction Slide 2

Understand control design (PID, lead-lag) in terms of shaping frequency response Introduce the use of stability margins and loop shaping to design controllers for more general systems See some extensions to higher-order and nonlinear systems Slide 3

In week 7 we looked at modifying the transient and steady state response of a system using root locus design techniques Gain adjustment (speed, steady state error) Lag (PI) compensation (steady-state error) Lead (PD) compensation (speed, stability) We will now examine methods for designing for a particular specification by examining the frequency response of a system We still rely to some extent approximating CL behaviour as 2 nd Order Slide 4

Control system performance generally judged by time domain response to certain test signals (step, etc.) Simple for < 3 OL poles or ~2 nd order CL systems. No unified methods for higher-order systems. Freq response easy for higher order systems Qualitatively related to time domain behaviour More natural for studying sensitivity and noise susceptibility Slide 5

Remember the root locus satisfies 1+ K(s)G(s) = 0 So poles cross over imaginary axis (stabilityinstability transition) when the frequency response satisfies or 1+ K( j")g( j") = 0 K( j")g( j") =1, #K( j")g( j") =180 Slide 6

Crossover frequency: Gain Margin: Phase Margin: Bandwidth (CL specification) Less intuitive than RL, but easier to draw for high order systems. Slide 7

The root locus demonstrated that we can often design controllers for a system via gain adjustment to meet a particular transient response We can effect a similar approach using the frequency response by examining the relationship between phase margin and damping Slide 8

20 db Slide 9

Recall that the Phase Margin is closely related to the damping ratio of the system For a unity feedback system with open-loop function We found that the relationship between PM and damping ratio is given by Slide 10

Slide 11

Phase Margin Slide 12

Given a desired overshoot, we can convert this to a required damping ratio and hence PM Examining the Bode plot we can find the frequency that gives the desired PM Slide 13

The design procedure therefore consists of Draw the Bode Magnitude and phase plots Determine the required phase margin from the percent overshoot Find the frequency on the Bode phase diagram that yields the desired phase margin Change the gain to force the magnitude curve to go through 0dB Slide 14

For the following position control system shown here, find the preamplifier gain K to yield a 9.5% overshoot in the transient response for a step input Slide 15

Draw the Bode plot For 9.5% overshoot,!=0.6 and PM must be 59.2 o Locate frequency with the required phase at 14.8 rad/s The magnitude must be raised by 55.3dB to yield the cross over point at this frequency This yields a K = 583.9 Slide 16

As we saw previously, not all specifications can be met via simple gain adjustment We examined a number of compensators that can bring the root locus to a desired design point A parallel design process exists in the frequency domain Slide 17

In particular, we will look at the frequency characteristics for the PD Controller Lead Controller PI Controller Lag Controller Understanding the frequency characteristics of these controllers allows us to select the appropriate version for a given design Slide 18

The ideal derivative compensator adds a pure differentiator, or zero, to the forward path of the control system The root locus showed that this will tend to stabilize the system by drawing the roots towards the zero location We saw that the pole and zero locations give rise to the break points in the Bode plot Slide 19

The Bode plot for a PD controller looks like this The stabilizing effect is seen by the increase in phase at frequencies above the break frequency However, the magnitude grows with increasing frequency and will tend to amplify high frequency noise Slide 20

Introducing a higher order pole yields the lead compensator This is often rewritten as where 1/" is the ratio between pole-zero break points The name Lead Compensation reflects the fact that this compensator imparts a phase lead Slide 21

The Bode plot for a Lead compensator looks like this The frequency of the phase increase can be designed to meet a particular phase margin requirement The high frequency magnitude is now limited Slide 22

The lead compensator can be used to change the damping ratio of the system by manipulating the Phase Margin The phase contribution consists of The peak occurs at with a phase shift and magnitude of Slide 23

This compensator allows the designer to raise the phase of the system in the vicinity of the crossover frequency Slide 24

The design procedure consists of the following steps: 1. Find the open-loop gain K to satisfy steady-state error or bandwidth requirements 2. Evaluate phase margin of the uncompensated system using the value of gain chosen above 3. Find the required phase lead to meet the damping requirements 4. Determine the value of " to yield the required increase in phase! Slide 25

5. Determine the new crossover frequency 6. Determine the value of T such that " max lies at the new crossover frequency 7. Draw the compensated frequency response and check the resulting phase margin. 8. Check that the bandwidth requirements have been met. 9. Simulate to be sure that the system meets the specifications (recall that the design criteria are based on a 2 nd order system). Slide 26

Returning to the previous example, we will now design a lead compensator to yield a 20% overshoot and K v =40, with a peak time of 0.1s Slide 27

From the specifications, we can determine the following requirements For a 20% overshoot we find!=0.456 and hence a Phase Margin of 48.1 o For peak time of 0.1s with the given!, we can find the required closed loop bandwidth to be 46.6rad/s To meet the steady state error specification Slide 28

From the Bode plot, we evaluate the PM to be 34 o for a gain of 1440 We can t simply increase the gain without violating the other design constraints We use a Lead Compensator to raise the PM Slide 29

We require a phase margin of 48.1 o The lead compensator will also increase the phase margin frequency so we add a correction factor to compensate for the lower uncompensated system s phase angle The total phase contribution required is therefore 48.1 o (34 o 10 o ) = 24.1 o Slide 30

Based on the phase requirement we find The resulting magnitude is Examining the Bode magnitude, we find that the frequency at which the magnitude is -3.77dB is # max =39rad/s The break frequencies can be found at 25.3 and 60.2 Slide 31

The compensator is The resulting system Bode plot shows the impact of the phase lead Slide 32

We need to verify the performance of the resulting design The simulation appears to validate our second order assumption Slide 33

We saw that the integral compensator takes on the form This results in infinite gain at low frequencies which reduces steady-state error A decrease in phase at frequencies lower than the break will also occur Slide 34

The Bode plot for a PI controller looks like this The break frequency is usually located at a frequency substantially lower than the crossover frequency to minimize the effect on the phase margin Slide 35

Lag compensation approximates PI control This is often rewritten as where " is the ratio between zero-pole break points The name Lag Compensation reflects the fact that this compensator imparts a phase lag Slide 36

The Bode plot for a Lag compensator looks like this This compensator effectively raises the magnitude for low frequencies The effect of the phase lag can be minimized by careful selection of the centre frequency Slide 37

In this case we are trying to raise the gain at low frequencies without affecting the stability of the system Slide 38

Nise suggests setting the gain K for s.s. error, then designing a lag network to attain desired PM. Franklin suggests setting the gain K for PM, then designing a lag network to raise the low freq. gain w/o affecting system stability. System without lag compensation, Franklin et al. procedure Slide 39

The design procedure (Franklin et al.) consists of the following steps: 1. Find the open-loop gain K to satisfy the phase margin specification without compensation 2. Draw the Bode plot and evaluate low frequency gain 3. Determine " to meet the low-frequency gain error requirement 4. Choose the corner frequency #=1/T to be one octave to one decade below the new crossover frequency 5. Evaluate the second corner frequency #=1/"T 6. Simulate to evaluate the design and iterate as required Slide 40

Returning again to the previous example, we will now design a lag compensator to yield a ten fold improvement in steady-state error over the gaincompensated system while keeping the overshoot at 9.5% Slide 41

In the first example, we found the gain K=583.9 would yield our desired 9.5% overshoot with a PM of 59.2 o at 14.8rad/s For this system we find that We therefore require a K v of 162.2 to meet our specification We need to raise the low frequency magnitude by a factor of 10 (or 20dB) without affecting the PM Slide 42

First we draw the Bode plot with K=583.9 Set the zero at one decade, 1.48rad/s, lower than the PM frequency The pole will be at 1/" relative to this so Slide 43

The resulting system has a low frequency gain K v of 162.2 as per the requirement The overshoot is slightly higher than the desired Iteration of the zero and pole locations will yield a lower overshoot if required Slide 44

As with the Root Locus designs we considered previously, we often require both lead and lag components to effect a particular design This provides simultaneous improvement in transient and steady-state responses In this case we are trading off three primary design parameters Crossover frequency # c which determines bandwidth, rise time and settling time Phase margin which determines the damping coefficient and hence overshoot Low frequency gain which determines steady state error characteristics Slide 45

Recall the Sensitivity Function S(s) = 1 1+ G(s)K(s) The Complementary Sensitivity Function T(s) = G(s)K(s) 1+ G(s)K(s) Complementary because S(s)+T(s)=1 for all s The system response to reference R, disturbance D and measurement noise N is Y(s) = T(s)R(s) + S(s)D(s) + T(s)N(s) Slide 46

Rate of gain roll-off is related to phase phase margin: each 20dB/decade corresponds to about 90 degrees of phase lag Slide 47

Also known as the waterbed effect. Recall S = For any system with any controller 1 1+ KG $ log S( j") d" = % Re(p i ) 0 # Where p i are open-loop unstable pole locations & i Slide 48

From Gunter Stein s article Respect the Unstable Slide 49

We examined several ways to assess CL stability from the OL transfer function. Root Locus (rules for plotting CL poles as a function of OL gain) Bode plots (GM and PM gave an indication of how much the gain or phase could change before instability) For some systems, GM and PM may be ambiguous or give contradictory stability information. Dr. Michael Jakuba Amme 3500 : Nyquist Slide 50

An OL unstable system: Slide 51

A conditionally stable system: GM = 14 db PM = -36.9 deg Slide 52

A complicated system: 0.7 rad/s 8.5 rad/s 9.8 rad/s Slide 53

To generate the Nyquist plot we will need a concept called contour mapping Given a series points on a contour A in the s-plane, the function F(s) will map these points to another contour B * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 54

Consider a complex-valued function of a complex variable, e.g. G(s).... Slide 55

We extend the contour to enclose the entire RHP Applying the contour mapping techniques we can determine if any closed loop poles exist in the RHP by examining the encirclements of the origin for 1+KG(s)H(s) Im(s) Contour at infinity Re(s) Slide 56

We can simplify this a bit by noticing that 1+KG(s)H(s) is just KG (s)h(s) shifted to the right by 1. Find the contour for KG (s)h(s) Count encirclements of -1. Im(s) Contour at infinity Re(s) Slide 57

The poles of the CL transfer function are given by the zeros of the characteristic equation. Let Z denote the number of zeros in the RHP of 1+KG(s)H (s) The number of encirclements of -1 depends on both the RHP zeros and poles of KG(s)H(s). Let P denote the number of OL unstable poles If N = ( # ccw - # cw ), then Z = P N. For stability Z = 0. Slide 58

. Im(s).. x x x Re(s). Slide 59

A typical system: Im(s).B N = 0 P = 0 Z = 0 D.. F.. x.a x Re(s) D E.. F.. B, C E.C A Slide 60

An OL unstable system: Im(s).B. A o. F...A x D E x Re(s).. E F. B,C.C. D Slide 61

F/A-18 on landing approach, 140 knots. G(s) = "(s) # e (s) = 0.072 (s + 23)(s 2 + 0.05s + 0.04) (s $ 0.7)(s +1.7)(s 2 + 0.08s + 0.04) Amme 3500 : Root Locus Design Slide 62

The gain and phase margins can also be read from the Nyquist plot * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 63

Time delay is common in controlled systems (e.g. computation or communication delay, transport of waves, fluid flow along pipes ) Although it is linear and time-invariant, it cannot be represented by a rational transfer function. Why not? What is its transfer function? Slide 64

Transfer function for a time delay of # is: G " (s) = e #"s So the frequency response is G " ( j#) = e $"j# Slide 65

G " ( j#) = e $"j# Slide 66

Slide 67

Slope = %! Slope = $! Slide 68

Slide 69

Nise Chapter 11 Franklin & Powell Chapter 6 Åström and Murray (free online) Chapter 9, 10, 11 Slide 70