Systems Analysis and Control


 Doris Payne
 1 years ago
 Views:
Transcription
1 Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 2: Drawing Bode Plots, Part 2
2 Overview In this Lecture, you will learn: Simple Plots Real Zeros Real Poles Complex Zeros Complex Poles M. Peet Lecture 2: Control Systems 2 / 33
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 2: Control Systems 3 / 33
4 Review Recall: Bode Plot Definition 1. The Bode Plot is a pair of loglog and semilog plots: 1. Magnitude Plot: 2 log 1 G(ıω) vs. log 1 ω 2. Phase Plot: G(ıω) vs. log 1 ω BiteSize Chucks: G(ıω) = i (ıωτ zi + 1) i (ıωτ pi + 1) 2 log G(ıω) = i 2 log ıωτ zi + 1 i 2 log ıωτ pi + 1 M. Peet Lecture 2: Control Systems 4 / 33
5 Plotting Simple Terms Plotting Normal Zeros Behaves as G 1 (s) = (τs + 1) { ıωτ ω << 1 τ = ıωτ ω >> 1 τ A constant at low ω. A zero at high ω. Break Point at ω = 1 τ. Magnitude: db ω << 1 τ 2 log ıωτ + 1 = 3.1 db ω = 1 τ 2 log ω ω >> 1 τ ω = τ M. Peet Lecture 2: Control Systems 5 / 33
6 Plotting Real Zeros Phase Geometry: (ıωτ + 1) = tan 1 ( ωτ 1 ω << 1 τ (ıωτ + 1) = 45 ω = 1 τ 9 ω >> 1 τ ) 1+ω 2 τ 2 1 Im(s) } ωτ } Re(s) ω = τ M. Peet Lecture 2: Control Systems 6 / 33
7 Plotting Real Zeros Phase We can improve our sketch using a line (ıωτ + 1) ω <.1 τ = 45.1 (1 + log ωτ) τ < ω < 1 τ 9 ω > 1 τ 9 45 ω =.1 τ 1 ω = τ 1 ω = 1 τ M. Peet Lecture 2: Control Systems 7 / 33
8 Plotting Real Poles Magnitude We can plot real zeros Now lets do the poles. Poles are the opposite of zeros G zero = G 1 pole 1 G pole (ıω) = ıωτ + 1 G zero (ıω) = ıωτ + 1 Magnitude: G pole (ıω) = 1 ıωτ log G pole (ıω) = 2 log ıωτ M. Peet Lecture 2: Control Systems 8 / 33
9 Plotting Real Poles Magnitude Our approximation is also a reflection of the zero 2 log 1 ıωτ + 1 = db ω << 1 τ 3.1 db ω = 1 τ 2 log ω ω >> 1 τ ω = τ ω = τ M. Peet Lecture 2: Control Systems 9 / 33
10 Plotting Real Poles Phase We can plot real zeros Now lets do the poles. Poles are the opposite of zeros G pole (ıω) = 1 ıωτ + 1 = G 1 zero(ıω) Phase: G pole (ıω) = tan 1 ωτ = G zero (ıω) M. Peet Lecture 2: Control Systems 1 / 33
11 Plotting Real Poles Phase Our approximation is also a reflection of the zero (ıωτ + 1) 1 ω <.1 τ = 45.1 (1 + log ωτ) τ < ω < 1 τ 9 ω > 1 τ ω =.1 τ 1 ω = τ 1 ω = 1 τ ω =.1 τ ω = τ 19 ω = 1 τ M. Peet Lecture 2: Control Systems 11 / 33
12 Plotting Real Poles and Zeros LeadLag We now have the pieces for a LeadLag Compensator. Example: Lead Compensator First Step: Standard Form G(s) = s + 1 s + 1 G(s) = s + 1 s + 1 = 1 s s + 1 Second Step: Plot the Constant c =.1 = 2dB M. Peet Lecture 2: Control Systems 12 / 33
13 Plotting Real Poles and Zeros LeadLag Third Step: Plot the Zero at ω = 1. Magnitude: db ω << 1 2 log ıω + 1 = 3.1 db ω = 1 2 log ω ω >> ω = 1 ω = Phase: (ıω + 1) ω <.1 = 45 (1 + log ω).1 < ω < 1 9 ω > ω = 1 zero M. Peet Lecture 2: Control Systems 13 / 33
14 Plotting Real Poles and Zeros LeadLag Fourth Step: Plot the Pole at ω = 1. Magnitude: 2 log ıω db ω << 1 = 3.1 db ω = 1 2 log ω ω >> ω = 1 ω = Phase: ( ıω ) ω < 1 = 45 ( ) 1 + log ω 1 1 < ω < 1 9 ω > ω = 1 ω = 145 pole M. Peet Lecture 2: Control Systems 14 / 33
15 Plotting Real Poles and Zeros LeadLag Add them all together 2 1 ω = ω = zero ω = 1 ω = 1 total pole M. Peet Lecture 2: Control Systems 15 / 33
16 Plotting Real Poles and Zeros LeadLag Compare with the True Plot 2 1 ω = True ω = True 45 ω = 1 ω = M. Peet Lecture 2: Control Systems 16 / 33
17 Plotting Real Poles and Zeros LeadLag Example: Lag Compensator G(s) = 1 s s + 1 = s s Pole at ω = 1 Zero at ω = 1 Output Phase Lags behind the input. M. Peet Lecture 2: Control Systems 17 / 33
18 Plotting Real Poles and Zeros LeadLag Put them together to get a LeadLag Compensator. Zero at ω = 1 Pole at ω = 1 Pole at ω = 1 Zero at ω = 1 Magnitude: Changes in Slope. ω = 1: +2 ω = 1: 2 ω = 1: 2 ω = 1: +2 G(s) = s + 1 s + 1 s + 1 s + 1 = s + 1 s s s M. Peet Lecture 2: Control Systems 18 / 33
19 Plotting Real Poles and Zeros LeadLag G(s) = s + 1 s + 1 s + 1 s + 1 Phase: Changes in Slope. ω =.1: +45 ω = 1: 45 ω = 1: 9 ω = 1: +9 ω = 1: +45 ω = 1: M. Peet Lecture 2: Control Systems 19 / 33
20 Complex Poles and Zeros Now we move on to the final topic: Complex Poles and Zeros Complex Zeros: G(s) = s 2 + 2ζω n s + ω 2 n First Step: Put in standard form ( ( s G(s) = ωn 2 ω n ) ) 2 + 2ζ s + 1 ω n So y ss = 1 Ignoring the constant G 1 (ıω) = ( ω ω n ) 2 + 2ζ ω ω n ı + 1 = 2ζı ( 1 ω ) 2 ω ω n ω n << 1 ω ω n = 1 ω ω n >> 1 M. Peet Lecture 2: Control Systems 2 / 33
21 Complex Poles and Zeros Magnitude: ( ) 2 ω 2 log G 1 (ıω) = 2 log + 2ζ sω ı + 1 ω n ω n ω db ω n << 1 ω = 2 log(2ζ) ω n = 1 ω 4(log ω log ω n ) ω n >> M. Peet Lecture 2: Control Systems 21 / 33
22 Complex Poles and Zeros The Behavior near ω = ω n depends on ζ. When ζ =, 2 log G 1 (ıω) = 2 log(2ζ) = When ζ =.5, When ζ = 1, 2 log G 1 (ıω) = 2 log 1 = 2 log G 1 (ıω) = 2 log 2 = 6.2dB M. Peet Lecture 2: Control Systems 22 / 33
23 Complex Poles and Zeros Phase: ( ( ) ) 2 ω G 1 (ıω) = 1 + 2ζ ω ı ω n ω n ( ) = tan 1 2ζωωn ωn 2 ω 2 ω ω n << 1 = 9 ω ω n = 1 18 ω ω n >> 1 Re(s) Im(s) } 2 ζ ω/ω n < G(iω) } 1(ω/ω n ) M. Peet Lecture 2: Control Systems 23 / 33
24 Complex Poles and Zeros The Behavior near ω = ω n depends on ζ. For ω = ω n = 1, When ζ =, d G 1(ıω) = dω When ζ =.1, d G 1(ıω) dω When ζ =.5, d G 1(ıω) dω When ζ = 1, d G 1(ıω) dω = 1 = 2 = 1 M. Peet Lecture 2: Control Systems 24 / 33
25 Complex Poles and Zeros Comparison vs. True for ω n = 1, ζ = Good for magnitude, bad for phase M. Peet Lecture 2: Control Systems 25 / 33
26 Complex Poles and Zeros Comparison vs. True for ω n = 1, ζ = M. Peet Lecture 2: Control Systems 26 / 33
27 Complex Poles Treatment of Complex Poles is similar Complex Poles: 1 G(s) = ( ( ) ) 2 s ω n + 2ζ s ω n Magnitude: ( ) 2 ω 2 log G 1 (ıω) = 2 log 1 + 2ζ sω ı ω n ω n ω db ω n << 1 ω = 2 log(2ζ) ω n = 1 ω 4(log ω log ω n ) ω n >> 1 M. Peet Lecture 2: Control Systems 27 / 33
28 Complex Poles The Behavior near ω = ω n depends on ζ. When ζ =, 2 log G 1 (ıω) = 2 log(2ζ) = + When ζ =.5, When ζ = 1, 2 log G 1 (ıω) = 2 log 1 = 2 log G 1 (ıω) = 2 log 2 = 6.2dB M. Peet Lecture 2: Control Systems 28 / 33
29 Complex Poles Phase: ( ( ω G 1 (ıω) = 1 ω n ω ω n << 1 = 9 ω ω n = 1 18 ω ω n >> 1 ) ) 2 + 2ζ ω ı ω n M. Peet Lecture 2: Control Systems 29 / 33
30 Complex Poles The Behavior near ω = ω n depends on ζ. For ω = ω n = 1, When ζ =, d G 1(ıω) = dω When ζ =.1, d G 1(ıω) dω When ζ =.5, d G 1(ıω) dω When ζ = 1, d G 1(ıω) dω = 1 = 2 = 1 M. Peet Lecture 2: Control Systems 3 / 33
31 Complex Poles Comparison vs. True for ω n = 1, ζ = Good for magnitude, bad for phase M. Peet Lecture 2: Control Systems 31 / 33
32 Complex Poles Comparison vs. True for ω n = 1, ζ = Phase improves, Magnitude gets worse. M. Peet Lecture 2: Control Systems 32 / 33
33 Summary What have we learned today? Simple Plots Real Zeros Real Poles Complex Zeros Complex Poles Next Lecture: Compensation in the Frequency Domain M. Peet Lecture 2: Control Systems 33 / 33
Systems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationSystems 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 informationSystems 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
More informationSystems 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
More informationStep 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
More informationSystems 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
More informationMASSACHUSETTS 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
More informationResponse 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
More informationSystems 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
More informationSystems 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
More informationSystems 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 informationPlan 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 frequencyresponse design method Goal: wrap up lead and
More informationSystems 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
More informationSystems 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
More informationSystems 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
More informationSystems 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
More information100 (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 informationFrequency 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
More informationHomework 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 informationControl 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 DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationReview: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Plan of the Lecture Review: transient and steadystate
More informationFrequency Response of Linear Time Invariant Systems
ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationTransient 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 informationEE 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
More informationME 375 EXAM #1 Friday, March 13, 2015 SOLUTION
ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned
More informationThe 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 informationEE 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
More informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop 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 closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationMAS107 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
More informationDESIGN 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)
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More information8.1.6 Quadratic pole response: resonance
8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Secondorder denominator, of the form 1+a 1 s + a s v 1 (s) + C R Twopole lowpass filter example v (s) with
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationEECS C128/ ME C134 Final Wed. Dec. 14, am. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C128/ ME C134 Final Wed. Dec. 14, 211 8111 am Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth 1 points total. Problem Points Score 1 16 2 12
More informationRoot 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 informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationModern Control Systems
Modern Control Systems Matthew M. Peet Illinois Institute of Technology Lecture 18: Linear Causal TimeInvariant Operators Operators L 2 and ˆL 2 space Because L 2 (, ) and ˆL 2 are isomorphic, so are
More informationSinusoidal Forcing of a FirstOrder Process. / τ
Frequency Response Analysis Chapter 3 Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
More information( ) Frequency Response Analysis. Sinusoidal Forcing of a FirstOrder Process. Chapter 13. ( ) sin ω () (
1 Frequency Response Analysis Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A tis: sin
More informationFrequency 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
More informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More informationStudio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter
Studio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC LowPass
More informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
More informationAsymptotic Bode Plot & LeadLag Compensator
Asymptotic Bode Plot & LeadLag Compensator. Introduction Consider a general transfer function Ang Man Shun 20225 G(s = n k=0 a ks k m k=0 b ks k = A n k=0 (s z k m k=0 (s p k m > n When s =, transfer
More informationProcess Control & Instrumentation (CH 3040)
Firstorder systems Process Control & Instrumentation (CH 3040) Arun K. Tangirala Department of Chemical Engineering, IIT Madras January  April 010 Lectures: Mon, Tue, Wed, Fri Extra class: Thu A firstorder
More informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More information12.7 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 informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informatione iωt dt and explained why δ(ω) = 0 for ω 0 but δ(0) =. A crucial property of the delta function, however, is that
Phys 531 Fourier Transforms In this handout, I will go through the derivations of some of the results I gave in class (Lecture 14, 1/11). I won t reintroduce the concepts though, so you ll want to refer
More informatione iωt dt and explained why δ(ω) = 0 for ω 0 but δ(0) =. A crucial property of the delta function, however, is that
Phys 53 Fourier Transforms In this handout, I will go through the derivations of some of the results I gave in class (Lecture 4, /). I won t reintroduce the concepts though, so if you haven t seen the
More informationDr 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 splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationAutomatic 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 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More informationDiscrete 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},
More informationINTRODUCTION 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 lineartimeinvariant
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
More informationAMME3500: 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
More informationTime Series Analysis. Solutions to problems in Chapter 7 IMM
Time Series Analysis Solutions to problems in Chapter 7 I Solution 7.1 Question 1. As we get by subtraction kx t = 1 + B + B +...B k 1 )ǫ t θb) = 1 + B +... + B k 1 ), and BθB) = B + B +... + B k ), 1
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationCHAPTER 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.
More informationChapter 6  Solved Problems
Chapter 6  Solved Problems Solved Problem 6.. Contributed by  James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the ZN tuning strategy for the nominal
More informationHomework 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 closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More information4.5. Applications of Trigonometry to Waves. Introduction. Prerequisites. Learning Outcomes
Applications of Trigonometry to Waves 4.5 Introduction Waves and vibrations occur in many contexts. The water waves on the sea and the vibrations of a stringed musical instrument are just two everyday
More informationAPPLICATIONS 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 informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationLecture 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 5: Calculating the Laplace Transform of a Signal Introduction In this Lecture, you will learn: Laplace Transform of Simple
More informationEE 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 Phaselead and Phaselag compensators using
More informationDynamic 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 informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationGATE EE Topic wise Questions SIGNALS & SYSTEMS
www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationMASSACHUSETTS 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 informationChapter 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 openloop system (a system without feedbac) and a closedloop
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control DMAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More information2.004 Dynamics and Control II Spring 2008
MIT OpeCourseWare http://ocw.mit.edu 2.004 Dyamics ad Cotrol II Sprig 2008 For iformatio about citig these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts Istitute of Techology
More information13.42 READING 6: SPECTRUM OF A RANDOM PROCESS 1. STATIONARY AND ERGODIC RANDOM PROCESSES
13.42 READING 6: SPECTRUM OF A RANDOM PROCESS SPRING 24 c A. H. TECHET & M.S. TRIANTAFYLLOU 1. STATIONARY AND ERGODIC RANDOM PROCESSES Given the random process y(ζ, t) we assume that the expected value
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
More informationBoise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain
Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type servomechanism:
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
More informationEE3CL4: 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 informationEE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation
EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginallystable
More information16.30/31, Fall 2010 Recitation # 1
6./, Fall Recitation # September, In this recitation we consiere the following problem. Given a plant with openloop transfer function.569s +.5 G p (s) = s +.7s +.97, esign a feeback control system such
More informationH(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 information1 (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
More informationRoot 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 informationControl 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
More informationParametric Functions and Vector Functions (BC Only)
Parametric Functions and Vector Functions (BC Only) Parametric Functions Parametric functions are another way of viewing functions. This time, the values of x and y are both dependent on another independent
More informationLecture 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
More informationControl Systems I. Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback. Readings: Emilio Frazzoli
Control Systems I Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH)
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
More informationPoles and Zeros and Transfer Functions
Poles and Zeros and Transfer Functions Transfer Function: Considerations: Factorization: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial
More informationChapter Eight. Transfer Functions. 8.1 Frequency Domain Modeling
Chapter Eight Transfer Functions The typical regulator system can frequently be described, in essentials, by differential equations of no more than perhaps the second, third or fourth order. In contrast,
More information