# Systems Analysis and Control

Size: px
Start display at page:

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 log-log and semi-log plots: 1. Magnitude Plot: 2 log 1 G(ıω) vs. log 1 ω 2. Phase Plot: G(ıω) vs. log 1 ω Bite-Size 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 -1-9 ω = 1 τ M. Peet Lecture 2: Control Systems 11 / 33

12 Plotting Real Poles and Zeros Lead-Lag We now have the pieces for a Lead-Lag 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 Lead-Lag 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 Lead-Lag 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 = 1-45 pole M. Peet Lecture 2: Control Systems 14 / 33

15 Plotting Real Poles and Zeros Lead-Lag 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 Lead-Lag 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 Lead-Lag 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 Lead-Lag Put them together to get a Lead-Lag 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 Lead-Lag 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

### 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 Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real

### 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

### 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 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using

### 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

### 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

### 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 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### Frequency 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

### 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

### 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 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

### ME 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

### 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

### 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

### 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

### 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)

### 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

### 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)

### Control Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017

### 8.1.6 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 + Second-order denominator, of the form 1+a 1 s + a s v 1 (s) + C R Two-pole low-pass filter example v (s) with

### Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design 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

### EECS 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 81-11 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

### Root Locus Design Example #3

Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll

### Radar 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

### Modern Control Systems

Modern Control Systems Matthew M. Peet Illinois Institute of Technology Lecture 18: Linear Causal Time-Invariant Operators Operators L 2 and ˆL 2 space Because L 2 (, ) and ˆL 2 are isomorphic, so are

### Sinusoidal Forcing of a First-Order Process. / τ

Frequency Response Analysis Chapter 3 Sinusoidal Forcing of a First-Order Process For a first-order transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A

### 16.30/31, 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

### ( ) Frequency Response Analysis. Sinusoidal Forcing of a First-Order Process. Chapter 13. ( ) sin ω () (

1 Frequency Response Analysis Sinusoidal Forcing of a First-Order Process For a first-order transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A tis: sin

### 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

### Outline. 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?

### Studio Exercise Time Response & Frequency Response 1 st -Order Dynamic System RC Low-Pass Filter

Studio Exercise Time Response & Frequency Response 1 st -Order Dynamic System RC Low-Pass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC Low-Pass

### Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros)

Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.

### Asymptotic Bode Plot & Lead-Lag Compensator

Asymptotic Bode Plot & Lead-Lag Compensator. Introduction Consider a general transfer function Ang Man Shun 202-2-5 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

### Process Control & Instrumentation (CH 3040)

First-order 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 first-order

### 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

### EECS 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 5-8 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

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

### 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,

### e 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

### e 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

### 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

### 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

### 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},

### 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

### 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

### Dynamic System Response. Dynamic System Response K. Craig 1

Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. Non-LTI Behavior Solution of Linear, Constant-Coefficient, Ordinary Differential Equations Classical

### 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

### Time 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

### 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

### 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.

### Chapter 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 Z-N tuning strategy for the nominal

### Homework 11 Solution - AME 30315, Spring 2015

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

### 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

### 4.5. 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

### APPLICATIONS FOR ROBOTICS

Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table

### ECE 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

### Frequency domain analysis

Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 2010-2011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 2010-2011

### 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.

### Systems 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

### 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

### Dynamic circuits: Frequency domain analysis

Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution

### 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

### GATE 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)

### 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)

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

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:

### 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

### Control Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:

### 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

### 13.42 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

### Control 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 =

### Boise 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:

### Control 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

### EE3CL4: Introduction to Linear Control Systems

1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We

### EE 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 marginally-stable

### 16.30/31, Fall 2010 Recitation # 1

6./, Fall Recitation # September, In this recitation we consiere the following problem. Given a plant with open-loop transfer function.569s +.5 G p (s) = s +.7s +.97, esign a feeback control system such

### H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )

.7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a

### 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

### Root Locus Design Example #4

Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is

### 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

### Parametric 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

### 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

### Control 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 D-MAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH)

### UNIVERSITY 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