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

Save this PDF as:

Size: px
Start display at page:

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

## Transcription

1 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: The open loop transfer function is K(s +)(s +) G(s) =. (s +5)(s +6) To find the break in and breakaway points, we apply rule 6 as follows: Taking the derivative, (σ +5)(σ +6) σ +σ + K(σ) = =. (σ +)(σ +) σ +σ + dk 8σ 56σ 68 σ σ 7 = =. dσ (σ +σ +) (σ +σ +) and setting dk/dσ =, we find σ =.565 and σ =.65.. Nise problem 7 from chapter 8, page 77. Answer: From rule about the real axis segment, we know that the root locus should exist between the two zeros in the right hand plane as well as the pole and zero in the left hand plane. Next step is to deal with the two poles with

2 imaginary parts. The root locus should depart from the two poles. It cannot go to the infinity because we already have the same number of poles and zeros. Therefore it has to be joined one of the two real axis segment. The root locus departing from the two poles eventually should arrive at the two zeros in the right hand plane. Here two possibilities arise: ) The root locus directly is joined to the real axis segment between the two zeros in right hand plane, or ) It first cuts through the real axis segment between the pole and zero in the left hand plane, and comes back to the right hand plane to join the real axis segment between the two zeros. Which option turns out to be the actual root locus depends on the relative location of the open loop poles and zeros. The two plots below show two examples obtained using Matlab Nise problem 9 from chapter 8, page 77. Answer: The open loop transfer function is K(s 9) K(s )(s +) G(s) = =, (s +) (s j)(s +j) and the root locus is shown below;

3 Clearly, the system becomes unstable for gain equal to or larger than the gain required to bring the closed loop poles together to a double pole at the origin. This critical gain is given by K = G(s) s= = s + s +j s j = = =. s 9 s + s 9 s= See the geometrical interpretation below. s= Alternatively, we can find K from the relationship KG(s) + =, which gives s = K(s 9), thus 9K s =. K + If 9K 9K > (or K>/9), then we get two real poles at s = ±, which make K+ K+ 9K the system unstable because of the pole in the right hand plane. If K+ < 9K (or K </9), then we obtain two pure imaginary poles at s = ±j, which K+ make the system marginally stable. If K =/9, then two double poles are at s =.. Nise problem 7 from chapter 8, page 79. Answer: The open loop transfer function is a. Sketch the root locus. K(s +) G(s) =. s(s +)(s +)(s +)

4 b. The asymptotes. By rule 5, ( ) ( ) 8 σ a = =, { } (m +)π π 5π θ a = =,π,. c. The value of gain that makes the system marginally stable. By rule 7, setting G(jω)=, we find K(jω +) =. (jω) +9(jω) + 6(jω) + (jω) Rearranging, ( ) K(jω +) = ω j9ω 6ω +jω, we end up with two equations for the real and imaginary parts, respectively: ω 6ω + K =, 9ω ω Kω =. From the second equation, ω =( + K)/9. Inserting this result to the first equation, we find K =.796. (The other value is K =.796, which is invalid because it is negative.)

5 5 Editor (C) Imag Axis Indeed, at gain K =.796, the root locus crosses the imaginary axis in the Matlab simulation as well. d. The value of gain for which the closed loop transfer function will have a pole on the real axis at.5. If a pole is on the real axis at.5, then it should be on the root locus, which means that s =.5 should satisfy the following relation: K(s +) + =. s(s +)(s +)(s +) s=.5 Hence, Imag Axis 5 (.5)(.5)(.5)(.5) K = =...5 Editor (C) Note that one pink dot is indeed at s =.5 for this value of the gain K =.. 5. Nise problem from chapter 8, page 8. Answer: 5

6 The open loop transfer function is a. Sketch the root locus. b. K for % overshoot 5 K G(s) =. (s +)(s +)(s +) To find K for % overshoot, we have to identify the location of poles that generates % overshoot. Then using the values of the poles and + KG(s) =, we can compute the gain K at the poles. For the first step, we have to find the location of the poles which produce % overshoot. The damping ratio ζ determines how much overshoot is generated. You can estimate the damping ratio ζ from the root locus, using the relation ζ =cos θ, where θ is the angle subtended from the pole to the origin of the s plane. (Please refer to pp. of lecture note.) Therefore, the poles that yield % overshoot should be the straight line cos θ = ζ for the value of ζ that yields % OS. From the relation between %OS and damping ratio, we find Also, cos θ = ζ θ = ±6.9. ln (%OS/) ζ = =.559. π +ln (%OS/) Next step is to find the location of the pole at ζ =.559. Matlab s sisotool provides you damping ratio, pole location, and gain K. (Also it gives the natural frequency ω n, which can be used in the next problem.) It turns out that ζ =.559, p = ±j.69, and K =9.7. You will find approximate values due to tuning error of the gain. If you want to find the exact location of the pole analytically, you can try as follows: Let s assume that a pole p = σ + jω. From the damping ratio, 6

7 θ =6.9, which means ω =tan(6.9 ) σ =.95σ. To be on the root locus, the pole σ + jω should satisfy + KG( σ + jω) =. We expand this equation as follows: s +6s +s +(6+ K) =. ( σ + jω) +6( σ + jω) + ( σ + jω)+(6 + K) =. ( σ +σ jω +σω ω j)+6(σ σωj ω ) σ +ωj +6+ K =. Separating real and imaginary parts as σ +σω +6σ 6ω σ +6+ K =, σ ω ω σω +ω =, and using ω =.95σ, we find σ =.866 and ω =.69 from the second equation. Using these values in the first equation, we find K = Editor (C) Imag Axis Note that the pink line indicates ζ =.559, cos θ = ζ(θ = ±6.9 ). c. The settling time, peak time for K found in (b) To estimate the settling time and peak time, we have to know the damping ratio and natural frequency. From the previous question, we found the damping ratio ζ (and natural frequency ω n as well if you used Matlab sisotool). If you solved (b) analytically, you know the pole p =.866 ± j.69. The imaginary part of the pole corresponds the damped natural frequency ω d, which is ω d = ω n ζ. Hence, the natural frequency ω n.9 (rad/s). The settling time T s /(ζω ( n )=.6(s). ) The peak time T p = π/ ω n ζ.(s). 7

8 .8 Step Response Amplitude Time (sec) Note that the setting time and peak time computed by the second order approximation well agree with the Matlab result plotted above. d. The locations of higher order poles for K found in (b) If the pole is purely real, + KG(s) = for s = σ. Solving +(σ +)(σ +)(σ +) = 9.7 numerically, we find σ =.68. Since , therefore the second order assumption is valid in this case. e. The range of K for stability We find the imaginary axis crossings by setting KG(jω) = (rule 7). Solving ω j 6ω +ωj +6+ K =, and separating to real and imaginary parts respectively, ω +ω =, 6ω +6+ K =, we find ω = and hence K = 6 6 = 6. At K = 6, the system is marginally stable and for K > 6 it becomes unstable. 8

### 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

### Lecture 1 Root Locus

Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the

### School of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:

Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See

### Example on Root Locus Sketching and Control Design

Example on Root Locus Sketching and Control Design MCE44 - Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We

### 7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)

### ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled

### 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

### 1 (20 pts) Nyquist Exercise

EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically

### 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

### Chapter 7 : Root Locus Technique

Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci

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

1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.

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

1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot

### Systems Analysis and Control

Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering

### 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)

### Root Locus Methods. The root locus procedure

Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain

### Chapter 5 HW Solution

Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, time-invariant system. Let s see, I

### PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada

PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop

### Chemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University

Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this

### Goals for today 2.004

Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation

### Alireza Mousavi Brunel University

Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched

### 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

### 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

### Solutions to Skill-Assessment Exercises

Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part

### ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]

ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative

### I What is root locus. I System analysis via root locus. I How to plot root locus. Root locus (RL) I Uses the poles and zeros of the OL TF

EE C28 / ME C34 Feedback Control Systems Lecture Chapter 8 Root Locus Techniques Lecture abstract Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley

### An Internal Stability Example

An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several pole-zero cancellations between the controller and

### 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.

### Notes for ECE-320. Winter by R. Throne

Notes for ECE-3 Winter 4-5 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................

### 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

### Performance of Feedback Control Systems

Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steady-state Error and Type 0, Type

. The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer

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

Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position

### Transient Response of a Second-Order System

Transient Response of a Second-Order System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a well-behaved closed-loop

### Lab # 4 Time Response Analysis

Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an

### EE3CL4: Introduction to Linear Control Systems

1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop

### 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

### Root locus 5. tw4 = 450. Root Locus S5-1 S O L U T I O N S

Root Locus S5-1 S O L U T I O N S Root locus 5 Note: All references to Figures and Equations whose numbers are not preceded by an "S" refer to the textbook. (a) Rule 2 is all that is required to find the

### 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.

### EXAMPLE PROBLEMS AND SOLUTIONS

Similarly, the program for the fourth-order transfer function approximation with T = 0.1 sec is [num,denl = pade(0.1, 4); printsys(num, den, 'st) numlden = sa4-2o0sa3 + 1 80O0sA2-840000~ + 16800000 sa4

### 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

### MODERN CONTROL SYSTEMS

MODERN CONTROL SYSTEMS Lecture 1 Root Locu Emam Fathy Department of Electrical and Control Engineering email: emfmz@aat.edu http://www.aat.edu/cv.php?dip_unit=346&er=68525 1 Introduction What i root locu?

### The Frequency-Response

6 The Frequency-Response Design Method A Perspective on the Frequency-Response Design Method The design of feedback control systems in industry is probably accomplished using frequency-response methods

### CDS 101: Lecture 8.2 Tools for PID & Loop Shaping

CDS : Lecture 8. Tools for PID & Loop Shapig Richard M. Murray 7 November 4 Goals: Show how to use loop shapig to achieve a performace specificatio Itroduce ew tools for loop shapig desig: Ziegler-Nichols,

### Lecture 15 Nyquist Criterion and Diagram

Lecture Notes of Control Systems I - ME 41/Analysis and Synthesis of Linear Control System - ME86 Lecture 15 Nyquist Criterion and Diagram Department of Mechanical Engineering, University Of Saskatchewan,

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

Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 First-Order Specs: Step : Pole Real inputs contain

### Dynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.

Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control

### Homework Assignment 3

ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full

### CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of

### 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

### PM diagram of the Transfer Function and its use in the Design of Controllers

PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude

### FEEDBACK CONTROL SYSTEMS

FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control

### 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 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

### Robust Performance Example #1

Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants

### Professor Fearing EE C128 / ME C134 Problem Set 2 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

Professor Fearing EE C128 / ME C134 Problem Set 2 Solution Fall 21 Jansen Sheng and Wenjie Chen, UC Berkeley 1. (15 pts) Partial fraction expansion (review) Find the inverse Laplace transform of the following

### MAK 391 System Dynamics & Control. Presentation Topic. The Root Locus Method. Student Number: Group: I-B. Name & Surname: Göksel CANSEVEN

MAK 391 System Dynamics & Control Presentation Topic The Root Locus Method Student Number: 9901.06047 Group: I-B Name & Surname: Göksel CANSEVEN Date: December 2001 The Root-Locus Method Göksel CANSEVEN

### The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location.

Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location. Mapping. If we take a complex number

### 16.31 Homework 2 Solution

16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p

### ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =

ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.

### 9/9/2011 Classical Control 1

MM11 Root Locus Design Method Reading material: FC pp.270-328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),

### EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8- am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total

### Transfer func+ons, block diagram algebra, and Bode plots. by Ania- Ariadna Bae+ca CDS Caltech 11/05/15

Transfer func+ons, block diagram algebra, and Bode plots by Ania- Ariadna Bae+ca CDS Caltech 11/05/15 Going back and forth between the +me and the frequency domain (1) Transfer func+ons exist only for

### 2.010 Fall 2000 Solution of Homework Assignment 8

2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)

### 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

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

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control DC Motor Control Trainer (DCMCT) Student Manual Table of Contents 1 Laboratory Objectives1 2 References1 3 DCMCT Plant

### 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

### 12.7 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

### Autonomous Mobile Robot Design

Autonomous Mobile Robot Design Topic: Guidance and Control Introduction and PID Loops Dr. Kostas Alexis (CSE) Autonomous Robot Challenges How do I control where to go? Autonomous Mobile Robot Design Topic:

### Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10

Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:

### 2.004 Dynamics and Control II Spring 2008

MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 8 I * * Massachusetts

### Richiami di Controlli Automatici

Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici

### Laboratory handouts, ME 340

Laboratory handouts, ME 340 This document contains summary theory, solved exercises, prelab assignments, lab instructions, and report assignments for Lab 4. 2014-2016 Harry Dankowicz, unless otherwise

### 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

### 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

### FREQUENCY-RESPONSE ANALYSIS

ECE450/550: Feedback Control Systems. 8 FREQUENCY-RESPONSE ANALYSIS 8.: Motivation to study frequency-response methods Advantages and disadvantages to root-locus design approach: ADVANTAGES: Good indicator

### Problem Set 5 Solutions 1

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski Problem Set 5 Solutions The problem set deals with Hankel

### Control Systems II. ETH, MAVT, IDSC, Lecture 4 17/03/2017. G. Ducard

Control Systems II ETH, MAVT, IDSC, Lecture 4 17/03/2017 Lecture plan: Control Systems II, IDSC, 2017 SISO Control Design 24.02 Lecture 1 Recalls, Introductory case study 03.03 Lecture 2 Cascaded Control

### Linear Control Systems Solution to Assignment #1

Linear Control Systems Solution to Assignment # Instructor: H. Karimi Issued: Mehr 0, 389 Due: Mehr 8, 389 Solution to Exercise. a) Using the superposition property of linear systems we can compute the

### Step Response for the Transfer Function of a Sensor

Step Response f the Transfer Function of a Sens G(s)=Y(s)/X(s) of a sens with X(s) input and Y(s) output A) First Order Instruments a) First der transfer function G(s)=k/(1+Ts), k=gain, T = time constant

### Proportional, Integral & Derivative Control Design. Raktim Bhattacharya

AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University

### LECTURE 21: Butterworh & Chebeyshev BP Filters. Part 1: Series and Parallel RLC Circuits On NOT Again

LECTURE : Butterworh & Chebeyshev BP Filters Part : Series and Parallel RLC Circuits On NOT Again. RLC Admittance/Impedance Transfer Functions EXAMPLE : Series RLC. H(s) I out (s) V in (s) Y in (s) R Ls

### Lecture 14 - Using the MATLAB Control System Toolbox and Simulink Friday, February 8, 2013

Today s Objectives ENGR 105: Feedback Control Design Winter 2013 Lecture 14 - Using the MATLAB Control System Toolbox and Simulink Friday, February 8, 2013 1. introduce the MATLAB Control System Toolbox

### Root Locus. 1 Review of related mathematics. Ang Man Shun. October 30, Complex Algebra in Polar Form. 1.2 Roots of a equation

Root Locus Ang Man Shun October 3, 212 1 Review of relate mathematics 1.1 Complex Algebra in Polar Form For a complex number z, it can be expresse in polar form as z = re jθ 1 Im z Where r = z, θ = tan.

### Course Outline. Higher Order Poles: Example. Higher Order Poles. Amme 3500 : System Dynamics & Control. State Space Design. 1 G(s) = s(s + 2)(s +10)

Amme 35 : System Dynamics Control State Space Design Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the s-plane

### Essence of the Root Locus Technique

Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general set-up, namely for the case when the closed-loop

### Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -

### Ver 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)

Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit.

### Pitch Rate CAS Design Project

Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability

### Analysis and Design of Control Systems in the Time Domain

Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.

### Reglerteknik Allmän Kurs. Del 2. Lösningar till Exempelsamling. Läsår 2015/16

Reglerteknik Allmän Kurs Del Lösningar till Exempelsamling Läsår 5/6 Avdelningen för Reglerteknik, KTH, SE 44 Stockholm, SWEDEN AUTOMATIC CONTROL COMMUNICATION SYSTEMS LINKÖPINGS UNIVERSITET Reglerteknik

### DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2

### Signals and Systems. Problem Set: The z-transform and DT Fourier Transform

Signals and Systems Problem Set: The z-transform and DT Fourier Transform Updated: October 9, 7 Problem Set Problem - Transfer functions in MATLAB A discrete-time, causal LTI system is described by the

### Compensator Design for Helicopter Stabilization

Available online at www.sciencedirect.com Procedia Technology 4 (212 ) 74 81 C3IT-212 Compensator Design for Helicopter Stabilization Raghupati Goswami a,sourish Sanyal b, Amar Nath Sanyal c a Chairman,

### Implementation of a Communication Satellite Orbit Controller Design Using State Space Techniques

ASEAN J Sci Technol Dev, 29(), 29 49 Implementation of a Communication Satellite Orbit Controller Design Using State Space Techniques M T Hla *, Y M Lae 2, S L Kyaw 3 and M N Zaw 4 Department of Electronic

### ME 375 System Modeling and Analysis. Homework 11 Solution. Out: 18 November 2011 Due: 30 November 2011 = + +

Out: 8 November Due: 3 November Problem : You are given the following system: Gs () =. s + s+ a) Using Lalace and Inverse Lalace, calculate the unit ste resonse of this system (assume zero initial conditions).

### Lab Experiment 2: Performance of First order and second order systems

Lab Experiment 2: Performance of First order and second order systems Objective: The objective of this exercise will be to study the performance characteristics of first and second order systems using

### School of Mechanical Engineering Purdue University. ME375 Feedback Control - 1

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