# Homework 11 Solution - AME 30315, Spring 2015

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 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 2 s 2 ` 2ζω n s ωn 2 We will use k, ζ, and ω n values for our system found from the hanging configuration so that the transfer function is: Our pole locations are: Θpsq Ipsq p6.23q 2 s 2 ` 2p0.039qp6.23qs p6.23q 2 s 1,2 5.99, 6.48 Rule The root locus starts at the open-loop poles pk 0q. By Rule 9.2 we know for our case (no zeros), as k Ñ 8, the root locus will grow unbounded for both poles.

2 2 Rule 9.3 gives the asymptotes along which the root locus will grow unbounded. θ n 180 ` n 360 n z n p θ ` p0q θ ` p1q Next, Rule 9.4 gives where the asymptotes intersect the real axis. s int ř nz i 1 z i řn p i 1 p i n z n p s int 0 p q By Rule 9.5 we recognize that on the real axis the root locus exists to the left of an odd number of poles and zeros. Lastly, Rule 9.6 gives the angle at which a branch of the root locus leaves a pole p j ÿn z =ps p j q =pp j z i q i 1 n ÿ p i 1,i j p r180 ` 0s =pp j p i q 180 p r0 s

3 3 We can make a complete root locus sketch [3 pts, must show steps] Thus, we can see that a proportional controller will be able to attain a stable closed-loop system. We will use the magnitude criterion to find the minimum k which will give us a stable closed loop system: k 1 Gpsq Therefore: Gpsq s 0 ˆk Gpsq s 0 p0.248qp q ś nz ś nz i 1 s z i i 1 s p i k 1 Gpsq For any k ą we will have a stable closed loop system [3 pts]. To determine if p-control is sufficient to meet our design criteria of %OS ă 25% and t r ă 1.5 sec, we want to map this region in the complex plane. We use the rise time approximation formula:

4 4 t r «1.8 ω n Ñ ω n ą 1.2 And using Fig on p. 361 of Goodwine, we find that the damping ratio must be: Plotting, in Matlab [2 pts] ζ ą 0.4. We can see from the trajectories of the root locus branches that there does not exist a value of k such that all the closed-loop poles fall within the design criteria region simultaneously (unshaded region) [2 pts]. 1 % AME HW 11 4/19/ % Problem G = zpk([],[5.99,-6.48],0.248*6.23ˆ2) 6 rltool

5 5 Problem 2 [25/25 pts] We choose a point within the specified region which satisfies the design criteria, knowing that: s desired ζω n iω n a 1 ζ2 ζω n iω d We choose ζ 0.7 and ω n 2, so that our desired point is: s «1.4 ` 1.43i which meets our design criteria [2 pts]. We want to design a lead controller to make the root locus go through the desired point in the complex plane. A lead controller has the form: Then for the root locus analysis, we define: And design k such that k P p0`, 8q. K lead psq k s ` z s ` p. Gpsq s ` z 0.248p6.23q 2 s ` p s 2 ` 2p0.039qp6.23qs p6.23q 2 We choose a fixed z-value, and we need to adjust p to make the root locus go through our desired point. Some rules of thumb include that z can be placed at z ω n ; however, this places the zero in between the poles of our plant. We chose z 8 to the left of the stable plant pole. n z n ÿ ÿ p =Gpsq =ps z i q =ps p i q 180 ` 360 pn 1q i 1 i 1 We find the angles referring to Fig. 1.

6 6 Im sdesired θ X -p 1 O -z θ θ θ4 2 3 X s2 X s1 Re Fig. 1. Vector representation of G(s). Then, θ 2 pθ 1 ` θ 3 ` θ 4 q 180 ` 360 pn 1q where θ 1 is an unknown function of p, so we solve for θ 1 : θ 1 θ 2 θ 3 θ pn 1q ˆ ˆ ˆ tan 1 ωd tan 1 ω d ˆ180 ` tan 1 ω d pn 1q z ζω n 6.48 ζω n pζω n ` 5.99q So From trig, Our controller has the form where k is still an unknown parameter. θ 1 «7.45. tan θ 1 ω d p ζω n p K lead psq k s ` 8 s ` Before finding k, we can plot the root locus. [10 pts]. By Rule 9.1, the root locus starts at the open-loop poles (k=0).

7 7 Rule 9.2, n z 1 and n p 3, therefore, the root locus will have one bounded branch and 2 unbounded branches. Rule 9.3, calculate the asymptotes to find that θ 0 90 and θ 1 `90 θ n 180 ` n360 n z n p Rule 9.4, Asymptote intersection with real axis: s int ř nz i 1 z i řn p i 1 p i n z n p z p p 6.48 ` 5.99q s int Rule 9.5, the root locus exists on the real axis to the left of an odd number of poles and zeros. Rule 9.6, we do not need because all of the open loop poles and zeros are on the real axis. Rule 9.7, Break away points from the real axis can be calculated where we let Gpsq Npsq{Dpsq ddpsq dnpsq Npsq Dpsq 0 (1) ds ds The components of this equation are: Npsq 0.248p qps ` 8q dnpsq 0.248p q ds Dpsq ps ` 12.33qps ` 6.48qps 5.99q ddpsq 3s 2 ` 12.82p2qs ds So that we find the roots of Eq. 1 using Mathematica: s i, Then, s is the only breakaway point that makes sense. We can plot our root locus by hand [5 pts, must show steps]:

8 8 Solve for k such that the root locus passes through a given location. We will use the magnitude criterion to find the minimum k which will give us a stable closed loop system: where Gpsq s 0 ˆk k 1 Gpsq ś nz ś nz i 1 s z i i 1 s p i 0.248p q a pz ζω n q 2 ` pω d q a 2 pp ζωn q 2a a p6.48 ζω n q 2 ` ωd 2 p5.99 ` ζωn q 2 ` ωd 2 Gpsq « Ñ k «6.73 [5 pts] Then, the controller by design that will achieve the desired design specifications is: K lead psq 6.73 s ` 8 s ` We verify our sketch using Matlab rltool [3 pts],

9 9 30 Root Locus Editor for Open Loop 1(OL1) Imag Axis Real Axis 1 %Construct the system transfer function 2 G=tf([0.248*6.23ˆ2],[1 2*0.039* ˆ2]) 3 4 % %Assign value for the compensator pole and zero 5 z=8; %Zero 6 p=12.33; %Pole 7 8 %Construct the lead compensator transfer function 9 K_lead=tf([1 z],[1 p]); %Plot the root locus 12 rltool(k_lead*g) We want to evaluate whether the design criteria is actually satisfied, first by evaluating our linear system. When we plot the step response of our system with the controller we designed, we see that:

10 10 14 Step Response Amplitude Time (seconds) Fig. 2. k=6.73, z=8, p=12.33 as designed following the root locus specifications. The overshoot is 4% and the rise time is 1.5 seconds which meets our design criteria. The steady-state value is off; for a step reference input, the output should fall to 1, but we did not set a design criteria for that parameter.

11 11 Problem 3 [10/10 pts] We now want to study the full nonlinear system for the inverted pendulum. Based on homework 9, we can determine the constant parameters for the plant in the provided simulink model: g l ω2 n b ml 2 2ζω n 2p0.039qp6.23q k τ ml 2 kω 2 n These are what we plug into the nonlinear plant diagram. Using our design parameters we get: [5 pts] Reference System Response 80 Theta [deg] Time [sec] This has an overshoot of: OS Ñ %OS 38% and a rise time of 0.25 seconds. This does not meet our overshoot design criteria [5 pts]. To bring the response closer to our desired gain values, we increase the gain and shift the compensator pole and zero. We tried gain k 15, z 7, and p 20.

12 Reference System Response Theta [deg] Time [sec] This decreases the overshoot so that %OS 17% and did not effect the rise time. Our design criteria is satisfied. Time Clock To Workspace2 Step K Gain s+z s+p Transfer Fcn Plant input: i(t) i(t) theta(t) Plant output: theta(t) Plant (double-click to edit) System Response Step1 Reference To Workspace1 System_Response To Workspace System Response1 1 close all 2 clear all 3 4 %Construct the system transfer function 5 G=tf([0.248*6.23ˆ2],[1 2*0.039* ˆ2]) 6 7 %Assign value for the compensator pole and zero 8 K = 6.73; 9 z = 8; %zero 10 p = 12.33; %pole 11

13 13 12 %Construct the lead compensator transfer function 13 K_lead=tf([1 z],[1 p]); rltool sim('problem3sim') %Plot system response from simulink data 20 figure(2) 21 plot(time,reference,'-b',time,system_response,'-r','linewidth',1.5) 22 legend('reference','system Response') 23 xlabel('time [sec]') 24 ylabel('theta [deg]')

### 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

### 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:

### 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)

### 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

### a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a

Root Locus Simple definition Locus of points on the s- plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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

### 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 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering

### 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

### 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

### 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

### 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

### 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

### 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)

### CHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System

CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages

### 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

### 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

### 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...................

### (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.

### 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

### 2.010 Fall 2000 Solution of Homework Assignment 7

. Fall Solution of Homework Assignment 7. Control of Hydraulic Servomechanism. We return to the Hydraulic Servomechanism of Problem in Homework Assignment 6 with additional data which permits quantitative

### 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

. 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 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

### 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

### 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

### 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

### 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

### 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

### KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open

### 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.

### KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS STAFF NAME: Mr. P.NARASIMMAN BRANCH : ECE Mr.K.R.VENKATESAN YEAR : II SEMESTER

### 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

### 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

### 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

### 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

### Chapter 6 Steady-State Analysis of Continuous-Time Systems

Chapter 6 Steady-State Analysis of Continuous-Time Systems 6.1 INTRODUCTION One of the objectives of a control systems engineer is to minimize the steady-state error of the closed-loop system response

### 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

### Analysis of SISO Control Loops

Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities

### Inverted Pendulum. Objectives

Inverted Pendulum Objectives The objective of this lab is to experiment with the stabilization of an unstable system. The inverted pendulum problem is taken as an example and the animation program gives

### Mechanical Systems Part A: State-Space Systems Lecture AL12

AL: 436-433 Mechanical Systems Part A: State-Space Systems Lecture AL Case study Case study AL: Design of a satellite attitude control system see Franklin, Powell & Emami-Naeini, Ch. 9. Requirements: accurate

### 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

### 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

### Power System Control

Power System Control Basic Control Engineering Prof. Wonhee Kim School of Energy Systems Engineering, Chung-Ang University 2 Contents Why feedback? System Modeling in Frequency Domain System Modeling in

### EEE 184: Introduction to feedback systems

EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)

### 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

### 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

### 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)

### 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

### 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

### Class 12 Root Locus part II

Class 12 Root Locus part II Revising (from part I): Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K Comple plane jω ais 0 real ais Thus,

### SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequency-domain analysis and control design (15 pt) Given is a

### 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

### 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

### Design via Root Locus

Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steady-state error (Sections

### Homework 7 Solution - AME 30315, Spring s 2 + 2s (s 2 + 2s + 4)(s + 20)

1 Homework 7 Solution - AME 30315, Spring 2015 Problem 1 [10/10 pt] Ue partial fraction expanion to compute x(t) when X 1 () = 4 2 + 2 + 4 Ue partial fraction expanion to compute x(t) when X 2 () = ( )

### 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

### D G 2 H + + D 2

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.302 Feedback Systems Final Exam May 21, 2007 180 minutes Johnson Ice Rink 1. This examination consists

### The dynamics of a Mobile Inverted Pendulum (MIP)

The dynamics of a Mobile Inverted Pendulum (MIP) 1 Introduction Saam Ostovari, Nick Morozovsky, Thomas Bewley UCSD Coordinated Robotics Lab In this document, a Mobile Inverted Pendulum (MIP) is a robotic

### 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.

### 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

### AN INTRODUCTION TO THE CONTROL THEORY

Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter

### Transfer function and linearization

Transfer function and linearization Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata Corso di Controlli Automatici, A.A. 24-25 Testo del corso:

### Vibrations: Second Order Systems with One Degree of Freedom, Free Response

Single Degree of Freedom System 1.003J/1.053J Dynamics and Control I, Spring 007 Professor Thomas Peacock 5//007 Lecture 0 Vibrations: Second Order Systems with One Degree of Freedom, Free Response Single

### Laboratory 11 Control Systems Laboratory ECE3557. State Feedback Controller for Position Control of a Flexible Joint

Laboratory 11 State Feedback Controller for Position Control of a Flexible Joint 11.1 Objective The objective of this laboratory is to design a full state feedback controller for endpoint position control

### Department of Mechanical Engineering

Department of Mechanical Engineering 2.14 ANALYSIS AND DESIGN OF FEEDBACK CONTROL SYSTEMS Fall Term 23 Problem Set 5: Solutions Problem 1: Nise, Ch. 6, Problem 2. Notice that there are sign changes in

### Represent this system in terms of a block diagram consisting only of. g From Newton s law: 2 : θ sin θ 9 θ ` T

Exercise (Block diagram decomposition). Consider a system P that maps each input to the solutions of 9 4 ` 3 9 Represent this system in terms of a block diagram consisting only of integrator systems, represented

### Digital Control Semester Project

Digital Control Semester Project Part I: Transform-Based Design 1 Introduction For this project you will be designing a digital controller for a system which consists of a DC motor driving a shaft with

### Feedback Control part 2

Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open- and closed-loop control Everything before chapter 7 are open-loop systems Transient response Design criteria Translate criteria

### 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.

### 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

### EE 370L Controls Laboratory. Laboratory Exercise #7 Root Locus. Department of Electrical and Computer Engineering University of Nevada, at Las Vegas

EE 370L Controls Laboratory Laboratory Exercise #7 Root Locus Department of Electrical an Computer Engineering University of Nevaa, at Las Vegas 1. Learning Objectives To emonstrate the concept of error

### 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.

### 1 (30 pts) Dominant Pole

EECS C8/ME C34 Fall Problem Set 9 Solutions (3 pts) Dominant Pole For the following transfer function: Y (s) U(s) = (s + )(s + ) a) Give state space description of the system in parallel form (ẋ = Ax +

### ESE319 Introduction to Microelectronics. Feedback Basics

Feedback Basics Stability Feedback concept Feedback in emitter follower One-pole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability

### 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

### Lab-Report Control Engineering. Proportional Control of a Liquid Level System

Lab-Report Control Engineering Proportional Control of a Liquid Level System Name: Dirk Becker Course: BEng 2 Group: A Student No.: 9801351 Date: 10/April/1999 1. Contents 1. CONTENTS... 2 2. INTRODUCTION...

### 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

### ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II

ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and

### Digital Pendulum Control Experiments

EE-341L CONTROL SYSTEMS LAB 2013 Digital Pendulum Control Experiments Ahmed Zia Sheikh 2010030 M. Salman Khalid 2010235 Suleman Belal Kazi 2010341 TABLE OF CONTENTS ABSTRACT...2 PENDULUM OVERVIEW...3 EXERCISE

### 1 Steady State Error (30 pts)

Professor Fearing EECS C28/ME C34 Problem Set Fall 2 Steady State Error (3 pts) Given the following continuous time (CT) system ] ẋ = A x + B u = x + 2 7 ] u(t), y = ] x () a) Given error e(t) = r(t) y(t)

### 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

### 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

### 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

### 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

### Professor Fearing EE C128 / ME C134 Problem Set 10 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

Professor Fearing EE C28 / ME C34 Problem Set Solution Fall 2 Jansen Sheng and Wenjie Chen, UC Berkeley. (5 pts) Final Value Given the following continuous time (CT) system ẋ = Ax+Bu = 5 9 7 x+ u(t), y

### 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

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

Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid

### Rotary Inverted Pendulum

Rotary Inverted Pendulum Eric Liu 1 Aug 2013 1 1 State Space Derivations 1.1 Electromechanical Derivation Consider the given diagram. We note that the voltage across the motor can be described by: e b

### 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