ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "ME 375 Final Examination Thursday, May 7, 2015 SOLUTION"

Transcription

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

2 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 electro-mechanical equations for an electric motor (ignoring armature inductance and armature damping) are given by: J!ω K T i a τ L i a + K T ω e i t where ω is the rotation rate of the motor, J is the mass moment of inertia of the armature, K T is the motor constant, i a is the armature current, τ L is the applied motor torque, e i ( t) is the input voltage applied to the armature and is the armature resistance. An electric motor governed by the above equations is used to propel a cart along a horizontal roadway. In this application, a disk attached to the output shaft of the motor drives one of the rear wheels of the cart through a no-slip frictional contact. The cart moves with no slipping of the wheels on the roadway and experiences a drag force of f D cv to the right as it moves to the left, where v is the speed of the cart to the left. Let M represent the total mass of the motor and cart. Ignore the mass of all of the wheels. For this problem: a) Draw free body diagrams of the driven rear wheel and of the total system. From these develop the differential equation of motion of the cart in terms of the speed v. b) Develop the transfer function relating the input voltage to the cart speed, G( s) V ( s) / E i ( s). c) The motor is given a step input voltage e i (t) e 0 h(t). Determine the steady-state speed attained by the cart as it moves along the roadway.

3 SOLUTION r F D O F M Oy O O x Let W be the radius of the rear wheels. From FBD of the wheel: f f M O F M W f W 0 f F M From the FBD of the cart/wheel: F f F D M!v F M τ L M!v + cv r (1) Taking LT of (1) gives: T L (s) r( Ms + c)v (s) (2) Taking LT of the motor equations gives: JsΩ K T I a T L (3) E i I a + K T Ω (4) For kinematics, we have: ω W r ω Ω W r Ω W W v W ω W V W Ω W rω Ω V / r (5) where ω W is the angular speed of the driven wheel. Combining above gives: (4) I a E i K T Ω (3),(5) & (6) Js V r K T ( E i r K T V ) r 2 ( Ms + c)v E i K T V / r r ( Ms + c )V JsV K T ( J + r 2 M )s + r 2 2 c + K T V K T re i J + Mr 2 And: G(s) V (s) E i (s) K T r / ( J + Mr 2 )s + r 2 c + K 2 T / From FVT: v ss lim sv (s) s 0 lim s e 0 G(s) K G(0) T re 0 s 0 s r 2 2 c + K T!v + r2 c + K T 2 / v K T re i (t) / (6)

4 POBLEM 2 (25%) plant ( s) K ( s) Y ( s) Consider the proportional feedback control system shown above. The plant transfer function s has a static gain of 10, has no zeros and has the three poles shown below. a) Determine the plant transfer function s. b) For the closed-loop root locus with K 0 : i. Determine the K asymptotes and their real-axis intersections. ii. Determine the breakaway/breakin (re-entry) points on the real axis. iii. Determine the departure angles from the open-loop poles. iv. Determine the values of K, if any, at which the root locus crosses into the right-half plane (unstable). v. Sketch the root locus on the plot provided below. vi. For what range of values for the gain K, if any, would the closed-loop response have a 2% settling time t s < 0.5AND a percent overshoot of %OS < 4% for a step input? K θ K 0 p2 p 2 11/ 3 p 3 K 0 θ p2 60 K 0 60 p K

5 SOLUTION A (s) ( s p 1 ) s p 2 A ( s +1 ) s + 5 Since G(0) 10, we have: ( s p 3 ) A ( s +1) ( s j) ( s j) (0) 10 A 29 A 290 A s 3 +11s s + 29 N(s) D(s) ule #2: The L has starting points at the three open-loop poles and end points at infinity. ule #4: The L exists on the real axis to the left of the real pole p 1 1. ule #5: The asymptotes have real axis intercepts and angles of: p i z i 1+ ( j) + ( 5 2 j) 0 σ 0 11 N P N z θ k ( 2k 1) ( 2k 1) 180 ( 2k 1)60 60,180,300 N P N z 3 0 ule #6: BA/E points on the real axis are found from: N dd ds D dn A( 3s s + 39) 0 ds s 1,2 22 ± 222 (4)(3)(39) (2)(3) 22 ± 4 6 3, 13 3 ule #7: From the general equation: ( s z 1 ) ( s p 1 ) + ( s p 2 ) 180 we can write the following for OL pole p 1 : j +1 +θ p2 + ( p 2 p 3 ) p 2 p 1 +θ p2 + ( j j) +θ p2 + ( 4 j) j 180 tan θ p2 + tan θ p θ p2 63.4

6 ule #8: The characteristic equation for the closed-loop system is: 0 D(s) + KN(s) s 3 +11s s K Substituting s jω into the CLCE gives: 3 +11( jω ) ( jω ) K 0 jω jω 3 11ω 2 + j(39ω ) K K 11ω 2 + j( 39 ω 2 )ω Imag : ( 39 ω 2 )ω 0 ω 2 0,39 eal : K 11ω 2 0 K ω ,1.38 For t s ζω n ζω n 8. It is not possible to have all of the roots to the left of 8. Therefore, it is not possible to meet the settling time criteria.

7 POBLEM 3 (25%) A plant with a transfer function s 1 s s + 1 is to be controlled with unity PD feedback control where the controller has a transfer function G C ( s) K P + K D s. a) Derive the closed-loop transfer function and closed-loop characteristic equation for this control system. b) Determine the gain values K P and K D corresponding to a 2% settling time of t s 4 and a %OS 4 for a step input. c) If K P 4, what value of the gain K D produces the smallest possible 2% settling time for a step input? SOLUTION The CL transfer function and characteristic equation are given by: G CL (s) G C 1+ G C D CL (s) 1+ G C 1+ K P + K D s s s +1 0 s 2 + ( 1+ K D )s + K P 0 respectively. The general form of the characteristic equation for a second-order system is: s 2 + 2ζω n s + ω n 2 0 For %OS 4 andt s 4, we have: %OS 100e ζπ / 1 ζ 2 ζ + π 2 ln %OS / 100 ln 2 %OS / 100 t s ζω ζω n 1 ω n n Balancing coefficients in the CLCE s: K P ω n K D 2ζω n K D 2ζω n 1 2 ( 1) π 2 ln 0.04 ln The smallest possible settling time occurs when ζ 1. Since ω n K P 4 2, we can write: K D 2ζω n

8 POBLEM 4 PAT A (10%) Consider the electrical circuit shown below. Develop an expression for the impedance V i ( s) I ( s) Z s. Express your final answer as a single fraction. a b i v i (t) + - L d e C SOLUTION Inductor be and capacitor de are connected in series. That connection is in parallel with resistor bd. In turn, that is in series with resistor ab. Therefore: Z be Ls and: Z de 1 Cs Z bd Z bde Z bd Z be + Z de + 1 Ls +1/ Cs 1 + Cs LCs 2 +1 LCs Cs LCs 2 + LCs 2 Z bde + LCs Cs which leads to: LCs 2 + Z(s) Z a + Z bde + LCs 2 + Cs +1 2LCs Cs + 2 LCs 2 + Cs +1

9 POBLEM 4 (continued) PAT B (10%) s plant K ( s) Y ( s) The closed-loop transfer function for the above system has the following characteristic equation: D CL (s) s 4 + 5Ks 3 + (3+ K)s 2 + 6s K 0 a) Determine the plant transfer function (s). b) Determine the magnitude of the steady-state response of the closed-loop system to a unit harmonic input r(t) sin5t for K 1. SOLUTION 0 s 4 + 5Ks 3 + (3+ K)s 2 + 6s K s 4 + 3s 2 + 6s K 5s 3 + s s 3 + s K s 4 + 3s 2 + 6s K (s) The CL transfer function is written as: For K 1: G CL (s) K (s) K 5s 3 + s K (s) s 4 + 3s 2 + 6s K 5s 3 + s s 3 + s 2 +1 G CL (s) s 4 + 5s 3 + 4s 2 + 6s + 5 Magnitude of steady-state response for harmonic input: G CL (5 j) 5(5 j) 3 + (5 j) 2 +1 (5 j) 4 + 5(5 j) 3 + 4(5 j) 2 + 6(5 j) j j

10 POBLEM 4 (continued) PAT C (5%) G 1 (s) s Y ( s) G 2 (s) Consider the feedback control system above, with: s 2 +1 G 1 (s) s 3 + 3s +1 G 2 (s) s s + 4 Determine the single input-output differential equation governing the output y(t) for an input r(t). What is the order of the closed-loop system? SOLUTION From block diagram: Y G 1 G 2 Y Y G 1 (s 2 +1) / (s 3 + 3s +1) 1+ G 1 G 2 1+ s(s 2 +1) / (s 3 + 3s +1)(s + 4) (s 2 +1)(s + 4) (s 3 + 3s +1)(s + 4) + s(s 2 +1) s 3 + 4s 2 + s + 4 s 4 + 5s 3 + 3s 2 +14s + 4 ( s 4 + 5s 3 + 3s 2 +14s + 4)Y ( s 3 + 4s 2 + s + 4) Performing an inverse LT on the above:!!!! y + 5!!! y + 3!!y +14!y + 4y!!! r (t) + 4!! r(t) +!r(t) + 4r(t) Fourth-order system.

ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION

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

More information

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

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

More information

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

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

More information

Homework 7 - Solutions

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

More information

Software Engineering 3DX3. Slides 8: Root Locus Techniques

Software Engineering 3DX3. Slides 8: Root Locus Techniques Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007

More information

If you need more room, use the backs of the pages and indicate that you have done so.

If you need more room, use the backs of the pages and indicate that you have done so. EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty

More information

FEEDBACK CONTROL SYSTEMS

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

More information

Module 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions

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

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

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

More information

R10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1

R10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1 Code No: R06 R0 SET - II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry

More information

Transient Response of a Second-Order System

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

More information

Example on Root Locus Sketching and Control Design

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

More information

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 K(s +1)(s +2) G(s) =. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:

More information

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

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

More information

Introduction to Feedback Control

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

More information

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

(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.

More information

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

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

More information

Control Systems. University Questions

Control Systems. University Questions University Questions UNIT-1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write

More information

AN INTRODUCTION TO THE CONTROL THEORY

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

More information

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 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

More information

EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =

EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) = 1. Pole Placement Given the following open-loop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the state-variable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback

More information

Root Locus Techniques

Root Locus Techniques 4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since

More information

Bangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory

Bangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system

More information

Course Summary. The course cannot be summarized in one lecture.

Course Summary. The course cannot be summarized in one lecture. Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: Steady-State Error Unit 7: Root Locus Techniques

More information

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

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)

More information

Outline. Classical Control. Lecture 5

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?

More information

ECEN 420 LINEAR CONTROL SYSTEMS. Lecture 6 Mathematical Representation of Physical Systems II 1/67

ECEN 420 LINEAR CONTROL SYSTEMS. Lecture 6 Mathematical Representation of Physical Systems II 1/67 1/67 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 6 Mathematical Representation of Physical Systems II State Variable Models for Dynamic Systems u 1 u 2 u ṙ. Internal Variables x 1, x 2 x n y 1 y 2. y m Figure

More information

Conventional Paper-I-2011 PART-A

Conventional Paper-I-2011 PART-A Conventional Paper-I-0 PART-A.a Give five properties of static magnetic field intensity. What are the different methods by which it can be calculated? Write a Maxwell s equation relating this in integral

More information

AMME3500: System Dynamics & Control

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

More information

Rotational Systems, Gears, and DC Servo Motors

Rotational Systems, Gears, and DC Servo Motors Rotational Systems Rotational Systems, Gears, and DC Servo Motors Rotational systems behave exactly like translational systems, except that The state (angle) is denoted with rather than x (position) Inertia

More information

Control of Manufacturing Processes

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 =

More information

Positioning Servo Design Example

Positioning Servo Design Example Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pick-and-place robot to move the link of a robot between two positions. Usually

More information

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27 1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect

More information

Time Response Analysis (Part II)

Time Response Analysis (Part II) Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary

More information

Discrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture

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},

More information

Performance of Feedback Control Systems

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

More information

Compensator Design to Improve Transient Performance Using Root Locus

Compensator Design to Improve Transient Performance Using Root Locus 1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning

More information

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

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

More information

Systems Analysis and Control

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

More information

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

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

More information

School of Engineering Faculty of Built Environment, Engineering, Technology & Design

School of Engineering Faculty of Built Environment, Engineering, Technology & Design Module Name and Code : ENG60803 Real Time Instrumentation Semester and Year : Semester 5/6, Year 3 Lecture Number/ Week : Lecture 3, Week 3 Learning Outcome (s) : LO5 Module Co-ordinator/Tutor : Dr. Phang

More information

Lab 3: Quanser Hardware and Proportional Control

Lab 3: Quanser Hardware and Proportional Control Lab 3: Quanser Hardware and Proportional Control The worst wheel of the cart makes the most noise. Benjamin Franklin 1 Objectives The goal of this lab is to: 1. familiarize you with Quanser s QuaRC tools

More information

Lecture 1 Root Locus

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

More information

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

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

More information

MAS107 Control Theory Exam Solutions 2008

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

More information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

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

More information

Chapter 12. Feedback Control Characteristics of Feedback Systems

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

More information

Introduction to Process Control

Introduction to Process Control Introduction to Process Control For more visit :- www.mpgirnari.in By: M. P. Girnari (SSEC, Bhavnagar) For more visit:- www.mpgirnari.in 1 Contents: Introduction Process control Dynamics Stability The

More information

Design of a Lead Compensator

Design of a Lead Compensator Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc - Funded by MHRD The Lecture Contains Standard Forms of

More information

Chapter three. Mathematical Modeling of mechanical end electrical systems. Laith Batarseh

Chapter three. Mathematical Modeling of mechanical end electrical systems. Laith Batarseh Chapter three Mathematical Modeling of mechanical end electrical systems Laith Batarseh 1 Next Previous Mathematical Modeling of mechanical end electrical systems Dynamic system modeling Definition of

More information

EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO

EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total

More information

Dynamic circuits: Frequency domain analysis

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

More information

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS

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

More information

Due Wednesday, February 6th EE/MFS 599 HW #5

Due Wednesday, February 6th EE/MFS 599 HW #5 Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]

More information

Exam. 135 minutes + 15 minutes reading time

Exam. 135 minutes + 15 minutes reading time Exam January 23, 27 Control Systems I (5-59-L) Prof. Emilio Frazzoli Exam Exam Duration: 35 minutes + 5 minutes reading time Number of Problems: 45 Number of Points: 53 Permitted aids: Important: 4 pages

More information

Systems Analysis and Control

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

More information

Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)

Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar

More information

Chapter 2 SDOF Vibration Control 2.1 Transfer Function

Chapter 2 SDOF Vibration Control 2.1 Transfer Function Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:

More information

Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax:

Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax: Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 02-760-4253 Fax:02-760-4435 Introduction In this lesson, you will learn the following : The

More information

Dr Ian R. Manchester

Dr Ian R. Manchester Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign

More information

Alireza Mousavi Brunel University

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

More information

Proportional plus Integral (PI) Controller

Proportional plus Integral (PI) Controller Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on

More information

INTRODUCTION TO DIGITAL CONTROL

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

More information

Root Locus Design Example #3

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

More information

Controls Problems for Qualifying Exam - Spring 2014

Controls Problems for Qualifying Exam - Spring 2014 Controls Problems for Qualifying Exam - Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function

More information

Lab 3: Model based Position Control of a Cart

Lab 3: Model based Position Control of a Cart I. Objective Lab 3: Model based Position Control of a Cart The goal of this lab is to help understand the methodology to design a controller using the given plant dynamics. Specifically, we would do position

More information

Introduction to Control (034040) lecture no. 2

Introduction to Control (034040) lecture no. 2 Introduction to Control (034040) lecture no. 2 Leonid Mirkin Faculty of Mechanical Engineering Technion IIT Setup: Abstract control problem to begin with y P(s) u where P is a plant u is a control signal

More information

Notes for ECE-320. Winter by R. Throne

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...................

More information

Linear Control Systems Solution to Assignment #1

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

More information

JRE SCHOOL OF Engineering

JRE SCHOOL OF Engineering JRE SCHOOL OF Engineering Class Test-1 Examinations September 2014 Subject Name Electromechanical Energy Conversion-II Subject Code EEE -501 Roll No. of Student Max Marks 30 Marks Max Duration 1 hour Date

More information

Chapter 5 HW Solution

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

More information

Mechatronics Engineering. Li Wen

Mechatronics Engineering. Li Wen Mechatronics Engineering Li Wen Bio-inspired robot-dc motor drive Unstable system Mirko Kovac,EPFL Modeling and simulation of the control system Problems 1. Why we establish mathematical model of the control

More information

Dynamic Compensation using root locus method

Dynamic Compensation using root locus method CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the

More information

Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current.

Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current. Inductance Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current. Basis of the electrical circuit element called an

More information

EE C128 / ME C134 Final Exam Fall 2014

EE C128 / ME C134 Final Exam Fall 2014 EE C128 / ME C134 Final Exam Fall 2014 December 19, 2014 Your PRINTED FULL NAME Your STUDENT ID NUMBER Number of additional sheets 1. No computers, no tablets, no connected device (phone etc.) 2. Pocket

More information

Linear Systems Theory

Linear Systems Theory ME 3253 Linear Systems Theory Review Class Overview and Introduction 1. How to build dynamic system model for physical system? 2. How to analyze the dynamic system? -- Time domain -- Frequency domain (Laplace

More information

Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits

Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the timevarying

More information

Chapter 7 : Root Locus Technique

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

More information

Control of Manufacturing Processes

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

More information

EE102 Homework 2, 3, and 4 Solutions

EE102 Homework 2, 3, and 4 Solutions EE12 Prof. S. Boyd EE12 Homework 2, 3, and 4 Solutions 7. Some convolution systems. Consider a convolution system, y(t) = + u(t τ)h(τ) dτ, where h is a function called the kernel or impulse response of

More information

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies. SET - 1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies..

More information

First and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015

First and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015 First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage

More information

Solutions to Skill-Assessment Exercises

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

More information

ME 230: Kinematics and Dynamics Spring 2014 Section AD. Final Exam Review: Rigid Body Dynamics Practice Problem

ME 230: Kinematics and Dynamics Spring 2014 Section AD. Final Exam Review: Rigid Body Dynamics Practice Problem ME 230: Kinematics and Dynamics Spring 2014 Section AD Final Exam Review: Rigid Body Dynamics Practice Problem 1. A rigid uniform flat disk of mass m, and radius R is moving in the plane towards a wall

More information

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

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.

More information

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONICS ENGINEERING

More information

Implementation Issues for the Virtual Spring

Implementation Issues for the Virtual Spring Implementation Issues for the Virtual Spring J. S. Freudenberg EECS 461 Embedded Control Systems 1 Introduction One of the tasks in Lab 4 is to attach the haptic wheel to a virtual reference position with

More information

2.010 Fall 2000 Solution of Homework Assignment 1

2.010 Fall 2000 Solution of Homework Assignment 1 2. Fall 2 Solution of Homework Assignment. Compact Disk Player. This is essentially a reprise of Problems and 2 from the Fall 999 2.3 Homework Assignment 7. t is included here to encourage you to review

More information

1 (20 pts) Nyquist Exercise

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

More information

Chapter 32. Inductance

Chapter 32. Inductance Chapter 32 Inductance Joseph Henry 1797 1878 American physicist First director of the Smithsonian Improved design of electromagnet Constructed one of the first motors Discovered self-inductance Unit of

More information

MECHATRONICS ENGINEERING TECHNOLOGY. Modeling a Servo Motor System

MECHATRONICS ENGINEERING TECHNOLOGY. Modeling a Servo Motor System Modeling a Servo Motor System Definitions Motor: A device that receives a continuous (Analog) signal and operates continuously in time. Digital Controller: Discretizes the amplitude of the signal and also

More information

PHYSICS 218 Final Exam Fall, 2014

PHYSICS 218 Final Exam Fall, 2014 PHYSICS 18 Final Exam Fall, 014 Name: Signature: E-mail: Section Number: No calculators are allowed in the test. Be sure to put a box around your final answers and clearly indicate your work to your grader.

More information

Introduction to Controls

Introduction to Controls EE 474 Review Exam 1 Name Answer each of the questions. Show your work. Note were essay-type answers are requested. Answer with complete sentences. Incomplete sentences will count heavily against the grade.

More information

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 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

More information

PID controllers. Laith Batarseh. PID controllers

PID controllers. Laith Batarseh. PID controllers Next Previous 24-Jan-15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time

More information

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0. 6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)

More information