# Linear Control Systems Solution to Assignment #1

Size: px
Start display at page:

## Transcription

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 transfer functions from each input the the output individually and then obtain the overall transfer function by adding the individual contributions. Let s start with the transfer function from r to c. Disregard the other two inputs, i.e. set Ds) = 0 and Ns) = 0. rearranging: Cs) = Ks)Gs)Es) = Ks)Gs)Rs) Bs)) = Ks)Gs)Rs) Ks)Gs)Hs)Cs) Cs) = Ks)Gs) + Ks)Gs)Hs) Rs) In a similar fashion, considering one input at the time, for disturbance we get: and for noise: Cs) = Cs) = Gs) + Ks)Gs)Hs) Ds) Ks)Gs)Hs) + Ks)Gs)Hs) Ns) As a final result, the overall transfer function is: Cs) = [Gs)Ds) + Ks)Gs)Rs) Ks)Gs)Hs)Ns)] + Ks)Gs)Hs) b) If there is no noise and no disturbance, and Hs) =, the error signal is given by: Es) = Rs) Bs) = Rs) Cs) = Rs) KGs) + KGs) Rs) = + KGs) Rs) Linear Control Systems Solution to Assignment # Page of 0

2 c) If our input is the unit step, and if the conditions of the theorem are met, then the steady-state error is the following: e ss = lim t et) = lim s 0 ses) = lim s 0 s + KGs) s = lim s 0 + KGs) Final Value Theorem doesn t apply to E s), since it has poles on the imaginary axis. Actually, e ss doesn t exist, and e t) oscillates sinusoidally about 0. e 2ss = lim + K s 0 ss+2) = lim s 0 ss + 2) s 2 + 2s + K = 0 Solution to Exercise 2. a) Input: switch-on; output: clean clothes. Open loop. The washing machine does not feed back the cleanliness of clothes. b) Input: electrical signal from readout device e.g. laser in a CD player); output: microphone vibration. Open loop. The output audio level and quality is not used to determine the input to the speaker. c) Input: desired temperature; output: temperature. Closed loop. The vast majority of air conditioners have temperature sensors often, a simple bimetal switch) so they turn themselves on when the temperature is outside a prescribed range, off when it is inside the prescribed range. d) Manual gear train in an automobile Input: lever location set by driver; output: wheel RPM. Open loop. The wheel speed is not used in determining the gear setting though a good driver would probably apply a sense of the vehicle s present speed to decide the proper gear to apply so as not to stall or rev up the engine.) Automatic gear train in an automobile Unlike the manual gearbox, in the automatic gearbox as the name implies!) there is a feedback mechanism: the wheel RPM heats up a special fluid causing it to expand or contract. The engine gear ratio is changed due to the fluid volume change. Solution to Exercise 3. a) From force balance, you can derive the equation of motion. For simplicity, the system variable θt) is chosen with polar coordinates. Then you don t need to care about tension on the rod and centrifugal force. Free-body diagram: Assuming that the length of the rod is l, we obtain ml 2 θt) = mgl sin θt) + ft)l cos θt) Note that inertia of the mass with respect to the rotation axis is ml 2. It is a nonlinear differential equation because it has sin θt) term. If θt) is small enough to be assumed sin θt) θt) and cos θt), then it is linear equation as ml θt) = mgθt) + ft). Linear Control Systems Solution to Assignment # Page 2 of 0

3 Figure : Problem 3a) b) You have one input ft) and two outputs xt) and θt), which are related to each other. Unlike rolling with no-slipping case, you have to consider slipping because viscous friction happens when the relative velocity at the interface of two surface is nonzero. Free-body diagram of the mass: Figure 2: Problem 3b), Free-body diagram of the mass Free-body diagram of the inertia: Figure 3: Problem 3b), Free-body diagram of the inertia The viscous friction between the mass and inertia is proportional to the relative velocity ẋ r θ). From free-body diagrams, the equations of motion are Mẍt) + 2f v ẋt) + Kxt) f v r θt) = ft) J θt) + r 2 f v θt) rfv ẋt) = 0 Linear Control Systems Solution to Assignment # Page 3 of 0

4 If you want to obtain the relation of single input ft) and single output xt), then you can eliminate xt) and combine the above two equations. From the second equation, ẋt) = J rf v θt) + r θt) This implies ẍt) = J rf v θ 3) t) + r θt) and xt) = J rf v θt) + rθt). Zero initial condition is assumed. If not, there is only offset θ 0 and it doesn t make any difference physically.) Using those relations, the equation of motion can be simplified to JMθ 3) t) + f v r 2 M + 2J) θt) + f v 2 r 2 + KJ) θt) + Kr 2 f v θt) = f v rft) It is a linear differential equation because it has only θt), θt), θt) and θ 3) t) terms. Solution to Exercise 4. a) Linear If the differential equation is linear and two functions x t) and x 2 t) are satisfied with ) 2π 7ẍ + 0.5ẋ + 5 sin 0 t x = f t) ) 2π 7ẍ ẋ sin 0 t x 2 = f 2 t) then x + x 2 also should be the solution of the differential equation with the input f t) + f 2 t), which is true because ) 2π 7ẍ + ẍ 2 ) + 0.5ẋ + ẋ 2 ) + 5 sin 0 t x + x 2 ) = f t) + f 2 t) b) Nonlinear With the same method as a), the two functions x t) and x 2 t) are the solutions of 7ẍ + 0.5ẋ x )x = f t) 7ẍ ẋ x 2 )x 2 = f 2 t) However, x +x 2 is not the solution of the differential equation with the input f t)+ f 2 t) because c) Nonlinear 7ẍ + ẍ 2 ) + 0.5ẋ + ẋ 2 ) x )x x 2 )x 2 7ẍ + ẍ 2 ) + 0.5ẋ + ẋ 2 ) x + x 2 ))x + x 2 ) d dt 2 mẋ2 + ) 2 kx2 = 0 ẋmẍ + kx) = 0 ẋ = 0 and/or mẍ + kx = 0 Linear Control Systems Solution to Assignment # Page 4 of 0

5 Let A and B denote the solution space of the homogeneous linear differential equations ẋ = 0 and mẍ + kx = 0, respectively. Clearly, A B. Choose x t) A A B and x 2 t) B A B; then both x t) and x 2 t) are solutions of d dt 2 mẋ2 + kx2) = 2 0, but x s t) = x t) + x 2 t) is not, since it satisfies none of the differential equations k ẋ = 0 and mẍ + kx = 0. A simple example is x t) = and x 2 t) = sin ). t m Some people may believe that the correct answer is Linear [under an assumption]. They may justify their answer either way:. By taking derivative first and inspecting. d dt 2 mẋ2 + ) 2 kx2 = ẋmẍ + kx) = 0 If we assume ẋ 0, then the equation is linear. The reason why the assumption is valid, if ẋ = 0 then x = cte and trivial.) 2. With the same method, we have two functions x and x 2 satisfied with d dt 2 mẋ2 + ) 2 kx 2 = 0 d dt 2 mẋ2 2 + ) 2 kx 2 2 = 0 From the first equation, and again assuming ẋ 0, ẋ 2 0 for nontrivial solution, From the second equation mẍ + kx = 0 mẍ 2 + kx 2 = 0 Now consider [ d dt 2 mẋ + ẋ 2 ) 2 + ] 2 kx + x 2 ) 2 = d [ dt 2 mẋ2 + ẋ ẋ ẋ 2 ) + ] 2 kx 2 + x x x 2 ) = mẋ ẍ + ẋ 2 ẍ 2 + ẋ ẍ 2 + ẍ ẋ 2 ) + kx ẋ + x 2 ẋ 2 + x ẋ 2 + ẋ x 2 ) = ẋ mẍ + kx ) + ẋ 2 mẍ 2 + kx 2 ) + ẋ mẍ 2 + kx 2 ) + ẋ 2 mẍ + kx ) = 0 Since the input x + x 2 also satisfies the equation of motion, the system is linear. Solution to Exercise 5. Let s define φt) as the rotation angle of the motor. From the relation of the gear pair, we know φn = θn 2 and φn = θn 2. T is the torque generated by the motor and it is scaled by N 2 /N at the inertia by the gear pair. Also the inertia of the motor is scaled by N 2 2 /N 2 by the gear pair. For the detail, please refer to section 2.7 of Nise) In my opinion, their assumption is not valid. Linear Control Systems Solution to Assignment # Page 5 of 0

6 From KCL at the node attached to the resistor, i s t) v e R T K m = 0. From torque balance at the inertia, J + N ) 2 2 J N 2 m θ = T N 2 b N θ Kθ. Converted torque from the motor drives the effective inertia inertia + converted inertia of the motor shaft), viscous friction and compliance.) Using the Laplace transform and v e = K v φt) = N Kv 2 N θt), the above two equations can be written as I s s) K v R J + N ) 2 2 J N 2 m s 2 + bs + K N 2 Θs)s T s) = 0 N K m ) Θs) = T s) N 2 N Solving the first equation with respect to T and plugging it into the second equation, we get the transfer function defined by Θs) I s s) = Solution to Exercise 6. df s) a) Use L{tft)} = ds ) J + N 2 2 J N 2 m s 2 + twice: N 2 /N )K m b + N 2 2 N 2 ). K v K m s + K R F s) = s + )3. 6 b) Note that sin t cos t = 2 sin2t).2 Then, F s) = s df s) c) Use the identity sint 3) = sin t cos 3 cos t sin 3 and L{tft)} = ds : F s) = cos 3 2s s 2 + ) 2 + sin 3 s2 + ) sin 3 2s 2 s 2 + ) 2. df s) d) Here you should use L{tft)} = first and then T > 0, L{ft T )t T )} = ds e T s F s): 2 avval ya dovome dabirestan! F s) = e 3s s ) 2. Linear Control Systems Solution to Assignment # Page 6 of 0

7 e) Since hence L{te at } = s + a) 2 L{t cos at} = s2 a 2 s 2 + a 2 ) 2 Solution to Exercise 7. a) Let F s) = L{te at + 2t cos t} = L{te at } + L{2t cos t} = s 2 + s + s + )s + 2)s + 3). Write s + a) + 2 s2 2 s 2 + ) 2 F s) = a s + + a 2 s a 3 s + 3 ; a = s2 + s + s + 2)s + 3) = s= 2, a 2 = s2 + s + s + )s + 3) = 3, s= 2 a 3 = s2 + s + s + )s + 2) = 7 s= 3 2. f t) = 2 e t 3e 2t + 7 ) 2 e 3t t). Since T > 0, L{xt T )t T )} = e T s Xs), hence ft) = f t )t ) = 2 e t ) 3e 2t ) + 7 ) 2 e 3t ) t ). b) { } { L } = L 4 ss + 2) 2 s 4 s s + 2) 2 = 4 { e 2t 2te 2t} t) c) It can be re-written as F s) = s 2 3s ) Hs). s Therefore its inverse Laplace transform is ft) = d2 ht) dt 3 dht) dt t + ht) + 2 hτ) dτ. 0 Linear Control Systems Solution to Assignment # Page 7 of 0

8 d) F s) = ω 2 n s s 2 + 2ζω n s + ω 2 n) = s F s) F s) = ω 2 n s 2 + 2ζω n s + ω 2 n For all 0 < ζ <, f t) = L {F s)} ω n ) = ζ 2 e ζω nt sin ω n ζ2 t ft) = ) ζ 2 e ζω nt sin ω n ζ2 t + cos ζ Hint: You should use integration by parts twice. e) F s) = e s s 2 + e s ) = s 2 F s) F s) = e s + e s = e s e s ) e s + = e s e s) e s + e 2s ) = e s 2e 2s + 2e 3s f t) = δt ) 2δt 2) + 2δt 3) = ft) = t )t ) 2t 2)t 2) + 2t 3)t 3) Note that expanding the term / + e s ) [directly] to a geometric series, leads to an invalid non-causal answer; we are using one-sided Laplace transform definition. Solution to Exercise 8. Linear Control Systems Solution to Assignment # Page 8 of 0

9 a) ) ) Y s) = s + s 2 + 2s + 5 [ ] [ ] = s + s + + 2j)s + 2j) = c s + + c 2 s + + 2j) + c 3 s + 2j) c = 2j. 2j = 4 c = 2j. 4j = 8 c = c 2 = [ 8 yt) = 4 e t 8 e 2j)t ] 8 e +2j)t t) = [ e t 4 2 e t e 2jt + e 2jt)] t) = 4 e t [ cos 2t] t) b) 3s + 4 Y s) = s 4 + 4s 3 + 8s 2 + 8s + 4 s = 3s + 4 s [s + ) 2 + ] 2 = k s + c s + + j + c 2 s + + j) 2 + d s + j + d 2 s + j) 2 k = c = 2 4 j d = c = j c 2 = 4 2 j d 2 = c 2 = j [ yt) = ) j e j)t ) j te j)t ) j e +j)t ) ] j te +j)t t) = [ e cos t t + 2 sin t + 2 )] t cos t + t sin t t) Solution to Exercise 9 Nise, 4th Edition, Chapter 2, Problem 53). The relationship between the nonlinear spring s displacement, x s t), and its force, f s t), is x s t) = e f st). Linear Control Systems Solution to Assignment # Page 9 of 0

10 Solving for the force, f s t) = ln x s t)) ) Writing the differential equation for the system by summing forces, d 2 xt) dt 2 + dxt) dt ln xt)) = ft). 2) Letting xt) = x 0 + δx and ft) = + δf, linearize ln xt)): d ln x) ln x) ln x 0 ) = dx δx. x=x0 Solving for ln x), ln x) = ln x 0 ) d ln x) dx δx = ln x 0 ) δx. 3) x=x0 x 0 When f =, δx = 0. Thus from ), = ln x 0 ). Solving for x 0, Substituting x 0 = into 3), x 0 = e ln x) = ln 0.632) x 0 = δx = eδx Placing this value into 2) along with xt) = x 0 + δx and ft) = + δf, yields the linearized differential equation or d 2 δx dt 2 + dδx dt d 2 δx dt 2 + dδx dt + + eδx = + δf + eδx = δf. Taking the Laplace transform and rearranging yield the transfer function Xs) F s) = s 2 + s + e Good Luck! Linear Control Systems Solution to Assignment # Page 0 of 0

### Electrical Machine & Automatic Control (EEE-409) (ME-II Yr) UNIT-3 Content: Signals u(t) = 1 when t 0 = 0 when t <0

Electrical Machine & Automatic Control (EEE-409) (ME-II Yr) UNIT-3 Content: Modeling of Mechanical : linear mechanical elements, force-voltage and force current analogy, and electrical analog of simple

### 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.

### Control System. Contents

Contents Chapter Topic Page Chapter- Chapter- Chapter-3 Chapter-4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of

### Raktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries

. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace

### 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

### 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

### MCE 366 System Dynamics, Spring Problem Set 2. Solutions to Set 2

MCE 366 System Dynamics, Spring 2012 Problem Set 2 Reading: Chapter 2, Sections 2.3 and 2.4, Chapter 3, Sections 3.1 and 3.2 Problems: 2.22, 2.24, 2.26, 2.31, 3.4(a, b, d), 3.5 Solutions to Set 2 2.22

### 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.

### MATH 251 Examination II April 7, 2014 FORM A. Name: Student Number: Section:

MATH 251 Examination II April 7, 2014 FORM A Name: Student Number: Section: This exam has 12 questions for a total of 100 points. In order to obtain full credit for partial credit problems, all work must

### 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

### 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:

### MATH 251 Examination II April 4, 2016 FORM A. Name: Student Number: Section:

MATH 251 Examination II April 4, 2016 FORM A Name: Student Number: Section: This exam has 12 questions for a total of 100 points. In order to obtain full credit for partial credit problems, all work must

### 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

### 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

### INC 341 Feedback Control Systems: Lecture 2 Transfer Function of Dynamic Systems I Asst. Prof. Dr.-Ing. Sudchai Boonto

INC 341 Feedback Control Systems: Lecture 2 Transfer Function of Dynamic Systems I Asst. Prof. Dr.-Ing. Sudchai Boonto Department of Control Systems and Instrumentation Engineering King Mongkut s University

### MODELING OF CONTROL SYSTEMS

1 MODELING OF CONTROL SYSTEMS Feb-15 Dr. Mohammed Morsy Outline Introduction Differential equations and Linearization of nonlinear mathematical models Transfer function and impulse response function Laplace

### MATH 251 Final Examination December 16, 2015 FORM A. Name: Student Number: Section:

MATH 5 Final Examination December 6, 5 FORM A Name: Student Number: Section: This exam has 7 questions for a total of 5 points. In order to obtain full credit for partial credit problems, all work must

### 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

### 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

### MATH 251 Final Examination December 16, 2014 FORM A. Name: Student Number: Section:

MATH 2 Final Examination December 6, 204 FORM A Name: Student Number: Section: This exam has 7 questions for a total of 0 points. In order to obtain full credit for partial credit problems, all work must

### Dynamic Modeling. For the mechanical translational system shown in Figure 1, determine a set of first order

QUESTION 1 For the mechanical translational system shown in, determine a set of first order differential equations describing the system dynamics. Identify the state variables and inputs. y(t) x(t) k m

### ( ) ( = ) = ( ) ( ) ( )

( ) Vρ C st s T t 0 wc Ti s T s Q s (8) K T ( s) Q ( s) + Ti ( s) (0) τs+ τs+ V ρ K and τ wc w T (s)g (s)q (s) + G (s)t(s) i G and G are transfer functions and independent of the inputs, Q and T i. Note

### 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

### MATH 251 Examination II April 3, 2017 FORM A. Name: Student Number: Section:

MATH 251 Examination II April 3, 2017 FORM A Name: Student Number: Section: This exam has 12 questions for a total of 100 points. In order to obtain full credit for partial credit problems, all work must

### Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response

.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........

### 20.6. Transfer Functions. Introduction. Prerequisites. Learning Outcomes

Transfer Functions 2.6 Introduction In this Section we introduce the concept of a transfer function and then use this to obtain a Laplace transform model of a linear engineering system. (A linear engineering

### Introduction & Laplace Transforms Lectures 1 & 2

Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Control System Definition of a Control System Group of components that collectively

### 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

### Name: Fall 2014 CLOSED BOOK

Name: Fall 2014 1. Rod AB with weight W = 40 lb is pinned at A to a vertical axle which rotates with constant angular velocity ω =15 rad/s. The rod position is maintained by a horizontal wire BC. Determine

### 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

### 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

### Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc.

Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc. Electrical Engineering Department University of Indonesia 2 Steady State Error How well can

### ENGIN 211, Engineering Math. Laplace Transforms

ENGIN 211, Engineering Math Laplace Transforms 1 Why Laplace Transform? Laplace transform converts a function in the time domain to its frequency domain. It is a powerful, systematic method in solving

### STABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable

ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Bounded-input bounded-output (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated

### An Introduction to Control Systems

An Introduction to Control Systems Signals and Systems: 3C1 Control Systems Handout 1 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie November 21, 2012 Recall the concept of a

### 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

### 1 x(k +1)=(Φ LH) x(k) = T 1 x 2 (k) x1 (0) 1 T x 2(0) T x 1 (0) x 2 (0) x(1) = x(2) = x(3) =

567 This is often referred to as Þnite settling time or deadbeat design because the dynamics will settle in a Þnite number of sample periods. This estimator always drives the error to zero in time 2T or

### MAE143 B - Linear Control - Spring 2018 Midterm, May 3rd

MAE143 B - Linear Control - Spring 2018 Midterm, May 3rd Instructions: 1. This exam is open book. You can consult any printed or written material of your liking. 2. You have 70 minutes. 3. Most questions

### 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 =

### MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions

MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noise-cancelling headphone system. 1a. Based on the low-pass filter given, design a high-pass filter,

### 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

### SAMPLE EXAMINATION PAPER (with numerical answers)

CID No: IMPERIAL COLLEGE LONDON Design Engineering MEng EXAMINATIONS For Internal Students of the Imperial College of Science, Technology and Medicine This paper is also taken for the relevant examination

### MATH 251 Final Examination December 19, 2012 FORM A. Name: Student Number: Section:

MATH 251 Final Examination December 19, 2012 FORM A Name: Student Number: Section: This exam has 17 questions for a total of 150 points. In order to obtain full credit for partial credit problems, all

### 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

### Dynamics of structures

Dynamics of structures 1.2 Viscous damping Luc St-Pierre October 30, 2017 1 / 22 Summary so far We analysed the spring-mass system and found that its motion is governed by: mẍ(t) + kx(t) = 0 k y m x x

### In the presence of viscous damping, a more generalized form of the Lagrange s equation of motion can be written as

2 MODELING Once the control target is identified, which includes the state variable to be controlled (ex. speed, position, temperature, flow rate, etc), and once the system drives are identified (ex. force,

### MATH 251 Final Examination May 4, 2015 FORM A. Name: Student Number: Section:

MATH 251 Final Examination May 4, 2015 FORM A Name: Student Number: Section: This exam has 16 questions for a total of 150 points. In order to obtain full credit for partial credit problems, all work must

### MATHEMATICAL MODELING OF CONTROL SYSTEMS

1 MATHEMATICAL MODELING OF CONTROL SYSTEMS Sep-14 Dr. Mohammed Morsy Outline Introduction Transfer function and impulse response function Laplace Transform Review Automatic control systems Signal Flow

### ADMISSION TEST INDUSTRIAL AUTOMATION ENGINEERING

UNIVERSITÀ DEGLI STUDI DI PAVIA ADMISSION TEST INDUSTRIAL AUTOMATION ENGINEERING September 26, 2016 The candidates are required to answer the following multiple choice test which includes 30 questions;

### Math 266 Midterm Exam 2

Math 266 Midterm Exam 2 March 2st 26 Name: Ground Rules. Calculator is NOT allowed. 2. Show your work for every problem unless otherwise stated (partial credits are available). 3. You may use one 4-by-6

### Topic # Feedback Control Systems

Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the

### Math 308 Exam II Practice Problems

Math 38 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture and all suggested homework problems..

### Last week: analysis of pinion-rack w velocity feedback

Last week: analysis of pinion-rack w velocity feedback Calculation of the steady state error Transfer function: V (s) V ref (s) = 0.362K s +2+0.362K Step input: V ref (s) = s Output: V (s) = s 0.362K s

### 2.161 Signal Processing: Continuous and Discrete Fall 2008

MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS

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

.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to

### University of Alberta ENGM 541: Modeling and Simulation of Engineering Systems Laboratory #7. M.G. Lipsett & M. Mashkournia 2011

ENG M 54 Laboratory #7 University of Alberta ENGM 54: Modeling and Simulation of Engineering Systems Laboratory #7 M.G. Lipsett & M. Mashkournia 2 Mixed Systems Modeling with MATLAB & SIMULINK Mixed systems

### Time Response of Systems

Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) s-plane Time response p =0 s p =0,p 2 =0 s 2 t p =

### Video 5.1 Vijay Kumar and Ani Hsieh

Video 5.1 Vijay Kumar and Ani Hsieh Robo3x-1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior

### Do not write below here. Question Score Question Score Question Score

MATH-2240 Friday, May 4, 2012, FINAL EXAMINATION 8:00AM-12:00NOON Your Instructor: Your Name: 1. Do not open this exam until you are told to do so. 2. This exam has 30 problems and 18 pages including this

### Final Exam April 30, 2013

Final Exam Instructions: You have 120 minutes to complete this exam. This is a closed-book, closed-notes exam. You are allowed to use a calculator during the exam. Usage of mobile phones and other electronic

### CITY UNIVERSITY SCHOOL OF ENGINEERING AND MATHEMATICAL SCIENCES

CITY UNIVERSITY SCHOOL OF ENGINEERING AND MATHEMATICAL SCIENCES AERONAUTICAL ENGINEERING MEng/BEng (Hons) AIR TRANSPORT ENGINEERING MEng/BEng (Hons) AIR TRANSPORT ENGINEERING BSc (Hons) AUTOMOTIVE AND

### Computer Problems for Methods of Solving Ordinary Differential Equations

Computer Problems for Methods of Solving Ordinary Differential Equations 1. Have a computer make a phase portrait for the system dx/dt = x + y, dy/dt = 2y. Clearly indicate critical points and separatrices.

### Basic Procedures for Common Problems

Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available

### Second Order Systems

Second Order Systems independent energy storage elements => Resonance: inertance & capacitance trade energy, kinetic to potential Example: Automobile Suspension x z vertical motions suspension spring shock

### Lab 6a: Pole Placement for the Inverted Pendulum

Lab 6a: Pole Placement for the Inverted Pendulum Idiot. Above her head was the only stable place in the cosmos, the only refuge from the damnation of the Panta Rei, and she guessed it was the Pendulum

### 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

### STABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse

SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential

### Mechatronics. MANE 4490 Fall 2002 Assignment # 1

Mechatronics MANE 4490 Fall 2002 Assignment # 1 1. For each of the physical models shown in Figure 1, derive the mathematical model (equation of motion). All displacements are measured from the static

### I Laplace transform. I Transfer function. I Conversion between systems in time-, frequency-domain, and transfer

EE C128 / ME C134 Feedback Control Systems Lecture Chapter 2 Modeling in the Frequency Domain Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley Lecture

### 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

### 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

### Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter

### 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. Massachusetts Institute

### Chapter 5 Design. D. J. Inman 1/51 Mechanical Engineering at Virginia Tech

Chapter 5 Design Acceptable vibration levels (ISO) Vibration isolation Vibration absorbers Effects of damping in absorbers Optimization Viscoelastic damping treatments Critical Speeds Design for vibration

### 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

### 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

### 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

### 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

### (Refer Slide Time: 00:01:30 min)

Control Engineering Prof. M. Gopal Department of Electrical Engineering Indian Institute of Technology, Delhi Lecture - 3 Introduction to Control Problem (Contd.) Well friends, I have been giving you various

### 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

### 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

### 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 -

### Spontaneous Speed Reversals in Stepper Motors

Spontaneous Speed Reversals in Stepper Motors Marc Bodson University of Utah Electrical & Computer Engineering 50 S Central Campus Dr Rm 3280 Salt Lake City, UT 84112, U.S.A. Jeffrey S. Sato & Stephen

### Unit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace

Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, 010 1 Pair Table Unit, Part : Unit,

### 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

### Differential Equations and Linear Algebra Exercises. Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS

Differential Equations and Linear Algebra Exercises Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS CHAPTER 1 Linear second order ODEs Exercises 1.1. (*) 1 The following differential

### DON T PANIC! If you get stuck, take a deep breath and go on to the next question. Come back to the question you left if you have time at the end.

Math 307, Midterm 2 Winter 2013 Name: Instructions. DON T PANIC! If you get stuck, take a deep breath and go on to the next question. Come back to the question you left if you have time at the end. There

### Given: We are given the drawing above and the assumptions associated with the schematic diagram.

PROBLEM 1: (30%) The schematic shown below represents a pulley-driven machine with a flexible support. The three coordinates shown are absolute coordinates drawn with respect to the static equilibrium

### Control of Electromechanical Systems

Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance

### 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

### 06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance.

Chapter 06 Feedback 06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Lesson of the Course Fondamenti di Controlli Automatici of

### MATH 251 Examination II November 5, 2018 FORM A. Name: Student Number: Section:

MATH 251 Examination II November 5, 2018 FORM A Name: Student Number: Section: This exam has 14 questions for a total of 100 points. In order to obtain full credit for partial credit problems, all work

### Lezione 9 30 March. Scribes: Arianna Marangon, Matteo Vitturi, Riccardo Prota

Control Laboratory: a.a. 2015/2016 Lezione 9 30 March Instructor: Luca Schenato Scribes: Arianna Marangon, Matteo Vitturi, Riccardo Prota What is left to do is how to design the low pass pole τ L for the

### 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

### 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

### Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.

ISS0031 Modeling and Identification Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design. Aleksei Tepljakov, Ph.D. September 30, 2015 Linear Dynamic Systems Definition

### Strauss PDEs 2e: Section Exercise 4 Page 1 of 6

Strauss PDEs 2e: Section 5.3 - Exercise 4 Page of 6 Exercise 4 Consider the problem u t = ku xx for < x < l, with the boundary conditions u(, t) = U, u x (l, t) =, and the initial condition u(x, ) =, where

### MATHEMATICAL MODELING OF DYNAMIC SYSTEMS

MTHEMTIL MODELIN OF DYNMI SYSTEMS Mechanical Translational System 1. Spring x(t) k F S (t) k x(t) x i (t) k x o (t) 2. Damper x(t) x i (t) x o (t) c c 3. Mass x(t) F(t) m EXMPLE I Produce the block diagram