Linear Control Systems Solution to Assignment #1

Save this PDF as:

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

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

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

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

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

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

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

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

MA 266 Review Topics - Exam # 2 (updated)

MA 66 Reiew Topics - Exam # updated Spring First Order Differential Equations Separable, st Order Linear, Homogeneous, Exact Second Order Linear Homogeneous with Equations Constant Coefficients The differential

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

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

Practice Problems For Test 3

Practice Problems For Test 3 Power Series Preliminary Material. Find the interval of convergence of the following. Be sure to determine the convergence at the endpoints. (a) ( ) k (x ) k (x 3) k= k (b)

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

Lecture Notes on Control Systems/D. Ghose/01 106 1.7 Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there

One-Sided Laplace Transform and Differential Equations

One-Sided Laplace Transform and Differential Equations As in the dcrete-time case, the one-sided transform allows us to take initial conditions into account. Preliminaries The one-sided Laplace transform

The basic principle to be used in mechanical systems to derive a mathematical model is Newton s law,

Chapter. DYNAMIC MODELING Understanding the nature of the process to be controlled is a central issue for a control engineer. Thus the engineer must construct a model of the process with whatever information

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

Lecture 12. AO Control Theory

Lecture 12 AO Control Theory Claire Max with many thanks to Don Gavel and Don Wiberg UC Santa Cruz February 18, 2016 Page 1 What are control systems? Control is the process of making a system variable

6.003 Homework #10 Solutions

6.3 Homework # Solutions Problems. DT Fourier Series Determine the Fourier Series coefficients for each of the following DT signals, which are periodic in N = 8. x [n] / n x [n] n x 3 [n] n x 4 [n] / n

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

Deterministic Dynamic Programming

Deterministic Dynamic Programming 1 Value Function Consider the following optimal control problem in Mayer s form: V (t 0, x 0 ) = inf u U J(t 1, x(t 1 )) (1) subject to ẋ(t) = f(t, x(t), u(t)), x(t 0

MEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 2017) Lecture DOF Modeling of Shock Absorbers. This lecture covers:

MEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 207) Lecture 4. 2-DOF Modeling of Shock Absorbers This lecture covers: Revisit 2-DOF Spring-Mass System Understand the den Hartog

Oscillations Simple Harmonic Motion

Oscillations Simple Harmonic Motion Lana Sheridan De Anza College Dec 1, 2017 Overview oscillations simple harmonic motion (SHM) spring systems energy in SHM pendula damped oscillations Oscillations and

EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 3 TUTORIAL 1 - TRIGONOMETRICAL GRAPHS

EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 3 TUTORIAL 1 - TRIGONOMETRICAL GRAPHS CONTENTS 3 Be able to understand how to manipulate trigonometric expressions and apply

EE 3054: Signals, Systems, and Transforms Summer It is observed of some continuous-time LTI system that the input signal.

EE 34: Signals, Systems, and Transforms Summer 7 Test No notes, closed book. Show your work. Simplify your answers. 3. It is observed of some continuous-time LTI system that the input signal = 3 u(t) produces

Lecture: Sampling. Automatic Control 2. Sampling. Prof. Alberto Bemporad. University of Trento. Academic year

Automatic Control 2 Sampling Prof. Alberto Bemporad University of rento Academic year 2010-2011 Prof. Alberto Bemporad (University of rento) Automatic Control 2 Academic year 2010-2011 1 / 31 ime-discretization

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

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

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

Frequency domain analysis

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

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

Angular velocity and angular acceleration CHAPTER 9 ROTATION. Angular velocity and angular acceleration. ! equations of rotational motion

Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for

Introduction to Vibration. Mike Brennan UNESP, Ilha Solteira São Paulo Brazil

Introduction to Vibration Mike Brennan UNESP, Ilha Solteira São Paulo Brazil Vibration Most vibrations are undesirable, but there are many instances where vibrations are useful Ultrasonic (very high

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

MEM04: Rotary Inverted Pendulum

MEM4: Rotary Inverted Pendulum Interdisciplinary Automatic Controls Laboratory - ME/ECE/CHE 389 April 8, 7 Contents Overview. Configure ELVIS and DC Motor................................ Goals..............................................3

Physics 351, Spring 2015, Homework #5. Due at start of class, Friday, February 20, 2015 Course info is at positron.hep.upenn.

Physics 351, Spring 2015, Homework #5. Due at start of class, Friday, February 20, 2015 Course info is at positron.hep.upenn.edu/p351 When you finish this homework, remember to visit the feedback page

6 OUTPUT FEEDBACK DESIGN

6 OUTPUT FEEDBACK DESIGN When the whole sate vector is not available for feedback, i.e, we can measure only y = Cx. 6.1 Review of observer design Recall from the first class in linear systems that a simple

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

z x = f x (x, y, a, b), z y = f y (x, y, a, b). F(x, y, z, z x, z y ) = 0. This is a PDE for the unknown function of two independent variables.

Chapter 2 First order PDE 2.1 How and Why First order PDE appear? 2.1.1 Physical origins Conservation laws form one of the two fundamental parts of any mathematical model of Continuum Mechanics. These

Exam 3 Practice Solutions

Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at

Modeling of Belt-Pulley and Flexible Coupling Effects on Submarine Driven System Electrical Motors

Modeling of Belt-Pulley and Flexible Coupling Effects on 319 JPE 11-3-10 Modeling of Belt-Pulley and Flexible Coupling Effects on Submarine Driven System Electrical Motors Mehrdad Jafarboland and Mahmoud

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

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

Lecture 9 Time-domain properties of convolution systems

EE 12 spring 21-22 Handout #18 Lecture 9 Time-domain properties of convolution systems impulse response step response fading memory DC gain peak gain stability 9 1 Impulse response if u = δ we have y(t)

Signal Correction of Load Cell Output Using Adaptive Method Rajesh Dey¹, Atreyee Biswas 2, Suman Kumar Laha 3, Amlan Pal 4 and Dr.Achintya Das 5 Asst. Prof., Dept. of ECE, SDET-BGI, Barasat, Kolkata, India

10 Measurement of Acceleration, Vibration and Shock Transducers

Chapter 10: Acceleration, Vibration and Shock Measurement Dr. Lufti Al-Sharif (Revision 1.0, 25/5/2008) 1. Introduction This chapter examines the measurement of acceleration, vibration and shock. It starts

RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS. Ryszard Gessing

RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS Ryszard Gessing Politechnika Śl aska Instytut Automatyki, ul. Akademicka 16, 44-101 Gliwice, Poland, fax: +4832 372127, email: gessing@ia.gliwice.edu.pl

2.152 Course Notes Contraction Analysis MIT, 2005

2.152 Course Notes Contraction Analysis MIT, 2005 Jean-Jacques Slotine Contraction Theory ẋ = f(x, t) If Θ(x, t) such that, uniformly x, t 0, F = ( Θ + Θ f x )Θ 1 < 0 Θ(x, t) T Θ(x, t) > 0 then all solutions

Physics 326 Lab 6 10/18/04 DAMPED SIMPLE HARMONIC MOTION

DAMPED SIMPLE HARMONIC MOTION PURPOSE To understand the relationships between force, acceleration, velocity, position, and period of a mass undergoing simple harmonic motion and to determine the effect

Study Material. CONTROL SYSTEM ENGINEERING (As per SCTE&VT,Odisha new syllabus) 4th Semester Electronics & Telecom Engineering

Study Material CONTROL SYSTEM ENGINEERING (As per SCTE&VT,Odisha new syllabus) 4th Semester Electronics & Telecom Engineering By Sri Asit Kumar Acharya, Lecturer ETC, Govt. Polytechnic Dhenkanal & Sri

Transfer Functions. Chapter Introduction. 6.2 The Transfer Function

Chapter 6 Transfer Functions As a matter of idle curiosity, I once counted to find out what the order of the set of equations in an amplifier I had just designed would have been, if I had worked with the

Solutions to Homework 5

Solutions to Homework 5 1. Let z = f(x, y) be a twice continuously differentiable function of x and y. Let x = r cos θ and y = r sin θ be the equations which transform polar coordinates into rectangular

Linear System Theory

Linear System Theory - Laplace Transform Prof. Robert X. Gao Department of Mechanical Engineering University of Connecticut Storrs, CT 06269 Outline What we ve learned so far: Setting up Modeling Equations

Examination paper for TMA4195 Mathematical Modeling

Department of Mathematical Sciences Examination paper for TMA4195 Mathematical Modeling Academic contact during examination: Elena Celledoni Phone: 48238584, 73593541 Examination date: 11th of December

Math Assignment 5

Math 2280 - Assignment 5 Dylan Zwick Fall 2013 Section 3.4-1, 5, 18, 21 Section 3.5-1, 11, 23, 28, 35, 47, 56 Section 3.6-1, 2, 9, 17, 24 1 Section 3.4 - Mechanical Vibrations 3.4.1 - Determine the period

DC Motor Position: System Modeling

1 of 7 01/03/2014 22:07 Tips Effects TIPS ABOUT BASICS INDEX NEXT INTRODUCTION CRUISE CONTROL MOTOR SPEED MOTOR POSITION SUSPENSION INVERTED PENDULUM SYSTEM MODELING ANALYSIS DC Motor Position: System

:= 0.75 Ns m K N m. := 0.05 kg. K N m. (a.1) FBD and forces. (a.2) derive EOM From the FBD diagram, Newton's 2nd law states:

P FA - Derive EO for simple mechanical system L San Andres (c) For the system shown in the figure, Z (t) Z o cos( t) is a periodic displacement input (known). Perform the following tasks: a) Set X(t) as

Physics 231 Lecture 18

Physics 31 ecture 18 τ = Fd;d is the lever arm Main points of today s lecture: Energy Pendulum T = π g ( ) θ = θmax cos πft + ϑ0 Damped Oscillations x x equibrium = Ae bt/(m) cos(ω damped t) ω damped =

Robotics. Dynamics. Marc Toussaint U Stuttgart

Robotics Dynamics 1D point mass, damping & oscillation, PID, dynamics of mechanical systems, Euler-Lagrange equation, Newton-Euler recursion, general robot dynamics, joint space control, reference trajectory

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

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

ECE 3793 Matlab Project 3 Solution

ECE 3793 Matlab Project 3 Solution Spring 27 Dr. Havlicek. (a) In text problem 9.22(d), we are given X(s) = s + 2 s 2 + 7s + 2 4 < Re {s} < 3. The following Matlab statements determine the partial fraction

Series RC and RL Time Domain Solutions

ECE2205: Circuits and Systems I 6 1 Series RC and RL Time Domain Solutions In the last chapter, we saw that capacitors and inductors had element relations that are differential equations: i c (t) = C d

Selection Calculations For Motorized Actuators

Selection Calculations/ Selection Calculations For Linear Slides and Cylinders Select from the EZS Series, EZS Series for Cleanroom Use, EZC Series First determine your series, then select your model.

VI. Transistor amplifiers: Biasing and Small Signal Model

VI. Transistor amplifiers: iasing and Small Signal Model 6.1 Introduction Transistor amplifiers utilizing JT or FET are similar in design and analysis. Accordingly we will discuss JT amplifiers thoroughly.

Polytechnic Institute of NYU MA 2132 Final Practice Answers Fall 2012

Polytechnic Institute of NYU MA Final Practice Answers Fall Studying from past or sample exams is NOT recommended. If you do, it should be only AFTER you know how to do all of the homework and worksheet

Dynamic Response of Structures With Frequency Dependent Damping

Dynamic Response of Structures With Frequency Dependent Damping Blanca Pascual & S Adhikari School of Engineering, Swansea University, Swansea, UK Email: S.Adhikari@swansea.ac.uk URL: http://engweb.swan.ac.uk/

On the Stability of Linear Systems

On the Stability of Linear Systems by Daniele Sasso * Abstract The criteria of stability defined in the standard theory of linear systems aren t exhaustive and show some inconsistencies. In this article

Generalized sources (Sect. 6.5). The Dirac delta generalized function. Definition Consider the sequence of functions for n 1, Remarks:

Generalized sources (Sect. 6.5). The Dirac delta generalized function. Definition Consider the sequence of functions for n, d n, t < δ n (t) = n, t 3 d3 d n, t > n. d t The Dirac delta generalized function

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 3. Periodic functions

Chapter 3. Periodic functions Why do lights flicker? For that matter, why do they give off light at all? They are fed by an alternating current which turns into heat because of the electrical resistance

Solving a RLC Circuit using Convolution with DERIVE for Windows

Solving a RLC Circuit using Convolution with DERIVE for Windows Michel Beaudin École de technologie supérieure, rue Notre-Dame Ouest Montréal (Québec) Canada, H3C K3 mbeaudin@seg.etsmtl.ca - Introduction

Robust Speed Controller Design for Permanent Magnet Synchronous Motor Drives Based on Sliding Mode Control

Available online at www.sciencedirect.com ScienceDirect Energy Procedia 88 (2016 ) 867 873 CUE2015-Applied Energy Symposium and Summit 2015: ow carbon cities and urban energy systems Robust Speed Controller

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work

Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational

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

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

Matlab Problem Sets. Math 246 Spring 2012 Sections 0112, 0122, 0132, R. Lipsman

Matlab Problem Sets 1 Math 246 Spring 2012 Sections 0112, 0122, 0132, 0142 R. Lipsman 2 Problem Set A Practice with MATLAB In this problem set, you will use MATLAB to do some basic calculations, and then

Optimal delayed control for an overhead crane

Optimal control for an overhead crane Carlos Vazquez Joaquin Collado Department of Automatic Control, CINVESTAV-IPN,Av. IPN 58, 736 Mexico, D.F., Mexico (e-mail: electroncvaitc@gmail.com) Department of

= m. 30 m. The angle that the tangent at B makes with the x axis is f = tan-1

1 11. When the roller coaster is at B, it has a speed of 5 m>s, which is increasing at at = 3 m>s. Determine the magnitude of the acceleration of the roller coaster at this instant and the direction angle

Goals for today 2.004

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

Parametric Equations, Function Composition and the Chain Rule: A Worksheet

Parametric Equations, Function Composition and the Chain Rule: A Worksheet Prof.Rebecca Goldin Oct. 8, 003 1 Parametric Equations We have seen that the graph of a function f(x) of one variable consists

Stability Analysis for ODEs

Stability Analysis for ODEs Marc R Roussel September 13, 2005 1 Linear stability analysis Equilibria are not always stable Since stable and unstable equilibria play quite different roles in the dynamics

Homework 5 EE235, Summer 2013 Solution

Homework 5 EE235, Summer 23 Solution. Fourier Series. Determine w and the non-zero Fourier series coefficients for the following functions: (a f(t 2 cos(3πt + sin(πt + π 3 w π f(t e j3πt + e j3πt + j2

6.241 Dynamic Systems and Control

6.241 Dynamic Systems and Control Lecture 12: I/O Stability Readings: DDV, Chapters 15, 16 Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology March 14, 2011 E. Frazzoli

3 Stability and Lyapunov Functions

CDS140a Nonlinear Systems: Local Theory 02/01/2011 3 Stability and Lyapunov Functions 3.1 Lyapunov Stability Denition: An equilibrium point x 0 of (1) is stable if for all ɛ > 0, there exists a δ > 0 such

Solutions to Problem Set 1

Due by :00pm sharp Fall 005 Friday, Sept. 16, 005 Solutions to Problem Set 1 Part I/Part II Part I(0 points) (a) ( points) p. 57, Section., Problem 1 (b) ( points) p. 6, Section.3, Problem 1 (c) ( points)

Acoustics-An An Overview. Lecture 1. Vibro-Acoustics. What? Why? How? Lecture 1

Vibro-Acoustics Acoustics-An An Overview 1 Vibro-Acoustics What? Why? How? 2 Linear Non-Linear Force Motion Arbitrary motion Harmonic Motion Mechanical Vibrations Sound (Acoustics) 3 Our heart beat, our

Active surge control of centrifugal compressors using drive torque

Active surge control of centrifugal compressors using drive torque Jan Tommy Gravdahl, Olav Egeland and Svein Ove Vatland Department of Engineering Cybernetics, NTNU, N-749 Trondheim, Norway ABB Corporate

2.004 Dynamics and Control II Spring 2008

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

APPENDIX. SELECTING THE SureServo SERVO SYSTEM. In This Appendix... Selecting the SureServo Servo System...B 2. Leadscrew - Example Calculations...

SELECTING THE SureServo SERVO SYSTEM APPENDIX B In This Appendix... Selecting the SureServo Servo System............B 2 The Selection Procedure......................................B 2 How many pulses

Part 1. The simple harmonic oscillator and the wave equation

Part 1 The simple harmonic oscillator and the wave equation In the first part of the course we revisit the simple harmonic oscillator, previously discussed in di erential equations class. We use the discussion

CALCULATION OF NONLINEAR VIBRATIONS OF PIECEWISE-LINEAR SYSTEMS USING THE SHOOTING METHOD

Vietnam Journal of Mechanics, VAST, Vol. 34, No. 3 (2012), pp. 157 167 CALCULATION OF NONLINEAR VIBRATIONS OF PIECEWISE-LINEAR SYSTEMS USING THE SHOOTING METHOD Nguyen Van Khang, Hoang Manh Cuong, Nguyen

Partial Differential Equations

Partial Differential Equations Xu Chen Assistant Professor United Technologies Engineering Build, Rm. 382 Department of Mechanical Engineering University of Connecticut xchen@engr.uconn.edu Contents 1

Study on re-adhesion control by monitoring excessive angular momentum in electric railway tractions

Study on re-adhesion control by monitoring excessive angular momentum in electric railway tractions Takafumi Hara and Takafumi Koseki Department of Electrical Engeneering and Information Systems Graduate

Concept Question: Normal Force

Concept Question: Normal Force Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical

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

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department. Physics 8.01 Fall Term 2006

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.01 Fall Term 2006 Momentum Demonstration Purpose of the Experiment: In this experiment you allow two carts to collide on a level track

Simple Harmonic Motion

3/5/07 Simple Harmonic Motion 0. The Ideal Spring and Simple Harmonic Motion HOOKE S AW: RESTORING FORCE OF AN IDEA SPRING The restoring force on an ideal spring is F x k x spring constant Units: N/m 3/5/07

TRANSIENT RESPONSE TO STEP AND PULSE FUNCTIONS*

CHAPTER 8 TRANSIENT RESPONSE TO STEP AND PULSE FUNCTIONS* Robert S. Ayre INTRODUCTION In analyses involving shock and transient vibration, it is essential in most instances to begin with the time-history

8 sin 3 V. For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0.

For the circuit given, determine the voltage v for all time t. Assume that no energy is stored in the circuit before t = 0. Spring 2015, Exam #5, Problem #1 4t Answer: e tut 8 sin 3 V 1 For the circuit

21.55 Worksheet 7 - preparation problems - question 1:

Dynamics 76. Worksheet 7 - preparation problems - question : A coupled oscillator with two masses m and positions x (t) and x (t) is described by the following equations of motion: ẍ x + 8x ẍ x +x A. Write

Lyapunov Stability Theory

Lyapunov Stability Theory Peter Al Hokayem and Eduardo Gallestey March 16, 2015 1 Introduction In this lecture we consider the stability of equilibrium points of autonomous nonlinear systems, both in continuous

High Frequency Variation Speed Control of Spindle Motor for Chatter Vibration Suppression in NC Machine Tools

High Frequency Variation Speed Control of Spindle Motor for Chatter Vibration Suppression in NC Machine Tools Teruaki Ishibashi, Hiroshi Fujimoto, Shinji Ishii, Kouji Yamamoto, and Yuki Terada Abstract