FEEDBACK CONTROL SYSTEMS


 Arleen Brooks
 2 years ago
 Views:
Transcription
1 FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control Systems. Control System Design. Establishing system goals (objectives), e.g. to control the velocity of a motor accurately What do you want to control? e.g. speed, position, level, temperature,... What do you want to achieve? e.g. fast response, less vibration,... How do you translate them into control terms? e.g. overshoot, rising time,.... Obtaining a model of the process, the actuator, and the sensor. E.g. G k/(ts) Differential equations Transfer functions Block diagrams State space equations 3. Designing a controller (PID, root locus, frequency response, state space) Proportional (P) control Integral (I) control ProportionalDerivative (PD) control ProportionalIntegral (PI) control ProportionalDerivativeIntegral (PID) control State space based control 4. Simulation (MATLAB) or experimental test.. Open and ClosedLoop Control Systems Design requirements and specification: An excellent transient response in terms of risetime, size of overshoot and settlingtime; An excellent steadystate response in terms of steadystate error; Acceptable stability margins; Robustness in terms of disturbance rejection; and Robustness in terms of sensitivity to parameter changes.
2 Openloop control system: utilises an actuating device to control the process (plant) directly without using feedback. Input Actuating device Process or plant Output Closedloop control system: uses a measurement of the output and feedback of this signal to compare it with the desired input (reference or command). E U G c _ G H Notation: : Laplace transform of the input signal; U: Laplace transform of the control or actuating signal; : Laplace transform of the output signal; E: Laplace transform of the error signal; G c : Transfer function of the controller; G: Transfer function of the process or plant; H: Transfer function of the sensor.
3 3. Why ClosedLoop Control? An advantage of the closedloop control system is the fact that the use of feedback makes the system response relatively insensitive to external disturbances (e.g. temperature and pressure) and internal variations in system parameters (e.g. component tolerances) which are not known or predicted. 3.. Sensitivity of control systems to parameter variations Suppose that G changes to G G due to the environment, ageing, etc. For small G << G, changes to. This can be illustrated by expanding to a firstorder Taylor series: d Y(G G) Y(G) G 443 dg Openloop: G G G Closedloop: G/[GH] G/[GH] _ G H The system sensitivity is defined as: the ratio of the percentage change in the system transfer function T to the percentage change of the process transfer function G: Openloop: S T G Closedloop: S T G GH dt / T S T G dg / G G T dt dg A system with zero sensitivity is ideal (i.e. changes in the system parameters have no effect on the transfer function). Exercise: A temperature control system of a chemical fluid in a tank is shown, where R is the resired temperature, Y the actual temperature and 'a' is a constant related to the mass in the tank. Determine the openloop and closedloop sensitivity of this sytem to changes in s a parameter 'a' due to chemical reactions an ageing. How would you reduce the sensitivity at low frequencies (i.e. )? 3
4 3.. Disturbance rejection Suppose that there is an input disturbance D applied between the controller and plant: D Openloop: U G c G c GGD G c U G Closedloop: U G c [ H] GcG G GH c G D G GH c G c D U H G Exercise: For the systems shown above, the process transfer function G /(0.5s), the controller G c 99, R 0, a stepunit disturbance D /s, and H. Determine the response to the disturbance for both the openloop and closedloop systems and sketch the outputs. 4
5 3.3. Transient response The transient response is the response of a system as a function of time. A motor speed control system: Objective: the actual speed y approaches the desired speed r quickly Openloop: G c k, G k /(Ts) Assume, for example, kk, T 0 Closedloop with a proportional control term k, H Assume, for example, kk 0, T 0. Unitstep response of openloop control Unitstep response of closedloop control Amplitude Amplitude Time (sec.) Time (sec.) 3.4. Steadystate error The steadystate error is the error after the transient response has decayed, leaving only the continuous response, that is for a unityfeedback system: e( ) lim[r(t) y(t)] t E . Considering a unitstep input, /s: Openloop: G, E [G]. By the finalvalue theorem, e( ) lim e(t) lim se G(0) t Closedloop: G/[GH]. For H, E /[G] By the finalvalue theorem, e( ) lim e(t) limse /[ G(0)] t 5
6 Exercise: Find the open and closedloop steadystate errors for the following system for unitstep and unitramp inputs. E 5 s Summary The advantages of closedloop control: Reduce effects of process (plant) disturbances Make system insensitive to process variations Stabilise an unstable system Create welldefined relations between output and reference The disadvantages: Risk of instability Complexity in analysis and implementation, and expensive 6
7 4. Case Study  Speed Control of a DC Motor The purpose of this case study is: to compare the openloop and closedloop control, and to provide a guide to the solution of a firstorder control design problem using a proportional (P) controller. Techniques used in the solution include: Mathematical modelling of physical systems Transfer function analysis Laplace transforms Feedback control 4.. DC motor  problem description R L i a (t) v a _ Armature circuit v b b ω _ J θ T m Bω T d For the armature controlled DC motor, we will use the following parameter values: R armature resistance Ω L armature inductance 0.5 H m motortorque constant 0 Nm/A b back emf constant 0. Vs/rad J moment of inertia of the motor kgm B viscousfriction coefficient of the motor 0.5 Nms/rad The motor torque, T m, is related to the armature current, i a, by a constant factor m : T m (t) m i a (t) The back emf, v b, is related to the rotational velocity, ω dθ/dt, by the following equation: v b (t) b ω(t) 4.. Design objectives (goals) The objective is to design a simple proportional controller (selecting the gain ) to achieve a fast response to the step input, a small steadystate error, and a reduction of the effect of disturbance. These requirements can be represented, for a unitstep input and unitstep disturbance, as: Steadystate error e ss 0.0 to a unitstep input The effect of a unitstep disturbance < rad/s Time constant 0. s. Can we meet these requirements using an openloop controller? If not, we will try the closedloop controller. 7
8 4.3. Mathematical model of the DC motor In this study, zero initial conditions are used. Time relation Laplace transform The motor torque: T m (t) m i a (t) T m m I a The back emf voltage: Electrical circuit equation: v b b ω(t) V b b Ω di a (t) va (t) bω(t) Ri a (t) L V a b Ω (slr)i a () dt The torque relation: T (t) T m (t) T d (t) T T m T d Mechanical system (rotor): dω(t) J Bω(t) T dt m (t) T d (t) (JsB)Ω m I a T d () Equations () and () are used to obtain the following block diagram: T d V a _ m Ls R T m T l Js B Ω b 4.4. Simplified model Transfer function from V a to Ω: Transfer function from T d to Ω: m G (Ls R)(Js B) G d (Ls R) (Ls R)(Js B) m m b b Neglecting L (assumed very small), we have the simplified firstorder model: m G RJs RB, R G d RJs RB m b m b Substituting the numerical values into G and G d : 0 G, s.5 G d s.5 8
9 T d V a _ 0 T m T l s 0.5 Ω 0. The block diagram can be simplified by moving the summing point ahead of the block with the constant gain of 0: T d /0 V a _ 0 s 0.5 Ω 0. This can be further reduced by representing the feedback loop with an equivalent block: /0 T d V a 0 s.5 Ω 4.5. Simplicity versus accuracy We must make a compromise between the simplicity of the model and the accuracy of the results of the analysis. In deriving a reasonably simplified mathematical model, we frequently find it necessary to ignore certain parameters Openloop control Introducing the constant gain (proportional),, block into the system (R desired velocity): T d Motor dynamics /0 V a 0 s.5 Ω 9
10 Steadystate error: T d 0, /s, 0 E  Ω s s.5 0 ess lim se. By choosing 3/0, e ss 0, O. s 0.5 Disturbance rejection: 0, T d /s, Ω s(s.5) Taking the inverse Laplace transform: ω(t)(/3)[ e 0.75t ] ω(t ) /3 rad/s, not O! Time constant: /.5 4/3 s, again not O! A simple openloop controller can not achieve the required goals Position and velocity sensors in control systems Potentiometers Converts mechanical energy to electrical energy. Input displacement can either be linear or rotational. Three terminal device: two fixed where voltage is applied, third is variable with the movable shaft to provide a voltage (with respect to ground) proportional to the displacement: e(t) e(t) Ly(t) Linear Rθ(t) Rotational Fixed terminals V s y(t) Variable terminal e(t) y(t) Rotary type available for single and multiturn rotational motion. Rotary potentiometer Linear potentiometer 0
11 4.7.. Incremental optical encoder Converts linear or rotary displacement of shaft into a digital pulse signal for position and velocity sensing. Consists of a rotary glass or plastic disk with evenly spaced opaque and transparent region each region represents a displacement increment. For example, the angular resolution of 000 lines disk is θ π/ radians per increment. Light passes through transparent lines as the disk rotates with the movement of the shaft, giving a sinusoidal or triangular waveform at the photodetector, followed by an amplifier/comparator to produce the pulse signal. Position information is obtained from counting the number of pulses using a counter; speed information is obtained from the frequency of the pulse train. Light source Photodetector Squaring Shaft Transparent Opaque For direction of rotation information (clockwise or anticlockwise), a second track of opaque/transparent lines at a different radius of the disk, that is offset 90 o with respect to the first track (quadrature tracks), is included. The pulse trains from the two channels are then compared for lead/lag to determine the direction of rotation Tachometer Works as a voltage generator with a DC voltage proportional to the angular velocity of the input shaft. Used in velocity indicators, velocity feedback control and stabilisation. dθ(t) Mathematical model: e(t) t tω(t) dt t is given as a parameter by the manufacturer in volts/000 rpm. E The transfer function is: Ω t
12 4.8. Closedloop control A tachometer is attached directly to the motor that measures the angular velocity and produces a proportional voltage that is fedback and compared with the reference voltage corresponding to the desired speed. e ref Controller DC motor amplifier M Load t ω Tachometer T ω 5V V REF R R Gain control Driver M ω R V REF t ω Summing amplifier t ω T
13 Assuming that t for simplicity: /0 T d V a 0 s.5 Ω t Steadystate error: T d 0, /s, s.5 E  Ω s s e ss limse (s ). By choosing 5, e ss < 0.0, O. s Disturbance rejection: 0, T d /s, Ω s(s.5 0) Taking the inverse Laplace transform (or from finalvalue theorem): ω( ) /(.50), by choosing 0, ω( ) < 0.005, O. Time constant: /(.50), by choosing, T 0.7 s, O. From the above, we choose a suitable value 0. From the above study, it can be seen that the larger, the better performance for a simple firstorder system. Can we choose very large? 4.9. Problems with a very large gain V a [  Ω] E, in the time domain v a (0) e(0), the small error e(0) at the beginning will lead to a large v a (0) since is large. When is big, the inductance L (which we ignored for simplicity) can not be neglected any more. It will lead to a problem. Assume L mh, then the transfer function from V a to Ω is: m (Ls R)(Js B) s s.5 G m b The closedloop transfer function with the proportional term is: 3
14 T 0.00s 0 s.5 0 For 0, two real poles: s 4 and s 886 (ζ > ) For 500, two complex poles: s 500j500 and s 500j500 (ζ < ) A compromise should be found between the need to increase and the need to avoid the above problems Summary Feedback (closedloop) control can be used to stabilise systems, speed up the transient response, improve the steadystate characteristics, provide disturbance rejection, and decrease the sensitivity to parameter variations. Proportional feedback control reduces errors and improves transient responses, but the high gain (large ) may lead to many problems. 4
15 5. SteadyState Errors in Unity Feedback Control Systems When a command input (desired output) is applied to a control system it is generally hoped that after any transient effects have died away the system output will settle down to the command value. The error with any system is the difference between the required (desired) output signal, i.e. the reference input signal which specifies what is required, and the actual output signal. For the unityfeedback control system shown in the figure, the error is: E G _ E G Using the finalvalue theorem (assuming the system is stable), the steadystate error is: e ss s e( ) lim e(t) limse lim t G A general representation of G is: G (s b s m m m N n n s (s a ns L b s b L a s a where is a constant and m and n are integers and neither a 0 nor b 0 is zero. N is an integer, the value of which is called the type of the system. Thus if N 0 then the system is said to be type 0, if N then type, if N then type and so on. The type number is thus the number of /s factors in the openloop transfer function G. Since /s represents integration, the type number is the number of integrators in the openloop transfer function. Exercise: What are the type numbers for the systems shown in the following figures? 0 0 ) ) (s ) (s )(s 6) (s ) s(s )(s 6) 5
16 5.. Static error constants Just as we defined the damping ratio, the natural frequency, settling time, percentage overshoot etc. as performance specifications for the transient response of a system, we can use static error constants to specify the steadystate error characteristics of control systems Static position error constant po The steadystate error of the unityfeedback control system for a unitstep input is: e ss s e( ) lim e(t) limse lim t s( G) G(0) The static position error constant po is defined by: lim G G(0) po s 0 Thus, the steadystate error in terms of the static position error constant po is; ess For a type 0 system, po C where C is a constant. For a type system, po. Hence, for a type 0 system, the static position error constant po is finite, while for a type or higher system, po is infinite. This is, e ss /(C) for a type 0 system, and e ss 0 for a type or higher system. Exercise: Find the static position error constants and steadystate errors, respectively, of the systems shown in the following figures for a unitstep input. po s s s s(s ) 6
17 5... Static velocity error constant v The steadystate error of a system with a unitramp (velocity) input is: e ss s e( ) lim e(t) lim se lim t s ( G) lim sg The static velocity error constant v is defined by: limsg v s 0 e ss For a type 0 system, limsg 0. For a type system, limsg C. For a type or higher system, v s 0 limsg. v s 0 v s Static acceleration error constant a The steadystate error of a system with a unitparabolic (acceleration) input, r(t) t /, is: s e( ) lim e(t) lim se lim lim t 3 s ( G) s G ess The static acceleration error constant a is defined by: a lims G s 0 e ss For a type 0 system, lims G 0. For a type system, lims G 0. For a type system, a s 0 lims G C a s 0 a s 0 lims G s 0.. For a type 3 or higher system, Exercise: A unity feedback system has the following forward transfer function: 000(s 8) G (s 7)(s 9) (a) Evaluate the system type, po, v, and a. (b) Use your answers to (a) to find the steadystate errors for the unitstep, unitramp, and unitparabolic inputs. 7
18 Exercise: For the unity feedback system shown, determine the value of to yield a 0% error in the steadystate when the input is r(t). s (s 4)(s 8) 8
19 5.. SteadyState Error for Disturbances D E G c G The error component when D 0 is: E  Y R GcG where the output due to the reference input, Y R, is given by: YR G G The steadystate error value due to is given by the final value theorem: e R s ( ) lim se lim G G c c When 0, the error in this case is given by: E 0  Y D where the output due to the disturbance, Y D, is: Y D G D G G The steadystate error value due to D is found using final value theorem: e D c sgd ( ) lim se lim G G The steadystate error due to the reference and disturbance inputs is: e( ) e R ( ) e D ( ) Exercise: For the block diagram shown below, determine the steadstate error component due to a unitstep disturbance. D c 00 s(s 5) 9
20 TUTORIAL PROBLEM SHEET 5. List the major advantages and disadvantages of closedloop control systems.. For the system shown in the figure, what are the steadystate errors when a unitstep input is applied to the following openloop transfer functions G: (a) (b) (c) 0 G (s )(s ) 6(s 3) G (s 6)(s ) 0 G s(s )(s ) E G 3. What are the type numbers for the systems shown in the following figures? (s ) (s )(s 6) (a) (s ) s(s )(s 6) 4. Determine the steadystate error for the system shown in question () above with: (s ) G s (s ) when subjected to the input r(t) t t. Plot the time response y(t) with MATLAB. (b) 5. Find the static position error constants and steadystate errors, respectively, of the systems shown in the following figures for a unitstep input. s s s s(s ) (a) (b) Plot the time response y(t) for a unitstep input using MATLAB. 0
21 6. For the system shown in the following figure: s(s 5)(s 0) (a) What value of will yield a steadystate error in position of 0.0 for an input of r(t) t/0? (b) What is the value of v for the value of found in (a)? (c) Plot the time response using MATLAB. 7. Find the total steadystate error due to a unitstep input and a unitstep disturbance in the system shown below. D 00 s 5 s 8. Consider the system shown in the following figure with G 5/(s ). What are the values of the gain which will achieve the following design specifications? D G (a) Steadystate error due to a unitstep input to be less than 0.. (b) Steadystate error due to a unitstep change in disturbance D to be less than 0.. (c) Validate the above results using MATLAB, that is to plot the time response for a unitstep input and a unitstep disturbance, respectively. 9. Redo question (8) for 0. s 0s 50 G
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 pickandplace robot to move the link of a robot between two positions. Usually
More informationManufacturing Equipment Control
QUESTION 1 An electric drive spindle has the following parameters: J m = 2 1 3 kg m 2, R a = 8 Ω, K t =.5 N m/a, K v =.5 V/(rad/s), K a = 2, J s = 4 1 2 kg m 2, and K s =.3. Ignore electrical dynamics
More informationDC 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
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BSC (HONS) MECHATRONICS TOPUP SEMESTER 1 EXAMINATION 2017/2018 ADVANCED MECHATRONIC SYSTEMS
ENG08 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BSC (HONS) MECHATRONICS TOPUP SEMESTER EXAMINATION 07/08 ADVANCED MECHATRONIC SYSTEMS MODULE NO: MEC600 Date: 7 January 08 Time: 0.00.00 INSTRUCTIONS TO
More informationQuanser NIELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual
Quanser NIELVIS 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 informationMECHATRONICS 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 informationME 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
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationCHAPTER 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 informationExample: Modeling DC Motor Position Physical Setup System Equations Design Requirements MATLAB Representation and OpenLoop Response
Page 1 of 5 Example: Modeling DC Motor Position Physical Setup System Equations Design Requirements MATLAB Representation and OpenLoop Response Physical Setup A common actuator in control systems is the
More information3 Lab 3: DC Motor Transfer Function Estimation by Explicit Measurement
3 Lab 3: DC Motor Transfer Function Estimation by Explicit Measurement 3.1 Introduction There are two common methods for determining a plant s transfer function. They are: 1. Measure all the physical parameters
More informationC(s) R(s) 1 C(s) C(s) C(s) = s  T. Ts + 1 = 1 s  1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain
analyses of the step response, ramp response, and impulse response of the secondorder systems are presented. Section 5 4 discusses the transientresponse analysis of higherorder systems. Section 5 5 gives
More informationMechatronics Engineering. Li Wen
Mechatronics Engineering Li Wen Bioinspired robotdc 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 informationControl 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 informationExample: DC Motor Speed Modeling
Page 1 of 5 Example: DC Motor Speed Modeling Physical setup and system equations Design requirements MATLAB representation and openloop response Physical setup and system equations A common actuator in
More informationR10 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 informationIndex. Index. More information. in this web service Cambridge University Press
Atype elements, 4 7, 18, 31, 168, 198, 202, 219, 220, 222, 225 Atype variables. See Across variable ac current, 172, 251 ac induction motor, 251 Acceleration rotational, 30 translational, 16 Accumulator,
More informationAppendix 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 informationDr 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 splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More information2.004 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
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 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 informationBangladesh 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 informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More informationSubject: BT6008 Process Measurement and Control. The General Control System
WALJAT COLLEGES OF APPLIED SCIENCES In academic partnership with BIRLA INSTITUTE OF TECHNOLOGY Question Bank Course: Biotechnology Session: 005006 Subject: BT6008 Process Measurement and Control Semester:
More informationSystem Modeling: Motor position, θ The physical parameters for the dc motor are:
Dept. of EEE, KUET, Sessional on EE 3202: Expt. # 2 2k15 Batch Experiment No. 02 Name of the experiment: Modeling of Physical systems and study of their closed loop response Objective: (i) (ii) (iii) (iv)
More informationENGG4420 LECTURE 7. CHAPTER 1 BY RADU MURESAN Page 1. September :29 PM
CHAPTER 1 BY RADU MURESAN Page 1 ENGG4420 LECTURE 7 September 21 10 2:29 PM MODELS OF ELECTRIC CIRCUITS Electric circuits contain sources of electric voltage and current and other electronic elements such
More informationLab 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 informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage 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, Forcecurrent and ForceVoltage analogies..
More informationAutomatic Control Systems. Lecture Note 15
Lecture Note 15 Modeling of Physical Systems 5 1/52 AC Motors AC Motors Classification i) Induction Motor (Asynchronous Motor) ii) Synchronous Motor 2/52 Advantages of AC Motors i) Costeffective ii)
More informationKINGS 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 informationPerformance 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 Steadystate Error and Type 0, Type
More informationsc Control Systems Design Q.1, Sem.1, Ac. Yr. 2010/11
sc46  Control Systems Design Q Sem Ac Yr / Mock Exam originally given November 5 9 Notes: Please be reminded that only an A4 paper with formulas may be used during the exam no other material is to be
More informationME 3210 Mechatronics II Laboratory Lab 4: DC Motor Characteristics
ME 3210 Mechatronics II Laboratory Lab 4: DC Motor Characteristics Introduction Often, due to budget constraints or convenience, engineers must use whatever tools are available to create new or improved
More information6.1 Sketch the zdomain root locus and find the critical gain for the following systems K., the closedloop characteristic equation is K + z 0.
6. Sketch the zdomain 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 informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE 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 informationThe 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
More information(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
More informationIntroduction 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 information06 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
More informationSchool of Mechanical Engineering Purdue University. ME375 Feedback Control  1
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop 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:
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 informationUNIVERSITY OF WASHINGTON Department of Aeronautics and Astronautics
UNIVERSITY OF WASHINGTON Department of Aeronautics and Astronautics Modeling and Control of a Flexishaft System March 19, 2003 Christopher Lum Travis Reisner Amanda Stephens Brian Hass AA/EE448 Controls
More informationFeedback Control Systems
ME Homework #0 Feedback Control Systems Last Updated November 06 Text problem 67 (Revised Chapter 6 Homework Problems attached) 65 Chapter 6 Homework Problems 65 Transient Response of a Second Order Model
More informationState Feedback Controller for Position Control of a Flexible Link
Laboratory 12 Control Systems Laboratory ECE3557 Laboratory 12 State Feedback Controller for Position Control of a Flexible Link 12.1 Objective The objective of this laboratory is to design a full state
More informationLaboratory Exercise 1 DC servo
Laboratory Exercise DC servo PerOlof Källén ø 0,8 POWER SAT. OVL.RESET POS.RESET Moment Reference ø 0,5 ø 0,5 ø 0,5 ø 0,65 ø 0,65 Int ø 0,8 ø 0,8 Σ k Js + d ø 0,8 s ø 0 8 Off Off ø 0,8 Ext. Int. + x0,
More informationLaboratory 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
More informationDepartment of Mechanical Engineering
Department of Mechanical Engineering 2.010 CONTROL SYSTEMS PRINCIPLES Laboratory 2: Characterization of the ElectroMechanical Plant Introduction: It is important (for future lab sessions) that we have
More informationMechatronic System Case Study: Rotary Inverted Pendulum Dynamic System Investigation
Mechatronic System Case Study: Rotary Inverted Pendulum Dynamic System Investigation Dr. Kevin Craig Greenheck Chair in Engineering Design & Professor of Mechanical Engineering Marquette University K.
More informationElectrical Machine & Automatic Control (EEE409) (MEII Yr) UNIT3 Content: Signals u(t) = 1 when t 0 = 0 when t <0
Electrical Machine & Automatic Control (EEE409) (MEII Yr) UNIT3 Content: Modeling of Mechanical : linear mechanical elements, forcevoltage and force current analogy, and electrical analog of simple
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationRadar 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
More informationControl 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 informationLecture 25: Tue Nov 27, 2018
Lecture 25: Tue Nov 27, 2018 Reminder: Lab 3 moved to Tuesday Dec 4 Lecture: review timedomain characteristics of 2ndorder systems intro to control: feedback openloop vs closedloop control intro to
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
More informationINC 341 Feedback Control Systems: Lecture 3 Transfer Function of Dynamic Systems II
INC 341 Feedback Control Systems: Lecture 3 Transfer Function of Dynamic Systems II Asst. Prof. Dr.Ing. Sudchai Boonto Department of Control Systems and Instrumentation Engineering King Mongkut s University
More informationSchool of Mechanical Engineering Purdue University. ME375 ElectroMechanical  1
ElectroMechanical Systems DC Motors Principles of Operation Modeling (Derivation of fg Governing Equations (EOM)) Block Diagram Representations Using Block Diagrams to Represent Equations in s  Domain
More informationLecture 12. Upcoming labs: Final Exam on 12/21/2015 (Monday)10:3012:30
289 Upcoming labs: Lecture 12 Lab 20: Internal model control (finish up) Lab 22: Force or Torque control experiments [Integrative] (23 sessions) Final Exam on 12/21/2015 (Monday)10:3012:30 Today: Recap
More informationPID Control. Objectives
PID Control Objectives The objective of this lab is to study basic design issues for proportionalintegralderivative control laws. Emphasis is placed on transient responses and steadystate errors. The
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN MECHANICAL ENGINEERING SEMESTER 1EXAMINATION 2017/2018
ENG00 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN MECHANICAL ENGINEERING SEMESTER EXAMINATION 07/08 ADVANCED THERMOFLUIDS & CONTROL SYSTEMS MODULE NO: AME6005 Date: 8 January 08 Time: 0.00.00
More informationOverview of motors and motion control
Overview of motors and motion control. Elements of a motioncontrol system Power upply Highlevel controller owlevel controller Driver Motor. Types of motors discussed here; Brushed, PM DC Motors Cheap,
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationVideo 5.1 Vijay Kumar and Ani Hsieh
Video 5.1 Vijay Kumar and Ani Hsieh Robo3x1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More information6) Motors and Encoders
6) Motors and Encoders Electric motors are by far the most common component to supply mechanical input to a linear motion system. Stepper motors and servo motors are the popular choices in linear motion
More informationControl Systems. University Questions
University Questions UNIT1 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 informationStepping Motors. Chapter 11 L E L F L D
Chapter 11 Stepping Motors In the synchronous motor, the combination of sinusoidally distributed windings and sinusoidally time varying current produces a smoothly rotating magnetic field. We can eliminate
More informationRobust Controller Design for Speed Control of an Indirect Field Oriented Induction Machine Drive
Leonardo Electronic Journal of Practices and Technologies ISSN 15831078 Issue 6, JanuaryJune 2005 p. 116 Robust Controller Design for Speed Control of an Indirect Field Oriented Induction Machine Drive
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationMechatronics Modeling and Analysis of Dynamic Systems CaseStudy Exercise
Mechatronics Modeling and Analysis of Dynamic Systems CaseStudy Exercise Goal: This exercise is designed to take a realworld problem and apply the modeling and analysis concepts discussed in class. As
More informationOutline. 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 informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
More informationSRV02Series Rotary Experiment # 1. Position Control. Student Handout
SRV02Series Rotary Experiment # 1 Position Control Student Handout SRV02Series Rotary Experiment # 1 Position Control Student Handout 1. Objectives The objective in this experiment is to introduce the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance
More information(a) Torsional springmass system. (b) Spring element.
m v s T s v a (a) T a (b) T a FIGURE 2.1 (a) Torsional springmass system. (b) Spring element. by ky Wall friction, b Mass M k y M y r(t) Force r(t) (a) (b) FIGURE 2.2 (a) Springmassdamper system. (b)
More informationBASIC PROPERTIES OF FEEDBACK
ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)
More informationCourse 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: SteadyState Error Unit 7: Root Locus Techniques
More informationEDEXCEL NATIONALS UNIT 5  ELECTRICAL AND ELECTRONIC PRINCIPLES. ASSIGNMENT No. 3  ELECTRO MAGNETIC INDUCTION
EDEXCEL NATIONALS UNIT 5  ELECTRICAL AND ELECTRONIC PRINCIPLES ASSIGNMENT No. 3  ELECTRO MAGNETIC INDUCTION NAME: I agree to the assessment as contained in this assignment. I confirm that the work submitted
More informationSECTION 4: STEADY STATE ERROR
SECTION 4: STEADY STATE ERROR MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Steady State Error Introduction 3 Consider a simple unity feedback system The error is the difference between
More informationFATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY
FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY Senkottai Village, Madurai Sivagangai Main Road, Madurai  625 020. An ISO 9001:2008 Certified Institution DEPARTMENT OF ELECTRONICS AND COMMUNICATION
More informationAutomatic 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 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More informationRoot Locus Design Example #4
Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is
More informationRotary Motion Servo Plant: SRV02. Rotary Experiment #01: Modeling. SRV02 Modeling using QuaRC. Student Manual
Rotary Motion Servo Plant: SRV02 Rotary Experiment #01: Modeling SRV02 Modeling using QuaRC Student Manual SRV02 Modeling Laboratory Student Manual Table of Contents 1. INTRODUCTION...1 2. PREREQUISITES...1
More informationSpontaneous 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
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequencydomain analysis and control design (15 pt) Given is a
More information1 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
More informationKing Saud University
motor speed (rad/sec) Closed Loop Step Response ing Saud University College of Engineering, Electrical Engineering Department Labwork Manual EE 356 Control and Instrumentation Laboratory (كهر 356 معمل
More informationSurvey of Methods of Combining Velocity Profiles with Position control
Survey of Methods of Combining Profiles with control Petter Karlsson Mälardalen University P.O. Box 883 713 Västerås, Sweden pkn91@student.mdh.se ABSTRACT In many applications where some kind of motion
More informationCoupled Drive Apparatus Modelling and Simulation
University of Ljubljana Faculty of Electrical Engineering Victor Centellas Gil Coupled Drive Apparatus Modelling and Simulation Diploma thesis Menthor: prof. dr. Maja AtanasijevićKunc Ljubljana, 2015
More informationRotary Motion Servo Plant: SRV02. Rotary Experiment #11: 1DOF Torsion. 1DOF Torsion Position Control using QuaRC. Student Manual
Rotary Motion Servo Plant: SRV02 Rotary Experiment #11: 1DOF Torsion 1DOF Torsion Position Control using QuaRC Student Manual Table of Contents 1. INTRODUCTION...1 2. PREREQUISITES...1 3. OVERVIEW OF
More informationLIAPUNOV S STABILITY THEORYBASED MODEL REFERENCE ADAPTIVE CONTROL FOR DC MOTOR
LIAPUNOV S STABILITY THEORYBASED MODEL REFERENCE ADAPTIVE CONTROL FOR DC MOTOR *Ganta Ramesh, # R. Hanumanth Nayak *#Assistant Professor in EEE, Gudlavalleru Engg College, JNTU, Kakinada University, Gudlavalleru
More informationEE 410/510: Electromechanical Systems Chapter 4
EE 410/510: Electromechanical Systems Chapter 4 Chapter 4. Direct Current Electric Machines and Motion Devices Permanent Magnet DC Electric Machines Radial Topology Simulation and Experimental Studies
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
More informationControl 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
More informationWHAT A SINGLE JOINT IS MADE OF RA
Anthropomorphic robotics WHAT A SINGLE JOINT IS MADE OF Notation d F ( mv) mx Since links are physical objects with mass dt J J f i i J = moment of inertia F r F r Moment of inertia Around an axis m3 m1
More informationLezione 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
More informationEC6405  CONTROL SYSTEM ENGINEERING Questions and Answers Unit  I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
More informationFast Seek Control for Flexible Disk Drive Systems
Fast Seek Control for Flexible Disk Drive Systems with Back EMF and Inductance Chanat Laorpacharapan and Lucy Y. Pao Department of Electrical and Computer Engineering niversity of Colorado, Boulder, CO
More informationTexas A & M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr. Alexander G. Parlos
Texas A & M University Department of Mechanical Engineering MEEN 364 Dynamic Systems and Controls Dr. Alexander G. Parlos Lecture 6: Modeling of Electromechanical Systems Principles of Motor Operation
More information(a) Torsional springmass system. (b) Spring element.
m v s T s v a (a) T a (b) T a FIGURE 2.1 (a) Torsional springmass system. (b) Spring element. by ky Wall friction, b Mass M k y M y r(t) Force r(t) (a) (b) FIGURE 2.2 (a) Springmassdamper system. (b)
More informationRigid Manipulator Control
Rigid Manipulator Control The control problem consists in the design of control algorithms for the robot motors, such that the TCP motion follows a specified task in the cartesian space Two types of task
More information