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) =


 Eleanor Lawrence
 3 years ago
 Views:
Transcription
1 567 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 less as veriþed below. The estimatorerror equation : x(k +)=(Φ LH) x(k) = T x (k) T x 2 (k) For k =,, 2, 3,... x() = x(2) = x(3) = x () T x 2() T x () x 2 (). (c) Yes, since one state element is measured directly, one can estimate the other element with a Þrstorder estimator following the development for reduced order estimators for contintuous systems in Section From the discrete statespace representation, x 2 (k +)=Tx (k)+x 2 (k)+ T 2 2 u(k) = x (k) = T x 2 (k +) x 2 (k) T 2 2 u(k) {z } known measurement and input For K =, x () = x 2 () x 2 () T 2 T 2 u() = y() y() T 2 T 2 u() x 2 () = x 2 () = y() 4. Singleaxis Satellite Attitude Control: Satellites often require attitude control for proper orientation of antennas and sensors with respect to Earth.
2 568 CHAPTER 8. DIGITAL CONTROL Figure 8.23: Satellite control schematic for Problem 4 Figure 2.6 shows a communication satellite with a threeaxis attitudecontrol system. To gain insight into the threeaxis problem we often consider one axis at a time. Figure 8.23 depicts this case where motion is only allowed about an axis perpendicular to the page. The equations of motion of the system are given by where I θ = M C + M D, I = moment of inertia of the satellite about its mass center, M C =control torque applied by the thrusters, M D =disturbance torques, θ =angle of the satellite axis with respect to an inertial reference with no angular acceleration. We normalize the equations of motion by deþning and obtain u = M C I, w d = M D I, θ = u + w d. Taking the Laplace transform yields which with no disturbance becomes θ(s) = s 2 [u(s)+w d(s)], θ(s) u(s) = s 2 = G (s).
3 569 Inthediscretecasewhereuis applied through a ZOH, we can use the methods described in this chapter to obtain the discrete transfer function G (z) = θ(z) u(z) = T 2 z + 2 (z ) 2. (a) Sketch the root locus of this system by hand assuming proportional control. (b) Draw the root locus using MATLAB to verify the hand sketch. (c) Add a lead network to your controller so that the dominant poles correspond to ζ =.5 andω n =3π/(T ). (d) What is the feedback gain if T = sec? If T =2sec. (e) Plot the closedloop step response and the associated control time history for T =sec. Solution (a) The hand sketch will show that the loci branches depart vertically from z = ; therefore, the system is marginally stable or unstable foranyvalueofgain. (b) The Matlab version below conþrms the situation. G (z) = T 2 (z +) 2 (z ) 2 = K (z +) (z ) 2 (c) To obtain the desired damping and frequency, Fig. 8.4 shows that a root locus branch should go through the desired poles at
4 57 CHAPTER 8. DIGITAL CONTROL z =.44 ±.44j. After some trial and error, you can Þnd that this can be accomplished with the lead compensation : (z.63) D(z) =K (z +.44) The speciþc value of K that yields the closedloop poles are at : z =.44 ±.44j,.3 is K =.692 K. The secondorder pair give : ω n =.97 T ζ =.59 rad/sec which is close enough to the desired speciþcations. locus for the compensated design is: The root (d) K =.692 T 2 2 ½ ¾.383 for T =sec =.3458 for T =2sec
5 u theta 57 (e) Closedloop step response : ClosedLoop Step Response (T=s) Time (sec).5 Time History of Control Effort (T=s) Time (sec) 5. In this problem you will show how to compute Φ by changing states so
6 a b (c) Let T = c d, TA = FT = 2a = c, b = d 573 = Choosing c = d =,thena =.5, b= c = d =.5 = T = This T satisþes TAT = F. Note : Let T = v v 2, TFT = A = v v 2 F v v 2 = 2 ½ ¾ Fv =( )v = Fv 2 =( 2)v 2 (d) Thus, the columns v and v 2 are the eigenvectors of F. e FT = Te AT T.5 = e T 2 2 e 2T 2 e = T +2e 2T e T e 2T 2e T +2e 2T 2e T e 2T 6. It is possible to suspend a mass of magnetic material by means of an electromagnet whose current is controlled by the position of the mass (Woodson and Melcher, 968). The schematic of a possible setup is shown in Fig. 8.24, and a photo of a working system at Stanford University is shown in Fig The equations of motion are mẍ = mg + f(x, I), where the force on the ball due to the electromagnet is given by f(x, I). At equilibrium the magnet force balances the gravity force. Suppose we let I represent the current at equilibrium. If we write I = I + i, expand f about x =andi = I, and neglect higherorder terms, we obtain the linearized equation mẍ = k x + k 2 i. () Reasonable values for the constants in Eq. () are m =.2 kg, k =2 N/m, and k 2 =.4 N/A. (a) Compute the transfer function from I tox, and draw the (continuous) root locus for the simple feedback i = Kx.
7 574 CHAPTER 8. DIGITAL CONTROL Figure 8.24: Schematic of magnetic levitation device for Problems6 and 7 (b) Assume the input is passed through a ZOH, and let the sampling period be.2 sec. Compute the transfer function of the equivalent discretetime plant. (c) Design a digital control for the magnetic levitation device so that the closedloop system meets the following speciþcations: t r.sec, t s.4 sec, and overshoot 2%. (d) Plot a root locus with respect to k foryourdesign,anddiscussthe possibility of using your closedloop system to balance balls of various masses. (e) Plot the step response of your design to an initial disturbance displacement on the ball, and show both x and the control current i. If the sensor can measure x only over a range of ±/4cm and the ampliþer can only provide a current of A, what is the maximum displacement possible for control, neglecting the nonlinear terms in f(x, I)? Solution: (a) G(s) = X(s) I(s) = k 2/m s 2 k /m 2 = s 2 (2)
8 575 (b) T =.2 sec, G(z) = ½ ¾ G(s) ( z )Z s = z (z.533)(z.8822) (c) The speciþcations imply that : t r. sec= ω n.8 = 8 rad/sec. t s.4 sec= σ 4.6 =.5 rad/sec.4 = r = z e.5.2 =.7945 ( z = e st ) M p 2% = ζ.48 Thus, the closedloop poles must be pulled into the unit circle near r =.8 andζ =.5. Using the template of Fig. 8.4, we experiment with lead compensation and select, D(z) = 6 z.533 z.93 The closedloop poles are : z =.75 ±.39j,.53 Performance : t r =.72 t s =.397 M p = 8.3% which meet all the speciþcations. The root locus is below.
9 x [m] 576 CHAPTER 8. DIGITAL CONTROL The step response shows M p 8 ClosedLoop Step Response Time (sec)
10 i [A] Time History of Control Effort Time (sec) (d) As can be seen from Eq. (2), the loop gain and the open loop pole locations depend on the mass of the ball. Changing the mass will therefore affect the dynamic characteristics of the system and may render it unstable. A root locus of the closedloop poles versus k shows how the locus changes as a function of the mass: The closedloop system becomes unstable for k 24. Since a small increase in k makes the system unstable and a decrease
11 i [A] x [m] 578 CHAPTER 8. DIGITAL CONTROL in m has the same effect on the system, it is difficult to balance balls of smaller masses. (e) The response to an initial x displacement is shown : ClosedLoop Initial Response Time (sec) Time History of Control Effort Time (sec)
12 579 The assumption here is that an allowable transient must stay in the range of the sensor and not require more than the limit of the current. From I(z) = D(z)( X(z)), we have a difference equation : i(k).93i(k ) = 6{x(k).533x(k )} For k =, i() = 6x(). We see that if x() = then i() = 6. Note that i() = D( )x(). Thus, if i is to be kept below A then x()mustbekeptbelow /6 =.862 m =.862 cm =.339 inch, which is greater than the sensor range. The current control can handle any displacement in the range of ±.25 inch. 7. In Problem 6 we described an experiment in magnetic levitation described by Eq. () which reduces to Let the sampling time be. sec. ẍ =x +2i. (a) Use pole placement to design a controller for the magnetic levitator so that the closedloop system meets the following speciþcations: settling time, t s.25 sec, and overshoot to an initial offset in x that is less than 2%. (b) Plot the step response of x, x, andi to an initial displacement in x. (c) Plot the root locus for changes in the plant gain, and mark the pole locations of your design. (d) Introduce a command reference input r (as discussed in Section 7.8) that does not excite the estimate of x. Measure or compute the frequency response from r to the system error r x and give the highest frequency for which the error amplitude is less than 2% of the command amplitude. Solution (Note: part (d) of this problem is a stretch for the students. The text doesn t cover the relevant discrete development and the student would be helped by referring to Digital Control of Dynamic Systems, by Franklin, Powell, and Workman.) (a) Statespace representation of the plant : x x x = =,y= x, u = i úx d x dt x 2 x 2 = = Fx + Gu x x u
13 y y 66 CHAPTER 8. DIGITAL CONTROL.2 Step Response (T=.).8 continuous tustin and MPZ time (sec).5 Step Response (T=.).5 Continuous Tustin and MPZ time (sec) 22. Repeat Problem 5.29 by constructing discrete root loci and performing the designs directly in the zplane. Assume that the output y is sampled,
14 67 the input u is passed through a ZOH as it enters the plant, and the sample rate is 5 Hz. Solution (a) The most effective discrete design method is to start with some idea what the continuous design looks like, then adjust that as necessary with the discrete model of the plant and compensation. We refer to the solution for Probelm 5.29 for the starting point. It shows that the specs can be met with a lead compensation, and a lag compensation, (s +) D (s) =K (s + 6) D 2 (s) = (s +.4) (s +.32). Although it is stated in the solution to Problem 5.29 that a gain, K = 24 will satisfy the constraints, in fact, a gain of about K = 27 is actually required to meet the rise time constraint of t r.4 sec. So we will assume here that our reference continuous design is (s +) (s +.4) D (s) = 27 (s + 6) (s +.32) It yields a rise time, t r =.38,Mp = 5%, and Kv = (27)(.4 6 )(.32 )= 56. So all specs are met. For interest, use of the damp function shows that ω n =6.4 rad/sec for the dominant roots, and those roots have a damping ratio, ζ =.7. For the discrete case with T =5Hz, we should expect some degradation in performance, especially the damping, because the sample rate is approximately 5 ω n. The discrete transfer function for the plant described by G(s) and preceeded by a ZOH is: ½ ¾ G(s) G(z) = ( z )Z s = z Z z This is most easily determined via Matlab, sysc = tf([],[ ]); T=/5; sysd = c2d(sysc,t, zoh ); which produces: ½ s(s +)(s + ) ¾
15 68 CHAPTER 8. DIGITAL CONTROL (z +3.36)(z +.22) G(z) =.4424 (z )(z.9355)(z.534 The essential elements of the compensation are that the lead provides velocity feedback with a T D = and the lag provides some high frequency gain. The discrete equivalent of the proportional plus lead would be (see Eq. 8.42): D (z) =K( + T D T ( z )) = K ( + T D/T )z T D /T z which for T =/5 =.667 and T D = reduces to 6z 5z D (z) = 27 z = 432 z z The lag equivalent is best introduced by use of one of the approximation techniques, such as the matched polezero: D 2 (z) = z e.4t z.9737 = z e.32t z.9979 as its whole function is to raise the gain at very low frequencies for error reduction. Examining the resulting discrete root locus and picking roots with rlocfind to yield the required damping shows that the gain, K = 6. While the use of damp indicates that the frequency and damping are acceptable, the time response shows an overshoot of about 2% and the rise time is slightly below spec. We therefore need to increase the lead (move the lead zero closer to z =+)to decrease the overshoot and increase gain to speed up the rise time. Several iterations on these two quantities indicates that moving the lead zero from z =.9375 to z =.96 and increasing the gain from K = 6toK = 65 meets both specs. The velocity coefficient is found from and is also satisþed. K v = lim (z )D(z)G(z) z Tz ThetimeresponseoftheÞnal design below shows that all specs are met.
16 y 69.2 Step Response time (sec) 23. Design a digital controller for the antenna servo system shown in Figs and3.64anddescribedinproblem3.3. Thedesignshouldprovideastep response with an overshoot of less than % and a rise time of less than 8 sec. (a) What should the sample rate be? (b) Use emulation design with the matched polezero method. (c) Use discrete design and the zplane root locus. Solution (a) The equation of motion is : J θ + Bθ = T c where J = 6 kg.m 2,B= 2 N.m.sec If we deþne : a = B J = 3,U= T c B after Laplace transform, we obtain : G(s) = θ(s) u(s) = s(3s +)
17 y 69.2 Step Response time (sec) 23. Design a digital controller for the antenna servo system shown in Figs and3.64anddescribedinproblem3.3. Thedesignshouldprovideastep response with an overshoot of less than % and a rise time of less than 8 sec. (a) What should the sample rate be? (b) Use emulation design with the matched polezero method. (c) Use discrete design and the zplane root locus. Solution (a) The equation of motion is : J θ + Bθ = T c where J = 6 kg.m 2,B= 2 N.m.sec If we deþne : a = B J = 3,U= T c B after Laplace transform, we obtain : G(s) = θ(s) u(s) = s(3s +)
18 6 CHAPTER 8. DIGITAL CONTROL From the speciþcations, µ M p < % = M P = ζ = ζ >.54.6 t r < 8 sec = t r.8 = < 8 = ω n = ωbw >.225 ω n Note that ω pn = /3 < ωn. If designing by emulation, a sample rate of 2 times the bandwidth is recommended. If using discrete design, the sample rate can be lowered somewhat to perhaps as slow as times the bandwidth. However, to reject random disturbances, best results are obtained by sampling at 2 times the closedloop bandwidth or faster. Thus, for both design methods, we choose : T = sec ω s =.628 rad/sec, which is > 2ω n =.45 rad/sec (b) Continuous design : Use a proportional controller : u(s) =D(s)(θ r (s) θ(s)) = K(θ r (s) θ(s)) Root locus : Choose K =.2. The closedloop pole location in splane : s =.67 ±.25j The corresponding natural frequency and damping : ω n =.265, ζ =.6299
19 6 Digitized the continuous controller with matched polezero method : Performance : D(z) =.2 T c (z) = Bu(z) = 42(θ r (z) θ(z)) M p =.9 t r = 67.3 sec (c) With u(k) applied through a ZOH, the transfer function for an equivalent discretetime system is : where z + b G(z) =K (z )(z e at ) K = at +e at,b= e at at e at a at +e at = z G(z) =.4959 (z )(z.765) Use a proportional control of the form : Root locus : D(z) =K
20 62 CHAPTER 8. DIGITAL CONTROL The speciþcation can be achieved with the proportional control. However, we try to achieve the same closedloop poles as the emulation design (part (b)) for comparison. These closedloop pole locations are denoted by + in the root locus. UseaPDcontroloftheform: D(z) =K z α z Root locus : Choose K =.294, α =.3. The resulting zplane roots : z =.828 ±.725j,.65 This corresponds to the splane roots : s =.67±.25j (the design point of emulation design),.44 which satisfy the speciþcation : The control law : ω n =.265,.44 ζ =.632,. D(z) =.294 z.3 z T c (z) = Bu(z) = 588 z.3 z
21 u (Tc/B) x (m) 63 Performance : M p =.79 t r = 7.3 sec Step response :.2 Step Response emulation design discrete design time (sec).4 Time History of Control Effort.3 discrete design.2 emulation design time (sec) 24. The system G(s) = (s +.)(s +3)
22 64 CHAPTER 8. DIGITAL CONTROL is to be controlled with a digital controller having a sampling period of T =. sec. Using a zplane root locus, design compensation that will respond to a step with a rise time t r secandanovershootm p 5%. What can be done to reduce the steadystate error? Solution (a) Continuous plant : G(s) =, Type system (s +.)(s +3) Discrete model of G(s) preceeded by a ZOH (T =. sec) : z +.99 G(z) =.45 (z.748)(z.99) SpeciÞcations : t r sec ω n.8 rad/sec M p 5% ζ.7 Discrete design : A simple proportional feedback, D(z) = K = 4., will bring the closedloop poles to : which are inside the specs region. Root locus : z =.8564 ±.278j ω n =2.7 rad/sec, ζ =.7
23 u y 65 Step response : ClosedLoop Step Response Time (sec) Time History of Control Effort Thestepresponseshowsthat: Time (sec) t r =.2 sec M p = 4.7% However, since the system is type, steadystate error exists and is 7% in this case. An integral control of the form, D(z) = K T I Tz z
24 66 CHAPTER 8. DIGITAL CONTROL can be added to the proportional control to reduce the steadystate error, but this typically occurs at the cost of reduced stability. 25. The transfer function for pure derivative control is z D(z) =KT D Tz, where the pole at z = adds some destabilizing phase lag. Can this phase lag be removed by using derivative control of the form (z ) D(z) =KT D? T Support your answer with the difference equation that would be required, and discuss the requirements to implement it. Solution: (825) (a) No, we cannot use derivative control of the form : D(z) =KT D z T to remove the phase lag. The difference equation corresponding to D(z) =K p T D z T = U(z) E(z) is e(k +) e(k) u(k) =K p T D T This is not a causal system since it needs the future error signal to compute the current control. In real time applications, it is not possible to implement a noncausal system.
6.1 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. 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 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 informationProblem Set 3: Solution Due on Mon. 7 th Oct. in class. Fall 2013
EE 56: Digital Control Systems Problem Set 3: Solution Due on Mon 7 th Oct in class Fall 23 Problem For the causal LTI system described by the difference equation y k + 2 y k = x k, () (a) By first finding
More informationMechatronics Assignment # 1
Problem # 1 Consider a closedloop, rotary, speedcontrol system with a proportional controller K p, as shown below. The inertia of the rotor is J. The damping coefficient B in mechanical systems is usually
More informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More informationPositioning 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 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 informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationFEEDBACK CONTROL SYSTEMS
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
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationDIGITAL CONTROLLER DESIGN
ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steadystate accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More informationAnalysis 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.
More informationDISTURBANCE ATTENUATION IN A MAGNETIC LEVITATION SYSTEM WITH ACCELERATION FEEDBACK
DISTURBANCE ATTENUATION IN A MAGNETIC LEVITATION SYSTEM WITH ACCELERATION FEEDBACK Feng Tian Department of Mechanical Engineering Marquette University Milwaukee, WI 53233 USA Email: feng.tian@mu.edu Kevin
More informationImplementation Issues for the Virtual Spring
Implementation Issues for the Virtual Spring J. S. Freudenberg EECS 461 Embedded Control Systems 1 Introduction One of the tasks in Lab 4 is to attach the haptic wheel to a virtual reference position with
More 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 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 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 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013 Problem Set #4 Posted: Thursday, Mar. 7, 13 Due: Thursday, Mar. 14, 13 1. Sketch the Root
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 informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus
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 informationK(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s
321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationController Design using Root Locus
Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA EXPERIMENT NO : CS II/ TITLE : FAMILIARIZATION
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 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 informationTopic # 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
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
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 informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More information(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 informationDepartment of Electronics and Instrumentation Engineering M. E CONTROL AND INSTRUMENTATION ENGINEERING CL7101 CONTROL SYSTEM DESIGN Unit I BASICS AND ROOTLOCUS DESIGN PARTA (2 marks) 1. What are the
More informationRoot Locus Design Example #3
Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll
More informationChapter 3. State Feedback  Pole Placement. Motivation
Chapter 3 State Feedback  Pole Placement Motivation Whereas classical control theory is based on output feedback, this course mainly deals with control system design by state feedback. This modelbased
More informationControl of SingleInput SingleOutput Systems
Control of SingleInput SingleOutput Systems Dimitrios HristuVarsakelis 1 and William S. Levine 2 1 Department of Applied Informatics, University of Macedonia, Thessaloniki, 546, Greece dcv@uom.gr 2
More informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More information1 An Overview and Brief History of Feedback Control 1. 2 Dynamic Models 23. Contents. Preface. xiii
Contents 1 An Overview and Brief History of Feedback Control 1 A Perspective on Feedback Control 1 Chapter Overview 2 1.1 A Simple Feedback System 3 1.2 A First Analysis of Feedback 6 1.3 Feedback System
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 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 informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationChapter 2 SDOF Vibration Control 2.1 Transfer Function
Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:
More informationEE 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 openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
More information9/9/2011 Classical Control 1
MM11 Root Locus Design Method Reading material: FC pp.270328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),
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 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 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 informationInverted Pendulum. Objectives
Inverted Pendulum Objectives The objective of this lab is to experiment with the stabilization of an unstable system. The inverted pendulum problem is taken as an example and the animation program gives
More informationDistributed RealTime Control Systems
Distributed RealTime Control Systems Chapter 9 Discrete PID Control 1 Computer Control 2 Approximation of Continuous Time Controllers Design Strategy: Design a continuous time controller C c (s) and then
More informationThe Frequencyresponse Design Method
Chapter 6 The Frequencyresponse Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)
More informationEngraving Machine Example
Engraving Machine Example MCE44  Fall 8 Dr. Richter November 24, 28 Basic Design The Xaxis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional
More informationControl System Design
ELEC4410 Control System Design Lecture 19: Feedback from Estimated States and DiscreteTime Control Design Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationControls Problems for Qualifying Exam  Spring 2014
Controls Problems for Qualifying Exam  Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
More informationEE402  Discrete Time Systems Spring Lecture 10
EE402  Discrete Time Systems Spring 208 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 0.. Root Locus For continuous time systems the root locus diagram illustrates the location of roots/poles of a closed
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationECEn 483 / ME 431 Case Studies. Randal W. Beard Brigham Young University
ECEn 483 / ME 431 Case Studies Randal W. Beard Brigham Young University Updated: December 2, 2014 ii Contents 1 Single Link Robot Arm 1 2 Pendulum on a Cart 9 3 Satellite Attitude Control 17 4 UUV Roll
More informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
More informationDepartment of Electrical and Computer Engineering. EE461: Digital Control  Lab Manual
Department of Electrical and Computer Engineering EE461: Digital Control  Lab Manual Winter 2011 EE 461 Experiment #1 Digital Control of DC Servomotor 1 Objectives The objective of this lab is to introduce
More informationEECS 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
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 informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More 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 Systems. Design of State Feedback Control.
Control Systems Design of State Feedback Control chibum@seoultech.ac.kr Outline Design of State feedback control Dominant pole design Symmetric root locus (linear quadratic regulation) 2 Selection of closedloop
More informationCL Digital Control
CL 692  Digital Control Kannan M. Moudgalya Department of Chemical Engineering Associate Faculty Member, Systems and Control IIT Bombay Autumn 26 CL 692 Digital Control, IIT Bombay 1 c Kannan M. Moudgalya,
More informationLinear Control Systems Solution to Assignment #1
Linear Control Systems Solution to Assignment # Instructor: H. Karimi Issued: Mehr 0, 389 Due: Mehr 8, 389 Solution to Exercise. a) Using the superposition property of linear systems we can compute the
More informationRoot Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus  1
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position
More informationSchool 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 Coordinator/Tutor : Dr. Phang
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationMechanical Systems Part A: StateSpace Systems Lecture AL12
AL: 436433 Mechanical Systems Part A: StateSpace Systems Lecture AL Case study Case study AL: Design of a satellite attitude control system see Franklin, Powell & EmamiNaeini, Ch. 9. Requirements: accurate
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
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 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 informationAcceleration Feedback
Acceleration Feedback Mechanical Engineer Modeling & Simulation Electro Mechanics Electrical Electronics Engineer Sensors Actuators Computer Systems Engineer Embedded Control Controls Engineer Mechatronic
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 informationDigital Control: Summary # 7
Digital Control: Summary # 7 Proportional, integral and derivative control where K i is controller parameter (gain). It defines the ratio of the control change to the control error. Note that e(k) 0 u(k)
More informationSatellite Attitude Control System Design Using Reaction Wheels Bhanu Gouda Brian Fast Dan Simon
Satellite Attitude Control System Design Using Reaction Wheels Bhanu Gouda Brian Fast Dan Simon Outline 1. Overview of Attitude Determination and Control system. Problem formulation 3. Control schemes
More informationRecursive, Infinite Impulse Response (IIR) Digital Filters:
Recursive, Infinite Impulse Response (IIR) Digital Filters: Filters defined by Laplace Domain transfer functions (analog devices) can be easily converted to Z domain transfer functions (digital, sampled
More informationEXAMPLE: MODELING THE PT326 PROCESS TRAINER
CHAPTER 1 By Radu Muresan University of Guelph Page 1 EXAMPLE: MODELING THE PT326 PROCESS TRAINER The PT326 apparatus models common industrial situations in which temperature control is required in the
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 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 informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
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 informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More informationSignal sampling techniques for data acquisition in process control Laan, Marten Derk van der
University of Groningen Signal sampling techniques for data acquisition in process control Laan, Marten Derk van der IMPORTANT NOTE: You are advised to consult the publisher's version (publisher's PDF)
More informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
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 informationHinfinity Model Reference Controller Design for Magnetic Levitation System
H.I. Ali Control and Systems Engineering Department, University of Technology Baghdad, Iraq 6043@uotechnology.edu.iq Hinfinity Model Reference Controller Design for Magnetic Levitation System Abstract
More informationDIGITAL CONTROL OF POWER CONVERTERS. 3 Digital controller design
DIGITAL CONTROL OF POWER CONVERTERS 3 Digital controller design Frequency response of discrete systems H(z) Properties: z e j T s 1 DC Gain z=1 H(1)=DC 2 Periodic nature j Ts z e jt e s cos( jt ) j sin(
More informationStep input, ramp input, parabolic input and impulse input signals. 2. What is the initial slope of a step response of a first order system?
IC6501 CONTROL SYSTEM UNITII TIME RESPONSE PARTA 1. What are the standard test signals employed for time domain studies?(or) List the standard test signals used in analysis of control systems? (April
More information