INTRODUCTION TO DIGITAL CONTROL


 Mavis Rich
 2 years ago
 Views:
Transcription
1 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 (LTI) plant G(s). We used feedback to: Reduce error due to disturbance, Reduce error due to model mismatch, Stabilize an unstable plant, Improve transient response. w(t) r(t) e(t) D(s) u(t) G(s) y(t) v(t) There are advantages to using a digital computer to replace D(s). Flexible: Modifications to design easy to implement. Accurate: Does not rely on imprecise R, C,... values. Advanced algorithms are possible (nonlinear, adaptive...). Logic statements may be included. There are problems with using a digital computer:
2 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 2 D(s) knows e(t) at each instant in time but digital computers cannot process input infinitely quickly. D(s) outputs u(t) at each instant of time but... e(t) is an infiniteprecision value, but digital computers must work with finiteprecision values. Digital computers are finite. The necessary compromises are to sample the input e(t) to get e[k] =e(kt), tocomputeanoutputu[k] that is held constant from u(kt) to u((k + )T ), andto quantize internalcomputervariables. e(t) r(t) A2D Q e[k] u[k] D(z) D2A zoh w(t) u(t) G(s) y(t) v(t) D(z) is the transfer function of the controller; discretetime. We need to know how to design D(z), howquicklytosample(/t ), and how finely to quantize. Fortunately, many of the techniques are similar either in principle or practice to analogcontrol techniques. Proceeding, we look at: Review. Examine impact A2D/D2A with approximate control method. Study ztransform. More indepth study of A2D/D2A using ztransform knowledge. Control design of D(z) using transform methods. Implementation using DSP. Adaptive form of digital control called Adaptive Inverse Control.
3 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 3.2: Review: System modeling Goals of feedback control Change dynamic response of a system to have desired properties. Ensure that closedloop system is stable. Constrain system output to track a reference input with acceptable transient and steadystate responses. Reject disturbances. These goals are accomplished via frequencydomain (Laplace) analysis and design tools for continuoustime systems. Dynamic response We wish to control linear timeinvariant (LTI) systems. These dynamics may be specified via linear, constantcoefficient ordinary differential equations (LCCODE). Examples include: Mechanical systems: Use Newton s laws. Electrical systems: Use Kirchoff s laws. Electromechanical systems (generator/motor). Thermodynamic and fluiddynamic systems. EXAMPLE: Secondorder system in standard form : ÿ(t) + 2ζω n ẏ(t) + ωn 2 y(t) = ω2 n u(t). u(t) istheinputand y(t) istheoutput.
4 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 4 ẏ(t) = dy(t) dt and ÿ(t) = d2 y(t) dt 2. The Laplace Transform is a tool to help analyze dynamic systems. Y (s) = H(s)U(s), where Y (s) is Laplace transform of output y(t); U(s) is Laplace transform of input u(t); H(s) is transfer function the Laplace tx of impulse response h(t). KEY IDENTITY: L{ẏ(t)} = sy(s) for system initially at rest. EXAMPLE: Secondorder system: s 2 Y (s) + 2ζω n sy(s) + ω 2 n Y (s) = ω2 n U(s) Y (s) = ω 2 n U(s). s 2 + 2ζω n s + ωn 2 Transforms for systems with LCCODE representations can be written as Y (s) = H(s)U(s), where H(s) = b ms m + b m s m + +b s + b 0 a n s n + a n s n + +a s + a 0, and where n m for physical systems. These can be represented in MATLAB using vectors of numerator and denominator polynomials: num=[bm b b0]; den=[an a a0]; sys=tf(num,den); Can also represent these systems by factoring the polynomials into zeropolegain form: H(s) = K m i= (s z i) n i= (s p i).
5 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 5 u sys=zpk(z,p,k); % in MATLAB Input signals of interest include the following: u(t) = k δ(t)... U(s) = k impulse u(t) = k (t)... U(s) = k/s step u(t) = kt (t)... U(s) = k/s 2 ramp u(t) = kt 2 /2(t)... U(s) = k/s 3 parabola kω u(t) = k sin(ωt) (t)... U(s) = sinusoid s 2 + ω 2 MATLAB s impulse, step, and lsim commandscanbeusedto find output time histories. The Final Value Theorem states that if a system is stable and has a final, constant value, lim x(t) = lim sx(s). t s 0 This is useful when investigating steadystate errors. Block diagrams Useful when analyzing systems comprised of a number of subunits. U(s) H(s) Y (s) Y (s) = H(s)U(s) U(s) H (s) H 2 (s) Y (s) Y (s) = [H (s)h 2 (s)] U(s)
6 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 6 i k H (s) U(s) Y (s) Y (s) = [H (s) + H 2 (s)] U(s) H 2 (s) R(s) U (s) Y 2 (s) H (s) H 2 (s) U 2 (s) Y (s) Y (s) = H (s) + H 2 (s)h (s) R(s) Blockdiagram algebra (or Mason s rule) may be used to reduce block diagrams to a single transfer function. U(s) H(s) Y (s) Y 2 (s) U(s) H(s) H(s) Y (s) Y 2 (s) U (s) U 2 (s) H(s) Y (s) U (s) U 2 (s) H(s) H(s) Y (s) R(s) H (s) Y (s) R(s) H 2 (s) H 2 (s) H (s) Y (s) H 2 (s) Unity Feedback
7 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 7.3: Review: Actual and desired dynamic response The poles of H(s) determine (qualitatively) the dynamic response of the system. The zeros of H(s) quantify the relationship. If the system has only real poles, each one has the form H(s) = s + σ. If σ>0, thesystemisstable,andh(t) = e σ t (t). Thetimeconstant is τ = /σ, and the response of the system to an impulse or step decays to steadystate in about 4 or 5 time constants. impulse([0 ],[ ]); step([0 ],[ ]); h(t) e σ t e y(t) K K ( e t/tau ) System response. K = DC gain Response to initial condition t = τ Time (sec τ) t = τ Time (sec τ) If a system has complexconjugate poles, each may be written as: H(s) = ω 2 n. s 2 + 2ζω n s + ωn 2 We can extract two more parameters from this equation: σ = ζω n and ω d = ω n ζ 2.
8 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 8 σ plays the same role as above it specifies decay rate of the response. I(s) θ = sin (ζ ) ωd is the oscillation frequency of the output. Note: ω d = ω n unless ζ = 0. ζ is the damping ratio and it also plays a ω n σ ω d R(s) role in decay rate and overshoot. Impulse response h(t) = ωn e σ t sin(ω d t) (t). ( Step response y(t) = e σ t cos(ω d t) + σ ) sin(ω d t). ω d y(t) Impulse Responses of 2ndOrder Systems ζ = 0.8 ζ = ω n t y(t) Step Responses of 2ndOrder Systems ζ = Asummarychartofimpulseresponsesandstepresponsesversus pole locations is: I(s) ω n t I(s) R(s) R(s) Impulse responses vs. pole locations Step responses vs. pole locations
9 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 9 Timedomain specifications determine where poles SHOULD be placed in the splane (stepresponse). t p M p t r t s t Rise time tr =timetogofrom0% to 90% of final value. Settling time ts =timeuntilpermanently within % of final value. Overshoot Mp =maximumper 0 CENT overshoot Some approximate relationships useful for initial design processes are: M p, % t r.8/ω n... ω n.8/t r t s 4.6/σ... σ 4.6/t s M p e πζ/ ζ 2... ζ fn(m p ) ζ
10 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 0.4: Review: Feedback control r(t) y(t) D(s) G(s) Y (s) R(s) = D(s)G(s) + D(s)G(s) = T (s). Stability depends on roots of denominator of T (s): + D(s)G(s) = 0. Routh test used to determine stability. Steadystate error found from (for unityfeedback case) E(s) R(s) = + D(s)G(s). ess = lim t e(t) = lim s 0 se(s) if the limit exists. System type = 0 iff e ss is finite for unitstep (t). System type = iff e ss is finite for unitramp r(t). System type = 2 iff e ss is finite for unitparabola p(t).. For unityfeedback systems, K p = lim D(s)G(s), s 0 K v = lim sd(s)g(s). s 0 K a = lim s 2 D(s)G(s). s 0 position error constant velocity error constant acceleration error constant Steadystate errors versus system type for unity feedback: Step input Ramp input Parabola input Type 0 + K p Type 0 K v Type K a
11 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL Some types of controllers Proportional ctrlr: u(t) = Ke(t). D(s) = K. Integral ctrlr u(t) = K t e(t) dt. D(s) = K T I T I s Derivative ctrlr. u(t) = KT D ė(t) D(s) = KT D s Combinations: PI: D(s) = K ( + T I s ); PD: D(s) = K ( ( + T D s) ; PID: D(s) = K + ) T I s + T Ds. Lead: D(s) = K Ts + αts +, Lag: D(s) = K Ts + αts +, α < (approx PD) α > (approx PI; often, K = α) Lead/Lag: D(s) = K (T s + )(T 2 s + ) (α T s + )(α 2 T 2 s + ), α <, α 2 >. Root locus Arootlocusplotshows(parametrically)thepossiblelocations of the roots of the equation + K b(s) a(s) = 0. For a unitygain feedback system, T (s) = D(s)G(s) + D(s)G(s). The poles of the closedloop system T (s) depend on the openloop transfer functions D(s)G(s). Suppose D(s) = KD 0 (s).
12 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 2 closedloop poles at + K (D 0 (s)g(s)) = 0 which is the rootlocus form. Drawing the root locus allows us to select K for good pole locations. Intuition into the rootlocus helps us design D 0 (s) with lead/ lag/ PI/ PID...controllers. Rootlocus drawing rules The steps in drawing a 80 root locus follow from the basic phase definition. This is the locus of + K b(s) a(s) = 0, K 0 They are ( phase of b(s) ) a(s) = 80 STEP : On the splane, mark poles (roots of a(s)) byan and zeros (roots of b(s)) withan. There will be a branch of the locus departing from every pole and a branch arriving at every zero. STEP 2: Draw the locus on the real axis to the left of an odd number of real poles plus zeros. STEP 3: Draw the asymptotes, centered at α and leaving at angles φ, where n m = number of asymptotes. n = order of a(s) m = order of b(s) pi z i α = n m = a + b n m φ l = 80 + (l )360, l =, 2,...n m n m For n m > 0, therewillbeabranchofthelocusapproachingeachasymptote and departing to infinity. For n m < 0, therewillbeabranchofthelocusarrivingfrominfinityalong each asymptote. STEP 4: Compare locus departure angles from the poles and arrival angles at the zeros where qφ dep = ψ i φ i 80 l360 qψ arr = φ i ψ i l360 where q is the order of the pole or zero and l takes on q integer values so that the angles are between ±80. ψ i is the angle of the line going from the i th zero to the pole or zero whose
13 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 3 angle of departure or arrival is being computed. Similarly, φ i is the angle of the line from the i th pole. STEP 5: If further refinement is required at the stability boundary, assume s o = jω o and compute the point(s) where the locus crosses the imaginary axis for positive K. STEP 6: For the case of multiple roots, two loci come together at 80 and break away at ±90. Three loci segments approach each other at angles of 20 and depart at angles rotated by 60. STEP 7 Complete the locus, using the facts developed in the previous steps and making reference to the illustrative loci for guidance. The loci branches start at poles and end at zeros or infinity. STEP 8 Select the desired point on the locus that meets the specifications (s o ), then use the magnitude condition to find that the value of K associated with that point is K = b (s o ) /a (s o ). When K is negative, the definition of the root locus in terms of the phase relationship changes. We need to plot a 0 locus instead. 0 locus definition: The root locus of b(s)/a(s) is the set of points in the splane where the phase of b(s)/a(s) is 0. For this case, the steps above are modified as follows: STEP 2: Draw the locus on the real axis to the left of an even number of real poles plus zeros. STEP 3: The asymptotes depart at φ l = (l )360, l =, 2,...n m. n m STEP 4: The locus departure and arrival angles are modified to Note that the 80 term has been removed. qφ dep = ψ i φ i l360 qψ arr = φ i ψ i + l360. STEP 8 Select the desired point on the locus that meets the specifications (s o ), then use the magnitude condition to find that the value of K associated with that point is K = b (s o ) /a (s o ).
14 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 4.5: Review: Frequency response The frequency response of a system tells us directly the relative magnitude and phase of a system s output sinusoid if the system input is a sinusoid. [What about output frequency?] If the plant s transfer function is G (s), theopenloopfrequency response is G ( jω) = G(s) s= jω. We can plot G( jω) as a function of ω in three ways: Bode Plot. Nyquist Plot. Nichols Plot (not covered in ECE450/550). Bode plots A Bodeplot isreallytwoplots:20 log0 G ( jω) versus ω and G( jω) versus ω. The Bodemagnitude plot is plotted on a loglog scale.the Bodephase plot is plotted linearlog scale. The transfer function G(s) is broken up into its component parts, and each part is plotted separately. The factorization must be of the form KG(s) = K o (s) n (sτ + )(sτ 2 + ) (sτ a + )(sτ b + ), which separates real zeros and poles, complex zeros and poles, zeros or poles at the origin, delay, and gains. The following dictionary is used when plotting a transfer function which has been factored into Bode form.
15 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 5 Dictionary of Bodeplot relationships (tempor Gain of K ;magnitude&phase db 0. 0 K > 0. 0 K < Delay; magnitude & phase db Zero at origin; magnitude & phase db Pole at origin; magnitude & phase db
16 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 6 Zero on real axis; magnitude & mpphase & nmpphase db ω n ωn 0ω n 0.ω n ωn 0ω n 0.ω n ωn 0ω n Pole on real axis; magnitude & mpphase & nmpphase 0.ω n ω n 0ω n 0.ω n ω n 0ω n db 0.ω n ωn 0ω n Complex zeros; magnitude & mpphase & nmpphase ω n ω n 0ω n 0.ω n ω n 0ω 0 n 0.ω n ω n 0ω Complex poles; magnitude & mpphase & nmpphase n ω n ω n 0ω n ω n ω n 0ω n 0.ω n ω n 0ω n 360 Bodeplot techniques It is useful to be able to plot the frequency response of a system by hand in order to Design simple systems without the aid of a computer, Check computerbased results, and
17 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 7 Understand the effect of compensation changes in design iterations. H.W. Bode developed plotting techniques in the 930s that enabled quick hand potting of the frequency response. His rules are: STEP : Manipulate the transfer function into the Bode form KG( jω) = K o ( jω) n ( jωτ + )( jωτ 2 + ) ( jωτ a + )( jωτ b + ) STEP 2: Determine the value of n for the K o ( jω) n term. Plot the lowfrequency magnitude asymptote through the point K o at ω = rad/sec with the slope of n (or n 20 db per decade). STEP 3: Determine the break points whereω = /τ i.completethecompositemagnitude asymptotes by extending the low frequency asymptote until the first frequency break point, then stepping the slope by ± or±2, dependingonwhetherthebreakpointisfromafirstorsecond order term in the numerator or denominator, and continuing through all break points in ascending order. STEP 4: Sketch in the approximate magnitude curve by increasing from the asymptote by a factor of.4 (+3dB) at first order numerator breaks and decreasing it by a factor of ( 3dB) at first order denominator breaks. At second order break points, sketch in the resonant peak (or valley) using the relation that G( jω) = /(2ζ) at the break. STEP 5: Plot the low frequency asymptote of the phase curve, φ = n 90. STEP 6: As a guide, sketch in the approximate phase curve by changing the phase gradually over two decades by ±90 or ±80 at each break point in ascending order. For first order terms in the numerator, the gradual change of phase is +90 ;inthedenominator,thechangeis±80. STEP 7: Locate the asymptotes for each individual phase curve so that their phase change corresponds to the steps in the phase from the approximate curve indicated by Step 6. Sketch in each individual phase. STEP 8 Graphically add each phase curve. Use dividers if an accuracy of about ±5 is desired. If lessor accuracy is acceptable, the composite curve can be done by eye, keeping in mind that the curve will start at the lowest frequency asymptote and end onthehighestfrequency asymptote, and will approach the intermediate asymptotes to an extent that is determined by the proximity of the break points to each other. Nyquist plots Nyquist plot is a mapping of loop transfer function D(s)G(s) along the Nyquist path to a polar plot.
18 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 8 Think of Nyquist path as four parts: I: Origin. Sometimes a special case. II: + jω axis. FREQUENCY response of O.L. system! Just plot it as a polar plot. III: For many physical systems, zero. Some counterexamples. IV: Complex conjugate of II. The test: I(s) III II R(s) I IV N =#CWencirclementsof /K point when F(s) = D(s)G(s). P =#ofopenloopunstablepoles. Z =#ofclosedloopunstablepoles. Z = N + P The system is stable iff Z = 0. For poles on the jωaxis, use modified Nyquist path. Bodeplot performance At low frequency, approximate Bode plot with KG( jω) = K0 ( jω) n. n =systemtype...slope = 20n db/decade. K 0 = K p for type 0, = K v for type,... Gain margin (for systems that become unstable with increasing gain) =factorbywhichgainislessthan0dbwhenphase= 80. Phase margin (for same systems) = amount phase is greater than 80 when gain =.
19 ECE4540/5540, INTRODUCTION TO DIGITAL CONTROL 9 PM related to damping. ζ PM/00 when PM < 70. k B PM therefore related to Mp. Damping ratio ζ Damping ratio versus PM Phase margin M p, overshoot Overshoot fraction versus PM Phase margin Gainphase relationship: For minimumphase systems, G( jω) n 90 where n =slopeofloglogplotofgain(n = if slope = 20 db/decade, n = 2 if slope = 40 db/decade...) Therefore want slope at gain crossover 20 db/decade for decent phase margin. Where from here? We have reviewed continuoustime control concepts. Can any of these be applied directly to the discretetime control problem? We look next at two different ways to emulate continuoustime controllers using discretetime methods, allowing a quick redesign and control implementation. However, we also find that these methods are not optimized and that we can do better by designing digital controllers from scratch.
Course Background. Loosely speaking, control is the process of getting something to do what you want it to do (or not do, as the case may be).
ECE4520/5520: Multivariable Control Systems I. 1 1 Course Background 1.1: From time to frequency domain Loosely speaking, control is the process of getting something to do what you want it to do (or not
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
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 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 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 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 informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
More 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 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 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 information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More information(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 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 informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
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 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 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 informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
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 informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
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 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 informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
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 informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
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 informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Boundedinput boundedoutput (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
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 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 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 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 informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
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 informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More informationEC CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING
EC 2255  CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING 1. What is meant by a system? It is an arrangement of physical components related in such a manner as to form an entire unit. 2. List the two types
More informationChemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University
Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this
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 informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
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 informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More 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 information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More 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 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 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 informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
More informationTime Response of Systems
Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) splane Time response p =0 s p =0,p 2 =0 s 2 t p =
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 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 informationEE3CL4: Introduction to Linear Control Systems
1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We
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 informationLecture 1 Root Locus
Root Locus ELEC304Alper Erdogan 1 1 Lecture 1 Root Locus What is RootLocus? : A graphical representation of closed loop poles as a system parameter varied. Based on RootLocus graph we can choose the
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More 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 informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
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 informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design
AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University Routh s Stability
More informationSECTION 5: ROOT LOCUS ANALYSIS
SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More 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 informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
More 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 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 informationReview: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Plan of the Lecture Review: transient and steadystate
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 informationSchool of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See
More informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More 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 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 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 informationFrequency Response Analysis
Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions
More informationLecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore A Fundamental Problem in Control Systems Poles of open
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 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 informationOutline. Classical Control. Lecture 2
Outline Outline Outline Review of Material from Lecture 2 New Stuff  Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
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 informationResponse to a pure sinusoid
Harvard University Division of Engineering and Applied Sciences ES 145/215  INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid
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 informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 5. 2. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
More informationControl System (ECE411) Lectures 13 & 14
Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 SteadyState Error Analysis Remark: For a unity feedback system
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 09Dec13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More informationMethods for analysis and control of dynamical systems Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.gipsalab.fr/ o.sename 5th February 2015 Outline
More information