Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types


 Shana Cunningham
 2 years ago
 Views:
Transcription
1 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 draft: March 2, 28 Introduction:Frequency domain design With lecture 4, we have completed what is considered the time domain method of system design. The reason being that time domain parameters, rise time, settling time, overshoot and steady state error were directly addressed by the PID controller which differentiates and integrates the time domain error. PID modifies effectively the pole positions and we can use the Root Locus method or Ziegler Nichols for finding the PID gains. The paramaters under the engineer s control are Kp, Ki and Kd, which using the method of Root Locus modify the pole positions and influence the final objective, viz., t r,t s,m p andsse. Let us say we model a system and then build it. We may find that the model differs quite a bit from reality. And so we have to upgrade the model and say we converge. Then we measure the t r,t s,m p etc. and we are not satisfied. Then we specify what we want and proceed with PID design. It seems simple enough. But, it is difficult to measure time domain characteristics. We need to give a step input which can be light and no energy input occurs. Or the input can be strong, then the system can become nonlinear. If the system is stiff, then the response can be very quick and die away. Measuring it is difficult. On the other hand the system may oscillate with a lot of low frequency dynamics. For every kind of system we cannot design a step inputu system. Locally, step inputs create nonlinear responses and deformations. On the other hand, the frequency response of a system can be very easily measured in the lab. Either a hammer impact is given or a sinusoid/random input is given and this method has been standardized. Now, does such a frequency response which captures sustained oscillation behavior capture information about transient response is the question. We will see that the frequency response is characterized by the resonance peak Mp, resonance frequency ω n, BW, Phase and gain margin. Indirectly, these indicate the time domain response parameters t r, etc. Here, the final requirement is still the time response. The tools here are compensators which use bode diagram and nyquist diagram methods.
2 Nyquist Criterion Nyquist Criterion: For a closed loop system to be stable, the Nyquist plot of G(s)H(s) must encircle the (,j) point as many times as the number of poles of G(s)H(s) that are in the right half of splane, and the encirclement, if any, must be made in the closkwise direction. (Read from Benjamin Kuo) 2 Nyquist or polar plots The characteristic equation is + or so that when the system is just unstable amplitude of and phase = 8 The Nyquist plot is the polar (amplitude phase) plot of a GH transfer function on the complex plane as s = jω, and ω takes values from to. The length of the arrow from the origin to the plot indicates the amplitude of GH and the angle of the arrow w.r.t the real axis indicates the phase of GH. The left figure (figure ) is for The right figure in figure is for s + s 2 + 3s + 2 s 3 + 4s 2 + 5s + 2 s(s + ) K s(s 2 + 3s + 2) () (2) where K takes values,3,6. At K = 6, the nyquist plot just passes through the point s = +j. Take for example the left figure and look at /(s 2 +3s+2). At ω = the amplitude is.5 and the phase is degrees. At ω = the amplitude is and the phase is dominated by /s 2 = /ω 2 and hence 8 degrees. This can be seen in the plot. Similarly in the right figure of figure the second order transfer function has phase 8 at high frequencies and the third order transfer function has 27 degrees. And as the gain in the third order transfer function is raised it gets closer to the point +j and finally crosses it going unstable. Thus, a polar plot of GH gives the relative stability of the closed loop system. If we know that at a given gain, the system is stable, then by increasing the gain and seeing which way the polar plot moves we can find out the relative stability of the system. How far is the system from going unstable. Note that we use the polar plot of GH in order to figure the relative stability of the closed loop system. How far the amplitude is from  and how far the phase is from 8 degrees are indicators of stability. These are called gain and phase margins respectively and we will elaborate more on these two in a little while. 2
3 Nyquist Diagrams Nyquist Diagrams Imaginary Axis [ 4 5 2] θ [ 3 2] Imaginary Axis [ 3 2 ] [ ] [ ] Real Axis Real Axis Figure : Nyquist plots 3 Bode Diagrams The same Nyquist information can be plotted with the amplitude and phase on separate plots. The amplitude is plotted after conversion to a db scale and the x axis on a log scale. Such plots are called Bode plots. Bode plot although very simple carries a lot of information such as: relative stability (gain and phase margins), steady state errors, damping coefficient etc. We begin with a simple example, 3. Example : /s the amplitude and phase are given by 2log ω s = jω = 2log ω and φ = jω = 9o The amplitude and phase are plotted in figure 9. Lets look at the amplitude plot. The x axis is on a log scale,.,.,,, etc. The amplitude goes through zero at ω = and has a negative drop of 2 db for every times increase in frequency, i.e., 2 db/decade. The phase plot is relatively simple. 3.2 Example 2: /(s + ) Amplitude is given by s + jω + ω 2 + 2log ω 2 + 3
4 2 Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 2: Bode plot for /s. Amplitude and phase. Low frequency ω =, the plot approaches asymptotically. Hence amplitude= is the low frequency asymptote. On the log scale as ω, the plot approaches db. High frequency ω, the plot approaches ω asymptotically. Hence amplitude=/ω is the high frequency asymptote. On the log scale as ω, the amplitude approaches 2log ω. At ω =, this asymptote goes through db and hence intersects the low frequency asymptote. The actual value of the curve at ω = is 2log + = 3dB. The curve lies 3dB below the db asymptote. Phase is given by jω+ as ω, the phase goes to as ω, the phase approaches 9 See figure Example 3: /(s +.) s+. jω+. s +. Magnitude ω 2 +. ω, reaches constant= asymptote ω, ω is the asymptote. db = 2log ω
5 Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 3: Bode plot for /s +. Note the asymptotes in the magnitude plot. ω, 2log. = 2dB ω, 2log ω 2 db/decade Phase as before. The meeting point of the asymptotes is the 2 db line at ω =. (note s+.). The actual value of the curve at ω =. is 2log = This value is 3 db below the 2 db asymptote. 3.4 Example 4: /(s + ) s+ jω+ s + Magnitude ω 2 + ω, asymptote. ω, ω asymptote. Following the earlier examples, the low frequency asymptote reaches the value 2 db. The high frequency asymptote goes as 2dB/decade. The two meet at ω =. The actual curve is 3 db below the 2 db asymptote at ω =. See figure 5. 5
6 2 Bode Diagram Phase (deg) Magnitude (db) Asymptote 9 2 Frequency (rad/sec) Figure 4: Bode plot for /s +.. Note the asymptotes in the magnitude plot meet at ω =.. 2 Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 5: Bode plot for /s +. Note the asymptotes in the magnitude plot. 6
7 3.5 Example 4: /((s +.)(s + )(s + )) We examine a complicated case. There are three terms and a (s +.)(s + )(s + ) ω,. = asymptote. ω, asymptote. ω 3 Following the earlier examples, the low frequency asymptote reaches the value db. The high frequency asymptote goes as 6dB/decade. The curve will follow /ω asymptote at ω =. with a slope of 2dB/decade. Then, the curve will follow /ω 2 asymptote at ω = with a slope of 4dB/decade. Then, the curve will follow /ω 3 asymptote at ω = with a slope of 6dB/decade. The actual curve is 3 db below the meeting points of asymptotes at the corresponding ω. See figure 6. Phase goes to at low frequencies and to 27 o at high frequencies. Bode Diagrams Phase (deg); Magnitude (db) w=. w=. w= Frequency (rad/sec) Figure 6: Bode plot for /((s +.)(s + )(s + )). 4 Second order system: ω n 2 s 2 +2ξω n s+ω n 2 Lets look at the closed loop Tr. Fn. M(s) = ω n 2 s 2 + 2ξω n s + ω n 2 + (2ξ/ω n )s + /ω n 2 s 2 7
8 We are interested in values of ξ <<. Substituting s = jω gives M(jω) = [ (ω/ω n ) 2 ] + 2jξω/ω n As can be seen, at low frequencies GH reaches the db asymptote. At high frequencies we get the 4 db/decade asymptote. These two meet at ω = ω n. The phase is degrees at low frequencies and at ω = ω n it is 9 and for higher frequencies it slowly approaches 8. The actual plots around ω = ω n are quite complicated depending on the actual value of ξ. Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 7: Bode plot of a second order transfer function. 4. M p, ω p and bandwidth of a second order system: ω n 2 M(s) = s 2 + 2ξω n s + ω 2 n { } { M(s) L = y(t) = e σt cosω d t + σ } sinω d t s ω d ω d = ω n ξ 2, σ = ξω n π y(t) =, ω d t p = π, t peak = ω n ξ 2 y(t p ) = + M p M p = e πξ = ξ.6. ξ.6. 8 ξ 2. ξ.
9 For a % settling time we know Looking at M(s) in the frequenc domain: M(jω) = C(jω) R(jω) = t s = 4.6 ξω n ω n t r =.8 ω 2 n (jω) 2 + 2ξω n (jω) + ω 2 n = ( ω ω n ) 2 + j2ξ ω ω n M(ju) = u 2 + j2uξ M(ju) = [( u 2 ) 2 + (2ξu) 2 ] 2 2ξu M(ju) = φ m (u) = tan u 2 In order to find out the frequency at which the peak transfer function value occurs in frequency: dm(u) du = 4u 3 4u + 8uξ 2 = u = u p = u p = 2ξ 2 So M p is purely a function of ξ. ω p = ω n 2ξ 2 valid for ξ <.77 M p = 2ξ ξ 2 For ξ.77,ω p =,M p =. Bandwidth of a second order system is frequency at which M(ω) drops to 7.7% of its value from its low frequency asymptote. as ξ, bandwidth and M p. M(u) = ( u 2 ) 2 + (2ξu) 2 =.77 u 2 = ( 2ξ 2 ) ± 4ξ 4 4ξ Bw = ω n [( 2ξ 2 ) + 4ξ 4 4ξ 2 + 2] 2. Maximum overshoot of unit step response depends on ξ only. 2. Resonance peak of closed loop frequency response M p depends only on ξ. 9
10 M P ω P ω n.77 ζ ζ ζ =.77 Figure 8: 3. Bandwidth is directly proportional to ω n. 4. High bandwidth = largerm p. 5. ω n =.8 t r. 6. t s = f(ξω n ) = 4.6 ξω n for % settling time. 7. Also, the following relations are relevant between bandwidth, ω n, t s, t r and ξ: ω Bw = ω Bw = ω n [( 2ξ 2 ) + 4ξ 4 4ξ 2 + 2] 2 ω n = 4.6 ξt s ω Bw = 4.6 ξt s [( 2ξ 2 ) + π ω n = t p ξ 2 π t p ξ 2 [( 2ξ2 ) + 8. Open loop system must be stable for Bode plot design. 4ξ 4 4ξ 2 + 2] 2 4ξ 4 4ξ 2 + 2] 2 9. For 2 nd order system closed loop damping ratio is approximately equal to PM, if phase margin is between and 6.. bandwidth ω n. We use these relations quite freely even for nonsecondorder systems...with caution. 5 System Types The type of the open loop system one is dealing with decides its steady state error nature. Hence, it is important to know. Also, the steady state error depends not only on the form of the transfer function but also on the input, like step or ramp. Here we will look at system types with regard to only a step input. More is available in Benjamin Kuo.
11 r(t) R(s) +  e(t) E(s) G(s) c(t) C(s) b(t) B(s) H(s) Figure 9: The error E(s) is given by E(s) = R(s) C(s)H(s) (3) C(s) = E(s)G(s) E(s) = R(s) E(s)G(s)H(s) R(s) E(s) = + G(s)H(s) E(s), the LT of e(t) is given by E(s) = = [ e st e(t) ] s = e() s + s se(s) = e() + e(t)e st dt. e st s de(t).dt dt e stde(t).dt dt e stde(t).dt dt lim se(s) = e() + de = e( ) e() = e( ) s Thus steady state error, i.e., e( ) is given by the above formula. Step response For step response R(s) = R s and let us denote lim k p, then s and if the steady state error then E(s) = R s( + k p ) e( ) = lim s se(s) = lim se(s) = s or k p =. If the loop transfer function has the form R ( + k p ) = s j F(s)
12 then the system is of type j. If j=, then lim F() = Kp s is a finite value. Thus, Kp is finite for type systems and hence the steady state error is not zero. For type and higher systems Kp is infinity. Hence, for step inputs steady state error is not for type systems and is for type and higher systems. A similar set of rules can be defined for ramp inputs. (Read Kuo). 2
Homework 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 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 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 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 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 informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
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 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 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 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 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 informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More 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 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 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 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 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 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 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationME 375 EXAM #1 Friday, March 13, 2015 SOLUTION
ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned
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 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 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 informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
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 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 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 informationAutomatic Control (TSRT15): Lecture 4
Automatic Control (TSRT15): Lecture 4 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13282226 Office: Bhouse extrance 2527 Review of the last
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 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 informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More 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 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 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 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 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 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 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 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 information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
More informationChapter 6  Solved Problems
Chapter 6  Solved Problems Solved Problem 6.. Contributed by  James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the ZN tuning strategy for the nominal
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 informationIntroduction to Root Locus. What is root locus?
Introduction to Root Locus What is root locus? A graphical representation of the closed loop poles as a system parameter (Gain K) is varied Method of analysis and design for stability and transient response
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More 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 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 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 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 informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
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 information10ES43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A
10ES43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative
More informationLecture 3: The Root Locus Method
Lecture 3: The Root Locus Method Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 56001 This draft: March 1, 008 1 The Root Locus method The Root Locus method,
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 informationCHAPTER # 9 ROOT LOCUS ANALYSES
F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closedloop system is closely related to the location of the closedloop poles. If the system
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 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 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 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 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 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 informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationActive Control? Contact : Website : Teaching
Active Control? Contact : bmokrani@ulb.ac.be Website : http://scmero.ulb.ac.be Teaching Active Control? Disturbances System Measurement Control Controler. Regulator.,,, Aims of an Active Control Disturbances
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 informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
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 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 informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
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 informationThe FrequencyResponse
6 The FrequencyResponse Design Method A Perspective on the FrequencyResponse Design Method The design of feedback control systems in industry is probably accomplished using frequencyresponse methods
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 informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
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 informationECE 388 Automatic Control
Lead Compensator and PID Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage: http://ece388.cankaya.edu.tr
More informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationThe Nyquist criterion relates the stability of a closed system to the openloop frequency response and open loop pole location.
Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the openloop frequency response and open loop pole location. Mapping. If we take a complex number
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 informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 9. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
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 informationData Based Design of 3Term Controllers. Data Based Design of 3 Term Controllers p. 1/10
Data Based Design of 3Term Controllers Data Based Design of 3 Term Controllers p. 1/10 Data Based Design of 3 Term Controllers p. 2/10 History Classical Control  single controller (PID, lead/lag) is designed
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 information8.1.6 Quadratic pole response: resonance
8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Secondorder denominator, of the form 1+a 1 s + a s v 1 (s) + C R Twopole lowpass filter example v (s) with
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 informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More informationDesign and Tuning of Fractionalorder PID Controllers for Timedelayed Processes
Design and Tuning of Fractionalorder PID Controllers for Timedelayed Processes Emmanuel Edet Technology and Innovation Centre University of Strathclyde 99 George Street Glasgow, United Kingdom emmanuel.edet@strath.ac.uk
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles
More informationH(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )
.7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a
More informationCollocated versus noncollocated control [H04Q7]
Collocated versus noncollocated control [H04Q7] Jan Swevers September 2008 00 Contents Some concepts of structural dynamics Collocated versus noncollocated control Summary This lecture is based on parts
More informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
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 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 information