Raktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries


 Kimberly Robertson
 2 years ago
 Views:
Transcription
1 . AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University.
2 Preliminaries Signals & Systems Laplace transforms.. Transfer functions from ordinary linear differential equations System interconnections Block diagram algebra simplification of interconnections General feedback control system interconnection. d n r + u + +. e C P y + + y m y m AERO 632, Instructor: Raktim Bhattacharya 2 / 46
3
4 Signals & Systems.. u(t). P y(t) Actuator applies u(t) Sensor provides y(t) Feedback controller takes y(t) and determines u(t) to achieve desired behavior The controller is typically implemented as software, running in a micro controller Imperfections exist in real world sensors have noise actuators have irregularities plant P is not fully known AERO 632, Instructor: Raktim Bhattacharya 4 / 46
5 System Response to u(t).. u(t). P y(t) Given plant P and input u(t), what is y(t)? P is defined in terms of ordinary differential equations y(t) is the forced + initial condition response. Linear Dynamics mẍ + cẋ + kx = u(t) dynamics y(t) = x(t) measurement Nonlinear Dynamics ẍ µ( x 2 )ẋ + x = u(t) dynamics y(t) = x(t) measurement In this class we focus on linear systems AERO 632, Instructor: Raktim Bhattacharya 5 / 46
6 Linear Systems... u(t) P y(t) Dynamics is defined by linear ordinary differential equation Super position principle applies u (t) y (t) u 2 (t) y 2 (t) = (u (t) + u 2 (t)) (y (t) + y 2 (t)) AERO 632, Instructor: Raktim Bhattacharya 6 / 46
7
8 Given signal u(t), Laplace transform is defined as Exists when L {u(t)} := u(t)e st dt lim t u(t)e σt =, for some σ >.. Very useful in studying linear dynamical systems and designing controllers AERO 632, Instructor: Raktim Bhattacharya 8 / 46
9 Properties Linear operator Additive L {u (t) + u 2 (t)} = Superposition = (u (t) + u 2 (t)) e st dt u (t)e st dt +.. = L {u (t)} + L {u 2 (t)} L {au(t)} = al {u(t)}, a is a constant u 2 (t)e st dt AERO 632, Instructor: Raktim Bhattacharya 9 / 46
10 Properties (contd.). U(s) := L {u(t)}.. 2. L {au (t) + bu 2 (t)} = al {u (t)} + bl {u 2 (t)} = au (s) + bu 2 (s) 3. t s U(s) u(τ)dτ 4. U (s)u 2 (s) u (t) u 2 (t) Convolution 5. lim su(s) lim s 6. lim su(s) u( + ) Initial value theorem s 7. du(s) tu(t) { ds } du 8. L su(s) su() dt t u(t) Final value theorem 9. L {ü} s 2 U(s) su() u() AERO 632, Instructor: Raktim Bhattacharya / 46
11 Important Signals... L {δ(t)} = δ(t) is impulse function 2. L {(t)} = s (t) is unit step function at t = 3. L {t} = s 2 ω 4. L {sin(ωt} = s 2 + ω 2 s 5. L {cos(ωt} = s 2 + ω 2 AERO 632, Instructor: Raktim Bhattacharya / 46
12
13 Spring Mass Damper System.. m u(t) Equation of Motion mẍ + cẋ + kx = u(t) Take L { } on both sides L {mẍ + cẋ + kx} = L {u(t)} ml {ẍ} + cl {ẋ} + kl {x} = L {u(t)} m ( s 2 X(s) sx() ẋ() ) + c (sx(s) x()) + kx(s) = U(s) (ms 2 + cs + k)x(s) = U(s) ẋ() and x() are assumed to be zero AERO 632, Instructor: Raktim Bhattacharya 3 / 46
14 Transfer Function.. m u(t). u(t) P y(t) (ms 2 + cs + k)x(s) = U(s) = X(s) U(s) = ms 2 + cs + k Choose output y(t) = x(t) = Y (s) = X(s). Therefore P (s) := Y (s) U(s) = ms 2 Transfer function + cs + k AERO 632, Instructor: Raktim Bhattacharya 4 / 46
15 Transfer Function (contd.) In general P (s) = N(s) D(s) where N(s) and D(s) are polynomials in s Roots of N(s) are the zeros Roots of D(s) are the poles determine stability.. AERO 632, Instructor: Raktim Bhattacharya 5 / 46
16 Response to u(t) Given input signal u(t) and transfer function P (s). Determine output response y(t)... Laplace transform U(s) := L {u(t)} 2. Determine Y (s) := P (s)u(s) 3. Laplace inverse y(t) := L {Y (s)} = L {P (s)u(s)}. u(t) P y(t) AERO 632, Instructor: Raktim Bhattacharya 6 / 46
17
18 Block Diagram Representation of s Series Parallel Feedback A simple example A complex example.. AERO 632, Instructor: Raktim Bhattacharya 8 / 46
19 Series Connection ụ. G G 2 y.. AERO 632, Instructor: Raktim Bhattacharya 9 / 46
20 Parallel Connection.. G 2 ụ. G. + y AERO 632, Instructor: Raktim Bhattacharya 2 / 46
21 Feedback Connection r. + G y.. AERO 632, Instructor: Raktim Bhattacharya 2 / 46
22 Simple Example.. r. + G G 2 G 3 + y G 4 AERO 632, Instructor: Raktim Bhattacharya 22 / 46
23 Complex Example.. r. + G G 2 G y G 4 AERO 632, Instructor: Raktim Bhattacharya 23 / 46
24 Frequency Response
25 Response to Sinusoidal Input u(t) y(t). P Let u(t) = A u sin(ωt) Vary ω from to.. A linear system s response to sinusoidal inputs is called the system s frequency response AERO 632, Instructor: Raktim Bhattacharya 25 / 46
26 Response to Sinusoidal Input Example Let P (s) = s+, u(t) = sin(t).. y(t) = 2 e t 2 cos(t) + 2 sin(t) = 2 e t + sin(t π }{{} 2 4 ) }{{} natural response forced response Forced response has form A y sin(ωt + ϕ) A y and ϕ are functions of ω AERO 632, Instructor: Raktim Bhattacharya 26 / 46
27 Response to Sinusoidal Input Generalization In general ω.. Y (s) = G(s) s 2 + ω 2 = α α n + + α + α s p s p n s + jω s jω = y(t) = α e pt + + α n e p nt + A }{{} y sin(ω + ϕ) }{{} natural forced Forced response has same frequency, different amplitude and phase. AERO 632, Instructor: Raktim Bhattacharya 27 / 46
28 Response to Sinusoidal Input Generalization (contd.) For a system P (s) and input forced response is where u(t) = A u sin(ω t), y(t) = A u M sin(ω t + ϕ), M(ω ) = P (s) s=jω = P (jω ), magnitude.. ϕ(ω ) = P (jω ) phase In polar form P (jω ) = Me jϕ. AERO 632, Instructor: Raktim Bhattacharya 28 / 46
29 Fourier Analysis
30 Fourier Series Expansion Given a signal y(t) with periodicity T, y(t) = a 2 + ( 2πnt a n cos T a = 2 T a n = 2 T b n = 2 T n=,2, T T T y(t)dt ( 2πnt y(t) cos T ( 2πnt y(t) sin T ) dt ) dt ) + b n sin.. ( ) 2πnt T AERO 632, Instructor: Raktim Bhattacharya 3 / 46
31 Fourier Series Expansion Approximation of step function...2 N=2.2 N= N=8.2 N= N=2.2 N= AERO 632, Instructor: Raktim Bhattacharya 3 / 46
32 Fourier Transform Step function.. Fourier transform reveals the frequency content of a signal AERO 632, Instructor: Raktim Bhattacharya 32 / 46
33 Fourier Transform Step function frequency content y(t) t ŷ(ω) ω AERO 632, Instructor: Raktim Bhattacharya 33 / 46
34 Signals & Systems.. Input Output. u(t) P y(t) Fourier Series Expansion superposition principle. i u i(t) P i y i(t) Fourier Transform Ụ(jω) Y (jω). P u i (t) = a i sin(ω i t) y iforced (t) = a i M sin(ω i t + ϕ) Y (jω) = P (jω)u(jω) Suffices to study P (jω) P (jω), P (jω) AERO 632, Instructor: Raktim Bhattacharya 34 / 46
35 Bode Plot
36 First Order System.. 5 Bode Diagram y(t) Freq =. rad/s Magnitude (db) y(t) y(t) Freq =. rad/s Freq = 5. rad/s Phase (deg) 45 y(t) Freq =. rad/s Frequency (rad/s) P (s) = /(s + ) loglog scale db = log ( ) 2dB = log (/) u(t) = A sin(ω t) y forced (t) = AM sin(ω t + ϕ) AERO 632, Instructor: Raktim Bhattacharya 36 / 46
37 Second Order System.. Magnitude (db) Bode Diagram y(t) y(t) 2 Freq =. rad/s Freq =. rad/s Freq = 5. rad/s Phase (deg) y(t) y(t) Freq =. rad/s 8 Frequency (rad/s) P (s) = /(s 2 +.5s + ) ω n = rad/s u(t) = A sin(ω t) y forced (t) = AM sin(ω t + ϕ) AERO 632, Instructor: Raktim Bhattacharya 37 / 46
38 S(jω) + T (jω) =.. d n r + u e y + y. m C P y m Magnitude S(jω) P (s) = C(s) = (s+)(s/2+) S = G er = +P C = +P T = G yr = P C +P C = P +P ω rad/s Bode Diagram S T S+T Magnitude T(jω) Magnitude (db) ω rad/s Frequency (rad/s) AERO 632, Instructor: Raktim Bhattacharya 38 / 46
39 All transfer functions With proportional controller.. G er G ed G en Magnitude (db) 2 Magnitude (db) Magnitude (db) Frequency (rad/s) 8 2 Frequency (rad/s) Frequency (rad/s) G yr G yd G yn 2 2 Magnitude (db) 2 4 Magnitude (db) Magnitude (db) Frequency (rad/s) 8 2 Frequency (rad/s) 6 2 Frequency (rad/s) AERO 632, Instructor: Raktim Bhattacharya 39 / 46
40 Controller Considerations
41 Using Bode Plot of P (jω)c(jω) Loop Shaping.. Develop conditions on the Bode plot of the open loop transfer function Sensitivity +P C Steadystate errors: slope and magnitude at lim ω Robust to sensor noise Disturbance rejection Controller roll off = not excite highfrequency modes of plant Robust to plant uncertainty Look at Bode plot of L(jω) := P (jω)c(jw) AERO 632, Instructor: Raktim Bhattacharya 4 / 46
42 Frequency Domain Specifications Constraints on the shape of L(jω).. P (j!)c(j!) Steadystate error boundary slope! c Sensor noise, plant uncertainty! Choose C(jω) to ensure L(jω) does not violate the constraints Slope at ω c ensures P M 9 stable if P M > = Sensor noise, disturbance Plant uncertainty P C > 8 AERO 632, Instructor: Raktim Bhattacharya 42 / 46
43 Plant Uncertainty P (jω) = P (jω)( + P (jω)).. Bode Diagram 5 5 True Model Unc+ Unc P (j!)c(j!) Steadystate error boundary slope!c Sensor noise, plant uncertainty! Magnitude (db) slope 5 2 P (j!)c(j!) Steadystate error boundary!c Sensor noise, disturbance Plant uncertainty! Frequency (rad/s) AERO 632, Instructor: Raktim Bhattacharya 43 / 46
44 Sensor Characteristics Noise spectrum.. P (j!)c(j!) Steadystate error boundary slope!c Sensor noise, plant uncertainty! P (j!)c(j!) slope!c Steadystate error boundary Sensor noise, disturbance Plant uncertainty! G yn = P C + P C AERO 632, Instructor: Raktim Bhattacharya 44 / 46
45 Reference Tracking Bandlimited else conflicts with noise rejection.. Spectrum of r(t) X(f) P (j!)c(j!) Steadystate error boundary slope!c Sensor noise, plant uncertainty! Frequency (Hz) X(f) Spectrum of n(t) slope Sensor noise, disturbance Plant uncertainty!!c Steadystate error boundary Frequency (Hz) P (j!)c(j!) G yr = G yn = P C +P C P C + P C AERO 632, Instructor: Raktim Bhattacharya 45 / 46
46 Disturbance Rejecton Bandlimited else conflicts with noise rejection.. Spectrum of d(t) X(f) P (j!)c(j!) Steadystate error boundary slope!c Sensor noise, plant uncertainty! Frequency (Hz) X(f) Spectrum of n(t) slope Sensor noise, disturbance Plant uncertainty!!c Steadystate error boundary Frequency (Hz) P (j!)c(j!) G yd = G yn = P C +P C P + P C AERO 632, Instructor: Raktim Bhattacharya 46 / 46
Raktim 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 informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
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 informationSingular Value Decomposition Analysis
Singular Value Decomposition Analysis Singular Value Decomposition Analysis Introduction Introduce a linear algebra tool: singular values of a matrix Motivation Why do we need singular values in MIMO control
More informationProportional, Integral & Derivative Control Design. Raktim Bhattacharya
AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University
More informationFrequency Response of Linear Time Invariant Systems
ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z
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 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 informationPrüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationH 2 Optimal State Feedback Control Synthesis. Raktim Bhattacharya Aerospace Engineering, Texas A&M University
H 2 Optimal State Feedback Control Synthesis Raktim Bhattacharya Aerospace Engineering, Texas A&M University Motivation Motivation w(t) u(t) G K y(t) z(t) w(t) are exogenous signals reference, process
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 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 informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Norms for Signals and Systems
. AERO 632: Design of Advance Flight Control System Norms for Signals and. Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Norms for Signals ...
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 informationExam. 135 minutes, 15 minutes reading time
Exam August 6, 208 Control Systems II (5059000) Dr. Jacopo Tani Exam Exam Duration: 35 minutes, 5 minutes reading time Number of Problems: 35 Number of Points: 47 Permitted aids: 0 pages (5 sheets) A4.
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 informationTransfer function and linearization
Transfer function and linearization Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata Corso di Controlli Automatici, A.A. 2425 Testo del corso:
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
MIT OpenCourseWare http://ocw.mit.edu 2.6 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS
More informationLecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.
ISS0031 Modeling and Identification Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design. Aleksei Tepljakov, Ph.D. September 30, 2015 Linear Dynamic Systems Definition
More informationDefinition of the Laplace transform. 0 x(t)e st dt
Definition of the Laplace transform Bilateral Laplace Transform: X(s) = x(t)e st dt Unilateral (or onesided) Laplace Transform: X(s) = 0 x(t)e st dt ECE352 1 Definition of the Laplace transform (cont.)
More informationMODELING OF CONTROL SYSTEMS
1 MODELING OF CONTROL SYSTEMS Feb15 Dr. Mohammed Morsy Outline Introduction Differential equations and Linearization of nonlinear mathematical models Transfer function and impulse response function Laplace
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #11 Wednesday, January 28, 2004 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Relative Stability: Stability
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 informationIntroduction & Laplace Transforms Lectures 1 & 2
Introduction & Lectures 1 & 2, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Control System Definition of a Control System Group of components that collectively
More informationCDS 101/110a: Lecture 81 Frequency Domain Design
CDS 11/11a: Lecture 81 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve
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 informationQuadratic Stability of Dynamical Systems. Raktim Bhattacharya Aerospace Engineering, Texas A&M University
.. Quadratic Stability of Dynamical Systems Raktim Bhattacharya Aerospace Engineering, Texas A&M University Quadratic Lyapunov Functions Quadratic Stability Dynamical system is quadratically stable if
More informationCDS 101/110a: Lecture 101 Robust Performance
CDS 11/11a: Lecture 11 Robust Performance Richard M. Murray 1 December 28 Goals: Describe how to represent uncertainty in process dynamics Describe how to analyze a system in the presence of uncertainty
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 informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
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 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: 30Apr14 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More informationECE 3084 QUIZ 2 SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING GEORGIA INSTITUTE OF TECHNOLOGY APRIL 2, Name:
ECE 3084 QUIZ 2 SCHOOL OF ELECTRICAL AND COMPUTER ENGINEERING GEORGIA INSTITUTE OF TECHNOLOGY APRIL 2, 205 Name:. The quiz is closed book, except for one 2sided sheet of handwritten notes. 2. Turn off
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 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 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 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 informationSchool of Mechanical Engineering Purdue University
Case Study ME375 Frequency Response  1 Case Study SUPPORT POWER WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power wire, which is suspended from a catenary. During
More informationLecture 1: Feedback Control Loop
Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure
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 informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servooriented
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 informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
More 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 informationControl Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/2017
More informationLinear Systems Theory
ME 3253 Linear Systems Theory Review Class Overview and Introduction 1. How to build dynamic system model for physical system? 2. How to analyze the dynamic system?  Time domain  Frequency domain (Laplace
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
More informationTransform Solutions to LTI Systems Part 3
Transform Solutions to LTI Systems Part 3 Example of second order system solution: Same example with increased damping: k=5 N/m, b=6 Ns/m, F=2 N, m=1 Kg Given x(0) = 0, x (0) = 0, find x(t). The revised
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 informationTopic # Feedback Control Systems
Topic #20 16.31 Feedback Control Systems Closedloop system analysis Bounded Gain Theorem Robust Stability Fall 2007 16.31 20 1 SISO Performance Objectives Basic setup: d i d o r u y G c (s) G(s) n control
More informationBasic Procedures for Common Problems
Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available
More information9.5 The Transfer Function
Lecture Notes on Control Systems/D. Ghose/2012 0 9.5 The Transfer Function Consider the nth order linear, timeinvariant dynamical system. dy a 0 y + a 1 dt + a d 2 y 2 dt + + a d n y 2 n dt b du 0u +
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 informationReview of Linear TimeInvariant Network Analysis
D1 APPENDIX D Review of Linear TimeInvariant Network Analysis Consider a network with input x(t) and output y(t) as shown in Figure D1. If an input x 1 (t) produces an output y 1 (t), and an input x
More informationELG 3150 Introduction to Control Systems. TA: Fouad Khalil, P.Eng., Ph.D. Student
ELG 350 Introduction to Control Systems TA: Fouad Khalil, P.Eng., Ph.D. Student fkhalil@site.uottawa.ca My agenda for this tutorial session I will introduce the Laplace Transforms as a useful tool for
More informationControl Systems. Frequency domain analysis. L. Lanari
Control Systems m i l e r p r a in r e v y n is o Frequency domain analysis L. Lanari outline introduce the Laplace unilateral transform define its properties show its advantages in turning ODEs to algebraic
More informationThe Laplace Transform
The Laplace Transform Introduction There are two common approaches to the developing and understanding the Laplace transform It can be viewed as a generalization of the CTFT to include some signals with
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
More informationGrades will be determined by the correctness of your answers (explanations are not required).
6.00 (Fall 20) Final Examination December 9, 20 Name: Kerberos Username: Please circle your section number: Section Time 2 am pm 4 2 pm Grades will be determined by the correctness of your answers (explanations
More informationControl System. Contents
Contents Chapter Topic Page Chapter Chapter Chapter3 Chapter4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of
More informationIntroduction to Modern Control MT 2016
CDT Autonomous and Intelligent Machines & Systems Introduction to Modern Control MT 2016 Alessandro Abate Lecture 2 Firstorder ordinary differential equations (ODE) Solution of a linear ODE Hints to nonlinear
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationagree w/input bond => + sign disagree w/input bond =>  sign
1 ME 344 REVIEW FOR FINAL EXAM LOCATION: CPE 2.204 M. D. BRYANT DATE: Wednesday, May 7, 2008 9noon Finals week office hours: May 6, 47 pm Permitted at final exam: 1 sheet of formulas & calculator I.
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 informationModule 4. Related web links and videos. 1. FT and ZT
Module 4 Laplace transforms, ROC, rational systems, Z transform, properties of LT and ZT, rational functions, system properties from ROC, inverse transforms Related web links and videos Sl no Web link
More informationECEN 420 LINEAR CONTROL SYSTEMS. Lecture 2 Laplace Transform I 1/52
1/52 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 2 Laplace Transform I Linear Time Invariant Systems A general LTI system may be described by the linear constant coefficient differential equation: a n d n
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 5: Calculating the Laplace Transform of a Signal Introduction In this Lecture, you will learn: Laplace Transform of Simple
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 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 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 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 informationLaplace Transform Part 1: Introduction (I&N Chap 13)
Laplace Transform Part 1: Introduction (I&N Chap 13) Definition of the L.T. L.T. of Singularity Functions L.T. Pairs Properties of the L.T. Inverse L.T. Convolution IVT(initial value theorem) & FVT (final
More informationLaplace Transforms and use in Automatic Control
Laplace Transforms and use in Automatic Control P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: P.Santosh Krishna, SYSCON Recap Fourier series Fourier transform: aperiodic Convolution integral
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 informationToday s goals So far Today 2.004
Today s goals So far Feedback as a means for specifying the dynamic response of a system Root Locus: from the openloop poles/zeros to the closedloop poles Moving the closedloop poles around Today Proportional
More informationTopic # Feedback Control Systems
Topic #17 16.31 Feedback Control Systems Deterministic LQR Optimal control and the Riccati equation Weight Selection Fall 2007 16.31 17 1 Linear Quadratic Regulator (LQR) Have seen the solutions to the
More informationCDS 101/110a: Lecture 102 Control Systems Implementation
CDS 101/110a: Lecture 102 Control Systems Implementation Richard M. Murray 5 December 2012 Goals Provide an overview of the key principles, concepts and tools from control theory  Classical control 
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 informationLecture 12. Upcoming labs: Final Exam on 12/21/2015 (Monday)10:3012:30
289 Upcoming labs: Lecture 12 Lab 20: Internal model control (finish up) Lab 22: Force or Torque control experiments [Integrative] (23 sessions) Final Exam on 12/21/2015 (Monday)10:3012:30 Today: Recap
More informationEE C128 / ME C134 Final Exam Fall 2014
EE C128 / ME C134 Final Exam Fall 2014 December 19, 2014 Your PRINTED FULL NAME Your STUDENT ID NUMBER Number of additional sheets 1. No computers, no tablets, no connected device (phone etc.) 2. Pocket
More informationEE Experiment 11 The Laplace Transform and Control System Characteristics
EE216:11 1 EE 216  Experiment 11 The Laplace Transform and Control System Characteristics Objectives: To illustrate computer usage in determining inverse Laplace transforms. Also to determine useful signal
More informationChapter 6: The Laplace Transform. ChihWei Liu
Chapter 6: The Laplace Transform ChihWei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace
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 informationEECS C128/ ME C134 Final Wed. Dec. 14, am. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C128/ ME C134 Final Wed. Dec. 14, 211 8111 am Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth 1 points total. Problem Points Score 1 16 2 12
More informationState Regulator. Advanced Control. design of controllers using pole placement and LQ design rules
Advanced Control State Regulator Scope design of controllers using pole placement and LQ design rules Keywords pole placement, optimal control, LQ regulator, weighting matrixes Prerequisites Contact state
More informationLINEAR SYSTEMS. J. Elder PSYC 6256 Principles of Neural Coding
LINEAR SYSTEMS Linear Systems 2 Neural coding and cognitive neuroscience in general concerns inputoutput relationships. Inputs Light intensity Presynaptic action potentials Number of items in display
More informationExplanations and Discussion of Some Laplace Methods: Transfer Functions and Frequency Response. Y(s) = b 1
Engs 22 p. 1 Explanations Discussion of Some Laplace Methods: Transfer Functions Frequency Response I. Anatomy of Differential Equations in the SDomain Parts of the sdomain solution. We will consider
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. 8 I * * Massachusetts
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 informationANALOG AND DIGITAL SIGNAL PROCESSING CHAPTER 3 : LINEAR SYSTEM RESPONSE (GENERAL CASE)
3. Linear System Response (general case) 3. INTRODUCTION In chapter 2, we determined that : a) If the system is linear (or operate in a linear domain) b) If the input signal can be assumed as periodic
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 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 informationLecture 9. Welcome back! Coming week labs: Today: Lab 16 System Identification (2 sessions)
232 Welcome back! Coming week labs: Lecture 9 Lab 16 System Identification (2 sessions) Today: Review of Lab 15 System identification (ala ME4232) Time domain Frequency domain 1 Future Labs To develop
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 informationAn Internal Stability Example
An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several polezero cancellations between the controller and
More informationIdentification Methods for Structural Systems
Prof. Dr. Eleni Chatzi System Stability Fundamentals Overview System Stability Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can be defined from
More informationENGIN 211, Engineering Math. Laplace Transforms
ENGIN 211, Engineering Math Laplace Transforms 1 Why Laplace Transform? Laplace transform converts a function in the time domain to its frequency domain. It is a powerful, systematic method in solving
More informationMath 221 Topics since the second exam
Laplace Transforms. Math 1 Topics since the second exam There is a whole different set of techniques for solving nth order linear equations, which are based on the Laplace transform of a function. For
More information