School of Mechanical Engineering Purdue University


 Buck Bradford
 3 years ago
 Views:
Transcription
1 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 highspeed runs between New Haven, CT and New York City, the train experiences intermittent power loss at 4 km/hr and 1 km/hr. ME375 Frequency Response  1
2 Case Study Pantograph Model F c (t) m z ( ) ( ) mz + b + b z + k + k z bz k z mz + bz + kz bz kz F () c t 1 1 k b z 1 For m 1 3. kg, b 1 15 N/(m/s), k 1 96 N/m, m 11.5 kg, b 75 N/(m/s), and k 958 N/m: k 1 m 1 b 1 Z( s).87( s s+ 834) 4 3 F () s s s + 179s s+ 347,75 c ME375 Frequency Response  3 Case Study Frequency Response 6 Magnitude (db) Phase (deg) ME375 Frequency Response  4
3 Frequency Response Forced Response to Sinusoidal Inputs Frequency Response of LTI Systems Bode Plots ME375 Frequency Response  5 Forced Response to Sinusoidal Inputs Ex: Let s s find the forced response of a stable first order system: y + 5y 1u to a sinusoidal input: ut () sin() t Forced response: Y() s G() s U() s [ ] where Gs ( ) and Us ( ) Lsin( t) Y( s) PFE: A1 A A3 Y() s + + Compare coefficients to find A 1, A and A 3 : ME375 Frequency Response  6 3
4 Forced Response to Sinusoidal Inputs Ex: (cont.) Use ILT to find y(t) ) : [ ] 1 1 yt ( ) L Y( s) L + + Useful Formula: Asin( ω t) + Bcos( ω t) A + B sin( ω t + φ) Where φ atan( BA, ) ( A+ jb) Using this formula, the forced response can be represented by 5t yt ( ) e + sin( t+ φ) ME375 Frequency Response  7 Forced Response of 1st Order System Output Input is sin(t) Input.5 Response Time (sec) ME375 Frequency Response  8 4
5 Forced Response to Sinusoidal Inputs Ex: Given the same system as in the previous example, find the forced response to u(t) ) sin(1 t). Y() s G() s U() s [ ] where Gs ( ) and Us ( ) Lsin(1 t) Y( s) ME375 Frequency Response  9 Forced Response of 1st Order Systems Input is sin(1t) 1 Output Response Input Time (sec) ME375 Frequency Response  1 5
6 Frequency Response Ex: Let s s revisit the same example where y + 5y 1u and the input is a general sinusoidal input: sin(ω t). 1 ω 1 ω Ys () Gs () Us () s+ 5 s + ω s+ 5 ( s jω)( s+ jω) A1 A A3 Ys () + + s+ 5 s jω s+ jω Instead of comparing coefficients, use the residue formula to find A i s: 1 ( 5) ( ) 5 ( 5) 1 ω A s+ Y s s+ s ( 5) s+ s + ω s 5 ω A ( s jω) Y( s) ( s jω) Gs ( ) s jω s + ω s jω ω A3 ( s+ jω) Y( s) ( s+ jω) G( s) s jω s + ω s jω ME375 Frequency Response  11 Frequency Response Ex: (Cont.) 1ω A1 5 + ω A G( jω ) j jω + 5 j j A3 G( jω ) j jω + 5 j j The steady state response Y SS (s)) is: A A3 YSS () s + s jω s+ jω 1 jω t jω t y () t L Y ( s) A e + A e SS [ ] SS 3 y ( t) G( jω ) sin( ω t+ φ) where φ G( jω) SS ME375 Frequency Response  1 6
7 Frequency Response Frequency Response ME375 Frequency Response  13 In Class Exercise For the current example, y + 5y 1u Calculate the magnitude and phase shift of the steady state response when the system is excited by (i) sin(t) ) and (ii) sin(1t). Compare your result with the steady state response calculated in the previous examples. Note: 1 1 Gs () G( jω ) s+ 5 jω G( jω ) and φ G( jω) atan( ω,5) ω + 5 ME375 Frequency Response
8 Frequency Response Frequency response is used to study the steady state output y SS (t)) of a stable system due to sinusoidal inputs at different frequencies. In general, given a stable system: ( n) ( n 1) ( m) ( m 1) any + an 1y + + a1y + ay bmu + bm 1u + + bu 1 + bu m m 1 bms + bm 1s + + bs 1 + b Ns () bm( s z1)( s z) ( s zm) Gs () n n 1 as n + an 1s + + as 1 + a Ds () an( s p1)( s p) ( s pn) If the input is a sinusoidal signal with frequency ω, i.e. ut () Asin( ω t) then the steady state output y SS (t)) is also a sinusoidal signal with the same frequency as the input signal but with different magnitude and phase: y () t G( jω ) A sin( ω t+ G( jω)) SS u where G(jω) ) is the complex number obtained by substitute jω for s in G(s) ), i.e. m m 1 bm( jω) + bm ( jω) + + b ( jω) + b G( jω ) G( s) s jω n n 1 a ( jω) + a ( jω) + + a ( jω) + a n u 1 1 n 1 1 ME375 Frequency Response  15 Frequency Response Input u(t) U(s) LTI System G(s) Output y(t) Y(s) u π/ω y SS π/ω t ut () Asin( ω t) u y () t G( jω ) A sin( ω t+ G( jω)) SS u t A different perspective of the role of the transfer function: Amplitude of the steady state sinusoidal output G( jω ) Amplitude of the sinusoidal input G( jω ) Phase difference (shift) between yss ( t) and the sinusoidal input ME375 Frequency Response
9 Frequency Response G Input u(t) Output y(t) G ME375 Frequency Response  17 In Class Exercise Ex: 1st Order System The motion of a piston in a cylinder can be modeled by a 1st order system with force as input and piston velocity as output: f(t) v The EOM is: Mv + Bv f() t (1) Let M.1 kg and B.5 N/(m/s), find the transfer function of the system: () Calculate the steady state output of the system when the input is Input f(t) Steady State Output v(t) sin(ω t) G(jω) sin(ω t + φ ) sin(t) sin(t + sin(1t) sin(1t + sin(t) sin(t + sin(3t) sin(3t + sin(4t) sin(4t + sin(5t) sin(5t + sin(6t) sin(6t + ME375 Frequency Response
10 In Class Exercise (3) Plot the frequency response plot Magnitude ((m/s)/n) Phase (deg) ME375 Frequency Response  19 Example  Vibration Absorber (I) Without vibration absorber: EOM: z 1 M z Bz K z f t () K 1 M 1 f(t) B 1 Let M 1 1 kg, K 1 1 N/m, B 1 4 N/(m/s). Find the steady state response of the system for f(t) ) (a) sin(8.5t) ) (b) sin(1t) ) (c) sin(11.7t). TF (from f(t) ) to z 1 ): Input f(t) Steady State Output z 1 (t) sin(ω t) G(jω) sin(ω t + φ ) sin(8.5t) sin(8.5t + sin(1t) sin(1t + sin(11.7t) sin(11.7t + ME375 Frequency Response  1
11 Example  Vibration Absorber (I).1 f(t) sin(8.5 t).5 z 1 (m) f(t) sin(1 t) z 1 (m) f(t) sin(11.7 t) z 1 (m) Time (sec) ME375 Frequency Response  1 Example  Vibration Absorber (II) With vibration absorber: K K 1 M M 1 f(t) z B B 1 z 1 TF (from f(t) ) to z 1 ): EOM: M 1z1+ ( B1+ B) z 1+ ( K1+ K) z1 Bz Kz f( t) M z + Bz + Kz Bz Kz 1 1 Let M 1 1 kg, K 1 1 N/m, B 1 4 N/(m/s), M 1 kg, K 1 N/m, and B.1 N/(m/s). Find the steady state response of the system for f(t) ) (a) sin(8.5t) ) (b) sin(1t) ) (c) sin(11.7t). Input f(t) Steady State Output z 1 (t) sin(ω t) G(jω) sin(ω t + φ ) sin(8.5t) sin(8.5t + sin(1t) sin(1t + sin(11.7t) sin(11.7t + ME375 Frequency Response  11
12 Example  Vibration Absorber (II).4 f(t) sin(8.5 t). z 1 (m) f(t) sin(1 t) z 1 (m) f(t) sin(11.7 t) z 1 (m) Time (sec) ME375 Frequency Response  3 Example  Vibration Absorber (II) Take a closer look at the poles of the transfer function: The characteristic equation 4 3 1s + 5.1s + 1.4s + 5s+ 1 Poles: p1,.1± 8.5 j p.155 ± 11.7 j 3,4 What part of the poles determines the rate of decay for the transient response? (Hint: when p σ ± jω the response is e σt e jω t ) ME375 Frequency Response  4 1
13 Example  Vibration Absorbers Frequency Response Plot No absorber added Frequency Response Plot Absorber tuned at 1 rad/sec added.5.5 Magnitude (m/n) Magnitude (m/n) Phase (deg) Phase (deg) ME375 Frequency Response  5 Example  Vibration Absorbers Bode Plot No absorber added Bode Plot Absorber tuned at 1 rad/sec added Phase (deg); Magnitude (db) Phase (deg); Magnitude (db) ME375 Frequency Response
14 Bode Diagrams (Plots) Bode Diagrams (Plots) A unique way of plotting the frequency response function, G(jω), ), w.r.t. frequency w of systems. Consists of two plots: Magnitude Plot : plots the magnitude of G(jω) ) in decibels w.r.t. logarithmic frequency, i.e. G( jω) log 1 G( jω) vs log ω db 1 Phase Plot : plots the linear phase angle of G(jω) ) w.r.t. logarithmic frequency, i.e. G( jω ) vs log To plot Bode diagrams, one needs to calculate the magnitude and phase of the corresponding transfer function. Ex: s + 1 Gs ( ) s + 1s 1 ω ME375 Frequency Response  7 Bode Diagrams Revisit the previous example: s+ 1 ( jω) + 1 Gs () G( jω) s + 1 s jω( jω + 1) 5 3 G( jω) G( jω) w Gj ( ω) log 1 Gjω ( ) G( jω) Phase (deg); Magnitude (db) ME375 Frequency Response
School of Mechanical Engineering Purdue University. ME375 Frequency Response  1
Case Study ME375 Frequecy Respose  Case Study SUPPORT POWER WIRE DROPPERS Electric trai derives power through a patograph, which cotacts the power wire, which is suspeded from a cateary. Durig highspeed
More informationSchool of Mechanical Engineering Purdue University. ME375 Dynamic Response  1
Dynamic Response of Linear Systems Linear System Response Superposition Principle Responses to Specific Inputs Dynamic Response of f1 1st to Order Systems Characteristic Equation  Free Response Stable
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 informationControl Systems Design
ELEC4410 Control Systems Design Lecture 13: Stability Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 13: Stability p.1/20 Outline InputOutput
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 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 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 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 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 informationProcess Control & Instrumentation (CH 3040)
Firstorder systems Process Control & Instrumentation (CH 3040) Arun K. Tangirala Department of Chemical Engineering, IIT Madras January  April 010 Lectures: Mon, Tue, Wed, Fri Extra class: Thu A firstorder
More informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
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 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 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 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 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 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 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 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 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 informationClass 13 Frequency domain analysis
Class 13 Frequency domain analysis The frequency response is the output of the system in steady state when the input of the system is sinusoidal Methods of system analysis by the frequency response, as
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 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 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 informationSignals and Systems. Problem Set: The ztransform and DT Fourier Transform
Signals and Systems Problem Set: The ztransform and DT Fourier Transform Updated: October 9, 7 Problem Set Problem  Transfer functions in MATLAB A discretetime, causal LTI system is described by the
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 informationEE/ME/AE324: Dynamical Systems. Chapter 7: Transform Solutions of Linear Models
EE/ME/AE324: Dynamical Systems Chapter 7: Transform Solutions of Linear Models The Laplace Transform Converts systems or signals from the real time domain, e.g., functions of the real variable t, to 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 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 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 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 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 information2.3 Oscillation. The harmonic oscillator equation is the differential equation. d 2 y dt 2 r y (r > 0). Its solutions have the form
2. Oscillation So far, we have used differential equations to describe functions that grow or decay over time. The next most common behavior for a function is to oscillate, meaning that it increases and
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 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 informationTransfer Function Analysis
Trasfer Fuctio Aalysis Free & Forced Resposes Trasfer Fuctio Syste Stability ME375 Trasfer Fuctios  Free & Forced Resposes Ex: Let s s look at a stable first order syste: τ y + y = Ku Take LT of the I/O
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
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 informationDynamic Response of Linear Systems
Dyamic Respose of Liear Systems Liear System Respose Superpositio Priciple Resposes to Specific Iputs Dyamic Respose of st Order Systems Characteristic Equatio  Free Respose Stable st Order System Respose
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 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 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 informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
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 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 informationBasic Electronics. Introductory Lecture Course for. Technology and Instrumentation in Particle Physics Chicago, Illinois June 914, 2011
Basic Electronics Introductory Lecture Course for Technology and Instrumentation in Particle Physics 2011 Chicago, Illinois June 914, 2011 Presented By Gary Drake Argonne National Laboratory Session 2
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 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 informationStep Response for the Transfer Function of a Sensor
Step Response f the Transfer Function of a Sens G(s)=Y(s)/X(s) of a sens with X(s) input and Y(s) output A) First Order Instruments a) First der transfer function G(s)=k/(1+Ts), k=gain, T = time constant
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 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 informationAdvanced Analog Building Blocks. Prof. Dr. Peter Fischer, Dr. Wei Shen, Dr. Albert Comerma, Dr. Johannes Schemmel, etc
Advanced Analog Building Blocks Prof. Dr. Peter Fischer, Dr. Wei Shen, Dr. Albert Comerma, Dr. Johannes Schemmel, etc 1 Topics 1. S domain and Laplace Transform Zeros and Poles 2. Basic and Advanced current
More informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
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 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 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 informationSchool of Mechanical Engineering Purdue University. ME375 Transfer Functions  1
Trasfer Fuctio Aalysis Free & Forced Resposes Trasfer Fuctio Syste Stability ME375 Trasfer Fuctios  1 Free & Forced Resposes Ex: Let s look at a stable first order syste: y y Ku Take LT of the I/O odel
More informationBode Diagrams School of Mechanical Engineering ME375 Frequency Response  29 Purdue University Example Ex:
ME375 Hadouts Bode Diagrams Recall that if m m bs m + bm s + + bs+ b Gs () as + a s + + as+ a The bm( j z)( j z) ( j zm) G( j ) a ( j p )( j p ) ( j p ) bm( s z)( s z) ( s zm) a ( s p )( s p ) ( s p )
More informationChapter 8: Frequency Domain Analysis
Chapter 8: Frequency Domain Analysis Samantha Ramirez Preview Questions 1. What is the steadystate response of a linear system excited by a cyclic or oscillatory input? 2. How does one characterize the
More informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10
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 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 informationGATE EE Topic wise Questions SIGNALS & SYSTEMS
www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)
More informationDr. Ian R. Manchester
Dr 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
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 information'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ. EGR 224 Spring Test II. Michael R. Gustafson II
'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ EGR 224 Spring 2018 Test II Michael R. Gustafson II Name (please print) In keeping with the Community Standard, I have neither provided nor received any
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 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 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 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 informationEE 3054: Signals, Systems, and Transforms Summer It is observed of some continuoustime LTI system that the input signal.
EE 34: Signals, Systems, and Transforms Summer 7 Test No notes, closed book. Show your work. Simplify your answers. 3. It is observed of some continuoustime LTI system that the input signal = 3 u(t) produces
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 informationEXPERIMENTALLY DETERMINING THE TRANSFER FUNCTION OF A SPRING MASS SYSTEM
EXPERIMENTALLY DETERMINING THE TRANSFER FUNCTION OF A SPRING MASS SYSTEM Lab 8 OBJECTIVES At the conclusion of this experiment, students should be able to: Experimentally determine the best fourth order
More informationCourse roadmap. ME451: Control Systems. Example of Laplace transform. Lecture 2 Laplace transform. Laplace transform
ME45: Control Systems Lecture 2 Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Transfer function Models for systems electrical mechanical electromechanical Block
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 informationCHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang
CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 51 Road Map of the Lecture V Laplace Transform and Transfer
More informationPower System Control
Power System Control Basic Control Engineering Prof. Wonhee Kim School of Energy Systems Engineering, ChungAng University 2 Contents Why feedback? System Modeling in Frequency Domain System Modeling in
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 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 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 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 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 information'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ. EGR 224 Spring Test II. Michael R. Gustafson II
'XNH8QLYHUVLW\ (GPXQG73UDWWU6FKRRORI(QJLQHHULQJ EGR 224 Spring 2017 Test II Michael R. Gustafson II Name (please print) In keeping with the Community Standard, I have neither provided nor received any
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 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 informationMAE 142 Homework #2 (Design Project) SOLUTIONS. (a) The main body s kinematic relationship is: φ θ ψ. + C 3 (ψ) 0 + C 3 (ψ)c 1 (θ)
MAE 42 Homework #2 (Design Project) SOLUTIONS. Top Dynamics (a) The main body s kinematic relationship is: ω b/a ω b/a + ω a /a + ω a /a ψâ 3 + θâ + â 3 ψˆb 3 + θâ + â 3 ψˆb 3 + C 3 (ψ) θâ ψ + C 3 (ψ)c
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 informationPart IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 4. The Frequency ResponseG(jω)
Part IB Paper 6: Information Engineering LINEAR SYSEMS AND CONROL Glenn Vinnicombe HANDOU 4 he Frequency ResponseG(jω) x(t) 2π ω y(t) G(jω) argg(jω) ω t t Asymptotically stable LI systemg(s) x(t)=cos(ωt)
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 informationSchool of Mechanical Engineering Purdue University. ME375 Feedback Control  1
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More 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 informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Stability RouthHurwitz stability criterion Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
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 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 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 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 informationSection 4.9; Section 5.6. June 30, Free Mechanical Vibrations/Couple MassSpring System
Section 4.9; Section 5.6 Free Mechanical Vibrations/Couple MassSpring System June 30, 2009 Today s Session Today s Session A Summary of This Session: Today s Session A Summary of This Session: (1) Free
More informationMATH2351 Introduction to Ordinary Differential Equations, Fall Hints to Week 07 Worksheet: Mechanical Vibrations
MATH351 Introduction to Ordinary Differential Equations, Fall 111 Hints to Week 7 Worksheet: Mechanical Vibrations 1. (Demonstration) ( 3.8, page 3, Q. 5) A mass weighing lb streches a spring by 6 in.
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 information