Lesson 16: Basic Control Modes

Similar documents
Chapter 6. Operational Amplifier. inputs can be defined as the average of the sum of the two signals.

Lecture 5: Operational Amplifiers and Op Amp Circuits

FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)

Department of Electrical and Computer Engineering FEEDBACK AMPLIFIERS

Prof. Paolo Colantonio a.a

Complex Numbers, Signals, and Circuits

MAE140 - Linear Circuits - Winter 16 Final, March 16, 2016

Digital PI Controller Equations

FE REVIEW OPERATIONAL AMPLIFIERS (OP-AMPS)( ) 8/25/2010

FEEDBACK AMPLIFIERS. v i or v s v 0

Boise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab

(b) i(t) for t 0. (c) υ 1 (t) and υ 2 (t) for t 0. Solution: υ 2 (0 ) = I 0 R 1 = = 10 V. υ 1 (0 ) = 0. (Given).

ENGR-4300 Electronic Instrumentation Quiz 4 Fall 2010 Name Section. Question Value Grade I 20 II 20 III 20 IV 20 V 20. Total (100 points)

Translational Equations of Motion for A Body Translational equations of motion (centroidal) for a body are m r = f.

Boise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab

We are IntechOpen, the world s leading publisher of Open Access books Built by scientists, for scientists. International authors and editors

Chapter 3 Differentiation and Integration

The Decibel and its Usage

OPTIMISATION. Introduction Single Variable Unconstrained Optimisation Multivariable Unconstrained Optimisation Linear Programming

ME2142/ME2142E Feedback Control Systems. Modelling of Physical Systems The Transfer Function

36.1 Why is it important to be able to find roots to systems of equations? Up to this point, we have discussed how to find the solution to

EECE 301 Signals & Systems Prof. Mark Fowler

Digital Signal Processing

Force = F Piston area = A

Combinational Circuit Design

Physics 4B. Question and 3 tie (clockwise), then 2 and 5 tie (zero), then 4 and 6 tie (counterclockwise) B i. ( T / s) = 1.74 V.

Energy Storage Elements: Capacitors and Inductors

Complete Variance Decomposition Methods. Cédric J. Sallaberry

Solution Set #3

Physics 4B. A positive value is obtained, so the current is counterclockwise around the circuit.

ELG 2135 ELECTRONICS I SECOND CHAPTER: OPERATIONAL AMPLIFIERS

ELG3336: Op Amp-based Active Filters

Advanced Circuits Topics - Part 1 by Dr. Colton (Fall 2017)

AGC Introduction

55:041 Electronic Circuits

General Tips on How to Do Well in Physics Exams. 1. Establish a good habit in keeping track of your steps. For example, when you use the equation

Mechanical Systems Part B: Digital Control Lecture BL4

Module 3. Process Control. Version 2 EE IIT, Kharagpur 1

Transfer Functions. Convenient representation of a linear, dynamic model. A transfer function (TF) relates one input and one output: ( ) system

Why working at higher frequencies?

6. Hamilton s Equations

Work is the change in energy of a system (neglecting heat transfer). To examine what could

A Tale of Friction Basic Rollercoaster Physics. Fahrenheit Rollercoaster, Hershey, PA max height = 121 ft max speed = 58 mph

Week3, Chapter 4. Position and Displacement. Motion in Two Dimensions. Instantaneous Velocity. Average Velocity

ORDINARY DIFFERENTIAL EQUATIONS EULER S METHOD

Single Variable Optimization

Descriptor and nonlinear eigenvalue problems in the analysis of large electrical power systems. Nelson Martins Sergio Gomes Jr. Sergio Luis Varricchio

PHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014

Advanced Topics in Optimization. Piecewise Linear Approximation of a Nonlinear Function

EE 221 Practice Problems for the Final Exam

55:041 Electronic Circuits

Model Reference Adaptive Temperature Control of the Electromagnetic Oven Process in Manufacturing Process

Electrical Circuits II (ECE233b)

The Fourier Transform

Physics 2A Chapter 3 HW Solutions

2 Simulation exercise 2

Lecture 2 Solution of Nonlinear Equations ( Root Finding Problems )

PHYS 705: Classical Mechanics. Calculus of Variations II

Electrical Circuits 2.1 INTRODUCTION CHAPTER

PHYS 1101 Practice problem set 12, Chapter 32: 21, 22, 24, 57, 61, 83 Chapter 33: 7, 12, 32, 38, 44, 49, 76

Math1110 (Spring 2009) Prelim 3 - Solutions

PHYSICS - CLUTCH 1E CH 28: INDUCTION AND INDUCTANCE.

Electrical Engineering Department Network Lab.

Complex Variables. Chapter 18 Integration in the Complex Plane. March 12, 2013 Lecturer: Shih-Yuan Chen

Design of Recursive Digital Filters IIR

UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:

Diode. Current HmAL Voltage HVL Simplified equivalent circuit. V γ. Reverse bias. Forward bias. Designation: Symbol:

6.3.7 Example with Runga Kutta 4 th order method

Solutions to Problem Set 6

Practical Newton s Method

II. PASSIVE FILTERS. H(j ω) Pass. Stop

Physics 1202: Lecture 11 Today s Agenda

n α j x j = 0 j=1 has a nontrivial solution. Here A is the n k matrix whose jth column is the vector for all t j=0

Homework 9 Solutions. 1. (Exercises from the book, 6 th edition, 6.6, 1-3.) Determine the number of distinct orderings of the letters given:

Lecture # 02: Pressure measurements and Measurement Uncertainties

Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 3A

Name: PHYS 110 Dr. McGovern Spring 2018 Exam 1. Multiple Choice: Circle the answer that best evaluates the statement or completes the statement.

: Numerical Analysis Topic 2: Solution of Nonlinear Equations Lectures 5-11:

( ) = ( ) + ( 0) ) ( )

EE C245 ME C218 Introduction to MEMS Design

G4023 Mid-Term Exam #1 Solutions

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

PHYSICS - CLUTCH CH 28: INDUCTION AND INDUCTANCE.

PID Controller Design Based on Second Order Model Approximation by Using Stability Boundary Locus Fitting

Graphical Analysis of a BJT Amplifier

Note 10. Modeling and Simulation of Dynamic Systems

Managing Capacity Through Reward Programs. on-line companion page. Byung-Do Kim Seoul National University College of Business Administration

Summary with Examples for Root finding Methods -Bisection -Newton Raphson -Secant

Chapter 5 rd Law of Thermodynamics

Lecture 3 Examples and Problems

PHYS 1441 Section 002 Lecture #15

Priority Queuing with Finite Buffer Size and Randomized Push-out Mechanism

Flyback Converter in DCM

Mathematics Intersection of Lines

Dr. Shalabh Department of Mathematics and Statistics Indian Institute of Technology Kanpur

Mathematical Economics MEMF e ME. Filomena Garcia. Topic 2 Calculus

THE VIBRATIONS OF MOLECULES II THE CARBON DIOXIDE MOLECULE Student Instructions

Consistency & Convergence

The Bellman Equation

Momentum. Momentum. Impulse. Momentum and Collisions

Transcription:

0/8/05 Lesson 6: Basc Control Modes ET 438a Automatc Control Systems Technology lesson6et438a.tx Learnng Objectves Ater ths resentaton you wll be able to: Descrbe the common control modes used n analog control systems Lst the characterstcs o common control modes Wrte the tme, Lalace and transer unctons o common control modes Identy the Bode lots o common control modes Desgn OP AMP crcuts that realze theoretcal control mode erormance. lesson6et438a.tx

Gan (db) 0/8/05 Control Modes-Proortonal Control Acton Process characterstcs or otmum results: ) Small rocess caactance ) Rad load changes Lmtatons: Small steady-state error may requre hgh gan to acheve accetable error levels Mathematcal reresentatons Tme uncton: v K e v Lalace uncton: (s) K E(s) Transer uncton: o (s) K E(s) Where: e = tme doman error sgnal K = roortonal gan v o = controller outut wth e=0 v = controller tme doman outut Note: Intal condton v o =0 on Lalace uncton lesson6et438a.tx 3 Proortonal Control Frequency Resonse Bode lots o three values o K : K =0., K =, and K 3 =00 60 K K 3 K 40 0 0 0 Note: gan s ndeendent o requency. Practcal realzaton: Non-nvertng OP AMP crcut 40 0 00 0 3 0 4 Frequency (rad/sec) K K K3 lesson6et438a.tx 4

0/8/05 Motor Seed Control Examle 6-: Determne the eect o alyng roortonal control to the block dagram shown below. The motor roduces the ollowng results wth the control loo oen: T L = 0.05 N-m T L = 0.075 N-m (50% ncrease n load) T = 9.4 dc T =9.4 I a =.033 A I a =.45 A w = 300 rad/sec w = 9.7 rad/sec r + e Controller K v Power T Motor & Suly G am Load w - c m Tach Generator, K tac lesson6et438a.tx 5 Motor Seed Control Motor Parameters: T = 0.0 N-m K T = 0.06 N-m/A K e = 0.06 -sec/rad R a =. ohms K tac = 0. -sec/rad K G am = 0 / r Soluton: Fnd the error roduced and the setont value, r. Then wrte equatons around control loo. + - e c m K G am =0 T = 9.4 K tac = 0. -s/rad e r c m T e K G w =300 rad/sec am c m K tac w 0.- s/rad (300 rad/s) 33 lesson6et438a.tx 6 3

0/8/05 Examle 6- Soluton () Combne equatons e r c m T e K G am T K G am r c m T K G am c m r Substtute values Smly Comute error Substtute values Smly 9.4 33 34.94 0/ r 34.94 e r c e 34.94 33.0 e.94 m The ntal setont value o r=34.94 wth an error o.94 at a seed o 300 rad/s and T L =0.05 N-m lesson6et438a.tx 7 Examle 6- Soluton (3) When torque ncrease to T L =0.075 N-m new T s dened by T (r c Substtute n Tachometer ormula T (r K m tac ) K G am w ) K G Snce setont, r does not change error must change due to measured seed change. am Motor equatons e b T I e a R K w m a e b Combne these equatons to get: T I a R a K e w Need two equatons to nd both T and w m. lesson6et438a.tx 8 4

0/8/05 Examle 6- Soluton (4) Substtute n know values and smly equaton to get rst relatonsh. r =34.94 K G am =0 K tac = 0. -s/rad T T T (r K tac w ) K G 34.94 0.w 349.4.w am 0 () Equaton Now use the motor armature crcut equaton and the armature current or T L =0.075 N-m to nd second ndeendent equaton. I a =.45 A K e = 0.06 -s/rad R+ =. ohms T T T I a R (.45 A) (. ) (0.06 -s/rad) w.74 0.06w a K w e () Equaton lesson6et438a.tx 9 Examle 6- Soluton (5) Place equatons () and ) nto standard orm and solve smultaneously usng sotware or calculator. T T T T 349.4.w.w.74 0.06w 0.06w 349.4.74 () () T 9.7 w 99.6 rad/sec Answers Now comute the error sgnal rom the new tachometer outut voltage c m. c c e m m K tac 0. s / rad 99.6 rad/sec (r c w m ) 34.94 3.956.968 3.956 Error ncreases e >e.968>.94 to rebalance system lesson6et438a.tx 0 5

0/8/05 Examle 6- Soluton (6) Now determne the ercentage seed changes or oen loo and eedback control. Setont, r=300 rad/sec Oen loo seed change wr w 00% %SE (SE seed error) wr 300 9.7 00% %SE 300.77% %SE Answers Feedback loo seed change 300 99.6 00% %SE 300 0.43% %SE Answers Feedback reduces seed error by actor o 9.35 lesson6et438a.tx Integral Control Mode Integral Mode characterstcs: ) Outut s ntegral o error over tme ) Drves steady-state error to zero 3) Adds ole to transer uncton at s=0 (nnte gan to constant) 4) Integrators tend to make systems less stable Equatons Tme: Lalace: v(t) K Transer Functon: I e(t) dt v 0 0 (s) KI E(s s (s) E(s) K I s Where K I = ntegral gan constant lesson6et438a.tx 6

0/8/05 OP AMP Realzatons o Integral Control Practcal OP AMP Integrator Ideal OP AMP Integrator Transer Functon o(s) (s) R C s One ole at s=0 KI R C Transer Functon o(s) R (s) R R Cs One ole at s=-/r C lesson6et438a.tx 3 Bode Plots o Integrator Crcuts Substtute jw or s and nd the magntude and hase sht o the transer uncton or derent values o w. Ideal Integrator o(jw) (jw) R C jw Practcal Integrator o(jw) R (jw) R R C jw Take magntude and hase sht o each o these unctons usng rules o comlex numbers. z=a+jb Magntude o z: z= z z a b Re(z) Im(z) Scale or db db 0log( ) Phase Sht tan b tan a Im(z) Re(z) lesson6et438a.tx 4 7

0/8/05 Bode Plots o Integrator Crcuts Practcal Integrator Crcut Takng magntude gves Phase Sht gves Ideal Integrator Crcut Magntude gves R R R C jw R R w) 80 tan R ( R db 0log( ) C w Cw Gjw db 0log( ) R C 80 degree hase sht s rom nvertng conguratons Phase sht j 90 j 0 80 90 90 Contant value lesson6et438a.tx 5 Integrator Bode Plots Usng MatLAB R R R C jw R C jw Use MatLAB scrt to generate Bode lots and transer uncton. Dene arameters: R = 0 k, R = 00 k, C = 0.0 mf MatLAB Scrt r=nut('enter value o nut resstance: '); c=nut('enter value o caactance: '); r=nut('enter value o eedback resstance: '); % comute transer uncton model arameters or % ractcal ntegrator Inut statement Comments begn wth % tau=r*c; k=-r./r; lesson6et438a.tx 6 8

Phase (deg) Magntude (db) 0/8/05 Integrator Bode Plots Usng MatLAB MatLAB Scrt (Contnued) % comute arameter or deal ntegrator tau = r*c; % buld transer unctons % denomnator orm s a*s^+as+a3 Av=t([k],[tau ]) Av=t([-],[tau 0]) %lot both on the same grahs bode(av,av); Create transer unctons Plot both grahs on same gure lesson6et438a.tx 7 Integrator Bode Plots Usng MatLAB 60 40 0 Integrator Bode Dagrams Practcal Ideal w=/r C 0-0 w=000 rad/s 80 35 90 0 0 0 3 0 4 0 5 Frequency (rad/sec) lesson6et438a.tx 8 9

0/8/05 Integral Acton on Tme aryng Error Sgnals Integral o constant, k, s lne wth sloe k. e e Integrator roduces a lnearly ncreasng outut or constant error nut -e 3 e 4 =0 (t)=-e 3 t Negatve error causes decreasng outut (t)=e t Zero error mantans last outut value (t)=e t lesson6et438a.tx 9 Estmatng Integrator Outut From Calculus, ntegral s sum o area below a uncton lot b a (t) dt (t) n 0 (t ) t Where t t (t + ) b a t n a t b t For lnear error lots, ntegral s the sum o the areas o lnear segments. Use trangle, traezod, and rectangle ormulas to aroxmate outut lesson6et438a.tx 0 0

Error Inut 0/8/05 Integrator Outut Examle 6-: An deal ntegrator has a gan o K I =0. /s. Its ntal outut s v=.5. Determne the ntegrator oututs the error has ste ncreases gven by the table below. Error Magntude () Tme Interval e(t)=0 0 t seconds e(t)=.5 <t seconds e(t)=4 <t 3 seconds e(t)=0 3<t 4 seconds e(t)=-.5 4<t 5 seconds lesson6et438a.tx Examle 6- Soluton () Plot the error uncton that s nut to the ntegrator 5 Integrator Error Inut 4 3 0 - - 0 3 4 5 6 Tme (seconds) lesson6et438a.tx

Error Inut Integrator Outut oltage 0/8/05 Examle 6- Soluton () Use the aroxmate ormula to nd the error at the end o each nterval 5 4 3 0 - v K A A I n 0 Integrator Error Inut A 0 e t - 0 3 4 5 6 Tme (seconds) A 4 A 3 Where t n 5 0 =.5 A 0 =K I t e 0 =0.()(0) =0 =.5+A 0 =.5+0=.5 A =K I t e =0.()(.5) =0.5 =.5+A =.5+0.5=.75 A =K I t e =0.()(4) =0.40 3 =.75+A =.75+0.40=.5 A 3 =K I t e 3 =0.()(0) =0 4 =.5+A 3 =.5+0.0=.5 A 4 =K I t e 4 =0.()(-.5) =-0.5 5 =.5+A 4 =.5+-0.5=.00 lesson6et438a.tx 3 Examle 6- Soluton () Integrator outut lot 3 Integrator Outut.5 X:.0 Y:.75.5 0.5 0 0 3 4 5 6 Tme (seconds) lesson6et438a.tx 4

0/8/05 Dervatve Control Mode Dervatve Control Characterstcs: ) Produces outut only when error s changng ) Outut s roortonal to rate o change n error 3) Dervatve control never used alone 4) Used wth roortonal and/or ntegral modes 5) Antcates error by observng the rate o change lesson6et438a.tx 5 Dervatve Mode Equatons Tme Equaton: v(t) K d de(t) dt Lalace Equaton: (s) Kd se(s) Transer Functon Equaton: (s) K E(s) d s Derentators are hgh-ass lters to snusodal sgnals. They ncrease senstvty to rad error changes when added to controllers. lesson6et438a.tx 6 3

0/8/05 OP AMP Realzatons o Derentators Ideal OP AMP Derentator Practcal OP AMP Derentator Transer Functon o(s) R Cs (s) Introduces one zero at s=0 Transer Functon o(s) R Cs (s) R Cs Introduces: zero at s=0 ole at s=-/r C lesson6et438a.tx 7 Bode Plots o Derentators Ideal Derentator Equatons o(s) R (s) R db 0log 90 Cw C jw Constant over all w Practcal Derentator Equatons o(jw) R C jw (jw) R C jw R Cw R db 0log[ ] - 70 - tan (R Cw) C w Use MatLAB scrt to generate Bode lots and transer uncton. Dene arameters: R = 0 k, R = 00 k, C = 0.0 mf lesson6et438a.tx 8 4

Magntude (db) Phase (deg) 0/8/05 Bode Plots o Derentators -00 Bode Dagram Derentator Frequency Resonse -0-40 Ideal Practcal -60-80 w=/r C -00 35 70 5 80 35 90 0-6 0-5 0-4 0-3 0 - Frequency (rad/s) lesson6et438a.tx 9 ET 438a Automatc Control Systems Technology END LESSON 6: BASIC CONTROL MODES lesson6et438a.tx 30 5