Control System Engineering

Similar documents
Numeric Matlab for Laplace Transforms

MAT 275 Laboratory 7 Laplace Transform and the Symbolic Math Toolbox

The Laplace Transform

EE -213 BASIC CIRCUIT ANALYSIS LAB MANUAL

Linear System Theory

OKLAHOMA STATE UNIVERSITY

ECE 203 LAB 1 MATLAB CONTROLS AND SIMULINK

MATHEMATICAL MODELING OF CONTROL SYSTEMS

SIGNALS AND LINEAR SYSTEMS LABORATORY EELE Experiment (2) Introduction to MATLAB - Part (2) Prepared by:

The Laplace Transform

Experiment 2: Introduction to MATLAB II

Tutorial 4 (Week 11): Matlab - Digital Control Systems

Matlab Controller Design. 1. Control system toolbox 2. Functions for model analysis 3. Linear system simulation 4. Biochemical reactor linearization

(a) Torsional spring-mass system. (b) Spring element.

University of Alberta ENGM 541: Modeling and Simulation of Engineering Systems Laboratory #7. M.G. Lipsett & M. Mashkournia 2011

Lecture 14 - Using the MATLAB Control System Toolbox and Simulink Friday, February 8, 2013

EE 4314 Lab 1 Handout Control Systems Simulation with MATLAB and SIMULINK Spring Lab Information

EE -213 BASIC CIRCUIT ANALYSIS LAB MANUAL

INTRODUCTION TO TRANSFER FUNCTIONS

LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593

INFE 5201 SIGNALS AND SYSTEMS

ECE 3793 Matlab Project 3 Solution

Appendix 3B MATLAB Functions for Modeling and Time-domain analysis

Course Background. Loosely speaking, control is the process of getting something to do what you want it to do (or not do, as the case may be).

ECE 3793 Matlab Project 3

INC 341 Feedback Control Systems: Lecture 2 Transfer Function of Dynamic Systems I Asst. Prof. Dr.-Ing. Sudchai Boonto

Solution to Homework Assignment 1

Representing Polynomials

University of Toronto Department of Electrical and Computer Engineering ECE410F Control Systems Problem Set #3 Solutions = Q o = CA.

Chapter 6 Steady-State Analysis of Continuous-Time Systems

Bangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory

Control System. Contents

LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693

APPLICATIONS FOR ROBOTICS

Time Response Analysis (Part II)

S I X SOLUTIONS TO CASE STUDIES CHALLENGES. Antenna Control: Stability Design via Gain K s s s+76.39K. T(s) =

ChE 400: Applied Chemical Engineering Calculations. Tutorial 5: Use of Matlab to Calculate Laplace Transforms

ENGG4420 LECTURE 7. CHAPTER 1 BY RADU MURESAN Page 1. September :29 PM

16.31 Homework 2 Solution

Outline. Classical Control. Lecture 2

ECE317 : Feedback and Control

Poles and Zeros and Transfer Functions

Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax:

a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a

SIGNALS AND SYSTEMS LABORATORY 4: Polynomials, Laplace Transforms and Analog Filters in MATLAB

Advanced Control Theory

Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system

Power System Control

butter butter Purpose Syntax Description Digital Domain Analog Domain

Dr. Ian R. Manchester

Stability and Frequency Response of Linear Systems

EE Experiment 11 The Laplace Transform and Control System Characteristics

Chapter 7: Time Domain Analysis

Digital Control System Models. M. Sami Fadali Professor of Electrical Engineering University of Nevada

Formulation of Linear Constant Coefficient ODEs

Chap. 3 Laplace Transforms and Applications

Q 1.5: 1/s Sol: If you try to do a Taylor expansion at s = 0, the first term, w(0). Thus, the Taylor series expansion in s does not exist.

3. Array and Matrix Operations

Multivariable Control. Lecture 05. Multivariable Poles and Zeros. John T. Wen. September 14, 2006

APPENDIX 1 MATLAB AND ANSYS PROGRAMS

MAE143A Signals & Systems - Homework 5, Winter 2013 due by the end of class Tuesday February 12, 2013.

Solution of ODEs using Laplace Transforms. Process Dynamics and Control

Alireza Mousavi Brunel University

ECE382/ME482 Spring 2005 Homework 1 Solution February 10,

Introduction to Root Locus. What is root locus?

Exercise 1a: Transfer functions

Department of Mechanical Engineering

Digital Signal Processing: Signal Transforms

7.1 Introduction. Apago PDF Enhancer. Definition and Test Inputs. 340 Chapter 7 Steady-State Errors

CHAPTER 2 TRANSFER FUNCTION ANALYSIS

EEL2216 Control Theory CT1: PID Controller Design

Solving Differential Equations Using MATLAB

Time Response of Systems

SOLUTIONS TO CASE STUDIES CHALLENGES

UNIVERSITI MALAYSIA PERLIS

Laplace Transforms Chapter 3

3.5 Undetermined Coefficients

Control Systems Engineering (Chapter 2. Modeling in the Frequency Domain) Prof. Kwang-Chun Ho Tel: Fax:

18.03 Class 23, Apr 2. Laplace Transform: Second order equations; completing the square; t-shift; step and delta signals. Rules:

These videos and handouts are supplemental documents of paper X. Li, Z. Huang. An Inverted Classroom Approach to Educate MATLAB in Chemical Process

CHEE 319 Tutorial 3 Solutions. 1. Using partial fraction expansions, find the causal function f whose Laplace transform. F (s) F (s) = C 1 s + C 2

37. f(t) sin 2t cos 2t 38. f(t) cos 2 t. 39. f(t) sin(4t 5) 40.

Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.

Some of the different forms of a signal, obtained by transformations, are shown in the figure. jwt e z. jwt z e

Massachusetts Institute of Technology Department of Mechanical Engineering Dynamics and Control II Design Project

Signals and Systems Laboratory with MATLAB

Outline. Classical Control. Lecture 5

Control Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich

A Note on Bode Plot Asymptotes based on Transfer Function Coefficients

SECTION 5: ROOT LOCUS ANALYSIS

DIGITAL CONTROL OF POWER CONVERTERS. 3 Digital controller design

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27

MATH 2050 Assignment 8 Fall [10] 1. Find the determinant by reducing to triangular form for the following matrices.

Math Exam 3 Solutions

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

Third In-Class Exam Solutions Math 246, Professor David Levermore Thursday, 3 December 2009 (1) [6] Given that 2 is an eigenvalue of the matrix

Universiti Malaysia Perlis EKT430: DIGITAL SIGNAL PROCESSING LAB ASSIGNMENT 5: DIGITAL FILTER STRUCTURES

MAE 143B - Homework 9

It is common to think and write in time domain. creating the mathematical description of the. Continuous systems- using Laplace or s-

Lab Experiment 2: Performance of First order and second order systems

Transcription:

Control System Engineering Matlab Exmaples Youngshik Kim, Ph.D. Mechanical Engineering youngshik@hanbat.ac.kr

Partial Fraction: residue >> help residue RESIDUE Partial-fraction expansion (residues). [R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of a partial fraction expansion of the ratio of two polynomials B(s)/A(s). If there are no multiple roots, B(s) R(1) R(2) R(n) ---- = -------- + -------- +... + -------- + K(s) A(s) s - P(1) s - P(2) s - P(n) Vectors B and A specify the coefficients of the numerator and denominator polynomials in descending powers of s. The residues are returned in the column vector R, the pole locations in column vector P, and the direct terms in row vector K.

Partial Fraction: residue Example: ( s 2)( s 4) 1/6 3 / 2 8 / 3 Y() s 2 ( s s)( s 3) s 3 s 1 s -0.1667-1.5000 2.6667 s 3 s 1 s num = conv ([1 2],[1 4]); % form numerator polynomial den = conv ([1 1 0],[1 3]); % form denominator polynomial [r,p,k] = residue (num,den) % compute the residues [ r : residues, p : poles, k : direct terms ] r = [-0.1667-1.5000 2.6667] ; p = [-3-1 0] ; k=[];

Partial Fraction: residue Example Y() s 2 1/3 1 2/3 ( s 2)( s 1)( s 4) s 4 s 2 s 1 num = 2; % form numerator den = poly([-2;-1;-4]); % form denominator polynomial from its roots [r, p, k] = residue(num, den) % compute the residues r 0.3333 1 0.6667 p 4 2 1 k

Transfer Function: tf >> help tf TF Constructs transfer function or converts to transfer function. Construction: SYS = TF(NUM,DEN) creates a continuous-time transfer function SYS with numerator NUM and denominator DEN. SYS is an object of class @tf. SYS = TF(NUM,DEN,TS) creates a discrete-time transfer function with sampling time TS (set TS=-1 if the sampling time is undetermined). >> sys = tf([1 2],[1 0 10]) Transfer function: s + 2 -------- s^2 + 10

Symbolic Objects: syms >> help syms SYMS Short-cut for constructing symbolic objects. SYMS arg1 arg2... is short-hand notation for arg1 = sym('arg1'); arg2 = sym('arg2');... Each input argument must begin with a letter and must contain only alphanumeric characters. By itself, SYMS lists the symbolic objects in the workspace. Examples: >> syms x beta real

Laplace Transforms: laplace >> help laplace --- help for sym/laplace --- LAPLACE Laplace transform. L = LAPLACE(F) is the Laplace transform of the scalar sym F with default independent variable t. The default return is a function of s. If F = F(s), then LAPLACE returns a function of t: L = L(t). By definition L(s) = int(f(t)*exp(-s*t),0,inf), where integration occurs with respect to t. Examples: syms a s t w x laplace(t^5) returns 120/s^6 laplace(exp(a*s)) returns 1/(t-a) laplace(sin(w*t)) returns w/(s^2 + w^2) laplace(diff(sym('f(t)'))) returns laplace(f(t),t,s)*s-f(0)

Inverse Laplace Transform: ilaplace >> help ilaplace --- help for sym/ilaplace --- ILAPLACE Inverse Laplace transform. F = ILAPLACE(L) is the inverse Laplace transform of the scalar sym L with default independent variable s. The default return is a function of t. If L = L(t), then ILAPLACE returns a function of x: F = F(x). Examples: syms s t w x y ilaplace(1/(s-1)) returns exp(t) ilaplace(1/(t^2+1)) returns sin(x)

Laplace Transforms Example >> syms t; f = t^4; F=laplace(f) F = 24/s^5 >> ilaplace(f) ans = t^4

Zeros, Poles, Gains: tf2zp >> help tf2zp TF2ZP Transfer function to zero-pole conversion. [Z,P,K] = TF2ZP(NUM,DEN) finds the zeros, poles, and gains: (s-z1)(s-z2)...(s-zn) H(s) = K --------------------- (s-p1)(s-p2)...(s-pn) from a SIMO transfer function in polynomial form: NUM(s) H(s) = -------- DEN(s) Vector DEN specifies the coefficients of the denominator in descending powers of s. Matrix NUM indicates the numerator coefficients with as many rows as there are outputs.

Zeros, Poles, Gains: tf2zp Example: Y() s 3s 9 y 6y 25y 9u 3u 2 U() s s 6s 25 numg = [3 9]; deng = [1 6 25]; [z, p, k] = tf2zp(numg, deng) [numg, deng] = zp2tf(z, p, k) z 3 3 4 3 4 k numg 0 3 9 deng 1 6 25 p j j 3

Step Response: step >> help step STEP Step response of dynamic systems. STEP(SYS) plots the step response of the dynamic system SYS. For multi-input models, independent step commands are applied to each input channel. The time range and number of points are chosen automatically. STEP(SYS,TFINAL) simulates the step response from t=0 to the final time t=tfinal. For discrete-time models with unspecified sampling time, TFINAL is interpreted as the number of samples. STEP(SYS,T) uses the user-supplied time vector T for simulation. For discrete-time models, T should be of the form Ti:Ts:Tf where Ts is the sample time. For continuous-time models, T should be of the form Ti:dt:Tf where dt will become the sample time for the discrete approximation to the continuous system. The step input is always assumed to start at t=0 (regardless of Ti).

Step Response: step DC motor: s () s 100s Gs () V s s s s m 3 2 a ( ) 10.1 101 s 2 100 10.1s 101 numb = [0 0 100]; denb = [1 10.1 101]; sysb = tf(numb, denb); t = 0 : 0.01 : 5; y = step(sysb, t); plot(t, y) Figure 3.5 Transient response for DC motor