Chapter 7. Root Locus Analysis

Similar documents
Lecture 12: Examples of Root Locus Plots. Dr. Kalyana Veluvolu. Lecture 12: Examples of Root Locus Plots Dr. Kalyana Veluvolu

CONTROL SYSTEMS. Chapter 5 : Root Locus Diagram. GATE Objective & Numerical Type Solutions. The transfer function of a closed loop system is

ME2142/ME2142E Feedback Control Systems

Root Locus Diagram. Root loci: The portion of root locus when k assume positive values: that is 0

MODERN CONTROL SYSTEMS

CISE302: Linear Control Systems

Automatic Control Systems. Part III: Root Locus Technique

Control Systems Analysis and Design by the Root-Locus Method

Root Locus Contents. Root locus, sketching algorithm. Root locus, examples. Root locus, proofs. Root locus, control examples

ROOT LOCUS. Poles and Zeros

MEM 355 Performance Enhancement of Dynamical Systems Root Locus Analysis

The Root Locus Method

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Chapter #4 EEE Automatic Control

Figure 1: Unity Feedback System

ME 375 FINAL EXAM SOLUTIONS Friday December 17, 2004

EE Control Systems LECTURE 14

Control Systems. Root locus.

Control Systems. Root locus.

Feedback Control Systems (FCS)

Chapter 13. Root Locus Introduction

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax:

Analysis of Stability &

NAME (pinyin/italian)... MATRICULATION NUMBER... SIGNATURE

CHAPTER 4 DESIGN OF STATE FEEDBACK CONTROLLERS AND STATE OBSERVERS USING REDUCED ORDER MODEL

ME 375 FINAL EXAM Wednesday, May 6, 2009

Stability. ME 344/144L Prof. R.G. Longoria Dynamic Systems and Controls/Lab. Department of Mechanical Engineering The University of Texas at Austin

CHAPTER 8 OBSERVER BASED REDUCED ORDER CONTROLLER DESIGN FOR LARGE SCALE LINEAR DISCRETE-TIME CONTROL SYSTEMS

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

Advanced D-Partitioning Analysis and its Comparison with the Kharitonov s Theorem Assessment

The state variable description of an LTI system is given by 3 1O. Statement for Linked Answer Questions 3 and 4 :

Chapter 7 : Root Locus Technique

Digital Control System

Linear System Fundamentals

Stability Criterion Routh Hurwitz

Figure 1 Siemens PSSE Web Site

Function and Impulse Response

Gain and Phase Margins Based Delay Dependent Stability Analysis of Two- Area LFC System with Communication Delays

Lecture 4. Chapter 11 Nise. Controller Design via Frequency Response. G. Hovland 2004

G(s) = 1 s by hand for! = 1, 2, 5, 10, 20, 50, and 100 rad/sec.

Lecture 10 Filtering: Applied Concepts

Introduction to Laplace Transform Techniques in Circuit Analysis

6.447 rad/sec and ln (% OS /100) tan Thus pc. the testing point is s 3.33 j5.519

SKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot

EE Control Systems LECTURE 6

State Space: Observer Design Lecture 11

Given the following circuit with unknown initial capacitor voltage v(0): X(s) Immediately, we know that the transfer function H(s) is

Module 4: Time Response of discrete time systems Lecture Note 1

Root locus Analysis. P.S. Gandhi Mechanical Engineering IIT Bombay. Acknowledgements: Mr Chaitanya, SYSCON 07

Lecture Notes II. As the reactor is well-mixed, the outlet stream concentration and temperature are identical with those in the tank.

into a discrete time function. Recall that the table of Laplace/z-transforms is constructed by (i) selecting to get

EE 4443/5329. LAB 3: Control of Industrial Systems. Simulation and Hardware Control (PID Design) The Inverted Pendulum. (ECP Systems-Model: 505)

Digital Control System

March 18, 2014 Academic Year 2013/14

Automatic Control Systems, 9th Edition

Chapter 8. Root Locus Techniques

CHAPTER # 9 ROOT LOCUS ANALYSES

EE C128 / ME C134 Problem Set 1 Solution (Fall 2010) Wenjie Chen and Jansen Sheng, UC Berkeley

Lecture #9 Continuous time filter

S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS

Lecture 12 - Non-isolated DC-DC Buck Converter

Homework 7 Solution - AME 30315, Spring s 2 + 2s (s 2 + 2s + 4)(s + 20)

MM1: Basic Concept (I): System and its Variables

Analysis and Design of a Third Order Phase-Lock Loop

Massachusetts Institute of Technology Dynamics and Control II

ECE-320 Linear Control Systems. Spring 2014, Exam 1. No calculators or computers allowed, you may leave your answers as fractions.

Chemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University

Homework 12 Solution - AME30315, Spring 2013

Solutions. Digital Control Systems ( ) 120 minutes examination time + 15 minutes reading time at the beginning of the exam

CONTROL OF INTEGRATING PROCESS WITH DEAD TIME USING AUTO-TUNING APPROACH

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

ROUTH HURWITZ ANALYSIS

EXTENDED STABILITY MARGINS ON CONTROLLER DESIGN FOR NONLINEAR INPUT DELAY SYSTEMS. Otto J. Roesch, Hubert Roth, Asif Iqbal

Mechanics. Free rotational oscillations. LD Physics Leaflets P Measuring with a hand-held stop-clock. Oscillations Torsion pendulum

Design of Digital Filters

Department of Mechanical Engineering Massachusetts Institute of Technology Modeling, Dynamics and Control III Spring 2002

Lecture 8. PID control. Industrial process control ( today) PID control. Insights about PID actions

1 Routh Array: 15 points

CALIFORNIA INSTITUTE OF TECHNOLOGY Control and Dynamical Systems

SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuits II. R 4 := 100 kohm

Question 1 Equivalent Circuits

A PLC BASED MIMO PID CONTROLLER FOR MULTIVARIABLE INDUSTRIAL PROCESSES

ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances

Evolutionary Algorithms Based Fixed Order Robust Controller Design and Robustness Performance Analysis

HOMEWORK ASSIGNMENT #2

Sliding Mode Control of a Dual-Fuel System Internal Combustion Engine

Stability regions in controller parameter space of DC motor speed control system with communication delays

Root Locus Methods. The root locus procedure

Chapter 9: Controller design. Controller design. Controller design

A Simple Approach to Synthesizing Naïve Quantized Control for Reference Tracking

Root Locus. Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering

General Topology of a single stage microwave amplifier

Then C pid (s) S h -stabilizes G(s) if and only if Ĉpid(ŝ) S 0 - stabilizes Ĝ(ŝ). For any ρ R +, an RCF of Ĉ pid (ŝ) is given by

A Simplified Methodology for the Synthesis of Adaptive Flight Control Systems

/University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2009

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

Singular Value Analysis of Linear- Quadratic Systems!

376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with :

An estimation approach for autotuning of event-based PI control systems

ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8

Transcription:

Chapter 7 Root Locu Analyi jw + KGH ( ) GH ( ) - K 0 z O 4 p 2 p 3 p

Root Locu Analyi The root of the cloed-loop characteritic equation define the ytem characteritic repone. Their location in the complex -plane lead to prediction of the characteritic of the time domain repone in term of: damping ratio, z natural frequency, w n econd-order mode damping contant, firt-order mode Conider how thee root change a the loop gain i varied from 0 to. Frequency Repone Deign

Root Locu Example R() + E() K ( + 2) C() The cloed-loop tranfer function i The characteritic equation i 2 + + K C( ) K 2 0 R( ) ( + 2) + K Conider the characteritic root a K 0. Frequency Repone Deign

Root Locu Example -± - K For K 0 the cloed-loop pole are at the open-loop pole. For 0 K the cloed-loop pole are on the real axi. For K the cloed-loop pole are complex, with a real value of and an imaginary value increaing with gain K. loci of cloed-loop root K K 2 jw K 0 Frequency Repone Deign

Amplitude Root Locu Example: Step Repone.6 K 50.0 Step Repone K jw.4 K 5.0.2 0.8 0.6 0.4 0.2 K 2.0 K.0 K 0.5 K 2 K 0 0 0 2 4 6 8 0 Time (ec.) Frequency Repone Deign

Root Locu Example: Some Obervation Thi i a econd-order ytem and there two loci. The root loci tart at the open-loop pole. The root loci tend toward the open-loop zero at infinity a K. (Note: the number of zero i equal to the number of pole, when the zero at infinity are included.) The relationhip between the characteritic repone and the increaing gain i een through the root loci. Frequency Repone Deign

The General Root Locu Method Conider the general ytem R() C() G() + where C( ) R( ) H() G( ) + GH ( ) The characteritic equation i + GH( ) 0 or GH( ) - or GH( ) GH( ) (2k k + ) p 0, ±, ± 2 L Frequency Repone Deign

The General Root Locu Method All value of which atify ; ; GH ( ) GH( ) (2k + ) p are root of the cloed-loop characteritic equation. Conider the following general form GH ( ) Note : pi K( + z)( + z2) L ( + z ( + p )( + p ) L ( + p may be zero. 2 m n ) ) k 0, ±, ± 2 L Frequency Repone Deign

Frequency Repone Deign Then The General Root Locu Method i ) ( + + p z K GH n i m i i 2 L,, 0 ± ± k ) (2 ) ( ) ( ) ( + + - + k p z GH n i i m i i p

Root Locu Method: Geometric Interpretation Conider the example K( + z) GH ( ) ( + p )( + p Then the value of which atify K + + p 2 + z + 3 2 p 3 ) z ) - ( + ( + p2) + ( + p )) (2k + )p ( 3 are on the loci and are root of the characteritic equation. q p3 O -z -p 3 A -p 2 q z B D q p2 C jw q p -p Frequency Repone Deign

Root Locu Method: Geometric Interpretation In term of the vector, the condition for to be on the root loci are K A A or BCD BCD K and q p3 A O -z -p 3 q z B D C jw q p -p q k z - ( q + q + q 3) p 0, ±, ± p2 2, L p (2k + ) p -p 2 q p2 Frequency Repone Deign

Root Locu Method When plotting the loci of the root a K 0, the magnitude condition i alway atified. Therefore, a value of that atifie the angle condition, i a point of the root loci. The magnitude condition may then be ued to determine the gain K correponding to that value. How can we eaily determine if the angle condition i atified? Frequency Repone Deign

Root Locu Contruction Rule. The loci tart (K 0) at the pole of the open-loop ytem. There are n loci. 2. The loci terminate (K ) at the zeroe of the open-loop ytem (include zeroe at infinity). For our example ytem + z GH( ) + p + p Therefore, a K 0, GH(), the pole of the loop tranfer function. A K, GH() 0, the zeroe of the loop tranfer function. 2 3 K Frequency Repone Deign

Root Locu Contruction Rule 3. The root loci are ymmetrical about the real axi. 4. A K the loci approach aymptote. There are q n m aymptote and they interect the real axi at angle defined by + p (2k ) q, k 0, ±, ± 2, L The root with imaginary part alway occur in conjugate complex pair. When the loci approach infinity, the angle from all the pole and zeroe are equal. The angle condition then i mq nq (2k + )p Frequency Repone Deign

Root Locu Contruction Rule 5. The aymptote interection point on the real i defined by a poleof GH( ) - 6. Real axi ection of the root loci exit only where there i an odd number of pole and zeroe to the right. q zeroe of GH( ) The angle from pole and zeroe to the left of are zero. The angle from pole and zeroe to the right are p. An odd number are required to atify the angle condition. Frequency Repone Deign

Root Locu Contruction Rule Example Conider our example with z 4, p 2 2j jw aymptote GH ( ) ( Aymptote: a angle K( + 4) + + 2 j)( + - 2 j) [- - + - - ]-[ - ] 0 (2k + ) p p ± 3-2 ( 2 j) ( 3-2 j) ( 4) real axi locu O 4 + 2 2j + 2j Frequency Repone Deign

Root Locu Contruction Rule 7. The angle of departure, q d from pole and arrival, q a to zeroe may be found by applying the angle condition to a point very near the pole or zero. The angle of arrival at the zero, -z i obtained from q + az ( -z + z ) i 2 i - n ( -z + p ) (2k + i )p i m Frequency Repone Deign

Root Locu Contruction Rule Example Departure angle from p 2. q z tan - (2/3) 33.7 q p tan - ( 2/) 6.6 q p3 90 Then -z O 4 33.7 (90 + 6.6 + q p2 ) 80 q p2 352.9 + 7. 33.7 2 q p2 -p 2 -p 3 90 jw 2j 6.6 -p 2j + Frequency Repone Deign

Root Locu Contruction Rule 8. The imaginary axi croing i obtained by applying the Routh- Hurwitz criterion and checking for the gain that reult in marginal tability. The imaginary component are found from the olution of the reulting auxiliary equation. Marginal tability refer to the point where the root of the cloed-loop ytem are on the tability boundary, i.e. the imaginary axi. Frequency Repone Deign

Root Locu Contruction Rule Example Imaginary axi croing: ( Characteritic equation + + 2 2 j)( (5 + + - K) 2 j) 3 2 + + + Routh table 3 5+K 0 2 2 4K 0 5 K 0 0 0 4K 0 4K + K( + 0 4) 0 For marginal tability, K 5 and the auxiliary equation i Therefore, the imaginary axi interection i ± 3.6 j 2 2 + 20 0 ± 0 j ± 3.6 Frequency Repone Deign j

Root Locu Contruction Rule Example Summary: There are three root loci. One on the real axi from -p to -z, and a pair of loci from -p 2 and -p 3 to zeroe at infinity along the aymptote. The departure angle from thee pole i 7. and an imaginary axi croing at 3.6j. -z O 4 2 7. -p 2 -p 3 jw 3.6 j 2j -p 2j + Frequency Repone Deign

Root Locu Contruction Rule Breakaway Point: Some example When two or more loci meet, they will breakaway from thi point at particular angle. The point i known a a breakaway point. It correpond to multiple root. x x x x x x o x x x o x 45 Frequency Repone Deign

Root Locu Contruction Rule 9. The angle of breakaway i 80 /k where k i the number of converging loci. The location of the breakaway point i found from dk d [ ] d GH( ) 0 or d 0 Note: K -2 [ ] d[ GH ( ) ] 0 dk GH ( ) d Alo, - [ ] GH ( ) d D( ) N( ) - N( ) D( ) - [ ] [ ] d GH( ) d N ( ) - D( ) d N( ) D( ) d N( ) D( ) 2 0 D( ) 0 Frequency Repone Deign

Root Locu Plot: Breakaway Point Example Conider the following loop tranfer function. ( ) K GH 2 ( + 3) Real axi loci exit for the full negative axi. Aymptote: angle (2k+)p p/3, p, 5p/3 a 3 (-3-3 - 0) - (0) 3-2 4 2 60 jw 2j + 2j Frequency Repone Deign aymptote

Root Locu Plot: Breakaway Point Example Determine the breakaway point from jw d d K ( + 3) 2 - K(3 + 2 + 9) 3 2 2 ( + 6 + 9) then 2 d d 3 + 0 K 6 2 + 9 4 2 2j 2j + 2 + 4 + 3 -, - 3 ( + )( + 3) 0 Frequency Repone Deign

Root Locu Contruction Rule 0. For a point on the root locu, calculate the gain, K from K + + p z + + p z 2 2 L L Alternately, K may be determined graphically from the root locu plot O A B D C jw K BCD A Frequency Repone Deign

Summary of Root Locu Plot Contruction Plot the pole and zero of the open-loop ytem. Find the ection of the loci on the real axi (odd number of pole an zeroe to the right). Determine the aymptote angle and intercept. + p (2k ) angle, q a pole- q q zeroe n - m, k 0, ±, ± 2, L Frequency Repone Deign

Summary of Root Locu Plot Contruction Determine departure angle. For a pole -p -p+ z ) +( -p + z ) + L-q -(-p + p ) - p L (2k )p ( + 2 2 Check for imaginary axi croing uing the Routh-Hurwitz criterion. Determine breakaway point. angle p / k location from Complete the plot., k d # of [ ] GH( ) d converging 0 loci Frequency Repone Deign

Root Locu Plot Example 3 Loop Tranfer function: ( K GH ) ( + 4)( 2 + 4 + Root: 0, 4, 2 4j Real axi egment: between 0 and 4. Aymptote: angle a 20) (2k + ) p p 3p 5p 7p,,, 4-0 4 4 4 4 (-4-2 -2-0) -2 4 aymptote 4 2 jw 45 4j 2j 2j 4j Frequency Repone Deign +

Root Locu Plot Example 3 Breakaway point: jw 4j d K 4 3 2 d + 8 + 36 + 80 3 2 K(4 + 24 + 72 + - 80) 4 3 2 2 ( + 8 + 36 + 80) 3 2 or + 6 + 8 + 20 0 olving, b -2, - 2 ± 2. 45 j 0 4 2 2j 45 2j + Three point that breakaway at 90. 4j Frequency Repone Deign

Root Locu Plot Example 3 The imaginary axi croing: Characteritic equation 3 2 + 8 + 36 + 80 + K Routh table 4 4 36 K 3 8 80 0 2 26 K 0 80-8K/26 0 0 0 K 0 0 0 Condition for critical tability 80-8K/26 > 0 or K<260 The auxiliary equation 26 2 + 260 0 olving ± 0 j ± 3. 6 Frequency Repone Deign j

Root Locu Plot Example 3 The final plot i hown on the right. jw 4j 3.6j 2j What i the value of the gain K correponding to the breakaway point at b 2 ± 2.45j? 4 2 2j 4j Frequency Repone Deign

Root Locu Plot Example 3 Gain Calculation From the general magnitude condition the gain correponding to the point on the loci i K n i i For the point 2 + 2.45j + p i K 2 + 2.45j 2 + 2.45j + 4 2 + 2.45j + 2 + 4j 2 + 2.45j + 2 4j /.0 3.63 3.63 6.45.55 00.0 m + z i Frequency Repone Deign

Root Locu Plot Example 3 I there a gain correponding to a damping ratio of 0.707 or more for all ytem mode? z 0.707 co(q ) q 45 4 2 45 jw 4j 3.6j 2j 2j 4j Frequency Repone Deign

Root Locu Plot Example 3 Time Repone 4 2 K260 K64 K00 jw 4j 3.6j 2j 2j 4j Examine the repone for the variou gain and relate them to the location of the cloed-loop root. K 64, root are 2, 2, 2±3.46j K 00, root are 2±2.45j, 2±2.45j K 260, root are ±3.6j, 4±3.6j Frequency Repone Deign

Amplitude Root Locu Plot Example 3 Time Repone Step Repone, K 64 2±3.46j 0.8 2, 2 0.6 whole repone 0.4 0.2 0 0 2 3 4 5 Time (ec.) Frequency Repone Deign

Amplitude Root Locu Plot Example 3 Time Repone Step Repone, K 00 0.8 2±2.45j (repeated) 0.6 0.4 whole repone 0.2 0 0 2 3 4 5 Time (ec.) Frequency Repone Deign

Amplitude Root Locu Plot Example 3 Time Repone 2.8.6.4 Step Repone K 260 ±3.6j whole repone.2 0.8 0.6 4±3.6j 0.4 0.2 0 0 0.5.5 2 2.5 3 3.5 4 4.5 5 Time (ec.) Frequency Repone Deign

Root Locu with Other Parameter Can we plot the root locu in term of the variation of a parameter other than gain, for example a time contant? To achieve thi we apply a mathematical trick. Conider the loop tranfer function ( + ) GH ( ) 0 ( + 2) characteritic equation t + 2 + 0 t + 0 2 regrouping the term t ( 2 + 2 + 0) + 0 equivalent loop tranfer function t 0 0 GH ( ) 2 + 2 + 0 0 Frequency Repone Deign

Root Locu with Other Parameter The open-loop pole are t 0 ± 3j and the zero i t 0.294 x 0. Aymptote: 80 t 0.4325 t 0.9 Angle of departure: x x 08 (90 + q d ) 80 4 2 q d 98 Break-in point b 3.6 d d t 0 0 2 + 2 + 0 ( -0) 0 2 x jw O 3j 3j + Frequency Repone Deign

Amplitude Root Locu with Other Parameter Step Repone t 0, ± 3j.2 0.8 0.6 0.4 0.2 0 0 2 3 4 5 6 Time (ec.) t 0.293, 2.5 ± 2j t 0.4325, 3.6, 3.6 t 0.90,, 0 Frequency Repone Deign

Root Locu with Two Parameter The root locu method focue on the root of the characteritic equation. There can be everal different loop tranfer function that have the ame cloed-loop characteritic equation. To contruct the root locu for a characteritic equation which ha two parameter, we contruct fictitiou loop tranfer function and apply the normal method. Frequency Repone Deign

Root Locu with Two Parameter Conider the following characteritic equation. 3 + 2 + b + a 0. Write thi in the general form of + GH() 0 with b a a multiplying gain. b + 0 3 2 + + a GH () Thi will allow the plotting of the root locu with repect to the gain b for any given value of a. The root of the characteritic equation of GH () define the tarting point for the root loci. Conider the loci of thee root. Frequency Repone Deign

Root Locu with Two Parameter The characteritic equation of GH () i 3 + 2 + a 0, which may be written a a + 0 2 ( + ) GH () Now contruct the root locu of GH () in term of the gain a. jw a 2 a a 2 a 2 a 0.3.2, 0.±0.5j a 2.8.66, 0.33±.0j Frequency Repone Deign

Root Locu with Two Parameter Now contruct the root locu for b GH () 3 2 + + a where the open-loop pole correpond to the previou root locu for varying a. Aymptote: ± 90 a 0.3 a (-.2 +0. +0.)/2-0.5 a.8 a (-.66 +0.33 +0.33)/2-0.5 Frequency Repone Deign

Root Locu with Two Parameter Imaginary axi croing: 3 + 2 + b + a 0 3 b 0 2 a 0 b-a 0 0 0 a 0 0 a b, 2 + a 0 - j a 2 jw a 2 a a 2 a O Frequency Repone Deign