Figure 1: Unity Feedback System

Size: px
Start display at page:

Download "Figure 1: Unity Feedback System"

Transcription

1 MEM 355 Sample Midterm Problem Stability 1 a) I the following ytem table? Solution: G() = Pole: -1, -2, -2, i, i 1 ( + 1)( )( + 2) 2 A you can ee, all pole are on the LHP. Thu, the ytem given above i table. 1b) Unity feedback ytem i a hown below Figure 1: Unity Feedback Sytem Cloed loop tranfer function: Y() R() = K()G p() 1 + K()G p ()

2 For what range of value of i the cloed loop ytem table? Characteritic equation; 1 + K()G p () = K ( 1 2 )2 ( + 1) = 0 Reorganize characteritic equation; ( + 1) + K( 1 2 )2 = 0 (1 + K) 2 + (1 K) + K 4 = 0 2 (1 + K) K 4 1 (1 K) 0 0 K 4 Table 1: Routh-Hurwitz Stability Criteria Hence, 0 < K < 1 When proportional Controller K i greater than 0 and le than 1, cloed loop ytem depicted in Figure 1 i table!

3 Ultimate State Error 2) What i the velocity error contant for the ytem? ( 2 + 2) G() = 5 ( + 2)( ) You may look at lecture note (Lecture 3: Ultimate State Error), or textbook Chapter 7.3, pg.350 (Nie, Control Sytem Engineering, 6th ed.) velocity error contant K v lim G() ( 2 + 2) K v lim 5 ( + 2)( ) = 10 2 = 5 3) For a unity feedback ytem, compute the teady-tate error in repone to: a) a unit tep and b) a unit ramp for a unity feedback ytem with the following forward loop tranfer function a) ( + 1)( 2 + 4) G() = 10 ( )( ) poition error contant K p lim G() velocity error contant K v lim G() K p lim 10 ( + 1)( 2 + 4) ( )( ) = 40 3 ( + 1)( 2 + 4) K v lim 10 ( )( ) = 0 Then, poition and velocity contant can be plugged in teady tate equation a follow.

4 For Step input > teady tate error: e( ) = 1 1+K p = 3 43 For Ramp input > teady tate error: e( ) = 1 K v = b) ( + 3) G() = 2 ( ) poition error contant K p lim G() velocity error contant K v lim G() K p lim 2 ( + 3) ( ) = ( + 3) K v lim 2 ( ) = 6 Then, poition and velocity contant can be plugged in teady tate equation a follow. For unity Step input ( 1 1 ) > teady tate error: e( ) = = 0 1+K p For unity Ramp input ( 1 2) > teady tate error: e( ) = 1 K v = 1 6

5 Solution#1: A we can ee, forward loop TF i G() = K ( + 5) velocity error contant K v lim G() K v lim K ( + 5) = K 5 We know that, For unity Ramp input ( 1 2) > teady tate error: e( ) = 1 K v = Therefore, we can redefine teady tate error For an arbitrary Ramp input, which i ( ) > teady tate error: e( ) = = K v K v = 10 K = 5 K v Then, K = 50

6 Solution #2: (Alternative olution) Error TF ----> E() = 1 1+G() R() In quetion, ramp input i given a R() = Steady tate error for ramp input > Thu, K = 50 e( ) = lim e( ) = lim E() = lim 1 K 1 + ( + 5) = G() R() Characteritic equation: K 1 + ( + 5) = 0 ( + 5) + K = K = K K Table 1: Routh-Hurwitz Stability Criteria Sytem given in quetion i table if K>0 Therefore, ytem i table for K=50!!!

7 Solution #1: Solution #2: (Alternative olution) Forward loop TF i G() = K( 2) ( + 1)( + 10) Error TF ----> E() = 1 1+G() R() In quetion, ramp input i given a R() = 20 Steady tate error for ramp input > e( ) = lim E() = lim 1 1+G() R()

8 e( ) = lim 1 K( 2) 1 + ( + 1)( + 10) 20 e( ) = lim ( + 1)( + 10) 20 ( + 1)( + 10) + K( 2) = K = K = Characteritic equation: K( 2) 1 + ( + 1)( + 10) = 0 ( + 1)( + 10) + K( 2) = (11 + K) K = K 1 (11 + K) K Sytem given in quetion i table if 11 < K < 5 Table 1: Routh-Hurwitz Stability Criteria And, we found previouly that K = Hence, the ytem i NOT table with K =

9 Solution: Let find forward loop TF to find error contant: a = K 2( ) J 2 ( K 1( ) G c ) a = K 1K 2 (( )) 2 ( 2 J + K K 2 ) G = a 1 = K 1 K 2 (( )) 2 2 ( 2 J + K K 2 )

10 After forward loop G i found, we can find error contant poition error contant K p lim G() velocity error contant K v lim G() accelarition error contant K v lim 2 G() K p lim K 1 K 2 (( )) 2 2 ( 2 J + K K 2 ) = K v lim K 1 K 2 (( )) 2 2 ( 2 J + K K 2 ) = K a lim 2 K 1 K 2 (( )) 2 2 ( 2 J + K K 2 ) = 0.01K 1 Error Contant Steady State Error K p = K p = 0 K v = 1 K v = 0 K a = 0.01K 1 1 K a = 100 K 1

11 Root Locu Solution: Forward loop TF i G() = K (2 + 1)( + 2) 2 ( + 1)( + 4)( + 5) Step#1: Pole: 0,0, 1, 4, 5 zero: 2, i, i Step#2: There are 5 branche becaue there are 5 pole! Step#3: Aymptote angle θ l = (2l + 1) π where l = 0,1,2 m n 1 m n

12 n: # of pole m: # of zero θ l = (2l + 1) 5 3 π θ l = π 2, 3π 2 for l = 0,1 Step#4: interection of aymptote with real axi (finite pole) (finite zero) σ = #pole #zero ( 1 4 5) ( i i) σ = 5 3 σ = Step 5: Break away-in point on the Root Loci 10 ( 1) 5 3 The breakaway point on the root loci of 1 + KG = 0 = 9 2 Satify, K = 1 G dk d = 0 If we want to imply the equation above, we can write dk = 0 d d d (1) = 0 G If define G() = N() D() where G() i open loop TF. d d (1) = d G d (D() d(n()) ) = D() N() d N() d(d()) =0 d N() = ( 2 + 1)( + 2) D() = 2 ( + 1)( + 4)( + 5)

13 ( ) = 0 There are a lot of olution but we know that ome of them are not on the loci. When we olve the equation above, we will ay that Break away point 0 (it i obviou from tep#1 ) and ( we peak thi becaue we know that break away point mut be between -4 and -5 in thi problem) G= zpk([-2, i, i],-[0,0,1,4,5],1) % Step2: Look at pole/zero of ytem [p,z]=pzmap(g) % Step3: Look aymptote angle i=1; for k=0:length(p)-length(z)-1 theta(i)=(2*k+1)*pi/( length(p)- length(z)) % angle for aymptote i=i+1; end theta=theta*180/pi % [deg] % theta= pi/2,3pi/2 % Step4: look interection of aymptote imga=(um(p)-um(z))/( length(p)- length(z)) % igma= = -7/3 in thi examle % Step5: Breakawway or break in if exit! ym N=((+2)*^2-+1); D=^2*(+1)*(+4)*(+5); DK=D*diff(N,)-N*diff(D,)

14 Breakaway=double(olve(DK==0,)) % Breakaway point are 0 and %Step6: Look at root loci rlocu(g) We have 5 branche and all are ymmetrical with real axi. Break away point are at 0 and at Aymptote interect with real axi at -4.5 and aymptote angle are 90 and 270 degree! Not to forget that, draw root locu tarting from pole to zero!

15 Solution: H i contant and aume that it i 0! Let find forward loop TF. G() K( ) a = ( 1 a) a = K( )0.04 ( ) G= 10 4 K(+0.01) 2 a = (+0.04) Pole 0,0, 0.04 Zero 0.01 σ θ l π 2, 3π 2 Break-away point 0 ( other are not on loci)

16 %Forward Loop TF G= zpk([-0.01],-[0,0,0.04],4e-6) % Step2: Look at pole/zero of ytem [p,z]=pzmap(g) % Step3: Look aymptote angle i=1; for k=0:length(p)-length(z)-1 theta(i)=(2*k+1)*pi/( length(p)- length(z)) % angle for aymptote i=i+1; end theta=theta*180/pi % [deg] % theta= pi/2,3pi/2 % Step4: look interection of aymptote imga=(um(p)-um(z))/( length(p)- length(z)) % igma= % Step5: Breakawway or break in if exit! ym N=(4e-06)*(+0.01); D=(^2)*(+0.04); DK=D*diff(N,)-N*diff(D,) Breakaway=double(olve(DK==0,)) % Breakaway point 0 %Step6: Look at root loci rlocu(g)

17

MODERN CONTROL SYSTEMS

MODERN CONTROL SYSTEMS MODERN CONTROL SYSTEMS Lecture 1 Root Locu Emam Fathy Department of Electrical and Control Engineering email: emfmz@aat.edu http://www.aat.edu/cv.php?dip_unit=346&er=68525 1 Introduction What i root locu?

More information

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

ECE-320 Linear Control Systems. Spring 2014, Exam 1. No calculators or computers allowed, you may leave your answers as fractions. ECE-0 Linear Control Sytem Spring 04, Exam No calculator or computer allowed, you may leave your anwer a fraction. All problem are worth point unle noted otherwie. Total /00 Problem - refer to the unit

More information

Chapter 7. Root Locus Analysis

Chapter 7. Root Locus Analysis 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

More information

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

Root Locus Contents. Root locus, sketching algorithm. Root locus, examples. Root locus, proofs. Root locus, control examples Root Locu Content Root locu, ketching algorithm Root locu, example Root locu, proof Root locu, control example Root locu, influence of zero and pole Root locu, lead lag controller deign 9 Spring ME45 -

More information

CISE302: Linear Control Systems

CISE302: Linear Control Systems Term 8 CISE: Linear Control Sytem Dr. Samir Al-Amer Chapter 7: Root locu CISE_ch 7 Al-Amer8 ١ Learning Objective Undertand the concept of root locu and it role in control ytem deign Be able to ketch root

More information

ME2142/ME2142E Feedback Control Systems

ME2142/ME2142E Feedback Control Systems Root Locu Analyi Root Locu Analyi Conider the cloed-loop ytem R + E - G C B H The tranient repone, and tability, of the cloed-loop ytem i determined by the value of the root of the characteritic equation

More information

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

CONTROL SYSTEMS. Chapter 5 : Root Locus Diagram. GATE Objective & Numerical Type Solutions. The transfer function of a closed loop system is CONTROL SYSTEMS Chapter 5 : Root Locu Diagram GATE Objective & Numerical Type Solution Quetion 1 [Work Book] [GATE EC 199 IISc-Bangalore : Mark] The tranfer function of a cloed loop ytem i T () where i

More information

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

Lecture 12: Examples of Root Locus Plots. Dr. Kalyana Veluvolu. Lecture 12: Examples of Root Locus Plots Dr. Kalyana Veluvolu ROOT-LOCUS ANALYSIS Example: Given that G( ) ( + )( + ) Dr. alyana Veluvolu Sketch the root locu of 1 + G() and compute the value of that will yield a dominant econd order behavior with a damping ratio,

More information

ME 375 FINAL EXAM SOLUTIONS Friday December 17, 2004

ME 375 FINAL EXAM SOLUTIONS Friday December 17, 2004 ME 375 FINAL EXAM SOLUTIONS Friday December 7, 004 Diviion Adam 0:30 / Yao :30 (circle one) Name Intruction () Thi i a cloed book eamination, but you are allowed three 8.5 crib heet. () You have two hour

More information

Automatic Control Systems. Part III: Root Locus Technique

Automatic Control Systems. Part III: Root Locus Technique www.pdhcenter.com PDH Coure E40 www.pdhonline.org Automatic Control Sytem Part III: Root Locu Technique By Shih-Min Hu, Ph.D., P.E. Page of 30 www.pdhcenter.com PDH Coure E40 www.pdhonline.org VI. Root

More information

Analysis of Stability &

Analysis of Stability & INC 34 Feedback Control Sytem Analyi of Stability & Steady-State Error S Wonga arawan.won@kmutt.ac.th Summary from previou cla Firt-order & econd order ytem repone τ ωn ζω ω n n.8.6.4. ζ ζ. ζ.5 ζ ζ.5 ct.8.6.4...4.6.8..4.6.8

More information

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

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial :. PT_EE_A+C_Control Sytem_798 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubanewar olkata Patna Web: E-mail: info@madeeay.in Ph: -4546 CLASS TEST 8-9 ELECTRICAL ENGINEERING Subject

More information

Feedback Control Systems (FCS)

Feedback Control Systems (FCS) Feedback Control Sytem (FCS) Lecture19-20 Routh-Herwitz Stability Criterion Dr. Imtiaz Huain email: imtiaz.huain@faculty.muet.edu.pk URL :http://imtiazhuainkalwar.weebly.com/ Stability of Higher Order

More information

ME 375 FINAL EXAM Wednesday, May 6, 2009

ME 375 FINAL EXAM Wednesday, May 6, 2009 ME 375 FINAL EXAM Wedneday, May 6, 9 Diviion Meckl :3 / Adam :3 (circle one) Name_ Intruction () Thi i a cloed book examination, but you are allowed three ingle-ided 8.5 crib heet. A calculator i NOT allowed.

More information

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

Root Locus Diagram. Root loci: The portion of root locus when k assume positive values: that is 0 Objective Root Locu Diagram Upon completion of thi chapter you will be able to: Plot the Root Locu for a given Tranfer Function by varying gain of the ytem, Analye the tability of the ytem from the root

More information

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

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax: Control Sytem Engineering ( Chapter 7. Steady-State Error Prof. Kwang-Chun Ho kwangho@hanung.ac.kr Tel: 0-760-453 Fax:0-760-4435 Introduction In thi leon, you will learn the following : How to find the

More information

1 Routh Array: 15 points

1 Routh Array: 15 points EE C28 / ME34 Problem Set 3 Solution Fall 2 Routh Array: 5 point Conider the ytem below, with D() k(+), w(t), G() +2, and H y() 2 ++2 2(+). Find the cloed loop tranfer function Y () R(), and range of k

More information

ROOT LOCUS. Poles and Zeros

ROOT LOCUS. Poles and Zeros Automatic Control Sytem, 343 Deartment of Mechatronic Engineering, German Jordanian Univerity ROOT LOCUS The Root Locu i the ath of the root of the characteritic equation traced out in the - lane a a ytem

More information

Stability Criterion Routh Hurwitz

Stability Criterion Routh Hurwitz EES404 Fundamental of Control Sytem Stability Criterion Routh Hurwitz DR. Ir. Wahidin Wahab M.Sc. Ir. Arie Subiantoro M.Sc. Stability A ytem i table if for a finite input the output i imilarly finite A

More information

Control Systems. Root locus.

Control Systems. Root locus. Control Sytem Root locu chibum@eoultech.ac.kr Outline Concet of Root Locu Contructing root locu Control Sytem Root Locu Stability and tranient reone i cloely related with the location of ole in -lane How

More information

MEM 355 Performance Enhancement of Dynamical Systems Root Locus Analysis

MEM 355 Performance Enhancement of Dynamical Systems Root Locus Analysis MEM 355 Performance Enhancement of Dynamical Sytem Root Locu Analyi Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity Outline The root locu method wa introduced by Evan in

More information

The Root Locus Method

The Root Locus Method The Root Locu Method MEM 355 Performance Enhancement of Dynamical Sytem Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity Outline The root locu method wa introduced by Evan

More information

Control Systems. Root locus.

Control Systems. Root locus. Control Sytem Root locu chibum@eoultech.ac.kr Outline Concet of Root Locu Contructing root locu Control Sytem Root Locu Stability and tranient reone i cloely related with the location of ole in -lane How

More information

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

376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with : 376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore by applying the lead compenator with ome gain adjutment : D() =.12 4.5 +1 9 +1 we get the compenated ytem with : PM =65, ω c = 22 rad/ec, o

More information

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

Module 4: Time Response of discrete time systems Lecture Note 1 Digital Control Module 4 Lecture Module 4: ime Repone of dicrete time ytem Lecture Note ime Repone of dicrete time ytem Abolute tability i a baic requirement of all control ytem. Apart from that, good

More information

EE Control Systems LECTURE 14

EE Control Systems LECTURE 14 Updated: Tueday, March 3, 999 EE 434 - Control Sytem LECTURE 4 Copyright FL Lewi 999 All right reerved ROOT LOCUS DESIGN TECHNIQUE Suppoe the cloed-loop tranfer function depend on a deign parameter k We

More information

6.302 Feedback Systems Recitation 6: Steady-State Errors Prof. Joel L. Dawson S -

6.302 Feedback Systems Recitation 6: Steady-State Errors Prof. Joel L. Dawson S - 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon A valid performance metric for any control ytem center around the final error when the ytem reache teadytate That i, after all initial

More information

Control Systems Analysis and Design by the Root-Locus Method

Control Systems Analysis and Design by the Root-Locus Method 6 Control Sytem Analyi and Deign by the Root-Locu Method 6 1 INTRODUCTION The baic characteritic of the tranient repone of a cloed-loop ytem i cloely related to the location of the cloed-loop pole. If

More information

Chapter 13. Root Locus Introduction

Chapter 13. Root Locus Introduction Chapter 13 Root Locu 13.1 Introduction In the previou chapter we had a glimpe of controller deign iue through ome imple example. Obviouly when we have higher order ytem, uch imple deign technique will

More information

Linear System Fundamentals

Linear System Fundamentals Linear Sytem Fundamental MEM 355 Performance Enhancement of Dynamical Sytem Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity Content Sytem Repreentation Stability Concept

More information

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

The state variable description of an LTI system is given by 3 1O. Statement for Linked Answer Questions 3 and 4 : CHAPTER 6 CONTROL SYSTEMS YEAR TO MARKS MCQ 6. The tate variable decription of an LTI ytem i given by Jxo N J a NJx N JN K O K OK O K O xo a x + u Kxo O K 3 a3 OKx O K 3 O L P L J PL P L P x N K O y _

More information

ME 375 EXAM #1 Tuesday February 21, 2006

ME 375 EXAM #1 Tuesday February 21, 2006 ME 375 EXAM #1 Tueday February 1, 006 Diviion Adam 11:30 / Savran :30 (circle one) Name Intruction (1) Thi i a cloed book examination, but you are allowed one 8.5x11 crib heet. () You have one hour to

More information

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

SKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot SKEE 3143 CONTROL SYSTEM DESIGN CHAPTER 3 Compenator Deign Uing the Bode Plot 1 Chapter Outline 3.1 Introduc4on Re- viit to Frequency Repone, ploang frequency repone, bode plot tability analyi. 3.2 Gain

More information

State Space: Observer Design Lecture 11

State Space: Observer Design Lecture 11 State Space: Oberver Deign Lecture Advanced Control Sytem Dr Eyad Radwan Dr Eyad Radwan/ACS/ State Space-L Controller deign relie upon acce to the tate variable for feedback through adjutable gain. Thi

More information

FRTN10 Exercise 3. Specifications and Disturbance Models

FRTN10 Exercise 3. Specifications and Disturbance Models FRTN0 Exercie 3. Specification and Diturbance Model 3. A feedback ytem i hown in Figure 3., in which a firt-order proce if controlled by an I controller. d v r u 2 z C() P() y n Figure 3. Sytem in Problem

More information

EE Control Systems LECTURE 6

EE Control Systems LECTURE 6 Copyright FL Lewi 999 All right reerved EE - Control Sytem LECTURE 6 Updated: Sunday, February, 999 BLOCK DIAGRAM AND MASON'S FORMULA A linear time-invariant (LTI) ytem can be repreented in many way, including:

More information

Chapter #4 EEE Automatic Control

Chapter #4 EEE Automatic Control Spring 008 EEE 00 Chapter #4 EEE 00 Automatic Control Root Locu Chapter 4 /4 Spring 008 EEE 00 Introduction Repone depend on ytem and controller parameter > Cloed loop pole location depend on ytem and

More information

Figure 1 Siemens PSSE Web Site

Figure 1 Siemens PSSE Web Site Stability Analyi of Dynamic Sytem. In the lat few lecture we have een how mall ignal Lalace domain model may be contructed of the dynamic erformance of ower ytem. The tability of uch ytem i a matter of

More information

CONTROL SYSTEMS. Chapter 2 : Block Diagram & Signal Flow Graphs GATE Objective & Numerical Type Questions

CONTROL SYSTEMS. Chapter 2 : Block Diagram & Signal Flow Graphs GATE Objective & Numerical Type Questions ONTOL SYSTEMS hapter : Bloc Diagram & Signal Flow Graph GATE Objective & Numerical Type Quetion Quetion 6 [Practice Boo] [GATE E 994 IIT-Kharagpur : 5 Mar] educe the ignal flow graph hown in figure below,

More information

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

MM1: Basic Concept (I): System and its Variables MM1: Baic Concept (I): Sytem and it Variable A ytem i a collection of component which are coordinated together to perform a function Sytem interact with their environment. The interaction i defined in

More information

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

Lecture 4. Chapter 11 Nise. Controller Design via Frequency Response. G. Hovland 2004 METR4200 Advanced Control Lecture 4 Chapter Nie Controller Deign via Frequency Repone G. Hovland 2004 Deign Goal Tranient repone via imple gain adjutment Cacade compenator to improve teady-tate error Cacade

More information

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

G(s) = 1 s by hand for! = 1, 2, 5, 10, 20, 50, and 100 rad/sec. 6003 where A = jg(j!)j ; = tan Im [G(j!)] Re [G(j!)] = \G(j!) 2. (a) Calculate the magnitude and phae of G() = + 0 by hand for! =, 2, 5, 0, 20, 50, and 00 rad/ec. (b) ketch the aymptote for G() according

More information

Digital Control System

Digital Control System Digital Control Sytem - A D D A Micro ADC DAC Proceor Correction Element Proce Clock Meaurement A: Analog D: Digital Continuou Controller and Digital Control Rt - c Plant yt Continuou Controller Digital

More information

Homework 12 Solution - AME30315, Spring 2013

Homework 12 Solution - AME30315, Spring 2013 Homework 2 Solution - AME335, Spring 23 Problem :[2 pt] The Aerotech AGS 5 i a linear motor driven XY poitioning ytem (ee attached product heet). A friend of mine, through careful experimentation, identified

More information

Mathematical modeling of control systems. Laith Batarseh. Mathematical modeling of control systems

Mathematical modeling of control systems. Laith Batarseh. Mathematical modeling of control systems Chapter two Laith Batareh Mathematical modeling The dynamic of many ytem, whether they are mechanical, electrical, thermal, economic, biological, and o on, may be decribed in term of differential equation

More information

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

Stability. ME 344/144L Prof. R.G. Longoria Dynamic Systems and Controls/Lab. Department of Mechanical Engineering The University of Texas at Austin Stability The tability of a ytem refer to it ability or tendency to eek a condition of tatic equilibrium after it ha been diturbed. If given a mall perturbation from the equilibrium, it i table if it return.

More information

SECTION 5: ROOT LOCUS ANALYSIS

SECTION 5: ROOT LOCUS ANALYSIS SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path

More information

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

NAME (pinyin/italian)... MATRICULATION NUMBER... SIGNATURE POLITONG SHANGHAI BASIC AUTOMATIC CONTROL June Academic Year / Exam grade NAME (pinyin/italian)... MATRICULATION NUMBER... SIGNATURE Ue only thee page (including the bac) for anwer. Do not ue additional

More information

Correction for Simple System Example and Notes on Laplace Transforms / Deviation Variables ECHE 550 Fall 2002

Correction for Simple System Example and Notes on Laplace Transforms / Deviation Variables ECHE 550 Fall 2002 Correction for Simple Sytem Example and Note on Laplace Tranform / Deviation Variable ECHE 55 Fall 22 Conider a tank draining from an initial height of h o at time t =. With no flow into the tank (F in

More information

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

EE 4443/5329. LAB 3: Control of Industrial Systems. Simulation and Hardware Control (PID Design) The Inverted Pendulum. (ECP Systems-Model: 505) EE 4443/5329 LAB 3: Control of Indutrial Sytem Simulation and Hardware Control (PID Deign) The Inverted Pendulum (ECP Sytem-Model: 505) Compiled by: Nitin Swamy Email: nwamy@lakehore.uta.edu Email: okuljaca@lakehore.uta.edu

More information

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)

More information

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

ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances ECE 350 Root Locu Deign Example Recall the imple crude ervo from lab G( ) 0 6.64 53.78 σ = = 3 23.473 PI To eliminate teady-tate error (for contant input) & perfect reection of contant diturbance Note:

More information

ECE382/ME482 Spring 2004 Homework 4 Solution November 14,

ECE382/ME482 Spring 2004 Homework 4 Solution November 14, ECE382/ME482 Spring 2004 Homework 4 Solution November 14, 2005 1 Solution to HW4 AP4.3 Intead of a contant or tep reference input, we are given, in thi problem, a more complicated reference path, r(t)

More information

Lecture 5 Introduction to control

Lecture 5 Introduction to control Lecture 5 Introduction to control Tranfer function reviited (Laplace tranform notation: ~jω) () i the Laplace tranform of v(t). Some rule: ) Proportionality: ()/ in () 0log log() v (t) *v in (t) () * in

More information

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

CHAPTER 4 DESIGN OF STATE FEEDBACK CONTROLLERS AND STATE OBSERVERS USING REDUCED ORDER MODEL 98 CHAPTER DESIGN OF STATE FEEDBACK CONTROLLERS AND STATE OBSERVERS USING REDUCED ORDER MODEL INTRODUCTION The deign of ytem uing tate pace model for the deign i called a modern control deign and it i

More information

Function and Impulse Response

Function and Impulse Response Tranfer Function and Impule Repone Solution of Selected Unolved Example. Tranfer Function Q.8 Solution : The -domain network i hown in the Fig... Applying VL to the two loop, R R R I () I () L I () L V()

More information

Digital Control System

Digital Control System Digital Control Sytem Summary # he -tranform play an important role in digital control and dicrete ignal proceing. he -tranform i defined a F () f(k) k () A. Example Conider the following equence: f(k)

More information

Chapter 8. Root Locus Techniques

Chapter 8. Root Locus Techniques Chapter 8 Rt Lcu Technique Intrductin Sytem perfrmance and tability dt determined dby cled-lp l ple Typical cled-lp feedback cntrl ytem G Open-lp TF KG H Zer -, - Ple 0, -, -4 K 4 Lcatin f ple eaily fund

More information

Chapter 10. Closed-Loop Control Systems

Chapter 10. Closed-Loop Control Systems hapter 0 loed-loop ontrol Sytem ontrol Diagram of a Typical ontrol Loop Actuator Sytem F F 2 T T 2 ontroller T Senor Sytem T TT omponent and Signal of a Typical ontrol Loop F F 2 T Air 3-5 pig 4-20 ma

More information

Chapter 6 Control Systems Design by Root-Locus Method. Lag-Lead Compensation. Lag lead Compensation Techniques Based on the Root-Locus Approach.

Chapter 6 Control Systems Design by Root-Locus Method. Lag-Lead Compensation. Lag lead Compensation Techniques Based on the Root-Locus Approach. hapter 6 ontrol Sytem Deign by Root-Lou Method Lag-Lead ompenation Lag lead ompenation ehnique Baed on the Root-Lou Approah. γ β K, ( γ >, β > ) In deigning lag lead ompenator, we onider two ae where γ

More information

March 18, 2014 Academic Year 2013/14

March 18, 2014 Academic Year 2013/14 POLITONG - SHANGHAI BASIC AUTOMATIC CONTROL Exam grade March 8, 4 Academic Year 3/4 NAME (Pinyin/Italian)... STUDENT ID Ue only thee page (including the back) for anwer. Do not ue additional heet. Ue of

More information

Exercises for lectures 19 Polynomial methods

Exercises for lectures 19 Polynomial methods Exercie for lecture 19 Polynomial method Michael Šebek Automatic control 016 15-4-17 Diviion of polynomial with and without remainder Polynomial form a circle, but not a body. (Circle alo form integer,

More information

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

ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27 1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system

More information

Lecture 1 Root Locus

Lecture 1 Root Locus Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the

More information

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

Department of Mechanical Engineering Massachusetts Institute of Technology Modeling, Dynamics and Control III Spring 2002 Department of Mechanical Engineering Maachuett Intitute of Technology 2.010 Modeling, Dynamic and Control III Spring 2002 SOLUTIONS: Problem Set # 10 Problem 1 Etimating tranfer function from Bode Plot.

More information

Systems Analysis. Prof. Cesar de Prada ISA-UVA

Systems Analysis. Prof. Cesar de Prada ISA-UVA Sytem Analyi Prof. Cear de Prada ISAUVA rada@autom.uva.e Aim Learn how to infer the dynamic behaviour of a cloed loo ytem from it model. Learn how to infer the change in the dynamic of a cloed loo ytem

More information

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

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 Root Locus Simple definition Locus of points on the s- plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation

More information

EE302 - Feedback Systems Spring Lecture KG(s)H(s) = KG(s)

EE302 - Feedback Systems Spring Lecture KG(s)H(s) = KG(s) EE3 - Feedback Systems Spring 19 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 1.. 1.1 Root Locus In control theory, root locus analysis is a graphical analysis method for investigating the change of

More information

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

Solutions. Digital Control Systems ( ) 120 minutes examination time + 15 minutes reading time at the beginning of the exam BSc - Sample Examination Digital Control Sytem (5-588-) Prof. L. Guzzella Solution Exam Duration: Number of Quetion: Rating: Permitted aid: minute examination time + 5 minute reading time at the beginning

More information

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

into a discrete time function. Recall that the table of Laplace/z-transforms is constructed by (i) selecting to get Lecture 25 Introduction to Some Matlab c2d Code in Relation to Sampled Sytem here are many way to convert a continuou time function, { h( t) ; t [0, )} into a dicrete time function { h ( k) ; k {0,,, }}

More information

Chapter 7 : Root Locus Technique

Chapter 7 : Root Locus Technique Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci

More information

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

Homework 7 Solution - AME 30315, Spring s 2 + 2s (s 2 + 2s + 4)(s + 20) 1 Homework 7 Solution - AME 30315, Spring 2015 Problem 1 [10/10 pt] Ue partial fraction expanion to compute x(t) when X 1 () = 4 2 + 2 + 4 Ue partial fraction expanion to compute x(t) when X 2 () = ( )

More information

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

ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8 Learning Objectives ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8 Dr. Oishi oishi@unm.edu November 2, 203 State the phase and gain properties of a root locus Sketch a root locus, by

More information

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

6.447 rad/sec and ln (% OS /100) tan Thus pc. the testing point is s 3.33 j5.519 9. a. 3.33, n T ln(% OS /100) 2 2 ln (% OS /100) 0.517. Thu n 6.7 rad/ec and the teting point i 3.33 j5.519. b. Summation of angle including the compenating zero i -106.691, The compenator pole mut contribute

More information

EE/ME/AE324: Dynamical Systems. Chapter 8: Transfer Function Analysis

EE/ME/AE324: Dynamical Systems. Chapter 8: Transfer Function Analysis EE/ME/AE34: Dynamical Sytem Chapter 8: Tranfer Function Analyi The Sytem Tranfer Function Conider the ytem decribed by the nth-order I/O eqn.: ( n) ( n 1) ( m) y + a y + + a y = b u + + bu n 1 0 m 0 Taking

More information

What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine?

What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine? A 2.0 Introduction In the lat et of note, we developed a model of the peed governing mechanim, which i given below: xˆ K ( Pˆ ˆ) E () In thee note, we want to extend thi model o that it relate the actual

More information

S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS

S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS by Michelle Gretzinger, Daniel Zyngier and Thoma Marlin INTRODUCTION One of the challenge to the engineer learning proce control i relating theoretical

More information

EE105 - Fall 2005 Microelectronic Devices and Circuits

EE105 - Fall 2005 Microelectronic Devices and Circuits EE5 - Fall 5 Microelectronic Device and ircuit Lecture 9 Second-Order ircuit Amplifier Frequency Repone Announcement Homework 8 due tomorrow noon Lab 7 next week Reading: hapter.,.3. Lecture Material Lat

More information

MM7. PID Control Design

MM7. PID Control Design MM7. PD Control Deign Reading Material: FC: p.79-200, DC: p.66-68. Propertie of PD control 2. uning Method of PD Control 3. Antiwindup echnique 4. A real cae tudy BO9000 0/9/2004 Proce Control . PD Feedback

More information

CHE302 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang

CHE302 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang CHE3 ECTURE V APACE TRANSFORM AND TRANSFER FUNCTION Profeor Dae Ryook Yang Fall Dept. of Chemical and Biological Engineering Korea Univerity CHE3 Proce Dynamic and Control Korea Univerity 5- SOUTION OF

More information

CALIFORNIA INSTITUTE OF TECHNOLOGY Control and Dynamical Systems

CALIFORNIA INSTITUTE OF TECHNOLOGY Control and Dynamical Systems Control and Dynamical Sytem CDS 0 Problem Set #5 Iued: 3 Nov 08 Due: 0 Nov 08 Note: In the upper left hand corner of the econd page of your homework et, pleae put the number of hour that you pent on thi

More information

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

Given the following circuit with unknown initial capacitor voltage v(0): X(s) Immediately, we know that the transfer function H(s) is EE 4G Note: Chapter 6 Intructor: Cheung More about ZSR and ZIR. Finding unknown initial condition: Given the following circuit with unknown initial capacitor voltage v0: F v0/ / Input xt 0Ω Output yt -

More information

Today s Lecture. Block Diagrams. Block Diagrams: Examples. Block Diagrams: Examples. Closed Loop System 06/03/2017

Today s Lecture. Block Diagrams. Block Diagrams: Examples. Block Diagrams: Examples. Closed Loop System 06/03/2017 06/0/07 UW Britol Indutrial ontrol UFMF6W-0- ontrol Sytem ngineering UFMUY-0- Lecture 5: Block Diagram and Steady State rror Today Lecture Block diagram to repreent control ytem Block diagram manipulation

More information

If you need more room, use the backs of the pages and indicate that you have done so.

If you need more room, use the backs of the pages and indicate that you have done so. EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty

More information

Massachusetts Institute of Technology Dynamics and Control II

Massachusetts Institute of Technology Dynamics and Control II I E Maachuett Intitute of Technology Department of Mechanical Engineering 2.004 Dynamic and Control II Laboratory Seion 5: Elimination of Steady-State Error Uing Integral Control Action 1 Laboratory Objective:

More information

Linear-Quadratic Control System Design

Linear-Quadratic Control System Design Linear-Quadratic Control Sytem Deign Robert Stengel Optimal Control and Etimation MAE 546 Princeton Univerity, 218! Control ytem configuration! Proportional-integral! Proportional-integral-filtering! Model

More information

Automatic Control Systems

Automatic Control Systems Automatic Control Sytem Lecture- Block Diagram Reduction Emam Fathy Department of Electrical and Control Engineering email: emfmz@yahoo.com Introduction A Block Diagram i a horthand pictorial repreentation

More information

Test 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010

Test 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010 Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the

More information

CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang

CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang CHBE3 ECTURE V APACE TRANSFORM AND TRANSFER FUNCTION Profeor Dae Ryook Yang Spring 8 Dept. of Chemical and Biological Engineering 5- Road Map of the ecture V aplace Tranform and Tranfer function Definition

More information

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

Advanced D-Partitioning Analysis and its Comparison with the Kharitonov s Theorem Assessment Journal of Multidiciplinary Engineering Science and Technology (JMEST) ISSN: 59- Vol. Iue, January - 5 Advanced D-Partitioning Analyi and it Comparion with the haritonov Theorem Aement amen M. Yanev Profeor,

More information

Analysis and Design of a Third Order Phase-Lock Loop

Analysis and Design of a Third Order Phase-Lock Loop Analyi Deign of a Third Order Phae-Lock Loop DANIEL Y. ABRAMOVITCH Ford Aeropace Corporation 3939 Fabian Way, MS: X- Palo Alto, CA 94303 Abtract Typical implementation of a phae-lock loop (PLL) are econd

More information

SECTION 8: ROOT-LOCUS ANALYSIS. ESE 499 Feedback Control Systems

SECTION 8: ROOT-LOCUS ANALYSIS. ESE 499 Feedback Control Systems SECTION 8: ROOT-LOCUS ANALYSIS ESE 499 Feedback Control Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed-loop transfer function is KKKK ss TT ss = 1 + KKKK ss HH ss GG ss

More information

CHAPTER # 9 ROOT LOCUS ANALYSES

CHAPTER # 9 ROOT LOCUS ANALYSES F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closed-loop system is closely related to the location of the closed-loop poles. If the system

More information

MAE140 Linear Circuits Fall 2012 Final, December 13th

MAE140 Linear Circuits Fall 2012 Final, December 13th MAE40 Linear Circuit Fall 202 Final, December 3th Intruction. Thi exam i open book. You may ue whatever written material you chooe, including your cla note and textbook. You may ue a hand calculator with

More information

A Simplified Methodology for the Synthesis of Adaptive Flight Control Systems

A Simplified Methodology for the Synthesis of Adaptive Flight Control Systems A Simplified Methodology for the Synthei of Adaptive Flight Control Sytem J.ROUSHANIAN, F.NADJAFI Department of Mechanical Engineering KNT Univerity of Technology 3Mirdamad St. Tehran IRAN Abtract- A implified

More information

Section Induction motor drives

Section Induction motor drives Section 5.1 - nduction motor drive Electric Drive Sytem 5.1.1. ntroduction he AC induction motor i by far the mot widely ued motor in the indutry. raditionally, it ha been ued in contant and lowly variable-peed

More information

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

Lecture 8. PID control. Industrial process control ( today) PID control. Insights about PID actions Lecture 8. PID control. The role of P, I, and D action 2. PID tuning Indutrial proce control (92... today) Feedback control i ued to improve the proce performance: tatic performance: for contant reference,

More information

Homework Assignment No. 3 - Solutions

Homework Assignment No. 3 - Solutions ECE 6440 Summer 2003 Page 1 Homework Aignment o. 3 Problem 1 (10 point) Aume an LPLL ha F() 1 and the PLL parameter are 0.8V/radian, K o 100 MHz/V, and the ocillation frequency, f oc 500MHz. Sketch the

More information

CHEAP CONTROL PERFORMANCE LIMITATIONS OF INPUT CONSTRAINED LINEAR SYSTEMS

CHEAP CONTROL PERFORMANCE LIMITATIONS OF INPUT CONSTRAINED LINEAR SYSTEMS Copyright 22 IFAC 5th Triennial World Congre, Barcelona, Spain CHEAP CONTROL PERFORMANCE LIMITATIONS OF INPUT CONSTRAINED LINEAR SYSTEMS Tritan Pérez Graham C. Goodwin Maria M. Serón Department of Electrical

More information

Lecture 7 CONTROL ENGINEERING. Ali Karimpour Associate Professor Ferdowsi University of Mashhad

Lecture 7 CONTROL ENGINEERING. Ali Karimpour Associate Professor Ferdowsi University of Mashhad CONTROL ENGINEERING Ali Karimpour Aociate Profeor Ferdowi Univerity of Mahhad Dr. Ali Karimpour Sep 5 Root Locu Criteria Topic to be covered include: Root locu criterion. Root loci (RL). Complement root

More information