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# Control Systems ME 451

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This 154 page Class Notes was uploaded by Princess Rolfson on Saturday September 19, 2015. The Class Notes belongs to ME 451 at Michigan State University taught by Staff in Fall. Since its upload, it has received 534 views. For similar materials see /class/207536/me-451-michigan-state-university in Mechanical Engineering at Michigan State University.

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Date Created: 09/19/15

ME451 Control Systems Lecture 1 Introduction Dr Jongeun Choi Department of Mechanical Engineering Michigan State University MICHIGAN STATE UNIVERSITY Fall 2008 1 l nstructor Class Instructor Dr Jongeun Choi Website httpwwwegrmsuedu39choi Assistant Professor at ME department 2459 Engineering Building Email 39choiegrmsuedu Office Hours 2459 EB MW 200300pm Extra hours by appointment Laboratory Instructor Dr C J Radcliffe 2445 Engineering Building Email radcliffegrmsuedu Fall 2008 Course information Lecture When MWF 1130am1220pm Where C103 McDonel Hall Class website httpwwwegrmsueduclassesme45139choi2008 Laboratory website httpwwwegrmsueduclassesme451radclifflab Required Text Feedback Control Systems C L Phillips and R D Harbor Prentice Hall 4th edition 2000 Fall 2008 3 Main components of the course Lectures about 40 lectures Midterm1 October 3rd Friday in class Midterm2 Final Final exam period Laboratory work Grading Homework 15 Exam 1 15 Exam 2 15 Final Exam comprehensive 30 Laboratory work 25 Homework will be due in one week from the day it is assigned Fall 2008 Tips to pass this course Come to the lectures as many times as you can Print out and bring lecture slides to the lecture Do Exercises given at the end of each lecture Do homework every week Read the textbook and the slides Make use of instructor s office hours If you want to get a very good grade Read the textbook thoroughly Read optional references too Do more than given Exercises Use and be familiar with Matlab Fall 2008 What is Control Make some object called system orplant behave as we desire Imagine control around you Room temperature control Carbicycle driving Voice volume control Control move the position ofthe pointer Cruise control or speed control Process control etc Fall 2008 What is Control Systems Why do we need control systems Convenient room temperature control laundry machine Dangerous hotcold places space bomb removal Impossible for human nanometer scale precision positioning work inside the small space that human cannot enter They exist in nature human body temperature control Lower cost high efficiency factory automation etc Many examples of control systems around us Fall 2008 OpenLoop Control Openloop Control System Toaster microwave oven shooting a basketball Signal Input input output yd Controller u y Actuator Calibration is the key Can be sensitive to disturbances Fall 2008 Example Toaster A toastertoasts bread by setting timer Setting of timer Toasted bread Objective make bread golden browned and crisp A toaster does not measure the color of bread during the toasting process For a fixed setting in winter the toast can be white and in summer the toast can be black Calibration A toaster would be more expensive with sensors to measure the color and actuators to adjust the timer based on the measured color Fall 2008 9 Example Laundry machine A laundry machine washes clothes by setting a program Program setting Washed clothes A laundry machine does not measure how clean the clothes become Control without measuring devices sensors are called openloop control Fall 2008 10 ClosedLoop Feedback Control Compare actual behavior with desired behavior Make corrections based on the error The sensor and the actuator are key elements of a feedback loop Design control algorithm Signal Input yd Fall 2008 Error output 6 Controller l I I Sensor I Ex Automobile direction control Attempts to change the direction of the automobile Desired Error Steering I I direction I Direction Manual closedloop feedback control Although the controlled system is Automobile the input and the output of the system can be different depending on control objectives Fall 2008 12 Ex Automobile cruise control Attempts to maintain the speed of the automobile Disturbance Desired Elmquot speed Cruise control can be both manual and automatic Note the similarity ofthe diagram above to the diagram in the previous slide Fall 2008 13 Basic elements in feedback control systems Error Disturbance Reference 1 mput Output Sensor Control system design objective To design a controller st the output follows the reference in a satisfactory manner even in the face of disturbances Fall 2008 14 Systematic controller design process 39 turbance Reference Controller 4 Implemena ion 1 Mo eling controller I Mathematical model 2 Analysis 3 DeSIgn Fall 2008 15 Goals of this course To learn basics of feedback control systems Modeling as a transfer function and a block diagram Laplace transform Mathematics Mechanical electrical electromechanical systems Analysis Step response frequency response Stability Routh HunNitz criterion Nyquist criterion Design Root locus technique frequency response technique PID control leadlag compensator Theonj simulation with Matlab practice in laboratories Fall 2008 16 ME451 Control Systems Lecture 3 Solution to ODEs via Laplace transform Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design Laplace transform Transfer function Models for systems electrical mechanical electromechanical Linearization El Time response Transient Steady state Frequency response Bode plot Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples t Matab simulations amp laboratories FaH2008 Laplace transform review One of most important math tools in the course Definition For a function ft ftO for tltO 5 complex variable W U gtt FS We denote Laplace transform of ft by Fs FaH2008 3 An advantage of Laplace transform We can transform an ordinary differential equation ODE into an algebraic equation AE t domain sdomain I 51 Partial fraction Solution to ODE lt FaH2008 4 Example 1 cont d unknowns 2 Partial fraction expan3ion J Similarly FaH2008 Example 1 cont d 3 Inverse Laplace transform B C u s l s 2 r A Lil iYCS Z gt If we are interested in only the nal value of yt apply Final Value Theorem s2 s5 5 lm yo I39m8Y8 WI M EH00 s gto S gtU ts i LMs i Z z FaH2008 ME451 Control Systems Lecture 19 Root locus Multiple parameter design Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems electrical mechanical Welectromechanical Block diagrams El I Time response I Transient W Steady state Frequency response Bode plot W Stability RouthHurwitz Nyquist Design specs Root locus El Frequency domain PID amp Leadlag Design examples W Linearization t Matab simulations amp laboratories FaH2008 What is Root Locus Review Consider a feedback system that has one parameter gain KgtO to be designed I Ls openloop TF Root locus graphically shows how poles of CL system varies as K varies from O to infinity Today multiple design parameters FaH2008 3 Example 1 II I Ys Kt s a Set Kt0 Draw root locus for KgtO b Set K10 Draw root locus for Ktgt0 c Set K5 Draw root locus for Ktgt0 FaH2008 4 Example 1 b 8 lm L 8 8358210 k x 017 l 136339 535 y gt 0 Re 0 By increasing Kt we can stabilize the CL system FaH2008 x017 136j Finding Kt for marginal stability Characteristic equation Kts Routh array 53 l1 When Kt2 O ltgt 53552Kts10 O Stability condition Kt gt 2 532IO02gtsi j FaH2008 Matlab command rlocusm Root Locus num 1 0 15 iiii rigv quot igi gogslwqIi if I gt12 den1 5 O 10 systf numden 10 I rlocussys 5 quot grid on Imagina Axis o Damping ratio 5f If K10 we cannot achieve C 08 40 1 quot1j31quot80 570Q4777quot I I for any Ktgt0 15 177 I A I I4 6 5 4 3 2 1 0 1 Real Axis FaH2008 9 Example 1 c K5 38gt 1 Characterlstlc eq 15 52545 I O K s K1 wit H 42 I L39 I 1 A rzn H n 5 A 5 5 3 I 5 3 U l X 444444444444 2 U KT T 75739 quotiquot ts35825 W FaH2008 10 Example 1 0 Root locus plot ImagimryAaxis amp 4 RealAaXis Fall 2008 11 Example 2 1Ts Ms a Set TO Draw root locus for KgtO 1 18 88 ls 2 b Vary T to see the effect of a zero on root locus FaH2008 12 Examle 2 a O O O O Root locus for 39m M 1 j s ss1s2 K 6 er Breakaway point 1 3 j K k K 6 FaH2008 3 13 Examle 2 When K is fixed and T is a positive parameter the characteristic equation can be written as 1 I Ts O ss 1s 2 H ss1s 2I K TK8O H v Term without T Term with T I 1 3939flj 39lK S 88182K Fall 2008 14 1 I K ME451 Control Systems Lecture 20 Root locus Lead compensator design Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems electrical mechanical Welectromechanical Block diagrams El I Time response I Transient W Steady state Frequency response Bode plot W Stability RouthHurwitz Nyquist Design specs Root locus El Frequency domain D PID amp Leadlag Design examples W Linearization t Matab simulations amp laboratories FaH2008 Closedloop design by root locus Designable Fixed am Controller Plant Place closedloop poles at desired location by tuning the gain CsK fortime domain specs If root locus does not pass the desired location then reshape the root locus by adding poleszeros to Cs How Compensation FaH2008 3 General effect of addition of poles Pulling root locus to the RIGHT Less stable Slow down the settling lm lm lm V V Add a pole Add a pole FaH2008 4 General effect of addition of zeros Pulling root locus to the LEFT 39m More stable Speed up the settling Re lm lm lm 9 Re 9 Re 9 Re FaH2008 5 Some remarks Adding only zero 03 3 z z gt O often problematic because such controller amplifies the highfrequency noise Adding only pole 08 18 10 P gt 0 often problematic because such controller generates a less stable system by moving the closedloop poles to the right These facts can be explained by using frequency response analysis Add both zero and pole FaH2008 6 Lead and lag compensators E l Controller Plant 5 z 0sK ltzgtopgtogt 8 T P Lead compensator Lag compensator lm Re Why these are called lead and lag We will see that from frequency response in this class FaH2008 7 Lead compensator Positive angle contribution Test point quot1 S ZOLead8 2 ZS FaH2008 8 Lag compensator Negative angle contribution Test point Im r e F r lt r3 eVLagvu VLag V s 2 chago 1 as 22 as 202 FaH2008 Roles of lead and lag compensators Lead compensator Today Improve transient response OLead5 K1 Improve stability sl zl 8P1 Lag compensator Next Reduce steady state error 822 C s 2K Lag 2Sp2 Leadlag compensator Next Take into account all the above issues CLL5 OLead5CLag5 FaH2008 10 aar trackin system Radar receiver Amplifier gt6 Plant 9A 9 POWfer I E Motor pedestal 63 amplifier and so forth T b a Power amplifier Plant 943 K 4 933 88 2 i m 11 Figure 71 Radar tracking system Lea cmensatr esin Consider a system Desired pole Analysis of CL system for Cs1 lm Damping ratio O5 Undamped natural freq mn2 rads 39 Performance specification Damping ratio QO5 Re I CL pole Undamped natural freq on 4 rads with Cs1 Fall 2008 12 Angle and magnitude conditions review A point s to be on root locus 69 it satisfies Angle condition Odd number ZLs18OO gtlt 2k 1 k 0 i1 i2 For a point on root locus gain K is obtained by Magnitude condition 1 L s lt l K FaH2008 13 Lead compensator design cont d Evaluate Gs at the desired pole 4 1 0 22 z 2 2mm 2 3 a o If angle condition is satisfied Desired pole lm compute the corresponding K A o In this example ZG 2 2 j 210 Angle condition is not satisfied 2 Angle deficiency 30 FaH2008 14 Lead compensator design cont d To compensate angle deficiency design a lead compensator Cs CsKSZ 81 satisfying Desired pole I 10 2 2W3 30 gb A gt ZGC 2 2 j 180 There are many ways to design such Cs FaH2008 15 Lead compensator Positive angle contribution Test point lm s QLead 9p lt2 0 gtRe Triangle quotp1 3921 6P QLead 77 92 77 ZCLead g QLead gt O FaH2008 16 How to select pole and zero Draw horizontal line PA Desired pole Im Draw line PO A P Draw biseotor PB 2 37 ll K 1 APB ZBPO EKAPO Draw PC and PD 4 B f 2 D ZCPB ZBPD p54 z29 Pole and zero of Cs are shown in the figure Re O Mia FaH2008 17 Comparison of root locus Gs GSCS Improved stability FaH2008 18 How to design the gain K Lead compensator 08 K8 i 9 s 54 Open loop transfer function GsC s B 48 2399 33 2s 54 Magnitude condition i Ki I V Ir 1 K4675 SKSl Ksro4s22 j FaH2008 19 Comparison of step responses Compensated system Uncompensated system Cs1 FaH2008 20 ME451 Control Systems Lecture 16 Root locus Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function W Models for systems Yelectrical mechanical electromechanical Block diagrams I Time response 9 Transient W Steady state Egt Frequency response Bode plot istability VRouthHurwitz Nyquist Design specs Root locus El Frequency domain PID amp Leadlag Design examples 7 Linearization f Matab simulations amp laboratories FaH2008 Lecture plan L16 Root locus sketching algorithm L17 Root locus examples L18 Root locus proofs L19 Root locus control examples L20 Root locus influence of zero and pole L21 Root locus lead lag controller design FaH2008 What is Root Locus W R Evans developed in 1948 Pole location of the feedback system characterizes stability and transient properties Consider a feedback system that has one parameter gain KgtO to be designed gt Ls openloop TF Root locus graphically shows how poles of CL system varies as K varies from O to infinity FaH2008 A simle examle MS 88 1 2 1 33 2 Closedloop poles gt 8228KO gt s 1i1 K KOs02 A I K1I S11 2 1 O gt Re Kgt1 complex numbers Characteristic eq 1 K lm FaH2008 5 A mre cmlicate examle s 1 88 2s l 3 gt ssI 2sl 3I K3l 1 0 gt S It is hard to solve this analytically for each K Is there some way to sketch roughly root locus by hand In Matlab use command rlocusm Characteristic eq 1 K O FaH2008 6 Root locus Step 0 Root locus is symmetric Wrt the real axis The number of branches order of Ls II I Mark poles of L with X and zeros of L with 0 51 M8 35 25 3 FaH2008 7 Root locus Step 1 RL includes all points on real axis to the left of an odd number of real poleszeros RL originates from the poles of L and terminates at the zeros of L including in nity zeros Re Indicate the direction with an arrowhead FaH2008 8 Root locus Step 2 Asymptotes Number of asymptotes relative degree r of L r 1 15 m Angles of asymptotes are deg den deg num Zgtlt 2k1 k01 T r 1 r 2 r 3 r 4 2 7r 7r 5 I A 5 3 4 FaH2008 9 Root locus Step 2 Asymptotes Intersections of asymptotes w s 1 gpoie Zzero OH2H 3k1 2 88 2s 3 Asymptotes Im Not root locus FaH2008 10 Root locus Step 3 Breakaway points are among roots of Points where two or more branches meet and break away 51 dL5 534s 553 L s 2 O 55 2s 3 d8 q s 24656 07672 i 07926i For each candidate 3 check the positivity of K 1 7907 T 0770 T Tle lLl FaH2008 11 Quotient rule KY N D ND D D2 31 J39I 552536 3 13821086 n 2 min I in I o PP T Aquot T 0 FaH2008 12 Root locus Step 3 lm lt Re 0 Breakaway oim 246 K 04186 FaH2008 13 Matlab command rlocusm RootLocus 8 l num1 1 6 L 81 den1560 5 Z 4 sgs2gs5 systfnumden M rlocussys 2 E n g V 3953 a 2 4 6 8 x W x V x 3 2 5 2 1 5 1 0 5 0 Realeis Fall 2008 14 A simple example revisited I 1 L 5 88 2 Asymptotes Relative degree 2 lm Intersection 0 2 1 2 L 472 0 Re Breakaway point LSy gtllt 1 0 gt s 1 FaH2008 15 Summary and exercises Root locus What is root locus How to roughly sketch root locus Sketching root locus relies heavily on experience PRACTICE To accurately draw root locus use Matlab Next more examples Exercises Read Chapter 7 FaH2008 16 a w w 4 g 4762171 26E797043a397l 49 X399 3 4M quotWav w zfr 734357 ow lj 1715 WWW W 77 a m 9 FA m a EUDGE a 1L1 mom I I ME451 Control Systems Lecture 5 Modeling of mechanical systems Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design J Laplace transform f Transfer function I if Models for systems W electrical mechanical electromechanical Block diagrams Linearization Time response Transient Steady state Frequency response Bode plot Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples t Matab simulations amp laboratories FaH2008 Timeinvariant amp timevarying A system is called timeinvariant timevarying if system parameters do not do change in time Example Mx tft amp Mtx tft For timeinvariant systems NW Nye Jmeshift JmeLshift ys ut 7 T T to tOT totoT This course deals with timeinvariant systems FaH2008 3 Newton s laws of motion lSt law A particle remains at rest or continues to move in a straight line with a constant velocity ifthere is no unbalancing force acting on it 2nd law 2 39 ZFZ OE mitt translational 2 39 27105 1 rotational 3rd law For every action has an equal and opposite reaction FaH2008 4 Translational mechanical elements constitutive equations Mass Spring Damper flttgt ft l FaH2008 5 Massspringdamper system FaH2008 6 Free body diagram 4444 fkt flit 1 I w2to 3 i to K J B 2 I I I I I ll I I I I ll Newton s law Fma FaH2008 Direction of actual force will be automatically determined by the relative values 1340 fbt W W Massspringdamper system f t Equation of motion at By Laplace transform with zero initial conditions FaH2008 2nd order system 8 Gravity I5 2 lil At rest y coordinate x coordinate i V FaH2008 5 m Ifsk36 El 1mg ZFima Automobile suspension system automobile suspension wheel I 11 tire fa w I M1 I t K I FaH2008 Automobile suspension system I Laplace transform with zero le Block diagram F X2 X1 E FaH2008 11 Rotational mechanical elements constitutive equations Moment of inertia Rotational spring Friction torque 7t 775 K 7t 70 3 t 61t 6 t 62 91t 9205 rotation angle 7t J9 t Ta mm 9205 7t BMW 650 elt gt 1 1 933 1 llgigtgl Te 19298 W K1s 928 w Bs1s 928 FaH2008 12 Torsional pendulum system Ex212 friction between bob and air FaH2008 13 Torsional pendulum system 4 322221 Kl friction between bob and air J B 9t 7t Equation of Motion By Laplace transform with zero ICs 2nd order system FaH2008 14 Example Tm K MOTOR 5 3 LOAD I I JL JmIIBm 9m 9L By Newton s law By Laplace transform with zero ICs FaH2008 15 Example cont d From second equation 2nd order system From first equation 4 order system Blockdiagram TMS Ll IeMSh I ed I 62 I I G1 I FaH2008 16 G 6 Th rustor Broadcasting 7t 2 J90 Weather forecast Communication 98 i DOUbe ePs etc m J32 integrator FaH2008 17 Summary amp Exercises Modeling of mechanical systems Translational Rotational Next block diagrams Exercises Read Sections 25 26 Derive equations for the automobile suspension problem FaH2008 18 gtSTEP4 COW9W 1 WC WWQ M gymMW As lt 00 g6 Q0 VaQas 4FL a X WL POM 9 GM Zerog CQL CvngrTF 71 0923 9 P 39m quot1 yw k z bw so U126 Tes r POW So 4441 3143 1 7 V A I M J izf lx 4 70 5 i 1 21 a 5 31 00 K So u fi PAM WI dam jig v A 2 4 46 92 4 1 quot 430 4 M 450492 M 45 Z inge 6aHD x L ateP qoa ilXQo 4ampW tmampyw2250 291 file 2 225 1 63 Edi6 A g u 9 Angle 01 Olaaaere n 4 Copr bait 2 Z 35670 OCZQH z gum af LMi 2L 70 V691 0 A oowaLX pok n quot 557LI5 mm WV 70 euw 91 Many kWer 4 aww zx p700 m qwzyvasho k 26 Anal M VO K 76 0V mm Zen 2 15m 9 ZOOMHO 2 1091 VanLvI A 0 1 4M W am f M71272 Jr eogt 40 73 mm 701207 H9174 WK QTQPS Ehd szmm 43 fm etaVia la N S use a 2001 ywm x y awemm 6 w w 2 lt ex 4 FEW i lt WAF 5 USN 7le 5237413 54 32 quot397 934 zs 3Q5Kso 3 L 3 Q 0 2 O 5 8 lt r 1 31 5 v I 0 5 L514 hngme 5 K O CVFFFMZ Valid ZgGKio C bl Cum zH b I M 00 W 3095 91 V ME451 Control Systems Lecture 23 Bode diagram of simple systems Dr Jongeun Choi Department of Mechanical Engineering Michigan State University FaH2008 Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems electrical mechanical Welectromechanical Block diagrams I Time response I Transient r Steady state DVFrequency response Bode plot W Stability RouthHurwitz Nyquist El Design specs 39Root locus Frequency domain W PID amp Leadlag Design examples W Linearization t Matab simulations amp laboratories FaH2008 Frequency response review Steady state output ysst A lGjoJl Sintwt GU12 Frequency is same as the input frequency w Amplitude is that ofinput A multiplied by Gjw Phase shifts away 6339 y 75 Stable y lquot quot 88 35 1 4 Frequency response function FRF Gjoa Bode plot Graphical representation of G003 l U05 2 Asinwt FaH2008 Bode plot of Gjco review Bode lot consists of gain plot amp phase plot 20nnA1n39yl mm V 39VUlU UJWl U 10 1 01 1 101oo 39w ZGjw deg 98 o11 10 100 39w l8O Logscale FaH2008 Sketching Bode plot Basic functions Today Constant gain Differentiator and integrator Double integrator First order system and its inverse Second order system Time delay Product of basic functions Next lecture 1 Sketch Bode plot of each factor and 2 Add the Bode plots graphically Main advantage of Bode plot FaH2008 5 Bode plot of a constant gain FaH2008 6 Bode plot of a differentiator FaH2008 Bode plot of an integrator 1 Mirror image of the bode plot of 15 with respect to oaaXiS FaH2008 d CZIC 5 5w r n 0 a 7 A4452 5 0a 0 1 rZaN 39 0 14 IVA 61917 M497 7091M KW 7N J x 7 00 675 8 N M45 3 20103 Ajin w d 5 w N 4 739 quot V 4 JV 40 A l 5 Paw 4 3 30m 0 Ill 1 as l a 3 w m3 m ARAS 5 gm a e m 3 Q HQ E XQ K a Qw x gm Nan U A3mvwxmo u i k wa me u LNQ H xmxww v v 0339 Am a3 if if 6 5 N g 2 p SHACH E ganggg mm V9 24 1165 iizalj Ka 39kfo Goof a v 7 514436 Wm rm ML it Pv u 97c Lg m Mfg215 7 9 La 39 Yr R g Q m K Fmvvh In wd K PRQ H mo mLmdb LCD 2 9 5 r Pv ragga 39W a r a yet 5 93 9T1 915 7 Kquot FIPD39 k a at V 6 s 3 921 9 9p 13 L15 r quot Ya quot Vinr1 ME451 Control Systems Lecture 9 Stability Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems Yelectrical mechanical electromechanical J Block diagrams El Time response Transient Steady state Frequency response Bode plot Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples W Linearization f Matab simulations amp laboratories FaH2008 Simple mechanical examples We want mass to stay at x0 but wind gave some initial speed Ft0 What will happen ft K ft Xsi ME E 1 Fs s2 Fs 52K Fmn Lmn K B 1 E 1 E 139 Fs 32Bs Fs52BsK rgtmo rgtMo How to characterize different behaviors with TF FaH2008 3 Stability Utmost important specification in control design Unstable systems have to be stabilized by feedback Unstable closedloop systems are useless What happens if a system is unstable may hit mechanicalelectrical stops saturation may break down or burn out FaH2008 4 What happens if a system is unstable Tacoma Narrows Bridge July 1Nov7 1940 Fall 2008 Mathematical definitions of stability BIBO Boundedlnput BoundedOutput stability Any bounded input generates a bounded output le0 t ut Y BIBO stable I Asymptotic stability Any le generates yt converging to zero ICS yt Asymp stable 00 system Fall 2008 Sme terminlies 8 728 Ex s 1sI 1 6 s Gs s 2s2 1 Zero roots of ns Zeros of a i1 Pole roots of ds Poles of G 2 ij Characteristic polynomial ds Characteristic equation ds0 FaH2008 7 Stability condition in sdomain Proof omitted and not required For a system represented by a transfer function Gs system is BIBO stable All the poles of Gs are in the open left half of the complex plane system is asymptotically stable FaHZOOB 8 Idea of stability condition HA I M A MIA mrm Example 9 w T Uyw aw yw yo nVn m l nVn rrn 1 Vn TTn J mm J O U o I ywu s or 71 all 1 l a i Asvm Stability git z YlsljFlC a i grow 0 110 we Rem gt0 Us0 lsio 1 quot 1 1 lH iff or BIBO Stabllltyi yl Ys yTlllTdfl e ulti Tu r y00 to V0 5 ft WW 3 lewlllult Tld73 leiwldi umaz 0 0 Bounded if Reocgt0 Fall2008 9 Remarks on stability For a general system nonlinear etc BIBO stability condition and asymptotic stability condition are different For linear timeinvariant LTI systems to which we can use Laplace transform and we can obtain a transfer function the conditions happen to be the same In this course we are interested in only LTl systems we use simply stable to mean both BIBO and asymptotic stability Fall 2008 10 Remarks on stability cont d Marginally stable if Gs has no pole in the open RHP Right Half Plane amp Gs has at least one simple pole on jwaXiS amp Gs has no multiple poles on jwaxis N x l GS 98 sls 4tsi 35 4 31 Marginally stable NO T marginally stable Unstable if a system is neither stable nor marginally stable Fall 2008 Examples Repeated poles 2s 39 Vo TF cw521 gt 1 Fle ut smt 1 Us 82 l I 1 71 l xrl n r r 23 1 L 1t5U51Ut59 I HOEgti E L we n im Fall 2008 Stability summary Let Si be poles of G Wm Then G is Stable lingmhlc BIBO asymptotically stable if gm 8 0quot ReSilt0 for all i I marginally stable if ReSilt0for all i and 0 0 simple root for ReSi0 I unstable if it is neither stable nor marginally stable Fail mm is Mechanical examples revisited Eli Xs1 K it Xlt 1 1quot3 is l t K X0 Poles X Poles stable stable K B m 1 KO w 1 EIquot Fit sz Bs quot m39 imm B xtt p es 39 x Poles stable stable Fail 2mm 14 gode FM WW ZQWoMe 3 Q 39K gmw i gage N 222mm y mm MIL 476 WfLawwW 4351mmst fm mf f fmlc I74 a a fhm 5a M 59 in rheaIx mlu 9 3 92 g 5399 he V26 7960 371 27 3 A8 79 f 9 5 997 r JiVZeJZGQJ3 39 gt Y Y 5 9Z39 glrgtg 6L3 6 2 lt9 1 VB We r W et 2 imam W I HILoLL in cwrm ar 1 7 Va 3 030 1 G970 Cato V E K7an Y2 Floafoh r JoaloVLw o 76 z quotW F V Fr v4 i f 9 Iquot A z E 95 T T V 3 Imammmm t39 We MUM 04 ower ax M Maids govt73905 oweV MW 10 a of f Plums a PI in WM iv DWIwt ameV pt dv 16 I m C 0 R T B O 310 V V1 9 Mi 399 6169 V 261001 SM 1 7 679742 v Growl G 2 J I 4 8 Zoba o V l 2 20 loam l 9140 Zoioalo l tgw W fl jg43956 K 29 QKS 6 QS39F st 51 Zgwmsjawm cam 079 Mm lt 2 5 tr S L 4 61Z 39g77 53 3 wh Q Cd ZISM lL gm r A 79 lt d8 IGQ W d c 2733 lt I 1147 1 gig 0 k70 N w ltltl 4 675 5 gm 0H Qlodydame M Ma 3 a r 39M QOIa D c 900620 0 1L U7 y 46797 ij 610 L ax quotquot31 C75 S i M F or d5 gala Om 0 03 w 4792 4600 z 6100 ME451 Control Systems Lecture 12 Timedomain specifications Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems electrical mechanical electromechanical W Block diagrams Time response Transient Steady state Egt Frequency response Bode plot W Stability VRouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples W Linearization t Matab simulations amp laboratories FaH2008 What we did and what we do next We have learned stability Definition in time domain Condition in sdomain RouthHunNitz criterion to check the condition Stability is a necessary requirement but not sufficient in most control problems Specifications other than stability How to evaluate a system quantitatively in time domain How to give specifications in time domain What are the corresponding conditions in sdomain FaH2008 Time response We would like to analyze a system property by applying a test input rt and observing a time response yt Time response is divided as W ytlttgt w Transient response Steadystate response tlim ytt 0 after yt dies out gtOO FaH2008 4 ME451 Control Systems Lecture 4 Modeling of electrical systems Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 5 Laplace transform Transfer function Models for systems electrical mechanical electromechanical Block diagrams Linearization Time response Transient Steady state Frequency response Bode plot Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples t Matab simulations amp laboratories FaH2008 4 Implemenalon Controller 3 Design 2 Analysis What is the mathematical model Transferfunction Modeling ofelectrical circuits FaH2008 3 Mathematical model Representation of the inputoutput signal relation of a physical system Physical Ot t In t upu pu system 1 Modeling A model is used for the analysis and design of control systems FaH2008 4 Important remarks on models Modeling is the most important and difficult task in control system design No mathematical model exactly represents a physical system Math model Physical system Math model z Physical system Do not confuse models with physical systems In this course we may use the term system to mean a mathematical model FaH2008 Transfer function A transfer function is defined by 8 Gltsgt EESL ms n5 A system is assumed to be at rest Zero initial condition FaH2008 Impulse response Suppose that ut is the unit impulse function and system is at rest ut so tem 9t US 1 The output gt for the unit impulse input is called impulse response Since Us1 the transfer function can also be defined as the Laplace transform of impulse response Gs E 905 FaH2008 7 Models of electrical elements constitutive equations Resistance Inductance Capacitance W W W VtA R vtA L vtA 1quot C ut Rz t ut 13 W Cal 5 Laplace transform 0 1000 2 0 FaH2008 8 Impedance Generalized resistance to a sinusoidal alternating current AC ls V Zs VsZsls Element Time domain Impedance Zs Resistance Inductance Capacitance Fall2008 Memorize 9 Kirchhoff s Voltage Law KVL The algebraic sum of voltage drops around any loop is 0 Fall 2008 10 Kirchhoff s Current Law KCL The algebraic sum of currents into any junction is zero FaH2008 11 Impedance computation Series connection I V1S I V2S 39 I I I N A Cl v N ll A C v N h A Cl v 5 A v H A U v A I N A U v VS Proof Ohm s law V518 Z2109 FaH2008 12 Impedance computation Parallel connection 13 28M 13 218 228 quotquot I Proof Ohm s law I2S Vs Vs 21118 I KCL Modeling example it 1 Input v1t R2 v2t Output C Kirchhoff voltage law with zero initial conditions By Laplace transform Fall 2008 14 Modeling example cont d it 1 Input v1t R2 v2t Output C Transfer function rstorder system FaH2008 15 Example Modeling of op amp MS Rule1 i390 Rule2 Vd0 L Impedance Zs VsZss Transfer function of the above op amp FaH2008 16 Modeling example op amp R2 C i0 Vd0 Kt R1 Input Vit I Iwa Output L By the formula in previous two pages firstorder system FaH2008 17 Modeling exercise op amp 1 2 AAA 39IV R2 0 R1 I Vd0 le Input vt I IVoG Output Find the transfer function I FaH2008 18 More exercises in the textbook I Find a transfer function from v1 to v2 Rt Fm Li C r t t l t l mu a villi quot 7 I Find a transfer function from Vito Vo R E E A 77 Ii quot 39 in l i L m w I Fall ZEIEIE lB Summary amp Exercises I Modeling I Modeling is an important task I Mathematical model I Transferfunction I Modeling of electrical systems I Next modeling of mechanical systems I Exercises I Read Sections 22 23 I Solve problems 21 22 and 23 in page 62 Fall ZEIEIE 2D ME451 Control Systems Lecture 22 Frequency response Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function I Models for systems Yelectrical mechanical Welectromechanical Block diagrams I Time response I Transient W Steady state Frequency response Bode plot W Stability RouthHurwitz Nyquist Design specs W Root locus Frequency domain W PID amp Leadlag Design examples W Linearization f Matab simulations amp laboratories FaH2008 An examle ont 1 TF RC1 39 ulttgt 1 06 yt g 1 14 utsint 039 412a Cs o4 416 418 1 A I I I I O 5 1O 15 20 25 3O 35 4O 45 50 At steadystate ut and yt has same frequency but different amplitude and phase FaH2008 5 An examle ont Derivation of yt 1 1 1 1 s1 ysGSUSs1s2I1 2lt81 321 Inverse Laplace Partial fraction expansion yt 16 7 cost sin 15 2 O as t goes to infinity 31880 cost I sint sint 450 Derivation for general Gs is given at the end of lecture slide FaH2008 6 esnse t sinusial inut How is the steady state output of a linear system when the input is sinusoidal yt A 93805 M I I I gt mt A Sm wt GS Steady state output yss A GJw Si wt 13000 Frequency is same as the input frequency w Amplitude is that of input A multiplied by Gjw Phase shifts 4Gjw Gain FaHZOOB 7 Freuency resnse functin For a stable system Gs G000 00 is positive is called frequency response function FRF FRF is a complex number and thus has an amplitude and a phase First order example In 1 1 1 W GSs1 9 1 5 Gjw 1 2 1 Re 10W 11 1m 1 tan1 w FaH2008 8 Another example of FRF Second order system FaH2008 First order example revisited 1 FRF IUKJWN LUUW O 1 O0 05 0894 266 10 0707 7450 oo 0 90 Two graphs representing FRF Bode diagram Bode plot Today Nyquist diagram Nyquist plot FaH2008 0 W4 0 Mpg 65ch g lt7 Ky 00 M 50 5 65512065 9 67 0 Jquot 2 a 5 E Ka 2ltA 1 6 C y 5amp0 002 VR39 2t m k w Na a Egg X8 E1 h m 5 Wivz 39 C 5 M S TJQXm was a g IM3i RC5 51 l 5 L W 83 6 m 9 T l w H 3 tar 3 X jwm 9 1 1 0 590 2 14 00 2 k 1 32 62 65 5 x5 5 6 52 2quot 3 9 raw 1 3 540 Iggy M j n kan wg Mirgig 55 50 gazi gigif A SWW 5 1136 g wk E 4 age WM Ob p W 9pr y TLIM M7 17 I 4 quot y4 gm 395 I 5 xix5 1 75 353 T 1 quot95 MS f gx39f wi 643 W3 W J 050 P amp Za 2 ww TC 5 i Ci wmfeml ifz b 65 i 32gt quotquot gym539v i K m 70 a Tquot 5quot 5 1 77 6 pmcmmL b S Fi 0W quotgt3 YR 7733 7973 m NC 3 quot n 3 37 s 30 W C S Pg 3 quotquot 1C 5C2 Zc M 5 r 678 5 If fig 2 Saw Y C 3quot viii 2 J gf Peg4mg g fa Jew 35 Mm Wavix39 a 404 109252 mg 2255 i wa e ASSUMG AM HM W m v 4 0 WPgai apk Veggm ww 0 f 907 50 i fg c lt x a k Pzt Wb M WWWu M 27 abMi 78W Waf e W F Vexaw 6F L165 4 21 V 4me Du C M 711 T 5 7g tap5amp5 a F gt e g A e P gum quotWm m x 392 quot r K L if f xiii 7 quot 9 w a W V 15 10556 f ziijjb fawdcizz wa Ma WW Kg L2 27 jquot mam 7 51 6 5 41 3 2 rf 1 w W265 Iii3 5 5 g quotWWquot 5gt L 3 W W A W f f frf ic a mg a5 saggy V Vi 959 11m 5 EX1 VW E EE pa ex 51 z magwe 7553 7 er v 393 quot r 6 AL 2 LP Q A 4 1 a bf NJ 3 EA g 5 94 71fo 95 6 V U W g i 2amp25 z E 21 w 3 5 s a lt j W wt 3 5amp5 x J A Wg j j 3 wf g a E 3 if Eva23 3 394 g hi 5 1 2 5 3lt E3 Qg39 g a W33 1 k 4 gray a an 5 3 1 y M3 gm my mg x M x Q j x l 5va V5 2 Vs Chg f R m 3 A saw g a d 6 1 MW 55 F j EHV25 a 9 a ig i 6amp5 853 3113 59 3 N W a a s K 41 23 lt 953 53va 545 25 fig 3 3M 5quot w K 1 m 7 2 WW 3 3M E O M a x I f5 w x 5 fi f 2 quot17 I g g 25 iP x 5 3 f if E323 W3 r 2 if 5 M m K51 a 5 x 1 E s M 2 j 5 3 39WW 7 23 s m if 5aquot i9 2 a 35quotM M M N m We My 7 4M in w 9ka f quot 3 g 3quot 5 M e W 1 3 9 2 2 5 M m w P 3 W 39 M Cisa 39xfx 2351 g r w a a s Q Q m N SM w v Mi 1 s V 3 w v y lt j quot 392 392 a 9quot ltf w J M Wi r W quot f C g m ZP 3 E5 six5m 6 3 5 3 4 5 f i I I 43 2amp2 L 59 a f M 3 W xing 5quot WW 3 lt4 6 g W 5 x P fquot M WW y w 2 my p a few mi 4 M W M m w V535 w 1 ow W m 4 3 2 a f 2 E w w m a quotin is to assay condition for H stability easessary and sufficient for stability in array Routh s Stability Criterion There are several methods of obtaining information about the locations of the roots of a polynomial without actually solving for the roots These methods were especially useful before the availability of CACSD software They are still useful for determining the ranges of coef cients of polynomials for stability especially when the coef cients are in symbolic nonnumerical form Consider the characteristic equation of an nthorder system8 assquot als 1 lazs 2 alsal 468 It is possible to make certain statements about the stability of the system without actually solving for the roots of the polynomial This is a classical problem and several methods exist for the solution A necessary condition for stability of the system is that all of the roots of Eq 468 have negative real parts which in turn requires that all the ai be positive9 If any of the coefficients are missing are zero or are negative then the system will have poles located outside the LHP This condition can be checked by inspection Once the elementary necessary conditions have been satis ed we need a more powerful test Equivalent tests were independently proposed by Routh in 1874 and Hurwitz in 1895 we will discuss the former version Routh s formulation requires the computation of a triangular array that is a function of the 0 He showed that a necessary and suf cient condition for stability is that all of the elements in the rst column of this array be positive To determine the Routh array we rst arrange the coef cients of the characteristic polynomial in two rows beginning with the rst and second coef cients and followed by the evennumbered and oddnumbered coef cients 3quot laz a4 nl S 01 a3 a5 8 Without loss of generality we can assume the polynomial to be monic that is the coef cient of the highest power of s is l 9 This is easy to see if we construct the polynomial as a product of rst and secondorder factors We then add subsequent rows to complete the Routh array Row 72 squot 1 a2 a4 Row n l 5quot a1 a3 a5 Row n 2 5quot b1 b2 173 Row 11 3 3quot 3 c1 c2 c3 Row 2 2 Row 1 s Row 0 so We compute the elements from the n 2th and n 3th rows as follows 1 a det 2 a1 as 102 03 b1 w 01 01 det 1 a4 b D m a1 5 a1a4 05 2 a quot 7 a a1 det 1 as a1 a7 alaa a7 b3 3 W w 11 a a a det 1 3 C 2 b1 b2 b1a3alb2 1 b1 121 a a det 1 5 c 3 legs xbxas albs 2 b1 b1 a a det 1 7 c b1 174 17107 0le 3 11 b1 Note that the elements of the n 2th row and the rows beneath it are formed from the two previous rows using determinants with the two elements in the rst column and other elements from successive columns Normally these are n 1 elements in the rst column when the array terminates If these are all positive then all the roots of the characteristic polynomial are in the LHP However if the elements of the rst column are not all positive then the number of roots in the RHP equals the number of sign changes in the column A pattern of is counted as two sign changes one change from to and another from to 629 5 3g3 C Si3quot 993 9 35 535 QS 3511 CisTro 3 3 2 S eqtzgs MO WP 577 M07 2 g f minis y y 762 9396 Mmsl F 45 klxl 9c3 I K AlXI mm 1 PH 4 9 pk 1 B Oh92 Mm 2 961 39 kX1rx 9TEF MAIk I7065 01 wa j m wryquot O lrf l N we V4902 axfs povl39 m Pmla W 120 cm Veal T6635 AXIS oJrN 6A oc V ahmm Ar 26403 lpemmmc Wk a59W 9655 5 2lt 9 00 yum a 3 2m m 1a madnm Wyn W WO Z In M Z W P Z 6 x 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32 3 739 974m As gfg39 9quot1 M 5 xaioovfg 7 1 Oq323 59 answer ME451 Control Systems Lecture 18 Root locus Sketch of proofs Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems electrical mechanical Welectromechanical Block diagrams El I Time response I Transient W Steady state Frequency response Bode plot W Stability RouthHurwitz Nyquist Design specs Root locus El Frequency domain PID amp Leadlag Design examples W Linearization t Matab simulations amp laboratories FaH2008 What is Root Locus Review Consider a feedback system that has one parameter gain KgtO to be designed Root locus graphically shows how poles of the closedloop system varies as K varies from O to infinity FaH2008 3 Characteristic equation amp root locus Characteristic equation 1 Ls Ls E 1I KL3O HK Root locus is obtained by for a fixed KgtO finding roots of the characteristic equa onand sweeping K over real positive numbers A point s is on the root locus if and only if Ls evaluated for that s is a negative real number FaH2008 4 Anle an manitue cnitins Characteristic eq can be split into two conditions Angle condition Odd number ZL3 1800 x 2k 1 k 0 i1 i2 Magnitude condition 1 For any points L8 E gt this condition holds for some positive K Fall 2008 5 A simle examle L 1 A lm S 53 2 gt Re Select a point s1j Select a point s2j 1 1 18 882 t l 2jj 1Jl17 2 1 2j 1 ZLS 180 s is on root locus ZLS 7amp 180 gt s is NOT on root locus Fall 2008 Ls 6 Root locus Step 0 Root locus is symmetric Wrt the real axis Characteristic equation is an equation with real coef cients Hence ifa complex number is a root its complex conjugate is also a root The number of branches order of Ls If Lsnsds then Ch eq is dsKns0 which has roots as many as the order ofds Mark poles of L With and zeros of L with o 5 Z1 I 39m 0 Re ts pliks ibi P T M5 X 2 X pl 21 FaH2008 7 Root locus Step 11 RL includes all points on real axis to the left of an odd number of real poleszeros lm Test point lr l l 8 L143 LKS le M5 Pll LKS PQl WOa Re P2 pl 21 0 0 0 Not satisfy angle condition lm Llslzls g1ll5n1lllsml 8 1 I ll ru Hoe Re LVi l LY J H4 P2 pl 21 180 0 0 Satisfy angle condition FaH2008 8 Root locus Step 11 cont d RL includes all points on real axis to the left of an odd number of real poleszeros lm Ls Z821829048192 99 x e Re H4 W4 W4 0 Z 1 180 180 Not satisfy angle condition lm 5 Z13882 1lM8101l18l2 X X e gt Re H4 H4 H4 102 101 21 180 180 180 Satisfy angle condition Fall 2008 9 Root locus Step 12 RL originates from the poles of L and terminates at the zeros of L including infinity zeros ns 1 ns 1 I K d8 O gt ds I Kns O ltgt EICKS O Ls K I O K I 00 V f n s ds O 8 O s Poles of Ls s Zeros of Ls Fall 2008 10 Root locus Step 21 Angles of asymptotes are 71 A14 1 A4 XKAlCii KUJ7 r r l 7 2 r 3 r 4 5 7r 7T E I A 5 z 3 4 FaH2008 11 Root locus Step 21 cont d Fora very large 5 138 Osnir i m 5n 8r Ch eq is approximately 1KLsOgt1K0gt3TKn020 gt 57quot Kn0 lt 0 we assume no gt O IgtZST7Tgtlt2kIl k012 Zsgtlt2k l k0l2 FaH2008 12 Root locus Step 22 Intersections of asymptotes Z pole 2 zero 7 Proof for this is omitted and not required in this COUFSG Interested students should read page 363 in the book by Dorf amp Bishop FaH2008 13 Root locus Step 3 dLs Breakaway points are among roots of d8 0 Suppose that sb is a breakaway point d b Kn b O ltgt ltgt sdbdltbgtnibo d b Kn b 0 711 dLs n bdb nbd b d5 85 db2 9 Kb d2b d b W W O FaH2008 14 Root locus Step 4 RL departs from a pole pj with angle of departure 9d 200939 Zi Z Pj Pi 180 75 73773j RL arrives at a zero Zj with angle of arrival 9a 2 19239 Z 29 373 180 z m j No need to memorize these formula FaH2008 15 Root locus Step 4 cont d Sketch of proof for angle of departure 6 39m For s to be on root locus 8 1 due to angle condition 191 1 1 91 6 2 180 e L R Z1 9 1 l5 pll O 92 Z101 Z191 1091 p2 180 FaH2008 16 69 gf QPIB Demmm m asympqLJFz 910 mn o 0 5 lt9 0039 M quot 6i2 Zm5mv ms 2 F Pr sjquot D39 I 5 I 398 5s PM W K B 47747101 agocee zypo eg We 3 WvMWry 0475 REV a LBu ltP 900 W1 5191 3 54 R6 5 Rea W D WS K 1 AG Mquot n Aquot g 9 Hgmri y d Silk ZPL 2ampSWM K 5177 w quot900 MPVW gm Mev 7aih 7P a5 W Fc 39eg K 5 ouwr M CD Suw PM MR 2 d lR 1 71 w cp lt lt1 M 3 a K Cf Swimwear 75 2 lt 5 2W9 3 quotf WqL y ss 3 W 1 7 97 0 L ZVL Wm W Jquot L acmw r M W m 4307 gx vlt JPwa i 31quot sm cg a 9Fi l 51 7 X J7amp6 1 quotWar 4 2L j qu M Wham1amp3 EL 3 C 0 m7 avg q O 3 W3 00X 72 Mask 6 K a farm 4quot Velma W 7613 e 9 M p I Liavgw X5 9 OS O 99 w 07 A J m y 3 m CI f s539 ir 95 mm w Wi m 2de 0M M m r M LMXM M 467 W 94 2 lt5 272 W2 72 2 i A i 5 23 3 399 I v 62 7O Vcs a ve E 2 jwn Q E V WCAXHJW Z 9 09 A9 pMmM 9 7F 1 A owe a Vi Lx1 n M2 c 2lt O 3 8 V w l M 0 2 94 A quot2 W 3704 3 x o 3 Manm3 WV 014 km J 2 e 9 5w W f v x m Gas ow 6W Fo y 6 Elny of g V A 595wlt V Z2112 A5 E9 C ikwfquot S a yeah 6 7 3928 5 jw jo 3 me gt lt l of 3 ng k jw5 F 441quot F SP A W 55 C50 2 M4 7 2 W 90 5 a 16 k2 e e 4 O aw do95 LH Hf p L 3466 7 KB a 4 Kg eazf 1 46 0 g 539 as quot 5C1 ijt lt2 giwq 9453 0 W gogmwn Mk La 1 I7 cf uwc M e 241 566 a fe 7f 5661 1 H s 1 bblm l 9 A AWE 146 Mer Lu LM A 5 My L4 may 4 1 RIII quotF2Ir 13o R2 CII IG L1513 13 I 0 AL 2 AIL I A3 x31 ML f O V3 333 V341 2123 4 39 5 KIKZ E Q212er RBR Ll Rmzb Aa bk 09 0th V 5 W kid 0 If 130016 g F kw 39 TF 2 EMFb 5 0th n 69 25 55 5 C71 9 9 y 39 4391 17 l V W K CS 03 z 5 N 399 F quot l zCS quot Ty 05 a 41739 r245 GD gummnna Juno mm C Pam d lanain 4quot a l E r9 44 at E a 27 ewe E gt L Z w 4 a 67 pawNoi cbmz by W 17C El 5439 E G I gt 7 V07 V917 prM w t Faibwol Comm PM WWW ECS 6 539 5 C 5 K 97 El W5 KCsCs 7751574 f 855 45 55 39thb39VPKcL m l 49C a Eat D 9 c s Q s 265 CSJCCSJJ 7 1 We ij 65 2 3283 C CT 65 5 F LzE 75 lt15 K L1 3 CS R B W 1W 5ng Y Y7 rz 9 lt Macaw 39 Y T J dv L Y Hf 9 Li O zLE lj Y1 a T Log c i Y 411 I f 39 FL s 1 612 TE QIM quot C 67 71 150 ME451 Control Systems Lecture 3 Solution to ODEs via Laplace transform Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design Laplace transform Transfer function Models for systems electrical mechanical electromechanical Block diagrams El Time response Transient Steady state Frequency response Bode plot Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples Linearization t Matab simulations amp laboratories FaH2008 Laplace transform review One of most important math tools in the course Definition For a function ft ftO for tltO 5 complex variable W U gtt FS We denote Laplace transform of ft by Fs FaH2008 3 An advantage of Laplace transform We can transform an ordinary differential equation ODE into an algebraic equation AE t domain sdomain I 51 Partial fraction Solution to ODE lt FaH2008 4 Example 1 cont d unknowns 2 Partial fraction expanSIon J n is 7895 A B I C y lSl z 4 4 ss l 1s l 2 5 39s139s2 Multiply both sides by s amp lets go to zero t B l C l SHSMHOA39Ss1l8ioss2ls0 39gt A swish 0 Similarly D DLHVAl i 11 OTL1Osgt1 3J O sl 2Yssgt2 E FaH2008 I Example 1 cont d 3 Inverse Laplace transform c l39Ylt A B Vquot l 8 5 51 52 Cgt ya I lt 5et E eZtlusa 2 HP 2 K7 B J If we are interested in only the nal value of yt apply Final Value Theorem ii 52 s 5 rY1 I l IA S gt0sis2 iim W iimsYs EH00 5 gtU Fall 2008 MlU I ME451 Control Systems Lecture 13 Steadystate error Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function W Models for systems Yelectrical mechanical Welectromechanical Block diagrams l WTime response Transient Steady state Egt Frequency response Bode plot W Stability VRouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples W Linearization f Matab simulations amp laboratories FaH2008 Performance measures review I Transient response A From next lecture I Peak value I Peak time 39PercenltoverShOOt Next we will connect I Delay line gt these measures Rise time with sdomain I Settling time I Steady state response I Steady state error 47 Today s lecture Fall 2008 3 Steadystate error unity feedback rll ll We assume that the w 39 ltd CL system is stable 4 Unity feedback 1 ill I Suppose that we want output yt to track rt I Error ell rt all I Steadystate error ass 2 lim lim 2 little R t gtoo s gt0 5 gtO Final value theorem Suppose CL system is stable Fall2008 4 1 Error constants Step error position error constant KP Iirn 05 SHO Ramp error velocity error constant Kc Iim 9CR 5H0 Parabolic error acceleration error constant K 132Gs Kp Kv Ka ability to reduce steady state error FaH 2mm Steadystate error for step rt Reference input rIDRHf llm s 5 1 1GO 3939 KP 0 FaH 2mm m m Steady state error for ramp rt Rcl cmuuc Illplll rl Rum 1 Pt Ilm 57 72 sac R1 s n Kv Fe 2005 Steady state error for parabolic rt 1 5 gt071G39A5 s3 A Reference inpm m um pg lim lim s239s 5vOV quot D 39 Ka Fe 2005 System type System type ofG is defined as the order number of poles of Gs at 50 Examples GU Kai 05 t 1 s e s1s12s1s 2 quot yp r7u Gse 1 gt type2 S Gs W t type 3 Zero steadystate error If error constant is infinite we can achieve zero steadystate error Accurate tracking For step rt Kp lin1Gs 2 oo gt Gs is of at least type 1 s U For ramp rt K ling 3615 2 oo gt Cs is of at least type 2 8H For parabolic rt FaH2008 10 Example 1 Gs of type2 rt 6t y s ts 14 Characteristic equation 1Gs 0 e 52s12K 0 a s31232K 0 CL system is NOT stable for any K et goes to infinity Don t use today s results if CL system is not stable FaH2008 11 Example 2 I GSOftype1 rt et 21 h A K 5 5 H U ss 153 05 By RouthHurwitz criterion CL is stable iff o lt K lt 1304 Step rt 655 R O 1 Kp R 315K Ramp rt 658 Ky 2 W 42K v Parabolic rt 585 7 00 K Iim 82Gs 2 Kg 80 FaH2008 12 2M WSW 3 a 647 L C f 6 63 WWquot mlz i n 11 Pi J 249 2 ma a 0quot 9 Jmmm quot I 431411 Mdgfyo 42 th WMl f 739 7 1 2900 Mdg 2002m 231 4 F w 1 1302 4 6 4 g w o an ao Z 5 F a ll Kai cl 5 7LM4T 70 l XOO 2an Esau 4 Gm 681 0quot C39 L sy m 13 WWW 9644 at kzgz w ij faa 00st 2 ya u atjco axis L Cann m c g Kcoz lk r5a 03 WM math Mk 4 jw 5 46 2 02 CE 0 9 1 K 6 13940 0 K rqw 4k J60 50 k 67067 i KWW W39 H Re 39 llt JV lg g Jule 18 blot I WHAMN A all ag 3 W9 6 D co quot39 OWquot Mquot 55 0 ms MU Mm m w W 0 of m 6 WSOlw Mul rd 0 bk 1 5W n fm wb Q y wmwg 4 n 992 1 so 2 500 graf ti W m w W mg mlok Q0 M yobs 9 W 9 1 H gt b 7 o 7th 77 W M k 7L NW 70 5800 Mae 07299 0 My 79 kg 0113 2093 9 3 20 Lag67014 W a 9Wa My qu phap 570 Z Macqu Aw mm m M 311171 414 6 hCW Mug 4916 SW a m 7 quot CT 69 W amm fudy m 987 oaf39 vdvyfwa K 0V 499 xQOCJ Q l OV 0 Ma 0 00 pola R0015 20015 Ted39mmw doseJ 25009 Chm t M l Quad391 7M1 1 KG S 0 j L 0pm eao TF dJJ LUJ db gallv 739 Wm gag 129 2 Ms 2 Ans 397 73911 Sam 5 open 402 095 isJm Pd 0 Zoo9 F01 banJ 7 1 llt gt CL EC C705 J AG K what Kapo 9 CL page 0L pom quot9 Poles 2 m 0L ms 39m megaw key ideas 247 S bc 4 CL polc m g5gt K To 5 Q C 9 51 G OICSJJ 7642 fart Ianma a part i0 1 I Go 5 J L39 M z W 4 70 s 5 0 x 7 397 Q a i 2 3600K J J J W 3 flit521 77 1547 lt1 l M h Q 0 12quot39 a z mmzu 24mm P m 1 365151139 9 27 quot EX Yo S pl SPz Sf Cir176127 Sr9v Mre J 4 S Pu Q S Zl24 S Pv 2 fL as wish F quot Ma quotIquot MP MP9 a a cgJ QTEPL Ffmol HE 1er cog PO Y JFo V S H 2 rmz 4 5quot 102 H cmID 45m 95 9 o 97 970 ZLqursn ls o iL 350 93 3 X So 2 lt7 5 9 9Lquot93 T o o 366 Q 06 C9 i 0 W50 Wee 319 Q PGL WIW QL I W lfo 5o gs e IQ CMwa 1 J Re 8Q 7 V 5 945 rimNJ P 44443 4 395q fr 3 60 gquot 0 39 xo u quot O SI 8quotD M O i O J 0 809 6 8G0 X 6 c 39 gtlt Yule Draw Jhe ows on WM 0063 0 44 Q 3 of M 0090 number 0 may po e PMS 4 39 MR ME451 Control Systems Lecture 10 RouthHurwitz stability criterion Dr Jongeun Choi Department of Mechanical Engineering Michigan State University FaH2008 1 Course roadmap Modeling Analysis Design 7 Laplace transform Time response Design specs I Transient W Transfer function Steady state I Root locus f Models for systems Egt Frequency response E electrical Bode plot 7 mechanical Welectromechanical istability I RouthHurwitz W Linearization Nyquist Frequency domain PID amp Leadlag Design examples Matab simulations amp laboratories FaH2008 2 Stability summary review Let si be poles of Wm rational G Then G is Smbk llugmhlc BIBO asymptotically stable if gm 8 0quot Resilt0 for all i marginally stable if Resilt0for all i and 0 0 simple root for Resi0 unstable if Stable Unskililr mgion region it is neither stable nor marginally stable Fall 2cm 3 RouthHunNitz criterion This is for LTl systems with a polynomial denominator without sin cos exponential etc lt determines if all the roots of a polynomial lie in the open LHP left halfplane or equivalently have negative real parts It also determines the number of roots of a polynomial in the open RHP right halfplane It does NOT explicitly compute the roots No proof is provided in any control textbook Fall 2cm 4 I m i Le 539 80 V a F v 15 V6 2 R R EJ gt 39 V55 RI Cy o W W K a CV aw W 15 e36 f 0 1 t 6 V V P J 1 Q G quot MM lt lDco VK 2 15 Imm al T 1 160 V63 V L J 6 L J l I 2 Va S L I6 V15 25 1 5 S A filling dam141112 A 7 7 7 a on L v L1H 3 v ffc39ott 265 Rt 11 RL t RL v1vw 2 W I LN C H a 2 J 1 Ab B Va I Kb 8 LL quot39 L CL V N R V r V g k LZL 41 31 N C Llt z R L LL47 51 2 RM L E 1 3 Rt EU b 942 fag 771 21 A 55 V425 4w 4 9A4 T 42 1 o 62Vo z a VHS zac VJ 576031 quot735 1 T76 1265 0 I V R 0 17ch A 2 h Ipi mfl 3 M II Q13 1 M A N v a x m U m 5 mm NM 6 KNATE mu T3AJN 1 MU 3N N N 30 4W3 g m l u a MM monuwx C 83 g S C flM T R 165 V0 3 9 V RCS 75 4 1 i2 83 ME451 Control Systems Lecture 4 Modeling of electrical systems Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 5 Laplace transform Transfer function Models for systems electrical mechanical electromechanical Block diagrams Linearization Time response Transient Steady state Frequency response Bode plot Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples t Matab simulations amp laboratories FaH2008 4 Implemenalon Controller 3 Design 2 Analysis What is the mathematical model Transferfunction Modeling ofelectrical circuits FaH2008 3 Mathematical model Representation of the inputoutput signal relation of a physical system Physical Ot t In t upu pu system 1 Modeling A model is used for the analysis and design of control systems FaH2008 4 Important remarks on models Modeling is the most important and difficult task in control system design No mathematical model exactly represents a physical system Math model Physical system Math model z Physical system Do not confuse models with physical systems In this course we may use the term system to mean a mathematical model FaH2008 5 Transfer function A transfer function is defined by La lace transform of s stem out ut Cs U 3 Laplace transform of system input ms n5 A system is assumed to be at rest Zero initial condition FaH2008 6 Impulse response Suppose that ut is the unit impulse function and system is at rest ut m ystem 9t W Z 1 The output gt for the unit impulse input is called impulse response Since Us1 the transfer function can also be defined as the Laplace transform of impulse response Gs L3 9WD FaH2008 7 Models of electrical elements constitutive equations Resistance Inductance Capacitance it it it WE R m L will C m Rit w 2 g it Caz29 Laplace transform Vs 18 R FaH2008 8 Impedance Generalized resistance to a sinusoidal alternating current AC s s W8 39 Zs VSZSIS Zs Element Time domain Impedance Zs Resistance 75 Riot 38 Inductance vt Ld i 3688 Capacitance z t 0 g FaH2008 Memorize 9 Kirchhoff s Voltae Law KVL The algebraic sum of voltage drops around any loop is 0 03 v2 v3 02 v3vgv3 3 C O I 03 FaH2008 10 Kirohhoff s Current Law KCL The algebraic sum of currents into any junction is zero 11 22 2320 Z391 i2 i3 FaH2008 11 Impedance computation V V Series oonneotlon lt 13 28 4 Z8 318 Z28 3 I A VS Proof Ohm s law V518 Z2109 W V1ltsgt V2ltsgt 4218 t Z2ltsgt21ltsgt 233 FaH2008 12 Impedance computation Parallel connection 13 28M 13 218 228 quotquot I Proof Ohm s law 39 392 39 I Vs I V8 21118 Vs Vs KltL Iltsgt11ltegt12ogt 7 Ul UZVDI ISA 1 l V5 l V3 zlcs 228 2a Modeling example it 1 Input V1 l R2 V2t Output C Kirchhoff voltage law with zero initial conditions 111 31 321713 l a I zir a idr l I 1 U un t Rq 217 1739 L I L l I L JU I By Laplace transform Trn unrI lr 1amp8 IE1 l n213 l 31 t3 Vafsl 2 Rays 4 Lung 39 4 V4 I sb39 I FaH2008 Modeling example cont d it 1 Input v1t R2 v2t Output C Transfer function 1 V2S R2m V1S R1R2sic RQCSl 31Rg081 rstorder system FaH2008 Example Modeling of op amp MS Rule1 i390 Rule2 Vd0 L Impedance Zs VsZss Transfer function of the above op amp FaH2008 Modeling exam ple op amp R2 C i0 Kt R1 Vd0 Input Via I IVoCt Output L By the formula in previous two pages V I39S I RA i i PU Us I l OK z 50 z T VXS R1 R108 firstorder system Fall 2008 17 Modeling exercise op amp C1 C2 AM 1quot R2 0 R1 I Vd0 le Input vt I IVoG Output L Find the transfer function Fall 2008 U av 3 CIquot L 12740 Scrfog EKPW M an Kgm 0697 H0145 13769 l iZAtl VL VA 96 c 75794 L WW 64m 0506 m 06 1AM meow Jam ac a 8 he mdm par ax er 966 XpCH xo 757 0 7354 was 5 765 u do 91970 3 LL 9C16rclu 9 39Km 1316uo i FWn Ul 5 9X wayulqsu UPIml 5r 6 30PM J Bade TWSW 2 7m CQWI Jogv 3 35 44m LilW wt ii Mo 96 A Momma MP0 q L156 uquot VestaM WWW K UL 45 SM 3976 a 7 393 a figIF m 7 JT I J S pm 5 9cu 9M L g wan m f J qgWla bLo 9 MT vac Ea 514 alC39Fiwe e 5723 g 397 b 7 The 7 39 9 39 V 2 I A MM a came39t sl ecu lt49 x V vvx ae by F34 2 9 I ha Ma Lg 5 71 1235 chgt La as be lmo39tmvx 4 1 W 39 MowwGC 5f Imam 9 a A gat 46 Mina rP 1 33 60562 391 2511831 fng WI 95 a r gt 399 a gt 9 7 3 56 elgtlt 0 2X es L9 4 4123 4 340 M3 MA 9 63 aquot Mia 9quot4 lakemg 11 a 0 ME451 Control Systems Lecture 14 Time response of 1storder systems Dr Jongeun Choi Department of Mechanical Engineering FaH2008 Michigan State University Course roadmap Modeling Analysis Design 7 Laplace transform Vi Transfer function f Models for systems electrical mechanical Welectromechanical Block diagrams El Time response Transient W Steady state Frequency response Bode plot W Stability RouthHurwitz Nyquist Design specs Root locus Frequency domain PID amp Leadlag Design examples W Linearization t Matab simulations amp laboratories FaH2008 Performance measures review Transient response Peak value Peak time Percent overshoot Delay time Rise time Settling time Steady state response Steady state error FaH2008 Today s lecture Next we will connect these measures with sdomain Done Firstorder system A standard form of the firstorder system DC motor example l irt JIM E 39Him nil E1 1 a sale 1quot u Fall 2008 DC mtr examle ont If LaltltRa we can obtain a 1storder system 97713 Ki N Ki Iltp h I i JFIH 1aJhn 39 T3 1 39 RaBm KbKi 39 RaBm ltsz TF from motor input voltage to motor speed is 1stoder motor position is 2ndorder Fall 2008 5 Ste res nse fr 1strer system Input a unit step function to a firstorder system Then what is the output ut 31t 1 p K V 0 Ts I l O KiTT l 1Ys 2 Z K1 e tT S s I lT W Partial fraotion expansion FaH2008 6 Hw t eliminate steaystate errr Make a feedback system with a controller having an integrator copy of Laplace transform of a unit step function Controller A ut 1 K13K2 K o gt I S I I One has to select controller parameters to stabilize the feedback system Suppose K T1 and obtain such parameters Fall 2008 7 Meanin f K an T K Gain Final steadystate value 15 tit W K T Time constant Time when response rises 63 of final value Indication of speed of 02 eeeeee eeeeee eeeee response convergence 39 39 39 Response is faster as T o 2 T3 4 5 a me becomes smaller Fall 2008 8 P m Amplitude DC ain fr a eneral system DC gain Final value of a unit step response For firstorder systems DC gain is K For a general stable system G DC gain is 30 Iim yt lim 3613 2 GO t oo I 8amp0 Final value theorem 39 Examples 3 3 KS 2s 5 Gm g 18 2 7 00 Z s I 23 3 3 FaH2008 9 Settlin time f 1strer systems ya Km at Relation between time and exponential decay t e tT O 1 T 03679 2T 01353 3T 00498 4 5 settling time is about 3T 4T 00183 O I I I 39 5T 00067 2 A Settllng tlme Is about 4T FaH2008 10 Step response for some K amp T K1 T1 K1 T2 2 0 0 o o a 3 3931 a1 E E lt lt O 0 5 10 0 5 10 Time Time K2T1 K2T2 2 w r a o o a 3 3931 a1 E E lt lt O 0 5 10 0 5 10 Time Time FaH2008 11 System identification Suppose that we have a blackbox system Obtain step response Can you obtain a transfer function How FaH2008 12

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