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# Feedback Systems ECE 4700

WMU

GPA 3.61

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This 36 page Class Notes was uploaded by Lizeth Hegmann on Wednesday September 30, 2015. The Class Notes belongs to ECE 4700 at Western Michigan University taught by Bradley Bazuin in Fall. Since its upload, it has received 62 views. For similar materials see /class/216789/ece-4700-western-michigan-university in Engineering Electrical & Compu at Western Michigan University.

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Date Created: 09/30/15

Exam 2 Review Lecture Chapter 9 Stability in the Frequency Domain Mapping Contours in the splane Nyquist plots The Nyquist Stability Criterion Relative Stability and the Nyquist Criterion 7 Gain and Phase Margins The Stability of Control Systems with Time Delay PID Controllers in the Frequency Domain Chapter 10 The Design of Feedback Control Systems Approaches to System Design Cascade Compensation Networks PhaseLead Networks Design Using Bode Diagrams and Root Locus PhaseLag Networks Design Using Bode Diagrams and Root Locus PhaseLeadLag Networks Design Using Bode Diagrams and Root Locus PID Network Design System Design with a Pre lter Design for a Deadbeat Response Chapter 3 State Variable Models The State Variables of a Dynamic System The State Differential Equations The Transfer Function from the State Equation The State Transition Matrix and Time Response The Exam Four 4 multipart questions approximately 125 min per question All my exams are very long you don t have enough time and are considered hard Open Book Open Notes Note 1 mark pages that might be important and 2 you won t have time to learn what you don t know in the exam so don t waste your time trying Dumb Calculators or No Calculator Only your choice I will provide a table of x pi radians to sin tan cot and cos If you see lots of variables and integers with a few simple fractions you are probably doing the problem correctly Note some things can t be easily made into integers or simple integer fractions so they aren t This is a course on design so elements that help you understand the performance and parameters of existing systems or that help you improve system performance are stressed Chapter 9 Stability in the Frequency Domain Mapping Contours in the s plane Nyquist plots The Nyquist Stability Criterion Relative Stability and the Nyquist Criterion Gain and Phase Margins The Stability of Control Systems with Time Delay PID Controllers in the Frequency Domain The Nyquist Stability Criterion ans Nyquist Plots The frequency response in fact Laplace domain deals with complex variables and values From complex variable theory we can use concepts developed by Cauchy concerning mapping contours in the complex splane This all starts by going back to the characteristic equation of feedback systems Ms E ltsgt G or T 1Gs 1GsHs Focusing on the denominator and characteristic equations Fs1Ls1Gs or Fs1Ls1GsHs What Nyquist proposed was a mapping of contours from the splane to the F 3 plane The result that this all get us are p 472 If a contour in the splane encircles Z zeros and P poles of Fs and does not pass through any poles or zeros of Fs and the traversal is in the clockwise direction along the contour the corresponding contour in the Fsplane encircles the origin NZP times in the clockwise direction Since FslLslGsHs Ifwe form F39sFs 1LsGsHs The new function simply described the open loop gain By doing this shift the above can be modified Nyquist stability criterion 0 A feedback system is stable if and only if the contour 113 in the Ls plane does not encircle the 10 point when the number of poles of Lsin the righthand s plane is zero P0 A feedback system is stable if and only if for the contour 113 the number of counterclockwise encirclements of the 10 point is equal to the number of poles of Ls with positive real parts Relative Stability For the system that may be either stable or unstable based on speci c variables or gains They are expressed in terms of a gain margin and a phase margin They describe how much the gain and phase of a stable system may change before it becomes unstable From Dorf The gain margin is the increase in the system gain when the phase 180 degrees that will result in a marginally stable system with intersection of the 10 or 1 j 0 point on the Nyquist diagram The phase margin is the amount of phase shift of the GHjw at unity magnitude that will result in a marginally stable system with intersection of the 10 or 1 j 0 point on the Nyquist diagram These margins are based on the gain and phase of Ls or GHs the open loop or loop gains It is easiest to see and measure gain and phase margin by plotting the Bode Plot of GHs Gain and Phase Margin based on frequency domain analysis The transfer function equations of feedback systems are ltsgt G T s G TS 1Gs SH 1GsHs Focusing on the denominator or characteristic equations Fs1Ls1Gs Fs1Ls1GsHs Using the frequency domain magnitude and phase of the open loop gain Fs 1 Gs expjtanzGs Fs 1 lGHs expjtanzGHs Marginal stability will occur at 0 1 Gjw eXpjtan4Gjw 0 1 lGHjw eXpjtanzGHjw or 1 IG 1W1 explj39tan4Giw 1 IGHJ39W1 explj39tan4GHiw Gain and Phase Margin Computation The gain margin where the phase 180 degrees is Solve wGM for tanzGijM i180 deg Then the gain margin is 1 KGM GijM expj7r 1 Lead1ng to KGM GM The phase margin where the magnitude 1 is Solve wPM for 1 GIWPM Then the phase margin is 1 1eXpjtant9PM 4G0pr Leading to 7139 9PM 460pr or 9 7r 4GijM PM PID Controllers in the Frequency Domain Control functions or processes may be placed in the openloop path or as a feedback element A proportional integral differential derivative controller is defined as GcsKP KD s s Proportional Integral PI GSSKP KP 35 KI Psl KP I Goo S S A pole at zero and a zero at KI KP with frequency wPI KI KP Proportional Derivative PD GcsKP KD s K GcsKDSKPKPsl D P No pole and a zero at KP KD with frequency wPD KP KD Proportional Integral Derivative PID K GcsKP IKD s s K K g32 Ps7IJ KSi1i1 K KD w1 w2 D acltsgt S s K K 2 K Apole at zero and two zeros at KD i KD ltD Now what are the Bode Plots for each of these contollers Chapter 10 The Design of Feedback Control Systems A suitable control system is o s able o results in an acceptable response to an input Command o is less sensitive to system parameter changes o results in less steady state error for input commands and o is able to reduce the effect ofdisturbances Compensation Often 39 quot 39 Luuuuuculu all 39 39 L inserted G b no0 C II Compensation types Cascade or series compensation a on b 9 equot quota Ea o o a 1 we a o Input compensation d A compensator PhaseLead Network Gog L S 1 Z for zltp jw Why is it referred to as phaselead In the frequency domain GAS where r and a The phase of this system is then 31 Kp z K1jwar1 jwr1 Kp w tan 1wa r tan 1wr 20 log a 2010gGcK1 dB a log scale 90 4 VJ O 00 M w PhaseLag Network GE 5 M for Zgtp sp jw 0 gt a f p Why is it referred to as phaselead K F 1 39 1 In the frequency domain GE s p Z 2 K1 jwT z s J jw0 l391 1 I7 K 17 1 z Where r and 0 and Z Z P The phase of this system is then w tan 1w r tan 1wot r I l m T I 39 T In 4 10 Invg Q 1 din 1 4M Y Maximum phase leadlag at the center frequency of the network The geometric mean ofthe pole and zem Resulting in Which is equivalently written as sin wm L a The gain is 39639me Phase Lead Using Root Locus 1 List the system speci cations and translate them into a desired root location for the dominant roots 2 Sketch the uncompensated root locus and determine whether the desired root locations can be realized with an uncompensated system If a compensator is necessary place the zero of the phaselead network directly below the desired root location or to the left of the rst two real poles desired Determine the pole location so that the total angle at the desired root location is 180 degrees and therefore is on the compensated root locus Evaluate the total system gain at the desired root location and then calculate the error constant Repeat the steps if the error constant is not satisfactory Phase Lag Design Compensation for DC gain due to steadystate error requirements For the location of the pole and zero place them where the gain contributes but the phase does not a factor of 10 below the desired circuit settling time Phase Lead Lag Design When both DC gain and additional phase margin is required Perform the PhaseLag design to include required gain and compensate for the phaselead gain reduction at DC This may require a design iteration Perform the PhaseLead design to achieve additional phase at the 0 dB openloop gain point Design trick not discussed in class used in homework solutions Zeros placed very near poles may result in cancellation in the system response Zeros can be placed tricky or large open loop gains may drive poles to zeros that were added for compensation Design Criteria Percent Overshoot Settling Time 2 criteria Steady State Error System Design Using Integration Networks Steady State Error Criteria szm E K E L Consider the open loop GOL s 0 SN39 3pk w Using the equation for steady state error negative unity gain feedback N1 Rs S S Rs 1 800 S GOL S 2 SN SN GOL Where N de nes the type number of the system For N 0 eoo Resulting in eoolsi 01ms fl RS i 01sw1 RS M Hzm Where KN lings 39GOLS K m Hltpk k1 De nition of Error Constants and cancellation of input response Position N0 KN0 KP liIIOI GOL 3 step Rs l H 3 Velocity Nl KN1 KV ling s GOL s ramp Rs Li H S Acceleration N2 KN2 KA lings2 GOL s parabola Rs is H S Jerk N3 KN 3 K li 0132 GOLS Rs i Deadbeat Response A deadbeat response is one that proceeds to the desired level and remains within a narrow bound with minimal overshoot As always we will use a 2 steadystate band Criteria 1 steadystate error 0 compensated for the appropriate openloop type 2 rapid response 7 minimum rise time and settling time 3 01 to 2 overshoot stays within the 2 bound 4 percent undershoot checked or monitored to insure compliance Chapter 3 State Variable Models x DJ yzg D IS Important aspects 0 Structures of canonical forms 0 Forming the state transition matrix or fundamental matrix 0 Solving for the time response of the state transition matrix Solving the state equation Phi 2s sg gl l quotlt 3 0 quota 32 X g m quota 32 II by H b h a 2 Often we are looking for l the transfer function based on a driven input where 0 Q and 2 an output that is a linear combination of the internal system states g 2 Characteristic Equation As detL j ECE 3710 Review Chap 6 to 7 Chapter 6 The Stability of Linear Feedback Systems The Concept of Stability The RouthHurwitz Stability Criterion The Relative Stability of Feedback Control Systems Chapter 7 The Root Locus Method The Root Locus Concept The Root Locus Procedure Parameter Design by the Root Locus Method Sensitivity and the Root Locus Three Term ProportionalIntegralDifferential PID Controllers Page 1 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Chapter 6 The Stability of Linear Feedback Systems The Concept of Stability The RouthHurwitz Stability Criterion The Relative Stability of Feedback Control Systems Poles and Zeros Poles and zeros provide knowledge about the system response TO 2 2 Pl 54 As For a system transfer function can be de ned as K Hltszm TSK39SZl39SZZquotquot39SZM quotHM 3p1 spz sp1v g wn Kmglszm or Ts H5Pk39l ISZ 261 39Sa2 w2 k1K PM The poles are then at pk and a iiwj The response to a step input is KHszm 1 Ys WW Hspk11sz2ajsaj2wjz S k1K k A B SC SEQ 2 2 j i 2 k1K pk Fm S 2ajsaj wj ytk 214k e39 2D e a12 sinwj tt9j k1K J1J Page 2 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 For a system transfer function how do we check for stability Look at the roots of the characteristic equation qS AS 3p1393p239quot 3p1v A necessary and sufficient condition for a system to be stable is that the poles of the transfer function have negative real parts A system with both negative and zero real parts is called marginally stable Finding Insights qltsgts ngtlts rgtlts rgt qs sN rl r2 rNSN391 r1r2 r2 r3 r1r3 FN7139FN39SN72 N73 r1 39r2 39r3r1 39rz 39r4r2 39rs39rzt quot39rN72 39rer 39rN39S k1Nr1rzrN If all roots are in the left halfplane a necessary but not suf cient condition is that all the signs in the polynomial must be the same either all positive or all negative RouthHurwitz Criterion The RouthHurwitz Criterion states that the number of roots of qs with positive real parts is equal to the number of changes in sign of the first column of the Routh Array For qsansquota squot391alsla0 quot71 Page 3 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Then compute additional rows until done Where b 1 1 anrl and CH 1 burl The Routh Array is the rst column of the above computations Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130 an and anrzt n71 anrl an73 anrs ml 5 burl ban bnrs n73 cnrl can can 0 anl a b 1 a n73 an73 anrl anrl an73 c 1 anrl n73 39 ban bnrl burl Page 4 50660 6 ECE 3710 Review Chap 6 to 7 Interpreting the Routh Array The number of roots of qs with positive real parts is equal to the number of ch anges in Sign of the rst column of the Routh Array Case 1 If no elements in the rst column of the Routh Array are zero Sign changes describe the positive real roots If there is a sign change the system is not stable Case 2 If any element in the rst column of the Routh Array is zero but the rest of the row has non zero elements You must replace the zero with a small number epsilon 8 and complete the computations as though performing case 1 Case 3 If an element in the rst column of the Routh Array is zero and the rest of the row is also zero The row prior to the all zero or the epsilon and zero row forms an auxiliary equation This identi es that the qs polynomial has two four etc roots on the jw axis at the zeros of the solution to the auxiliary equation If there is only one row and the auxiliary equation de nes one pair of complex roots the system is marginally stable the steady state is a marginally stable oscillation Case 4 Repeated roots of the characteristic equation on the jw axis The row prior to the all zero or the epsilon and zero row forms an auxiliary equation If the auxiliary equation results in repeated imaginary roots the system is unstable Only instances of one conjugate pair of imaginary roots is marginally stable Page 5 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Using the RouthHurwitz Stability to determine if a settling time specification is met Axis shifting the splane 7 Meeting Settling Time Specs Also called Relative Stability If you know that a certain settling time is required for a control system the splane s jw axis can be offset by performing a change in the variable Let snsa Then solve for the shifted sdomain If there are poles to the right the system does not meet the settling time Question If we have software tools that solve for the poles and zeros of a system Why do we still need to know the RouthHurwitz criterion The RouthHurwitz criterion solves for stability boundaries when one or more variables appear in the system transfer function Instead of guessing at different values for a controller design using a trial and error approach until you get lucky you can derive the stability bound and know the range of selected variables Other approaches do not allow you to work with multiple variables all at the same time Page 6 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Chapter 7 The Root Locus Method The Root Locus Concept The Root Locus Procedure Parameter Design by the Root Locus Method Sensitivity and the Root Locus Thr ee Term Pr oportionaleIntegraleDi erential PID C ontr oller s w cL ILL 4 rr system parameter is changed The locus or system poles tor a varying parameter For a simple closed 100 system Transfer Functi on characteristic 43 Ms 1 K 03 Roots ofeharaeteristie eqn 0 1 K 03 siuee s is a complex variable to solve for the roots we have 71 KGs motelAme We require K Gk l and zKGs7ru2n P ge 7 a Nutes and gures are based rm ur taken 39um matenals m the ECE 371 euurse textbuuk RFDerdRHthn whip HH nm39mquot ECE 3710 Review Chap 6 to 7 Generalized Root Locus Format Roots of characteristic eqn 0 1Fs Find the roots of Fs lj0 Where FltSgtK szlszzszM 3p139sp2uiusp1v Or FltSgtK Sle39SZzlquotquot39SZM39eXpj6Z1 jt9z2 jt92M lsplspzlspNexpjt9p1 119p2 jt9pN Which requires Fs K Sp1Sp2SpN And zFs eXpf7r r 172 j 1 p1 p2 pN nin27rltt921t922 t9zM 9P1 9P2 9PN All values of s in the splane that meet the above criteria can be part of one solution to the characteristic equation However once one root is selected out of all possible roots the rest of the roots may then be determined Further note Letting 01KG31K lt16 0 qsK ps Then for K0 the roots correspond to the characteristic equation or poles of the openloop response And for K 00 the roots correspond to the numerator equation or zeros of the openloop response Therefore we often say that the root locus travels from the poles to the zeros of the open loop response But what if my openloop system has no zeros Assume that there are zeros at in nity using as many zeros as needed so that the number of poles and zeros are equal Page 8 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710Remew Chap 6to7 Using RnnthcnstLhm ls e srplanE 1oeus coef cient can be se1eeted take as 12 steps In the 10quot 1 a through s followed by 21 In the 9quot Edmon they are nets 1 z nMustosetmmooueeoe poem ui udEuuumnanure toutoutttettonooeotmwomen s mm o powwow monotth X Let my New my tom to hummus mo mm votoeosttotmmem w who Page 9 Notes and gures are oosee on oetokeo mm materials mm 505 71 cause textbank RC DarfandRH ExshnpMademCammlS gemggmedPxenuceIIHJLZUUI ISENUell z e ECE 3710 Review Chap 6 to 7 Root Locus a 12 or 7 with 1ae Step Program Step 1 1a Using the characteristic equation write it in the root locus form Root Locus Form 0 1 Fs 1 K Polys Where K is the variable that is allowed to vary thereby moving the roots Step 2 1b Factor Polys into poles and zeros 01 K ltpgtltspzgtltsPN Step 3 1c Locate the Open Loop poles and zeros of Fs in the s plane Define their locations with x poles and 0 zeros 0spspzspNKszls22SZM Note For K0 the poles of the characteristic equation are the poles of Fs For Kin nity the poles of the characteristic equation are the zeros of Fs As a result the root locus plot always moves from the poles to the zeros of Fs Step 4 2 Locate the segments of the root loci that are on the real axis Loci lie to the left of odd numbers of poles and zeros This is a result of requiring irt27239 61 62 9ZM 6p 19p2 6pM Step 5 1d Determine the number of separate Loci Count the number of separate nite Loci determining poles nP and zeros n Step 6 1e For a real system the root loci are symmetric with respect to the real axis Since complex poles and zeros occur as conjugate pairs the root loci describing their positions must be symmetric about the real axis Page 10 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Step 7 3 For systems with multiple zeros at in nity the loci proceed to in nity along asymptotes centered at 039 A and with angles A Where M 0A 71 n p 2 And A Mforq01pnz1 np nz Az 1180 1 nZ n 180 z 5 1 324 Step 8 4 Use Routh Hurwitz Criteria to determine if and where the loci become marginally stable by touching or crossing the jw axis Zero rows in the Routh array expose crossings of the jw axis By solving the auxiliary equation the jw axis points of the root locus can be computed Step 9 5 Determine and plot the break away points from the real axis 1 a Form K ds m b Obtain i0 0 63 c Determine the roots one root will be the breakaway point tofrom the real axis d Breakaway angles are based on the angle criteria nin27rltt921t922 t9zM 9P1 9P2 9PN Using the number of poles 2n at the breakaway point the angles are 2 7 1 BA Page 11 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Step 10 6 Determine and plot the angle of the departure of the loci from poles and the arrival at zeros nin27rltt9z1 622 t9ZM t9p1 9p2 6FM Solve for the unknown pole or zero departure angle given the vectors from all the other poles and zeros The optional steps Step 11 not listed For expected root locations that must be on the root locus determine the angle phase criteria to help plot the path The angles are based on the angle criteria again irt27239 921 622 t9ZM t9p1 19p2 9FN Examples 45 deg from the origin the line of a design criteria Step 12 not listed For the expected root locations determine the gain K criteria for the root The gain is based on the magnitude again FSKWzl SP1 SP2 quot39Sle K I lszll393zzl 39lsZMAw Examples Table 77 on pp 466498 11Lh ed or pp 406408 10111 ed show a number of generic root locus plots For most problems once you have completed steps 1 and 2 you can probably nd the generic form of the problem you are solving in the Tables Page 12 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 610 7 Dort s Textbook Steps Table 72 Seven Steps for Sketching 3 Root Locus Step 1 Prepare the Iool iocns skeich a Viiie llie cin pammetei of n e 39siic equation so mat the imcresl K nppenrs as n nnrhipiier b Encior 175 in Iei ms of n poies and M zeros c Locnre he operrioop poies and zeros 017715 r d Deiennine the nnrnher ofsepm39me ioci5L e The roor ioci nre symmetrical with iesptcl Io rhe horizomal rcal ans Related Equaiion or Rule 1 KPir o gtlt poles o zeros n 2610 gt W171 Inunbsrof finite poles 71 when I x i ilimber of finite zeios r zeros 3 The loci proceed 0 the zeros at iii niiy along EVE 21 asymplmesccmcred a an and wihauglcs 4 Lquot n M 39 2k i U 7 a n 7 M180 k 012n M 1 r 39 Use D 39 39 39 39 imaginary axis it does so 5 Determine ihe breakaway poim on ihereai ms iinny a St K 7 r b Deermine 10015 of dpdx 7 0 or use graphical merhod io iind mnxirnnnr oi pm 6 Determine the ting of locus depmluic from compiex 175 1m 1660 al 7p or 72 poies nnd rhe angle of Jews nrrivni m complex zeros using ihe phase criierron 7 Complexe rhe rooi iocns sketch Page 13 F K F 391 00k RC Dorfand RH Bishop Modem Control Systems 9 h ed Prentice Hall 2001 ISBN 0130 39 39 1 coursetextb 306606 ECE 3710 Review Chap 6 to 7 Design Example 76 Varying Two Parameters Gs Kl Rm bO bO V m 39Y 2 HM K25 Steadystate error for ramp input 5 35 A Type 1 openloop system Damping ratio of dominant roots 2 0707 implies 5 2 X5 or arccos lt 45 Settling time to Within 2 for 3 sec implies roots are lpxl 2 Note Page 14 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 K 1 ss2 K 0 en Loo ustthe 1nner loo G s 1 p M p W0 1 ng 322SSK1K2 ss2 ClosedLoop Ys GIWS sz2ssK1K2 K1 Rs 1GWs 1 ss2sK1K2K1 sz2ssK1K2 YS K1 W3 Rs 32239SS39K139K2K1 322gwnsw RouthHurwitz stability is K1 gt0 and K1K2 gt 2 2K K Steady state errortoaramp essi 1 1 2 035 K sKl K1 limz HO 3 2ssK1K2 Note that K2 must be small less than 035 and that K1 must be large enough so that 2K1 is smaller than 035 Damping ratio and natural frequency 2K1K2 2 wH K1 w 2K1K2 2K1K2 gt 1 9quot 2w 241 5 0354K gt 2K1K2 1 gt 2 2JK1 E 2 2 K12 1632 and w 24041 03545 Page 15 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 For root locus purposes we will use Assz 2sBsa First assume one value is zero and perform the root locus for the other variable Try it both ways to see which roots on the root locus would be better to let the second variable vary from 0 to in nity with Since K 0 is the starting point for all variable ranges in a root locus you are setting one to zero and varying the other Then once you are happy with a point on one of these curves it becomes the startingpoint for a root locus involving the variable previously set to zero If you like it ne otherwise pick another value from the original root locus and try again Form the root locus for B 0 1 5 S sZ2sa Form the root locus fora 01aZ s 2sBs See if you like the rst one with 0L0 as a root locus or the second one with 30 For01B 15 S 7 sz2s0 s2 the root locus is trivial but doesn t say much For 01aZ s 2s 77 1136 4 Page 16 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 If we pick an 0c the root locus in 5 becomes 0 1 s 2 s all The book picks 0c 20K1 which meets the criteria already established 3 sz2s20 Thenfor 01 R001 Locus 1 System sysb can 0 443 gt x Frequencyradsec 447 Pole 74 4361 c Damping o 224 Overshool 48 6 Frequency radsec 447 L lmagmary Awe Real Axis The book picks l 43 which is approximately if exact match at l 4325 0315 S 035 2K K This nishes the design as Kz0215 and egg f 1 Note start with the variable with the lowest power in s rst If that doesn t work start with the next highest power Page 17 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 6 to 7 Table 77 Root Locus Plots for Typical Transfer Functions Table 77 H00 Locus Plots for Typical Transler Funclions G s G13 Root Locus 17 J 45 w m 1 Rumle L 39 m 1 m my 1n1 Page 18 course textbook RC Duff and RH Bishop Modem Control Systems 9quoth ed Prentice Hall 2001 ISBN 0130306606 ECE 3710 Review Chap 610 7 Table 77 continued 5151 R001 Locus mm 1 sm my 1 7 sign l Root Locus N S r a Tnpk P u 7 quot Tnple 391 pale 7 r a re Table 11 continued Gm Root Locus 513 R001 Locm tam mm 1 V my 1 N quot s quot asquot mm 11 1 Dmvhle l u r v H l K1 mm 1 1 n mm mm 1 mm k Notes and gures are based on or taken from RC DOIf and RH Bishop Modem Control Systems 9 h ed Prentice Hall 2001 ISBN 0130306606 maten 39als in the ECE 371 course textbook ECE 3710 Review Chap 6 to 7 Proportional Integral Derivative PID Controller Control functions or processes may be placed in the openloop path or as a feedback element A proportional integral differential derivative controller is defined as GcsKP KD s s For on 663 Rn KP IKD sjm The response is Ce KP rtKI J39rtdtKD 5 ProportionalIntegral Pl A pole at zero and a zero at KI KP ProportionalDerivative PD No pole and a zero at KP KD ProportionallntegralDerivative PID Gcs KP KD s S K K KDs2 Ps I G S KD KD KDszlszz C s s 2 K K K Apoleatzeroandtwozerosat KDi KD ltD Page 20 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control S stems 93911 ed Prentice Hall 2001 ISBN 0130306606

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