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# Feedback Systems ECE 4700

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This 142 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 43 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

The Root Locus Method for Two Variables Avv31v2 xzz Let Form the root locus for c 0 1 1 f x 1s 134 Form the root locus for 3 0 1 x 1x 11 1 First assume one Value is zero and perform the root locus for the second Variable 2 Picking speci c Values of the root locus then de ne the root locus with respect to the original Variable that was zeroe 3 If this doesn t give you some insight into picking the Variables which Variable you use in step one and repeat 1 and 2 Perform the root locus operation for 3 0 AHG 3 142 0 Thenewrootlocusforc 01z 3 1 21a x 1x x s1 Ramlacus in a erh bu lmagAxls 5 71 Realm The marks are at steps of 0120 in 0 Note there are no Values that would make the system stable yet Notes and figures are based on or taken from materials In the ECE 371 course textbook R c Dorf and R H Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 0130306606 Performing the root locus in 3 for c 0 results in Form the root locus for 3 Ramlacus in a erh 33 imam us a RealAxis The marks are at steps of 0 20 in 3 For this approach it appears that stable room can be found So maybe by using speci c Values of c the root locus in 3 will show us a useful design curve Notes and flgures are based on or taken from materials in the ECE 371 course textbook R c Dorf and R H Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 0130306606 Now assume that o 1 and perform the root locus in 3 Form the root locus for 3 0 1 1 3 15 2 i39 1 roots1 1 01 1465 02328 07926i 02328 07926i Ramlacus in a WM 31 a 5 Real Axls The marks are at steps of014 for the root locus of a when 30 Notice that the root locus begins at one the 11 locus points and follows a characteristic pattern This appears to be a useful approach for understanding Variations in c an 3 01 7 s31x2zz Notes and figures are based on or taken from materials in the ECE 371 course textbook c Dorf and R H Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 071370306606 7 Performing in the opposite order Assume that 3 4 and perform the root locus in 0 Formtherootlocusforc 01a1a 2 1 roots1140 0 0500019365i 0500019365i Ramlacus in a erh a4 lmagAxls 2 Real Axls The marks are at steps of028 for the root locus of 3 when 10 Notice size the root locus in a must start atthe 10 root locus of 3 it appears that similar curves will exist for all values of 3 selected and that the useful values of a must be relatively small the root locus rapidly goes toward the jw axis and becomes unstable as 0 increases While this approach provide information the previous sequence is likely to be m ore useful in design 1 01 7 01 13 SS1 1534442434 WWW l 3 S liiii i Notes and flgures are based on or taken from materials In the ECE 371 course textbook R c Dorf and R H Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 0130306606 Matlab script for the preceding curves and analysis clear clfl clf2 clf3 clf4 numal denal l 0 0 numbl 0 denbl l 0 0 sysatfnumadena sysbminrealtfnumbdenb figurel rlocussysa0 l 20 axisl axis hold on rlocussysa hold off axisaxisl titlevarsprintf39Rootlocus in a with b039 titletitlevar figure2 rlocussysb0 l 20 axis2axis hold on rlocussysb hold off axisaxis2 titlevarsprintf39Rootlocus in b with a039 titletitlevar Set al and do the rlocus for b again num 2num denb2l l 01 sysb2minrealtfnumb2denb2 figure3 rlocussysa0 l 4 axis3axis rlocussysb2 hold off axisaxis3 titlevarsprintf39Rootlocus in b with al 39 titletitlevar Set b4 and do the rlocus for a again numa2numa dena2l l 4 0 sysa2tfnuma2dena2 figure4 rlocussysb028 axis4axis hold on rlocussysa2 hold off axisaxis4 axis2 6 4 4 titlevarsprintf39Rootlocus in a with b439 titletitlevar 5 0f 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 9th ed Prentice Hall 2001 ISBN 013030660 6 Design Example 75 Walding Head Desxgn cmena 1 Steady state enorto a 2 Damp m d 3 amp mput of lt 0 35 ofthe mputramp ng no of the ommantroots gt7 0 707 Setdxng time to wahm 2 of the fmal value m e 3 sec K1 5 52 K 11lt 52 2 5 K1 K1 5 s21ltl MK2 5 5 s21lt K21lt 551775 as 52 K1 K2 5 5 5 2K K21lt Note For stability each ofthe coef cients must be posmve A s 525 2K K21lt 2 m21lt 1lt 220 320 6 of 12 Nmes and gures are based an a lakenfmm maunals m39he ECE 371 cums mka RC Dmme x Exshap Mademcanknls items 9quot ed Pranks Hall 2mm xsENmmmm szs2K1K2 1 1 R zum amPermr 8a H0 s2s2K1KZK1 s2 8a s2K1K2 22K1K2lt035 H032s2K1K2K1 K1 39 2 Damping factor 4 2 f g 2a ZJK1 4 3 Settling time T5 S 3 7 2K1K2 The two variable root locus allowing X K1 and 8 K1 K2 Assz 2Bs0 The rootlocus force is 0lX 322 s The root locus for is 01 s 2SX 7of12 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 9th ed Prentice Hall 2001 ISBN 013030660 6 Perform the root locus operation for at when for l 0 Therootlocusforocis 01a21a s 2 s s s 2 K K 2 t 1 5 i E i 1 2 K K K K 3 2395 1395 075 05 1395 2 Real Axis Note This does not meet the design criteria gm 7 Z What are other stalting locations for 5 8 of 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 9th ed Prentice Hall 2001 ISBN 013030660 6 The root locus for with 0c0 0lBZl s 2s is2 Imag Axis 1 3 25 2 15 1 05 Real Axis As 5 increases the starting point of the pole moves toward in nity and the root locus will begin to meet the design constraints Therefore the root locus for 0c is probably an adequate starting point Form the root locus for l with x 0 1 s 2 s X Perform the root locus operation for the equation for 0c 4 Imag Axis 2 n u r 1 05 Real Axis 9 0f 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 9th ed Prentice Hall 2001 ISBN 013030660 6 This looks better since the roots move to the left of a S 43 but let s look at a few different values for ac Perform the root locus operation for the equation for 0c 4 and 8 Imag Axis 15 Real Axis Now we can see the structure of the solutions and pick values that are appropriate for the design As at increases changes in i will drive the roots so that a faster settling time is achieved The steady state error consideration 4w QKVK5 QDlt035 1 lt035a 2 The upper bound on i is set by the magnitude of at if we are to meet the steady state error criteria So we will select a value for ac and let i change within the allowable range From As 322Bs06s2s2K1K2K1 Fora201et 0ltBlt035oc 25 s us1ng 0l 2 s 2SX 100f12 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 9th ed Prentice Hall 2001 ISBN 013030660 6 mums m h was 32El mews 4 Rea Axxs a eonunuous curve forthe 0 1 5 when 0720 Thu plot shows roots on the rootlocus of a0 rootlocus of when 120 and 5 251m the speeme roots for Nome that as 13 approaches 5 me desxgq emena can be meet 2K1lt 2 2 lt035 K1 URampen39or a 2 Dampmg factor 4 amp 21lt 21lt K 22 4 4 3 5m u 71 7 27 e mg me 2 2 a T 3 Subsmuung 0220 and 135 forthe ongmal vanable ac1lt 20 and K2 K 520025 s21lt K21lt 527 s20 A5 roots17 20 e3 5000 2 7839 V3 5000 r 2 7839 From lof12 Nmesmd gmes arebasedanmtakenfmmmaunalsm39he 71 cums RC Dmme x Exshap Mademcanknls items 9quot ed Pranks ECE 3 mm 2115 mm ISENEHLEIZEI EL Note from these conditions it can be derived that 035ocgt 22J2oc which leads to a gt A3sjz E 163 Matlab script for the preceding curves and analysis clear clf numa1 dena1 2 0 numb1 0 denb1 2 0 sysatfnumadena sysbminrealtfnumbdenb figure1 rlocussysa0120 axisl axis hold on rlocussys a hold off axisaxisl titlevarsprintf39Rootlocus in a with b039 titletitlevar figure2 rlocus sysb0120 axis2axis hold on rlocussysb hold off axisaxis2 titlevarsprintf39Rootlocus in b with a039 titletitlevar Set a4 and do the rlocus for b again numb2numb denb21 2 4 sysb2minrealtfnumb2denb2 figure3 rlocussysa 0 2 8 hold on rlocussysb2 hold off axis5 1 3 3 titlevarsprintf39Rootlocus in b with a439 titletitlevar Set b4 and do the rlocus for a again numb3numb denb31 2 8 sysb3minrealtfnumb3denb3 figure4 rlocussysa0 2 12 rlocussysb2 rlocussysb3 hold off axis7 1 4 4 titlevarsprintf39Rootlocus in b with a4839 titletitlevar Set a20 and do the rlocus for b again numb4numb denb41 2 20 sysb4minrealtfnumb4denb4 figure5 rlocussysa0 5 25 old on rlocussysb40 1 5 rlocussysb4 hold off axis10 2 6 6 titlevarsprintf39Rootlocus in b with a2039 titletitlevar grid 12 of 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 9t h ed Prentice Hall 2001 ISBN 013030660 6 Chapter 3 Lecture 1 State Variable Models Timedomain techniques can be utilized for nonlinear timevarying and multivariable systems A timevarying control system is a system for Which one or more of the parameters ofthe system may vary as a function of time We are interested in reconsidering the timedomain description of dynamic systems as they are represented by the system differential equation The timedomain analysis and design of control systems utilize the concept of the state of a system The state ofa system is a set ofvariables such that the knowledge of these variables and the input functions will with the equations describing the dynamics provide the future state and output of the system This is also the equivalent of a multipleinput multipleoutput MIMO system For a dynamic system the state of a system is described in terms of a set of state variables X1 t X2 txn t Input signals Output signals With the output as a combination of states and inputs represented as yQ z The state variables are those variables that determine the future behavior of a system When the present state of the system and the excitation signals are known Page 1 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modem Control Systems 9quot ed Prentice Hall 2001 ISBN 0130306606 An example to demonstrate the concept A springmass damper system dzy alt2 M bgkyu Wall friction b in 11ml p Js2 YsbsYskYsUs De ne a set of state variables as x1 x2 position and velocity where x1yt 2 dt Substituting this into the system equation dx M7bx2 kx1 ut Then we can describe the system using two equations having two state variables dx1 x2 dt dx k 7 Mxl M x2 ut The statevariable matrix equation is m 0 1 m k b MFu amp E M x2 1 dt yh Mou Page 2 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 9 11 ed Prentice Hall 2001 ISBN 013030660 6 Annther Exampletn demnhserztethe cnnczpt P2 3 A coupled spnngrmass system rs shown m Frg P2 3 FIGURE F23 Twormass system The masses and spnngs are assumedto be equal Obtam the dAfferenual equahorrs descnbmg the system d r 112 15 dz m Athkux2 rm 0 M kxkuxrx2Ft M Therefore we wru de ne a set ofstate yarrables as rrzr3r where Substrtutmg thrs mto the system equahor MTEk1x1kux rrzFz Mbx rum 1 0 Put thxs m a mamx form a x um x F0 011 M M M M 011 M M M Page 3 Nates mehgeres are based m mtrrm mm materials mthe EcE 371 cause terthmk RC DarrmaRxx ExshnpMademCammlS gemggmedPrenuceIIHJLZUUI ISENEIJLEIZEI EIr 1223 cunt am 7 X 11x20 7 K d 3 d M M They can be put thto a mathx form The statervanable mamx equation is dz 0 1 0 0 7 1 0 1t X2 0 i 0 0 112 2 14 Jm 0 i x 0 M t wtth the two outputs bethg x and he wutteh as outputs 1h y as 7 1 1 0 0 0 x 0 1 e 2 ha y 0 1 0 0 x3 0 7 5 OK now that the form looks ught so whst7 Lthear Algebra has a purpose Lthear A1gebhats usedto solve systems of equauOhs control system problems Page 4 Nates and gures he based uh mm mm muteuus mthe ECE 71 cause texthuuk RC Darf andRH ExshnpMademCanlmlS gem ameu Prenuce Hall 2mm SEN Dell z e State Variable Equations In general we are interested in developing a set of state differential equations of the form where 9 dx dt 39 XI aux1 aux2 a1nxn bnu1 b12u2 b1mum xi aux1 azzx2 a2nxn b21u1 b22u2 b2mum 5cquot amx1 anzx2 amxn bmu1 bnzu2 bmum using vectormatrix notation we have x A E 2 x1 an all 39 39 39 aln x1 bu 1m u 1 1 x2 an an 39 39 39 an x2 b21 39 39 39 blm dt z z um xn am anl arm xn bnl brim and y1 cllxl cllx2 quot39cinxn d11u1 d12u2 quot39d1mum y2 621x1 czzx2 quot3962nxn d21u1 d22u2 quot39d2mum ykck1x1ck2x2cknxndk1u1d u d u k2 2 km m yl 611 612 clquot 1 6111 lm u 1 yZ 621 622 Clquot x2 6121 dim um yk ckl ckz 61m xn dkl diam 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 Solving for the State Variable System Equations When we have a scalar equation we can solve it using Laplace Transform Methods as 5c a x b u SXS x0 aXs b Us s aXs x0 bUs Then we should be able to de ne the solutions as 0 exp A t 0jexpEt r grdr 0 Note The matrix exponential function exists and is de ned as Azjz 4k eileprtIAt 2 k It looks like a Taylor series expansion of the exponential Page 6 Notes and figures are based on or taken from materials in the ECE 371 course textboo RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 Another way of thinking about this problem involves the inverse function If we let Phi 33 g ll which is the Laplace transform of exp A t Then Xs SI 41 new st 41 gm Us becomes IIUU skgw gw The matrix exponential function describes the unforced response of the system and for control systems is called the fundamental or state transition matrix Therefore the result exp A t 0jexpEt d 0 can be written as at qgtltrgt ltogtigltr agar d The solution to the unforced system that is when 20 Q is simply x10 x0 x2 I 21 22 39 39 39 2n x2 0 xr W Page 7 Notes and figures are based on or taken from materials in the ECE 371 course textboo 39 RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 To determine the state transition matrix we can go through a tedious procedure x10 11 12 1n x10 x20 2 21 22 2n x20 xn I n1 n2 rm xn 0 1 Set the initial conditions for all state variables to zero 2 Select one state variable make it nonzero and evaluate the time response of all the state variables The resulting time response provides one column of the time domain state transition matrix That is the term 4 ij t is the response of the ith state variable due to an initial condition on the jth state variable when there are zero initial conditions on all the other states 3n Repeat for every state variable until the state transition matrix is complete If you don t have access to set all of the initial conditions of the state your in the real world There are ways to perform state estimation more advanced classes 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 If the solution to the state equations is 2ltsgt 2ltsgtgtfltsgt The solution for the multiple system outputs becomes zsg sggs 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 Q 2 Under these assumptions the solution becomes Zs Q2s ys B gsg2s where 23 Q ll Page 9 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 The de nition of an eigenvalue from Linear Algebra is liMl39rg We have been dwelling on the characteristic equation Can we de ne the characteristic equation based on the state transition matrix Sure detls j 0 Recognize something familiar The eigenvalues are the roots of the characteristic equation Most of the really far out stuff in control systems advanced signal processing image processing etc involves vectors matrices eigenvalues and eigenvectors 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 Example FIGURE 34 An FILC cwcmt d v L R x d Rx Let x v and x2 1 and M v39 we get 0 7 x dz 1 R alt L and V1sz age 11 Nates and gures are we an mm mm mum m39he ECE 71 cause mum RC Dmme x ExshnpMademCammlS gemggmedPxenuceHalll l xsaummmm Then we have 1 2 M ASl CR L L l R l s s L C L C l l S S gs L L Solving for the transfer function R 1 s L C 1 S 1 9sg 2s 0 R LR 1 E sz s 0 L LC l RJI s L C L i LC LC gltsgt 1o R1 2 1 2 R 1 35 L piji L C L C 1 L S E X 0 U 0 2 R 1 2 R 1 S s s s s L LC L LC PagelZ 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 R 3L 1andC 05 i 1 93 LC 0 5 6 2 S SL sz3s2 s2sl L LC 2s S is 12 2 S2 SL s 3s2 L LC R l sZ 1 33 2 33 2 s L l s l 3 23 R 1 S2 S sz3s2 sls2 L LC 20 2zt aerial Initial Condition Response t 2tr0 2et e 2e t 2e A0 0 6quot e39 equot 2639 Page 13 Notes and figures are based on or taken from materials in the ECE 371 course textboo RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 IVIaking z SignalFlnw Graph mm the State Mndels to de ne a smgle transfer function um mm 0 snumc FIGURE 34 An FILC cvcun URE 35 new grap m 5L0 netwam Usmg Mason39s signalr ow gam formula we obtam the transferfuncuon Page 14 Nules and gures are based un ur uken mm rra39anals m LheECE 371 cuurse lmbuuk R C DurfandR H Exshup Madam Cuntml Systems 9391 ed Prenuce Hall Z l EN Delir l r EXAMPLE Spread of an epidemic disease 39 I 7 and 1 39 39 The 39 39 and in IIquot 391 immunization dea39h 0i isolation from XI The feedback systemcan be represented by me following equations 7 m m gt um ell I Iquot e mi 1 use m T tuxws i i i and F12 A population we have u1t u2t a i Thephysical m aiia 39 x2andx3w 39 dial represents mis set ofdifferential equations is shown in Fig 317 epidemic disease The vector differential equation is equal to n u a H A i u a a 7y i i um Y a y n n u n quot1 By mininnF n L 39 In 1L H a depmdent on x1 and x2 and does not affect die variables x1 and x2 Page 15 Nutes and gures are based un ur taken 39nm materials in the ECE 371 euurse textbuuk RC Durfand RH Bishup Madam Cuntxul SysLems 9391 ed Prentice Hall ZUEIl ISBN Irlir l r A closed population no inputs at the steady state Let us consider a closed population so that u1t u2t 0 The equilibrium point in the state space for this system is obtained by setting dXdt 0 The equilibrium point in the state space is the point to which the system settles to the steady state or rest condition Wehave x1 0 a b 0 x1 ix2 0 b g 0 x2 dt x3 0 a g 0 x3 0 ax1 bx2 0 bxl gx2 0 ax1gx2 Which requires that 0 g bx2 x1 0 0 abx1 x2 0 0 ax1gx2 x3 To determine whether the epidemic disease is eliminated from the population we must obtain the characteristic equation of the system and figure out if it is stable or unstable Stable and the disease is eliminated unstable and the disease wipes out the population 3 0 0 a b 0 36 b 0 detsI Adet 0 s 0 b g 0 det b sg 0 0 0 s a g 0 a g s detsI Asasgs00 b bs detsI As3agszagbZs Page 16 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 Or from Mason s Flow there are three loops tWo ofWhich are nontouching Thus the characteristic equation is 2 Ar1a g Arr2 xagagb2 S S For stability 11 ggt 0 and ag b2gt 0 When roots are in the le hand plane We expect the unforced response to decay to zero as t 9 in nity Page 17 39 39 39 39 39 FCF1 1 coursetextbook RC Dorfand RH Bishop Modem Control Systems 9 ed Prentice Hall 2001 ISBN 0130306606 EXAMILE 33 Inverted pendulum cnntrnl uJ 4 39 MN nauns 349 A can me an lmar aa mm m m I Fnuumku mmta md w mm mm m m mm Nana r h Wu w F V upnghtposmon The state vanables mustbe expressedm terms ofthe angular rotation the system can be obtamed by wnhng the sum ofthe forces m the horizontal dArecuon A L We mu horizontal dArecnon is M mi 7 um pomt The sum of the torques abounhe vaotpomtxs my mI39H 7 mlgH o The state vanables for the two 52 s are chosen cond7order equahon as x1 x2 x3 x4 7 y dydt 6 016011 Then Eqs 3 58 and 3 59 are wnuen m terms of the state vanables as 1115quot um 1 n 153 7 gx 70 age 18 Nuns And gures he based ah mum Flam materials mm 505 71 cause mum RC DarfandRH ExshnpMademCammlS gemggmedPxenuceHalll l ISENEIJLEIZEI EL To obtain the necessary firstorder differential equations we solve for dX4dt and substitute to obtain Mir mgr MU since Mgtgtm Substituting dX2dt we have Mug Mgr milquot 1 Therefore the four firstorder differential equations can be written as was I v 3M a 13gt 1 9 Mr 33 quot will 1 1 M I E 1 T E 1 l i 3 39 r M 39 Thus the system matrices are U l U H y 7 ll 7 1 El mgquot1 139 l 539 M A E V fquot B E 1 IE I 391 i U Rn El 3 L J 39LquotML At steady state y x1 undertermined d yX20 dt t9x30 zxazo dt Anyone what to check stability It isn t Poles at 0 0 and i E Page 19 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 Note Ins o en a dAf cult taskto determine a senes of rstrorder dAfferenual equations funcuon Page 20 Nates and gures are we an mm mm mum m39he ECE 71 cause mum RC Dmme x ExshnpMademCammlS gemggmedPxenuceHalll l xsaummmm Deriving a canonical form for the state variable models of a 8180 system The general transfer function for a single input single output system can be Wrinen as new 771 LNV bm39 U Uv 1anr squota s bv 1 170 139 an 139 We which to show What looks like a sequence of integrators or squot l 1 r m Y S A j m r O 0 H O 0 o e O o 4 A 39 V 1 Then there are multiple canonical forms of the result 0 Phase Variable Canonical form 0 Input feedforward form 0 Physical variable form 0 Jordan Canonical form Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modem Control S tems 9 h ed Prentice Hall 2001 ISBN 0130305505 Phase Variable Canonical form Um x 7 xua xquotquotax anXS Then S m SHJW 91 WM 40 r m MSMyawanw s The mm x form for a 4 node system ls men x o 1 o o 1x20010xzo 39o 1 dtxg 0 0 0 1 a X 41 42 4 X X x yrbn b b bsl39 x Page 22 and gures are based nn nrtaken fmm materials m the ECE 371 cmursetextbnnk Nntes RC DnrfandRH BxshnpMndeanntml Systems SmedFrenuceHall 2001 ISBN 0713703066076 Input Feed Forward Canonical form Xs s 5 Jam5quotquot a anYs m squot 2 s39quot bm4SWquot b bnUs Xs1aHsquot a ans 175 Y s Xsr aH 4quot asquot H an s39quotYs Xs Q7quot 539 quot bH 39quot quotquot b 539 bn Us V 3 H Qym W bH 5Wquot b 539 bn s39quot U 7aHsquot a 2 as39quot ans39 Ys H Then de ne the equivalent levels ofintegration as the appropriate value of states The matrix x form for a4 node system is then x 7 a3 1 0 0 x b3 x r a 0 1 0 x b g 2 2 2 2 W dz x3 7 a 0 0 1 X b x4 7 an 0 0 0 x4 bu x 2041 o o 01 2 x4 Fag Notes and gures are based on or taken from matenals m the ECE 371 course textbook RCDrFandRHR hm h Dranh nu am Hm Physical variable form Conlrollcr Molm39 and load Field Field vollngc Velocin 515 l t 5JD cuncnl 7 I R gt 7 7 m cm 1392t 1xh m Gltsgtt tmtt De ne the node variables as shown and build the matrix form x1 73 6 0 x1 0 x2 0 72 720xz5ut x3 0 0 75 x3 1 x1 x2 NH 0 o 01 x x4 Page 24 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control S tems 93911 ed Prentice Hall 2001 ISBN 0130306606 Jordan Canonical form Also known as the diagonal form when all the poles are real Conlrol Icr Mulm39 21ml luzlrl Field Field vollngc cuncnl 7 1x3r Velocin 515 It m Rm gt C m 55t 7Ys7 30sl 7A B C W W s25355 Gs 72071030 52 53 Page 25 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control S tems 93911 ed Prentice Hall 2001 ISBN 0130306606 Department of Electrical and Computer Engineering 0119 of Engineering and Applied Suences ECE 47005950 Feedback Systems Westem Mchxgan University College ofEngmeenng and Applied Sciences panment of Elecmcal and Computer Engmeenng 1 chxgan Ave Kalamazoo MI 4900875329 De Course Objectives an ability to analyze design simulate and experimentally validate linear control systems While taking into account practical limitations of operations an ability to utilize circuit simumton mm or mammmtua W we tools for control system design and analysis an understanding of feedback systems and their application to circuit analysis and control system design an understanding of frequency compensation and its application to control system design a knowledge of compensation for disturbances in control systems ECE 47005950 2 Content Review Review and ex and on our 1 reVious knowledve of Linear System Transfer functions steadystate error 2nd order systems characteristic equation stathy loot locus methods Frequency Response What the transfer functions means in the frequency domain Filters as described by a Laplace transfer function Linear Control System feedback design methods Root locus based Frequency response based Exposure to statespace control systems Exposure to digital control systems ECE 47005950 Feedback Desiun A L roaches Design 5 ecifications Settling time Percent overshoot Rise time Tracking errors system type and gain Using the openloop responses Root locus design methods Bode plot 0 dB frequency and the gain and phase margin Compensators Proportional integral and differential PID Phaselead phaselag and leadlag networks ECE 47005950 Final Exam Part 1 Inclass exam type questions With fewer points per answer Basic design elements for a linear system 39 Part 2 New material since exam 2 P13 10 Digital control system proportional controller root locus step response Chap 11 oscillator state space 2 X 2 oscillator state feedback control state estimator and observer state feedback When feedback used there is no input only initial conditions When only the system or observer there is a step response and no initial conditions you may want to try initial conditions particularly with the observer states different than system states ECE 47005950 Final Exam Part 3a Controller Design PI and PID DP108 PI and PID Controllers for a system With time delay Analytically set up as though no delay exists then add the additional phase or overshoot at a frequency due to the time delay Simulation use PADE to simulate the time delay You should check that in the fre Luencd domain the anal tical answer and PADE simulation are approximately the same if not increase the order of PADE The specs can be metrwith a PI controller you should be able to do better with a PID ECE 47005950 6 Final Exam Part 3b Controller Design Lead Lag and LeadLag P1021 P1022 and P1023 Dual phaselead phaselag and lead lag compensators Multiple phaselead compensators provide gt90 degrees phase but the 0 dB crossing can only happen in a region where the original open 100 s stem was at 60 dBdecade Compare the system with K20 and no compensation to the compensated system 0 For P1021 if you use 2 leads that maintain the 0 dB frequency you will need a phaselag to restore the lost DC gain 0 The 0 dB frequencies and therefore step response settling times are different for P1021 P1022 and P1023 Expect WP1021gtWP1023gtWP1022 ECE 47005950 7 P1310 Go 1 ss10 Digital proportional control Root locus 30 overshoot Compute zdomain pole locations ECE 47005950 Chap 11 Observer controller Similar to examl 1e 1110 1 rovided in class and MATLAB You will need to construct an augmented matrix for the observer an error is helpful to see convergence but it was not part or the questlon ECE 47005950 DP108 TillV Controller Ys39 l T RU GLS 26 gt Engine 0215 1X45 1 spccd PI and PID compensation for T0066 10 overshoot 10 sec settling time 0 Zero steadystate error to a step ECE 47005950 10 P102122 and 23 K G 39 A sis2lHS6li Velocity constant Kv20 Phase margin 45 P1021 Closed loop bandwidth greater than n radsec open loop 0 dB crossing point P1022 Closed loop bandwidth greater than w2 radsec P1023 Closed loop bandwidth greater than 4ltwlt10 ECE 47005950 11 Special Note Filter Design Methods Spectral Power Responses For Tjw Tjw expzTjw What is T jw Magnitude must be the same Phase 2 4 w M N 4Tjw i atanl R Z atanl Z atan 2 11 21 2 m4 Pm n1 1 w W 2 4quot w Q M N 4T jw Zatan WR Zatan WJ Zatan wquot2 11 21 2 m4 Pm n1 1 w W Therefore T jw Tjw and AT jw 4Tjw 0 Then what is TjwT jw Tjw eXpzTjwTjw eXp 4Tjw or TJIW T 1W ITOW12 This is a power term notice the square so we use 10log to create decibels It is the same result Note this works for ss too This is the generic form for de ning the magnitude response of a lter Why The poles and zeros are symmetric about the jw axis of the splane Therefore the LHP and RHP elements can be separated into Ts and Ts and guarantee marginal stability Page 1 The Butterworth Lowpass Filter ROW7km MN 7 1an2 1r Swn 171rnj Eunerwunh Huey Characteristic Eq Frequency normalized As AG 3 1 71 3quot 0 an Valkenburg Analog Filter Design Oxford Univ Press1982 Reference ME V ISBN 0195107349 Page 2 Solving for the Butterworth Filter poles Filter in jw Tn jw Tn jw 1 2 1 V 1 WO Laplace Tn Tn s 1 2n 1 2n 1 2n 1 woj 1 2quot 1 1quotV0 2n Characteristic equation 1 01 j As A s 0 0 Normalize AsA s1 1quotSZ 0 For 11 odd AsA s1 s2 1s 1 s 0 Rootsat sz 1expj2m 9 Szexpm n Let As be the LHP poles and A s be the RHP poles For 11 even As A 3 1 32quot 1 jsquot 1 1393quot 0 Roots at 32quot 1 eXpj2m 7r 17239 9 s eXp mj n Let As be the LHP poles and A s be the RHP poles Page 3 Matlab Code BW Filter generation demonstration mo close all clear all Rinl Rloadl Rmatchl PBfregl PiWlog5paceloglOPBfreg72loglOPBfreg21024 I r I I muucr colorseg b ii0 PolesRange6ell for BWnPole5Range iimodii6l denProot571PBfreg 2 BWn zerosl2BWnel 1 Y sortrealdenP denPsortdenPD denpolydenPsortlBWn figurel plotrealdenPimagdenP5printf39cx39color5egii title39Power Magnitude Poles39 grid on hold on num PBfreg BWn zpiab5root5num ppiab5root5den BW5y5tfnumden PiMAG PiPHASEbodeBW5y5PiW figure2 5emilogxPiW dBV5gueezePiMAGcolor5egii title39Power vs Frequency39 xlabel39Freg rad5ec39 ylabel39Magnitude dB39 plotvaxi5 axi5plotvl plotv2 7120 10 figure3 5emilogxPiW 5gueezePiPHASEcolor5egii grid on hold on title39Pha5e v5 Freguency39 xlabel39Freg rad5ec39 ylabel39Pha5e39 axi5plotvl plotv2 emaxPole5Range90 15 pause end Page 4 Magmtude dB Phage Results PoweFMagmtude Po es Powervs Frequency Freq Fadsec Phase v5 Frequency Freq radsec Page 5 What if we want to change the frequency mom st 1 Hm WM Just change the natural frequency w0 27f0 the center frequency is simply scaled w TnsTVl S 2n 0 n in We 1 S Design approach 1 Determine the order of the lter you want What attenuation do you need at the 10 w0 point There are plenty of curves like those above if the value you need comes before t 10X the cutoff frequency 2 Generate the Butterworth Coefficients on the unit circle for wl 3 Scale the poles by the desired frequency remember that wl is in radianssec therefore multiply by w0 27f0 Page 6 The Chebyshev Type 1 Lowpass Filter 4 182 Cn 2 C w cosn cos391w for M 31 coshn cosh391w for lwl gt 1 Tn139wTn M Chebyshev Type 1 Filter n15 Buttervv unh Fmer n1 5 xx gtlt X X gtlt gtlt X 05 gtlt 3 E x x 35 x X X XXX gt0 x 1 1 1 2 715 1 1 15 2 Rea1Ax1s Reference ME Van Valkenburg Analog Filter Design Oxford UniV Press 1982 ISBN 0195107349 Page 7 Active Audio Frequency Filters An active lowpass lter implementation of a lSt order Butterworth lter CZ II II R2 WAVAVL R1 Vin W The transfer function for this circuit is Vouts R 2 Vins R1 1 sRZCZ MaxGaz39rzG R1 To tune the circuit Page 8 An alternate approach Vdc Vouts R0 Rb 1 Vins Rb 1 sRlCl R R MaxGainGa b w0 L Rb R101 To tune the circuit Page 9 A Second Order Butterworth Lowpass Filter Let s do the math for the second order system for n2 and w0 1 1 1 T2ST2 S 22 22 1y 1 12j J39wo WO 1 1 T T M 2 s 1H4 1 1 T d T 2S 32 2s1i an 2 S isZ s1i For 713 T23 1 1 32 2s1s22 s1 A second order underdamped system with 2 if xE or if XE 0707 i1 2 1 r 31 32 9 4 J5 J5 After frequency scaling 3132 gw0iw0 g2 1W ijw5 How to make a second order LPF for audio Page 10 SallenKey Circuit Lowpass Filter An active lowpass lter implementation of a unity gain Friend Circuit also referred to as a SallenKey circuit as described in Walter G Jung IC OPAmp Cookbook Howard W Sams Co Inc Indianapoli IN 1974 R1 V1 am The transfer function for this circuit is a generic second order f11ter equation is also shown Rum 1 Vout R3 C1C2R1R2 Kw2 Vls 1 1 R3 I 1 322 wsw2 2 S S C2R1 C2R2 C1R2 C1C2R1R2 VoutsR3R4 1 V1s R3 1sC1R2C1R1 R3C2R1s2 C1C2R1R2 Letting C1C2C and R1R2R and 6 MaxGaz39n G R3R4 1 R3 w R Page 1 1 Function Derivation The circuit derivation assumes a perfect opamp with in nite gain in nite input impedance and zero output impedance nonlimiting power supplies and voltage drops and no frequency response considerations The circuit derivation follows V2 iLsC2 EQV04CZ R1 R2 R1 R2 V2 VP LSC1 R2 R2 R3 Vo R3R4 Vn Letting VpVn VOLLHC1Q R3R4 R2 R2 R31sC1R2 V2 Vo R3R4 V2 iLsC2 ELVOSCZ R1 R2 R1 R21sC1R2 V2 iS39 C1SC2 EVosC2 R1 1sC1R2 R1 R31sC1R2 1 sCl V1 V0 SC2 VosC2 R3R4 R1 1sC1R2 R1 R3 1sC1R2sC1R1sC2R1szC1C2R1R2 V1 0 V0sC2 R3R4 R1 R1 R4 2 1sClR2sC1R1 sC2R1 s C1C2R1R2 R3 R3 V1 V0 R3R4 R1 R1 Page 12 V0R3R4 1 R3 Letting Resulting in Note that for a stable system Implying that V1 13C1R2sC1R1 sC2R1s2 C1C2R1R2 R3R4 1 R3 C1C2R1R2 1 L 1 Re3 L 1 C2R139C2R2 C1R2 39C1C2R1R2 amp V1 szs C1C2CandR1R2RandGM R3 1 E CRY V1 2 3 6 1 s 2 CR CR MaxGainszm 1 R3 w CR 3 G CI T 1 Glt3 0SR4lt2R3 Page 13 Multiple Feedback MFB Circuit Lowpass Filter An active lowpass lter implementation of a multiple feedback circuit MFB that is may also be referred to as a derivative of the SallenKey Filter R3 Vout Vdc V Figure 1 Sallen Key Lowpass Filter The transfer function for this circuit is E l yClC2R2R3 V1 R1 32 S391R11R21R31C2R2R3 E j 1 V1 R1 1sC2 RZ R1C2R3C2R2szC1C2R2R3 Resulting in R3 MaxGam G E W 1C2R2R3 Page 14 Function Derivation The circuit derivation assumes a perfect opamp with in nite gain in nite input impedance and zero output impedance nonlimiting power supplies and voltage drops and no frequency response considerations The circuit derivation follows V2 LLLSC1 EE R1 R2 R3 R1 R3 sC2VoQ0 R2 Combining Vo sC2R2iiisClZi2 R1 R2 R3 VOSCZR2R1R2R1R3R2R3sC1R1R2R3E V1 R1R2R3 R3E V0 R1sC2R1R2R1R3R2R3s2 C1C2R1R2R3 R1R3 R1 V0 R3 V1 R1sC2R1R2R1R3R2R3s2C1C2R1R2R3 E l yCLCZRZR3 V1 R1 325391R11R21R31C2R2R3 Resulting in R3 l MaxGam G R1 W C1C2 R2R3 And 4 W391R11R2139R3 Page 15 Higher Order Butterworth Lowpass Filters Take multiple stages and cascade them Remember 39 39 stagem As a ruleofthumb you should select the order for the stages ofyour lter Ifyou look at the output of each stage it will be the product ofthe transfer functions to that location So possible use those with damping factors closest to one before the smaller ones ColnpleeronjugmerPale Pairs LuwestQ gt Higheer Optional Figure 3 Building EvenOrder Filters by cascading SecondOrder Stages Opricnnl Complex 0ujugarerpole Pairs I R St 2 39 V39 A T C Lowest gt Highestu Optional SecondOrder Stages and e 7 Real Pole Figure 4 Building OddOrder Filters by Cascading Audi 9 a Single Real Pol 11m Karla Texas Instruments nnlw at m n w r 04 n m ml r on Note 1 Real elements may not exactly match the values you select 2 Components have a tolerance they are within some 3 rcr u r r What do RF designers do Why might it be different Page 16 Low Pass Filter 3ml Order The classic 3rd order LC Ladder Low Pass Filter L1 L3 Vin n M M A Vout CZ V Figure 2 LC Ladder 3rel Order Low Pass Filter The circuit derivation assumes a source and load resistance The source resistance is placed prior to the input voltage and the load is placed on the output For RSRLR and L1L2L M Vin L CR 2 CL 1s 1s s R 2 2 Page 17 Theoretical Derivation The circuit derivation assumes a source and load resistance The source resistance is placed prior to the input voltage and the load is placed on the output The circuit node equations follow VoutL 1 V2 1 RL sL3 sL3 V2 sC2 1 Vout 1 Vz39n RssLl sL3 sL3 RssLl Solving for V2 and substituting V2 Vow RL Vow M c V0m Wm 4 RL RssLl sL3 sL3 RssLl 1 2m L3 RssLl VWRLsL3RssL1sL3s2 C2L3RssL17RssL1RLVm RLsL3 RssL1sL3s2C2L3RssLl Vout Vout RL RssLlsL3 s sL3RL Vouti sL3RL Vin 7RLsL3RssL1sL3s2 C2L3RssL17RssL1RL Vouti sL3RL Vin RLsL3HZ CzL3RssL1sL3RssL1sL3s2 C2L3RssL1 Vouti RL Vin RLSC2RLRSS2 CzL1RLRssL1sL3s2 C2L3RssL1 Vouti RL Vin RLRssL1L3C2RLRssz C2L1RLC2L3Rss3 C2L3L1 Page 18 1 Vouti RL Vin RLRSJ L1L3 RLRs 2 C2L1RLC2L3Rs 3 C2L3L1 1s 7C2 3 RLRS RLRS RLRS RLRS ForRsRLR Vouti R Vin 2RsL1L3CzRZSZ C2L1L3Rs3 C2L3L1 0r Vouti 1 V 2s gczR s2CzR g s3C2R g R R R R R ForL1L2L Vouti RL Vin RLRss2LCRLRss2CLRLRss3CL2 RL Vout AL FRS 2 V 1s 2 L CRL39RS s2CLH3L RLRS RLRS RLRS ForRsRLRandL1L2L Vouti 2 7 7 2 Vm 1s C szCLsjCIL 1s 1sC Rsz 2 2R R 2 2 Page 19 Theoretical Derivation Pi Filter L2 Vin O m 0 Vout C1 CZ Figure 3 LC Ladder 3rel Order Low Pass Filter The circuit derivation assumes a source and load resistance The source resistance is placed prior to the input voltage and the load is placed on the output ForRsRLR and ClC3C Vour1 1 Vin 2 lsCRlsisz 2R 2 Page 20 Theoretical Derivation Pi Filter The circuit derivation assumes a source and load resistance The source resistance is placed prior to the input voltage and the load is placed on the output The circuit node equations follow Vout L 1 SC3 V2 1 RL 3 L2 sL2 V2 LSCl 1 Vout 1 Vz39n i Rs sL2 sL2 Rs Solving for V2 and substituting RLsL2szL2C3RLJ V2 Vout RL 2 Vow LHCH 1 V0m 1 V L RL Rs sL2 sL2 Rs RLsL2szL2C3RL RssL2szC1L2Rs l l V0ut 2 Vm R s s Vout RL 3 L2 Rs RLRssL2RsL2RLs2L2C3RLRsL2C1RsRLL22 33L22C3RLL22C1Rssquot L22C1C3RLRs RLRS Vm 1 Vout sLZRsRL Rs RsRLsC3RLRsC1RsRLL2 32L2C3RLL2C1Rss3L2C1C3RLRs Vm1 Vout Rs RL Rs Vout RL Vin RsRLsC3RLRsC1RsRLL2 s2 L2C3RLL2C1RSS3 L2C1C3RLRs Page 21 Vout RL Vin 1 RHRL 1sC3C1 RS39 RL L L2 LS2 L2C3RLL2C1RS RSRL 39 RsRL39 RSRL 33L2C1C3 RsRL For RSRLRI Vout 1 Vin 2 J1s RC3RC1E s 2R 1 2 L2C3C1s3 L2C1C35 For C1C3C Vout RL 1 Viquot RHRL 1s2C39RS39RLL2s2L2Cs3L2CC RS39RL RsRL RSRL ForRsRLR and C1C3C Voutl 1 Vm 21s CR Js2LCs3LC2 Vour1 1 Vm 2lsCR1sisz 2R 2 Page 22 15 MHz Low Pass Filter 73911 Order Cuilcra P7LP155 The 7m order elhpucal Lc Ladder Low Pass Frlter Figure 4 Cuilcran LC Ladder Luszss Film Manufacture frequmcy39 response Anenualiun ma Alummian Ida Amnuilian ma Fraqnency MHz ma Frequency MHz Cnilcmft Lc Ladder Luszss Finer Page 23 Test Analysis A test circuit was built and tested using the network analyzer R is assumed to be 50 ohms L1 and L3 were variable inductors in the range of0578 to 095 uH and C2 was 4 parallel 100pF 101K capacitors or 400 pF Network Analyzer Measurements 5mm 5mm LEYYER LEYYER spans spans am am spans spans msz msz um um pm pm smx DEV smx DEV mm mm mm mm 5mm 5mm LEYYER LEYYER spans spans am am spans spans msz msz um um pm pm smx DEV smx DEV mm mm mm mm Page 24 Selecting RMI Filters based on frequency and bandwidth me Vemmn mamaunnaL GmEml techmml mfmmunn mm WWer Lmn cmmgmdmssawgdf gg TECHYN39FO gdf Page 25 1 2 3 4 References Walter G Jung IC OPAmp Cookbook Howard W Sams Co Inc Indianapoli IN 1974 ME Van Valkenburg Analog Filter Design Oxford 1982 ISBN 0195107349 httpwww 39 quot mm lterhtml httn39 fnm ti t n h nntentT phionr quot T 78ampnavSectionaDD notes TI Application Notes Slod006b Sloa093 TI Application Notes on Filtering Active Filter Design Techniques SLOA088 Analysis of the Sallen Key Architecture Rev B SLOA024 FilterPro MFB and SallenKey LowPass Filter Design Program SBFAOOlA Active LowPass Filter Design Rev A SLOA049 Using the Texas Instruments Filter Design Database SLOA062 Filter Design in Thirty Seconds SLOA093 Filter Design on a Budget SLOA065 More Filter Design on a Budget SLOA096 Page 26 Chapter 3 Lecture 2 State Variable Models Mltiple input multiple output MIMO system Input signals Output signals ulm MU System 113 Dir The state variable equations are de ned using vectormatrix notation as jg quotD lb 2 with the output as a combination of states and inputs represented as y Q 96 Q a The state variables are those variables that determine the future behavior of a system when the present state of the system and the excitation signals are known Solving for the State Variable System Equations and Zs lS ll out Still gems The fundamental matrix or state transition matrix is de ned as Phi 0 2 ll 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 For a more compact notation 2s ggw h Eghlg ms The time domain solutions for the states are then 2 t 2t 0 I20 r1ar dr 0 ct exp At 0jexpEt r grdr zt grt2at Page 2 Notes and figures are based on or taken from materials in the ECE 371 course textboo RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 Farming a sum Space Systems Note Ins o en a dAf culttaskto determine a senes of rstrorder differential equations The above transfer function 5 usmg Mason39s Flow Page 3 Nates and gures are we an mm mm mum m39he ECE 71 cause mum RC Dmme x ExshnpMademCammlS gemggmedPxenuceHalll l xsaummmm If we add feed forward paths x 0 1 0 0 x 0 0 0 1 0 0 1 2 w 1 altv3 0 0 0 1 x3 0 x fan 7a 7a 7a x 1 X1 x A0451 5 b 5 7 7 Thxs 15 called the phase variable cannnical am Page 4 Nates and gures are based an mm mm materials m39he ECE 71 cause mum RC Darf andRH ExshnpMademCammlS gem aha mum Hall 2mm SEN Urlzr z r There are multiple canonical forms that can be used for an arbitrary transfer function 0 Phase Variable Canonical form a derivative ofthe controller or controllability canonical form 0 Input Feed Forward form similar to the observer or observability canonical form Physical variable form 0 Jordan Canonical form Deriving a canonical form for the state variable models of a 8180 system The general transfer function for a single input single output system can be written as r39quot bm4rmquot b v bn 7Us7 s aHs ar 11 So reformat the transfer mction to show negative powers of s G6 Yr 7 r Us s L Ys 7 bquot v39 quotquot bH HM b 439 bn 439 Ur 1aHrquota1r39 anr39 we which to show what looks like a sequence of integrators or rquot l l Mn 1 l r m x g 1 x O 4 H I O O 39 J 7 H 7 O 4 u y r J J H The numerator looks like multiple paths from Masson s Flow The denominator looks like multiple touching loops from Mason s Flow Then we can draw equivalent Mason s Flow Diagrams as Page 5 course textbook RC Dorfand RH Bishop Modem Control stems 9quoth ed Prentice Hall 2001 ISBN 0130306606 The Input Feed Forward Canomeal Form 00 The Phase Vanable Canomeal Form Page 6 Nates and gures are me an mm mm materials mm 505 71 cause textbank RC Dmme x ExshnpMademCammlS gemggmedPxenuceIIalll l xsaummmm Input Feed Forward Canonical form observer canonical form state space s as quotas a Y 711 I n Qym quotbms39quotquotbs bnUs i 3 p Xs1aHquot a 439 ans39quotYs 175 Xsr aH aquot a 539 an s39quotYs Xs 27quot s39 39quot bH 39quot quotquot b 539 bn s Us 175 Qym W bH PMquot b 5 bn s U 7aH squot aH 4392 a 539 an s39quotYs Then de ne the equivalent levels of integration as the appropriate value of states The matrix x form for a4 node system is then x rag 1 0 0 x b d x 7a 0 1 0 x b 7 2 2 2 2 ut dz x 7a 0 0 1 x b x4 ran 0 0 0 x4 bu Xi yr1 o o 01 2 X4 Page 7 Notes and figures are based on or taken from materials In the ECE 371 course textbook RCDrFandeT ihm ma Dranh M n w i Phase Variablz Canonical form YxbSquotbsquotquot hx bnXx U11 Sua qm a x anXs xquot 411 xHa nM s anXx 211a 494aawan5 S The mm x form for a 4 node system 15 then x o 1 o x o d x O O 1 O O I XI 111 drxg o o o 1 x3 0 x ran 5 7a fag x 1 X1 X2 yrb1 b1 b1 b3 gt95 X5 Page 8 Nate and gures are based nn nr taken hm matenals m LheECE 371 nurse tmbnnk R C DnrfandR H Blshnp Mndem Cnntml stsms 9 ed Prentice Hall 2001 ISBN 0713703066076 Deriving the Controller or Controllability Canonical Form Equations If we take the phase variable form and reverse the order of the states The state space matrices become x1 a3 2 a1 610 x1 1 x l 0 0 x 0 i 2 2 ut dt x3 0 1 0 0 x3 0 x4 0 0 1 0 x4 0 x1 x ytb3 b2 b1 bol39 2 x3 x Take the phase variable form and start ipping A ips leftright and updown B ips up down and C ips leftright The states in X are also ip updown Why explain this mode that s not in the textbook MATLAB When converting from a transfer function notation the tf2ss returns the result in the controller canonical form Page 9 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 Physical variable form Make the matrix re ect the physical implementation of the ow graph Conlrullcl Mulor and luzlil F CM FICM Velocily 59 It quot01mgD 1 current 7 Km 71m 7 1 Ym 55 Um 5Zr m 1xu Gltsgtis fiiiii ii i De ne the node variables as shown and build the matrix form x1 73 6 0 x1 0 Exz 0 72 720xz 5ut x3 0 0 75 x3 1 WM 0 o 01 2 This form is o en called a banded matrix form The A matrix has lots of zeros with the values appearing in blocks along the diagonal a band along the diagonal Linear algebra likes bidiagonal and tridiagonal matrix forms Page 10 Note and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control S tems 93911 ed Prentice Hall 2001 ISBN 0130306606 Jordan Canonical form Also known as the diagonal form when all the poles are real Conlrullcl Mulor 21ml luzlrl Ficltl FlL lll Vollagc current Velocily Km elm rm Jslt ts 7Ys7 30sl 7A B C GMm s25355 GsL720 J710 jJr30 Page 11 Notes and gures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control S tems 93911 ed Prentice Hall 2001 ISBN 0130306606 1mm fm nn example 3A MU Currcnl source and The nulpulls Tharefnre we have The sum translth mam Gaye 7 1 RC Dnn m andRHEu Pthrl r Page 53rd guex mm mmtakenfmm m W i E 12 E m L L y0 Rx0 ur 2 manual mm ECE 371 came mmqu 939 ed Puma Han mm mm momma Let R3 Ll and C05 EH i lllultrgt m y ou dets s2 s32sls2 s3 2 gs s1ls2 s1ss2 hnnn QUQ3 Solving for the state transition matrix I 2 eXp I eXp 2 I 2 eXp I 2eXp 2 eXp I eXp 2 eXp I 2eXp 2 I Using to solve for the states we have zltrgt rz0l2r rgardr t 2 eXp I eXp 2 I 2 eXp I 2eXp 2 I x l eXPIexp2I explt rgt2explt 2rgtgt l 4 2 eXp I r eXp 2I 2139 2 eXp I r 2eXp 2I 2139 2 u T T 1 exp I r eXp 2I 2139 eXp I r 2 eXp 2I 2139 0 d Resulting in 2 eXp I eXp 2 I 2 eXp I 2eXp 2 I x eXp I eXp 2 I eXp I 2eXp 2 I 0 2 eXp I r eXp 2I 2139 2 eXp I r eXp 2I 2139 d1 y h ago Page 13 Notes and figures are based on or taken from materials in the ECE 371 course textboo RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 For no input and initial conditions both equal to 1 0 2 2 eXp I eXp 2 I 2 eXp I 2exp 2 9 1 exp I eXp 2 I eXp I 2 eXp 2 i 1 t 2 eXp I eXp 2 I 2 eXp I 2eXp 2 I A l lt 1 exp I p 2 I eXp I 2 eXp 2 I 0 eXp 22 eXp 2I yltrgto 316XPE 239 3explt 2rgt I eXp 2I Matlab Chapter 3 example AO 2l 3 B2O CO 3 DO sysssABC D sys a x1 x2 x1 0 2 x2 1 3 b ul x1 2 x2 0 C x1 x2 yl O 3 d ul yl O Page 14 Notes and figures are based on or taken from materials in the ECE 371 course textboo RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 a1 conditions only 11 l 0121 yTx11simltsysutxogt figurel subplot211plotTxlH rid xlabel39Time sec39 ylabel39xl39 subplot212plotTx2H grid xlabel39Time sec39 ylabel39x239 1 g 05 XRH 0 0 01 02 03 04 05 06 07 08 09 1 Time sec 1 g 05 KKK 0 0 01 02 03 04 05 06 07 08 09 1 Time see 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 0l30 50660 6 figure2 XO2 l yTx11simltsysutx0gt subplot211plotTxlH rid xlabel39Time sec39 ylabel39xl39 subplot212plotTx2H grid xlabel39Time sec39 ylabel39x239 25 0 01 02 03 04 05 06 07 08 09 1 Timeseo 0 01 02 03 04 05 06 07 08 09 1 Timeseo Page 16 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 0l30306606 step reponse figure3 yTXJStepsys p10tTyTX xlabel 39Time sec 39 legend 39Output39 39Xl39 39X239 3 I 1 Output X1 2 5 7 7 X2 7 2 a 15 7 7 1 05 7 7 0 I I I I I I I I I 0 1 2 3 4 5 6 7 8 9 10 Time see 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 0713703066076 o 6 Conveting 55 to numden and dealing the MATLAB precision AlBlClDl ssdatasys numtfdentfss2tfAlBlClDl numtfle 6roundle6numtf dentfle 6round le6dentf systftfnumtfdentf Al O 2 l 3 Bl 2 0 Cl O 3 Bl O numtf O O 6 dentf l 3 2 Transfer function 322 3 s 2 Page18 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 o 6 Conveting numden to ss A2B2C2D2 tf255numtfdentf sy52ssA2B2C2D2 Converted State Space Original State Space A2 3 2 1 B2 l 0 C2 O 6 D2 O Al O 2 l 3 Bl 2 0 Cl O 3 D1 O Notes and figures are based on or taken from materials in the ECE 371 course textbook39 RC Dorf and RH Bishop Modern Control Systems 93911 ed Prentice Hall 2001 ISBN 0130 Page19 50660 6 State Variable Models DiscreteTime Evaluation The state equations are de ned using vectormatrix notation as iv A E 5 2 with the output as a combination of states and inputs represented as y Q E Q 2 Using the basic de nition of a derivative xt AI xt Atao A For a sampled system at a sample rate of UT m xt 72 xt Rearranging we have IT lttgtTA tTBgt IT T tl3Tgt Letting I k T k1T TA kT TgkT This is also written as k1 T39 39 k 39T 206 This is de ned as a solution by recursion 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 Systems 93911 ed Prentice Hall 2001 ISBN 0130306606 Extending this to k2 k2 QT kl Tgkl T39 T39 39 k Tzki T39zk1 g k2T TZ Letting T 2T k2 T T k T Tgk Tgk1 v2 v2 v 1TVALAZ kABLZk 2 4 2 Nb tltkgtg tltklgt Remember the matrix exponential function exists and is de ned as A Alf Aktk e eprtIAt 2 k We have approximated the rst three terms of the expansion estimating further it may look like k eXpEKT 0 trg0 grz glt0gt 2 T3 g0 trgl rz u1 213T3 gl tTzltk 2gtggwtoe n5 g Moe 2 239Tzk1 39E39TZMic1 239T2k This result is based on the matrix exponential as was the original solution and driVing terms based on B times the powers of A This recursive methodology is a form of an In nite Impulse Response IIR Filter from digital signal processing We are also going to the driVing function form shortly Page 21 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 The Root Locus Method The root locus is the path of the roots of the characteristic equation traced out in the s plane as a system parameter is changed Generalized Root Locus Format Structure from the char eqn 0 1 F s Findthe roots of Fs 1j0 321s22SZM Wh F K S ltspgtltspgtltspNgt Or FltSgtK lsles22sZMeXpj921j922j92M s p1sp2sleexpj9P1j9P2 j9PN Which requires Fs K Sp1lsp2lspN exp 1921 jazz 39 39 39 jezM And 4Fltsgtexpltxnifnmgtm p1 p2 F 71 in 271 021 922 02M 9p1 0p2 9pN 1 of 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 9th ed Prentice Hall 2001 ISBN 013030660 6 Root Locus 12Steps Step 1 Using the characteristic equation write it in the root locus form Root Locus Form 01Fs1KPs K is the root locus parameter of interest for 0 S K S 00 Step 2 Factor Ps into poles and zeros 3Zl39322 393zM 0 1 K ltspgtltspgtltspgt Step 3 Locate the Open Loop poles and zeros of Fs in the s plane Define their locations with x poles and 0 zeros 0 3plsp2 pNK 39SZl39SZZquotquot39SZM The root locus begins at the poles K 0 and moves to the zeros K 00 Step 4 Locate the segments of the root loci that are on the real axis Hint Loci lie to the left of odd numbers of poles and zeros This is a result of the angle criteria requiring 71 in27r 91 92 92M 0p1 9p2 0pM Step 5 Determine the number of separate Loci Count the number of separate nite Loci determining the number of poles np and number of zeros n2 The number of separate Loci is max np n2 typically np Step 6 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 The two 2nd order system root loci demonstrated this concept 20f5 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 9th ed Prentice Hall 2001 ISBN 013030660 6 Step 7 For systens with multiple zeros at in nity the loci proceed to in nity along asymptotes centered at 7A and with angles A 22920292 Where nP nz And AnMforq01np nZ 1 nP nz Ass s gt00 71 in 271 01 92 02M 9p1 0p2 9FN win2n 0 0 720Z A A N M A A 727in2727 Step 8 Use Routh Hurwitz Criteria to determine if and where the loci become marginally stable by touching 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 3 of 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 9th ed Prentice Hall 2001 ISBN 013030660 6 Step 9 Determine and plot the break away points from the real axis l a Form K d s 133 b Obtain MS 0 is c Determine the roots One root will be the break away Break away angles are base on the angle criteria 71 in 271 01 92 02M 9p1 0p2 9FN Since you are on the aXis a small amount above or below should not affect the angles from other poles and zeros For the pair or quadruple or 2n poles breaking away the angle criteria must be locally satis ed Therefore they are 2ql 2717 BA 2quot BA7Z Step 10 Determine and plot the angle of the departure of the loci from poles and the arrival at zeros The angles are base on the angle criteria again 71 in 271 021 922 02M 9p1 0p2 9pN This ends the basic portion Two additional steps can be included to provide additional knowledge for the root locus 4of5 Notes and figures are based on or taken from materials in the ECE 371 course textbook RC Dorf and RH Bishop Modern Control Systems 9th ed Prentice Hall 2001 ISBN 013030660 6 Step 11 For expected root locations determine the angle phase criteria to plot the path The angles are base on the angle criteria again 71 in 271 921 922 92M epl 0p2 9pN Examples 45 deg from the origin the line of a design criteria Step 12 For the expected root locations determine the gain K criteria for the root The gain is based on the magnitude again szlls22szM 1 Sp1lsp2lspN Fs1lt K Sp1Sp2spN X szllls22 sZM 5 5 of 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 9th ed Prentice Hall 2001 ISBN 013030660 6 Chapter 10 Lecture 3 The Design of Feedback Control Systems The performance of a feedback control system is of primary importance This subject has been discussed and quantitative measures of performance have been developed defined A suitable control system is 0 stable 0 results in an acceptable response to an input command is less sensitive to system parameter changes results in less steady state error for input commands and is able to reduce the effect of disturbances A feedback control system that provides an optimum performance without any necessary adjustments is rare Page 1 Using a PI Controller for a type 0 system Example 105 Frequency Response K 63 G 2s 105s 1 Resulting Transfer function K K Ts G G 23105s1KG s2 250sKG 1 Roots spsl 425311252 KG 1 125r jJKG 05625 Design goals 1 Settling time of approximately 163 sec poles to the left of TS4 lpl or 075 2 Percent overshot of 10 or less zeta N 05912 3 Steady state error to a step of zero require a type 1 system Estimating the phase from damping chap 9 PM 100 if m 59 We must use a PI controller or PID in order to achieve a zero steady state step error by making the openloop system into a type 1 G SKPI Using phaselead techniques to assign the zero A Make the system type 1 by including the integrator Bode Dagram Gm 7 96 dB at 1 radsec R39n 22 5 deg at 0608 radsec Magnitude dB in o 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 V 1 1 1 1 1 Phase deg Frequency radsec Page 2 B Can this system work if the gain is changed to achieve the phase margin K025 ammm Gm 2D dEaH van520 Pm SBZHEE a UZZBvadsec Step Respunse l l a WWW E SEMHETVVEGEE 177 E 7 7 7 77 7 7 7 7 7 7 7 7 8quot quotquot7 7777 77 7 7 7 7 7 7 7 7 7 7 W 1 g7 m i 7 1 7 i i t 1 1 n 3 TVYESEE Phase Margin achieved with K025 Settling time isn t close An integrator alone will not work C Use the zero and pole setting the zero at the pole boundary for the steady state error z075 and determine the phase margin without gain Bode Dagram Gm hf dB at hf radsec R39n 62 deg at 0591 radsec an H 7 7 E n 7 7 g g 51 m 5o n E a m 3 m w m D Frequency radsec Page 3 It appears to work check the step response Step Response 1 I Sstem sys2c Peak arrplitude 108 l l 12 7 7 7 7 7 H Overshootquot 23 7 7 7 7 V 7 7 7 7 74 t W At time sec 509 Sstem sys2c i i Settling39l39lmesec798 1 7 w w w E 3971 i E M w 0quot n IO 12 Time sec We can add some gain to still maintain the phase margin Kc125 Bode Dagram Gm Inf dB at Inf radsec Rn 592 deg at 0703 radsec M agnitude dB Phase deg t t 1 t 4357quot Hquot quotquot7quotquot HHH quotifquot n t t t t t 2 1oquot 10D 10 10 Frequency radsec Page 4 The step response is now Step Response 1394 7 System sys2c l l Peak arrplitude 1 11 l l 01ershoot 106 1 1 1 Amplitude Tlrre sec If we apply more gain the overshoot will get worse while the settling time decreases To deal with the high overshoot the transfer function can be compensated by a prefilter if the prefilter delay will not be signi cant The text adds a prefilter and suggests concerns if the prefilter does not perfectly cancel the zero Step Response 11 Sstem sys3f3c Peak arrplitude 1 04 1 139 Overshoot 449L Attimesec 06 1 Amplitude Tlrre sec The settling time does increase but the overshoot is reduced Page 5 Using a PI Controller for a type 0 system Example 105 Root Locus Design goals 1 Settling time of approximately 163 sec poles to the left of TS4 lpl or 075 2 Percent overshot of 10 or less zeta N 05912 3 Steady state error to a step of zero require a type 1 system For the PI controller as before place he root at the settling time estimate and make a root locus Root Locus cam 2 02 i Bowen 75 o 977 Demp g 0 609 4va sgr FFeqnenemraasee 1 23 L 7 r magmarymws 0 5 RealM s From the root locus a gain of approximately 202 text 208 looks good Sygtem WEN Step Response Peak amphtude 1 17 quot 1 Overshoot 16 6 x Amme sec 2 82 1 1 i 7 fKi irSystem sys2c 7 7 7 7 7 7 7 7 7 7 H Setthng T me sec 4 84 w 1 l l a U 8 J J U l l S g t t lt 06 e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e l l t t 04 e e e e e e e e e 7 e e e e e e 7 e e e e e e e e e e e e e e e e e e e e e e e H t t o 2 e e e e e e e e e 7 e e e e e e 7 e e e e e e e e e e e e e e e e e e e e e e e e e t t l l 0 2 4 6 8 10 12 Page 6 Using a pre lter Step Response System sys3f3c Amplitude Tlrre sec We did much better with this technique Page 7 Typical Compensators Phase Lead Network Poles Zeros and Gain Form GE 3 for 39239 lt lpl S P 39 1 Frequency Domain Form GE 3 K1 M jwr1 where ri and 013 and K1 p z z Phase Lag Network Poles Zeros and Gain Form GE 3 for 39239 gt lpl S P 39 1 Frequency Domain Form GE S 2 K1 JW 139 jw 01 139 1 where 71 and 013 and KlzK P z p z An Analytical Method An analytical method to determine phaselead or phaselag compensation is presented in Section 109 p 700702 To use this method you must know 1 The desired OdB crossover frequency wc possibly the same as the uncompensated system 2 The phase margin desired in order to provide a phase value 3 The magnitude of the compensation required at we Then check the conditions and solve the equations provided Page 8 A deadbeat response a c at that level with minimal overshoot cntena 1 39 39 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 4 andholds 1 Trad 05T l TS l 5 r 27 Nornmlrzed time See Table 102 This is similar to the optimal coef cients relatedto performance indices from chapter 5 p 303312 Practical definition 1 It has a zero steadystate error a quotL l 39 econuorue W Page 9 78 b Make sure it doesn t nndershoot the 4 boundary once it peaks 1 amp 2 above Example 1012 Deadbeat response as L s s 1 For a third order deadbeat response include a leadlagnetwork W This will require a pre lter to cancel the zero of the compensated system Therefore M K Ksz 3p ssl sslsp G0LSGCS39GS Resulting in the closed loop system of Gs z Ksz S acltsgtlt TS GP1GCSGS 32 ss1s p1ltsz Haz s3 lps2 KpsKz Normalize by s C K and for w 34K 2 z Then solve for a See MATLAB Example ChaplOiEX12m Page 10 Phase LeadiLag Network R2 C2 T Tl gsM sRcgtRzltscz T H 1JR C 15R2C2 Lag 1RC122czR C2CH2R C R2C2 mf BUdeg LeadLa Page 11 Example 1013 Rotor Winder control system Air supply Stepper I motor Ii Air chuck I Q 39M DC motor Input Rotor mmnmnd Winding loop Annature wire a Rm gt H xi b Gltsgt s s51s10 Design goals 1 Steady state error to a step is zero 2 Steady state error to a ramp equal to 5 3 Percent overshot of 5 Therefore 4 m 069 Estimating the phase from damping chap 9 PM 100 4 m 69 This is a type 1 system therefore goal 1 is met and goal 2 can be met with gain KG KG When N21 KV 5GSZSWZEZ This would require KG 2 1000 Page 12 Initial System Characteristics Bade Damam Gm 575dEat7U7vadsec Pm BB7dE aWUZvadsec Maunnuue as Phase neg newech vansec Rum Luc us mamnaN Am 72 4n RialM s Phase Margin Gain Margin 575dB 707 radsec Phase Margin 897 deg 002 radsec Root Locus Becomes unstable at Kgt750 Phase Lead Apply the required gain and determine the amount of phase lead needed Gain Margin 25 dB 707 radsec ammm Gm 725dEat7EI7vadSEE Pm r752 HEB 8 8 EvanSEE Phase Margin 752 deg 1813 radsec g m g Ann M 1 g H V H g 7225 n 1D 1D 1U 1U3 FVEWEHEV vansec o a 1 Therefore sm69 75 l l sin 0 a 716 l s1n0 This is not very practical particularly as the 0 dB point will shift a long ways to the right Page 13 The results aeee Damam aeee Damam Gm maeamawvaesecy Pm 267d2 atlaavawsec Gm anaeaemamaesecy Pm 316d2 aHUvadsec 2 2 m m m m m Fvequencv vansee Fvequencv vansee The step response is terrible Step Respunse E lt me see Page 14 Phase Lag Using the original curve to set a compensation gain and then using gain from the phaselag network The phase margin could be achieved at wl2 radsec Subtracting gain to get to 12 The frequency shift is 589 The new open loop gain is 1000589 1697 The system with gain is then Magnitude dB We will not be able to Magnitude dB Pnase deg Pnase deg Bode Gm 12 9 dB at 7 07 radSec Diagrarn Rn 44 6 deg at 2 84 radsec 71007 72m 71 10 i i i e i 40 Frequency radsec make our phase margin but can add the lagnetwork anyway Bode Diagrarn Gm i2 8 dB at 7 02 radsec Rn 43 6 deg am 84 radsec iiooe 5o 7507 riooe i i i 7 Phaseer i0 Frequency 40 radsec Page 15 The time responses are Step Response Amphtude Now what happens if we add a PhaseLead to the PhaseLag solution Lead Lag Network Therefore sin30 a 1 a 1 BodeEIagram 9n 149 dB at 972 radsec R39n 736 deg at 284 radsec Magnitude dB 1 1nn x x 1357 7 225 7 7 Phase deg 270 t t W W 10 10393 1039 10quot 10B 10 1o 10 Frequency radsec Page 16 And the step response is Step Response Amplitude 25 3 35 4 45 Time sec Page 17 Chapter 13 Digital Control Systems Discrete Time Steps Continuous in Time Page 1 of 19 ZDomain Operations for the Closed Loop From the previous example we see an input to output relationship with a corresponding transfo 39 n as seen in continuous time systems Therefore we use b to represent processing ofthe digitized signal rt from a below rlll 0 bit Ytzl Note Ifa discrete system is driving a continuous time system the goal of a discrete nwillnot in I between the sampling points Page 2 of19 Example 134 Response of a closed loop system an m r39l It gt 0 Model the system as a discrete time implementation using Example 133 for T1 using Table 131 with aZOH inction Eu Rm gt zze391172e39l zen17271 zziz 12l 271 ze3914r172e39l ze3914r172e39l Yz7 zlizlJre39lJre39l 7 ze3914r172e39l Rz71 ze3914r172e39l 722izlJre39lyre39lJrze39l172e39l 2272 1e391 Jre39l Page 3 of19 Determmmg the output response For a step response Rz L 2 equot z2 rzlre equot1 e z 172 equot 2 z2zzrlrz equote z2 24 172 2quot Perform longrhand polynomlal orvrsron and me solutmn wru be ofthe form we no no in The output sequence observed wru then be me coef ments m ume Yk7o Usmg MATLAB Chap137Ex4 Pagezzmg MATALB Tsl numl denl l 0 systfnumden sysCLfeedbacksysh sysdc2dsysTs39ZOH39 systLfeedbacksysdl stepsysCLsystL legend39Cont3939c2d39 grid Result Transfer function Transfer function 03679 Z 02642 ZAZ 1368 Z 03679 Sampling time 1 Transfer function 03679 Z 02642 ZA2 Z 06321 Sampling time 1 Amplitude Step Response Time sec Page 5 of 19 Why is the tracking so poor Taking the phase margin of the open loop equations Bode Dagram Gm 2 87 dBaM 32 radsec Fm12 7 deg am 19 radsec Magmtude dB Phase deg Frequency radsec The phase margin changed from almost 90 degrees to 127 degrees The gain margin went from in nity to 287 dB The step responses must be signi cantly different Improving the discrete time performance increasing the sample rate Step Response Com c2d15ec Amphtude Twesec Page 6 of 19 and Step Response Amplitude The gain and phase margins for the open loop systems are Bode Dagram Gm 23 7 dB at 3 96 radsec Fm 79 2 deg at 1 06 radsec Magnitude dB Phase deg Frequency radsec The sample rate is everything in digital and discrete time systems The Nyquist sampling rate and theory are valid but normal analog approaches that are transformed to digital will not perform in a similar fashion to the analog system The sample rate must be significantly higher than the Nyquist rate Digital systems and controls can be designed to perform at ar near the Nyquist rate but they should be designed using digital techniques not analog techniques Page 7 of 19 Using the element 2 digital filters 2391 represents a time period delay in a sampled system So b b z 1bz 2 Y 1 2 3 R Z ala2z39la3z392 Z Yza1 a2 z391 a3 z392 b1 b2 z391 b3 z392 Rz alyka2yk la3yk 2b1rkb2rk lb3rk 2 a1yk a2yk l a3yk 2 b1rkb2rk lb3rk 2 ykyk1yk2 rk rk1b3 rk 2 1 a1 a1 al The next output is based on the weighted sum of previous outputs and the weighted sum of the inputs PS We just repeated the first order difference estimate using 2 transforms Special Case If the denominator coefficients are zero except for a1 l then the output has a finite impulse response FIR yk 1rkb Zrk 1rk 2 611 Meaning that if rk is an impulse one for only one sample then zero the output will last for a finite period of time and go to zero This is where an FIR filter comes from Page 8 of 19 Stability Analysis in the zPlane The relationship we de ned is z e5T 6W T equotT eW39T We return to the concept of mapping from the splane to the zplane 1 For the magnitude of z Izl equotW39T I m Tl For the entire left hand plane of s we see that 039 lt 0 which means that 0 lt lzl lt 1 2 For the phase ofz 42 AAWW Lew wT Note that for the imaginary axis 039 0 and lzl yew 1 and 42 wT This describes a unit circle in the zplane Note that all poles in the LHP de nes stability in the splan The mapping of splane LHP to the zplane is a mapping into a unit circle Therefore stability in the z plane is de ned as all poles inside a unit circle Splane Zplane Page 9 of 19 Example Section 136 Roots to roots comparison GFS ssl1 GZ zTle 7 2 Te T e 7l zTle T 2 Te T e 7 1 z lZ e397 zz zle397e397 The zplane root are based on Az z2 z 1 e39T 6397 which have roots Z Z 1ei 1 2 2 2 7 77 77 2122 12i Hangar leTelillZLJ 77 2122 le The system is marginally stable one of the roots is on the unit circle Therefore marginally stable in the splane and marginally stable in the zplane Page 10 of 19 What if this was a closed loop system with gain K Yz K Gz Rz 1KGz 63 1 ef l sw Gz l wiT 2 6 2 1 Ze Gz XV Z 1 2 z 12 62 22 2H1JZ 11WTZWTeM GZA le WTWTZ11WTew7 z l 2 639 The characteristic equation would become AZ1KV le w397 7ZZle l74 WTemy M22 1 Z 7W39TK39Vle39w397 WTZl lwTe W397 Azz2 zle39W39T 1e39W39T wTe39W39T l lwTe39w397 w w The conditions for all roots to be inside the unity circle are given without proof A0lt1A1gt 0 and A 1gt 0 Page 11 of 19 Checking the conditions Az z2 z1e39W39T 1e39W39T wTe39W39T 1 1wTe39w397 w w A0lt1 AO e 1 1WTeWT1 lt1 K 1 w 1 1wTe39w397 Algt0 W w AlK T KTe39w gt0 KT1 e w397gt0 A 1gt 0 A 1111e w397 1e wTe W39T 1 1WTe w397gt0 W w Therefore Kgt0 and Tgt0 Page 12 of 19 For performance of a second order system From text w therefore K 239 1 and 7 wT Table 133 Maximum Gain for a SecondOrder Sampled System TT o 01 05 1 Maximum Kr 00 204 40 232 145 Figure 1319 The maximum overshoot I yl for a secondorder sampled system for a unit step input 025 050 075 LO 125 150 175 20 Figure 1321 The steadystate error of a secondorder sampled system for a unit ramp inputrll zgt 0 Nunxruhlu mnu 0 025 050 075 H 125 150 75 20 T T Page 13 0f19 The ZDomain Root Locus Az1KGz 0 Repeating the same old problem Example 135 for T1 Wm Gltzgt 15533 23 135 Yz KGz Rz 1KGz 1 1 Az1KGz1K Azzz z1e391e391Kze3911 2e391 A0lt1 A0e 1K1 2e 1lt1 ltKlt1121 0r 51766ltKlt23922 Algt0 Al12 11e 1e 1K1e 11 2e 1gt0 Kl e 1gt0 A 1gt0 A 1111e 1e 1K e 11 2e 1gt0 Klt 2 or Klt26397 36 1 Therefore 0ltKlt23922 Page 14 of 19 The MATLAB root locus plot is System sysd Root Locus Gain 238 15 Pole 0244 0965i Damping 000336 Overshoot 989 Frequency radsec 132 1 7 t 05 9 a 1 2 9 I E I I 05 7 1 15 V l i r r 395 4 3 2 1 o 1 2 Real Axis 71 71 Az1KGz1K 2 6 1 239e z2 zlle391le391 Poles and zeros zero 07183 pole 10000 and 03679 Note do all the normal root locus stuff Nothing is different except for the de nition of stability the unit circle Page 15 of 19 The ZDomain Root Locus Example 138 I39 1 D Az1KDzGz0 Repeating the same old problem Example 135 for Tl 71 7271 TZzzl iTzzl Ma s Z Wlth 62 2 21271 lizzaf Note this ZTransform with a ZOH is not in Table 131 I have other books M KDzGz Rz 1K DzGz Z Az 1KDzGz 1K1LZHZ 2271 Calculate real breakaway points Let 2 039 l2o10391703971ZJ0 0712039170397107103930 rhoz 13 Page 16 ofl9 From MATLAB Root Locus 25 i i Imaginary Axis O 25 i i i i 7 6 5 4 3 2 1 0 Real Axis Page 17 of 19 Digital Control andor Compensation Method 1 Perform everything in the sdomain and then convert to the zdomain Derive and de ne for example sa sb GE sK Then convert to a digital controller of the form I Dz C Z by Z Where A 6 and B e b39T and computing the value of C for a steady state response a G 0 K Cltgt b l A Dl C 1B Therefore GC0D1 and KgC1 b 1 3 CKg b l Page 18 of 19 Method 2 Use a zdomain root locus method in the digital domain Rn The characteristic equation is Az 1 K GzDz and all conditions andor requirements for forming the root locus now in the zdomain are identical to the continuous time case See Table 134 on p 922 Evaluation of the root locus must now be done using the zdomain unit circle stability Example 137 Method 1 Plant compensated with a phaselead network 1740 G Pm s025sl Phase lead compensation to provide 45 deg margin at w125 radsec s50 G S539639s312 Converting the compensation to a digital compensator for T0001 27A Dz C H Where Ae 095 and Be 073 and C56 1 03995 485 312 17073 Dz 48527095 27073 Page 19 of 19 Developing the Equations for a Physical System l39 2 From Figure 22 we have Md72yb ky 7t a dt or dv 1 1 2 fromFi re23wehave Ci7v7 Vt rt gu dt R at 0 Performing the Laplace Transform Mfszm ylto 7 7 lg j were yltorgtgt ms Ru m 2Mibk RiiMby07M The rst desired structure for the Laplace system equation is m M E Rr so Rr y m y s qs qs 10 D where the natural system response based on initial conditions is I10 qt and the transfer function relating the output to the input is Ynltsgt RltsgtsltsgtRltxgt 16 ECE 371 Notes Page 1 of 7 Solving for the example results in M 13 y0 M 0 1 Ys dt Rs szMsbk szMsbk d 0 qs szMsbk and mm Rn 32M Sb k i The time response can then be solved as ye y1ty0t Transient response consists of y1 t and the steadystate response is y0 The second desired structure for the Laplace equation is de ned for Rs as s quot13 3 quot3 quot13 f1s fzs 9757 mimime Y1 ZSY3S is the natural response based on initial conditions q s is the partial fraction expansion of the terms involving factors of qs the q s characteristic equation and Y s fl S is the partial fraction expansion of terms involving factors of ds the 3 ds denominator of the driving function ECE 371 Notes Page 2 of7 Solving for the example where Rs 1 results in 3 M 13 y0 Mdy0 S dt 1 l szMsbk szMsbk s dy0 S sMb y0 M dt SMA szMsbk szMsbk s d 1s qs szMsbk S39A Y2S 2 s Msbk and Y3samp s The time response can then be solved as yt y1t y2t Mt Transient and characteristic responses consist of y1t y2 t and the steadystate response is Mt ECE 371 Notes Page 3 of7 Second Order Equations 2 M d y b dy From Fi re 22 wehave k rt gu dtz dt y Performing the Laplace Transform 143 yo dy 23mg yokysRs dt Ys 32M Sb k RssM by0M let rt 0 and y0 yo and g 0 0 Mb Then YsW39yo Letting a 1 kM and Q bZ lkM change the structure to s4 Sagan YS 321s1 yo 322answ yo Where we de ne Q as the dimensionless damping ratio and aquot as the natural frequency of the system 2 06 06 Remember th1s s2 as 3 3 3132 i J 3 Solving for the roots of the characteristic equation s1s2 Qcon i 0144 1 The splane plot ofthe poles and zeros of Ys is shown in Fig 210 As Q varies with wquot constant the roots follow along the real aXis for Q gt1 are both real and equal for Q 1 and are complex conjugate following a circular path to the imaginary aXis for Q lt1 as shown in Fig ECE 371 Notes Page 4 of7 Figure 210 frnm Dnrl Seeond order entena When 5 gt 1 the roots are real and the condluon ls ealled nverdamped When 51 the roots are repeated and real andthe eondrtron ls ealled critical damping When lt l the roots are complex and conjugates and the response ls nnderdamped or ezw tawn 4 Snlving fur and Underdamped Seennd Order System 5 24 y s2 w 5w lt1LA 74w ham174 The pamal fraehon expmslon ls ut we know that s2 ls the complex conjugate of s therefore the resldue k2 ls the complex conjugate of It so that we obtam ECE 371 Notes Page 50f7

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