Solutions Manual Linear Control of Manipulators

Chapter 9 Linear Control of Manipulators Exercises 9.1) In the general solution, The constants с'1 and с2 must Ье (in general) imaginary since the S; are and yet x(t) must of course Ье real. Substituting for the S,: Using Euler's relation (9.13) we have То satisfy the constraint that x(t) Ьс rcal, wc have: where с1 and с2 are real. Тhis leads (о 9.2) From (9.5) 5 -6 ✓з6 - 4 х 2 х 4 + ' = 2х2 2х2 = -1.5+0.5 = -1.0 s, = -1.5 - 0.5 = -2.0 х(О) =О= -с, - 2с2 Adding [!] and [2] gives 1 =-с, so:c2=-l andc1 =2 :.1 x(t) = 2,-, - ,-" 1 [2] 9.3) 1 x(I) = 4(] + l)e-• 1 9.4) 1 x(t) = 2e-ir cos t + 4е-21 sint 1 9.6) Minimum when с,= -1 => ЦМ2 -2L1L2M2 + L;(M, + М2) Maximum when с,= 1 => ЦМ, +2L1L2M2 + Lj(M, + М2) :, nun =0.5 - 1 + 1.5 = 1; max = 0.5 + 1 + \.5 = 3; 9.7) Minimum effective inertia: Im;" + п'Iт = 1.0 + (400)(0.01) = 5.0 Maximum effect·ive inertia: '= - n 2 lm = 3,0 + (400)(0.01) = 7,0 VariaLion as % of maximum is � х НЮ= 128.57% 1 9.8) Closed loop system is: МХ"+В'Х"+К'Х=О (9.38) where В'= В+ К" and К'= К+ КР and 8'=2.../lvГl(' То achieve critical damping. Using rule (9.72), if \VRES = 6 rad/sec then we should design servo 10 havc WN = 3 rad/sec. From (9.20) WN = ,/1rТМ (in our case К') so, З=✓К'/М; 3=./К'; К'=9 К, = К' - К = 9 - 5 = 41 К, = 41 В'= 2.../lvГl(' = 2./9 = 6; К,= В' - В= 6 - 4 = 2 9.9) Min. effect. lnertia = 4 + (1()0)(0.01) = 5 + (HXJ)(0.01) = 6 Мах. effect. lnertia = 5 Use lowest resonancc. so WN =, (1/2)(8), or \Vн ,:::: 4 racVsec to insure never \Шderdamped, desigп using max effecr. lner(ia, so use М = 6 Sincc we didn't provide info about апу damping or stiffness iп tl1c open-loop system, we'II assume its zero. Therefore: Since а = 6 wc now cffcctivcly have а М = l systtm: = ..fl(;lМ; = .,;к;тr; К, = Wн К, = 2,/М7Г, = 2v']хТ6 = 8; 1К, 4 1 К, = = 161 81 9.10) Using (8.24) and assuming aluminum: К= (0.333)(2 х 10 11 )(0 054 -0.044 ) (4)(О.5О) = 123, ООО.О Using info from figure 9. 13, the equivalent mass is (0.23)(5) = 1. 15 kg so, IVRES rvm 123, ООО.О = vntm = -- -- = 1_ 1 5 1 327.04 rad/se� 1 This is very high- so the designer is рrоЬаЫу wrong in thinking that this Hnk vibrating rcprcsents the lowest ,inmodeled resonance ! 9.11) WREs = Jl000/1 = 31.62 rad/sec=I 5.03 Hz 1 9.12) Shaft appears stiffer due to gears: К = 500 х 82 = 32000; = 178.88 rad/sec = WRES = ,/З2ООО7f 128.47 Hz 1 9.13) As in proЫcm 9.12, the effective stiffness is К= 32000. Now, the effective inertia is / = 1 + (0. 1)(64) = 7.4 :. WRES = ✓32000/7.4 = 65.76 rad/sec = 1 10.47 Hz 1 9.14) = 612.5 " = 61 1 кр = 49.5 fJ = О 9.15) Stiffness of transmission is, as in Exercise 8.4: к, 1 1 1 = К п1Ко.2 + Ко.з; Wmax К = 1.843 пtn,/rad =� = 1.35 7 rad/sec ~г;;-;;-;:;::� = у/ТШ~ -Т-- w..., = ~ 0.679 rad/sec = ~� У/ТШ -;i- = � Vегу low because 0.2 and 0.3 cn, shaft are floppy ! Chapter 10 Nonlinear Control of Manipulators Exercises 10.1) Let < = ,п' + f3 f3 = 30 2 - siп8 then: where •' = iiv + Кwё + KfJe е = 80 - В For closed Ioop stiffness, КCL, use: Кв= KcL = \О 10.2) Let < = rп' + f3 rx=2f3 {3=580-\30 3 +5 and : r' = ёа +к,;+ К,е whcre е = Во - 0 апd: К,= 10 к, =2✓i'o 10.3) Whcre М(8), V(8, 0), and G(8) are as found in section 6.7. о к• - [К" к \V = [ Kw, Q Where Kwi = 2� 10.4) Where M,(IJ), �,(8, 0), G,(IJ) are as found in example 6.6. KIN(O) is the kinematics i,1 (fonvard) for this simple two-\ink. Also: Кр1 кр -[ о with К v; = 2,,fl(p; кV _ [Kv, Q Ann 10.5) r = М0 + V М = [::�: /Lj] So, as usual: r =ar' +{) with tx = М f) = V r' = 00 + к ve + к re and: е = 0v -0 к _ [ KPI р- о with Kvi = 2..[Тr'i:; Kv -[Kv, о Since the mass matrix, М, is not symmetric this can not rcprcscnt а coup)cd mechanical system. 10.6) In steady state, the system error equation is: 1 К,е = -.--(v,т Lgcos0) ML 2 so: v,gcos0 e=--.K,ML Note that it is а runction of М as well as the other variaЫcs mentioncd iп proЫem. It is max.imum when 0 = 0° or 180° , since this епоr comes from gravity loading, tbls is not surprising. 10.7) Not stated in proЬlem, but we wШ assume than an activator (а gas rocket thruster) сап :зррlу а force F; оп each mass. Positive F; tends to increase Xi , That is: Drawing free body diagrams leads to equations ! of motion: !, = м,х., +(В,+ В2)Х1 - В,х, + КХ,- КХ 1 F2 = М2Х2 + В2(Х2 - Х,)- К(Х2 - Х1) Q= [(В1 +В2)Х1-: В2Х2+К(Х2-Х1)] В2(Х2 - Х,) - К(Х2 - Х 1 ) 10.7) (Co11ti11ued) Then let F = aF' + р with а= М, Р = Q •• + К, Е· F, =Хо + Кр Е IF = [[1!21 7 7 Х = [Х 1 Х2] Е = [,,,,Jт еi =Хт-Х; First, лоtе (from sectioл 6. 7) that the only 01 depcndcnce is in G(0), so \Ve have: Тhе second column of [В] аЬоvе is given Ьу: ,vhich is ЭВ(0). . ао;- 0,е, + Э(С(0)[0 2 ]) ае, + м, м, 10.8) We would like ю say something about Э(М(0)0) -::j§!i )( ,vith Кщ = 2Jl{;; �+ -:3aJ!�F, __ F2 дС(0) ао;- )( )( )( )( 10.8) (Co11tir111ed) То say more we need to make some assuшp­ tions: Lets say average velocity is OA,.�nd l��s also assume that average acceleration 0А. = 0;;.. For many robots tl1is is about right (units are rads). At this speed 80 = В,Лt. where Лt is update iпterval of configuration-dependent terms. We then have; From here things are quite heuristic, perhaps the шaximum value of t" is found for "avcrage" trajectories, and �'t" is requircd to Ье Iess than 5% of this max. Then а Лt might bell found based оп ап expected 0А and other params. 10.9) Like exercise 10.8. Result will Ье in terms of average cartesian velocity, rather than average joint velocity. These might Ье converted using son1c average norm of the JacoЬian. 10.10) Let f = c,f' + /3 witha=2, f3f3=5XX-12 алdf'=Хо+к,;+к,,, К, = 20, К, = е=Хо-Х 2Jw. 10.11) Closed loop system, given Ьу: M(0)ii + Vм(0, 0)0 + G(0) = -к,0 - M(0)K,il + G(0) V= ½О' М(0)0 + ½&' К,0 \i = ½0' М(0)В (Lyapuпov candidate) + в т М(0)ё + в т к,0 \i = ½0' М(0)В + 0'[-Vм(О, 0)0 - G(0) - К,О - М(8)К,0 + G(O)] \i = _tJT М(0)К,В This is non-positive if М(8)К 11 is positive defi­ nite. This matrix product is positive def. if К u = ku l/1/ , where К 11 is positive sca)ar. Q.E.D. 10.12) Completely analogous to exercise 10.11. 10.13) Use 1. • V = В'М(0)0 + 2 То yield: 1'' ,, M(0)K,0 .в d5 > О 85 V = -0'М(О)К,8 < О if К,= K,IN , К,> О 10.14) A11alogous to exercise 10.13. 10.15) Use V = ½О' М(О)О + ½В' к,е + Р(О) where Р (0) is the potential energy sюre in mechanism. v = -8' G(O) - О' К/1 + j,(0) Now, G(O) = а p(O)(from Lagrange) д0 :. сап sho,v О'С(О) = j,(0) :. V=-O'K,i! From which stabllity follows. Do steady state analysis (set iJ = ё = 0) to gct к,о + G(O) = О or 0 = -к;;'С(О) steady statc erтor. 10.17 ' ,,, =ах х 1 /J = ЬХ 2 + csinxO 1 k, = 2ft, J0.18 мё + 80 2 + ci! = M[iio +к,;+ к,,]+ sinO Thats it. Jt сап Ье re-writtcn many ways.