final_f05

final_f05 - BB 428 FINAL EXAM 12 December 2005 Last Name...

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Unformatted text preview: BB 428 FINAL EXAM 12 December 2005 Last Name (Print): SOL/l3 l o Qé First Name (Print): ID number (Last 4 digits): Section: ewe? DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO 1 2 3 4 INSTRUCTIONS . You have 2 hours to complete this exam. This is a closed book exam. You may use one 8.5” x 11” note sheet. Calculators, protractors, and rulers are allowed. Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example, Problem 1.2 Continued. NO credit will be given to solutions that do not meet this requirement. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. V . The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. Problem 1: (25 Points) Consider the closed-loop system in Figure 1 with plant transfer function _ 8+1 (8—1)? GP(3) and proportional controller Gc(s) = K. Y(s) Figure 1: Feedback control system using cascade compensation. 1. (5 points) Write a m—file that generates the Bode and Nyquist plots of the loop transfer function Gc(s) Gp(s). K= 1', (gap: 2PK([—IJJ 51):], Cid); bor&£. ( Ray’s“: (6‘, GP\) 2. (10 points) Sketch the Bode magnitude and phase plot of Gp(s) using the semilog paper provided in Figure 2 on page 3. mm -— b” +' (aw/r! )‘ .4.” - ___L——-v- l‘Nfia Saint ma—a’n'b"& “'5 duh/[+1 _. -l 2—- . A ___L...L = - 2_ mm ~| 94H) 4(«9 3 w '2. Tom /"" <o\ 4411903- '360: l — 2.(+135)- ' 27° >10 -zc9a) = "3’°° Term; : Magnitude (dB) Phase (deg) ANDOA 0000 A 5 A l/(fwll-OL Campos, £2— Bode Diagram Frequency (rad/sec) Figure 2: Bode magnitude and phase plot of Gp(s). 3. (10 points) Sketch the Nyquist plot of the loop transfer function for K = 1 in Figure 37 and indicate the direction of travel as w is increases from —00 to 00. 1 Imaginary Axis O 0.2 -0.4 -0.2 0 0.4 0.6 Real Axis -0.6 Figure 3: Nyquist plot of K Gp(s) for K = 1. GP(’.W\)= 9W_l)z {GPC’vAl '7 (m); m ) 4L, leeqwfl 2L eplzw) pmtm .n ma 0 ] '3GO" :1 0" fl 0.2.7 0,77 3 -— 315° = ago 8, 6’ O.S] 0 37307 - 270° :: 700 c, c’ LO 0 71 30.7 - 27.50 —- [3:0 I). D’ [-7 0.5; 30.5' - 180° -: 130° E 3.7 016 350.3 v 135° :— 2256 F, F’ Problem 2: (25 Points) The closed-loop system in Figure 4 has a loop transfer function whose Nyquist plot is shown in Figure 5 for unity proportional control K = 1. a When en ti rpieaQ, ——) when and. -—§ When OWLQ, Figure 4: Closed-loop system with proportional control gain K. Nyquisi Diagrams 0.2 Nyquisi Pioi oi lhe Open-me Tmnsier Function for K - 1 0.15 0.1 53 o 01 Imaginary Axis 2 —0.1 —0.15 —1 —0.9 —0.8 -0.7 —O.6 -0.5 Real Axis -0.4 -0.3 -0.2 —0.1 0 Figure 5: Nyquist plot of the loop transfer function for K = 1. o<K 4| (Afr—o). léKC’Z. the [Don't the. Cc-w Jireuélmv K> Z. the. pom/E -—I Ehg Cw J/ree—ézok 5 In ‘ [n the pom": "’ +io '3 ,1-1-7‘0 [Ix/c— v21, ‘: Y(s) 1. (4 points) What is the loop transfer function ? How many poles of the loop transfer function lie in the right half of the s—plane 7 95;) K 3+1 5 Eta) 41: Raw-7(5) (s—t)($"+25+2) (S) ‘ =+\ one open Qeop po,QL(S<-—l3 Qmu m the, RH’P (P 2. (10 points) Using the Nyquist plot, for what range of K > 0 is the closed-loop system stable ? Z=~+P= N-l-I '1) a; K/ :— ch zncnrolemené C WO- fle 0‘9. ‘50 have. onQ. -l 4—?0 ‘F‘Or clokfl—Qogf; [3/00 S‘ISOL‘DI)¢£)(' 3. (6 points) Using the Nyquist plot, for what range of K is there one closed—loop pole in the right half of the s-plane ? For what range of K are there two closed-loop poles in the right-half plane ? . To ob‘laqfn ana- cloud/00 ole ,‘r, the RH? we, neat. 2' '2 N 1— P = U I : TS Occur) when + [_ IV: 0) 25o1wlen‘kg/1 we, Aeag. 50 M "l akfo is net chum/Pl, ' To alo'lra'm {3w scam—1’6" 9'95 r we neq’o’ Z; Nkp; M +1 :2. 74175 Dec/Ur.) wken o the V31, QVIUqlvnt(/) we nfieoa E] 5 t "l +/c9 75 encu-Uéa, $1162. In 4. (5 points) Verify your answers in parts 2 and 3 by using the Routh-Hurwitz stability criterion. ' Chm?c~c't2r€5£:<. Ebuoctl‘on % the. o’oseéZ—fioup Sda'hemt l + K (5")(57'1'2—5423 + KCS+1)= 53+ 57' 1- KS + 2.\<-2. = o ' 0.? Pay a 33 l K 5" I 2197, 5 2"IQ 0 O S ZK-z here fin/5": [9g - WM; [559.11-12.99 : 3km, t For 6”.” Sb b lrst COL/MK)? 00 Sign Changes I}! 2.10170 4) K 71 ’1 n 4 g a ‘ FOP one. unsbwbhz QOIQJ thaw. muff? be W Sln‘dlg’ ‘5’” a 3 Us the, 43me (Jump: 2—1<>o='> KLL .2 1K<ll 244-2. <0 => K 41 ' {2h f/woro, mqn’: b9. ‘lrwo Syn C‘WWWQ‘S '" 2' . Fa- ‘EWO \JQS has ‘FWS‘L‘ ca‘vm A 1 Z’K‘O” “2% E223 2K‘7v >0 :9 IL?! _ ’ a Z T’qeée’ W475 W Conan-Barre Wit“ tkaJ¢ a‘a‘billngng' '0 {7 ffs mi, 3. Problem 3: (25 Points) Consider the closed-loop system in Figure 6 with plant transfer function 20 Gp(8) = 3(3 +2) whose Bode magnitude and phase plots are shown in Figure 7 on page 9. Y(s) Figure 6: Feedback control system using cascade compensation. The closed-loop system specifications are: 0 The steady-state error to a unit-ramp input must be 0.1. o The peak overshoot to unit—step input must be 15%. 1. (5 points) What is the minimum value of proportional gain Gc(s) = K0 for which the steady-state accuracy requirement is achieved ? _ 655 = it; = 00K :9 KV =10. The m'nlfl'Ium Um‘:% kc resume.“- . ‘Ko A = k ‘0 {7° qclMeva 6;":- KV 7' 30k. "' 0‘ KV '3 Qom S 66 GP : 5;: S 5.6+.” o 5-)0 “ 2. (3 points) Using the value of K0 determined in part 1, estimate the phase margin and gain crossover frequency using the Bode magnitude plot of Gp(s). [:95 KO -_-_ 1) 4:1le “’6; anti pm 'FW‘ 6’; C5) . wait, «5, $3 '“JL/Seo PM x 2.5"” } gee Pom—b #- ” Faye 6" 3. (3 points) Using the phase margin in part 2, estimate the peak overshoot to a unit-step input. 7’ liq n Bacau“ gr“) has .5“; .Corm __£a...—._ we. can vac the, fin .M } scum“) ’ P resul'l: Lrom fut,le Set 8; (m «Hoof. 14mg, so ‘1’: 731; 6-11 WWW : 0.7.5: mP 2 7:; = .70 4. (2 points) Based on your results in parts 1 and 2, can both design specifications be met using proportional control 7 In grief 'to £ecy¢uw mp) we. new}, “[29 chFeuSza the. PM, USU-n; Frofgor'launfi can'Er-ol) (flawed-5y the, Pm mgv,pg) 0v UGJUQ’ % KO smalcr Ftkun !‘ 8d: PC we, flea-ease k0) ‘l’VUfh Pfirt l we. knew thwfi €55 WI“ QKCéefla 0.\ , price So bctk fizeS‘l3f1 Speci'FlCu'Llo/u CCU’"58£ 60.. va/‘quewsiy mt p Gnérol. Bode Diagram Magnitude (dB) 0 f Q 9.“ ‘K 0 1| P Frequency (rad/sec) Figure 7: Bode magnitude and phase plot of Gp(s). 5. (12 points) Design a phase-lead controller 1 +1 Go (.9) = K a w; k wa that achieves both design specifications. Show all work required to determine the value of the parameters K a, k, and Lola. +1’ v Choose Ko to mac-l: S‘Eeuuflj—S'Eu‘ée. anew/ram? fégv‘retflefl‘é: .5... #1., .__z_‘o___. = 'oke kv. = gym 5 59 GP : 2m 5 K. 4—H saw.) 3090 S 90 Sflo I .-.-. —— = o. = 655 loKa I :9 K° /- 0 From Part 2) «1., Know that ‘gr Kc CS) 5,‘ [) rm: (250qu K 4:0 Achéuo al— feu’c Phase. % tI-IO/ Con/2050564” ¢mz Pmoo-esweJ- "' pmj, kochSW + T" flefirm“fl¢ Fm JeSamrflJ Link (wlcwlw'Lz, the 70 {pr mic ‘LYZ f: M = V?L+ WNW]; P 5.16 FmSL an affronmwbz, uwlve % pm V5,? —Fm:100 —- . A’mfl. so ¢m= SZ°—2_s"+s" = 3z° SOL/Q, Lur- K 0 1+ Sm¢m = K- list/Wm l— §|n(32. ’ 7flo, 30‘ In C 1-05.50sz {7225.420 (2‘ ¢ tho Carafe/6‘, 'Fufl o'ekd‘ Sw‘ElJ £65 I 6.6 (aw?) a? (fur/3] : «be; 900/ ’b‘un) (9V zaooflprz + lfioép‘iwya\\&e”o ‘RGP’jwalM= ’wygwfia ” '50'3 10 Ff‘am the. Sofie magm'bofe. rig-L 0-; K0 Gr (/w))‘<a:l of) Wage 7) ‘ k9 G10 " "5A6 when Wfio :- 6 "ff—0L“ 5a., (Sen. Fonqfi 6 an 0427’ 7)_ fing— So flan. you}; cro‘SSo/er‘ fivjdnc/ ¢ gig» Cdfl/QHSa/£PL £0” é)"dflf-ér fiflc’éro/i I»: ' “ku- t‘xo. phat/1.2.. peak callbfléufiei % b“? Fk‘de’ fled Cantu)!er a£ Wm S i K We 5'- Wfi c, $CMUSQ. K :— qnfl Wflc/ :_ 6‘ “(9/ known) W; Can SoluQ ‘F-vr‘ W0 J 6 r a4; '—‘- Wfi :- __________,. TE liar 'I-C numcr} [Va 5 ‘m-J‘Jéw” raw.on m P umbs-Blf nor/43 Yepew‘e' the, £670 proceA'j OSI’7— 0" QM 0”- Vu’ua % PM Jamey. 7/f‘7o-Qrw 11 Problem 4: (25 Points) Using a dynamic signal analyzer (DSA), the Bode magnitude and phase plots in Figure 9 on page 13 were obtained for the plant Gp(s) in the closed—loop system shown in Figure 8. Y(S) Figure 8: Close-loop system with plant 0,,(3). The closed—loop system specifications are: 0 The steady-state error to a unit—step input must be 0.1. o The phase margin of the loop transfer function Gc(s)Gp(s) must be 60°. 1. (8 points) Using the Bode magnitude and phase plots, estimate the transfer function Gp(s). (Parts 2 and 3 can be answered without part 1). . Bacqwa 3.- GF flecreuses ‘Frdm 0" 4:0 — 180°) ‘Ukare are, "54/9 poie_f_ . Q Slh&|€. eer-nev- ‘crcbuancg. J. w=. [o {afi/gec Lg ng-JI'QJ qnfldth‘ls Garner H at [6 ( W93[ 5 - 6&8) anLCw'E two pics at s: 4:? M} P ? y“ (lo/Lev) ' 7R9, be gruff) l5 [(7 7' r1: Bis-seat an tines-e. obsewcuglafls 2. (5 points) Can both design specifications be met using proportional control Gc(s) = K o. Justify your answer. - Fmfl- kg £4 Swim-51 the J'ize4%—.r‘irw'ifl- 45M“? re? (/lfz-CM-efl’é. 353 = ———l in? :5 0.| é = KP; chgépa)’; kvh—o- Wo/ neeg- kg =1 Kf/I'T; = 7/fi'3 = 2.. 35 - 4: . Settm K, :- 2-8’5‘1 the. your) Crasswer '25 Van? — — trams‘iQ’: fivno'elan k0 6p Say-EIK‘FIQJ i 6p(d with ‘5 'Zagg‘o K0 " .3) lepzawfim: 4.1428 «£2 w o:- 27 Rug/sea (Pamfm, 04. the, Comp-enSaL'ée'L’ From raver when), PM 411 35" 75 Qex; than t‘m— :QeSH‘eaa e-—) P Qn'brai 00'1th SIMUI‘EUMwSl/ $06195? Lot!“ jaw-n rafiulVe/fleffffi‘ 12 Bode Diagram 20 10 wme ._ E'I'iépjéah} —20 ~ - - ~ - - - - - -{mac-:--3-:--:-:-i»:-~- E E f E Eaéiiiaurwllb éréxsiuéz+ifi . . _ . Magnitude (dB) Phase (deg) Frequency (rad/sec) wen kfo . , ; iiiiliépr+->fl= 9133mm Ikzéégwiléjoéelfi Figure 9: Bode magnitude and phase plot of Gp(s). l3 3. (12 points) Design a phase-lag controller 3 676(3) 2 KO [(3 £410 __ + 1 7 we that achieves both design specifications. Show all work required to determine the value of the parameters K o, k, and mo. ' Ch°°5¢ Kg 4:0 a-ohléve, Elno— erJnrcfl g'Leaogr-I—s-Lw‘lze “Cchy: - l =_ .L.’ 50.! kp = Ge 6,, = Ko '75; €55 = l+l<p H-kngE . The. clash-ma awn}?! cross, over ‘Frafiyenf; Wflc g...-I:.5-I[¢eJ +1 46plgw&°)= Ago + pmfleéma + 5° Chg-(e?) '4 'IQO°+60°+5'° :« "llf" FNm “tha— Bogflb phage, a": 6? an page/13 Céee Pomt 8) A Gptatwio) ’3’.— —-//50 ,1]: lwzc = If fwd/55f; ' Plane the, zero 0-; the. Quét Cuqflgn‘sw-L—yp i Way/[o so thuf the phage, many}; 7.: not all‘éreoar %_ the, cumflcnjwbrpde. c144. zero Choc“, K so they“: W30 3 'rrw’e/Sec— l) the act/)0 cmsfouev- 4r%,¢n7, 9%: Mac =, [OKWOB [6“ ko/k,’ dflfl So 14 Gwen ng 0mg. K Solvc‘gA/‘Wo‘. 1. “HQ ’g " 53¢ 2. 5‘1 . me CamflIIW-LQ’ Qai améroller «i 15 ...
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final_f05 - BB 428 FINAL EXAM 12 December 2005 Last Name...

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