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examII_f06 - EE 428 EXAM II 30 November 2006 Last...

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Unformatted text preview: EE 428 EXAM II 30 November 2006 Last Name (Print): 5 blv'lii Q a; First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score “- INSTRUCTIONS 1. You have 2 hours to complete this exam. 2. This is a closed book exam. You may use two 8.5” X 11” note sheet. 3. Calculators are allowed. 4. 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. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. 6. 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) Figure 1 shows a closed—loop system with nonunity feedback, where ‘ 10(8 + 10) 0(8) — 8(S + 2) and H (8) : .9 + 4. U6) R(s) Y(S) + Figure 1: Closed—loop system with nonunity feedback . Determine 1 5 points the system type, 5 points the value of the finite, nonzero error constant, - ( ) 2. ( ) 3. (5 points) the input waveform that yields a constant, nonzero steady—state error, 4. (5 points) the steady—state error for the waveform in part 3, and 5. < > 5 points the steady—value of the actuating signal. The. system is flettrmnnexjs Lg bot), 6C5) €4an H(5)_ Regatta“, th-‘L 5(7‘5tem Em Is the. ole Snoo— —- gowf gagiem GTflo J‘qugle 7 Commie»- thO. doch/Q ~floaf) 335%“ Y Z“ lOLS'f‘lo) ._.__:._ _ K _. .... W [4—5 lléz+l325+300+ [0.5+I00 Be‘C-QVW- Y/fk LS SQConfi-v—or‘oet" «uni «wag the. C°e¥¥mmb are, P°SI+4VLJ “H“?— 6554.2”! F5 6180 :‘EqHQ. l. gquuse, (5.0) ha.) rm po‘es act the away”) the saskm 2_' For a. tape 0 Sdfi'tev‘", on, the, position QYV‘W’ CvnS'éIU’J-é I5 ”“0 73am ungL 'f'tnt‘éQI P 513265 elem +132.)/+3°° 3 k? 5 FaIDf" ck. tflI’Q‘ O 5(75 {ZR/"J an; cu Conjtan'b Inpyt (J ‘bep WKJQLdfm) 54.9%} d— Cons'Eun‘E zvro r: 3. l l ’3 ‘1. e, ::. --—L-—-- = —- 3 C :: —' 35 l+Kp [+15 11., 5‘) V 3 S. 7/99, abet/w‘br’ 5tyrlMQJ UCS); affleang “It" HQ, l"/)~/b% 633 M Skew/n n7) Pyvr'z, I 7;: aokleua, a, S'EeqsLan-r5fi& emf 3““) Mt 7/ math 0‘, “Mimi EQCMQ 6C>\ has J o. '9an “A {55% (av-gym, it 6911190) 61—) ctr) Inféyrdiévfl I45 A ram/Ha) g667 CW7 Giff ar/promJu a. Coast‘aot— 17C the "M Wm 0-. Narmada.“ at!“ : F55 -—QSS = 1—- 3.7 3-, V7. Beam/3.2. b Problem 2: (25 Points) 1. (12 points) Find the sensitivity of the steady—state error to changes in the parameter K for the system in Figure 2 with an input r(t) : 1(t). fl (:1 C 3) ___.__K_____ (s + a)(s + b) Y(S) F0“ d~ Steadj’sbwbo. err-eh to eJusfiJ qu, Mus": En": Pesuire, tut- the paramefirs R a, card, (7 1r; oko‘en 30 M tLQ‘ °(°“&‘fl°afl éas'bem Is {5160 Sficb‘e, 6|uen _\/__— 6 _ K __ —- _. ) 9‘ 1+ 6 51+ (CL-P533 + «5+ k— R(s) we maul? have, as 1' b>0 aux, 6L5 +K§q OWL?" 5kg Cane/return Cola/4.2 we. an a / 5A6 hey/f6} oéflEVMQJ J flp/ '14» see“! deem MWWL I." gevfiwp / fl kp = an QC“ :- i... S59 ab PEP W unL'E’5Ee’l7 1“,?th a.“ .__L____ _ l = ,________._.__—. 655 = (“<9 ‘ he 5?; ab + K TKQ BQnSVEIvf‘Lfi 0‘? (-255 (JUL-t}? m)ch/~E "E0 "the. PauY‘wm-Qféir‘ 1‘ "6 gK Ben __’<__, '2. ‘9“ I< as; a K €51 (4,04, KVZM ”316:, 2. (13 points) A certain system has the transfer function representation 200 54 + 653 + 1153 + 6.9 + 200' GAS) — Find the number of poles in the left—half plane. the right half—plane, and on the flu axis. ”PLQ. charger/Ecru 5—579 abdaiflon “gar flu), ore/inflow [dang (:5 I 6 :10 ago _ 2w _. 5‘ 60—1200 ‘4“! to I" $0 2.00 Li‘s”: COL/(Jinn m (“£25 '[3vvo FWL‘E-‘M/‘IL - 7200 Man ckou‘laoa in tk“ eta“, (7 0‘95. Problem 3: (25 Points) 1. (15 points) Consider the unitx gain feedl shows the pole— 7ero map 101‘ the pl a1e both monic following details: ploys p1 oport1onal (ontrol. Figure 4 whose numerator and denominator polynomials > O. You1 analx sis and sketrh must include the ant transfe1 function G p(9) Sketch the 1oot locus in Fitune 4 101‘I1 o calculation of the angle of departine o dete1 n11nat1on of the Jw axis cussing 0 the 1ange 01 I1 point fo1 which the closed—loop system is BIBO stable, and 0 solution for the breakin point location. controller p Iant Figure 3: Feedback cont1ol system with proportional cont1oller GC( 211' and plant G ,,.(9 ). 'Ff‘om Flagrq 9 40a the 4119+ tkqi the numer¢'1:er~quL jeflOMInM—Ey “TQ- bdfl'} montg‘ C1 (3): W __._ Sz+$3+g ' Then an; t‘eQQ Q‘Q no zeros not 14-FlnI-%. 45 K-aoo t’w— 10°10; aw"? brawl—”’5 fit out.) bathe!“ -2. anQ, - 3J1Jecm/Je, this 75 the. gun/(7 reywn ./ the +0 411's when, the. 405057. Qua? Pol“ Jap'élb'F' the «.1712, (Ufiom In I an‘LEfmlflQ UV MU c7055! 0N ”Le/fie TLS‘ELC- ebvafécon "fl pom'EJ cmoQ uSacm'beflaq/‘nu % (unbugzrutfl 2- 1+ K 69(5) :-__ o :3) 51"3114-1. + kPC64’ 5—6 +6) “Hos?“ + (SK-2.55 + c1: + 2- =0 ~ 1 ., 'l: uath.) 7‘0 .btqm (>0le ya the aw q 711)} the. coexHflucw é: 5L 5 m ) Qéxvu? _a (1+K35‘7—‘r 6k+110 {‘Wk—S :21; +41% ‘tkafl; 3: I‘f-63§__:: :. t'vl7/7 1+2. ... 2.. $00 61,3115- croSSIA «j: '4. [17/7 05%: K- fl ’2. K7”". 6 H i/ 1.f€ Pole—Zero Map #7 g i E E j z i J I I j ' S : I 1.55 § J: i Wu. ' 5+ I : .(W 1 1 ' ' K V) ' i i// ' L // ' 3 g o 5 // a} g a / /” / ‘ ‘ a ,, g a 0 ”0/. ’79); 0/ /5\ .‘l‘L‘ ............. ‘ j E 2‘31 22.. . I .0 (J1 TW _,. ____J._ ~ = c.— l x lice-z. 3 ‘lh H a» > . ,4 “1:" W1: 2L CR ‘75) x 4(St—iL) XV): _— "4; ; 5°“ “"‘m ’° “W ¢S¥ : 2S (S 'b _pt) 3 . 1 02 QMWuwwmmxmmmwwq‘E,“m.WC.“,wm..»;gw_wi~_/v__m4 “MM“..LWNMJMWJJ —3 ~25 '2. —1.5 —C.5 O 0.5 1. 1.5 2 Real Axis Figure 4: Pole—zero map of the open—loop transfer function Gp(s). ' Defifm’aa tko bre*k‘n VOIHLE Ea (onihgefl :77 24—; é—p’ = 0 AL S"—2.J+2 _(25-2)(J’-+s’s+g)_g59-,7_,_,7_~)(7_5+S) a? SZ+SS+6 (52.4.55 +631 7. 2,53 +1051 +. {2.5 "ZJ " [OS “[2,... 2.53 -562' 1- ‘15?” + 10$ .—‘-15 “IO :0 7.57- +~Bs - 7.2. :0 :9 g: —§__1-_"(_3__>z +;_7-_ .—_ Lzrnz) 1—7., q’gl (“I H 7 IbquEm CEOIAZ «SE 5: -Z.‘!3j TRe anb ‘5;— oVe/zartwo. ¢§L Ja'E'S‘F'eS (‘99 Ske'b‘)’ “5‘343 4 GP (54;): ”50+ 360/0. = 46+. ‘2D+i(S£-2._)-2{_(5+.-fl) «ta-pl) 7' Tan—[0%) + Tau? (i?) “@Q -’ 70° For Q: '1) ¢fl/;|12.\10 See. Tout Qacub rkE‘EJ‘ on page/5’. 7 Imaginary Axis Pole—Zero Map W'WWW' “1“" ”WW”fi—j S —{ E i, ~i i 1 1 .1 i i i E a i E 4 i i I J ! i i » i J- .5 t . 1 —— L7.— até : 2- ! 9L 7 K ”s" } _2 4%». ' = W_,4-,,,.L_H.._MWLN”WMMJWWEMWJ ...... MAMMJA.WW_.L_J —3 115 2 ~1~GS C‘ 05 1 1.5 2 Real Axis Figure 5: Spare pole—zero map of the opcn~loop transfer function GAS). 2. (10 points) Neatly write a m—file that o generates the root locus diagram for the feedback system in part 1, 0 indicates curves of constant C = 0.69 and w" : 2.25, 0 allows the user to find the gain corresponding to a particular point on the root locus, o simulates the closed—loop unit—step response for the chosen proportional gain7 and 0 determines the steady-state value and time to peak of the unit-step response. 6? =.E+=(L"I,§, 63) [1, “2,1135 Y‘l0(u:> L67); 586$ ( {0,6611} [2.253)] lg: vioaémi (39)) 6t ‘- Sene$< £19030) (SP); H = #:‘c (1)0)- Gcfl. -— 1:20ng (GU H); “Le-P (Goa); [3mg :— sief (6&9; Vk_\h19>97k €- 41.»:9. Q; :7- ”‘Wkégfll lip = iGCk-IJ-eQ-l as: 5 5t<em®j 9 Problem 4: (25 Points) Figure 6 shows a feedback control system where the plant transfer function is Y(s) Figure 6: Feedback control system with controller 62(3) and plant Gp(s). 1. (20 points) Design a compensator so that the closed—loop system meets the following specifications: 0 A rise—time of 450 ms for a unit—step input 7'(t). o A complex pole pair with a dimensionless damping ratio of 1/\/§ 2. (5 points) Using your compensator design, estimate the value of the peak overshoot for a unit—step input. Is your estimate likely to be larger or smaller than the observed value 7 Justify your answer with one or two short sentences. As t.» x 1.3/wn, in- = Vs’oms é Con= Val/Sec, ’45 ‘F: 1/52 eolebnreuq- W’a ‘59.: -fwn + WAVI-f’z _ 2-5:;2'5- gagging/7533:”) SI Confiuaer the rod: gem); {of Propw‘kloflcu-Q CO-O'EPU‘ GLCS) = K “”91 PI) can-kml 6c: KCS+232 PDI-Z<z¢z_ P Control Pb : .3 > 2 coflJ'bY'u/o 4.; h P or PO glihoa MIG can r to obtain a— flmgo 542-3)" (on trul’Q/ c1050}. e000,; gas-12m} we canno‘lll achieve. ‘b‘fl— tfeuv-gua— cJOJQfl’floap flak! flan‘évufi) 5:1. ang- $11? on the. otherfldnfl a, P1: (wirelley- place) at; at the Ora/01’ “Mt thereLm W a. break.qu pa/n‘b bah-ween 0 “mg" +1 on 10 s+a 5+1 (5'30“?) a bin xrfiAi-i) Man”) 4-6.1 -0) In 5L:-;r2+/ 2/2 Choose. 7? 5,; vii: fine. ang‘q. COWJL‘EWD is so» SCWQ: - awn» wag-o) 49M?) 46(5).) :: 190 + 3602. = §Cs£+23+ «91+» hwy—J I Pa Q3 2cm) XS!“ \ : e 4 2r - C 2 ~\ Zf‘z-I‘ ~12o Kai-u): 70 +an(37‘?) «135° 5439.44): 70 -I- Tom '37:)" 3 XCSJL‘JD = ?0 + Tap (#2543; 2 [50° A— (skwb = 70 1— Tqu' (23%:292106" 2r; 09'! @- -.-. l80+§fi91 ~123° {.1060 1-13?" +Iro> : 88.10 fir X‘ Ernie = 0-"‘?‘t7 =>‘?= “Lim‘o'ow x Choose K So that the. mam-Luk— cai'éwn [J Sw'élbxcleflgi (55ng K” { lSJL-zlts.,1«01 lSA+zl :_ Q, 92 03 - 3L2 5'” _ [Say-$1 iqu 9-5‘ Q3 .5 (s-l-LXS-z) 5;”!- Ql: (mmhnmnm 9,1: W =z7r .Qz: VCL 1201+ (21137— = ‘f 95-: (2.72 —2J23"*(Z’3)L 22‘? 9'3 : VCZB—Qz +(zr237- 3 3'7 2_, For a. S€60ng-’drrgér yg‘bem “at" no ‘fimfgfiz beta: [7; = e'flf/VI—fz P f= //I2 3 0'0”” :9 game m 40541-2(”, was”. :s tarrawfiflar «w €0,252. zeros, fin. dwewoflu peak— wenhw-é UM“ Jr‘c‘ceo/ 4-er "/3 '29. 11 ...
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