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examII_f98 - BB 428 EXAM II 7 December 1998 Name 3...

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Unformatted text preview: BB 428 EXAM II 7 December 1998 Name: 3 olu'lzigas ID#: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO 50 “Weight 25 - - 2s - - "100 [\3 U1 [\D II This test consists of four problems. Answer each problem on the exam itself; if you use additional paper, repeat the identifying information above, and staple it to the rest of your exam when you hand it in. 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. Problem 1: (25 points) The closed-loop system in Figure 1 contains the plant 32—23+2 s(s+1) U G43) 2 Kg. Gal”) = and proportional controller controller ulant Y(s) Figure 1: Closed-loop system with cascade compensation. 1. (4 points) Express the characteristic equation of the closedwloop system in a form suitable for con- structing the root locus of the closed-loop system as the proportional gain K0 is varied from zero to infinity. 2. (4 points) Calculate the location of the break away point. 3. (4 points) Determine the value of the control gain Kc for which the closed-loop poles cross the go) axis. Specify the location of the crossing point on the imaginary axis. 4. (4 points) Determine the angles of arrival at the complex zeros. 5. (4 points) Sketch the complete root locus using the graph in Figure 2. Appropriately label the break away point, 3w axis crossings, and angles of arrival. Use arrows to indicate the direction of travel along U the loci as K9 increases towards infinity. Root Locus of Gp(s) lmag Axis Figure 2: Root locus of the closed-loop system as the proportional gain K0 is varied from zero towards infinity. 3 From Part C0.) tha. ohm-Ezrta'mo 9%. Vodka can 59-- efif’reSSQcEL (:5 514's + kc (SZ‘ZS +2.) 22.0 “(04")51 1" (I ‘LKAS 4. Zko _-... 0 TL ROV'HN a.“ I“? 1.5 5"" K0 + I Zko 5‘ [-2 ko . s" 2 k. $76!: ko = J- 55 the]: the, 52mg— row LJ 1:271). 2- The Ow x“ I out? ebvccbon 1‘s Problem 2: (25 points) The root locus of the closed-system in Figure 3 is shown in Figure 4 for 0 < Kg < so. The numerator and denominator polynomials of the plant transfer function Gp(s) are monic. Answer the following questions using the root locus in Figure 4 and the ruler provided on the last page of the exam. controller - lam Y(s) Figure 3: Closed-loop system with cascade compensation. 1. (8 points) For what range of K, is the system BIBO stable ? 2. (8 points) Determine the gain K0 for which the dominant pole pair of the closed-loop system has C = 0.8. 3. (9 points) For the gain determined in part (2), what is the steady-state error of the closed-loop system in response to a unit-step input ? l. Wham k0 -.-. 0’ the. glosea-Qoop poles gnu-e, In the. 19,-“: le‘F plane, A; Ko :ocrmesJ the. Pale; cram- 'HIQ. dw ”L: at Peifl'i'fi S. qua Sf ( walked-Ag cm Flaw-q. '1). Use. fl'lg "Imam‘lD-Wle— Confll'lzign £9 Jz'EerMIPI-a. i512. .29.ng # k, for “5“ch - P° le. «av-l tto. cJoSeJ- - peep rarity-n. U 5‘ is k .. t 3 I (5| 'fhl lcsl"?fll = Q; Q}... . .. [5960‘ l (5. ~23! l(s.- em 25 Q1 k _ (‘3.I)(Z-6) = 033 ° ‘ (3.8)(55) T'- system is 8180 .siaafle fif- o<K. (0.39 it 2.. 711.. ruled 52...,5; gmflhgglm (7") an on she. T‘oa‘i‘a flows ang— au- algt-Eafl. .44. mm‘hnat 19:0.3. The. cued-('93Ponglfla~ can'éruQ again K. B {:0st any 19m. muyntflbé. Imag Axis Root Locus of K o Gp(s) (5-3)(S—S) cs+ LX544) 6, (53 =- Real Axis Figure 4: Root locus of the closed-loop system as the proportional gain K0 is varied from zero towards infinity. (o-3)Co-s a». acts) 6,6.) = k. ”"L 3. as -- kP= S6, (o+2.)(o+l) —m S: (0.0'1‘0 (int—:9 : 0‘33 I 1+ KP Problem 3: (25 points) Figure 6 shows the Bode magnitude and phase plot of the plant Gp(s) in Figure 5. controller -lant U Figure 5: Closed-loop system with cascade compensation. 1. (6 points) Based on the Bode magnitude and phase plots, write an expression for the plant transfer function G,(s). 2. (6 points) What is the gain margin and phase crossover frequency of the uncompensated open—loop transfer function B(s)/E(s) : Gp(s) ? 3. (6 points) What is the phase margin and gain crossover frequency of the uncompensated open-loop transfer function B(s)/E(s) = G901) ? 4. (7 points) Suppose proportional control Gc(s) = K9 is used. What is the steady-state error due to a ramp input r(t) 2: tu,(t) if the gain margin of the compensated open 100p transfer function B(s)/E(s) = Koapp) is 6 dB ? l. Frau-n 'H'c. muvnpbflfll. put, 'For' w < 10% the. Slope. ls - 5° dfifle‘. . -2,od6[&c3 51mg. Cor w > lof..;.""..9.h the. 510p; 35 a“ ,0 seal U This 54332513 ,. total 7L M ”has; with 2m 0'- 01: lo "Wm. c _ e_ P‘om the phase. response.) i gov-D frebvencief ‘5 70 T915 Jagger's: 0-— Pale. “£3 the. “$7!"- 3GJOQ on this Cu '30” (Q the finsfir aflc‘élofl IS 10 U Phase (deg); Magnitude (dB) Bode Plots of Gp(s) Frequency (rad/sec) Figure 6: Bode magnitude and phase plot of Gp(3). (lo-€94- at IO "WE/52¢.) 7’12 Manam'blfi- #- 6,;(55 “76' ~ -643, the We .ae 0%c.¢r.1...w was, Qmok P 0‘9. 03‘: uvpc. can'brt Evie.) - 3 OQ- B From the EDE— nuancEJeQ outca- IIVUO'O. flo'ES Phat-52. mama tfl C-T‘OSSOUEY“ ;'7- inC. 000’) fianS‘QP ‘CNL'LEM i3 6 a! 6) than we, neeir '50 82% k0 ll KV" a flax. 5 69(536fC5) 5+9 _ ,,__LQ._~—- 5-. lo K9 “ 9:07“ "° Maw? 12 13 I. Problem 4: (25 points) Consider a. plant with transfer function s + 1 GP(3) : 0.5.9 _1. 1. (8 points) Neatly sketch the Bode magnitude and phase plots of Gp(-’)- 2. (8 points) Sketch the Nyquist plot of the open-loop transfer function Gp(s). Indicate the direction of travel along the contour [‘9' using arrows. 3. (4 points) Based on the Nyquist plot obtained in part (2), determine the range of proportional control Gc(s) z K,J for which the closed-loop system in Figure 7 is BIBO stable. Y(s) Figure 7: Closed-loop system with proportioanl control. 4. (5 points) Verify your answer in part (3) by applying the Routh—Hurwitz stability criterion. d‘” "1 ‘, r2 onoawldel ”fl .__'J.+Zoc15{£e‘ 14 \J % rag: MS oi. the. oriain. 3. ThQ. P‘ant 6? CS) “946 an. Pole. in the. RH PJ aria SO =1. Far Btfio «5.1%; we. naval ‘2': = u+ P —..-.-. 0. To db‘bl-lfl N =- ‘1} the. “a Euus‘l: FIJI: MUTE CnCtr‘t—Q ti?— 01'9”; once. It!" the. Coun‘bzr plank w\$e.. GQUEL‘EforL For “HID ‘50 OCC—ur“) we. nooJL kc >1. 15 Supposo. k, is mam-Ewe; this w.” aDQ. + {Zoo "£0 “50' 509'. Pink.» Pitt: shewn cm Pafl‘e‘ IS.- The. Nabmt Flo-i; ’3 to 6p (avg) 43.,— \<° < aJ .5 ma», below: "'_ kg 6??) 0<W<.¢° -—-~ k9 apt-ya 04:» (m Re. ikoéfl k0 <0 flew. reaJ'E?) when Qttl’lQr' ...
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