examII_s01

examII_s01 - EE 428 EXAM II 12 April 2001 Name:...

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Unformatted text preview: EE 428 EXAM II 12 April 2001 Name: golv‘lfio 95 ID#: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Weight Score - - - - 4 - loo - 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) Skecth Root Locus 1. (10 points) Consider the closed-loop system shown in Figure 1. Using the angle condition, determine if the point .5 = ——1 + 3 is a point on the loci of roots of the characteristic equation for some K 2 0. If it is a point on the root locus, determine the value of K at this point using the magnitude condition. Y(s) Fran the ska-trek Kiowa .|_L, _ s- q}! = T“ 1 -HSO -_-_ tic-F3600“ '9’ Q in 1' $100 {75" (.PL '1. 70° «1- l n it- : [35" "' J Usmj— the mmeinP-v Comp'b'w'i K 1693'“ a K_ I = 5W).— larsm l 3,9,+2.\ ’ 1—3 r7: K=| 2 1 P K) N \1 \m \l 2. (10 points) Once again consider the closed-loop system shown in Figure 2. Sketch the root loci of roots of the characteristic equation as the parameter K is varied from zero to infinity. Show the start points, end points, breakaway, and and breakin points. brewkdh {)0 U0 b break-tn pom-i: " 3 . '4112, \ _Z+E -_; 3. (5 points) The closed-loop system in Figure 3 uses derivative feedback with gain Kg. In order to sketch the root locus of the closed-loop system it is necessary to express the characteristic equation in the form 1+KG(3):0, where both the numerator and denominator polynomials of G(s) are monic. Specify K and G(s) for the closed—loop system shown in Figure 3. Do not sketch the root locus. 4-69» Y(s) so) " 52+4KQ5+ " The charactensf‘tc. @649. ton % due. ("1039-0 “poof ~é7Jtem LS 52+qm3+ L1 :0 Problem 2: (25 points) As discussed in lecture, the ability to quickly sketch a. root locus will allow you to choose between difi'erent controller designs. As an example, the following three controllers were suggested for use in the feedback control system shown in Figure 4 Controller A G43) 2 K Controller B G43) : K s + 4 s + 1 Controller C Gc(s) : K a + I s + 4 Y(s) Figure 4: Closed-loop system with controller Gc(s). 1. (18 points) For each controller, make a rough sketch of the root locus. Con‘bholler H 2 clown 260, l + K "SL1 1: D —___——-_H—u——'IH' K > 0 K 4 0 Insist :m 253 (Z) w [has I mas; Coh'b'wnew‘ 6 char- can; I .l. K 2. (7 points) Use the root locus sketches in part 1 to qualitatively evaluate the transient performance of each controller. Which is the best controller ? Explain your answer. 505*“ 15 009 13/80 c-talo‘o. «Jan we use, Confimflqr— C out") K .70. RegarJ‘oSS % 15110.. vw‘vg ;2 KJ COO-EPJJPLS H” amfl. 33 on” no'b firm/Owe. av c/osxo—Qaaf 6:30 stalk, SVSLem. f}an so the $251.4 Con‘A'r-oflxzr‘ 7S C (“It‘ll-TA 1950). Problem 3: (25 points) 1. (9 points) Consider the feedback control system in Figure 5 where the plant transfer function G(s) is a Type 1 system. You are asked to design a compensator G'C(s) so that the closed—loop system achieves zero steady-state error for the input 1-(t) : t 1(t). You are also told that the transient response of the open—loop system to a unit step input is satisfactory. Specify Whether a P, PI, PD1 phase-lag or phase—lag controller should be used to satisfy the design specifications. Justify your answer. 4. as» Figure 5: Closed-loop system with plant transfer function 0(3) and compensator Gc(.9). Y(s) If. orfler *0 wn‘J‘le-UQ; "Zero 92W 4N or... raun p "WV—b J GCCS) G (5) moJl: 5Q- jargz BQ—CmvSQ 6C5) is Tny, [J we, macaq- 66557 l3” bo— Type .’L-. Fa.» tho mucky/1J Gets) Shoulcg. lfiQ. av PI canfvoller, 2. (16 points) For each of the following compensators 3+4 Gc1(s) : K's-+1 1 G¢2(8) :. (a) (8 points) Determine the low and high frequency gains. (b) (8 point) State whether the compensator approximates a. PD or PI controller. Justify your answer. (“3 W I ac, (as) = K 15615:»)! = K .QOW £refiuemg gums le (0)1 =- HK lac-brow: i;— va Sam 15 a. PhasearOQQO, medley amuse 'GQ(°6))> JGQCDD) o Pfiuse ~0qu— con'éwuefs affrw‘m’dte or. PD minty—gun’s which M ,0 "9,192 5cm, ai [max gaauenuej- ’79. C one/SQ 6C, (0 75 a, Phase '9? (on-Emile?- {Gm Co)! > 16C, (00)}. Phase—Q7 Minnow hpprmmaf’a 0. pr Cantata wine)! flue [Ix-EMS: 0940‘ i Dc. Problem 4: (25 points) The closed-loop system in Figure 6 contains a. plant with transfer function G(s). Figure 7 showa the Bode magnitude and phase plots of the plant transfer function C(s). + R(s) —>®—> G(s) Y(s) Figure 6: Closed-loop system with plant transfer function G (s). 1. (5 points) Determine the steady—state error of the closed-loop system to a unit—step input. 323? the darker; 7.5 3 I60 S'iZLLiQJ they, k? :fiy’n. 6C5) :: “'10, I 5‘90 However-J {roan put, Ft) 3 amp- ‘1 we &LJCo-uor' 1"le @3th '8 unstable) and}. So 19-55 I :— co. 2. (10 points) Based on the magnitude and phase plots shown in Figure 7, determine the transfer function G(s) of the system. ‘rfio. Peak in the, majorbuog, Flo-I: cmgdciw Mt», 180° change. in Phase, lJle'Eéflf a.- xmflwr Th“ Ml“ “WV-5 «3‘3 “n =ic>¢=~J amQ 2490?“, 11- 5:19:96 =9 30-70-05- Th; DC any.) Km.) 75 -IO (at WcoJ 151; 2x03) 56,—.—180°) 7- 'IO“. = Kbcwn L L 1 51- +2fwnJ +wn .s + £08 +10 3. (5 points) Based on your result in part 2, find an ordinary differential equation that describes the response y(t) of the closed-loop system to the reference input 1-(t). ch) : G(§\ : -|05_ ets‘) 1+6m glues - affine; 4. (5 points) Using your result from part 3, do you expect the unit-step response of the closed-loop system to be underdamped, critically damped, or overdamped ? Explain your answer in a. single sentence. ‘ERQ. sawiem Bgfitfio with! “In resfonge ‘tso a. (JAKE-1642’) Input) laC£)l-—3°° “'45 10 Phase (deg):Magni1ude (dB) 100 50 Bode Diagrams 102 Frequency (rad/sec) Figure 7: Exact magnitude and phase plots of (1(3). 11 10 ...
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examII_s01 - EE 428 EXAM II 12 April 2001 Name:...

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