final_f98

final_f98 - FINAL EXAM 14 DECEMBER 1998 Name: \3 lg 0115 DO...

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Unformatted text preview: FINAL EXAM 14 DECEMBER 1998 Name: \3 lg 0115 DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO 1 30 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: (30 points) Figure 1 shows the block diagram of a position control system. The plant is a DC motor with input armature voltage u(t) and output angular displacement 3(t). The Bode magnitude and phase response of 6(s)/U(s) are shown in Figure 2. In this problem you will design a cascade controller so that the closed-loop system meets the following design specifications. i. A velocity error constant K1,. 2 10. ii. A phase margin greater than 40° for the compensated open-loop transfer function Gc(s)Gp(s). Figure 1: Position control system with cascade compensation. 1. (5 points) What is the phase margin of the uncompensated open-loop transfer function G'p(s) '1‘ 2. (5 points) Suppose proportional control Gc(s) : K0 is used. Specify the value of K.J required so that the design specification is met. 3. (5 points) Can both design specifications be met simultaneously using proportional control Gc(s) : K, ? Justify your answar. 4. (15 points) Design a phase lead controller that satisfies both design specifications. Clearly show how the parameters KO, la, and we were obtained. I. Fm, Ham. 2, wag = 0.839; , pm 3: 55° Zuess": "‘l—' Kr=fllh~kofila=ko '1.) Kw- 59. 35+!) 3. 13-? Kg: to) "59— fecal-lanai 7’“ 5d KoCPCQ i5 20‘? (The gain CM.Ver % lo GPCS\ )3 3 Fugue) ChOOSIy K) =‘—- [o 43., Sagas-E; Kr.- FQ‘SVH: 0’0 l/‘q, e i FM. Profur'lawnng {S COfl‘Bol (4” flat S'rmvl‘buf‘PQ} c “15% baa) you - Phase (deg); Magnitude (dB) Bode Diagrams Frequency (radlsec) Figure 2: Bode magnitude and phase 01: of Gp(3). gala C‘ICIIS‘ 1749)” 9:25.36 % 611m Crass w-Pr -Frezuerg % 10 Gr 65) Problem 2: (30 points) Consider the closed-loop system shown in Figure 3. Y(s) HIS) Figure 3: Closeddoop system with variable parameter a. 1. (5 points) Express the characteristic equation in a form suitable for constructing the root locus as the parameter a is varied from zero towards infinity. 2. (5 points) Determine the value of the parameter a for which the closed-loop poles cross the 3w axis. Specify the location of the crossing point on the imaginary axis. (5 points) Calculate the location of the break in point. (5 points) Determine the angles of departure at the complex poles. (5 points) Sketch the root locus in Figure 4. EDS-"99° (5 points) Using your root locus sketch and the magnitude condition, determine the value of the parameter a for which the step response of the closed-loop system is critically damped. A ruler has been provided on the last page of the exam. I. Chara-c‘iZQnJ'Lic. 55,9;me t+ amnion-:0 H. 9.7—5 ._'__ =- o ..) 51' + r51.2.51» 3 + a.(.$-D =0 ObSQTvQ thud; 6C8) “15 - mzl ‘cml‘i‘fi- £9“: “'4: 8:"! , n :7... poles «£0: 552, = + t i- J‘T '7'2‘5- 11 + N l— \t as o4 Imag Axis Root Locus for 0 < a < co Real Axis Figure 4: Root locus as a. is varied from zero towards infinity. 2-. qhe. LRmt‘chi—erer-ic grog.» non :3 57' + S(O\—"‘I\ + °f.2$"cc, =0 11‘9- RaJH-x our Nana. is 57‘ I Ci.2.§"‘0~ \ S m~l 0 é———> 01:24:44?! row % Zens 4;" ("Q-l wrfigx Pies. 5.655% 57- + 8.75:0 D s 7‘ "GI- ‘ 3'5 s: ta T G $042.37 I “the (“‘65 cvass 'U‘P— dw 0"“3 Id; t 2-37 when “:44 T he, breakfin F 0) n4: Sw‘LuS C165 fl (I It?“ : O __.—.—u———'—""_“""- —- .- 3. E" 51—5 + fizz: _, (25—06-13 "' CSZ'SH‘“) a o :1 7?. 5° Problem 3: (20 points) The closed-loop system shown in Figure 5 contains a plant G,(s) whose polar plot is shown in Figure 6. Using the ruler provided on the last page of the exam, answer the following questions. Figure 5: Closed-loop system with cascade compensation. 1. (5 points) What is the gain margin, in dB, of the uncompensated open-loop transfer function G,(s) ? 2. (5 points) What is the phase margin of the uncompensated opennloop transfer function ? 3. (5 points) Using only the polar plot, choose the value of K a so that the phase margin of the compensated open-loop transfer function is 45“. 4. (5 points) Is the open—loop plant BIBO stable ? Justify your answer. I. fit we, lGPCj—Wiochl: OJHS‘ Polar Plot of Gp(s) -1 —0.5 0 Re G(s) Figure 6: Polar plot of the plant transfer function Gp(s). 11 0.5 12 13 Problem 4: (20 points) A 8150 plant is described by the state space representation (-3 33)*+(3)“ (1 0):c. ll w) y“) 1. (5 points) Is the open-loop system BIBS stable ? Justify your answer. 2. (5 points) Suppose full state feedback i used, where u(t) z ( k1 kg ) a:(t) +gr(t). Can we choose in and kg to arbitrarily place the poles of the closed-loop transfer function Y(s) / 12(5) ? Justify your answer. 3. (5 points) Choose the controller parameters ’61, 512 so that the eigenvalues of the closed-loop matrix are located at "2 :l: 2]. 4. (5 points) Choose 9 so that for the unit step input r(t) = 11.05), y(t) = r(t) under steady-state conditions. 1,=-I I, Jgt(7t1‘.-fl)= ’2 -i )5 73%.321-2. = (3+I\(7~+23=o Kalgi 2. 2+3 62.64052, loo‘H-r Quflenwas are. negativeJ the Jlj‘Lem is 6185 «We. 2 P s (3 “53 = o a) lat“) = (o)(—63—c23m =~v #0 C 2 _ “(Limrul Plum? IS Oar-Sm ulcer qu. C‘?“ ¢h°°°e k') A in P 3 ‘ E35,”; (Jae-Jourth I'IH‘C- Eb uw‘Eton '. 2 (7x+2.--\(7H-z+1'3= 7\ + + 8 ‘-'-° Cl a PAL—'1 “Q t Undvfloop characteristic. egaocbicn m berm: % k‘l 6* 7- -=A K =. (ea-smxcsa“. X 3‘ + Y +a'r3 Glflfi’" 35km mmlmk. o u. kg 0 I 9+ [5K = (.02 'l" (I) ‘ =- (.2+2K‘ 6+3) --I out (m: - (6+er = 7‘ a z—zk. a—rs-zl‘; Cor-path? the “burg v.1 Jerlsnpfi. E KL?- ‘17: K. = '3 14 all: 3—LkL': "t a) 7‘": Z‘ZF‘I = 9 ‘9 YCS) ’1 —-— = c: sr-L’Awn [3 am ( > 3 I Th9. bc. 00.54. is 5:24:43 my. 501 chm! “it 09 as -l DC em; = c(—CH+GI~1)85L - l _. __i____,..__. a c (-59»: 5:13" (5 o I ° 1 3 Q 5 a BB = \ :3 fl _ .- + (-24-2F1 -3+2.F—2. ELL“); 3 '1 L1 ‘ (‘72. ’19 -\ .- o -l .- fl—f“ g (_ '- ( 9 l1 ‘- (°‘(q\‘(33L—n -3 o —I O ‘l _ V7. [/9 (’3 z (1 o§(1)= V7 = <1 03(4 0X2 o I .F—. :3 _...-————-—-*-—'—_T'" f g cf-mmn‘jb /7 H 15 16 17 ...
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This note was uploaded on 07/23/2008 for the course EE 428 taught by Professor Schiano during the Fall '07 term at Penn State.

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final_f98 - FINAL EXAM 14 DECEMBER 1998 Name: \3 lg 0115 DO...

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