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examI_f98 - EE‘428 EX‘AM I 19 October 1998 Name Sci n5...

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Unformatted text preview: EE‘428 EX‘AM I 19 October 1998 Name: Sci ' n5 ID#: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO 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) Consider the block diagram representation of a dynamic system shown in Figure 1. Figure 1: Block diagram representation of a. dynamic system. 1. (7 points) Determine the transfer function Y(s)/U(s). Express your answer as the ratio of two poly- nomials in s such that the denominator polynomial is monic. 2. (5 points) Represent the system as a single ordinary differential equation. 3. (7' points) Draw an allnintegrator block diagram that describes the system dynamics. 4. (6 points) Obtain a state-space representation of the system. 2.5‘ 9. CS» YCS) 9 2. From pm'L' Ch SZ'YC-S) ‘t’ 6.5 YCSE 'l' 2‘5 Qb‘rwolon'uy) . I I ‘4. \ Problem 2: (25 points) A 5130 system is described by the state space model no: we» y = (0 2):. l. (5 points) Is the system BIBS stable ? 2. (5 points) Determine the transfer function Y(s)/ U (s) 3. (5 points) Is the system BIBO stable ? 4. (5 points) Find the state transition matrix fit). 5. (5 points) Suppose that u(t) = 0 for t 2 0 and that Find z(t) and y(t) for t 2 0. 5+3 0 o I (o 2) (S) 2: ,____l_———- (0 2-3( 3( l = (5—0015) ' (you-+3) S + 3 -1 _ A; (51—40 " cx—msm) 2. __.'___. '2 5-1 0 Reedm. an}. 5:60 $51.47 as Jefinafl £w {N4 1&1‘0 - S r0 onSL . [tuba-94 can "Elm can 1%,, 0,6!(30 s—bab‘e 55549,“! 5 Problem 3: (25 points) Attractive magnetic suspension systems have many applications including magnetic bearings and maglev trains. Figure 2 shows a ferromagnetic sphere with mass M suspended by the attractive force of an electro- magnet. The input to the system is the current u(t) in the eiectromagnet, and the output of the system is the position y(t) of the sphere with respect to the surface of the electromagnet. - electromagnet current 5011 ['66 W) gravity Figure 2: Attractive magnetic suspension system. The inductance of the electromagnet depends on the position y and has been found to take the form Lo y! 1+— (1. 141:) = L1 + where 15;, InJ and a are positive constants. The electromagnet is driven by a current source with strength u(t). The force of electric origin f'(y, in.) acting on the sphere is given by f‘(y.u)=—-1- L“ , um), 2“ (1+ 3) a where f‘ is directed as shown in Figure 2. Note that the force of electric origin always acts to decrease the displacement 1; because f8 (y, u) 5 0. In addition to the force of electric origin, gravity exerts a force M g on the sphere, where g is the acceleration due to gravity. The equation of motion for the system is .. 1 La My:—— 2 u3(t)+Mg. . (5 points) Using :1 = y and :3 z 3) as state variables, show that the equations of motion can be described by a nonlinear state-space model of the form i1 = f1(=1,=2.u) £2 = f2(31:321u) I! ‘-= 9(=1,32)- Specify the functions f1, f2 and g. . (7 points) The coil is driven by a constant current u(t) = no. Calculate the physically meaningful (the displacement y must be positive) static equilibrium state 11,. Express :5: and a; in terms of the parameters L”, a, M , g, and the constant input no. . (7 points) Find a linear state-space model that describes the dynamic behavior of the magnetic sus— pension system for small perturbations from the static equilibrium state x. and constant input up. Let a = a, + 6n and u = u, + 6n, and for the state-space representation 65: = A63+B§u 6y = 065:, determine the matrices A, B, and C in terms of the parameters La, 0, M , g, equilibrium state :1, and constant input no. Do not substitute the expression for :5 obtained in part (2). . (6 points) Is the static equilibrium state a; stable or unstable '? Justify your answer. .. _ _ _L_ L" MPG-e) + fl; ”' (”—2.331 __L_ Lo uC'L-b} "' j— : "PI/(”HM”) 3"“ (”ea-WY“ Problem 4: (25 points) Consider the closed-loop system shown in Figure 3. ’ controller Figure 3: Feedback control system. 1. (5 points) Suppose that a proportional controller 06(3) = K. is used. For what range of values K, is the closed-loop system BIBO stable ? 2. (5 points) Determine the range of K, for which the closed-loop step response has less than 10% overshoot. 3. (5 points) Is it possible chooae a. value of proportional gain Kg for which (1) the closed-loop step- response has less than 10% overshoot M (2) the steady-state error due to a ramp input r(i) = tu.(t) is less than 0.02 ? If so, state the required range of Kc. If not, explain why. 4. (7 points) Suppose that the proportional-plus-derivative controller G‘,(s) = K a + st is used. Choose the smallest values of of K, and K5 for which both design specifications in part (3) are satisfied. As a starting point, neglect the affect of any finite zeros of the closed-loop transfer function on the transient response. 5. (3 points) Using the gains determined in part (4), draw a pole-zero plot of the closed-loop transfer function. Clearly indicate numeric values of the pole and zero locations on your sketch. If there are any finite zeros1 discuss their potential affect on the unit step response of the closed~loop system. (as)? 5L+cz+kms+Ko ' A: r'n Pa‘t (a), flu. Mlnlmvm :fl:Q-:— 0% )fia -: (DO gIVQS #1 6mm, 05"? <. PL... , Lo kickwkfismm -—-7. 5-30 L 0.02. we, “099' k") >’00 (07¢ LL comm, all. The gait-”8' 91W..- 5‘ YCS) __ k0 4, has '— ‘7.%’ C5 +' 'qu'g) 3 a ——-—'———"___—I— -—u m _ _ Ufi 52+ C2+h935+l<o 3“ + 11.85 + 100 There, 7.: one, QWHEEL zero 9% S: —!°O/c,'p: “#332. Thar-0.. ‘ID a.- Cgmflbfi ConJua—‘yé ng, (ON-fr Sui = ._ “.2? pm = ~59 + 9.) '2— ”d T I Fig- T’DQ’ ‘9‘“ £6 32m um” mc‘rpuA-n m 07;: 0-49— rsLoJE % {Ll-L chflw loop U n FL «5432? MSW gage ' I b 12 ...
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