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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 closedloop system
meets the following design speciﬁcations. i. A velocity error constant K1,. 2 10. ii. A phase margin greater than 40° for the compensated openloop 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 openloop 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— fecallanai 7’“ 5d KoCPCQ i5 20‘?
(The gain CM.Ver % lo GPCS\ )3 3 Fugue) ChOOSIy K) =‘— [o 43., SagasE; 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 wPr Frezuerg % 10 Gr 65) Problem 2: (30 points) Consider the closedloop 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 closedloop 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 closedloop system is critically damped. A ruler has
been provided on the last page of the exam. I. Charac‘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:"!
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s: ta T G $042.37 I “the (“‘65 cvass 'U‘P— dw 0"“3 Id; t 237 when “:44 T he, breakﬁn F 0) n4: Sw‘LuS C165 ﬂ (I It?“ : O __.—.—u———'—""_“"" — . 3. E" 51—5 + ﬁzz: _, (25—0613 "' CSZ'SH‘“) a o :1 7?. 5° Problem 3: (20 points) The closedloop 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: Closedloop system with cascade compensation. 1. (5 points) What is the gain margin, in dB, of the uncompensated openloop 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
openloop 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 openloop 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 closedloop transfer function Y(s) / 12(5) ?
Justify your answer. 3. (5 points) Choose the controller parameters ’61, 512 so that the eigenvalues of the closedloop 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 steadystate
conditions. 1,=I I, Jgt(7t1‘.ﬂ)= ’2 i )5 73%.3212. = (3+I\(7~+23=o Kalgi 2. 2+3 62.64052, loo‘Hr Quﬂenwas are. negativeJ the Jlj‘Lem is 6185 «We. 2 P s (3 “53 = o a) lat“) = (o)(—63—c23m =~v #0
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 Fall '07
 SCHIANO
 Bode diagrams, openloop transfer function

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