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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


 
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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 closedloop 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 skatrek Kiowa ._L, _ s
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\l 2. (10 points) Once again consider the closedloop 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 breaktn pomi:
" 3 . '4112, \ _Z+E _; 3. (5 points) The closedloop system in Figure 3 uses derivative feedback with gain Kg. In order to sketch the root locus of the closedloop 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. 469» Y(s) so) " 52+4KQ5+ " The charactensf‘tc. @649. ton % due. ("10390 “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: Closedloop 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
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(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 ctalo‘o. «Jan we use, Conﬁmﬂqr— C out") K .70. RegarJ‘oSS % 15110.. vw‘vg ;2 KJ COOEPJJPLS
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Con‘A'roﬂxzr‘ 7S C (“It‘llTA 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 steadystate 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 phaselag or
phase—lag controller should be used to satisfy the design speciﬁcations. Justify your answer. 4. as» Figure 5: Closedloop system with plant transfer function 0(3) and compensator Gc(.9). Y(s) If. orﬂer *0 wn‘J‘leUQ; "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
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[IxEMS: 0940‘ i Dc. Problem 4: (25 points) The closedloop 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: Closedloop system with plant transfer function G (s). 1. (5 points) Determine the steady—state error of the closedloop system to a unit—step input. 323? the darker; 7.5 3 I60 S'iZLLiQJ they, k? :ﬁy’n. 6C5) :: “'10,
I 5‘90 HoweverJ {roan put, Ft) 3 amp ‘1 we &LJCouor' 1"le @3th '8
unstable) and}. So 1955 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, FloI: cmgdciw Mt»,
180° change. in Phase, lJle'Eéﬂf a. xmﬂwr Th“ Ml“ “WV5 «3‘3 “n =ic>¢=~J amQ 2490?“, 11 5:19:96 =9 307005 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 closedloop 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 unitstep response of the closedloop system
to be underdamped, critically damped, or overdamped ? Explain your answer in a. single sentence. ‘ERQ. sawiem Bgﬁtﬁo with! “In resfonge ‘tso
a. (JAKE1642’) 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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