final-so1

final-so1 - Final Exam EEL 3657 (Spring 2003 Instructions...

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Unformatted text preview: Final Exam EEL 3657 (Spring 2003 Instructions 1. There are totally 4 problems in this exam. 2. Show all work for partial credit. 3. Assemble your work for each problem in logical order. 4. Justify your conclusion. I cannot read minds. Problem 1. (25 points) For the unity feedback system sho . below, G6) = (50035) , s s + _R(S) + % E(S) (3(5) 1). What is the expected percent overshoot for a unit step in‘ut? 1 2). What is the settling time for a unit step input? i 3). What is the steady-state error for an input of 5u(t), wher u(t) is a unit step fimction. 4). What is the steady-state error for an input of 5tu(t), whe - u(t) is a unit step function. §§vuiknnz The closed-loop transfer function is, “)' T _ 5000 “)"s2+75s+smm from which, a)" = V5000 and 2(0),, = 75. Thus, 4' = 0.5 ‘ and %OS = e—W‘IITFXIOO = 14.01%. 4 = —-§4—=0.107 second. 29‘ T‘= cub 75/2 3) , Since system is Type 1, eSS for 5u( t) is zero. i 5000 _ _5__ Q) Since KV is = 66.67, e33 - Ky — 0.075. I i - Problem 2. (20 points) 1 1 1. 10 oints Consider H s =Gs , H s =Gs a, )( p ) l() ()SZJFZS+2 () ()S2_2S+2 At a frequency of a) = 2 rad/s, G(ja)) has a magnitude of A and an angle of ¢. What are the magnitude and angle of H I ( jw) and H 2 ( jw) at the sam‘ frequency? Ke-Z: S 2). (10 points) Consider G(s)H(s) = , determine th frequency a) at which 4G(ja))H(ja)) = —l810°. = +a-C'I +-+a-"3 =“7e, 1 [swan] -.= {£479 = 2r? 0 ' . ' f . HA2» A Lt” . “S "7 2r?" : 011/} 415:“): . _________.___._.\_—_._..—- -3 w. I I . . __ 6 J 5 _~_ fim __%__/_i- ___.'_-_‘—.J.‘-Z_—:_____-._ .-_..___.__._. ________.._,___._______tv_4g'~___w 4______«~_______..__~__..____ .__._.—r——-—— Problem 3. (20 points) 1). (10 points) Sketch the straight-line Bode magnitude p101 for G(s) = 2) (10 points) Identify G(s) from the magni shown below (G(s) has no. complex poles): dB 20 2 s2(s+2)' tude plot of th straight-line Bode diagram as 1/0005 Problem 4. (35 points) The figure shown below is the blo k diagram of the servo-control system for one of the joints of a robot. 1). (10 points) Show the plant (“Motor and arm”) transfer ction is given by 49L (5) 0.15 Ea (s) s(s + 1)(s + 5) ' 2). (10 points) With the compensator replaced with a gain I , sketch the root locus of the system, and specify asymptotes, break-in/break-out points. 3). (15 points) Given the compensator Gc(s) = 30(K1D + Ds) as shown in the diagram, calculate the compensator gains K p,K D to put the dminate closed-loop poles at s = ~1i j, and sketch root locus of the compensated syste . Compensator Motor and arm r_____.. 12 lg 4514-2451911 =__’~ * l + ~L— 305% #54245 +20 W“ #59245“: = I I? _ 3/20 1.7.05 5"+é§+5 ‘ 5154-1)“ 5) (2D°T’>/>°c£ “=0 2 (pa: 3 ’9 " O 0 U flames“ gee ,mzsw, __’__$.~ out: ~=~1 ...
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final-so1 - Final Exam EEL 3657 (Spring 2003 Instructions...

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