Exam 2 solution

# Exam 2 solution - EML4312: Spring 2011 Test 2...

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Unformatted text preview: EML4312: Spring 2011 Test 2 Name—Ke¥____—_——__ Instructions. T 115 is a closed-book, closed notes exam. 1. (25 points ) Consider the following block diagram Block diagram. __ 2 _ s(s+3) GP Using the MC / AC, design a Lead Compensator so that a closed—loop pole is —5 + 5j. For the degree of design freedom, place a zero at s = —5. - lie; ZZeros- i‘polcs r 3/30 4 8+5 ~zs—z\$+3 —A\$+b} =i/3O sz'SJST Ag: a. 1—9:? «- z -a+{'5 ’ zb—\$+\$7 70° - 4u5'[=—D ~— Jm;'(:%> - 47;]323 =+IZO qo-C—A/q» («M933 — 4mm"(.D—§§.§ :HYSO E 3 5- .— Fs' ' -43 /S}u( Meal] (7‘ \ 2. (25 points )Design a PD controller to place a closed—loop pole to yield a percent overshoot of 9.5% and a settling time of 1.3 sec to reach a tolerance of 2% of the ﬁnal value. Note that 100 _ (57f P.O. r m (fawn We. 0 . 03 ll H -wéﬂ ) j“ (ac/5'» —' ﬁ /6 VF???) e"'(’w“(“37 «902 4,3 = .. gum/['33 ,th: 3. cl “2 5. (F31 : §(I.333 [2:1 ® 3. (25 points) For the block diagram in question 1, let G’p be redeﬁned as l G = —.—— p (s+1)(s+2)(s+4) Where GC is deﬁned as ( > s + 3 Ge _ (s + 5)' Modify the controller so that it yields a steady state error of 0.101 for a step input. For the degree of design freedom, place a zero at s = —O.1. Rl 9 - l a E (S\ : (’2 (\$7 ’ {S l + a? c. c 5+ 0(5), zY :4 q\(sh’3 I . r _L M £55: ‘: lislgos(s\ '+_£§i§}_———-——-—"" gfd l ¢-—- l+ “"3 5 HZWMYﬂcl A O 0|: I : 'Jgié’ 1+ '3 LIOJ+.3 403 \$+ ’1. \$7 X/o's- ,atg Comlyngalzr: \$+.-l . 4. (25 points) Design a controller that gives a closed—loop pole location of s = —3 — 3j for the following transfer function. Q K ﬁzzy COC'C'FlC‘ICn‘l Wovlclthz : Db: [s +3+3’S>(5 +3, 333 AA: (\$+D[S+S3+Q¥ = iz:s’+a\< \k: ...
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## This note was uploaded on 09/05/2011 for the course EML 4312 taught by Professor Dixon during the Spring '07 term at University of Florida.

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Exam 2 solution - EML4312: Spring 2011 Test 2...

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