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ExamIISolutionSP05

ExamIISolutionSP05 - ME 375 EXAM#2 Tuesday April 5 2005...

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Unformatted text preview: ME 375 EXAM #2 Tuesday, April 5, 2005 Division King 11:30 / Cunningham 2:30 (circle one) Name SOL—UT 10H Instructions (1) This is a closed book examination, but you are allowed two 8.5Xll crib sheets. (2) You have one hour to work all three problems on the exam. (3) Use the solution procedure we have discussed: what are you given, what are you asked to find, what are your assumptions, what is your solution, does your solution make sense. You must show all of your work to receive any credit. (4) Circle or box-in your answers. (5) You must write neatly and should use a logical format to solve the problems. You are encouraged to really “think” about the problems before you start to solve them. Please write your name in the top right-hand corner of each page. (6) A table of Laplace transform pairs and properties of Laplace transforms is attached at the end of this exam set. Problem No. 1 (30%) 30 Problem No. 2 (30%) 30 Problem No. 3 (40%) 4a l 00 TOTAL (***1100%) April 5, 2005 2 Name 6mMT ION PROBLEM N0. 1 (30%) Consider the second-order transfer function given below. Assume y(0) = 0 and 52(0) = 0. Ys Kama): () G(S) sz+2§'a)"-s+a): (a) Given the input f (t)= Au(z), where 11(1) is the unit step function, solve for the dynamic response 32(1), and clearly identify the transient and steady-state portions of the response. YB} - V‘: w“ I) (-2151) _I:I_\+ A? (:5 + Sham) * A?) L44 - ‘1 1- P—T 1. 51+ 2—3 wm- g .i coal {“9“} + Na @SW“) «- had A (91+Zgwn‘$+w1)+ Azsfiggwdt— A studs) : _~’___’,_____3____ 4": Aim». ~——-7Az.——Air-t,A {A “(Rabbit + 028%” +(wDA-s: o % AVE (5%” 43m" :v-Eiijw R'A (b) Using the result from part (a), find the numerical value of the product (g'wn) to ensure that y - YO) < -005 r :21 d. : as: «A M—r—x ) ‘:im = KsA- vsA 59““ [Cbstwd)+ film. H (o -u:[‘] 40.05 April 5, 2005 3 Name 59 LUTI o g (0) Consider a different input to 6(5), f(t) = 3.5 +11.71cos(4.26t)— 6.87sin[15.33t +3. When K x = 7.56 9' = 0.479 , and (on = 4.26 , find the steady-state response, y“ , to f (t) Grljw) 1L 1 ‘L—__‘—“ quZ‘ 4'25”“ (ind) +W} [1" "JO-z]. +d (255$) 3951+: = 3‘5 (5150” + (1'5“) 15(54-Zbfl to: (4'2“ “LXFG‘J ”“0 _ Gum) “1“? I533), 51‘“ < ,5.B++ 1;: + gtdqan) 355m : 25.45 + 92.41 Cog (4.2:, 3171:") _ 4.:5 5m (“5:33 + 13‘: — 'L%) April 5, 2005 4 Name 0 Lu PROBLEM NO. 2 (30%) Draw the asymptotic bode plot approximation for the transfer function, 31 l C(s)_ 0.0255-s+o.0332 __ ' (0. tab? 5+ 1 ) " 3+0.0380. 2+0.0036- ’ . S S S 5 (51+ e-o'seo 5+ enoo'se , . . t___..-—.r-—'J L/\_——/ usmg the followmg steps. 2,314,. bowl (a) Find the poles, zeros, and the static gain of the transfer function 6(5). Zero I ‘ l'S poles: .— 0.0!. 6113 0. 0'5ch 3 o 5541‘; Sam 2 61°) ——)00 (131:. (7; 43v“. 3m+cjva+u1F) (b) Extract the elemental transfer functions writing all first-order terms in the form (2'- 3 +1) (except for s or 1/ s) and all the second—order terms in the form a): /(s2 + 2(a)” - s + a): ). (so = K5 : {3% => W~zroQaWWu> 6* = ('75- +1) ___. OIDO—Sb (:3 52' + 0.0330 5 {— moon, (c) Identify the break frequencies of each of the elemental transfer functions comprising 6(5). WE». _> 0‘9 win .90 thE-mal but 1'- I-S N3: 0' 0"] rLzbufr-ed (d) Plot the asymptotic magnitude and phase plots for each of the elemental transfer functions on the graphs on the next page. Be sure to clearly label each of the elemental transfer functions on the graph along with all break frequencies and slopes. (e) Plot the total transfer function, 6(5), from the composite of the elemental transfer functions in part ((1). hkune SSOl_LA [C3 TOTAL — Gld‘w) BoDE DI AC—nzAM Ap ' s, 2005 IIIII LIIIIF'OI _ I‘ll—‘1' .l.||. _ittlJIIII~IIII _ _ _ 1:1:TIIIJIIIITIII . _ _ _ _ _ . . _ HHHHHHHHUHHHHHHHH Illtlel;1unnTI|| April 5, 2005 6 Name ’30 LUT ION PROBLEM N0. 3: (40%) In lecture the following Input / Output Model for a DC electric motor was developed. E1” In 7' EA ls; 9(5) : KrEi-(S) __ (LAs+RA)TL L,,J,,s2 + (LAB +R,J,)s+ (R,B+ KbKT) LAM2 + (L,B+R,J,)s + (RAB-t KbKT) where J A is the armature moment of inertia, KT is the torque constant, Kb is the back emf constant, RA is armature resistance, LA is the armature inductance, TL is the load torque, B is the armature friction, and Q is the armature speed. (a) Assuming that B is small develop the following: “For: Tfiau (STALL. szQuE) 11:0 () Kr Tim” AND W15 ‘5 WAD? STAT-E I —= ' RA ea 51:0 3 KT EA M147 (gWs + 'EAgl'th‘r) O o 0 AT 5.5- $=0 '1‘} — QZ+KA>T~mu KT EA: m; KT ' (inK — e“ “For. no (”Mm no—Loael (no Low gees) T._=o MD smarts/«617m; ( 5— wine-4034 :: KT EA J'bc‘é—L‘l' (LAEVA/JA35 4- RB High We’lcaél : y4EA ': EA ___> KL... _‘ E 135%“ o'l/B‘g iHCbK/T Kb If the transfer function for an electric motor loaded through a transmission is given by the following: KT R J 9’" (S) = ——-A—“’— where qu and Beq are the effective moment of inertia and E (S) 1 KTKb " (s + — (B, + )) J“! q R, friction for the motor, transmission, and load system. Find the following from the given motor speed map data: April 5, 2005 7 Name Sat—UT ION NI = 100 3 91.0) N; = 1000 .10 = 5 kg-rn2 ., JL = 700 kg-m2 8,, = 2 N-m slrad (a) Bx. = 800 N-rn Slrad Torque Speed (rad/s) ‘L (b) "L fa, =JA ”gym 3” = 3,, ”(£3 N 2 2 (b) Find the transfer function Qm(s)/Ea(s), where 0m is the motor speed. (Put values for the terms into you TF for credit .) Jab: JA—FJL N‘V': 5+7OO(T;)1 : {2. 7r), April 5,2005 8 Name sot-LL I [32$ (0) Find the transfer function 9L(s)/Ea(s), where 91, is the load position. (Put values for the terms into you TF for credit .) N29}, : N1 e'IM 1E9“ : QM 5 film _L 9m 1*. 5 N1, ~eL. April 5, 2005 9 Name Lanlace Transform Pairs m) F(s) Comment 1. 8(1) 1 Unit impulse 1 . 2. l for DO — Unit step 8 1 . 3. t for t>0 2 Unit ramp S -at 1 - 4. e — Exponential S 'l' a 5 ' t a) S' . sm a) 1116 s2 + 012 6 t S C ' . cos a) — osme s2 + w2 7. f( t) F(s) Function (if t _ 8. —d(—) SF(S)-f(0 ) First Derivative t dnf t ._ .. _ (if 0— ( ) SnF(S)- 8“ [HO )—Sn 2 ( ) -- - 11m Derivative dt“ dt n! 10. t" SrH-l 11. e'a‘fa) F(s + a) _. 1 12. te “1 —-2- (s + a) 13 Inc'“ 11 l ' (S + a ) n ‘i' l . [U 14. 6"“ sin mt —2—2 (s + a) + a) . S + a 15. 6"“ COS (1)! '—(2)—2 (s + a) + a} 16. Initial Value Theorem: f(0+) = 511,12 SF(S) 17. Final Value Theorem: lim f(t) == Sig?) SF(S) 1—)00 ...
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