{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw 5 - ‘ For ea'ch of the unit step function of the...

Info iconThis preview shows pages 1–12. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 12
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ‘ For ea'ch- of the unit step function of the system. l g o 5 ' i ? O 0.05 0.1 0.15 0.2 0.25 ‘ Time (seconds) ‘ (a) ., a, 1&2 1% I .49 . . - - l _ 4 5 Time. (seconds) ~ " (b) ’ i ‘ 9 : j _, f For the system shown in Figure P4.8, a step torque is applied at 61(1)qu a; a. The transfe‘f fujiéfiéfiTfls) = 92(S)/ T (S)- . . .' h. The percent Overshoot, settling time, and pea]; time for 92(0- ' ' 150:) 910:) ' 92 (t) ‘ ' ; . ~ "- - lN-m-slrad ‘ ' 3 For the system shown in Figure P4.16, find N1 / N2 so that the settling lime ' for a step torque input is 16 seconds. _ T(t) . ‘ e— _ 1 N-m/rad 1 N-m-s/rad ? ,_._'.— - '. .':‘ I'- ’5 w :. its responSes shown inFigure P43, find the ”WEI ii ' ’ ;.. 10 15 20 Time (seconds) ' (c) x mien 2/0 ewes: syfse‘rsmwwmmlw We” MW I . “(I-l4) Find M and K, shown' 1n the system of Figure P417, to yield zit) with 10% overshoat and 1.0 seconds settling time for a _st_ep input in mom: berque , , Tmm. I! 1 ' Tmft) J=1kg-m2 Forfllemoton J, =1kg-m2 hI-Da‘: INnmslmd R: 19 Kb=1V—"sfrad hkw fi=1N-sfm= L , ' Kt=1leA :K-r I ! MECH 310 Second Hourly Examination ._ ' 30 Mar 01 CLOSED BOOK AND NOTES NAME A railroad car coasts into a massless bumper at 4 mlsec and couples with the bumper without any loss of energy. The bumper position is shown recorded versus time. Suppose the viscous dashpot is removed and the coupling is repeated - - what will be the maximum bumper excursion? snail-l 3“, ,.. NM *5 sol—um”: 1 La) ovuonnpso Wfl'im T a 0" 3a. 314le oz Hm an .3qu 005:7 ..,. Lomc at! 1:55., 70%!) 21(0390 ., , THIS (,3_ mum's—raw wcm MP *5 __ Kama: UNI—L55 THE SYJTE,._, ,, - ‘0 ~ 0 0.05 0.1 0.15 _. w , (5 Hm 02,953,. Time (seconds) H...” Mummy»... M..‘.- MM m; ”Mum,“ ;‘. che. x q Tm -—.-. warm, 4- = 4a FINAI- VALW. (.3, 2-4 ”M‘- 1742 fiML-WLII‘ Twins». 3 .,.. Time (seconds) .5. WM 4‘1 Var 6» WHICUIJMYIC” “‘7" i 49:1 , {~09 can!” Tyuk‘ 1 J“ Hum. anuzzl a”: .- 0 5 10 15 20 2 Time (seconds) ' ' (0.) :0: Y9; AND Tut FVF WILL (we; A“! (.5106: yawn) —-) I ' . V , For the system shown' m Figure P4. 8, a step torque is applied at 61(1‘)Find§ , a. The transfer function, G(s)= 92(s)/ m). i b. The percent overshoot, settling time, and peak time for 62(1‘). 12'0) 910‘) I For the system shown in Figure P416, find N1 / N2 so that the settling time for a step torque input is 16 seconds .- __ _fi____#;-————f”' " Find M and K, shown m the system of Figure P417, to yield x(t) with 10% ., overshoot and 10 seconds settling time for a step mPut 111 motor torque, Tm(t) _____.-____.___‘-_.___.. _.-._- . - TmO‘) For the motor: Ja=1kg—m2 = b‘ ‘0“: lN—ms/rad XO‘) Ra=19 7 ‘-H- .. WEI: . MW. - . 7 . ‘ “m WNW-“W s I = k P = -s m - -_ (Mm TWQdi. - ‘ flit-7mg IF wM‘ -. ' We‘re“. HA 0. DEW SPchF’tD .1116!) m: - Panama; 7-7- ff] ”aim. I" ‘com’b "7o 05 -' "V 9 ”1’8th (0“ =' 7—; % W '9 , n: UC5’) ' 51.2 mm M+l 1, _ “/9? a? 591, 4 , 9999 .; (959(M-l-pei) K 7, ~ "*9? W = ”2“ 5:2- Kc va‘fl’x’ L23 0 510.52. we 94’»; Wu; :5 auto, mu 1mm Has no, qu 4w :1 ~chz, E+1jw+ng§iaic cmbo MD If: Mnmohmfim Jo firm—F yH) Be“ $90,, f .Sfin (Wt-r0 é" = Be: flail-.835) .rux(w+,_+{) DW'DWP Tm, 1“” aw By we 93" v. A- gififluwww) M‘Bw d , fl , B".- ”In“. +§iJhlv6+9 =2, . 3-- 8’ “WW“ (‘33?) AND, {wag 49st 3= 6” VF??? -—>§- 0.059 Shunt w“: Lg“ L%E=,fl.fl 311 Rulfw -> b= 3.0539621Qwéflw~ £146 4) A railroad car coasts into a massless bumper .11 4 m/sec and couples with the bumper without any loss of energy. The bumper position is shown recorded versus time. Suppose the viscous dashpot is removed and the coupling is repeated - - what will be the maximum bumper excursion? H ' o 0.1 a: 03 0.4 05 as 0.7 as ..., . ‘ , 7' . - fw “1' ti econds , t 7 .. ,, . amt-ammo: x '= A + B c J... W‘W __ _ che Fm; Mot-UL u x (5 Jam) nun-o ._ mm Mama» at Tm. (Mm-t an _ PSKMI): eve—:09: 045cc. = mew-=51!- 00 g Howwm.‘w£.TP.,K “Elam f—o _ 71413 n gr-l‘ A $72! Runner... ---Wm°.N-‘rz run-r _ (3&(Wt'f015- ' 9 £; '08’019) '1'? ...
View Full Document

{[ snackBarMessage ]}