Hw_3_w06_soln - l—tH—El—Elfilfib 1 l i 3 lYllzL'H 3 HtHLI IzNLillelelNLi b'l’j titi4 blfi‘jlfi l—‘[£11 L’i‘ll

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Unformatted text preview: l—tH—El—Elfilfib 1 l i 3':- lYllzL'H 3: HtHLI IzNLillelelNLi b'l’j titi4 blfi‘jlfi l—‘ . [£11 L’i‘ll Homewurit‘Set #3: MAE 3600 + System Dynamics, Winter 2006 Due date: Friday, February 17, 2006 42 Pgoblem 1. A~stensiitive instrument platform for microgravity experiments cribon the Space Shuttle is mounted on shock mounts in order to reduce vibrational motion. The system (shovm belotN) can be modeled as a simple mass-spring-damper system. The mass is m = 0.8 kg, total stiffness for the shock mount is k = 230 film, and‘to‘tal damping coefficient for the mount is b = 4 N-s/m. Vertical displacement a is measured from the static equilibrium position. The various system inputs are base motion vibration (so) from pmnps, fans, and astronaut motion; disturbance force (Fain) from electrical umbilical cords or motors mounted on the payload; and an active control force (Fm) produced by an electromagnetic actuator. For this problem, assume that the two input forces Fain = Fm = 0. The input base motion is 20(1‘) = 0.004 sin 47:! m (input frequency is 2 Hz). I Base motion, 30(1) J}; a) Derive the transfer function for the instnunent platform system. ‘ ' b) Simulate ‘thqfiynamic response using Simulink. Assume that the system is L, initially at rest! Hand in a copy of your Simulinlc model and plot the response 20) in mm. Describe the dynamic response of the platform with the sinusoidal input {base motion. {0 c) Repeat part b3; but now use the Matlab M-file LSIMM to computetthe dynamic response. Hand .in a copy of your Mfile, and plot the response 2,0) in mm. (1) Now suppose that the vibration of the input base motion is variable. Run nine f separate trials” with input frequencies f = 0.1, 0.5, 1, 1.5, 2.7, 5, 10, 20, and 50 Hz ' and record the m of the amplitudes of the output motion to the, input (base) motion at steady-state. Plot the amplitude ratios vs. input frequency on a log-log scale (logibaiie .10). Comment on the general trend that you observe from the amplitude ratiio- plot (note responses to both low and high frequency input). htH—El—EUUE 5T5 HH4 DUHU 11:2e NtEH & HtHU tNUthtHlNU Problem 2. A flywheelwith moment of inertia J = 0.5 kg-mz is initially spinning at 320 rpm with zero arigplar acceleration. At time I = 0, a brake (modeled by viscous friction coefficient 29 = 0.04 Nnrn-s/rad) is applied directly to the flywheel. [Z a) Obtain the. analytical solution for the dynamic response of the angular ____,_._.--' position, 60‘) ,ru'sing Laplace methods. Assume r9 = O at r = 0. Sket’ch’the angular position response. Compute the total number of revolutions of the flywheel after time r= 0. I 4 b) Simulatetlieidynamic response using Simulink. Use your simulation to compute the rate of eneggy change (power) and total energy of the system“ Hand in a copy of yourSimrilink model and plotthe angular position, angular velocity, power loss, andtotal energy vs. time. Problem 3. In» many; aerospace applications, a “notch” filter is used to separate vibration “noise” from real input signals (such as gyroscope measurements). In'the’“old days,” notch filters were constructed from an electrical network composed~of «operational amplifiers, resistors, and capacitors. Today, notch filters are digital filters that reside in the flight control softivare. In either case, the transfer function of a unity-gain notch filter is show below:’ 32 +62): _ Y(s) .92 +1253+wj [1(3) "(0(3) = where 140‘) is the inpit’t'énoisy) raw measurement signal from a gyroscope (in degis) and ya) is the output of the filter, or “filtered” gyroscope measurement (alsu‘in deg/s). This particular filter, is “tuned” to frequency a), = 10027 rad/s (or, 50 Hz), which represents the expected “bending mode” or rigid—body vibration of the vehicle frame. Suppose the gyroscope is sensing a steady 1 cler rotation plus a harmonicrotation rate (which may be real vehicle motion, or may be vibration of the vehicle frame). Using Simulink, obtain the, filter output for three input test signals: i) 110‘) = 1+ 0.5 sin 2: , ii) 1.10) =1 + 0.5sin 201", and iii) n(!) = 1 + 0.5 sin 1 00m (all inputs are in deg/s). Show a copy of \‘younQiSimulink model, and plot the filter input 1:0“) and- filter response y(z‘) on the same plotfor all three input signals. Set the simulation time this; sec for signals, i) and ii); andpse 0.5 sec simulation time for signal iii). Describe theateady-state responses for all three input signals. Can you explain why this filter is called a “notch” filter? a E. (Simona. + lifters ) l2. (disca.55r‘on~ WSW“) F. ME I-tH-El-Elfllflb 1 l : Eb _ MtLl'H E: HtHLI IzNLillelelNLi b'r’d 55-4 521521 |-‘ . 1:15 wag “ 'M/ng. 3m: ("3 Pm 11" ‘ 5-256”— md'fiMm a", :: 0-0m4-5r'n 71;“? m a) Fan f‘“ w W/ flew ICC, . ‘ MSE‘EZfi) +IB";§‘E(5) + k?(5) = 193298) I'+V‘k"v‘za(5) zax'figrsz + b: + k] : 2390659 k] s - « J 4. i” wimp”? ,IZm) " f k ~ 1” ' i Gar.) -; i -: 45 + 23c: ‘ J '2 n 5' -r- 55 4- 237.5 W, PtH—El—EUUE llifib ‘ NtEH & HtHU tNUthtHlNU - w - ‘ r. ‘. i ij HH4 DUHU F.U4 ..—-‘—.-......._._,d_, - ‘ “mm—L» on’.’__,...,m_,.__,.....‘ I “ )EH), : ’4 R *"7 I 2:. ml C” mm? lH-«JL Fri-1.0;!) I. 0354 /-/563 1.42é3 A 3. 533 3 6* "figghmm “ \ '5" 0-4570 C I? ' 5'. H59 3"?" 0.0445” {zFf/a 0.0162. €- ‘ mm»? Edi » ’09 (AR) Us '_ SE: CD4? ’03 ) — C: M mean": a r) 39E, fingumj, Mg: (aw; ampufimag) ’ Inpupf “V HyifiHfliex' 5' .“- E: ‘4 ‘ 731%”, ‘ Pasonant -F at.“ {M _fitflpicf_ (J‘ V by parrfijr (“c u‘SPeru) (. 2') At Jekffuencfl f: 2-7 HE, '3) ’45 fifluEnca '3’ 5 Hi Ffl-leha‘ h E 0-H: ‘4‘!“ Ar (chm mfg an Iv-/bj fad“ (£19 W‘éfiafibn) Mb P. 5Y5 554 bUHU 1112b FtH—El-EUUb MtCH & HtHU tNUthtHlNU “mumnmut 055 o... mumnmfiog a... Em n; EwEEfiE ommfifmmmd amme . a Eyfifig EHEEEAaBm‘, mg ., .. 9mg 35 FtH—El-Ewwb 10 Mass diapfaoement. 2, mm lliab MtEH & HtHU tNUthtHlNU (w/ Si’MulInk > MAE 3600. W06, Problem 1b ‘ de 554 bUHU P. Uh ' ' ' P.UT FtH-El-EUUb 11:2? MtEH & HtHU ththtHle 5H5 554 bUHU ( ‘ \ p \ y #H 2/l4(96 11:;} AM C:\Doo" nts and Setting s\klUEVero\Dosktop\M Doi‘ MAE 3600 , W06 HW#3, Problem lo o'FIfiPflFn‘fla'F'tF' oonstanto = 0.8; :4; = 236; w U a w % transfer function J ‘ x f V, numG = [ b k ]: denG = [ m b k 1; sysG = tf(numG,denG) % build input (basa‘nntion) dt = 53-3: %; time step tf = 4: %3 final time t = D:dt:tf; ‘%2 time vootor w = 4*pi; ‘ 5 :input froquenoy, rad/s 20 = 0.004*oin(w.*t); % base motion, m a % Simulate response d “w [2,tJ w lsim(syBG,z@,t);"" % plot result figure(1) plot(t,lo3*z) \,' grid ‘ “ ) xlabel('Time, 5') _ ylabel('Maso diaplaoement,‘o, mm') titlo('MAE 3600; wos: Hw#3ppion1em 10') < p” ‘. FtH—El-EUUb 10 Mass diSpIaceme , 2. mm 11:2? MtCH & HtHU tNUthtHlNU (w/ 1.9”“) MAE 3600, W06: HW#3 Problem 10 Time, s 5Y5 554 bUHU P. UH I-tH-El-Elfllflb 1 l : Hr“ MtL'H E: HtHLI IzNLillelelNLi fifij 55-4 521521 |-‘ . HE! 101 : MAE 3600. was: HW#3 Problem 1d 10 ‘% m ‘ . 1: I Z a . . a : : In: I I _ . . . . ‘ . . ‘ . . . . . . ‘ . . . . . ..; . \ . . . . . ‘ . . . . .. 1 J I 1" 10 : , ‘ k k . . ..; g | . ‘ . . l . . . . . ‘ . . . . . ‘ . . , . . . ‘ ..l ' . . ‘ . , . , . . . 1U . . . . .. . :a‘ _l_‘_ Input frequency, Hz ' rd I-tH-El-Elfllfib 11;;3. _‘ MtL'H kg HtHLI IzNLn'llelelNLi Ema 7339,94 i a ’ FEM? "baa +9427: gar, [03 37: ;b.o+s:l @(5) 4-2 . 935,! , . W5.) CD'S 3“ + 0.0%: r] .1; f b'l’j titi4 blfi‘jlfi (0‘ ‘3 J ( 3 5' 97%?“222‘3? * ‘ 535-5.'§§i5£.f 3' 3' -, 21-2973 ‘. 4‘» H - Wig} 4 I-‘. 1E1 5T5 HH4 DUHU F.11 . ..L_-..-- ., M-.. ____..m._~.._... ‘_ . . . é \ i r ‘ . FtH—El—EUUb 11:38 MtEH & HtHU tNUlHEtHlNU PM 2. \ ( aw 61-) SWy'SfiJa : §(m) m M7337" {(‘Q’L M”); 1‘ - ‘ i i 2' ;, « ' I I '1 i I? , yWCP- 34- FEMS: be flare Mac-.513 R‘L‘g (res-r) 1 m if PCS/Paar: __ SE: ,Mdi' ‘ ’ pafa'r‘: i F0 :‘EHV v P +v g Fwfir 3&5 r 6 WE,- ':= C (7.43 ) LBJ-3 ) :3??? = +~(b¢9')éi F- 495': ‘ W...) m (flgwer loss ) 3- 1‘ ‘F ‘Q‘f'CfiV-‘M \* I I; \ [ > 71M emit-33 E = T + L1 : k5, 1»- PE r-w I E : 19'11“ U '-“- 0 (always) I i . g: T -— g“ {hf = g cm" a 4 $ [Up-(u r Eflayljvfiqu E; = 7: 1 Cré'“ HI I I 3 but a = 'Ebs (1%”. mm) i; g _ r: ' FE ‘ — I- Z ‘ F I J E U— 9 ( 3'... a ) 7- E e. 7" ,Piolwér [.955 1 fl 1 E: i "5 .5 t' 5' I; g ‘ 4. ; f? g 13 F. 5T5 HH4 DUHU MtEH & HtHU tNUthtHlNU 11:38 FtH—El—EUUb $30,363 oh 1m. 0 :26: 2“. 5m: ,._ou..flm£ Evflmfi.9 535% ,N 533 $3: 8am EEEEE mi: I-tH-El-Elfllflb 11 : EH ‘ MtL'H E: HtHLI IzNLillelelNLi b'r’d 55-4 521521 |-'. 15 1200 :MAE 3600. W00 HW#3. Problem 2: Theta vs time 1000 800 Angular position. theta, rad m 0 CI 0 10 2o 30 4o 50 60 70 <36 ‘ »90 100 Time,s ‘ FtH—El—EUUb 90 BO 70 Angular velocity, theta-ch rada's (o O 20 10 11:38 MtEH & HtHU tNUthtHlNU i.” x Time, s 5T5 HH4 DUHU 90 F. 14 100 I-tH-El-Elfllflb 11 : EH MI;L'H E: HtHLI IzNLillelelNLi Ib'r'd 55-4 521521 |-'. 1::- 2000 MAE 3600. W06 HW#3. Problem 2: Energy vs time 1300 1600 1400 —.L M 0 CI Total enemy. E, Joules a; ‘ O D 400 200 o 10 20 f ‘ 30 40 50 60 70 do ‘,’* 90 100 i 1 Time, s a I-tH-El-Elfildb 11 : EH MtL'H #5: HtHLI IzNLillelelNLi 5'35 554 3115111 |-'. it: E r Mka 3600, W06 HW#3, Problem 2: Power loss VS time ” 100 —100 -15C| Power Jess. W -200 4250 -300 —350 4000 ' ‘ ‘ 10 20 30 4o 50 50 70 30 ' 90 100 » » Time, s ‘ afi“ FtH—El—EUUb 11:33 MtEH & HtHU tNUthtHlNU Mm!) {fifi(af' 7715, Has (or K. p , 3"L:Hu.lfnlt_ Marga! I K i ~ mva ‘ S 3;“ 94%??? tin-«1 in pug!- MHA “cwicfiéfl : $34.55 50 ) l+‘ 5-;an bflefl 51141.1; 5T5 HH4 DUHU H mm». .y.._.d_.,_,w..,.m‘ ‘ \ .1? firm! Fla-f5 M 1135;» ‘ 5 [Do rug, ) “flack: u exec-Hal ; (9‘33, Lfflw"#€t{kmg j + 0.5: 5;" EM 71‘; ,_ d a brig—E”, Sway-94+ Ff“ ,, I + 0- 5 sin: tad-1&5sz ‘1 Show: {Cg-gng‘f'fiav" 18 F. 5T5 HH4 DUHU MtEH & HtHU tNUthtHlNU 11:33 FtH—El—EUUb Engage}: n.._. u mama BEE . 8890?. E. :35 H 0 m E2321 fig: m8“ L353 3% mi: ii 532 magicazfimmrmm mEPDETNm .5456 99m EE magnumfioa E. m Emu—mace m.me me Hztl—El—Eldldb 1 l : EH MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUHU |-‘ . 1'5 54/15): /+ 0.5% 275 MAE 3600, W06 HW#3, Problem 3: Input signal P ' Filter input and output, degfs o 0.2 0.24 ‘ 0.5 0.3 1 1.2 1.4 ‘115 1 1.3 2 Time,s ' I-Izlzi-El-Elfildb 1 1 : EH MtL'H i5: HtHLI IzNLillelelNLi 5'35 554 3115111 |-‘ . EU 14(19): / + 0.551112% 1 a j ~ -MAE 3600. W06 HW#3. Problem 3: Input signal ii 1.5 Filter input and output. degi's .0 a: 0.6 0.4 ‘ 0 0.2 0.4 0.5 0.8 1 1.2 1.4 1.6" 1.3 2 Time, s Hztl—El—Eldldb 1 l : EH MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUHU |-‘ . 31 .‘~ 0/0) = I + 0.5.501/0000’ g: _ MAE 3600. W06 HW#3, Problem 3: Input signal iii“ “0000 1.5 7‘ O} .-‘ a _I. M _'L F} a: Filter input and output. degfs F3 c:- v H 5 J F’ .p. 0 0.05 ‘0.1' 0.15 0.2 0.25 0.3" 0.05 ‘ 014 “0.45 05 1 ~ Time-,5 ‘ J - E TEITFIL F' . 21 ...
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This note was uploaded on 02/11/2010 for the course MAE 3600 taught by Professor Pai during the Summer '09 term at Missouri (Mizzou).

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Hw_3_w06_soln - l—tH—El—Elfilfib 1 l i 3 lYllzL'H 3 HtHLI IzNLillelelNLi b'l’j titi4 blfi‘jlfi l—‘[£11 L’i‘ll

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