ME80 - Homework 4 Soln

ME80 - Homework 4 Soln - NES’O 2.007 W ii ‘f Jag/«Hon...

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Unformatted text preview: NES’O 2.007 W ii ‘f Jag/«Hon W {afi/(m I Look a! (4 7Q? 5001), 07F “I4Q fgr'o/ gag/y Conn-{fig 0% 44! “MK/{27 mo/ “40/ 4’46. M72— por‘fl’f‘ 9”?an I . ‘05 [-35:66 -* F{a‘)'l3Co.S'9 m In I/ ‘ ' ‘ qkf a 443/6 VF¢XI®QI704 {/Mt‘an'ze «(at/IL 9 =0) .' 51‘!) 92.“ 9 C319”! _ ‘- I) : ‘- 1 2 I ft“!!! 4’45”!- flf 5‘ fldihf ”?¢/f/ 0431.2,” t‘fifr‘fizl aloof fi/Oza% {‘r 3;: ML: ah") 7C0”? gfomcflryj :: - 353M? 4-: ~L39 (4’1/04” 43/24). I 2. , ” ‘ij 9: " F’X qu//a:J/dl) flJJJl‘W‘J} “11-3 (“:34 = «LIVE-ix _ 3 ' a 2 l at: (. L 75) Wig-113 x) _ Fang o- L ’ _ 2-. ml.3 x +0 a ’X +(kL: TmJLs)’X * L3 56") £21m” org/2.; 804/9,” I {coqfib/M) (CL) “12¢ MM?» ,chfizn (omr/ 7‘4? Lap/fl? 712,154? +w§x 1 V mt} 5" X + CZ:$X maf‘x =th 0(5) = 35-" : L31 F 3 CLJS f'kLIi MJL3 “gr G4<l (L) To Natale um I. IMF!) 3 >> m: k :(ftoob >7L-1=,w'=,. ~_ _ 00‘s.) 007/ L3‘O;IJ3= $1.3! A 77 mm: LGzz; j I” >> ‘6: +r(flamla(pal)j 3) {nipu/jf(g) [we a¢hc4ed {4%) ?/'D$/€m l _.._-—-—'—‘ (a) fee f/“r‘n'flO-Jf OI m—fi/e I 1/90 7Cbr'1 Cl") “1420f V‘afl/r/ word" (-14% M man), om o/ofi'am 11w): — N I kp -—5000 7;;— Nolr 01:47,” {’9' janjf H a £f(e frofol‘hbfla/ Ki 2 10000 +0 Jar/(«(me 079/; hence N/m. k; 3:1?! 4 flee/verbal 40 44c 47>»: a‘méjm/ OF AKme (rm/‘1 VIE/Ire - Impulse Response Time (sec) ‘/( {'0 It‘d: u/ak’ /O , (a) m; gyro/M6 W; ’9: WI 10/5/07 2:00 PM Q:\rwhite07\CLASSES\ME 80\Fall 2007\HW4\prob2.m l of l %First, define some constants; 1:40; W713 k=4000; g29.8; c:1000: L1:0.05; L2:0.07; L3=O.1: num=l; %Here is the numerator polynomial for G(s) den:[m c*(L2/L3)“2 k*(Ll/L3)”2+m*g/L3]; %Here is the denomenator polynomial for G(s) Gatf(num,den); %Now we have defined our plant transfer function aPI Controller: KF&+}(F Kp:5000; %Here is the proportional gain or _rruzi~ I.) Ki220000; %Here is the integral gain GC=tf([Kp Ki],[l 0]); %This defines the PI controller transfer function %Close loop step response: Gcl=feedback(series(Gc,G),1); %Now we have built the closed loop transfer function proL/Fm 5" (5) ‘_______.—-—I-- Cluck! [00/ W/D/i/C . ' N ' 5 102300 N Wr‘MI Kp' $000775 “1 A5 9.10) (Xéde 1’0 72K SlepResponse af— lJt’( M M J . g a : 3‘] ! %’/- : «VIC Cfm‘i‘ d iy/«Mvd' 3‘3. § 9... Amplitude {15% 1 é a i 1 072 Time (sec) I New (“Hog-S I L'—-»— Vow} = <5 V I 9x + I—— (5 : I VL QC: +1 S0 Vow} : \h____________4 V.‘,, [(R,+4;)LC5:+ 2, ac; +L5 + 1%] (1365+!) flog/(m 3 ((00"') if (L) Final Valve ‘l'ét'orem 5.7} 444. Jim s-Gb)-V,a,(s) = la'r», 15.1fm) 5-90 ¥~>ao go {II/’7 S' 3 If”? fag—Hf At) 'kr a 042.713.”? JA/v}; .(->o 1—300 a 1—)00 5-30 g“? {v meek” far... a [pa/7- bC v-o/fye u)” Jr'f’ MC 2.0Jvc‘1r0/ qjfi 4 Igor] 'A‘ijbvag/j' a" coma! wiilf/ow Maya mafia» (1+ 13C. 0an 1M? (2,94Q56/ L~I|// Mam), 94.36.3k/J (C) >> RI: locoj' £23510“); C = lle)’ L: /e_3/. >7 nam: ELika‘lz-l L 0])- ?7 dm:(onv(]:(fll+22)iL-isc RMQUKC+L @1133.“ 77 G : +£(num/ 59,7); 1 >7 54:7; (G) (W 41446" 0/ //°0 Step Response flak/€41 3 far} (Q 0.25 _ 3 O A ‘3‘? Ac. *\0>\ >¥ €235 0.8 0.6 Time (sec) 0.4 02 T Jam; 5, 67; .S 5. 0:6X/3' n.— h— Now/l UJ" C.) - 1‘0 Cflhfba-‘k Vi) Va, ‘4? w") 00/7 ‘4'0/ “9} ' Maftzcagl do 44¢ abféfa. fag ch/ (aJ~/nc{{/) Problem 4 Part A Given (Vin — V1) + (V2 ' V1) + (Vout ’ VI) =0 R4 R3 1 C2-s (V1 ’ V2) _ V2) =0 R3 l Cl-s 7V3 Vout _ V3 — + = 0 R1 R2 Find(vl , V2 , V3 , Vout) —’ RI'Vin + CI'RI'R3'Vin'S R] + + — + RI 'Vin 2 R1 "Vin 2 R1 'Vin + RZ'Vin 2 R1 + + 'RI’R4‘S — + Problem 4 part 0: >> R1=2.7Se3;R2:2Se3;R3=2.27e3;R4=22.7e3;C1=1&:-9;C2:le-10; >> num=R1+R2; >> den=[C1*C2*R1*R3*R4 C1*R1*R4-C2*R2*R4+C1*R1*R3 R1] >> G=tf(num,den) >> step(G) Step Response Time (sec) ...
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ME80 - Homework 4 Soln - NES’O 2.007 W ii ‘f Jag/«Hon...

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