ME80 - Homework 3 Soln

ME80 - Homework 3 Soln - HEXO 4-1007 Aim/#3 504,475,) A T...

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Unformatted text preview: HEXO 4-1007 Aim/#3 504,475,) A T 'T 6:44 fit Vie/45W ‘h‘a 3%? meé/é’m l / ‘V/hpmfixre = T 0"”. fp— a|~ LfimeM-flf :mc (a) RfiCa” £17m ’IYEMIIG‘ 'n 017“ MC __ — _ r P ’ an C20“! 012‘ me? if 6%,, “ 147- 50 L670 {ace Wig/“n1 Mei—(5) + 14775) c at), (s) w; M 28/0 I‘m 0‘ C¢O¢/J"J7(3nj .' 50 (3(5): :12 __ I OMS) ‘ I’MFSJ-l, I: 1': () Mat)“: Ki «(rm + k6, iii/flaw» L6? /QC(’ Mfg/m 2&0 2‘0 rhhté/ (0r) .' st>= ’2 L WM 50 Maw 2667 Him-tr? so/m‘m gob-aw) If fiégém 2 (a) If 52(31): I - ‘L (Ly/(«Q 74%)74077 071‘ <4»er Luge) -t ,r 77*): OZ”— f'm): =02 (mgwsg fw‘fa/ 4%«5‘0’1 exfwfofl: 4;, z ,4 is (mcps+A)S mg”) + J 1 A5 + 507925”) emwe 4% Pas/P i=0 33""3: I: 317 => LET 4‘ 1, "06 ll Pdvawk 4+ fwk 5:: - _f ‘mco ‘ TVS Ta): 4 m95+h S -l __l/A : of M’L 75 ij—):// U(S):_§l__ Em flag/m, [(4) X8): 6(5) (/6) = (kF+K0/5+_If£),4 S 5 _ l< $0 X(5)" {+Ka,+ {2 wk” 50);) %e ole/Ja—fiflcfibn. Pnai/em 3 ‘_____—.. (a) 101' confine/{tr fC’J/owf Jo a um‘f ramp (>1 error .' f.— wok-1k? + f3 .1135) I 6(5) : £§15§= Z;- ..L {I 53 k 1 -I 1" 44m; A”): x {Um kpt*;}’t 50 L18) : :16 + (Gabi/c" +0 4 am“!— amf )4 {’F/o/ .’ “(9" {'8“(kr+ke§s>';‘a§= z" ‘3}? 44m VHF kft + 5,7; l “p .7} 3“ 1 B-w-3—1 . W ,,,,w__________ W“——_m__—— 3—3—5- when D(s) is zero, the closed—loop transfer function CR(s)/R(s) is cm) _ 62 (r) w!) 2m 2+ 4:0“) 6;. (I) When Ris) = 0, the closed—loop transfer function CD(s)/D(s} is C3M5) n / ms) ‘ 1+ 6d!) (us) When both the reference input and disturbance input are present, the output C(s) is the sum of CR(S) and CD(5). Hence ((5):: CR {5) +— Cp(;) 2.: Mg m6 3) [66(5) Gina) kg) «4- 19(5)] c f .__......_._.______...M,._._______._.__.______..____....._._.-.__..__..—.._——.—___...-m..———.—__—————_.....-—.—.——- 9—3—7- When NS} = 0, the block diagram of the system can be simplified as follows: The closeddoop transfer function CR(s)/R(S) can be given by 6: (if (73 C71 CRIS) __ (+5, (51 #1 fl. (5:: 5t 52 6; W 1+ _____.._‘?c ‘9'62 ‘5: #2 2+ 6,6: m + 5m, (72 4mg "1" 51 61 HI when Me) = 0. the block diagram of the system shown in Figure 3—76 can be modified as follows: Hence CD 6’ 63 (51 C73 1i ...
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ME80 - Homework 3 Soln - HEXO 4-1007 Aim/#3 504,475,) A T...

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