ME410_HWSoln7

# ME410_HWSoln7 - I 511(5 9(“3 ‘ ssh—libs Em ‘ 31130...

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Unformatted text preview: _ . I _ . 511(5) 9 (“3) ‘ ssh—libs Em ‘ 31130 1+G(s) where (3(5) 2 , 3(5 + 38)(s‘ + 25 + 28) 37 For step. e = 0. For 37tu(t) . R(S) = —,_ . Thus. e = 6.0?5x10'2. For parabolic input. 5_ 13(33): 'I. 2. a. F1‘0111t11e ﬁgure {7‘55 2 1‘55 — (‘55 = 5. — 3 = 2 I). Since the system is linear. and because the original input was J‘O‘) = 2.5fi.‘(f) . the new steady 3' state elror is 655 = - = 0.8. 2.5. 3. . . s R(s) e 7.:- = lnnsEs = Inn ‘ () S—>0 () s—>01+CT{S) . 5(603’33) a 213%” 20(s +3](s+ 4)(s + 8) 2 09375 §@+DB+U) 6. E( ) RC5) DIS1 5 = =* 1+ G65) 1+ 200(5 + 2X5 + 5)(S + 7X5 + 9) 515+ 3)(s +10)(s +15) Thus. "I" — 1' 'E ' ——72 —02§71 ‘1' '9 — 511? (*5) — (200)(2)(5)(7)(9) ‘ ' * (3)(10}(15) 8. 1:. 1000 13 35 T2 ace) 2 a v : K = M = 24.78 . Therefore. e(=:C-} = 0.582. 1 +12” p (61)(73)(87) 13. 5 s§s+lggs+21 5 a. Gag): — 1+ﬂ — \$3 +3132+ +7s+15 s(s+])(s+2) Therefore. KP = 133'. Key = G: and Ka = O. ... 1+Kp b. F01'5011(t). 611:3?) = = 3?.5: For 50mm. 61:93} = 93'. F01‘50t3u(t). €{r'eC-j = 1:. Type 0 16. The transfer function from connnand input to error signal can be found using Mason‘s rule or any other method: 20 G 13(3) _ _ 5(5 W 3) 3 _ .s-{s- + 3) — 2062 RB) 1+ 20 a GI 3(5' —: 3} + 2061 SL5 —— 3) Letting RU) : — and by the final value theorem: 5. G, . 955 = Lfm.sE(.9) : _ Hm :(Si 5—)0 5&0 (71 a.If G1 is type 0. it is required that (32 (3) = 0 b.If G1 is type 1. it is required that G2 (5') must be type 0 c. If G1 is type 2. it is required that 621111153th type 1 10 .y 7,, a. e(r.o)=& = 0.01: where K‘. = =10.Thus. K = 685.71. Kv 5x8x12 b.Kv=10 e. The minimum error will occur for the maximum gain before instability. Using the Routh—Hum'itz 310+?) _ 34+3553 +19612 +(480+K}.;+7K' Criterion along with IL?) — s4 1 196 71: For Stability s3 25 480+K a? 44204: 1751: K 4420 51 —K3—435K+2121600 .1690.2<:K«-: 1255.2 5" 1.751; K > 0 , . . , , 7K , Thus. for stabﬂity and 11111111111011 elror K = 1255.2. Thus. iv 2 — I 18.3 and 5x8x12 10 10 - - docs): r = a = 0.0033. 1K1, 18.3 22. 10 1 30K , , a. 6033) =I-— = — . But. KV = = 60.000. Hence. K = 10.000. For inute error for a ramp av 6000 input. 11 = 1. . . 10000 :3 3 50 b. K1,=11111 (3(3) = 11111 2 ac 55200 SEED ' . . 10000 33+3 +3.0 K\.=11111 sG(s) =11m 3% = 60.000 s®0 SQJO ' . . . .100001 33+3 +30} Ka=11111 53G(s) = 11111 s-’ SSH-J s = 0 SQJO s®0 a. For 10% overshoot q = 0.456. Also. RV = 1000 =: . Since Ks} = . 2:03:11 = a. and t s as can = V? . Hence. a = 0.912. .Solviug fora and K. K = 831.?44.a11d a = 831.744. : .' h K = 0.01. Thus. K‘- = 100 =— . Since T15) = x. a 1]. For 10% overshoot. -: = 0.591. Also. — . s +as+Ei Egan = a. and (on = \fI—C .He11ce.a =1.181‘\IE . Solving fora and K. K =1397’1aud a =139.71. Progranu numgl=[i T];deng1=poly([0 —4 —8 —12]); 'Glis)=' Gi=tftnumgl,dengl) numg2=5*poly([—9 —13]);deng2=poly{[—lﬂ —32 —64]); 'GE(5)=' GC=tftnumg2,deng2) numhl=10;denhl=1: TH1{s)=' Hi=tftnumhl,denhl) numh2=1:denh2=[1 3}; 'H2(s)=' HC=tffnumh2,denh3) %Close loop with H1 and form G3 'GS(s)=GZ(s)/{1+G2Es}H1{s]’ G3=feedbackEGZ,HlJ %Form Gé=GlG3 'G4(s)=Gl(s)GB[s)' G4=series(G1,GS] %Form Ge=Géf1+G4HZ 'Gets)=34(s)/{l+Gd[s)H2{s})' Ge=feedbacktGé,E2) %Form T{s}=Ge{s]f[l+Ge(sJJ to test stability 'Tis)=Ge[s}/(1+Ge(s}}' T=fesdback{Ge,i] 'Poles of TIs)’ pole{TJ %Computer response shows that system is stable. Now find error specs. Kp=dcgain(Ge] 'sGe{s}=' sGe=tf([1 0],l)*5e; 'sGe{s}' sGe=minrealtsGe1 Kv=dcgain(sGe} 's"2<3e(s)=T sEGe=tf{{1 01,1}TsGe: 's“ZGe{s)' siGe=minrea1(sZGe) Ka=dcgain(sZGej essstep=30ft1+Kp} essramp=30lKv essparabola=60£Ka Computer response: Transfer function: ans 2 Transfer function: 5 S”: + 118 S + 595 5A3 + 186 5‘3 + 3888 S + 28488 Transfer function: G3 (s)=-S: (s) / (1+6: ts)H1{s} Transfer function: 5 S": + 118 S + 585 S“3 + 156 5‘3 + 4188 S + 26338 Transfer function: 5 5‘3 + 145 S“: + 1355 5 + 4895 S“? + 188 5‘6 + 8828 5A5 + 152?62 5A4 + 1.4156886 5A3 + 6.2126886 SAZ + 1.811688? 3 ans = Ge(s)=G4(s)f{1+Gé(s)H2{s]) Transfer function: 5 5A4 + 168 5A3 + 1798 SAE + 8168 S + 12285 588 + 183 5‘7 + 8568 5A6 + 176846 5A5 + 1.8738886 5A4 + 1.8468807 5‘3 + 2.8758807 S"2 + 3.8338807 S + 4095 ans = T(s)=Ge(s)f(1+Ge{SJ) Transfer function: 5 5A4 + 168 5A3 + 1798 SAZ + 8168 S + 12285 SAB + 183 5‘7 + 8568 5A6 + 176846 5A5 + 1.8736886 5A4 + 1.8468807 S"3 + 2.8758807 S"Z + 3.8348807 S + 16388 ans = Poles of T(s) —21.3495 —1:.0001 —9.8847 —7.9999 —4.0000 2.9994 —D.ODGS ans = 5Ge{sl= ans 2 sGe{s] Transfer func:ion: 5 5A5 + 160 5A4 + 1790 s”3 + 8160 s“: + 1.229e00é 5 s“8 + 183 s”? + 8568 SAE + 1.?68e005 5A5 + 1.8?3e006 5A4 + 1.046e00? s"3 + 2.8?5e00? s"2 + 3.033e00? s + é095 RV = s“SGe(s) Transfer function: 5 S“E + 160 S"5 + 1799 S“4 + 3150 S“3 " i.229€004 S”: 588 + 183 S”? + 8568 586 + 1.?686085 585 " 1.8?36086 584 + 1.0466007 S"3 + 2.8?5e00? S"Z + 3.033e00? S + €095 Ka = ESSSZED = E.— '1“- ? I ‘JLIIJIUI Warning: Divide by ze_:::. [Type "warning off MATLABzdivideByZ-ero" t::: suppress :his warning.) > In :J:\My :3:;cument5\c::+ntrol Systems: Engineering Bonk—MESS 4th ed‘xSolutions Manual'\CIiap 7’ Refer“ences\p'?_32.:r. at. line 42] essramp = Inf Warning: Divide by zern. (Type "warning off FﬁTLABzdivideByZero“ to suppress this warning.) :> In 3-:"\.My ::;:.~cument5:\C:::ntrol Systems Engineering 30:::k\.¢383 4th ecl\Solutions Manual‘xchap " References\p?_32.m at line 41 essparabola = Inf 39. Error due only to disnu‘hance: Rearranging the block diagram to ShOV.‘ D05) as the input. Therefore. __ s(s+4) 31305—3) '43) ‘ DC“) K1K2(s+2) ‘ D“) s(s—3)(s+4) + K1K3(s+2) + s(s+3)(s—4) '301‘D(s) =é .eD(:x:-) = lim 523(3) = - 2:1 SCZ‘O Trror due 0111 Fto in m e = 1 = 6 " 3 p ' R Ki.- K1K2 3le2 - 6 Design: 3 613(0)) = - 0.000012 = - 2K1 . 01‘K1=135.000. eR(-:c-) = 0.003 = .or K2 = 0.016 6 Bile ...
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ME410_HWSoln7 - I 511(5 9(“3 ‘ ssh—libs Em ‘ 31130...

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