Chapter 11 - - 3 llllfliéiifl'llllfll .

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Unformatted text preview: - 3 llllfliéiifl'llllfll . IIIIIHIJIllllll“Illlllllliiflfllllllllllm .. CHAPTER 11 mm Illumlllmm mmvllmw 1o"- 1 I I I I I I I | F-IllllllI-IIIIIHI-IIIIIIH _ IIImmllumulllmmsll mmIumInmiumnllum 'iiiiiimiiiiiiiiiilIinmuuumlllmm i'!| Illlllllllll!Illlllllllll .JI . ' IIIIIEIIIIII|Iii!IIIIIIIIIIIIIHIIIIIIIIH — IIIIIIIIIIIIIIIIIIIllllllllllllllIllllllll IIIII}IIIIII||fIIfilllllllfllllllllllllllll IIIIIIIIIIllIIJIIII#IHEIIIIIIIIIIIIIIHI 3-11-3. A Bode diagram of the PI con roller is shown in Figure (3). _ a * _———- —---——— ——-—_~ - _ u ' u -——__ _"--—_ _-—-I-——._ ———_..___ ! Ill"!!! ......um.u|mn II!!!” n h a |-—-— h I--————--—. - — * h h —--—___ -—-.__.__ _— I——_ —..._._ _.__.._ ..._.._. — — *- * - - - —' _ ‘- u.— = - _ — — 2": - — _ .— "‘_".. “ “ E m in *- — w" 10' 10‘ 10’ w IlllllllflIIIHHIIIIHHIiiiiifllllllllm - IImmilIllumlwaimIlumullllmu I 7» .... .-as;.:::§!!!!!mHM!!!EH!!! IIIIIWIIIIlmlfIlllWl!!!..[HIIHHW 6' 10‘ 10' ; !!!!l!!!!l!HH{iiiiiiflélllllllllllll Fl ' fl}!fl!!!I!!!llllllfllllflfllflfllfll!!!!!!!! # Inmlnu uuwllm 1 F B—II—fi. The equatim of Mid: for the system is 1:6: «46) = 1:19 lé+—%—19=i The .6. transform of this equatim, using zero initial auditions, gives- ' (I8 4' -E-*I)O(a} I 3X03) LBJ—3L a x“) 1 2+ (m) Notice that this systun is a diner-matting system. Sins: =M=J~__‘LL— 30*” x Jeri-(Vb) “obtain .L W IGUa-fl' “ 1 W {€4ij = 90“ - tan"1 .529 x The steady-state output Gnu-J is therefore 91m by and a (1:): -1- “’x 91mm 1:. + 90’- tan-1 -——‘”b) 83 . ’ J(—*—)"+w* " b - Nam. substituting 1 = 0.1 m. x a 2 N/m. and b =- 0-2 N—efm into thw) gives - GU60} -- 10 j“) - 190+ 10 A Bode diagram 01' GUM) is shown an next: page. GUOU) = Bod. Ulnar-m -——-. "unu- ‘ lllllllllllllllllllll‘n. llllllllllllIllllllllllll Illlllllllllllllllllll ‘ Fm (Induce) - _ _ — = = - - —' = E I Ii!!!!lfl!lll|ll|llllIIIIHIIIIIIIHIIIIIIIII IIIIIllllli!!!!!!!!!!!!!!llllllllllllllllllll III|llllllllllllllllllllllllIlliil!!!l!!|llll IllllllllllIlllllIIIllllllllllllllllllllfllfl I'Junllln...‘l" Illllll,.!.llllllIllllll‘Infill!!! 1o" 10" 10° 1 ‘ 1o2 0 ‘ — _ * _— _.__..._... .___ u - _ H — _— _._— _._. _ _ _ m _— _— _— _ — _ — .— _.__— .___...— __..—__-_.._ .___—_.- pH—u—— ——_—__—__.._ _—....-.-_._.__..---___-______. !Ill!!!H!!!||l||llllll|l|lIIIIIIIIIIIIIIIIII IIllllIllIniflifl!!!IIIIHIIIIIIIIIIIIIIIIII Illllllllllllllllllllllllllilllllflllllllllll _ IIIIIIHIIIIIllllllllllllllIIIIIIInE!Illllll IlllllmlllllllIIIIIIIIIHIIlllllllIIlII llllllllllllllllllllllllll .. _ * -___ _— _— _— uh == _— _— ‘— _.__— _.__...— _.__.. Illi'llflllfllllllllll to“ 10’ 3-11-9. A possible mm program to obtain a Bode diagram Of the given GEE) is morn below. The resulting Bode diagram is also shown below. 13-11-10. A possible HMIAB program for obtaining a Nyquiet plot of the given Gts} is shown below. Note that. to plot thw) locus only for a; ) 0, we use the following mad: [redmw] = nyquieflmmfiemw); plucked!!!) The resulting Nyquist plot is sin-an below the mm program. nun [O O 0 1] .' den - [1 0.8 1 0]; w—= O.1:O.1:100; [re,im,w] = nyquist{num,den,w); 'plotire,im) ' v = [—3 3 -4 2],- axistv) ; axisf'e'qual') grid titlet'Nyquist Plot'] xlahe1('Rea1 Axis') ylabelt'lmaginary Axis') -Since nine of the Open—loop poles lie in the right—half a plane and the thw)1ocus encircles the -1 + :10 point. twice clodcvise 1f GU60) locus is plotted Erma): «do 1104):“, the closed-loop system is unstable. - 3-11-11. A possible new program for obtaining a Moist plot. out the given 9(a) is shown below. The resulting Nyquist plot is also shown. balm. since none of the Glam-loop poles lie in the right-bait 3 plane and tron the Nyquiet plot it can be seen that the GUw) locus does not encircle the -1 + )0 point:' the mm is stable. num = [0 0 0 20 20]: den 8 [1 7 20 50 0}; w = 0.1:0.1:100; [re,1m,w] = nyquist{num,den,w); plottre,im} v = [—1.5 1.5 -2.5 0.5]; axistv]; axist'equal') grid ; ' titlet‘flyquist Plot') xlabelt'Real Axis') ylabelt'Imaginary Axis'} :3 415 5.1 9 .E 5 0 05 1 L5 '-2_ —1.5 -1 43. Real Aids 3—11-12. A possible MM‘LAB program for obtaining a Nyquist plot of the given 6(a) is shown on next page.. The resulting Nyquist plot is also shown on next page. Note that there are two open—loop poles in the right—half a plane, because s3+O.252-I-s+1 = (s + 0.7246)“ — 0.2623 +' 51.1451)“: - 0.2623 - 31.1451) num = [0 1 2 1]) >) den = [1 0.2 1 1]; >> w =0:0.005:10; >> [re.im:WJ - nyquist(num,den,w); >> plotjre, 1111) >> V = [-3 3 -3 3]; axistv}; axisl'equal'j >> grid >> title('Nyquist .Plot'} xlabelt'Real Axis') ylabelt'Imnginary Axis') From the plot. it is seen that the GU och-locus encircles the —1 + :10 point twice as a) is varied tram w: —oo to a): 0 to (a: 00. Referring to the N'yquist stability criterion, we have N=nmheroalclodcuiumc1rclmtoftlud+10pointna P=mnmaro£po1esofc(s)intheright~ln1£splam-2 z=mnbero£zaerosofl'+G'(s)intlieright-mlfaplane =N+P=—2+2?0 Thus, there are no closed-loop poles in the right-half a plane and the closed-100p system is stable. ‘ —-——u—- Bull—13. A closed—loop systan with the following Open-loop We: fmictim G(s)H(sJ = (r1 >0) (1) 32011: +. 1) is unstable. while a closed—loop system with the foucving open-loop transfer function is stable. K('I'23 + 1) 82(1‘15 "’ 1) I G(3)H(B) a (T2 7 T1 ? U) (2) Wu plot; a; the pzjeceding'twc systems are shown below. Note that - omeuw) we: start mi mum infinity on thc run and: (w: 0) autism thori'gin (wanna); The sweatithtlaopen-locp trends-r functicn given by_ Equation (1} encircles the -1 + 10 point. twice clockwise. The when is unstable. The system with oped-1009 transfer flu-action given by mum-(2) does not encircle the -1 +10 point. Hence. this m1: stable. K Unstabl 520115 + 1) ( a) etch-1(8) = 3(3)}.“3’ = 52(1'13 + 1) (T2>'I‘1> o) —— ————-—n——-——— ——-— n- —--—--——u—-——_——— —_ 3—11-14. For this system (2-5” 8060} =K——— ja)+1. By setting K = 1. we draw a Nyquist diagram as shown on next page. Note that ( e-Jw = -w (rad) = - 57.3%) The Nyquist locus crosses the ncgative real axis at 0': 43.442. Hence for stability, we require 1 0.442 >K>O or 2.262) K 70 Since .jw - Gm» 3w)” (1+.1'wM1—jg) K . . a 1+w2 [(MM'NBHWJ*j(sinaJ-ra)mw)] by setting the imaginary part of thw) equal to zero, we obtain sinw-I-wcosal = 0 w=-—tanw Solving this equation for the manage. positive value of an we obtain Cd 2 2.029 substituting w = 2.029 into thw) yields K (cos 2.029 - 2.029 sin 2.029) 1 + 2.02292 G(j2.029) = = -0.4421 K Tin critical value of K fur stability can be obtained by letting G(j2.029) = —1: 01‘ ' 0.4421 R = 1_ Thus. the range of gain K for stability is 2.262 > K > O Ell-'15. 6(8) = K I '= 0.251! _ Ms2 4- s + 4) 303.2552 1- 0.258 + 1) 'msquadraticteminthsdencminatorhasthsmdupsdmtunl frequency 01' 2 rad/s the damping ratio of 0.25. Define the {W corres- Ind pcnding to the angle of 430‘ to' be (01. {00.01) = - [jw - [1 - 0.25.012 + 10.254211 2 —90. — tan‘l &_= -130‘ 1 - 0.250212 Solving this last aquatic: for all, us find a); = 1.491. um. 1:11: phase angle beauties 8:11.31 to -130' at 4)- 1.491 rad/s. At this fre— quency. the mgnitude must be unity. 0:160:91)! = 1. The required gain 3‘. can be from [001.491)’ c 0'25“ 20.2090: = 1 (ji.491)(-0.555 +10.3725 + 1) from uhichse get K=3.46 that tbsphasa crossover frequency is aha): 2 rad/s. since Zcqz -» - (52 - (-0.25 x 22 + 0.25 x 12 + '1 = -90‘- 90": 490' 'me magnitude [602)] with K s 3.46 bananas ' 0.865 {j2)(-1 + 50.5 + 1) 10(12)[= = 0.865 a 4.25 an Thus. the gain margin is 1.26 dB. The Bode diagram of GUw) with k = 3.45 is shown on next page. MIQI‘IItude (as) _ .IIIIIIIIflllllllll-IIIIIIII _ .llllllllHIIIIllll-IIIIIIII IIIIlllll-llllllll-IIIIIIII .IIIIIIll-IE!!!!I!-Illllll| 10° 1 1o -n—p—m——_——-———uflumu——n—H wumflh——————.——--—--—_—— 13—11-16 . Not. that K jun 0.1 - 10 muogwa- 1) jw+0.5 JWUW+1)-ja’(2ja}+-1)(jé’+1) malnllplotflnnodediagrmmeK-l. 'I‘hatispwplottheaode diagram of =.—_J'OJ£+_1___.__ 140 (2:1 w + 1M1 ah 1) and. m .IIIIIIIl-IIIII .IIIIIIII -""lllllll-Illl"" ---.-Ilu—__ _--..II .Illlllll-llllllllh.'l'lll" .IIIlllll-IIIIIIll-i"llll .IIIIllll-llllllll-Iiiill" ___..-ullll-II.."!"-III I'll. ‘ 1 is —J.3D' at w= 1.438 rad/s. Since we requi 50'. the magnitude of (301.433) must be equal Bode diagram indicates thatIGULdaaN is 5.43 —5.43 dB. or K = 0.266 Since the phase elm lies above the 480' line for allto' . the gain margin is +0063. ' 3—11-17. Let: us' use the following lead cmpansawr: Ta+1 8++ 'Kc «Ts-+1 9+ Gem) = Kc“ . or T Since K... is apacifiad as 4.0 5-1, we haw.- - + chlimsfico: T6 1 K - =chxu'4 _ 5-”. «Ts-+1 s(0.13+1)(s+1) . Iat-fissetk=1anddeflnexco(=fi. mm _ . A x24 Next, plot a Bade diagram of I 4 _ 4 s(o.1s + 1H: + 1) 0.193 + 1.152 + s The following mum produces the Bode dingmshwn an page. >>num=l0 0 >> den = [0.1 1.1 >> bode (nun, den) 3 >> ggid >> titlet'Bode Diagram of Gta} = 4/[s{0.15 + 1}(5 + 1)]') Fran this plot. the phase and gun mm; are 17'and 3.7 dB. respectively. Since the specification-Is can for a. plane margin of 45’, let. us choose I’m: 45. - 17"? 12'2400 (This mans that 12' has been added to mate for the shift. in the gain crossover Hemmer.) me mximn phau lead is 40‘. since Bode Diagram of 6(3) 2 49MB.“ +1)(s 4 1)] lllflfl!!!IIIIIIIIIIIIIIIHIIIIIIIHIIIIIIIH .IIllllIl-=l!!...._ IIIIIllllllllllIllIIIlilflllllllllflllllllm IIIIIIIllIIIlllflllIIIIIIHIIIlill!!!!llllm IIIlllI".IIIIIIHIIllllIHIIIIIIHIIIEI!!!H _ IlliIlliiiii!!!l|llllll|llllIlllllHIllllllH - IIIIIHIIIIlllllHi!!!lllllllllllllllllllllll llllllllllllllllllllllil!!!IIlllllllllllllm _ — — — m __ - _ _ _ _ fl =—_. — _ fi — _ = Pm- (den) § - - .— — .— —' _ _ — _ —— ——._ .— IIIIIIIHIIIIIIHI Illlllli' _27° .II In...__ 10‘3 10“ 10° 10’ 10‘ 1o3 anunncy trachea) 1 - W sin 2 ( = 40') ‘m 1 +04 ‘5" d is determined as 0.2174. Instead of d= 0.2174. let us choose «to be 0.21, or c( = 0.21 Next step is to determine the corner frequencies k) I 1/‘1‘ and a). 1/(ot T) of the lead cunpensabar. Note that. the mximuu phase—lead angle aim occurs at the geanetric man of the two comer frequencies; are): 1/(57'1‘). The amount. of the nodifieation in the magnitude me at. can 1/(JE'rT) due to the inclusim of the term (T3 4- 1J/(ol T3 1- 1} is .__1.._ IE 1+ij 1&1qu 1 w- —— fiz'r Note that '1 = 2.1822 = 6.7778 63 Fina Weneedtofindthefrequencypointwhere, Mtheleadcmpensator is added, the total magnitude 0 dB. The magnitude thjaJJI is 4.7778 dB corresdeds to 60: 2.81 rad/s. We select this frequency to bettemgainmtrflqmywc. 'I'henweobbain'_ 7:. figwc = 0.21:: 2.31 = 1.2877 Thus a + 1.2877 %“)=E s+anw_ A 3K - 4 Kc cf ' 0.21 4 s + 1.2877 Go‘s) = . = 4 0.77663 + 1 0.21 s + 6.1319- 0.153033 4- 1 The open-loop transfer fmctim beocima as GC(B)G(3) = 4 0.77665 + 1 1 0.163083 + 1 Build: 1- 1M8 + 1) me closed-loop transfer fmctim 15 Mtonwingmmpmgrmproducesthemit-steprespmsemaa c! I _ 3.10643 + 4 RIB) 0.0163134 + 0.279433 + 1.263152 + 4.10643 + 4 sham below. 3.1064 0.2794 . num = [0 U 0 >> den = [0.01631 >> t C 0:0.01:4; >> c = ateptnum,den,t): >> plottt,C) >> grid >> t1t1e('Unit—Step Response of Compensated System') >> xlabelt't (necl') ylabelf'Output cltl'} 4]: 1.2631 4.1064 4]: Uni-Step Response of Compensated System similarly. the following MATLAB program produces the wit-ramp response curve as shcnm below the mm program. >> num = [0 0 0 0 3.1064 4]; >> den = [0.01631 0.2794 1.2631 4.1064 4 0}; >> t = :0.01:6; >> 0 = step£num,den.t}; >> plotIt,c,t,t} >> grid b> titlel'UnitrRamp Response of Compensated System'} >> xlabelt't {sec} ') >2 ylabelf'InPUt and Output c(t)') Unit—Ramp Response of Compensated System Input and Output c(t) M :- 3—11-18. 'Ib satisfy the requimts. try a lead War Gets) of the form . .L 1 . Ts+1 _ 5+ 'r ‘ GE‘s I:ch «TS'Hl -Kc 3+..L Define OZT K 61(s)—KG(B)= “8+1, winteK=Kc0(. Isincetheataticvelocityarrormtmtfiqis givenas 503‘1 mhave 'I‘s-i-l K =1imaG€B)Gs)‘-=lima =x=50 xv 5+0 _c ( 5-9-0 “Ts-F1 B(s+1} We shauncnr plot a Bode diagram of 50 91(3) = The rouwing new program produces the Bode diagram sham below the program. _>>num=[o 0 50]; >>den=£1 1 0]: >> w = logspacei-1,2,1001: >> bode(num.den,w); >> grid >> title{'Bode Diagram of G_1{s) ') Bade Ow of 61(5) =Efllllll-IIIIIIII-IIIIIIIl -IIIIIiii=!III|lll-IIIIIIII .IIIIIIIl-Iillllll-IIIIIIII IIIIIIIII-IIIlllllallllllll .IIIIIIll-IIIIIIII- i“- - - — — _ — H |||||||||i'llllllllllllllll _1 m I ....IIII——. _ 1o“ 10" 10‘ ' wa Fran this piot, the pause margin is found to be 7.3“. The gain margin is suede. Since the specifications can for a phase margin of 50’. the additional phase lead angle necessary to satisfy the phase margin requiranent is 42.2'. We may asm‘tm mum phase lead required _ toha48’. Thisneanstlnt 5.3’hasbeenedded to mate for-the shift in the gain crossover frequency. Since 1-4: 31n¢”='1+c< 95m = 48" corresponds to of = 0.14735. (Note that oi = 0.15'oorreepouds todn=47.657'.) . Wmdmae =48'or¢m=47.657'doesnot make much difference in the final solut on. Hence, we dance or = 0.15. _ 'menextstepistodeteminethecomerfreqmiesw=1/Tand m=1/(O<T}o£theleadcompensator. Notethattheuximplnse leadmgledmoocursatthegeemetricmeenoftmmeomrfrequencies: orw=1/(J;'T). Themmtofuaemdificatiouinthemgnitudecurve .It wr— 1/(a'1') due to theimluei‘m of the term (TB + ij/(ams + 1113 Note that 1 1 = 2.5320 = 3.239 as J3? 0.15 ‘ We need to find the frequency point where. when the lead Wat-,0: is added. thatatalmgnitudebeomom. 'Ihefqumcyatvhid-l 'the magnitude of G1(jw)is equal ton-8.239 a comm between a): 10 and 100rad/s. FrcmtheBodediagruuefindtJmtthofl-oquency point where [611(ij = —8.239.dB m'atw= 11.4 rad/s. Noting that this frequency correspmda to 1/(Jk-T). Or: we obtain 1 =w_¢ =&4_=29.4347 «‘1‘ Jo: 0.15 Mleadcmpensatorflmsdetemined is ' .1. E” 'r _Kc s+4.4152 Gem-Kc a... 1 s+29.4347 WT 0.22658 + 1 0.033973 + 1 The following mm program produces the Bode diagram of the leaf: oun- peneator just designed. It is shown on next page. >> num = [11.325 50] : >> den - [0.03397 1]; >> w - logspacet-1,3,100}; >> bodaInum,den,w}; >> grid >> titlef'Bode Diagram of Gusts} = 50(0.2265s + 1}/{_¢.03397s + 1] ') acci- Dhgrlm dean) - 50mm + mamas?" 1) IIIIIIIIIIII lllllllllllllllfllllfll IIIIIIIIIII "Ill-Eilllllllllllllll 8 Myra-innate!) B B '8 a Pl'lle(@g) I lllllllll-II filllnlllllllllllllllll _-_!!!!!!II!I! IIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIiIIIll-IIIIIIIIIIIIIIIII Ill||Hlllll|"""'ll||l a, 4 Illlllh.‘ IIIIIIIIIIil ||l|I|||"""||||| l o __..._-III“|‘ lllll......_._ 10" 10° 10' 10" 10’ Fm (m) mm-lmmmimdmduwm 18 100° 5 + 4.4152 '1 scams) = 3 (T, + 29.4347 ) so: + 1) The following mm program produces the Dada diagram of Getsmts) which is shown on next page. >> nun - [0 O 1000 4415.2]; >> den - [3 91.3041 88.3041 0]; >> w = logapacet-1,3,JOOJ: >3 bodetnum,den,w); 2) grid I >> titlet'Bode Diagram of Gwcis)G{s) = 1000i: + 4.4152)/[3(s + 29.4347)sts + 111'} Pm this Bode diagram, 11: is clearly seen tint the phase margin is appro- ximately 52'. the gain margin is +90 dB, and RV = 50 5'1; all spacit‘icaticru are'net. Time. the designed system is satisfactory. Both Diagram of GJSJGE) = 10000: + 4.415234%: + 3.434% + 1}] Iiiiiiiii:l!!!llllllllllllllllllllll """" IIIIII" Ill-IIIIIIII!_ _ "I iiifll!!!IIIIIHHIIIIIIWIIIIIll" "um III!!!" - _ _ — _ = _ _ _ _ w - _ _ u — _ — - — — _ .— _ _- - _ _ _ _ - - - _ — _ * - _ _ — — fl - - — — I“ W W — _ —— _— ——_— 01-1 immune! mmmmted system ‘meoriginal wipe-mated systan has the following closed-loop transfer mum: 0(8) .1, 11(3) 52+a+1 The closed—loop poles of the uncompensated system are as follows: a = —O.5 a; j0.866 'Ihe closed—loop transfer fmmtim of the mud mm is 0(3) 1000(3 + 4.4152) — .- Rta) 3(3 + 29.4347)s(s + 1) + 1000(s + 4.4152) 10005 + 4415.2 333 + 91.304132 + 103030415 + 4415.2 The closed-loop poles of the mated are as follows: memmmogmgimbelowmoduoesthemt-stepmesorun unmnpensated and Wanted systess. The- resulting zespmse curves ' are shown belowr the mm program. ' ' num = [0 0 1]; den = [1 1 1]; numc = [0 O 1000 4415.2]; deno = [3 91.3041 l083;3041 4415.2]; = 0:0.01:3; ' o1 - step£num,den,t}; o2 = steptnume.denc,t}; plottt,c1,t,o21 grid _' titlet'Unit-Step Response of Unoompensated and Compensated Systems'J xlabelt't (aeci'l ylabelt'Outputs') text (1, 1.25, 'Compensated system'] textt1.7,0.25,'Uncompensated system'} Una-Step Response of Unoompensated and Compensated Systems ' 1.4 The new program given on next page produces the unit-ramp responses of the uncompensated system and compensated system. The response curves are shown on next page. Fran the wit-ramp response curves it can be seen that the output of the. ocmpensated- systan follows the input ramp very closely. For the compensated system, the error in following the ramp input can be seen for 0 < t< 0.5, but it is almost zero for 0.5< t. ('me steady-state error in the unit-ramp response is 0.02.) 1]: 1 0}; 0 1000 4415.2}: 91.3041 1088.3041 4415.2 0]; t = 0:0.01:8; c1 = step£num,den,t}; c2 = steptnumc,denc,t); plot t,cl,t,c2,t,t} tit1e('Unit—Ramp Response of Uncompensated and Compensated Systems') xlabelt't (sec)'] ylabelt'Input and Outputs'} text[1,7,'Compensated system') thxtl3.5,1.7,'Uncompensated') textt3.5,1,'system‘) text(1,4,'Input') Unit-Ran Response of Uncompensatad and Compensated Systems Input and O ...
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This note was uploaded on 04/15/2008 for the course MECE 3338 taught by Professor Song during the Spring '08 term at University of Houston.

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Chapter 11 - - 3 llllfliéiifl'llllfll .

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