Chapter 10 - CHAPTER 10 3—10-1 A simplifiedwa diagram of the Bystan is shown to the right The transfer flmctim C(s/R(s is cm 3.61” 12(3 1 4(cl

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Unformatted text preview: CHAPTER 10 3—10-1. A simplifiedwa diagram- of the Bystan is shown to the right. The transfer flmctim C(s)/R(s} is cm 3 .61”: 12(3) 1 4- (cl 4- (mantisI - G4" 3610-2. lsinpnflad block diagrams for the system an aim balav. C “a ‘oll IIIQfim+G. +6Wfll| R ' ¢fid§+M) c “fim+finfi*fi&%+9fi6*¢fifi The ma: Win: C(s}/R(s) is 0‘53 _ G163w? + H1) R(s J 1 + GZHZ + GzGafla‘ + G3”1"3 + G16253 + G16:3”1 B—lO-B. — t:tnum1,den1): = [10]; == {1 2 0]: = tf‘num23den2); series(sysl,syszi; feedbackisysg.{lll Transfer function: The closed—loop transfer function obtained is cs3! = 53 + 10 N3) 32 + 73 + 10 13—10—4. A MATLAB program to » solve this prob]. is sham to >> the right. 30 The closed—loop transfer ,, fmctim obtained is ' >> >> >) C(s) 105 + 50 ,9 3(3) 53 + 83'2 + 375 + 50 >> >> >> A MA'I‘IAB program to solve this problem is given below. "Hmw___m____-._m_-fi— tftnum1,den1); 10; [1 3 2 0]: tf(num2,den2); seriestsysl,sy32); [0.5 1); 1; tftnum3.den3): sys = feedbacktsysg,syah) Transfex function: 5‘3 + 8 5‘2 + 37 s + 50 mun-u— ——————-—--—-p- —— —-————.—-——.—..._ .——.——__._. H-_ ____ _ - .- B—IO—S. Define the input impedance and feedback impedance as Z and z , respectively. as sham in the figure below. ’ 1 2 'men Because each Operational amplifier of the system involves negative feedv back. the differential_ input voltage of each amplifier is zero. or e' = 0 and e" = 0. The voltages 151(3) and £05) are given by 51(8) = ZINE) = 1211(5) 28 Renal E‘s) - RZCZI + 1 M80: ‘0‘“ “.51 3(3) R3 Therefore. E (a) R R + he 2’328 lifetu 1, E (3) R3. 12102: 121123 R2023 The control action is proportional plus integral. Tali-a2 +1: Since a speed ombroller is usually design-d lllzh that 1: a1 ha s elf-a2 baa-k )1 the transfer function Y(s)/E(s) become ——.."3‘_———-—‘1:2 see-sine 13-10-6. If the engine speed increases. the sleeve of the fly-ban gamma: moves upward. his mt acts as the input; to the hydraulic cmtrqnsr. A positive error signal (upward motion of the sleeve) causes the pane: pistm tomvedomward. reducesthe fuel valve Opening. anddecreasestheengine‘ speed. Referring to Figure (a) shown below, a block diagram for the system canbedrawnasslminflgure (b)mnextpage. Figure (I) - Figure (b) mus. the control action of this speed controller is proportimu-plus-mteg— r‘al. 3-10-7. For the first—order systeni 913i, 1 . Obts) 'ra,+1 theateprosponsecurveisanmoxmtialcuno. SomtimIIamstantTm_ be determined from such an expmential curve easily. From Figure III-98' the time constant '1‘ 13- 2 a. If this mm is placed in a bath. the taparature of which is increasing at a rate of lo'c/min = 1/6'0/3. or ' 1 9b ' 6 t + a more a is a constant. than the steady-state error can be determined as follows: Noting that , ' _ = _. .flEL 13(3) obts) 0(a) 613(5) [1 ob“) -_- .- y= --—29 Gb(s_)(1 4—25 + J-) Obcs) 23 + 1 we obtain e 11111 SH ) 11m 252 = s = as 3+0 94-0 26 +- 1 813(5) where = _L_1_. a _ l + 6!! 819(8) 6 92 + s 652 'marafore, , ______28"’ ____1+6as=_2_.'_1_=_1_. 833 $210 25 + 1 652. 1 6 3 C Thus. the steady-state error is 1/3'C. For a mama-order. system: £14. 1 s =’ Gibbs) (1-15 + 1)(Tzs + 1) A typical response curve; when this thermoeta: is placed in a bath held at a constant temperatum, i ; .J_- J . . .r. 9.214. The closed-loop transfer tum-tic: or the system is C(s) 3 10(5 + 1) R(B) 32 + 105 + 10 10(3 + J.) (a + 1.1270)(s + 8.8730) For the unit-step input, we have at) g _ 10{s + 1) (a + 1.12701“ + 8.8730) 3 1 0.1455 1.1455 + s a + 1.1270 ' a :i- 8.8730 Mil—“mm 3:10—9. Since Hp is specifieé! as 0.05,; we have n H a e-Jf-x-t = 0.05 P II 2.99.5 "‘5 t;ch = (2.99920 -;2) Solving for the damping ratio 5 we obtain ' 01' Rani ting, 5 a 0.69 The settling‘tina 1.3 is specified as 2 seconds. So we hum t. = 4 = 2 01'. “man-prom . *Hfl- -fimnnm- mu—wn-uu— -q-n-m— - — ——-._-_ — n- -_--—__-‘m 3-10—10. The closed—loop transfer mum's of the system is 0(a) 100 Rts)' s3 4- 232+ 10: + 100 100 F (s + 4.5815)(s - 1.2907 + j4.4901)(s - 1.2907 — 34.4901) This systa: is unstable became two music—conjugate closed—loop poles are in the right half plane. '1!) visualize the unstable response. we my enter the following mm program into the mater._ The resulting unstable respcnse curve is shown below. . 'I'ommesystamstnbls. it isnsosmrytondmstM-mnofthe system or add an appropriate cmpsnsator. >> num [0 0 D 100]; >> den [1 2 10 100]: >> t = 0:0.01:6: >> y = steptnum,den,tJ: >> plot{t,y) >> grid >> title(’Unit-3tep Response of Unstable System'l >> xlabe1{'t (seCJ') ' )) ylsba1('c{t}'l Unit-Step Response of Unstable System 5-10—11. I C(B) 16 . Rte) .52 + (0.8 + 16105 4; 16 I Fran the chancteristic polyncmial, we find (On-=4. 2$wn=2xo.5x4=o.a+16k 'Hence k=0.2 The rise tunaY tr ie'obtained from . 7(_ Wu Since . . wd-wn/1-52-f-4h-o.25=3.46 41d 4:: = 1-1—=31n"10.866=-—- [6 en fin 3 . wehave ..L. tr=_72_fl.c_=o.5053 3.46 ' mepeaktimetpisobtainedaa ' =£'=_3_°.1_4._=o,9075 (dd 3.46 _ Hp—e "3453 the settling time 115 is ta=__£_=__$___=25 _5wn 0.51:4 13-1042. The closed-loop transfer function for the system 13 ' cm) K 0.51! 3(3) 232 + a + Exhs + K 32 + 0.50. + lath): + 0.5K Fran this equation. we obtain can =J0.5K, 2 3'4)“ -_- 0.5(1 + KKh) Since the damping ratio 5' is specified as 0.5.. we get 6d,, = 0.5(1 + xxh) 'I‘hérefore. we have 0.5(1 + fifth] =1/o.sx The settling time is Specified as 4 4 16 t =—-—-= 2 3 Scan 0.25(1+Rfih) 1+axh 52 Since the feede transfer {met-Am 6(a) is K mm= 23+1 _L= K 1 1 +_ ER]: 3 23+1+I<Kh a 28+ 1 the static velocity error constant RV is Kv=1imsG(s)=1ims K ...1__= K' '9' 5+0 284-14-th 5 11-“ This value‘ must be equal to' or greater than 50. Hanna, ' K 1+K'Kh a 50 Thus. the cmditions to be satisfied can be mined as follow: 0.5(1 + min) = 10-51! (1) 16 $2 (2) 1 +KKh I g} 50 (3} 1 +KKh o<xh<1 ‘4’ From Equations (1)_and (2).r we gwt B$1+KKh=JEI 32 IA a: Prom Equation (3) we obtain K "56214-1311 =‘/2K xgsooo Ifuectmaeli=500mthenweget 1+KKh=fZK=100 99 Kh = .= 0.0193 Ms,uadetem1mdaaetofva1uesofkandxhasfonm: K = 5000. Rh = 0.0193 with these values of K and Rh. all spacificntims are satisfied. 3:10:91- c‘vs) . x (1 +133) I 4“— RM le + xpa + Tda) Since Ms) - 1/52, the output C(s) is obtained as K + K Ta: 1 1 1 as) _ J z 2 52 2 K Td 5 s + KpTda + KP a a + J2 s + J Since the system is urflerdanped, C(s) can be written as C(S) ' "12 "' T 2 l K T _ 2.1 J a (=+-%:d—)"+JJ 5:; )5 d _ - rd ’Ihe inverse Laplace transform of C(a) gives {I ‘ c(t)=ti- 1 e’gfitsin IR-lfia- t‘ K 2T 2 J é—fil— =_1_.- B2 The steady—state error a” for a. unit mp input. 1: as, a: 11m [rtt] ~ c(t)] = m: [t — c(t)] 13"" t-Mu ‘ 2 2 - _§j x 1' =11. 1 a JJtain éi-fit-o t'” x xz'rz ' J “2 The steady-state error can also be obtained by use of the final value theorem Since the error 31ml 3(3) is ' I 1 '1' H6) = Rm - C(s) = Ma) [1 - cm]: .35 1 -. .E( + :1" 12(3) Jo + xp(1 4. 1d.) we obtain the study-auto error a“ as e“: 11-: e(t)'z 11: 355(3) = 11:: TEE-£5— : 0 15-". s-oo s-oo 3 (Jo + xp 4- Tarps) WH‘-—-- ham—unh— —- - --——— —- nu—flnmu—I-I— -.. --—w-—-—u-—n -n—“m 3-10—14. The characteristic 'equ'atioen is ——.—.—§—.——.‘—I+ 1 I: 0 3(3 + 1H: + 5) s3+632+55+x=0 ‘meRoubh array for this equation is 1 5 6 x 30—1: 0 6 K Forthesystamtobestable.flareshmndbenosignchangesintrnflrst collm.‘1hiarcquires 30-K>0. K>O mnee.weg'ettherangeofgainxforstabilitytobe 30>K>0 3-10-15. Since the system is of higher order (5th order), it is easier to find the range of gain K for stability by first plotting the root loci and then finding critical points (for stability) on the root loci. The open—loop transfer fmotim Gts) can be written as Mez + 2; + 4) s(s.+ 4103+ tins:2 + 1.43 + 1) ' 1((82 + 25 + 4) s5 + 11.454 + 3933 + 43.632 + 245 Gts) = The MM'IAB program given next will generate in plot of the root loci for the system. The resulting root-locus plot is shown below the 1mm program . >> nun: [0 0 O 1 2 4] ; >> den [1 11.4 39 43.6 24 0] ; >> rlocustnum,den) >} v = [—B 2 —5 5}; axistv}; axist'equal'} >> title ( ' Root—Locus Plot ’) - Based on thisplot. it can be seen that the systh is umditionally stable. A11 critical points for stability lie on the jwaxie. To determine the crossing points of the root loci with the jwaxis substitute a = ju) into the characteristic equation which is - 35+11.4s4+3933+43.632+24s+K(sZ+25-I-4)=0 Then 1 (MN + 11.4(ij4 + 39(‘jw)3 + (43.5 + mum)? + (24 + 2mm +m=o This equation can be rewritten as [11.4504 — (43.5 + xij + 4K] + jth — 39:03 + (24 + now] =- o By equating the real part and imaginary part equal to zero. respectively. we obtain ' . - 11.4w4—(43.6+K)w2+4l(=0 (1) 505539w3+(24+2xm=o (2) Equation (2) can be written as w =0 or _ w4-39w2+24+2x=o (3) From Equation' (3) we obtain K=.—w4'+:9w2-24 I m By substituting Equation (4) into Equation (-1); we get 11.4 :04 - [43.6 + 5(-w4‘+ 39402 — 24mm? — 2:04. + 78.102 - 48 = 0 which can be simplified to 4:6 — 20.2414 +92.3a52 -96 =0 The roots of this last equation can be easily obtained by use of. the ram program as given below. >>a=[1 o 420.2 o 92.5 o >>roots(a) ans 3 3.7553 v3.7553 2.1509 1.2130 -2.1509 —1.2130 The root-lows branch in the upper half plane that goes to infinity crosses. the jw axis at u; = 1.2130, w = 2.1509, and a): 3.7553. The gain values at these crossing points are obtained as follows: ———————————-——————-—‘1'21304 + 39 x 131302 ‘ 24 = 15.61 for a) = 1 2130 K: 2 K - = 57.51 for w = 2.1509 2 = 163.55 mm: 3.7553 K = 2 Based on the K wlues above. we obtain the range of gain K for stability as follows: The system is stame if - ‘ 15.61 > K > 0 163.56 > K > 67.51 __-_____m—n__-—-———u———- fin".- 3-10—16. A mums program to plot the root loci and asymptotes {01' the following system K G(s)H(s) = -—-—-——-'—""_‘—"'—" 3(5 + 0.5)(52 + 0.63 + 10) K s4 + 1.153 + 10.332 + 53 is given below and the resulting root-locus plots-are sham on the next page. Note that the agitation for tha W 1- K (s + 0.275)4 Ga(s)Hh(s) = K ‘34 + 1.133 + 0.453852 + 0.08319: 4- 0.005719 8' >> am [0 D O 0 1]; I >> den [1 1.1 10.3 5 0]; >> numa = [0 0 0_ 0 1]; >> dena = [1 1.1 0.4538 0.08319 0.005719]; >> subplottZZl); zlocusInum,den); ' >> v = [-5 5 —5 5]; axislv); axist'equal') 3) subplot1222); :locusinum,den) >> hold Current plot held >> rlocu5{numa,dena]; >> v = {-5 5 -5 5]; axistv}; axisl'equal'} >3» titlet'Plot _of Root 1.021 and Asynptotes') Root Locus 3—10-17. The following mm program will generate ai root-locus plot. Gfs) = R resulting.plot is shown below. >> num= [0 >> den - [1 0 7 >> zlocus {nun-1, den] 2») v = [—4 2 -3 PM of Root Loci and Asyrnplalns The open-loop transfer.function 6(a) is m 2 ' s + 5 52(5 + 2) Kzts + 1) s4 + 733 + 1032 The 3] ; axis [v] J axis {'equal') >> title ( 'Root-Locus Plot: ' } Fran the plot we find that the critic'ai value of gain _K for stability oorrespmdstotheerossingpointoftherootlocusbnnditMtgoesto infinity and the imaginary axis. Hence, we first find the crossing frequency and then find the corresponding gain value. The characteristic equation for this system is s4+7s3+1052+2KB+ZK=0 By substituting a = 160 into the characteristic equation. we obtain- (J'wfl + 7(Jwi3 + 10(jw)2 + mij + 23 = 0 which can be rewritten as (w4-iow2+2x)+Jw(—7w2+2x) =0 By equating the real part and imaginary part of this last equation to zero, reSpectively, we get w4-1.0w2+2fi=0 (-1) cut-7422+2K) =0 (2) Equation (2) can be rewritten as w = 0 or —7w2+2x=o ‘ (3) By substituting Equation (3) into Equation (1); we rim! 6.)“ - ma}2 4- “dz = 0 or w4—3wzéo which yields #:0: (d‘o: 60:]; W"’J? Sinoew-J-JS—is the crossing frequency with the jwazda. by substituting a) I: E into Equation (3) we obtain the critical value of. gain K for stability. k=3.5w2= 3.5x 3=1o.5 Hence the stability range for K is m5>K>o ——--—_— —_--———-_———-_m 5-10—13. The angle deficiency is 130' — 120' -. 120" = — 60’ A lead compensator can contribute 60". Let us choose the zero of the lead compensator at e = -1. Then, to obtain phase lead angle of 60': the pole of the ompeneator must be located at a = -4. mos, 8+1 Ge“) = “Tr;- 'I'he gain K can be detehnined from the magnitude oouditim. JR 5+1 1 ' _1 8+4 2 n a s=—1+jf§- :8 +1 3 5=—1+j.l3 Heme the lead oomensator beams as follows: 8+1 Gc(5)=8 8+4 The feedfbrward transfer function is Bs+8 334-452 The following me program will gamma a root-10cm plot.- The resulting plot is Brawn below the new program. came) = >> num = [O 0 >> deaf [1 4 0 0]; a») rlocusinum,den) 2 -4 4]; axis {v}; axisanual 'J Note that the closed-loop transfer funotim‘is IC(s) - 83 + 8 12(8) 53 + 452 + 83 + 8 The closed-loop poles are located at B = -1 1 jfi and s = ~2. ———————.———.———_—____-'_—________— _—...——.——.—..—-—-——---...—__ 13—10-19. The mm program giVen below generates a root—locus plot for the given system. me resulting plot is sham below the program. >> num [0 0 1]: >> den [1 5 4 0]: >> zlocustnum,den} >> 1; = [-6 4 -5 5]; axis Iv}; axist'equal') Root LOO". Ram Note that mutant-g points (0< 5(1) lie on a straight line having angle 8 frcn the 1a) axis as sham in the figure below. Fran the figure we obtain Noteuoothat§=0.61ineoanbedefimdby s a: —O.75a +ja where a is a miable (o (a (co). "no find the wine of K- out that the damping ratio 3 of the dcminant closed—loco poles is 0.6 can he foumd by finding the intersection of the line a - 43.753. + 3: , The intersection point can be determined by solving the following aim].- taneoue equations .for a. ' s = -O.7Sa + ja (1) s{s + 1)(s + 41+ K = 0 (2) By substituting Equation (1) into Equation (2). (—0.75a + ja)(—O.75a + ja + 1)(7-0.75a + ja + 4) + K = 0 which can be rewritten as I (1.020103 - 2.1375212 — 3a + K} + 5(0.6075a3 — 7.52:2 4- 4a) '= 0 By equating the red. part and imaginary part of this last equation to zero. respectively. we obtain _ 1.8281a3 —2.1075a2 —3a 4 K = 0 . - (3) 0.63733 -— 7.5.12 + 431 = 0 _ (4) Equation (4) can be rewritten as a = 0 0.687532 — 7.5a + 4 = 0 much can be writbm as a2 _ 10.9091a + 5.8182 = 0 or (a - 0.5623)(a - 10.3468) = D a = 0.5623 ' or. a = 10.3403 Fran Equation (3) we find at = 4.323133 + 2.107332 + 30. a 2.0535 for a = 0.5623 x = -1.8281a3 + 2.1875122 + 3: 3 4759.74 for a = 10.3468 - since the R «an. is positive for a = 0.5523 Iandflnagultgive-for a = 10.3453, we choose a a 0.5623. The required 90,111 K is 2.0535. since the characteristic equetiom with K = 2.0535 18 s(s + 1M5 + 4) ' + 2.0535 = 0 01' s3 + 552 +43 + 2.0535 s 0 the closed-loop poles can be obtained by use of the following mm program. >> p = {1 .5 4 2.0535]; >> rootsip} afis = "4.1565 —0.4217 + 0.5623i —0.4217 — 0.56231 Thus. the cloud-loop poles are located at B = -0.4217 t j0.5623, a = -4.1565 The unit—step response of the systemwithx a 2.0535 can be obtained by entering the following mama pct-09m into the camber. The resulting wait-step response curve is shown below the MATIAB program. ' >> num n [0 0 0 2.0535}: >> den = [1 5 4 2.0535]; >> = 0:0.01I20; >> y - steplnum,den,t}; >> plotttryl >> grid >> titlet'Unit-Step Response‘} >> xlabell't {5ec)') >> ylabel('0utput'l Unit-Step Response —-————+n—.‘ "nw——m_wm—— 3-10—20. The closed—loop transfer function of the system 13 C(s) Ms + ms + b) R(s) 5(52 + 1) + K(s + 8003 + b) = K(az+as+'bs+ab2 s3+s+K(52+as+bs+ab) Since the daninant closed-loop poles need be located at s = -1 i jl, the Claracteristic equation must be divisible by (s+1+j1)(s+l-j1)=sz+2s+2 Hence 83+Ksz+(l+aK+ti{)s+abl{=(32+2s+2)(s+o() where s = -D( is the unknown third pole. By dividing the left side of this last equation by 52 + 23 + 2, we obtain - s3+Ksz+(1+aK+bK_)s+abK=-'(52+23+2)(s+K—2) +(axaibK-2K+3)s+xab—2(K—2) There-minder of divisionmustbezero. Hence we sat aK-ihK-2K‘+3=O Kab-ZtK-2)=O since a is specified as 0.5, by substituting a = 0.5 into these two equations; we obtain - bx = 1.5K — 3 (1) 0.5m: — 2(x — 2) = o (2) By substituting Equation {1) into nation (2); we have o.5(1.5x -‘3) — 2(K - 2) = o 01' K = 2 men. by substituting K = 2 into Equation (1). we get I 21: = 1.5 x 2 — 3 = 0 Hence b=0 The PID controller: with K = 2 and b I 0 beacon; (+1 +1: I . Gc(3)=K 5 a: J=K-m-—:—§E—=K(s+0.5) Thus. the controller beeanes a PI) controller. The open-loop transfer function bit-canes K(s + 0.5) 32 + l Gamma) = The closed—loop transfer fmctian (with b = 0 and K I 2) banana-s as follows: C(s) 2(5 + 0-5) 3‘5) '32 + 23 + 2 23+1 (s+1+j1)(s+1-j1) mam—Implotforthedesigxwdsystmcanbeobtaimdbymberring the following mmm into the when >> num = [0 1 0.5}; - >> den = [l 0 1]; >> rlocuatnum,denl >> v = [-2 1 -1.5 1.5]: axistv); axist'equal'l The resulting root-1m plot is 31m below. menal 15 1 3 Q5 E O = E .05 4 45 a 4 O I ...
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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 10 - CHAPTER 10 3—10-1 A simplifiedwa diagram of the Bystan is shown to the right The transfer flmctim C(s/R(s is cm 3.61” 12(3 1 4(cl

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