BALL & BEAM

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<p>LI TAM hnh qu bng v thanh c s dng trong cc th nghim v iu khin hu ht cc trng i hc k thut v n l mt m hnh d xy dng v l mt v d in hnh v k thut iu khin. H thng gm c mt qu bng, mt thanh , mt ng c v mt s sensor. Mt tay quay c gn vo mt u ca thanh , u cn li c gn vo ng c. Khi ng c quay mt gc teta ( ), thng qua tay quay s lm thanh quay mt gc anpha (). Khi thanh thay i gc t v tr cn bng, di tc dng ca trng lc s lm cho qu bng ln t do trn thanh . Cng vic ca ngi iu khin l phi thit k mt b iu khin kim sot c v tr ca qu bng trn thanh . Bng cch s dng cc sensor, thng tin v v tr ca qu bng c gi v v c so snh vi gi tr t vo trong b iu khin, t b iu khin s gi tn hiu iu khin gc quay ca ng sao cho qu bng s t ng v tr mong mun trn thanh . M hnh ton hc ca h thng thuc loi phi tuyn, tuy nhin ta c th coi h thng l h tuyn tnh quanh khong nm ngang. M hnh tuyn tnh n gin ny vn c p dng cho rt nhiu cc h thc nh h thng cn bng my bay theo phng nm ngang khi h cnh di nh hng ca cc dng kh chuyn ng hn lon. Trong n s dng cc nh lut vt l v phng php xp x tuyn tnh phn tch m hnh vt l ca h bng v thanh . T , ta xc nh c hm truyn ca h thng, l c s kho st tnh n nh ca h thng h. B iu khin c tng hp v thit k cng da trn cc kt qu thu c. Tt c cc cng on m hnh ha v tng hp b iu khin u s dng phm mm MATLAB &amp; SIMULINK.</p> <p>CHNG 1 GII THIUGII THIU CHUNG H qu bng v thanh cn c gi l h cn bng ca qu bng trn thanh . H thng ny c s dng nh mt bi th nghim hu ht cc trng i hc k thut trn th gii v n kh gn gi vi cc h thng iu khin thc nh vic n nh h thng cn bng my bay theo phng ngang khi h cnh di tc ng hn lon ca cc dng kh. Mc ch ca h thng l iu khin v tr ca qu bng trn thanh sao cho n t ng gi tr mong mun di cc nh hng t nhiu bn ngoi nh lc y t tay chng ta. Tn hiu iu khin c th nhn c t tn hiu phn h v tr ca qu bng qua cc b sensor. Tn hiu iu khin (in p) c a ti ng c mt chiu DC thng qua b khuych i cng sut s iu khin chnh xc gc quay ca ng c. T , qu bng s t n ng v v mong mun. Mt iu tt nhin l h thng ca chng ta l h h, phi tuyn v khng n nh. gii quyt vn khng n nh, ta cn s dng mt mch phn hi. Ngoi ra, ta c th xem h s tuyn tnh nu thanh ch dao ng vi mt gc anpha () tng i nh (-300300). Nu gc anpha ln hn gi tr trn th kt qu thu c t phng php xp x tuyn tnh s khng cn chnh xc. Trong trng hp bt buc ta phi dng n cc phng php iu khin phi tuyn. CC TI NGHIN CU V H THNG H qu bng v thanh dm c rt nhiu cc c nhn v t chc trn th gii nghin cu v c nhng thnh cng nht nh. Sau y s l mt s v d in hnh: Vo nm 2005, Arroyo xy dng mt h thng c tn Qu bng cn bng trn thanh nh hnh 1.1. H thng s dng mt cm bin in tr dy xc nh v tr ca qu bng. Tn hiu t cm bin c x l trong mt b DSP v xut tn hiu iu khin ng c mt chiu vi hp gim tc. H thng ny s dng lut iu khin PD.</p> <p>Hnh 1.1: Qu bng cn bng trn thanh dm c xy dng bi Berkeley Robotics Laboratory (Arroyo 2005) H thng trn tng i d thc hin v b iu khin PD cng kh n gin. Thc t, mc d v tr ca qu bng c iu khin bng b iu khin PD, tuy vy gc nghing ca thanh dm li khng c o v iu khin. Do , h thng hot ng cha hon ton n nh. Nm 2006, Quanser to ra mu qu bng v thanh dm mang tnh thng mi u tin nh trong hnh 1.2. Mu qu bng v thanh dm ny cng gm c mt cm bin in tr dy xc nh v tr qu bng, mt ng c servo vi hp gim tc. H thng c iu khin bng b PID. So vi mu qu bng cn bng trn thanh dm trn th mu ny hon thin hn rt nhiu.</p> <p>Hnh 1.2: Mu qu bng v thanh dm do Quanser ch to (nm 2006) Ngoi nhng mu in hnh trn cn c rt nhiu nhng mu tng t nh vy, chng ch khc nhau loi cm bin c s dng, cch iu khin (c th l iu khin qua gc quay theta ca ng c hoc iu khin bng mmen quay ca thanh khi gn trc c c vo tm thanh ). NHIM V CA N Mc ch chnh ca n l phn tch, m hnh ha v tng hp thit k b iu khin cho h thng qu bng v thanh dm. C th: - Xy dng phng trnh vi phn chuyn ng v thit lp hm truyn ca h thng. - Kho st tnh n nh ca h thng h - Tng hp b iu khin (t c cc gi tr p ng mong mun) bng phng php p ng tn s. - Kt lun v hng pht trin ca n. GI THIT CHO BI TON n gin ha m vn mang tnh khch quan, ta c th gi s cc iu kin sau y cho m hnh: - Qu bng ln khng trt trn thanh . - Gc quay ca thanh -300300.</p> <p>KT LUN Chng 1 gii thiu tng quan v n qu bng v thanh dm, cc cng trnh lin quan n n ny cng nh mc tiu, nhim v v cc iu kin cn thit ca n. Trong chng 2 s gii thiu v cch thit lp m hnh ton ca h thng, phn tch h thng h theo phng php p ng tn s.</p> <p>CHNG 2 PHN TCH V XY DNG M HNH TONHnh v 2.1 th hin chi tit m hnh ca bi ton:</p> <p>Hnh 2.1: M hnh qu bng v thanh dm. Vi bi ton ny, cc tham s v cc bin c cho trong bng sau: Bng 2.1: Cc tham s ca h thng Th t 1 2 3 4 5 6 7 8 Tham s Trng lng bng Bn knh bng Mmen qun tnh Chiu di cnh tay n Khong cch Ta ca bng Ta gc ca thanh Gc quay ca ng c K hiu m r I l d P n v kg m kg.m2 m m m rad rad Gi tr 0,11 0,005 9,9.10-6 1 0,15</p> <p>Mc tiu iu khin c cho trong bng sau:</p> <p>Bng 2.2: Mc tiu iu khin ca h thng Th t 1 2 Tham s cn iu khin Thi gian qu qu iu chnh Gi tr 3 (s) 5 (%)</p> <p>1. PHN TCH M HNH H THNG Phn tch m hnh qu bng trn thanh Xt chuyn ng ca qu bng trn thanh .</p> <p>0mgco s</p> <p>r</p> <p>mgsi n</p> <p>mgHnh 2.2: Cc lc tc dng ln qu bng Hot lc tc dng ln h ch gm c trng lc P ca qu bng (trng lc ca thanh khng ng k). Trng lc P l lc c th. Ch h ta suy rng ca h l q1=x, q2= (hnh 2.2). Ta thy h c 2 bc t do v phng trnh Lagrange c th vit di dng:</p> <p> d T . dt q iTrong : T = ng nng ca h.</p> <p> T = qi qi </p> <p>Fqt</p> <p>x</p> <p>(2.1)</p> <p>qi= ta suy rng tng ng i= (1,2).</p> <p>-</p> <p>= th nng ca h ng vi P. Th nng ca h:</p> <p> ( P ) = mg sin ( ) x + CSuy ra:</p> <p>(2.2)</p> <p> = mg sin ( ) x = mg cos( ) x </p> <p>(2.3)</p> <p>ng nng ca h bao gm ng nng ca bng v ng nng ca thanh : ng nng ca qu bng va chuyn ng quay va chuyn ng tnh tin nn ta c:</p> <p>T1 =</p> <p>1 2 1 2 mv1 + I1 2 2</p> <p>(2.4)</p> <p>ng nng ca thanh ch bao gm chuyn ng quay nn ta c:</p> <p>T2 =</p> <p>1 I thanh22 2</p> <p>(2.5)</p> <p>Thay (2.4) v (2.5) vo biu thc ng nng ca h ta c:</p> <p>T = T1 + T2</p> <p>T =</p> <p>1 2 1 2 1 2 mv1 + I1 + Ithanh 2 2 2 22 2</p> <p>2 . . 1 1 x 1 . T = m x + I + Ithanh 2 2 r 2 </p> <p>(2.6)</p> <p>Suy ra:</p> <p>T x.</p> <p>= m x,</p> <p>.</p> <p>T .</p> <p>= Ithanh ,..</p> <p>.</p> <p>T T = 0, = 0. x .. = Ithanh </p> <p>.. d T x d T . = m x+ I 2 , . dt x r dt </p> <p>Phng trnh vi phn chuyn ng ca h ca h Lagrange loi 2 nhn c l:..</p> <p>m x+ I</p> <p>x = mg sin( ) 2 r .. I thanh = mg cos( ) x ..</p> <p>(2.7)</p> <p>Gc quay ca thanh ( ) t l vi gc quay ca ng c ( ) theo biu thc sau:</p> <p>=</p> <p>d l</p> <p>(2.7)</p> <p>Thay (2.7) vo phng trnh (2.6) ta c phng trnh chuyn ng l hm v tr ca qu bng (x) v gc quay ca ng c ( ):</p> <p>d I .. 2 + m x = mg l r </p> <p>(2.8)</p> <p>Phn tch m hnh ng c DC ng c l thit b bin i nng lng in thnh nng lng c di dng chuyn ng quay hay tnh tin. ng c l mt phn khng th thiu trong h thng qu bng v thanh . Trn th trng hin nay c rt nhiu chng loi ng c: ng c c hp gim tc, ng c bc, ng c servo, ng c xoay chiu mt, hai hay ba pha. Tt c cc ng c k trn u c th c s dng trong h thng qu bng v thanh dm. Tuy nhin, trong rt nhiu cc ng dng thc t, ngi ta thng la chn ng c mt chiu bi kt cu n gin v d iu khin.</p> <p>ng c mt chiu ca hng Harmonic Driver nh hnh 2.3 c s dng lm b phn dn ng trong n. T hnh v ta c th d dng nhn thy cc b phn ca n, bao gm: hp gim tc, rotor v stato, enconder. Tt c c lp ghp li thnh mt chic ng c hon chnh.</p> <p>Hnh 2.3: ng c mt chiu Harmonic Driver S mch tng ng ca ng c c dng:</p> <p>Hnh 2.4: S mch tng ng ca ng c in mt chiu Cc tham s ca ng c gm c: Bng 2.3: Tham s ng c Harmonic Driver Th t Tham s K hiu n v Gi tr</p> <p>1 2 3 4 5 6 7</p> <p>M men qun tnh ca rotor T s gim chn in tr rotor in cm rotor Hng s sc in ng phn hi u vo in p Gc quay theta</p> <p>J b R L K V</p> <p>kg</p> <p>m2 s2</p> <p>0,043 1,5279 4,7 0,016 4,91 (-2020)</p> <p>Nms Ohm H</p> <p>Nm AVolt rad</p> <p>M men T ca ng c t l vi dng in phn ng I theo hng s Kt v sc in ng phn hi t l vi vn tc gc ca rotor theo hng s Ke. Ta c biu thc:T = Kt i e = Ke .</p> <p>(2.9)</p> <p>T s mch tng ng hnh 2.4, theo nh lut Kirchoff 2 ta c:</p> <p>J + b = K ti. di L + Ri = V K e dt</p> <p>..</p> <p>.</p> <p>(2.10)</p> <p>Mt khc ta c:</p> <p>K t = KeThay (2.11) vo (2.10) ta c:</p> <p>(2.11)</p> <p>J + b = Ki . di L + Ri = V K dt</p> <p>..</p> <p>.</p> <p>(2.12)</p> <p>XY DNG HM TRUYN CA H THNG Bin i Laplace cc phng trnh (2.8) v (2.12), ta c: Hm truyn gia v tr ca qu bng trn thanh v gc quay ca ng c:</p> <p>d I 2 2 + m Y (s )s = mg X (s ) l r G (s) = Y (s) mgd 1 = X (s) I s2 l 2 + m r (2.13)</p> <p>-</p> <p>Hm truyn ca ng c DC (gia in p t vo v gc quay):</p> <p>s( Js + b)(s ) = KI ( s ) ( Ls + R) I ( s) = V Ks( s)Thu gn ta c:</p> <p>H ( s) =-</p> <p>( s ) K = V ( s) s(( Js + b)( Ls + R ) + K 2 )</p> <p>(2.14)</p> <p>Hm truyn ca c h thng chnh l hm truyn ca ng c c mc ni tip vi thanh qua tay n d. Gi T(s) l hm truyn ca h thng, ta c:</p> <p>T ( s) = H ( s)G ( s)</p> <p>Kmgdr 2 T ( s) = 3 s (( Js + b)( Ls + R ) + K 2 )(lI + m)Kmgdr2 T ( s) = 5 4 ( JLlI + JLm ) s + ( JRlI +JRm + bLlI bLM ) s + (bRlI +2 bRm K +lI +</p> <p>2 K + m</p> <p>KHO ST P NG CA H THNG H Vi hm truyn thu c v cc tham s cho, s dng phn mm MATLAB ta c th xy dng ng c tnh h ca h thng bng m-file (xem Ph lc 1). Hm truyn cui cng ca h thng c dng: Transfer function:</p> <p>1.985e-005 ---------------------------------------7.569e-005 s^5 + 0.02492 s^4 + 3.442 s^3 V c tnh ca h h theo thi gian trn hnh 2.5:</p> <p>Hnh 2.5: c tnh theo thi gian ca h h</p>