C S T NG HCCHNG I

NHP MNNI DUNG : I. i cng. II.Cc nh ngha. III.Cc loi h thng iu khin t ng I. I CNG Hi tip (feedback) l mt trong nhng tin trnh cn bn nht trong t nhin. N hin din trong hu ht cc h thng ng, k c trong bn thn sinh vt, trong my mc, gia con ngi v my mc Tuy nhin, khi nim v hi tip c dng nhiu trong k thut. Do , l thuyt v cc h thng t iu khin (automatic control systems) c pht trin nh l mt ngnh hc k thut cho vic phn tch, thit k cc h thng c iu khin t ng v kim sot t ng. Rng hn, l thuyt cng c th p dng trc tip cho vic thit lp v gii quyt cc vn thuc nhiu lnh vc khc nhau, khng nhng cho vt l hc, ton hc m cn cho c cc ngnh khc nh: sinh vt hc, kinh t hc, x hi hc, Hin nay, h thng t iu khin m ng mt vai tr quan trng trong s pht trin v tin b ca cng ngh mi. Thc t, mi tnh hung trong sinh hot hng ngy ca chng ta u c lin quan n mt vi loi iu khin t ng: my nng bnh, my git, h thng audio-video ... Trong nhng c quan ln hay cc xng sn xut, t hiu sut ti a trong vic tiu th in nng, cc l si v cc my iu ho khng kh u c kim sot bng computer. H thng t iu khin c thy mt cch phong ph trong tt c cc phn xng sn xut : Kim tra cht lng sn phm, dy chuyn t ng, kim sot my cng c. L thuyt iu khin khng th thiu trong cc ngnh i hi tnh t ng cao nh : k thut khng gian v v kh, ngi my v rt nhiu th khc na. Ngoi ra, c th thy con ngi l mt h thng iu khin rt phc tp v th v. Ngay c vic n gin nh a tay ly ng mt vt, l mt tin trnh t iu khin xy ra. Quy lut cung cu trong kinh t hc, cng l mt tin trnh t iu khin II. CC NH NGHA. II.1 H thng iu khin II.2 H iu khin vng h II.3 H iu khin vng kn

II.4 Hi tip v cc hiu qu ca n 1- H thng iu khin: L mt s sp xp cc b phn vt l, phi hp, lin kt nhau, cch sao iu khin, kim sot, hiu chnh v sa sai chnh bn thn n hoc n iu khin mt h thng khc. Mt h thng iu khin c th c miu t bi cc thnh phn c bn (H.1_1). i tng iu khin (ch ch). B phn iu khin. Kt qu.

Ba thnh phn c bn c th c nhn dng nh ( H.1_1). Cc inputs ca h thng cn c gi l tn hiu tc ng (actuating signals ) v cc outputs c hiu nh l cc bin c kim sot (controlled variables ). Mt th d n gin, c th m t nh (H.1_1) l s li xe t. Hng ca hai bnh trc c xem nh l bin c kim sot c, hay outputs. Gc quay ca tay li l tn hiu tc ng u, hay input. H thng iu khin trong trng hp ny bao gm cc c phn li v s chuyn dch ca ton th chic xe, k c s tham gia ca ngi li xe. Tuy nhin, nu i tng iu khin l vn tc xe, th p sut tc ng tng ln b gia tc l input v vn tc xe l output. Ni chung, c th xem h thng iu khin xe t l mt h thng iu khin hai inputs (li v gia tc) v hai outputs (hng v vn tc). Trong trng hp ny, hai inputs v hai outputs th c lp nhau. Nhng mt cch tng qut, c nhng h thng m chng lin quan nhau.

Cc h thng c nhiu hn mt input v mt output c gi l h thng nhiu bin. 2.H iu khin vng h (open_loop control system). Cn gi l h khng hi tip (Nonfeedback System), l mt h thng trong s kim sot khng tu thuc vo output. Nhng thnh phn ca h iu khin vng h thng c th chia lm hai b phn: b iu khin (controller) v thit b x l nh (H.1_2).

Hnh H.1_2 : Cc b phn ca mt h iu khin vng h. Mt tn hiu vo, hay lnh iu khin hay tn hiu tham kho (Reference) r a vo controller. Tn hiu ra ca n l tn hiu tc ng u, s kim sot tin trnh x l sao cho bin c s hon tt c vi tiu chun t trc ng vo. Trong nhng trng hp n gin, controller c th l mt mch khuch i, nhng c phn ni tip hoc nhng th khc, tu thuc vo loi h thng. Trong cc b iu khin in t, controller c th l mt microprocessor. Th d : Mt my nng bnh c gn timer n nh thi gian tt v m my.Vi mt lng bnh no , ngi dng phi lu?ng nh thi gian nng cn thit bnh chn, bng cch chn la thi gian trn timer. n thi im chn trc, timer iu khin tt b nung.

Hnh H.1_3: Th d v h iu khin vng h. D thy ngay rng mt h thng iu khin nh vy c tin cy khng cao.Tn hiu tham kho c t trc, cn p ng ng ra th c th thay i theo iu kin xung quanh, hoc nhiu. Mun a p ng c n tr gi tham kho r, ngi dng phi qui chun li bng cch chn timer li. 3. H iu khin vng kn (closed loop control system). Cn gi l h iu khin hi tip (feedback control system). iu khin c chnh xc, tn hiu p ng c(t) s c hi tip v so snh vi tn hiu tham kho r ng vo.

Mt tn hiu sai s (error) t l vi s sai bit gia c v r s c a n controller sa sai. Mt h thng vi mt hoc nhiu ng hi tip nh vy gi l h iu khin vng kn. (Hnh H.1_4)

H.1_4 : H iu khin vng kn. Tr li v d v my nng bnh. Gi s b nung cp nhit u cc pha ca bnh v cht lng ca bnh c th xc nh bng mu sc ca n. Mt s c n gin ho p dng nguyn tc hi tip cho my nng bnh t ng trnh by nh (H.1_5).

Ban u, my nng c qui chun vi cht lng bnh, bng cch t nt chnh mu. Khng cn phi chnh li nu nh khng mun thay i tiu chun nng. Khi SW ng, bnh s c nng, cho n khi b phn tch mu "thy" c mu mong mun. Khi SW t ng m, do tc ng ca ng hi tip

(mch in t iu khin relay hay n gin l mt b phn c kh). H.1_6. l s khi m t h thng trn.

Mt th d khc v h thng iu khin vng kn nh hnh H.1_7: h thng iu khin my nh ch in t (Electronic Typewriter).

H.1_7: H thng iu khin my nh ch in t. Bnh xe in (printwheel) c khong 96 hay 100 k t, c motor quay,t v tr ca k t mong mun n trc ba g in. S chn la k t do ngi s dng g ln bn phm. Khi mt phm no c g, mt lnh cho bnh xe in quay t v tr hin hnh n v tr k tip c bt u. B vi x l tnh chiu v khong cch phi vt qua ca bnh xe, v gi mt tn hiu iu khin n mch khuch i cng sut. Mch ny iu khin motor quay thc bnh xe in. V tr bnh xe in c phn tch bi mt b cm bin v tr (position sensor). Tn hiu ra c m ha ca n c so snh vi v tr mong mun trong b vi x l. Nh vy motor c iu khin sao cho n thc bnh xe in quay n ng v tr mong mun. Trong thc t, nhng tn hiu iu khin pht ra bi vi x l s c th thc bnh xe in t mt v tr ny n v tr khc nhanh c th in mt cch chnh xc v ng thi gian.

H.1_8: Input v output ca s iu khin bnh xe in. Hnh H.1_8 trnh by input v output tiu biu ca h thng. Khi mt lnh tham kho c a vo (g bn phm), tn hiu c trnh by nh mt hm nc (step function). V mch in ca motor c cm khng v ti c hc c qun tnh, bnh xe in khng th chuyn ng n v tr mong mun ngay tc khc. N s p ng nh hnh v v n v tr mi sau thi im t1. T 0 n t1 l thi gian nh v. T t1 n t2 l thi gian in. Sau thi im t2, h thng sn sng nhn mt lnh mi. 4. Hi tip v cc hiu qu ca n : a)Hiu qu ca hi tip vi li ton th b)Hiu qu ca hi tip i vi tnh n nh c)Hiu qu ca hi tip i vi nhy d)Hiu qu ca hi tip i vi nhiu ph ri t bn ngoi Trong nhng th d trn, vic s dng hi tip ch vi ch ch tht n gin, gim thiu s sai bit gia tiu chun tham kho a vo v tn hiu ra ca h thng. Nhng, nhng hiu qu c ngha ca hi tip trong cc h thng iu khin th su xa hn nhiu. S gim thiu sai s cho h thng ch l mt trong cc hiu qu quan trng m hi tip c tc ng ln h thng. Phn sau y, ta s thy hi tip cn tc ng ln nhng tnh cht ca h thng nh tnh n nh, nhy, li, rng bng tn, tng tr.

H.1_9: H thng c hi tip. Xem mt h thng c hi tip tiu biu nh (H.1_9). Trong r l tn hiu vo. C l tn hiu ra. G v H l cc li.

a) Hiu qu ca hi tip i vi li ton th (overall Gain). So vi li ca h vng h (G), li ton th ca h vng kn (c hi tip) c thm h s 1+GH. Hnh H.1_9 l h thng hi tip m, tn hiu hi tip b c du (-). Lng GH t n c th bao gm du tr. Do , hiu qu tng qut ca hi tip l lm tng hoc gim li. Trong mt h iu khin thc t, G v H l cc hm ca tn s f. Sut c th ln hn 1 trong mt khong tn s no v nh hn 1 mt khong tn s khc . Nh vy, hi tip s lm tng li h thng trong mt khong tn s nhng lm gim n khong tn s khc. b) Hiu qu ca hi tip i vi tnh n nh. Ni mt cch khc khng cht ch lm, mt h thng gi l bt n khi output ca n thot khi s kim sot hoc l tng khng gii hn. Xem phng trnh (1.1). nu GH = -1, output ca h thng s tng n v hn i vi bt k input hu hn no. Nh vy, c th ni rng hi tip c th lm mt h thng (m lc u n nh) tr nn bt n. Hi tip l mt thanh gm 2 li. Nu dng khng ng cch, n s tr nn tai hi. Nhng cng c th chng t c rng, mi li ca hi tip li l to c s n nh cho mt h thng bt n. Gi s h thng hi tip (H.1_9) bt n v GH = -1. By gi, nu ta a vo mt vng hi tip m na, nh (H.1_10) .

li ton th ca h thng by gi s l :

(1.2) Nu do nhng tn cht ca G v H lm cho vng hi tip trong bt n, v G.H = -1. nhng ton th h thng c th vn n nh bng cch chn la li F ca vng hi tip ngoi. c) Hiu qu ca hi tip i vi nhy. (Sensibility) nhy thng gi mt vai tr quan trng trong vic thit k cc h thng iu khin. V cc thnh phn vt l c nhng tn cht thay i i vi mi trng xung quanh v vi tng thi k , ta khng th lun lun xem cc thng s ca h thng hon ton khng i trong sut ton b i sng hot ng ca h thng. Th d, in tr dy qun ca mt ng c in thay i khi nhit tng trong lc vn hnh. Mt cch tng qut, mt h iu khin tt s phi rt nhy i vi s bin i ca cc thng s ny c th gi vng p ng ra. Xem li h thng (H.1_9). Ta xem G nh l mt thng s c th thay i. nhy ton h thng c nh ngha nh sau:

M: li ton h thng. Trong : (M ch s thay i thm ca M G.(M/M v (G/G ch phn trm thay i ca M v G. Ta c:

(1.4) H thc ny chng t hm nhy c th lm nh tu bng cch tng GH, min sao h thng vn gi c s n nh. Trong mt h vng h, li ca n s p ng kiu mt - i - mt i vi s bin thin ca G. Mt cch tng qut, nhy ton h thng ca mt h hi tip i vi nhng bin thin ca thng s th tu thuc vo ni ca thng s . Ngi c c th khai trin nhy ca h thng (H.1_9) theo s bin thin ca H. d) Hiu qu hi tip i vi nhiu ph ri t bn ngoi. Trong sut thi gian hot ng, cc h thng iu khin vt l chu s ph ri ca vi loi nhiu t bn ngoi. Th d, nhiu nhit (thermal noise) trong cc mch khuch i in t, nhiu do tia la in sinh t chi v c gp trong cc ng c in Hiu qu ca hi tip i vi nhiu th tu thuc nhiu vo ni m nhiu tc ng vo h thng. Khng c kt lun tng qut no. Tuy nhin, trong nhiu v tr, hi tip c th gim thiu hu qu ca nhiu. Xem h thng (H.1_11)

Hnh H.1_11

Ouput ca h c th c xc nh bng nguyn l chng cht (super position) (hinh 1.5) - Nu khng c hi tip, H = 0 th output e = r T s tn hiu trn nhiu (signal to noise ratio) c nh ngha:

(1.6) tng t s S/N hin nhin l phi tng G1 hoc e/n. S thay i G2 khng nh hng n t s. - Nu c hi tip, output ca h thng khi r v n tc ng ng thi s l :

(1.7) So snh (1.5) v (1.7), ta thy thnh phn do nhiu ca (1.7) b gim bi h s 1+ G1G2 H. Nhng thnh phn do tn hiu vo cng b gim cng mt lng. T s S/N by gi l:

(1.8) V cng bng nh khi khng c hi tip. Trong trng hp ny, hi tip khng c hiu qu trc tip i vi t s S/N ca h thng. Tuy nhin , s p dng hi tip lm ny ra kh nng lm tng t s S/N di vi iu kin. Gi s rng sut G1 tng n G1v r n r, cc thng s khc khng thay i , output do tn hiu vo tc ng ring (mt mnh) th cng bng nh khi khng c hi tip. Ni cch khc ta c :

(1.9) Vi s tng G1, G1 output do nhiu tc ng ring mt mnh s l:

(1.10) Nh hn so vi khi G1 khng tng. By gi t s S/N s la:

(1.11). Nhn thy n ln hn h thng khng hi tip bi h s (1+ G1G2H) Mt cch tng qut, hi tip cng gy hiu qu trn cc tnh cht ca h thng, nh rng dy tn, tng tr, p ng qu ( Transient Response) v p ng tn s. III.CC LOI H THNG IU KHIN T NG. III.1 H t iu khin tuyn tnh v phi tuyn tnh III.2 H thng c thong s thay i v khng thay i theo thi gian III.3 H iu khin d liu lien tc III.4 H iu khin d liu gin on III.5 Chnh c t ng C nhiu cch phn loi h thng iu khin. Nu da vo phng php phn tch , thit k th chng gm cc loi tuyn tnh, phi tuyn thay i theo thi gian (time varying ), khng thay i theo thi gian (time invariant). Nu da vo loi tn hiu trong h thng th chng gm cc loi d liu lin tc( continous data), d liu gin on (discrete data), bin iu v khng bin iu. Nu da vo loi ca cc thnh phn ca h thng , th chng gm c cc loi in c , thy lc, kh ng . Ty vo mc ch chnh ca h m ngi ta xp loi chng nh kiu no . 1. H t iu khin tuyn tnh v phi tuyn. Ni mt cch cht ch, cc h thng tuyn tnh u khng c trong thc t . V mi h thng vt l u phi tuyn. H iu khin hi tip tuyn tnh ch l m hnh l tng ha lm n gin vic phn tch v thit k. Khi ln ca tn hiu ca h c gii hn trong mt vng m cc thnh phn biu l tnh thng ( ngha l nguyn l chng cht p dng c ) th h thng c xem l tuyn tnh . Nhng khi tn hiu vt qu vng hot ng tuyn tnh, ty vo s nghim ngt ca tnh phi tuyn, h thng s khng c xem l

tuyn tnh na. Th d : cc mch khuch i c dng trong h iu khin thng bo ha khi tn hiu a vo chng tr nn qu ln. T trng ca mt motor thng c tnh bo ha. Hiu ng phi tuyn thng gp trong cc h iu khin l vng cht (dead zone ) gia cc bnh rng ; tnh phi tuyn ca l xo ; lc ma st phi tuyn . Vi cc h tuyn tnh, c mt s phong ph v cc k thut gii tch v ha gip cho vic thit k c d dng. Cn trong cc h phi tuyn , mt liu php(treat ) ton hc thng l rt kh. V khng c phng php tng qut c th gii quyt mt s ln cc h phi tuyn. 2. H thng c thng s thay i v khng thay i theo thi gian. Khi cc thng s ca mt h iu khin c gi nguyn khng thay i trong sut thi gian hot ng ca n, th h c gi l h khng thay i theo thi gian ( time invariant). Trong thc t , hu ht cc h thng vt l u cha nhng thnh phn c thng s b tri, hay thay i theo thi gian. Th d : in tr dy qun ca mt ng c in s thay i khi t0 gia tng. Th d khc, h thng iu khin ng i ca ha tin, trong khi lng ca ha tin gim do s tiu th trn ng bay. Mc d mt h c thng s thay i theo thi gian khng phi tuyn th vn l mt h tuyn tnh, nhng s phn tch v thit k loi h ny thng l rt phc tp so vi cc h tuyn tnh c thng s khng thay i . 3. H iu khin d liu lin tc . Mt h iu khin s liu lin tc l mt h trong cc tn hiu nhng thnh phn khc ca h l cc hm lin tc ca bin s thi gian t. Trong cc h iu khin s liu lin tc, cc tn hiu c th l AC hoc DC. Khng ging trong nh ngha tng qut ca AC v DC dng trong k thut in, AC v DC ca h iu khin mang ngha chuyn bit. Khi ni mt h iu khin AC, c ngha l cc tn hiu trong c bin iu bi mt kiu bin iu no , v khi ni mt h iu khin DC, c ngha l tn hiu ca n khng bin iu nhng chng vn l tn hiu AC. 4. H iu khin d liu gin on. L h c tn hiu khng lin tc . a) Nu tn hiu c dng mt lot chui xung (pulse train ), th h c gi l h d liu mu ha ( sample data system ). b) Nu tn hiu l xung c m ha s thch hp cho vic s dng digital computer th gi l h iu khin digital. Th d: H iu khin my nh ch in t l mt h iu khin digital, v b x l nhn v cho ra cc s liu digital. Mt cch tng qut, mt h d liu mu ha ch nhn s liu v thng tin mt cch ngt qung ti nhng thi im ring. Th d: tn hiu sai s trong h c th

c cung cp ngt qung di dng xung. Nh vy h s khng nhn thng tin v sai s sut trong giai on gia hai xung lin tip.

H.1_12 : S khi mt h iu khin d liu mu ha. Mt tn hiu vo lin tc r(t) c a vo h thng. Tn hiu sai s e(t) c ly mu ( sampling). Ng ra ca b phn ly mu ( sampler) l m?t lot xung. Tn s ly mu c th u hay l khng. Hnh H.1_13 l s khi c bn ca h thng iu khin digital hng dn qu o tn la autopilot t tm mc tiu.

H.1_13 : S khi c bn ca h thng iu khin qu o tn la t tm mc tiu. 5. Chnh c t ng ( servomechanism). Mt loi h thng iu khin ng c c bit lu tm do tnh thnh hnh ca n trong k ngh v ngn ng iu khin hc. l servomechanism. Mt servomechanism l mt h iu khin t ng, trong bin s kim sot C l v tr c hc, hoc o hm theo thi gian ca v tr( vn tc hay gia tc). Th d : Xem mt b iu khin t ng ng m van nc.

H.1_14: Servo mechanism iu khin van. Ng vo ca h thng l mt bin tr loi quay P1, c u vi ngun in. Chn th 3( con chy) c quy chun theo v tr gc ( radians) v u vo mt ng vo ca mch khuch i servo. Mch khuch i ny cung cp in th cho mt ng c in gi l servo motor. Trc ca motor c truyn ( c kh ) n mt van m hay kha nc. Nu trc motor quay 3600 th van m hon ton. P2 gi l bin tr hi tip. Chn th 3 c ni ( c kh ) vi trc motor nh mt bnh rng v u ( in ) vi ng vo th hai ca mch khuch i servo. Ty v tr con chy ca hai bin tr, m in th sai bit e c th dng, m hay bng zero. in th ny c khuch i, sau t vo motor iu khin motor quay theo chiu m van, ng van hay vn gi van v tr c ( e= 0; khi o motor khng quay). Gi s van ang ng, ta quay P1 mt gc ( t mt tiu chun tham kho ng vo ). in th e mt cn bng ( khc 0), lm cho motor quay mt gc (thch ng vi gc quay ca con chy P1 ) lm van m. ng thi, qua b bnh rng truyn ng , con chy P2 cng quay mt gc sao cho in th sai bit e tr v 0 (motor khng quay ). Van c gi m y. H thng trn c trnh by bng s khi nh sau :

H.1_15 : S khi servomechamism iu khin van.

Mt s th d : 1. Xem mt cu phn th nh hnh v. Output l v2 v input l v1. Mch th ng ny c th m hnh ha nh l mt h vng h hoc nh mt h vng kn.

H.1_16 a. T cc nh lut Kirchhoff, ta c : v2 = R2. i i= v1/ (R1 + R2 ) Vy v2 =( R2 / (R1 + R2 )).v1= f(v1,R1,R2)

b. Nu bit dng i di dng khc hn: i = ( v1-v2 ) / R1 th: v2 = R2 ( v1 v2 ) / R1 = v1 . R2 / R1 v2 .R2 /R1 = f (v1, v2, R1, R2 )

2. H thng t iu khin tay ngi chm n mt vt, c th nhn dng nh sau : cc b phn chnh ca h l c, cnh tay, bn tay v mt.

B c gi tn hiu thn kinh n cnh tay. Tn hiu ny c khuch i trong cc bp tht ca cnh tay v bn tay, v xem nh cc tn hiu tc ng ca h thng. Mt dng nh b cm bin, hi tip lin tc v tr ca cnh tay v v tr vt n c. V tr tay l output ca h, v tr vt l input. Mc ch ca h iu khin l thu nh khong cch ca v tr tay v v tr vt n zero.

H.1_20 3. nh lut cung cu ca kinh t hc c th c xem nh mt h iu khin t ng. Gi bn ( gi th trng ) ca mt hng ha no l output ca h. Mc tiu ca h l gi cho gi n nh. nh lut cung cu cho rng gi th trng n nh nu v ch nu cung bng cu. Ta chn 4 b phn chnh ca h thng l ngi cung, ngi cu, ngi nh gi th trng, hng ha c mua v bn. Input l s n nh ca vt gi, hay tin li hn, l s nhiu lon gi bng zero. Output l gi thc t ca th trng. S hot ng ca h thng c giai thch nh sau : Ngi nh gi nhn mt tn hiu (zero) khi vt gi n nh. ng ta nh mt gi bn vi s gip ca nhng thng tin t tr nh hay gi biu ca s giao dch trc . Gi ny lm ngi cung sn xut a vo th trng mt lng hng ha no , v ngi cu mua mt s trong s . S chnh lch (sai s ) gia cung v cu c iu chnh bi h thng ny. Nu cung khng bng cu, ngi nh gi s thay i gi th trng theo hng sau cho cung bng vi cu. Vy c cung v cu u c th xem l hi tip v chng xc nh tc ng kim sot . H thng c biu din nh H.1_21.

H.1_21 ************* Ging vin: Phm Vn Tn

C S T NG HCChng II

HM CHUYN V S KHI CA H THNGNI DUNG: 2.1) i cng. 2.2) p ng xung lc v hm chuyn 2.3) S khi

I. I CNG Bc quan trng th nht trong vic thit k mt h iu khin l vic miu t ton hc v m hnh ha (modeling) cho thit b c kim sot. Mt cch tng qut, nhng c tnh ng ca thit b ny s c xc nh trc bng mt tp hp cc bin. Th d, xem mt ng c in trong h thng iu khin. Ta phi xc nh in p t vo, dng in trong cun dy qun, moment c khai trin trn trc, gc di v vn tc ca rotor, v nhng thng s khc na nu cn thit .Tt c nhng thng s y c xem nh cc bin ca h. Chng lin h nhau thng qua nhng nh lut vt l c thit lp v a n cc phng trnh ton hc di nhiu dng khc nhau. Ty bn cht ca thit b, cng nh iu kin hot ng ca h, mt vi hoc tt c cc phng trnh y l tuyn tnh hay khng, thay i theo thi gian hay khng, chng cng c th l cc phng trnh i s, phng trnh vi phn hoc tng hp. Cc nh lut vt l khng ch nguyn tc hot ng ca h iu khin trong thc t thng l rt phc tp. S c trng ha h thng c th i hi cc phng trnh phi tuyn v/hoc thay i theo thi gian rt kh gii. Vi nhng l do thc t, ngi ta c th s dng nhng gi nh v nhng php tnh xp x , nghin cu cc h ny vi l thuyt h tuyn tnh. C hai phng cch tng qut tip cn vi h tuyn tnh. Th nht, h cn bn l tuyn tnh, hoc n hot ng trong vng tuyn tnh sao cho cc iu kin v s tuyn tnh c tha. Th hai, h cn bn l phi tuyn, nhng c tuyn tnh ha xung quanh im hot ng nh mc. Nhng nn nh rng, s phn tch cc h nh th ch kh dng trong khong cc bin m s tuyn tnh cn gi tr. II. P NG XUNG LC V HM CHUYN. II.1) p ng xung lc(impulse). II.2) Hm chuyn ca h n bin II.3) Hm chuyn ca h a bin. 1. p ng xung lc(impulse). Mt h tuyn tnh, khng i theo thi gian c th c c trng bng p ng xung lc g(t) ca n. chnh l output ca h khi cho input l mt hm xung lc n v ((t). Hm xung lc d (t) = 0 ; t 0 .

d (t) ; t = 0 .

Tnh cht th ba l tng din tch trn xung lc l mt. V tt c din tch ca xung lc th tp trung ti mt im, cc gii hn ca tch phn c th di v gc m khng lm thay i tr gi ca n.

C th thy rng tch phn ca ((t) l u(t) (hm nc).

Mt khi p ng xung lc ca h c bit, th output c(t) ca n vi mt input r(t) bt k no c th c xc nh bng cch dng hm chuyn. 2. Hm chuyn ca h n bin. Hm chuyn (transfer function) ca mt h tuyn tnh khng thay i theo thi gian, c nh ngha nh l bin i Laplace ca p ng xung lc ca n,

vi cc iu kin u l zero. t G(s) l hm chuyn vi r(t) l input v c(t) l output. G(s)= L [g(t)] (2.1)

(2.2) Trong : #9; R(s)= L [r(t)] (2.3) &&C(s)= L [c(t)] (2.4) Vi tt c cc iu kin u t zero. Mc d hm chuyn c nh ngha t p ng xung lc, trong thc t s tng quan gia input v output ca h tuyn tnh khng thay i theo thi gian vi d liu vo lin tc, thng c miu t bng phng trnh vi phn thch hp, v dng tng qut ca hm chuyn c suy trc tip t phng trnh vi phn . Xem phng trnh vi phn vi h s thc hng, m t s tng quan gia input v output ca h tuyn tnh khng thay i theo thi gian.

Cc h s a1,a2,..an v b1, b2bn l hng thc vn(m. Mt khi r(t) vi t(to v nhng iu kin u ca c(t) v cc o hm ca n c xc nh ti thi im u t=t0, th output c(t) vi t(t0 s c xc nh bi phng trnh (2.5). Nhng, trn quan im phn gii v thit k h thng, phng php dng phng trnh vi phn m t h thng th rt tr ngi. Do , phng trnh (2.5) t khi c dng trong dng ban u phn tch v thit k. Thc quan trng nh rng, mc d nhng chng trnh c hi?u qu trn my tnh digital th cn thit gii cc phng trnh vi phn bc cao, nhng trit l cn bn ca l thuyt iu khin h tuyn tnh l: cc k thut phn gii v thit k s trnh cc li gii chnh xc ca h phng trnh vi phn, tr khi cc li gii trn my tnh m phng c i hi.

c hm chuyn ca h tuyn tnh m t bi phng trnh (2.5) , ta ly bin i Laplace c hai v, vi s gi nh cc iu kin u l zero. (Sn+anSn-1++a2S+a1)C(S)=(bm+1Sm+bmSm-1++b2S+b1)R(S) (2.6) Hm chuyn: (2.7) ( C th tm tt cc tnh cht ca hm chuyn nh sau: *Hm chuyn ch c nh ngha cho h tuyn tnh khng thay i theo thi gian. * Hm chuyn gia mt bin vo v mt bin ra ca h c nh ngha l bin i Laplace ca p ng xung lc. Mt khc, hm chuyn l t s ca bin i Laplace ca output v input. * Khi xc nh hm chuyn, tt c iu kin u u t zero. * Hm chuyn th c lp vi input ca h. * Hm chuyn l mt hm bin phc S. N khng l hm bin thc theo thi gian, hoc bt k mt bin no c dng nh mt bin c lp. * Khi mt h thuc loi d liu vo digital, vic m t n bng cc phng trnh vi phn s tin li hn. V hm chuyn tr thnh mt hm bin phc Z. Khi , bin i Z s c s dng. 3. Hm chuyn ca h a bin. nh ngha ca hm chuyn d c m rng cho mt h thng vi nhiu input v nhiu output. Mt h nh vy c xem l h a bin. Phng trnh (2.5) cng c m t s tng quan gia cc input v output ca n. Khi xt s tng quan gia mt input v mt output, ta gi s cc input khc l zero. Ri dng nguyn l chng cht (super position) cho mt h tuyn tnh, xc nh mt bin s ra no do hu qu ca tt c cc bin vo tc ng ng thi, bng cch cng tt c cc output do tng input tc ng ring l. Mt cch tng qut, nu mt h tuyn tnh c p input v c q output, hm chuyn gia output th i v input th j c nh ngha l:

Gij(s) =

(2.8)

Vi Rk(s)=0 ; k=1,2...p ; k (j Lu :phng trnh (2.8) ch c nh ngha vi input th j, cc input khc u zero. Nu cc input tc ng ng thi, bin i Laplace ca output th i lin h vi bin i Laplace ca tt c cc input theo h thc . Ci(s) =Gi1(s).R1(s)+ Gi2(s).R2(s)+....+Gip(s).Rp(s)

v Gij(s) xc nh bi phng trnh (2.8) Tht tin li, nu din t phng trnh (2.9) bng mt phng trnh ma trn: C(s) = G(s). R(s) (2.10) Trong (2.11) L mt ma trn qx1, gi l vector output.

L mt ma trn px1, gi l vector input.

L mt ma trn qxp, gi l ma trn chuyn (transfer matrix) Xem mt th d v mt h a bin n gin ca mt b iu khin ng c DC

Cc phng trnh cho bi :

Trong : v(t): in p t vo rotor i(t) : Dng in tng ng ca rotor. R : in tr ni cun dy qun rotor. L : in cm ca rotor. J : Qun tnh ca rotor. B : H s ma st. T(t): moment quay. TL(t): moment ph ri, hoc ti (moment cn). ((t): Vn tc ca trc motor. Moment ca motor lin h vi dng rotor bi h thc : T(t)=Ki.i(t) (2.16) Trong , Ki : l hng s moment tm hm chuyn gia cc input (l v(t) v TL(t)) v output (l ((t)), ta ly bin i Laplace hai v cc phng trnh (2.14) n (2.16). Gi s iu kin u l zero. V(s) = (R + LS) I(s) (2.17) T(s)= (B + JS) W (s) + TL(s) (2.18) T(s)= KI .I(s) (2.19)

Phng trnh ny c th vit li : C(s)= G11(s).R1(s) + G12(s).R2(s) (2.21) Trong C(s) = ((s) ; R1(s) = V(s) ; R2(s) = TL(s)

9; 9; G11(s) c xem nh hm chuyn gia in th vo v vn tc motor khi moment ti l zero. G12(s) c xem l hm chuyn gi moment cn v vn tc motor khi in th vo l 0 . III. S KHI ( block diagram ) II.1) . S khi ca mt h thng iu khin . III.2). S khi v hm chuyn ca h thng a bin. III.3) Nhng nh l bin i s khi. III.4) Thu gn cc s khi phc tp. Trong cc h iu khin phc tp, vic v s chi tit i hi nhiu thi gian. V vy, ngi ta hay dng mt k hiu gn gng gi l s khi. S t hp s khi v hm chuyn ca h s trnh by bng hnh v s tng quan nhn qu gia input v output. Chn hn, s khi H.2_1 biu din phng trnh: C(s)= G(s)R(s).

Mi tn trn s khi minh th rng, s khi c tnh nht hng (unilateral), tn hiu ch c th truyn theo chiu mi tn. Mc d mi h thng n bin c th trnh by bng mt khi duy nht gia input v output, nhng s tin li ca nim v s khi nm ch: n c th din t nhng h a bin v gm nhiu b phn m hm chuyn ca chng c xc nh. Khi ton b h thng c trnh by bi s ghp nhiu khi ca cc b phn ring r, sao cho s tham gia ca chng vo hnh trng chung ca h c lng gi . Nu cc h thc ton hc ca cc b phn y c bit, th s khi c th c dng tham kho cho li gii gii tch hoc cho my tnh. Xa hn na, nu tt c cc b phn ca h u tuyn tnh, hm chuyn cho ton b h thng c th tm c bng cch dng nhng php tnh i s v s khi. Mt im rt cn bn cn lu , s khi c th dng biu din cho cc h tuyn tnh cng nh phi tuyn. Hy tr li th d v ng c DC trn. H.2_2a: b phn khuch i th phi tuyn. Motor c gi s tuyn tnh hay hot ng vng tuyn tnh. Nhng tnh cht ng ca n biu din bng phng trnh (2.20). H.2_2b: cng h thng trn nhng b phn khuch i th tuyn tnh.

Lu l H.2_2a, v b khuch i l phi tuyn, nn khng c hm chuyn gia ng vo v ng ra ca n. Gi s chng ch c th xc nh bng h thc lin h gia hai bin vi(t) v v(t) m thi. Ngc li, H2_2b, hm chuyn gia ng vo v ng ra ca b khuch i l K. V , V(s)=K.Vi(s). 1. S khi ca mt h thng iu khin . Mt thnh phn c dng nhiu trong cc s khi ca h iu khin, l b cm bin (sensing device), n ng vai tr so snh tn hiu v thc hin vi thut ton n gin nh cng, tr, nhn v i khi t hp ca chng. B cm bin c th l mt bin tr, mt nhit tr hoc mt linh kin chuyn nng khc (transducer), cng c th l mt mch khuch i vi sai, mch nhn ...

S khi ca cm bin trnh by H.2_3a,b,c,d. + H.2_3a,b,c: mch cng tr th tuyn tnh. Nn cc bin ng vo v ra c th l bin theo t hoc s ( bin i Laplace ). e(t) = r(t) -c(t) (2.22) hoc E(s)=R(s)-C(s) (2.23)

H.2_3d, mch nhn th phi tuyn, nn lin h gia input v output ch c th phm vi thi gian (Time domain). Ngha l, e(t)=r(t).c(t) (2.24) Trong trng hp ny s khng a n E(s)=R(s) .C(s). C th dng nh l chp phc (complexe_convolution) ca bin i Laplace a (2.24) n : E(s)=R(s)*C(s) (2.25)

( Mt h t iu khin tuyn tnh c th c trnh by bng s khi chnh tc nh H.2_4. Trong : r(t), R(s): tn hiu tham kho vo. c(t), C(s): bin s c kim sot ng ra. b(t), B(s): tn hiu hi tip. e(t), E(s): tn hiu sai bit ( error ). : Hm chuyn vng h hoc hm chuyn ng trc tip (forward path). : Hm chuyn vng kn, hoc t s iu khin . 9; H(s): Hm chuyn hi tip (feedback transfer ) G(s).H(s): Hm chuyn ng vng (loop transfer)

T H.2_4 ta c : C(s)=G(s).E(s) (2.26) E(s)=R(s) B(s) (2.27) B(s)=H(s).C(s) (2.28) Th (2.27) vo (2.26):

C(s)=G(s).R(s)-G(s).B(s) (2.29) Thay (2.28) vo (2.29): C(s)=G(s)R(s)-G(s).H(s)C(s) (2.30) T phng trnh cui cng suy ra hm chuyn li vng kn:

2. S khi v hm chuyn ca h thng a bin. H.2_5 trnh by s khi nhiu bin, vi p input v q output.

H.2_5b c dng nhiu v n gin. S nhiu input v output c biu din bng vector . H.2_6 ch s khi dng chnh tc ca h thng a bin.

Hm chuyn c suy bng cch dng php tnh i s cc ma trn. C(s) = G(s). E(s) (2.32)

E(s) = R(s) - B(s) (2.33) B(s) = H(s). C(s) (2.34) : C(s) l ma trn qx1: vector output E(s), B(s), R(s): u l ma trn px1 G(s) v H(s) l ma trn qxp v pxq : ma trn chuyn. Thay (2.34) vo (2.33) v ri thay (2.33) vo (2.32) : C(s)=G(s). R(s) G(s). H(s).C(s) (2.35) Gii C(s) t (2.35) : C(s)=[ I + G(s). H(s)]-1. G(s). R(s) (2.36) Gi s I + G(s). H(s) khng k d (non singular). Nhn thy rng s khai trin tng quan vo ra y cng tng t nh h n bin. Nhng y khng th ni v t s C(s)/ R(s), v chng u l cc ma trn. Tuy nhin, vn c th nh ngha ma trn chuyn vng kn nh sau: M(s) = [ I + G(s). H(s)]-1. G(s) (2.37) Phng trnh (2.36) c vit li : C(s) = M(s). R(s) (2.38) Th d 2.1: Xem ma trn hm chuyn ng trc tip v ma trn hm chuyn hi tip ca h H.2_6 l :

Ma trn hm chuyn vng kn c cho bi phng trnh (2.37) v c tnh nh sau:

3.Nhng nh l bin i s khi. a. Cc khi ni tip. Mt s hu hn bt k cc khi ni tip c th kt hp bi mt php nhn i s. l, n khi vi hm chuyn tng ng G1,G2,..Gn mc ni tip th tng ng mt khi duy nht c hm chuyn l G cho bi:

Th d 2.2:

Php nhn ca hm chuyn th giao hon : Gi.Gj=Gj.Gi (2.45) Vi mi i,j.

b. Cc khi song song: n khi vi hm chuyn tng ng G1,G2,,Gn mc song song th tng ng mt khi duy nht c hm chuyn G cho bi:

c. Bng bin i s khi . S khi ca h iu khin phc tp c th n gin ha bng cch dng cc bin i. Trong bng sau y, ch P c dng ch mt hm chuyn bt k v W, X, Y, Z ch nhng tn hiu trong phm vi tn s s.

4. Thu gn cc s khi phc tp. S khi ca cc h t iu khin thc t th thng rt phc tp. c th a v dng chnh tc, cn thu gn chng li. K thut thu gn, c th theo cc bc sau y : - Bc 1: kt hp tt c cc khi ni tip, dng bin i 1. - Bc 2: kt hp tt c cc khi song song, dng bin i 2. - Bc 3: gim bt cc vng hi tip ph, dng bin i 4. - Bc 4: di cc im tng v bn tri v cac im ly v bn phi vng chnh, dng bin i 7, 10 v 12. - Bc 5: lp li cc bc t 1-> 4, cho n khi c dng chnh tc i vi mt input no . - Bc 6: lp li cc bc t 1-> 5 i vi cc input khc nu cn . Cc bin i 3, 5, 6, 8, 9 v 11 i khi cng cn n . Th d 2.3 : Hy thu gn s khi sau y v dng chnh tc.

Th d 2.4 : Hy thu gn s khi th d trn bng cch c lp H1 ( H1 ring)

.

Th d 2.5 : Hy thu gn h sau y v dng h iu khin hi tip n v.

Mt thnh phn phi tuyn ( trn ng truyn thng ) khng th thu gn nh bin i 5 c. Khi tuyn tnh trn ng hi tip c th kt hp v khi tuyn tnh ca ng truyn thng. Kt qu l:

Th d 2.6 : Hy xc nh output C ca h nhiu input sau y :

Cc b phn trong h u tuyn tnh, nn c th p dng nguyn l chng cht . - Cho u1=u2=0. S khi tr nn.

CR l output ch do s tc ng ring ca R. t phng trnh (2.31

- Cho R=u2=0, S khi tr nn :

C1 l p ng ch do s tc ng ring ca u1. Sp xp li cc khi :

Vy:

Cho R=u1=0. S khi tr nn :

C2 l p ng do tc ng ring ca u2 .

Vy:

Th d 2.7: S khi sau y l mt v d v h nhiu input v nhiu output. Hy xc nh C1 v C2.

a)Trc ht b qua C2. Xt h thng vi 2 input R1 ,R2 v output C1.

Nh vy, C11 l output C1, ch do R1 gy ra.

t R1=0:

Vy: b. By gi, b qua C1. Xt h thng vi 2 input R1,R2 v output C2.

Vy : - t R2=0.

Cui cng: C2 =C21+C22 . *****************

Ging vin: Phm Vn Tn

C S T NG HCCHNG III

HNH TRUYN TN HIUNI DUNG: I. I CNG II. NHNG NH NGHA III. TM LC NHNG TNH CHT C BN CA TTTH IV. I S HC V HNH TRUYN TN HIU V.CCH V HNH TRUYN TN HIU VI. CNG THC MASON VII. P DNG CNG THC MASON VO S KHI I. I CNG. hnh truyn tn hiu ( signal flow graph - HTTH) c gii thiu u tin bi S.J. MASON c xem nh l k hiu n gin ha ca s khi, trnh by mi tng quan nhn qu ca mt h tuyn tnh. Bn cnh s khc bit v hnh trng vt l gia HTTH v s khi, ta c th thy HTTH chc ch hn v nhng lin h ton hc. Nhng nhng nh lut dng cho s khi th mm do hn nhiu v km r rng hn. Mt HTTH c nh ngha nh l mt phng php ha miu t nhng lin h vo - ra gia cc bin ca mt tp hp nhng phng trnh i s. Xem mt h tuyn tnh c din t bi tp hp N phng trnh i s.

Hay n gin hn: Output =( ( li).(input) (3.3) hnh truyn tn hiu c v da vo tin quan trng nht ny. Trng hp h thng c m t bng cc phng trnh vi tch phn, trc nht ta phi bin i chng thnh cc phng trnh bin i Laplace v sp xp chng theo dng phng trnh (3.1).

(3.4) j=1,2,...N

Khi v HTTH , cc im ni hay l nt dng biu din cc bin yj hay yk . Cc nt c ni vi nhau bi cc on thng gi l nhnh, tu thuc vo cc phng trnh nhn qu. Cc nhnh c c trng bi li nhnh v chiu. Mt tn hiu ch c th truyn ngang qua nhnh theo chiu mi tn. Th d, xem mt h tuyn tnh c trnh bi bi phng trnh n gin. y2 =a12 .y1 (3.5) Trong , y1 l bin s vo , y2 l bin ra v a12 l li hay truyn dn (transmittansce) gia hai bin s. hnh truyn tn hiu biu din cho phng trnh (3.5) c v hnh H.3_1.

Chiu ca nhnh t nt y1 n nt y2 ch s ph thuc ca bin ra vi bin vo, v khng c ngc li. V th, mc d phng trnh(3.5) c th vit li:

(3.6) Nhng HTTH hnh H.3_1 khng a n mt tng quan nh vy. Nu phng trnh (3.6) c gi tr nh l mt tng quan nhn qu theo ngha vt l, th phi v mt HTTH khc. Mt th d khc, xem tp hp cc phng trnh i s : y2 = a12 y1 + a32 y3 y3 = a23 y2 + a43 y4 y4 = a24 y2 + a34 y3 + a44 y4 (3.7) y5 = a25 y2 + a45 y4 HTTH cho cc phng trnh ny c v tng bc nh hnh H.3_2. Cc nt biu din cc bin y1 , y2 , y3 , y4 v y5 c t theo th t t tri sang phi.

H.3_2. : HTTH ca h phng trnh (3.7) . II . NHNG NH NGHA.

1) Nt vo (ngun ) : Nt vo l mt nt ch c nhng nhnh ra. Th d nt y1 H.3_2 . 2) Nt ra : Nt ra l nt ch c nhng nhnh vo. Th d nt y5 H.3_2. Tuy nhin khng phi lc no cng c sn nt ra tha nh ngha trn. Th d HTTH hnh H.3_3a. khng c nt no ph hp nh ngha. Tuy nhin, c th xem y3 v/hoc y2 l nt ra nu ta a vo cc nhnh vi li n v cho cc bin y3 v y2 nh H.3_3b. Cc nt a thm vo gi l nt gi (dummy node).

. Mt cch tng qut ta c th thy rng, bt k mt nt no khng phi l nt vo u c th lm mt nt ra theo cch trn. Tuy nhin, ta khng th i mt nt khng phi l nt vo thnh mt nt vo theo cch tng t. Th d, nt y2 trong hnh H.3_3a khng phi l nt vo. Nhng nu ta c i n thnh nt vo bng cch thm nt gi nh H.3_4 th phng trnh m t tng quan ti nt y2 s l:

y2 = y2 + a12y1 + a32 y3 (3.8) Phng trnh ny khc vi phng trnh gc, c vit t hnh H.3_3a: y2 = a12 y1 + a32 y3 (3.9) Trng hp mun chn y2 l nt vo, ta phi vit li phng trnh nhn qu, vi kiu xp t : cc nguyn nhn nm bn v phi v hu qu nm bn v tri. Sp xp phng trnh (3.9) li, ta c hai phng trnh gc cho HTTH hnh H. 3_3 nh sau:

y3 = a32 y2 (3.11) HTTH cho hai phng trnh trn, v hnh H.3_5.

H.3_5: HTTH vi y2 l nt vo. 3) ng(path): L s ni tip lin tc theo mt hng ca cc nhnh , m dc theo n khng c mt nt no c i qua qu mt ln.

4) ng trc tip (forward path): L ng t nt vo n nt ra. Th d HTTH hnh H.3_2d, y1 l nt vo, v c 4 nt ra kh d : y2 , y3 , y4 v y5 . ng trc tip gia y1 v y2: l nhnh gia y1 v y2. C hai ng trc tip gia y1 v y3: ng 1, gm cc nhnh t y1 n y2 n y3. ng 2, gm cc nhnh t y1 n y2 n y4 (ngang qua nhnh c li a24) v ri tr v y3(ngang qua nhnh c li a43). Ngi c c th xc nh 2 ng trc tip t y1 n y4. Tng t, c 3 ng trc tip t y1 n y5. 5) Vng(loop): L mt ng xut pht v chm dt ti cng mt nt, dc theo n khng c nt no khc c bao qu mt ln. Th d, c 4 vng HTTH hnh H.3_2d.

6) li ng (path Gain) : Tch s li cc nhnh c nm trn mt ng. Th d, li ng ca ng y1- y2- y3- y4 trong hnh H.3_2d l a12 a23 a34. 7) li ng trc tip ( forward_path Gain) : li ng ca ng trc tip. 8) li vng (loop Gain) : li ng ca mt vng. Th du, li vng ca vng y2 - y3 - y4 - y2 trong hnh H.3_2d l a24 a43 a32. III. TM LC NHNG TNH CHT C BN CA HTTH. 1. HTTH ch p dng cho cc h tuyn tnh .

2. Cc phng trnh, m da vo v HTTH, phi l cc phng trnh i s theo dng hu qu l hm ca nguyn nhn. 3. Cc nt biu din cc bin. Thng thng, cc nt c sp xp t tri sang phi, ni tip nhng nguyn nhn v hu qu ngang qua h thng. 4. Tn hiu truyn dc theo nhnh, ch theo chiu mi tn ca nhnh. 5. Chiu ca nhnh t nt yk n yj biu din s ph thuc ca bin yj vo yk, nhng khng ngc li. 6. Tn hiu yk truyn dc mt nhnh gia nt yk v yj th c nhn bi li ca nhnh akj sao cho mt tn hiu akjyk nhn c ti nt yj . IV. I S HC V HNH TRUYN TN HIU. Da trn nhng tnh cht ca HTTH, ta c th tm lc nh sau: 1) Tr gi cu bin c biu din bng mt nt th bng tng ca tt c tn hiu i vo nt. Nh vy, i vi HTTH H.3_7, tr gi ca y1 bng tng ca cc tn hiu c truyn ngang qua mi nhnh vo : y1= a21 y2 + a31 y3 + a41 y4 + a51 y5 (3.12)

H.3_7: Nt nh l mt im tng, v nh l mt im pht .

2) Tr gi ca bin s c biu din bi mt nt th c truyn ngang qua tt c cc nhnh ri khi nt. Trong HTTH hnh H.3_7 , ta c : y6 = a16 y1 y7 = a17 y1 (3.13) y8 = a18 y1 3) Cc nhnh song song theo cng mt chiu gia hai nt c th c thay bi mt nhnh duy nht vi li bng tng cc li ca cc nhnh y. Th d hnh H.3_8.

4) S ni tip nhiu nhnh, nh hnh H.3_9, c th c thay bi mt nhnh duy nht vi li bng tch cc li nhnh.

V. CCH V HNH TRUYN TN HIU. 1) HTTH ca mt h t kim tuyn tnh m cc thnh phn ca n ch r bi cc hm chuyn th c th c v mt cch trc tip bng cch tham kho s khi ca h. Mi mt bin ca s khi s l mt nt. Mi khi s l mt nhnh. Th d 3.1: T s khi di dng chnh tc ca mt h thng t kim nh hnh H.3_10, ta c th v HTTH tng ng hnh H.3_11.

H.3_11 : HTTH tng ng ca h. Nh l du - hay + ca im tng th c kt hp vi H. T H.3_11, vit phng trnh cho tn hiu ti cc nt E v C :

v C(s) = G(s).E(s) (3.15) Hm chuyn vng kn : (hay t s iu khin)

(3.16) 2) i vi cc h c m t bng phng trnh vi phn, ta v HTTH theo cch sau y: a.Vit h phng trnh vi phn di dng : X1 = A11` X1 + A 12X2 + ... + A 1nXn X2 = A21X1 + A22X2 + ... + A2nXn (3.17)

. . . . . . . . . .. . X m= Am1 X1 + Am2X2 + ... + AmnXn Nu X1 l nt vo, th khng cn mt phng trnh cho n. b. Sp xp cc nt t tri sang phi sao cho khng gy tr ngi cho cc vng cn thit . c. Ni cc nt vi nhau bng cc nhnh A11, A12 ... d. Nu mun v mt nt ra, th thm nt gi c li nhnh bng 1 . e. Sp xp li cc nt v /hoc cc vng c mt hnh r rng nht. Th d 3.2 : Hy v HTTH cho mt mch in v hnh H.3_12 :

C 5 bin s : v1, v2, v3, i1 v i2 . Trong v1 bit. Ta c th vit 4 phng trnh c lp t cc nh lut Kirchhoff v th v dng.

(3.18)

t 5 nt nm ngang nhau vi v1 l mt nt vo, ni cc nt bng nhng nhnh. Nu mun v3 l mt nt ra, ta phi thm vo mt nt gi v li nhnh bng 1.

VI. CNG THC MASON. chng trc, ta c th rt gn cc s khi ca nhng mch phc tp v dng chnh tc v sau tnh li ca h thng bng cng thc:

V phn trn, ta cng c th dng hnh truyn tn hiu t tn th gi hn. V y, ta li c th dng cng thc Mason, nh l cng thc tnh li tng qut cho bt k mt hnh truyn tn hiu no.

9; 9;

(3.19)

li : yout/yin ; yout: bin ra, yin: bin vo. pi : li ng trc tip th i.

=1-( tng cc li vng)+(tng cc tch li 2 vng khng chm) - (tng cc tch li ca 3 vng khng chm)+.. (I = tr ca ( tnh vi cc vng khng chm vi cc ng trc tip th i. ( Hai vng, hai ng hoc 1 vng v 1 ng gi l khng chm (non_touching) nu chng khng c nt chung).

Th d : xem li HTTH ca 1 h iu khin dnh chnh tc H.3_11. Ch c mt ng trc tip gia R(s) v C(s). Vy : P1=G(s) P2=P3=...=0. - Ch c 1 vng . Vy: P11= G(s).H(s) Pjk=0; j 1, k 1. Vy, D =1-P11=1 G(s).H(s), Va, D 1=1-0=1 Cui cng,

9; R rng, ta tm li c phng trnh (3.16).

(3.20)

Th d : Xem li mch in VD3.2, m HTTH ca n v hnh H.3_13. Dng cng thc mason tnh li in th T= v3/v1.

. - Ch c mt ng trc tip. li ng trc tip:

9;- Ch c 3 vng hi tip. Cc li vng:

- C hai vng khng chm nhau (vng 1 v vng 3). Vy: P12 = tch li ca 2 vng khng chm nhau:

-Khng c 3 vng no khng chm nhau. Do : D =1- ( P11+ P21+ P31)+ P12 D=

V tt c cc vng u chm cc ng trc tip ( duy nht), nn:

Cui cng (3.21) VII. P DNG CNG THC MASON VO S KHI. Do s tng t gia S khi v HTTH, cng thc li tng qut c th c dng xc nh s lin h vo ra ca chng. Mt cch tng qut, t s khi ca 1 h tuyn tnh cho, ta c th p dng cng thc li tng qut MASON trc tip vo . Tuy nhin, c th nhn dng tt c cc vng v cc phn khng chm mt cch r rng, i khi cn n s gip ca HTTH. Vy cn v HTTH cho s khi trc khi p dng cng thc. Nu G(s) v H(s) l mt thnh phn ca dng chnh tc, th t cng thc Mason ta suy ra:

Hm chuyn ng trc tip G(s)= (3.22) Hm chuyn ng vng G(s).H(s) = ( - 1 (3.23) Th d: Xc nh t s iu khin C/R v dng chnh tc ca mt h iu kim th d 2.1.

- C 2 ng trc tip : P1 = G1.G2.G4 P2 = G1.G3.G4 - C 3 vng hi tip:

P11 = G1.G4.H1 P21 = - G1.G4.G2.H2 P31 = - G1.G3.G4.H2 D = 1 - ( G1.G4.H1 - G1.G2.H4.H2 - G1.G3.G4.H2) Khng c vng khng chm nhau, v tt c cc vng u chm vi cc ng trc tip. Vy : D 1= 1 ; D 2= 1 Do t s iu khin l:

(3.24) T phng trnh (3.23) v (3.24), ta c: G=G1G4(G2+G3) V: GH=G1G4(G3H2+G2H2-H1) (3.25) Vy: (3.26) S dng chnh tc c v hnh H.3_17.

Du tr im tng, l kt qu vic dng du cng trong cng thc tnh GH trn.

Th d: Xc nh t s iu khin (hoc hm chuyn vng kn) C/R ca mt h c s khi nh hnh H.3_18.

hnh truyn tn hiu ca h c v hnh H.3_19:

. C hai ng trc tip: P1= G1G2G3 ; P2= G1G4. C 5 vng hi tip: P11= - G1G2H1 ; P21= - G2G3H2 ; P31= - G4H2 ; P41= - G1G2G3 ; P51= - G1G4. Vy: D = 1- ( P11+ P21+ P31+ P41+ P51)

V (1 = (2 = 1. =>

BI TP CHNG III

3.1 : Hy xc nh t s C/R v dng s khi chnh tc ca mt h iu khin sau y: sau y: ;

3.2 : Xc nh hm chuyn cho s khi sau y, bng k thut dng HTTH:

3.3 : Xem TD2.4, gii bi ton bng HTTH.

3.4 : Tm hm chuyn C/R ca h thng sau y, vi k l hng s.

3.6 : Dng k thut HTTH gii bi tp 2.13. 3.7 : Tm C/R cho h iu khin sau y:

3.8 : V HTTH cho mch in sau:

3.9 : V HTTH cho mch in sau:

3.10 : V HTTH cho mch in sau, tnh li:

Gi : 5 bin v1, i1, v2, i2, v3. Vi v1 l input. Cn 4 phng trnh c lp.

GII BI TP CHNG III 3.1 : hnh truyn tn hiu:

Dng cng thc Mason xc nh C/R. C hai ng trc tip: P1= G1G2G4 ; P2=G1G3G4 C 3 vng: P11=G1G4H1; P21= - G1G2G4H2 ; P31= - G1G3 G4H2 Khng c vng khng chm. V tt c cc vng u chm c hai ng trc tip. Vy: D 1= 1 ; D 2= 1 Do , t s C/R:

Vi (= 1 - (P11+P21+P31). Suy ra:

T ( 3.25 ) v (3.26) , ta c: G = G1G4(G2 + G3) V : GH = G1G4(G3H2 +G2H2 - H1)

Dng chnh tc ca s khi ca h thng :

Du tr ti im tng l do vic dng du cng trong cng thc tnh GH trn. S khi trn c th a v dng cui cng nh trong VD2.1 bng cch dng cc nh l bin i khi. 3.2 : hnh truyn tn hiu v trc tip t s khi:

C hai ng trc tip, li l : P1 = G1G2G3 ; P2 = G4 C 3 vng hi tip, li vng l: P11 = - G2H1 ; P21 = G1G2H1 ; P31 = - G2G3H2 Khng c vng no khng chm, vy: ( = 1 - (P11 + P21 + P31) + 0 V (1 = 1 V c 3 vng u chm vi ng 1. V khng c vng no chm vi cc nt ng trc tip th nh, nn: (2= ( ( C 3 vng u khng chm vi ng trc tip th 2). Vy:

3.3 : HTTH v trc tip t s khi.

P1 = G1G2 ; P11 = G1G2H1H2 D = 1- P11 ; D 1 = 1 Vy:

Vi u2 = R =0, Ta c:

P1 = G2 ; P11 = G1G2H1H2 D = 1 - G1G2H1H2 ; D1=1

Vi R = u1 = 0

P1 = G1G2H1 ; P11 = G1G2H1H2 D = 1 - P11 ; D 1 = 1

3.4 :

3.5 : HTTH v trc tip t s khi:

-

3.6 :

3.7 : HTTH v t s khi:

c hai ng trc tip: P1= G1G2G3 ; P2 = G1G4 c 5 vng hi tip: P11 = G1G2H1 ; P21 = G2G3H2 ; P31 = - G1G2G3 P41 = G4H2 ; P51 = - G1G4 D = 1 - (P11 + P21 + P31 + P41 + P51) ; D 1 = D 2 = 1 Cui cng:

3.10 : 5 bin v1, i1, v2, i2, v3. Vi v1 l input, cn 4 phng trnh c lp.

li: 9; Tnh theo cng thc Mason. *********** Ging vin: Phm Vn Tn

C S T NG HCChng IV

TRNG THI CA H THNGNI DUNG : 4.1) i cng 4.2) Phng trnh trng thi v phng trnh output 4.3) S din bin bng ma trn ca phng trnh trng thi 4.4) Vi v d 4.5) M hnh trng thi

I.

I CNG. Trong cc chng trc, ta kho st vi phng php thng dng phn gii cc h t kim. Php bin i Laplace c dng chuyn cc phng trnh vi phn m t h thng thnh cc phng trnh i s theo bin phc S. Dng phng trnh i s ny ta c th tm c hm chuyn m t tng quan nhn qu gia ng vo v ng ra. Tuy nhin, vic phn gii h thng trong min tn s, vi bin phc, d l k thut rt thng dng trong t ng hc, nhng c rt nhiu gii hn. S bt li ln nht, l cc iu kin u b b qua. Hn na, phng php y ch c p dng cho cc h tuyn tnh, khng i theo thi gian. V n c bit b gii hn khi dng phn gii cc h a bin. Ngy nay, vi s pht trin ca my tnh, cc iu khin thng c phn gii trong min thi gian. V v vy, cn thit phi c mt phng php khc c trng ha cho h thng. Phng php mi, l s dngbin s trng thi (state variable) c trng cho h thng. Mt h thng c th c phn gii v thit k da vo mt tp hp cc phng trnh vi phn cp mt s tin li hn so vi mt phng trnh c nht cp cao. Vn s c n gin ha rt nhiu v tht tin li nu dng my tnh gii. Gi s mt tp hp cc bin x1(t), x2(t)...xn(t) c chn m t trng thi ng ca h thng ti bt k thi im cho sn t=t0 no, cc bin ny m t hon ton trng thi qu kh ( past history ) ca h cho n thi im t0. Ngha l cc bin x1(t0), x2(t0) . . . xn(t0), xc nh trng thi u ca h ti t=t0. Vy khi c nhng tn hiu vo ti t >= t0 c ch r, th trng thi tng lai ca h thng s hon ton c xc nh . Vy, mt cch vt l, bin trng thi ca mt h tuyn tnh c th c nh ngha nh l mt tp hp nh nht cc bin x1(t),x2(t),... xn(t), sao cho s hiu bit cc bin ny ti thi im t0 bt k no cng thm d kin v s kch thch (excitation) ng vo c p dng theo sau, th xc nh trng thi ca h ti bt k thi im t >=t0 no.

Ci ngt in, c l l mt th d n gin nht v bin trng thi. Ngt in c th v tr hoc ON hoc OFF, vy trng thi ca n c th l mt trong hai tr gi kh hu . Nn, nu ta bit trng thi hin ti (v tr) ca ngt in ti t0 v nu c mt tn hiu t ng vo, ta s c th xc nh c tr gi tng lai trng thi ca n. II. PHNG TRNH TRNG THI V PHNG TRNH OUTPUT. Xem li s khi hnh H.4_1, din t mt h thng tuyn tnh vi p input v q output. Ta gi s h thng c t trng bi tp hp sau y ca n phng trnh vi phn cp 1, gi l nhng phng trnh trng thi.

Trong :, , l cc bin trng thi ., , l cc input : hm tuyn tnh th i. Cc output ca h thng lin h vi cc bin trng thi v cc input qua biu thc sau.

: hm tuyn tnh th k .

Phng trnh (4.2) gi l phng trnh output ca h. Phng trnh trng thi v phng trnh output gi chung l cc phng trnh ng ca h. Th d, xem mt h tuyn tnh vi mt input v mt output c m t bi phng trnh vi phn :

Hm chuyn m t h thng d dng c c bng cch ly bin i Laplace hai v, vi gi s cc iu kin u bng 0.

Ta s chng t rng h thng cn c th m t bi mt tp hp cc phng trnh ng nh sau : Trc nht, ta nh ngha cc bin trng thi

Phng trnh 4.3 c sp xp li sau cho o hm bc cao nht v tri:

By gi phng trnh 4.6 v 4.7, thay th cc h thc nh ngha ca bin trng thi vo 4.8 . Ta s c nhng phng trnh trng thi:

Ch c phng trnh (4.9c) l tng ng phng trnh ban u (4.3). cn hai phng trnh kia ch l phng trnh nh ngha bin trng thi. Trong trng hp ny, output c(t) cng c nh ngha nh l bin trng thi x1(t), (khng phi lun lun nh vy). Vy phng trnh (4.5) l phng trnh output. Tng qut hn, nu p dng phng phng php m t trn, th phng trnh vi phn cp n:

S c trnh by bi cc phng trnh trng thi sau :

V phng trnh output gin d l :

Phng php nh ngha cc bin trng thi c m t trn khng thch hp khi v phi ca (4.10) c cha nhng o hm ca r(t).

Trong trng hp ny, nhng h thc ca cc bin trng thi cng phi cha r(t). Cc bin trng thi c nh ngha nh sau:

Vi cc gi tr :

Dng (14) v (15) ta a phng trnh vi phn cp n(4.13) vo n phng trnh trng thi sau y di dng bnh thng :

Phng trnh output, c c t biu thc th nht ca(4.14):

III. S BIU DIN BNG MA TRN CA PHNG TRNH TRNG THI . Nhng phng trnh trng thi ca mt h thng ng c th c vit di dng ma trn, s dng ma trn trnh by trong cc h phc tp lm cho cc phng trnh c dng c ng hn. Phng trnh (4.1) vit di dng ma trn th n gin sau:

Trong X(t) l ma trn ct biu din cc bin s trng thi gi l cc vct trng thi. R(t) l ma trn ct, biu din input gi l cc vct input.

A l ma trn vung n x n :

B l ma trn n x p

(v c p input r )

Tng t nh vy, q phng trnh trong (4.2) cng c th c trnh by bng mt ma trn duy nht

Trong D l ma trn q x n v E l ma trn q x p. Th d, cc phng trnh trng thi ca phng trnh (4.11) c vit di dng ma trn:

Khi so snh phng trnh (4.23) vi phng trnh (4.18), cc ma trn A v B s c ng nht d dng. Trng hp ny, phng trnh output (4.22) l mt phng trnh v hng.

IV. VI TH D.

Th d 4.1: Xem mt h thng tuyn tnh, c hm chuyn cho bi:

Cc bin s trng thi c nh ngha:

Do h thng c th c din t bng ma trn:

Th d 4.2:

Xem mt h thng iu khin nh H.4.2. Hm chuyn vng kn ca h l:

Phng trnh vi phn tng ng

Cc bin trng thi:

Vy h thng c th din t bng h thng vct:

Trong :

Th d 4.3 : Xem mt mch RLC nh H. 4.3

Trng thi ca h c th m t bi tp hp cc bin trng thi x1 = vc(t) ( 4.39)

x2 = iL(t) ( 4.40) i vi mch RLC th ng, s cc bin s trng thi cn thit th bng vi s cc b phn tch tr nng lng c lp. Cc nh lut Kirchhoff cho:

Vit li(4.41) v (4.42) nh tp hp cc phng trnh vi phn cp 1:

Dng cc phng trnh (4.44), (4.45), (4.46) v cc iu kin u ca mch x1(t0), x2(t0) ta c th xc nh trng thi tng lai ca mch v tn hiu ra ca n. Di dng vct, trng thi ca h c trnh by:

Lu l cc bin trng thi ca h thng khng phi l duy nht. Ty theo cch chn la, c th c nhng tp hp khc ca cc bin trng thi. V. HNH TRNG THI . hnh truyn tn hiu m ta ni chng 3 ch p dng cho cc phng trnh i s. y, ta s a vo cc phng php hnh trng thi, nh l mt s m rng cho hnh truyn tn hiu m t cc phng trnh trng thi ,v cc phng trnh vi phn. ngha quan trng ca hnh trng thi l n to c mt s lin h kn gia phng trnh trng thi, s m phng trn my tnh v hm chuyn. Mt hnh trng thi c xy dng theo tt c cc qui tc ca hnh truyn tn hiu. Nhng hnh trng thi c th c dng gii cc h tuyn tnh hoc bng gii tch hoc bng my tnh. Tr li mch RLC v d 4.3. din t ng lc 3 phng trnh (4.44) (4.45), (4.46), ta c th dng gin hnh trng thi nh hnh H.4_4 sau y :

Nhng ri thay, hu ht cc mch in, cc h thng in c hay nhng h iu khin u khng n gin nh mch RLC trn y, v thng kh xc nh mt tp hp cc phng trnh vi phn cp 1 din t h thng.V vy, n gin hn ,ta thng chuyn ha kiu mu trng thi t hm chuyn. Mt cch tng qut mt h c m t bng hm chuyn nh sau:

n>=m v mi h s a u thc dng. Nu nhn t v mu cho S-n ta c:

Cng thc Mason quen thuc gip ta tha nhn d dng rng t s l tng li trc tip, v mu s l tng li vng hi tip. Ta vit li cng thc Mason.

Nu tt c cc vng hi tip u chm nhau v tt c cc ng trc tip u chm vng hi tip th (4.51) thu li

Th d 4.4 : Trc ht xem hm chuyn ca h thng cp 4:

V h thng cp 4, ta s nh ngha 4 bin trng thi (x1,x2,x3,x4). Gi t cng thc Mason, ta c th thy rng mu s ca (4.53) c th c xem nh l 1 cng vi li vng, v t s ca hm chuyn th bng vi li ng trc tip ca hnh. hnh trng thi phi dng s ln ly tch phn bng vi cp s ca h thong.Vy cn ly tch phn 4 ln.

Ghp cc nt li .Nh rng Ta c hnh trng ca (4.53)

Th d4.5 : .By gi ta xem hm chuyn cp 4 khi t s l mt a thc theo S:

T s ca G(s) l tng li cc ng trc tip trong cng thc Mason . hnh trng thi (HTT) v hnh H.4_7.Trong li cc ng trc tuyn l b3/s; b2/s2; b1/s3 v b0/s4.

H4_7

H4.8 Ngoi ra phng trnh output l : C(t) = b0 x1 + b1 x2 + b2 x3 + b3 x4 Di dng ma trn ta c:

(4.57)

L u : din t phng trnh (4.54), HTT v hnh H.4_7 khng phi l duy nht.ta hy xem hnh H.4_8.

T HTT hnh H.4_8a, ta c mt tp hp phng trnh trng thi :

vit phng trnh (4.61a), ta hy tham kho hnh H.4_8b. Gia hai nt x2 v x1 , ta thm nt mi x2. Cc phng trnh kht cng lm tng t. hnh H.4_8a trnh trnh by cng mt hm chuyn nh hnh H.4_7.nhng cc bin trng thi ca mi hnh th khng ging nhau.. Th d4.6 : Ta hy xem mt h thng iu khin nh hnh H.4_9 c th dng HTT xc nh trng thi ca h.

Hm chuyn vng kn ca h :

Nhn t v mu vi s-3 :

hnh trng thi cho bi H.4_10

T hnh suy ra cc phng trnh trng thi..

V phng trnh output :

Di dng ma trn :

V Vi

Ging vin: Phm Vn Tn

C S T NG HCChng V M HNH HO CC H THNG VT LNI DUNG:5.1) I CNG

5.2) PHNG TRNH CA CC MCH IN 5.3) M HNH HO CC B PHN CA H THNG C.

5.4) PHNG TRNH CA CA CC H THNG C KH

5.5) M HNH HA NG C DC.

I. I CNG. Mt trong nhng cng vic quan trng nht trong vic phn gii v thit k cc h t kim l m hnh ho h thng. nhng chng trc, ta a vo mt s phng php m hnh ho h thng thng dng. Hai phng php chung nht l hm chuyn v phng trnh trng thi. Phng php hm chuyn ch c gi tr i vi cc h tuyn tnh, khng i theo thi gian. Trong khi cc phng trnh trng thi, l nhng phng trnh vi phn cp mt c th dng m t cc h tuyn tnh v c phi tuyn. V trong thc t, tt c cc h vt l u phi tuyn trong mt vi phm vi hot ng. Nn c th s dng ham chuyn v cc phng trnh trng thi tuyn tnh, h thng phi c tuyn tnh ho, hoc l hot ng ca n phi c hn ch trong vng tuyn tnh. D s phn gii v thit k cc h iu khin tuyn tnh c pht trin tt, nhng bn sao ca n cho cc h phi tuyn th thng rt phc tp. K thut iu khin thng phi xc nh khng ch vic lm sao m t chnh xc h thng mt cch ton hc, m con phi, quan trng hn, lm sao t cc gi thuyt ng, v php tnh xp x (nu cn thit) sao cho h thng c th c c trng ha mt cch tng xng bi mt m hnh ton hc tuyn tnh. Tht quan trng thy rng, k thut iu khin hin i phi da trn s m hnh ho h thng sao cho vn phn gii v thit k c th ph hp vi cc li gii nh my tnh. Nh vy, ch ch ca chng ny l: - chng t s m hnh ho ton hc ca cc h thng iu khin v cc b phn. - chng t bng cch no s m hnh ho s dn n cc li gii trn my tnh II. PHNG TRNH CA CC MCH IN.

Phng php c in vit cc phng trnh ca mch in c t trn c s hai nh lut v nt v vng ca kirchhoff. Tuy hai nh lut ny th n gin nhng cc phng trnh kt qu th khng t nhin i vi my tnh.

Mt phng php mi vit cc phng trnh mch in l phng php bin trng thi. V cc mch in trong phn ln cc h t kim th khng phc tp lm, ta s trnh by y ch mc gii thiu. Nhng l gii chi tit v cc phng trnh trng thi cho mch in c th tm cc gio trnh l thuyt mch.

H.5_1. Xem mch RLC nh hnh H.5_1. Phng cch thc hnh l xem dng in trong cun cm L v in th ngang qua t C l cc bin trng thi (tc i(t) v ec(t)). L do ca s chn la ny l v cc bin trng thi th lin h trc tip vi b phn tch tr nng lng ca mt h thng. Trong trng hp ny, cun cm tch tr ng nng v t tch tr th nng. Bng cch chn i(t) v ec(t) l cc bin trng thi, ta c mt s m t hon ton v qu kh (tc tr gi u ca chng) hin ti v trng thi tng lai ca mch. Ta c: Dng in trong t C : (5.1) in th ngang qua L : (5.2)

Cc phng trnh trng thi di dng ma trn, c vit:

9; 9; 9; 9; Th d5_1 : Xem mch in nh hnh H.5_2.

(5.3)

H.5_2 in th ngang qua t ec(t), cc dng in trong cc cun cm i1(t) v i2(t) c xem nh l cc bin s trng thi. Cc phng trnh trng thi c c bng cch vit in th ngang qua cc cun cm v dng trong t.

Xp xp li cc h s hng, cc phng trnh trng thi c vit di dng chnh tc nh sau:

III. M HNH HO CC B PHN CA H THNG C. III.1 Chuyn ng tnh tin III.2 Lc ma st trong chuyn ng tnh tin. III.3 Chuyn ng quay.

III.4 S tng quan gia chuyn ng tnh tin v chuyn ng quay. III.5 C nng v cng sut. III.6 Bnh rng - n by dy courroir. Hu ht cc h t kim u c cha cc b phn c kh cng nh cc b phn in. Trn quan im ton hc, s m t cc b phn c v in th tng ng nhau. Tht vy, ta c th chng minh rng mt b phn c kh thng l mt bn sao ca mt b phn in tng ng, v ngc li. D nhin, s tng ng ch trn ngha ton hc. Hai h thng th tng ng nhau nu chng c din t bng cc phng trnh ging nhau. S chuyn ng ca cc b phn c c th l tnh tin, quay hoc phi hp c hai. Cc phng trnh din ra chuyn ng ca cc h c th thng c vit mt cch trc tip hay gin tip t nh lut chuyn ng ca Newton. 1. Chuyn ng tnh tin. Chuyn ng tnh tin c nh ngha nh l mt chuyn ng di ch dc theo mt ng thng. Cc bin c dng m t chuyn ng tnh tin l gia tc, vn tc v di. nh lut Newton chng t rng tng i s cc lc tc ng ln mt c th theo mt phng cho th bng tch s ca khi lng ca c th v gia tc ca n theo cng phng . ( lc = Ma (5.8) Trong : M l khi lng v a l gia tc. Trong chuyn ng tnh tin, cc b phn sau y thng c a vo: a. Khi lng. Khi lng c xem nh l mt c trng ca mt b phn tch tr ng nng trong chuyn ng tnh tin. N tng ng vi cun cm ca mch in. Nu W l trng lng ca c th, th M c cho bi: (5.9) g: Gia tc trng trng.

Trong h thng SI, n v ca M l kg, ca g l m/s2; ca lc l Newton(N).

Hnh H.5_3: H thng lc- khi lng. HnhH. 5_3 m t v tr m mt lc tc ng ln mt c th c khi lng M. Phng trnh c vit:

9; 9; 9; 9; Trong y(t) ch di; v(t): vn tc; a(t): gia tc.

(5.10)

Tt c c tham chiu theo hng ca lc p dng. b. L xo tuyn tnh. Mt cch tng qut, l xo c xem nh l mt b phn tch tr th nng. N tng ng vi t in trong cc mch in. Trong thc t, l xo tuyn tnh c th l mt l xo thc s, hoc mt dy courroir. D tt c cc l xo u phi tuyn vi vng hot ng. Nhng, nu s bin dng ca l xo nh, trng thi ca n c th c xp x ho (approximated) bng mt h thc tuyn tnh: 9; 9; 9; f(t)= Ky(t) (5.11) Vi K l hng s l xo, hoc hng s n hi (Stifness) n v ca K: N/m Phng trnh (5.11) cho thy lc tc ng ln l xo th t l trc tip vi di ( bin dng) ca l xo. M hnh biu din mt b phn l xo tuyn tnh v hnh H.5_4.

H.5_4: H thng lc-l xo.

Nu l xo c mang trc mt sc cng T th (5.12) s c ci bin thnh:

9; 9; 9; f(t)-T= Ky(t) (5.12)2. Lc ma st trong chuyn ng tnh tin. Mi khi c s chuyn ng hoc khuynh hng chuyn ng gia hai vt, lc ma st s xut hin. Lc ma st gp trong cc h vt l thng l phi tuyn. Nhng c tnh ca cc loi lc ma st gia hai b mt tip xc thng ph thuc vo cc h s nh l s phi hp b mt, p sut gia cc b mt, vn tc tng i ca chng v nhng th khc, lm cho vic m t ton hc mt cch chnh xc lc ma st th rt kh. Tuy nhin, vi ch ch thc hnh, lc ma st c th chia thnh ba loi nh sau: Ma st trt, ma st ngh v ma st coulomb. a. Ma st trt ( ma st nht-Vicous Friction) Ma st trt biu din mt lc cn c lin h tuyn tnh gia lc tc dng v vn tc. Lc ma st trt thng c m hnh ho bng mt dashpot (ng m), c k hiu nh hnh H.5_5.

Phng trnh biu din lc ma st trt:

(5.13) Trong : B l h s ma st trt. (N/m/sec) Hnh H.5_5a, trnh by s tng quan gia lc ma st trt v vn tc. b. Ma st ngh (Static Friction). Ma st ngh biu din mt lc cn, c khuynh hng ngn cn chuyn ng lc va bt u (khi chuyn ng bt u ma st ngh c tr cc i bng ma st trt). Ma st ngh c biu din bi biu thc: f(t) = (Fs)y=0 (5.14) Trong : (Fs)y = 0 c nh ngha nh l lc ma st ngh tn ti ch khi vt ng yn nhng ang c khuynh hng chuyn ng. Du ca lc ty thuc v chiu chuyn ng hoc chiu ban u ca vn tc. S tng quan gia lc v vn tc v hnh H.5_5b. Nh l mt khi chuyn ng bt u, lc ma st ngh bin mt, v loi lc ma st khc xut hin. c. Ma st coulomb. Lc ma st coulomb l mt lc cn, c ln khng i i vi s bin thin ca vn tc. Du ca lc th thay i khi vn tc i chiu. Phng trnh ton hc ca lc ma st coulomb:

9;

(5.15)

Trong Fc l h s ma st coulomb. S tng quan gia lc v vn tc v hnh H.5_5c.

3. Chuyn ng quay. Chuyn ng quay ca mt vt c th c nh ngha nh l chuyn ng ca vt quanh mt trc c nh. Cc bin s thng dng m t chuyn ng quay l moment; gia tc gc (; vn tc gc (; v gc di (. Cc b phn sau y thng c a vo m hnh ho chuyn ng quay. Qun tnh (Inertia). Qun tnh J, c xem nh l ch th tnh cht ca mt b phn tch tr ng nng trong chuyn ng quay. Qun tnh ca vt ph thuc vo s tng hp hnh hc quanh trc quay v khi lng ca n. J cn gi l moment qun tnh. Th d: qun tnh ca mt da trn hoc mt trc trn quay quanh trc hnh hc l:

(5.16) Trong , M l khi lng ca da hoc ca trc v r l bn knh ca chng. Khi mt moment c p dng vo mt c th vi qun tnh J, nh hnh H.5_7, th phng trnh moment c vit:

(5.17)

J : Kg.m2 ; T :N.m ; q :radian.

H.5_7: H thng moment _qun tnh. b. L xo xon (torsional spring). Khi p dng mt moment ln mt thanh hay mt trc quay c khi lng khng ng k, trc quay mt gc (. Nu k l hng s xon, moment trn mt n v gc di, th h thng c th biu din bng hnh H.5_8 v phng trnh: T(t)=Kq (t) (5.18)

H.5_8: H thng moment- l xo xon. Nu l xo xon c mang trc mt moment Tp, th phng trnh trn c ci tin. T(t) TP =Kq (t) (5.19) c. Ma st trong chuyn ng quay. C ba loi ma st m t trong chuyn ng tnh tin u c th p dng cho chuyn ng quay. Do cc phng trnh (5.13), (5.14) v (5.15) c th vit li trong trng hp ny nh sau:

; ; (5.20) T(t)= (Fs)q =0 9; 9; (5.21)

(5.22) Trong , B :H s ma st nht, moment trn mt n v vn tc gc. (Fs)(=0 l ma st ngh. Fc : l ma st coulomb. 4. S tng quan gia chuyn ng tnh tin v chuyn ng quay. Trong vn iu khin chuyn ng, thng khi ta cn i mt chuyn ng quay thnh mt chuyn ng tnh tin. Th d, Hnh H.5_9 : b iu khin i mt chuyn ng quay thnh mt chuyn ng thng nh motor v b screw (Vis Faraday) Hnh H.5_10: cng c chc nng tng t, nhng s chuyn i thc hin nh thanh rng (rack) v pinion(nhng)./ Hnh H.5_11: Mt b iu khin chuyn ng thng dng khc, dng pulley (rng rc) v dy couroir .

Cc h thng trn iu c th c biu din bng mt h thng n gin vi mt qun tnh tng ng mc trc tip vo mt motor thc. Th d, khi lng hnh H.5_11, c th xem nh l mt khi im (point mass) chuyn ng quanh rong roc, bn knh r. B qua qun tnh ca rong roc, th qun tnh tng ng do motor l (5.23) Nu bn knh ca pinion hnh H.5_10 l r, qun tnh tng ng do motor cho bi phng trnh (5.23). By gi ta xem h thng hnh H.5_9. Gi L l khong di chuyn thng ca khi lng khi khoang cch space convis xoay mt vng. V nguyn tc, hai h thng hnh H.5_10 v H.5_11 th tng ng. hnh H.5_10 khong di chuyn thng ca khi lng trn mi vng quay ca pinion lL=2(r. Do , dng phng trnh (5.23) tnh qun tnh tng ng ca h hnh H.5_9.

(5.24) 5.C nng v cng sut. Nng lng v cng sut gi vai tr quan trng trong vic thit k cc h thng in c. Nng lng c tch tr di dng ng nng v th nng iu khin tnh "ng" ca h thng. Tuy nhin, nng lng tiu tn thng dng nhit, cng cn c kim sot. * Khi lng hoc qun tnh ca mt vt ch kh nng tch tr ng nng. ng nng ca mt khi lng di chuyn vi vn tc v l:

(5.25) Wk: Joule, hoc Nm ; M: N/m/sec2 ;v: m/s. i vi mt h thng quay, ng nng c vit:

(5.26) J: moment qun tnh Kg.m2 (: vn tc gc rad/s. *l xo tuyn tnh b bin dng mt chiu di y , s tch tr mt th nng: (5.27) * l xo xon, tch tr th nng: (5.28) ( : Gc xon. i vi mt b phn ma st, nng lng biu din mt s mt hoc tiu hao bi h thng khi i khng vi lc ma st. Cng sut tiu tn trong b phn c ma st l tch s ca lc v vn tc. P=f.v (5.29) V f= B.v, vi B l h s ma st, nn: P=B.v2 (5.30) ( P: N.m/s2 hoc watt (w)). Vy nng lng tiu tn trong b phn ma st la: (5.31) 6.Bnh rng - n by dy courroir. Bnh rng, n by hoc dy courroir v pu-li l nhng c phn truyn nng lng t mt b phn ny n mt b phn khc ca h thng thay i lc, moment, vn tc v di. Chng cng c xem nh l nhng b phn phi hp nhm t n s truyn cng sut ti a. Hai bnh rng ni nhau nh hnh H.5_12. Qun tnh v ma st ca chng c xem nh khng ng k trong trng hp l tng.

Nhng h thc gia moment T1 v T2, gc di (1 v(2 , s rng N1 v N2 ca b bnh rng c dn xut t cc s kin sau y: 1_ S rng trn b mt cc bnh rng t l vi bn knh r1v r2 ca bnh rng: r1N2=r2N1 (5.32) 2_ Khong dch dc theo b mt ca mi bnh rng th bng nhau. q 1r1=q 2r2 (5.33) 3_ Gi s khng c s mt nng lng, cng to bi bnh rng ny bng cng ca bnh rng kia. T1q 1=T2q 2 (5.34) Nu (1 v (2 l vn tc gc ca chng th:

(5.35)

Thc t, cc bnh rng u c qun tnh v lc ma st thng khng b qua.

T= moment p dng (1, (2: gc di. T1, T2: moment c truyn n bnh rng J1, J2; qun tnh ca bnh rng N1, N2: s rng Fc1,Fc2: H s ma st coulomb. B1, B2: H s ma st nht (trt). Phng trnh moment ca bnh rng 2 c vit:

(5.36) Phng trnh moment ca bnh rng 1 l:

(5.37) Dng (5.35), phng trnh (5.36) i thnh:

(5.38) Phng trnh (5.38) chng t rng c th phn x qun tnh, ma st,momen,vn tc v di t pha na sang pha kia ca b bnh rng. Nh vy, cc i lng sau y s c c khi phn x t bnh rng 2 sang bnh rng 1 : Qun tnh : H s ma st nht :

Momen : Gc di : Vn tc gc : Momen ma st coulomb : Nu c s hin din ca l xo xon, hng s l xo cng c nhn bi, khi phn x t bnh rng 2 sang bnh rng 1. By gi, thay (5.38) vo (5.37) :

Dy courroir v dy chain c dng cng mc ch nh b bnh rng. Nhng n cho php chuyn nng lng vi khong cch xa hn m khng dng cc bnh rng vi s rng qu ln. Hnh H.5_14 v s ca mt dy courroir (hoc chain) gia hai rng rc (pulley). Gi s khng c s trt gia chng. D thy rng phng trnh (5.41) vn cn c p dng trong trng hp ny. Tht vy, s phn x (hay s truyn dn) ca momen, qun tnh ma st th tng t nh trong mt b bnh rng.

n by (lever) nh trong hnh H.5_15 truyn chuyn ng thng v lc tng t cch thc m b bnh rng truyn chuyn ng quay. H thc gia lc v khong cch l :

(5.43)

IV. PHNG TRNH CA CA CC H THNG C KH. vit cc phng trnh ca mt h c tuyn tnh , trc nht phi xy dng trc mt m hnh ca h, bao gm cc b phn tuyn tnh ni nhau. Sau p dng nh lut Newton. Th d 5.2 :

Xem mt h thng v hnh H. 5_16a . S vt th t do ca h v hnh H.5_16b. Phng trnh lc ca h c vit :

(5.44)

Phng trnh cp 2 (5.44) c th phn thnh hai phng trnh trng thi cp mt. t=y v= nh l cc bin s trng thi.

h thng c trn y tng ng vi mch RLC ni tip ca mch in. Vi s tng ng gia mt h thng c v mt h thng in, vic thnh lp trc tip cc phng trnh trng thi cho mt h thng c s tr nn n gin. Nu ta xem khi lng th tng ng vi in cm, hng s l xo K th tng ng vi nghch o ca in dung 1/C . Vy c th ch nh v(t): vn tc v fk(t): lc tc ng ln l xo nh l cc bin s trng thi. L do l ci trc tng t dng in trong cun cm, v ci sau tng t nh in th ngang qua t. Do phng trnh trng thi ca h c vit bng: Lc trn khi lng:

(5.47) Vn tc ca l xo :

(5.48) Phng trnh trn th ging nh cch vit phng trnh in th ngang qua 1 cun cm. Cn phng trnh di ging nh phng trnh ngang qua t. Th d n gin trn cho thy cc phng trnh trng thi v bin s trng thi ca 1 h thng ng th khng duy nht. Th d 5.3: Xem 1 h thng nh hnh H.5_17a. V l xo b bin dng khi chu tc dng ca lc f(t) hai di y1 v y2 phi c ch nh cho 2 u mt ca l xo. S vt th t do ca h v hnh H.5_17b.

9;

T H.5_17b, cc phng trnh lc c vit : f(t)=K[y1(t)-y2(t)] (5.49)

(5.50)

vit cc phng trnh trng thi ca h thng, ta t: 9; X1(t)=y2(t)

9; Th cc phng trnh (5.49) v (5.50) c vit li:

Nu ta ch nh vn tc v(t) ca khi lng M l 1 trng thi bin s , lc fk(t) trn l xo l 1 bin s, th:

(5.53) fk(t)=f(t) (5.54) Mch in tng ng vi h c trn c v hnh H.5_18.

Nu mun tm di y1(t) ti in m y(t) p dng vo, ta dng h thc:

9;

(5.55)

Trong y2(0) l di ban u ca khi lng M . Mt khc, c th gii cho y2(t) t 2 phng trnh trng thi (5.51) v (5.52) v y1(t) c xc nh bng (5.49).

Th d 5.4: H thng quay v hnh H.5_19 gm 1 u th c nh. Moment qun tnh ca da quanh trc l J. Ra ca da c lt trn mt phng v h s ma st trt l B. B qua qun tnh ca trc. Hng s xon l K.

Gi s 1 moment p dng vo h thng nh hnh v: Phng trnh momen quanh trc c vit t hnh H.5_19b

H thng ny tng t nh h thng chuyn ng tnh tin H.5_16. Cc phng trnh trng thi c th vit bng cc nh ngha cc bin. x1(t) V Ng c c th thc hin cc bc tip theo vit phng trnh trng thi nh l 1 bi tp. V.M HNH HA NG C DC. V.1) S lc v cc lai ng c DC: V.2) M hnh ha ng c DC: 1/ S lc v cc lai ng c DC:

Motor DC c th c xp thnh 2 loi : loi c t thng thay i c v loi khng c t thng thay i c. -Trong loi th nht: T trng c to bi cun cm. M cun cm th u vi 1 t trng ngoi. Loi ng c ny li c c th chia lm 2 loi: kch t ni tip v kch t ring.

H.5_19a, k hiu ca ng c DC kch t ni tip. Cun cm u ni tip vi phn ng. H.5_19b ng c ni tip kch t ring. Cun cm cch ly vi phn ng v c cp in bi 1 ngun in khc. + Trong loi kch t ni tip, t thng trong ng c th t l vi dng in cm, m dng ny th thay i, s lin h gia moment v vn tc thng l phi tuyn. Nh vy loi ng c ny ch dng trong nhng ng dng t bit cn n moment ln vi vn tc thp. Momen ca motor gim rt nhanh khi vn tc tng. + i vi loi kch t ring t thng th c lp vi dng in ng. V vy n c th c iu khin t bn ngoi trong 1 phm vi rng. -Trong loi th 2 motor DC c t thng khng i, t trng phn cm l do 1 nam chm vnh cu v khng thay i . Loi ny gi l PM motor. iu ny khin c tuyn moment-vn tc tng i tuyn tnh. Cc ng c DC qui c u c chi v c gp. Nhng hin nay c loi ng c DC m c gp c thay bng b phn in t . Loi ny c gi l ng c DC khng chi(DC brushless motor).

2/ M hnh ha ng c DC: V cc ng c DC c dng rt nhiu trong cc h iu khin ta cn quan tm ti vic thip lp 1 m hnh ton hc cho chng. Sau y ta khai trin m hnh ton hc cho 2 lai ng c DC kch t ring v loi kch t bng nam chm vnh cu (PM.motor). c. ng c DC kch t ring:

Phn ng c m hnh ha nh l 1 mch vi in tr Ra, ni tip vi 1 cun cm La. Mt ngun in th Eb biu din cho sc in ng sinh ra trong phn ng khi rotor quay.Phn cm c biu din bng 1 in tr Rf ni tip vi 1 cun in cm Lf . T thng trong khe t l rng. Cc bin s v thng s tm tt nh sau: Ea(t): in th phn ng. Ef(t): in th phn cm. Ra: in tr phn ng. Eb(t): sut in ng trong phn ng. Rf: in tr phn cm. La: in cm phn ng.

Lf: in cm phn cm. I a(t): dng in phn ng. I f(t): dng in phn cm. 9; 9; Ki: hng s moment. Kb: hng s sut in ng phn ng. Tm(t): moment c khai trin bi ng c. 9; 9; Jm: qun tnh ca rotor. Bm: h s ma st trt. gc di ca rotor. vn tc di ca rotor. TL(t): moment ti. Gi s ef(t) c cung cp 1 cch hiu qu cho if(t) khng i. S iu khin c t ln 2 u phn ng di dng in th ea(t). V phn gii tuyn tnh ta gi s thm: 1- T thng khe t th t l vi dng in cm. 2- Moment khai trin bi ng c th t l vi t thng trong khe t v dng in ng . V K mKf If l hng s, nn: Tm(t)=Ki ia(t) (5.65) Ki l hng s moment. Bt u vi in th iu khin ng vo cc phng trnh nhn qu ca h c vit li:

(5.66) Tm(t)=Ki ia(t) (5.67)

Trong o, TL(t) l moment ti(cn). Mt cch tng qut TL(t) biu din 1 moment m ng c phi vut qu mi c th thay i c. TL(t) cng c th l moment ma st khng i th d ma st culomb. * Cc phng trnh (5.66) n (5.69) l nguyn nhn ca cc nguyn nhn. Phng trnh (5.56) xem diat)/dt l hu qu trung gian do ea(t) gy ra. Trong phng trnh (5.57) ia(t) to nn moment Tm(t). Phng trnh (5.68) nh ngha sut in ng phn ng v cui cng trong phng trnh (5.69) moment gy ra gc di (m. Cc bin s trng thi ca h c th c nh nhga l (m , Wm v ia. Cc phng trnh trng thi ca ng c DC , c vit di dng ma trn (5.70):

(5.70) Nh l trong trng hp ny TL(t) l input th 2 trong cc phng trnh trng thi. hnh trng thi ca h c v hnh H.5_27, bng cch dng phng trnh (5.70). Hm chuyn gia di v in th suy c t hnh trng thi.

(5.71)

Trong TL t Zero.

************* Ging vin: Phm Vn Tn

C S T NG HCChng VI

TNH N NH CA H THNGNI DUNG: 6.1) ai cng 6.2) nh ngha tnh n nh 6.3) Khai trin phn b tng phn 6.4) Mt phng phc v s n nh ca h thng 6.5) Cc phng php xc nh tnh n nh ca h thng 6.6) Tiu chun n inh ROUTH 6.7) Tiu chun HURWITZ I. I CNG. C nhiu c tnh c dng trong thit k h thng t kim. Nhng yu cu quan trng nht, l h thng c n nh theo thi gian hay khng? Ni chung, tnh n nh c dng phn bit hai loi h thng: Hu dng v v dng. Trn quan im thc t, ta xem mt h thng n nh th hu dng, trong khi mt h thng bt n th v dng. i vi nhiu h thng khc nhau: tuyn tnh, phi tuyn, khng i theo thi gian v thay i theo thi gian, tnh n nh c th c nh ngha theo nhiu hnh thc khc nhau. Trong chng ny, ta s ch xt tnh n nh ca nhng h tuyn tnh, khng i theo thi gian. Mt cch trc gic, tnh n nh ca mt h l kh nng quay tr v trng thi ban u sau khi lch khi trng thi ny, khi tc ng ca cc ngun kch thch t bn ngoi(hay cc nhiu) chm dt. II. NH NGHA TNH N NH

Mt h thng l n nh nu p ng xung lc gim ti zero khi thi gian tin ti v cc. * Th d 6.1: cho p ng xung lc ca vi h iu khin sau y. Trong mi trng hp, hy xc nh tnh n nh ca h thng. a) g(t) = e-t. b) g(t) = t.e-t. c) g(t) = 1. d) g(t) = e-t.sin3t. e) g(t) = sinw t.

H.6_1. Theo nh ngha, h thng: a) n nh. b) n nh. c) bt n. d) n nh. e)bt n III.KHAI TRIN PHN B TNG PHN (Parial Fraction

expansion) C th tm p ng xung lc ca mt h thng bng cch ly bin i laplace ngc hm chuyn ca h.

V khng phi dng n tch phn bin i laplace ngc.

ta c th dng phng php khai trin phn s tng phn Xem hm chuyn G(s) = C(s)/ R(s). (6.1) Trong , C(s) v R(s) l nhng a thc theo s. Gi s R(s) c bc ln hn C(s). a thc R(s) gi l a thc c trng v c th vit: R(s) = sn + a1sn-1 +....+an-1s +an. (6.2) Trong , a1,...an l nhng h s thc. Nhng nghim ca phng trnh c trng R(s) = 0 c th l thc, hay nhng cp phc lin hp n hay a cp (c ly tha hay khng). Ta xem trng hp nhng nghim ny thc v n cp, phng trnh (6.1) c th c vit:

(6.3) Trong , -s1, -s2,....-sn l nhng nghim ca phng trnh c trng zero ca R(s) hay l nhng cc ca G(s).

(6.4) Nhng h s Ksi (i=1, 2, 3,...n) c xc nh bng cch nhm 2 v ca (6.3) hoc (6.4) cho (s+si) ri t s = -si. Th d, tm h s Ks1, ta nhm c hai v (6.3) cho (s+s1) v t s = -s1.

(6.5) * th d 6.2: xem hm chuyn ca mt h thng.

(6.6). Hy tm p ng xung lc ca h. Trc ht, ta p dng k thut khai trin phn s tng phn.

(6.7) cc h s K-1, K-2, K-3 c xc nh nh sau:

Vy (6.7) tr thnh:

(6.8). By gi ta c th dng bng bin i tnh p ng xung lc ca h thng. g(t) =L-1[G(s)].

g(t) = -e-t + 7e-2t -6e-3t. (6.10) * Th d 6.3: bi ton tng t nh trn, vi hm chuyn nh sau:

(6.13)* Th d 6.4:

Khai trin phn s tng phn:

Bin i Laplace ngc : g(t) = - e-t + t e-t + e-2t. IV. MT PHNG PHC V S N NH CA H THNG 1. Hm chuyn l mt hm hu t, bao gm t s ca nhng a thc theo bin s phc s. 2. trn ta thy p ng xung lc ca mt h thng tuyn tnh khng thay i theo thi gian th gm tng cc hm expo theo thi gian, m cc s m ca chng l nghim ca phng trnh c trng.

1. Hm chuyn l mt hm hu t, bao gm t s ca nhng a thc theo bin s phc s.

9; 9;

(6.14)

Trong c c (s+zi ) l nh g th atha so cua a thc t va ( s+pi ) la nhng tha s ca a thc mu. a) Nhng gi tr ca s lm cho tr tuyt i ca |G(s)| bng zero th gi l cc zero ca G(s). b) Nhng gi tr ca s lm cho tr tuyt i ca |G(s)| tin ti v cc th gi l cc cc (pole) ca G(s). * Th d 6.5 : Xem mt h thng c hm chuyn

(6.16) G(s) c cc zero ti s = -1 v s = 2 G(s) c cc cc ti s = -3 ; s = -1-j v s = -1+j Cc v zero l nhng s phc, c xc nh bi hai bin s s = ? + j?. Mt biu din phn thc v mt biu din phn o cho s phc. Mt cc hay mt zero c th c biu din trong ta vung gc. Trc honh ch trc thc v trc tung ch trc o. Mt phng xc nhbi h trc ny gi l mt phng phc hoc mt phng s.

H.6-2 Na mt phng m trong ( < 0 gi l na tri ca mt phng s. v na kia trong ( > 0 gi l na phi ca mt phng s. V tr ca mt cc trong mt phng s c k hiu bng du (X) v v tr mt zero bng du (o). 2. trn ta thy p ng xung lc ca mt h thng tuyn tnh khng thay i theo thi gian th gm tng cc hm expo theo thi gian, m cc s m ca chng l nghim ca phng trnh c trng. Vy m bo hm xung lc gim theo hm expo theo thi gian th cc nghim ca phng trnh c trng phi c phn thc m. Nghim ca phng trnh c trng ca h thng cng l cc ca hm chuyn. Vy c th kt lun rng, iu kin cn mt h n nh l cc cc ca hm chuyn phi nm na tri ca mt phng s. Trc o, bao gm gc ta , th thuc v vng bt n.

H.6-3 * Th d 6.5 : Xem mt h thng c hm chuyn m cc cc ti -1 v -5 v cc zero ti 1 v -2

H.6-4 Cc cc u nm na tri mt phng s. vy h thng n nh. Mc d c mt zero nm na phi, nhng u khng tc ng ln tnh n nh ca h thng. V. CC PHNG PHP XC NH TNH N NH CA H THNG Ta thy tnh n nh ca mt h t kim tuyn tnh khng i theo thi gian c th xt bng cch kho st p ng xung lc, hoc tm v tr cc nghim ca phng trnh c trng trong mt phng s. Nhng cc tiu chun y thng l kh thc hin trong thc t. Th d, p ng xung lc c c bng cch ly bin i Laplace ngc ca hm chuyn, nhng khng phi lc no cng n gin. Cn vic tm nghim ca phng trnh bc cao ch c th nh vo my tnh. V vy, trong thc t phn gii tnh n nh cho h thng, ngi ta c th dng phng php sau y m khng cn n vic gii cc phng trnh c trng. Tiu chun ROUTH v HURWITZ : l mt phng php i s, cho d kin v tnh n nh tuyt i ca mt h tuyn tnh khng i theo thi gian. Cc tiu chun ny s th ch c bao nhiu nghim ca phng trnh c trng nm na tri, na phi v trn trc o. hnh qu tch nghim s (Root Locus Plot): trnh by mt hnh ca qu tch cc nghim ca phng trnh c trng khi mt thng s no ca h thng b thay i. Khi qu tch nghim s nm trn na phi mt phng s, h thng vng knh b bt n. Tiu chun NYQUIST : l mt phng php bn - - ha (Semi graphical), cho d kin trn s khc bit gia s cc v zero ca hm chuyn

vng kn bng cch quan st hnh trng ca hnh NYQUIST. Phng php ny cn bit v tr tng i ca cc zero. S Bode : s Bode ca hm chuyn vng kn G(s) H(s) c th c dng xc nh tnh n nh ca h vng kn. Tuy nhin, ch c th dng khi G(s) H(s) khng c cc cc v zero trong na phi mt phng s. Tiu chun LYAPUNOV : l phng php xc nh tnh n nh ca h phi tuyn, nhng vn c th p dng cho cc h tuyn tnh. S n nh ca h c xc nh bng cch kim tra cc tnh cht ca hm Lyapunov. VI. TIU CHN N NH ROUTH 9; Tiu chun Routh c th xc nh tnh n nh ca h m phng trnh c trng n bc n. 9; 9; ansn + an-1sn-1 + .. + a1s + a0 = 0 Tiu chun ny c p dng bng cch dng bng Routh nh ngha nh sau :

9;

Trong an , an-1 , , a0 l cc h s ca phng trnh c trng, v :

Bng c tip tc theo chiu ngang chiu dc cho n khi c ton zero. Tc c nghim ca phng trnh c trng c phn thc m nu v ch nu cc phn t ct th nht ca bng Routh c cng du (khng i du). Ni cch khc s nghim c phn thc dng bng vi s ln i du.

* Th d 6 -6 : H thng c phng trnh c trng 9; 9; s3 + 6s2 + 12s + 8 = 0 Xt tnh n nh Bng Routh :

v khng c i du ct th nht, nn tt c cc nghim ca phng trnh c trng u c phn thc m. Vy h n nh. * Th d 6 -7 : Phng trnh c trng ca mt h thng l : 9; 9; s3 + 3s2 + 3s + 1 + k = 0 Hy xc nh iu kin h n nh Bng Routh :

9; 9; h n nh, cn c s khng i du ct 1. Vy cc iu kin l : 8-k > 0 v 1+k > 0 vy phng trnh c trng c cc nghim vi phn thc m nu :9; -1 < k < 8

* Th d 6 -8 : Lp bng Routh v xc nh s nghim c phn thc dng ca phng trnh c trng 2s3 + 4s2 + 4s + 12 = 0 Bng Routh : s3 ; 2 4 0 Hng s2 c chia 4 trc khi s2 1 3 0 tnh hng s1. Hng s1 c chia9; 9; s1 -1 0 2 trc khi tnh hng s0

s0 3 V c hai ln i du ct 1, nn phng trnh trn c hai nghim c phn thc dng. * Th d 6 -9 : Xt tnh n nh ca h thng c phng trnh c trng : 9; 9; 9; s4 + s3 - s - 1 = 0 Bng Routh :

H s hng s0 c tnh bng cch thay 0 hng s1 bng (, ri tnh h s ca hng s0 nh sau :

cn phng cch ny khi c mt zero ct mt. V c mt ln i du ct mt, nn phng trnh c trng c mt nghim c phn thc dng. Do , h thng khng n nh. VII. TIU CHUN HURWITZ

9; Tiu chun n nh Hurwitz l phng php khc xc nh tt c nghim ca phng trnh c trng c phn thc m hay khng . Tiu chun ny c p dng thng qua vic s dng cc nh thc to bi nhng h s ca phng trnh c trng. Gi s h s th nht, an dng. Cc nh thc Ai vi i = 1, 2, .... , n-1 c to ra nh l cc nh thc con (minor determinant) ca nh thc :

Cc nh thc con c lp nn nh sau :

V tng dn n ?n Tt c cc nghim ca phng trnh c trng c phn thc m nu v ch nu ?i > 0 vi i = 1 , 2 , .. , n.

* Th d 6 -10: Vi n = 3

Tt c cc nghim ca phng trnh c trng c phn thc m nu 9; 9; a2 > 0 , a2 a1 a0 a3 > 0 a2 a1 a0 a02 a3 > 0 * Th d 6 -11 : Xt s n nh ca h thng c phng trnh c trng 9; 9; 9; s3 + 8s2 + 14s + 24 = 0 Lp cc nh thc Hurwitz

9; 9;

Cc nh thc u ln hn khng, cc nghim ca phng trnh c trng u c phn thc m, nn h thng n nh. * Th d 6 .12 : Vi khong gi tr no ca k th h thng sau y n nh :9; 9; s2 + ks + ( 2k 1 ) = 0

h n nh, cn c : Vy * Th d 6 .13 : Mt h thng thit k t yu cu khi mch khuch i ca n c li k = 2 . Hy xc nh xem li ny c th thay i bao nhiu trc khi h thng tr nn bt n, nu phng trnh c trng ca h l : s3+ s2 (4+k) + 6s + 16 + 8k = 0 Thay cc tham s ca phng trnh cho vo iu kin Hurwitz tng qut th d 6 .10. Ta c nhng iu kin h n nh : 9; 4 + k > 0 , (4+k)6 (16+8k) > 0 (4+k) 6 (16+8k) (16 + 8k)2 > 0 Gi s li k khng th m, nn iu kin th nht tha. iu kin th nh v th ba tha nu k < 4 Vy vi mt li thit k c gi tr l 2, h thng c th tng li ln gp i trc khi n tr nn bt n. li cng c th gim xung khng m khng gy ra s mt n nh. ****************** Ging vin: Phm Vn Tn

C S T NG HCChng VII

PHNG PHP QI TCH NGHIM SNI DUNG: 7.1) i cng. 7.2) Qu tch nghim s 7.3) Tiu chun v gc pha v sut 7.4) S ng qu tch 7.5) Qu tch trn trc thc 7.6) Cc ng tim cn 7.7) im tch 7.8) Gc phn x v gc n 7.9) Phng php v QTNS 7.10) Hm chuyn vng kn v p ng trong min thi gian 7.11) Ngng li v ngng pha t QTNS BI TP CHNG VII I.I CNG Trong vic thit k v phn gii cc h iu khin, ngi ta thng cn phi quan st trng thi ca h khi mt hay nhiu thng s ca n thay i trong mt khang cho sn no . Nh , ta c th chn mt cch xp x tr gn ng cho thng s (chng hn, chn li cho h, hoc kho st nhng bin i thng s do s la ha ca cc b phn ca h). thc hin mc ch y, ta c th dng k thut qu tch nghim s (Root locus). Ta bit, cc cc ca hm chuyn l nghim ca phng trnh c trng, c th hin th trn mt phng S.

Hm chuyn vng kn ca h: l mt hm ca li vng h K. Khi K thay i, cc cc ca hm chuyn vng kn di chuyn trn mt qi o gi l qi tch nghim s (QTNS). Trong chng ny, ta a vo nhng tch cht c bn ca QTNS v phng php v qi tch da vo vi nh lut n gin. K thut QTNS khng ch hn ch trong vic kho st cc h t kim. Phng trnh kho st khng nht thit l phng trnh c trng ca h tuyn tnh. N c th c dng kho st nghim ca bt k mt phng trnh i s no. V ngy nay, vic kho st . thit k mt h t iu khin (trong c k thut QTNS) tr nn d dng, nhanh chng v thun tin nhiu nh cc phn mm chuyn dng trn my tnh, chng hn Matlab. II.QU TCH NGHIM S Xem mt h t iu khin chnh tc:

- Hm chuyn vng kn:

- Hm chuyn vng h:

N(S) v D(S) l cc a thc hu hn theo bin phc S m(n ; K l li vng h. Cc cc ca hm chuyn vng kn l nghim ca phng trnh c trng:

D(S) + KN(S) = 0 (7.1) V tr ca cc nghim ny trn mt phng S s thay i khi K thay i. Qi o ca chng v trn mt phng s l mt hm ca K. Nu K = 0, nghim ca (7.1) l nghim ca a thc D(S), cng l cc ca hm chuyn vng h GH. Vy cc cc ca hm chuyn vng h l cc cc ca hm chuyn vng kn. Nu K tr nn rt ln, nghim ca (7.1), nghim ca (7.1) l nghim ca a thc N(S), l cc zero ca hm chuyn vng h GH. Vy khi K tng t 0 n (, qi tch ca cc cc vng kn bt u t cc cc vng h v tin n chm dt cc zeroca vng h. V l do , ta quan tm n hm chuyn vng h G(S).H(S) khi v QTNS ca cc h vng kn. Th d 7.1: Xem hm chuyn vng h ca mt h hi tip n v:

Vi H=1, hm chuyn vng kn: Cc cc vng kn:

- Khi K=0 ; S1=0 ; S2= -2 - Khi K= ; S1= -1 ; S2= - Qi tch cc nghim ny c v nh l mt hm ca K (vi K > 0)

QTNS gm hai nhnh: Nhnh 1: di chuyn t cc vng h ti gc ta (ng vi K=0) n zero vng h ti -1 (ng vi K=(). Nhnh 2: di chuyn t cc vng h ti -2 (ng vi K=0) n zero vng h ti -( (ng vi K=().

III.TIU CHUN V GC PHA V SUT mt nhnh ca QTNS i ngang qua mt im S1 trong mt phng S, iu kin cn l S1 phi l nghim ca phng trnh (7.1) vi vi tr gia thc ca K. D(S1) + KN(S1) = 0 (7.2)

Suy ra: Phng trnh (7.3) chng t: - Sut: - Gc pha: arg G(S1).H(S1) = 1800 + 3600l ; l = 0, (1, (2 .. arg G(S1).H(S1) = (2l + 1)( ra 9; 9; (7.5)

; (7.6) Phng trnh (7.4) gi l tiu chun ca sut v (7.6) gi l tiu chun v gc mt im S1 nm trn QTNS. Gc v sut ca G(S).H(S) ti mt im bt k no trong mt phng S u c th xc nh c bng hnh v. Vi cch y, c th xy dng QTNS theo phng php th v sa sai (Trial and error) nhiu im trn mt phng S. * Th d 7.2: Xem hm chuyn vng h ca th d 7.1, chng t S1=-0,5 l mt im nm trn QTNS, khi K=1.5

Vy tha tiu chn v sut v pha, nn S1 nm trn QTNS. H.7.1, im S1=-0.5 nm trn QTNS, l mt cc ca vng kn vi K=1.5. * Th d 7.3: Hm chuyn vng h ca h l. Tm arg GH(j2) v. Tr gi no ca K lm j2 nm trn QTNS?

arg GH(j2) = -900-450-450 = -1800

im j2 nm trn QTNS, th khi K=16 * Th d7.4: Chng t im nm trn QTNS. Cho

vi K > 0, v xc nh tr K ta im .

tha tiu chun sut, th:

IV.S NG QU TCH S ng qu tch, hay l s nhnh QTNS, bng vi s cc ca hm chuyn vng h GH. Th d 7.4: Vi, QTNS s c 3 nhnh.

V.QU TCH TRN TRC THC Nhnh ca QTNS nm trn trc thc ca mt phng S c xc nh bng cch m ton b s cc hu hn v s zero ca GH. Nu K>0: Nhnh ca QTNS trn trc thc nm bn tri ca mt s l cc cc v zero. Nu K0 nm trn trc thc. iu tng t cng ng vi K0 Phn cn li ca trc thc, t -4 n -2 v t -0 n +( l QTNS vi K 0

H. 7-4 VII. IM TCH (Break away point, saddle point). 9; im tch (b l mt im trn trc thc, ti hai hay nhiu nhnh QTNS i khi (hoc n) trc thc.

im tch l nghim ca phng trnh :

9; 9; Trong : - p i : cc cc ; -zi : cc zero * Th d 7-7 : Xc nh im tch ca :

(7.8)

9; 9; ( 3(b2 + 6(b + 2 = 0 . Phng trnh c hai nghim : s b1 = -0.423 ; k > 0 s b2 = -1,577 ; k < 0

9; 9; 9; 9;VIII. GC XUT PHT V GC N VIII.1). Gc xut pht ca QTNS t mt cc phc cho bi : VIII. 2). Gc n mt zero phc ca QTNS cho bi : 1). Gc xut pht ca QTNS t mt cc phc cho bi : q D = 1800 + arg GH (7.9) Trong arg GH l gc pha ca GH c tnh ti cc phc, nhng b qua s tham gia ca cc ny. * Th d 7-8 : Xem hm chuyn vng h :

,k>0

Gc xut pht ca QTNS ti cc phc s = -1 +j tnh nh sau : arg GH = 450 900 = -450 q D = 1800 450 = 1350 Gc xut pht ca QTNS ti cc phc s = -1 -j tnh nh sau : arg GH = 3150 2700 = 450 q D = 1800 + 450 = 2250 2). Gc n mt zero phc ca QTNS cho bi : q A = 1800 - arg GH (7.10) Trong GH l gc pha ca GH c tnh ti zero phc , nhng b qua s tham gia ca zero ny. * Th d 7-9 : Xem :

9; 9;

;k>0 Gc n ti zero phc s = j tnh nh sau : arg GH = 900 900 - 450= - 450 q A = 1800 (- 450 ) = 2250

H.7-8 IX. PHNG PHP V QTNS . ve QTNS chnh xc v d dng, c th theo cc bc sau :

Xc nh cc nhnh nm trn trc thc. Tnh tm, gc tim cn. V cc ng tim cn. Xc nh cc gc xut pht t cc cc phc v gc n cc zero phc ( nu c). Xc nh im tch. V cc nhnh sao cho mi nhnh xut pht ti 1 cc ri chm dt ti mt zero, hoc tin v ( dc theo mt ng tim cn. Ap dng tiu chun v gc pha cho cc im nm trn QTNS hnh v c chnh xc. Tiu chun v sut dng xc nh cc tr gi ca k dc theo cc nhnh. V cc cc phc ca h xut hin tng cp phc lin hp, nn QTNS th i xng qua trc thc. Vy ch cn v na trn ca QTNS. Tuy nhin, cn nh l cc cc phc v zero phc na di ca QTNS cng phi tha iu kin v sut v gc pha.

Thng thng, vi ch ch phn tch v thit k, mt QTNS chnh xc ch cn thit mt vi vng ca mt phng s. Khi , tiu chun v gc v sut ch p dng cho nhng vng ny c th v dng chnh xc ca qu tch. Th d 7-10 : QTNS ca h kn c hm chuyn vng h l :

, k >0 c v nh sau : - Nhnh trn trc thc nm t 0 n -2 v t -4 n -( - Tm tim cn, c xc nh bi phng trnh (7.6). s c = - (2+4) /3 = -2 C 3 ng tim cn, nh v bng cc gc ( c xc nh bi (7.7) : 9; 9; ( = 600 , 1800 v 3000 - V c hai nhnh cng nm trn trc thc gia 0 v 2, nn c mt im tch tn ti trong on ny. V tr im tch xc nh bi :

- Tiu chun v gc v sut c p dng ln tng im ln cn ca ng qu tch v phng, xc nh v tr chnh xc ca cc nhnh trong phn phc ca mt phng s.

V QTNS cho th d 7-10 trong trng hp k < 0

Cch v cng tng t mh trng hp k>0. s b = -3.115 ; b = 00 ; 1200 ; 2400 X. HM CHUYN VNG KN V P NG TRONG MIN THI GIAN 9; Hm chuyn vng kn C/R c xc nh d dng t QTNS vi mt tr gi ring ca k. T , ta c th tm c p ng ca h min thi gian C(t) bng cch ly bin i laplace ngc C(s) Xem hm chuyn vng kn C/R ca mt h hi tip n v :

9; 9; 9; 9; 9; 9;

(7.9)

Hm chuyn vng h l biu thc hu t

9; 9; -zi l cc zero ; -pi l cc cc ca G

(7.10)

(7.11) R rng C/R v G c cng zero, nhng khng cng cc ( tr khi k=0 ).

(7.12) vi l n cc vng kn. V tr cc cc ny c xc nh trc tip t QTNS vi v tr gi ring ca li vng h k. Th d 7.11: Xem h thng c hm chuyn vng h l

; k>0 QTNS c v hnh 7.11 Vi tr gi ca k c ch ti nhng im k hiu bng mt tam gic nh. y l cc cc vng kn tng ng vi nhng tr ring ca k. Vi k=2, cc cc l v

H.7.11

Vy Khi h c hi tip n v: (7.13) X. NGNG LI V NGNG PHA T QTNS . Ngng li l h s m tr thit k ca k c th nhn vo trc khi h vng kn tr nn bt n. N c th c xc nh t QTNS. tr ca k ti giao im ca QTNS vi trc o Ngng li=---------------------------------------------------

Nu QTNS khng ct trc o, ngng l li ca Th d 7.12: Xem h hnh 7.12. Tr thit k ca k l 8. Ti giao im ca QTNS v trc o, k = 64. Vy ngng li l 64/8 = 8.

Ngng pha ca h cng c xc nh t QTNS. Cn thit phi tm im j(1 trn trc o cho, vi tr thit k ca k thit k Thng cn n phng php th- v-sa sai nh vi j(1. Vy ngng pha c tnh t argGH(j() l: w PM =1800 +argGH(jw 1) (7.15) Th d 7.13: m h nh hnh 7.14. QTNS v hnh H.7.15.

im trn trc o l lm cho = 1. vi (1 = 1.35 Gc pha ca GH(j1.35) l 129.60 Vy ngng pha l (PM =1800 - 129.60 = 50.40 Lu : xc nh tn s v li ti giao im ca trc o vi QTNS, c th dng bng Routh. Ta bit rng mt hng cc zero trong hng s1 ca bng Routh cho bit a thc ca mt cp nghim tho phng trnh h tr : AS2 + B = 0 (7.16). Trong A, B l phn t th nht v th hai ca hng S2. Nu A v B cng du, nghim ca phng trnh (7.16) l o ( nm trn trc j( ) Vy nu bng Routh c vit cho hm c trng ca h, cc tr ca k v ( ng vi giao in QTNS v trc o c th c xc nh. Th d : Xem h vi GH nh sau Phng trnh c trng vng kn l: S3 + 4 S2 + 4S + k = 0I Bng Routh:

Hng S1 th bng khng ng vi k=16. Vy phng trnh h tr tr nn: 4 S2 + 16 = 0. Vy vi k=16 phng trnh c trng c cc nghim v QTNS ct trc o ti j2 *********************** Ging vin: Phm Vn Tn

C S T NG HCBI TP CHNG II

2.1: Tm hm chun ca 1 h thng m input v output ca n lin h bng phng trnh vi phn:

. 2.2 : Mt h thng cha thi tr c phng trnh vi phn:

Tm hm chuyn ca h. 2.3 : V tr Y ca 1 vt c khi lng khng i M lin h vi lc f t ln n bi phng trnh vi phn:

Xc nh hm chuyn tng quan gia v tr v lc. 2.4 : Mt ng c dc mang ti cho 1 moment t l vi dng in vo i. Nu phng trnh vi phn i vi ng c v ti l:

Trong J l qun tnh rotor, B l h s ma st. Xc nh hm chuyn gia dng in vo v v tr trc rotor. 2.5 : Mt xung lc c t vo ng vo ca 1 h thng v ng ra c 1 hm thi gian e-2t . Tm hm chuyn ca h. 2.6 : p ng xung lc ca 1 h l tn hiu hnh sin. Xc nh hm chuyn ca h v phng trnh vi phn. 2.7 : p ng nc ca h thng l:

. Tm hm chuyn. 2.8 : Tm hm chuyn ca cc mch b chnh sau y:

2.9 : Tm hm chuyn ca mch in gm 2 mch v bi tp 2.8f ni tip. 2.10 : Xc nh p ng dc (ramp) ca 1 h c hm chuyn:

2.11 : Xem 2 Mch in v bi tp 2.8d v 2.8e. Hm chuyn ca mch 2.9d l: P(s ) = ; vi a=1/RC. Hi hm chuyn ca mch 2.9e c bng khng? Ti sao?

II.12 : S khi chnh tc ca 1 h t kim c v nh sau :

Xc nh : a) Hm chuyn ng vng GH. b) Hm chuyn vng kn C/R. c) T s sai bit E/R. d) T s B/R. e) Phng trnh c trng. 2.13 : Thu gn s sau y v dng chnh tc v tm output C. Cho k l hng so.

II.14 : Xc nh hm chuyn ca h thng trong s khi sau y ri c H1 =1/G1 ; H2 =1/G2 .

II.15 : Xc nh C/R cho mi h sau y :

2.16 : Thu gn cc s khi sau y v dng chnh tc:

2.17 : Xem s khi ca 1 h nh sau . Xc nh p ng ng ra.

LI GII CHNG II

2.1 : Ly bin i laplace phng trnh trn, b qua cc s hng do iu kin u. S2 Y(s)+3SY(s) +2Y(s)=X(s)+SX(s)

Hm chuyn ca h : 2.2 : Ly bin i laplace phng trnh trn, b qua iu kin u: SY(s)+Y(s)=e-STX(s).

Hm chuyn ca h l:

2.3 : Ly laplace phng trnh: Ms2Y(s)=F(s) Hm chuyn : 2.4 : Bin i laplace ca phng trnh: (JS2+BS).((s)=KI(s) Hm chuyn: 2.5 : Hm chuyn l : P(s)=C(s)/R(s). V R(S) =1, khi r(t)=((t). Vy: II.6 : Hm chuyn ca h l phng trnh laplace ca p ng xung lc ca n:

Dng ton t D: D2c+c=r hoc : 2.7 :V o hm ca hm nc l 1 xung lc, nn p ng xung lc ca h l

Bin i laplace ca P(t) v hm chuyn:

2.8 : a) ; vi v b) vi v c) vi v

2.9 :

2.10 :

2.11 : Sinh vin t gii.

2.12 :

e) Phng trnh c trng ca h c xc nh bi: 1( GH=0 Trng hp ny v l hi tip dng nn :1-GH=0 =>s+p-K1K2 = 0 2.13 :

2.14 : Thu gn cc vng trong.

2.15 : Sinh vin t gii. 2.16 :

II.17 : y(t)=5(cost-2sin2t t2). Ging vin: Phm Vn Tn

C S T NG HCBI TP CHNG III 3.1 : Hy xc nh t s C/R v dng s khi chnh tc ca mt h iu khin sau y: sau y: ;

3.2 : Xc nh hm chuyn cho s khi sau y, bng k thut dng HTTH:

3.3 : Xem TD2.4, gii bi ton bng HTTH.

3.4 : Tm hm chuyn C/R ca h thng sau y, vi k l hng s.

3.6 : Dng k thut HTTH gii bi tp 2.13. 3.7 : Tm C/R cho h iu khin sau y:

3.8 : V HTTH cho mch in sau:

3.9 : V HTTH cho mch in sau:

3.10 : V HTTH cho mch in sau, tnh li:

Gi : 5 bin v1, i1, v2, i2, v3. Vi v1 l input. Cn 4 phng trnh c lp.

GII BI TP CHNG III 3.1 : hnh truyn tn hiu:

Dng cng thc Mason xc nh C/R. C hai ng trc tip: P1= G1G2G4 ; P2=G1G3G4 C 3 vng: P11=G1G4H1; P21= - G1G2G4H2 ; P31= - G1G3 G4H2 Khng c vng khng chm. V tt c cc vng u chm c hai ng trc tip. Vy: D 1= 1 ; D 2= 1 Do , t s C/R:

Vi (= 1 - (P11+P21+P31). Suy ra:

T ( 3.25 ) v (3.26) , ta c: G = G1G4(G2 + G3)

V : GH = G1G4(G3H2 +G2H2 - H1)

Dng chnh tc ca s khi ca h thng :

Du tr ti im tng l do vic dng du cng trong cng thc tnh GH trn. S khi trn c th a v dng cui cng nh trong VD2.1 bng cch dng cc nh l bin i khi. 3.2 : hnh truyn tn hiu v trc tip t s khi:

C hai ng trc tip, li l : P1 = G1G2G3 ; P2 = G4

C 3 vng hi tip, li vng l: P11 = - G2H1 ; P21 = G1G2H1 ; P31 = - G2G3H2 Khng c vng no khng chm, vy: ( = 1 - (P11 + P21 + P31) + 0 V (1 = 1 V c 3 vng u chm vi ng 1. V khng c vng no chm vi cc nt ng trc tip th nh, nn: (2= ( ( C 3 vng u khng chm vi ng trc tip th 2). Vy:

3.3 : HTTH v trc tip t s khi.

P1 = G1G2 ; P11 = G1G2H1H2 D = 1- P11 ; D 1 = 1 Vy:

Vi u2 = R =0, Ta c:

P1 = G2 ; P11 = G1G2H1H2 D = 1 - G1G2H1H2 ; D1=1

Vi R = u1 = 0

P1 = G1G2H1 ; P11 = G1G2H1H2 D = 1 - P11 ; D 1 = 1

3.4 :

3.5 : HTTH v trc tip t s khi:

-

3.6 :

3.7 : HTTH v t s khi:

c h