K Thut iu Khin T ng

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K Thut iu Khin T ng

Bin tp bi:Khoa CNTT HSP KT Hng Yn

Cc tc gi:Khoa CNTT HSP KT Hng Yn

Phin bn trc tuyn:http://voer.edu.vn/c/0d7fe2e0

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1. Bi 1: C bn v h thng iu khin t ng1.1. Khi nim iu khin1.2. Cc nguyn tc iu khin1.3. Phn loi iu khin1.4. Lch s pht trin l thuyt iu khin1.5. Mt s v d v cc phn t v h thng t ng

2. Bi 2: M t ton hc h thng iu khin lin tc3. Bi 3: c tnh ng hc ca h thng

3.1. Khi nim v c tnh ng hc3.2. Cc khu ng hc in hnh3.3. c tnh ng hc ca h thng t ng3.4. Kho st c tnh ng hc ca h thng

4. Bi 4: Kho st tnh n nh ca h thng4.1. Khi nim v n nh4.2. Tiu chun n nh i s4.3. Phng php qu o nghim s4.4. Tiu chun n nh tn s

5. Bi 5: nh gi cht lng ca h thng iu khin5.1. Cc tiu chun cht lng5.2. Sai s xc lp5.3. p ng qu 5.4. Cc tiu chun ti u ha p ng qu

6. Bi 6: Thit k h thng iu khin lin tc7. Bi 7: M t ton hc h thng iu khin ri rc

7.1. H thng iu khin ri rc7.2. Php bin i Z7.3. M t h thng ri rc bng hm truyn7.4. M t h thng ri rc bng phng trnh trng thi

8. Bi 8: Phn tch v thit k h thng iu khin ri rc8.1. H thng iu khin ri rc :Khi nim chung8.2. Cc tiu chun n nh8.3. nh gi cht lng ca h thng8.4. Cc phng php tng hp h thng iu khin ri rc

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9. Bi 9: ng dng thit k h thng iu khin t ng10. Bi 10: Lp trnh iu khin h thng t ng t my tnh11. Bi 11: Tho lun v tng kt

11.1. M T H RI RC DNG MATLABTham gia ng gp

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Bi 1: C bn v h thng iu khin tngKhi nim iu khin

Khi nim iu khin

Khi nim

Mt cu hi kh ph bin vi nhng ngi mi lm quen vi l thuyt iu khin liu khin l g?. c khi nim v iu khin chng ta xt v d sau. Gi s chngta ang li xe trn ng, chng ta mun xe chy vi tc c nh 40km/h. tc iu ny mt chng ta phi quan st ng h o tc bit c tc ca xeang chy. Nu tc xe di 40km/h th ta tng ga, nu tc xe trn 40km/h th tagim ga. Kt qu ca qu trnh trn l xe s chy vi tc gn bng tc mongmun. Qu trnh li xe nh vy chnh l qu trnh iu khin. Trong qu trnh iu khinchng ta cn thu thp thng tin v i tng cn iu khin (quan st ng h o tc thu thp thng tin v tc xe), ty theo thng tin thu thp c v mc ch iukhin m chng ta c cch x l thch hp (quyt nh tng hay gim ga), cui cng taphi tc ng vo i tng (tc ng vo tay ga) hot ng ca i tng theo ngyu cu mong mun.

iu khin l qu trnh thu thp thng tin, x l thng tin v tc ng ln h thng p ng ca h thng gn vi mc ch nh trc. iu khin t ng l qu trnhiu khin khng cn s tc ng ca con ngi.

Cu hi th hai cng thng gp i vi nhng ngi mi lm quen vi l thuyt iukhin l Ti sao cn phi iu khin?. Cu tr li ty thuc vo tng trng hp cth, tuy nhin c hai l do chnh l con ngi khng tha mn vi p ng ca h thnghay mun h thng hot ng tng chnh xc, tng nng sut, tng hiu qu kinh t.V d trong lnh vc dn dng, chng ta cn iu chnh nhit v m cho cc cnh v cc cao c to ra s tin nghi trong cuc sng. Trong vn ti cn iu khin cc xehay my bay t ni ny n ni khc mt cch an ton v chnh xc. Trong cng nghip,cc qu trnh sn xut bao gm v s mc tiu sn xut tha mn cc i hi v s anton, chnh xc v hiu qu kinh t.

Trong nhng nm gn y, cc h thng iu khin (HTK) cng c vai tr quan trngtrong vic pht trin v s tin b ca k thut cng ngh v vn minh hin i. Thct mi kha cnh ca hot ng hng ngy u b chi phi bi mt vi loi h thngiu khin. D dng tm thy h thng iu khin my cng c, k thut khng gian v

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h thng v kh, iu khin my tnh, cc h thng giao thng, h thng nng lng,robot,...

Ngay c cc vn nh kim ton v h thng kinh t x hi cng p dng t l thuytiu khin t ng. Khi nim iu khin tht s l mt khi nim rt rng, ni dungquyn sch ny ch cp n l thuyt iu khin cc h thng k thut.

Cc thnh phn c bn ca h thng iu khin

S khi h thng iu khin

Ch thch cc k hiu vit tt:

- r(t) (reference input): tn hiu vo, tn hiu chun

- c(t) (controlled output): tn hiu ra

- cht(t): tn hiu hi tip

- e(t) (error): sai s

- u(t) : tn hiu iu khin.

thc hin c qu trnh iu khin nh nh ngha trn, mt h thng iu khinbt buc gm c ba thnh phn c bn l thit b o lng (cm bin), b iu khin vi tng iu khin. Thit b o lng c chc nng thu thp thng tin, b iu khinthc hin chc nng x l thng tin, ra quyt nh iu khin v i tng iu khinchu s tc ng ca tn hiu iu khin. H thng iu khin trong thc t rt a dng,s khi hnh 1.1 l cu hnh ca h thng iu khin thng gp nht.

Tr li v d li xe trnh by trn ta thy i tng iu khin chnh l chic xe,thit b o lng l ng h o tc v i mt ca ngi li xe, b iu khin l bno ngi li xe, c cu chp hnh l tay ngi li xe. Tn hiu vo r(t) l tc xemong mun (40km/h), tn hiu ra c(t) l tc xe hin ti ca xe, tn hiu hi tip cht(t)l v tr kim trn ng h o tc , sai s e(t) l sai lch gia tc mong mun v tc hin ti, tn hiu iu khin u(t) l gc quay ca tay ga.

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Mt v d khc nh h thng iu khin mc cht lng hnh 1.2 d rt n gin nhngcng c y ba thnh phn c bn k trn. Thit b o lng chnh l ci phao, v trca phao cho bit mc cht lng trong bn. B iu khin chnh l cnh tay n m vanty theo v tr hin ti ca phao, sai lch cng ln th gc m van cng ln. i tngiu khin l bn cha, tn hiu ra c(t) l mc cht lng trong bn, tn hiu vo r(t) lmc cht lng mong mun. Mun thay i mc cht lng mong mun ta thay i dica on ni t phao n cnh tay n.

H thng iu khin mc cht lng

Cc bi ton c bn trong lnh vc iu khin t ng

Trong lnh vc iu khin t ng c rt nhiu bi ton cn gii quyt, tuy nhin cc biton iu khin trong thc t c th quy vo ba bi ton c bn sau:

Phn tch h thng: Cho h thng t ng bit cu trc v thng s. Bi ton t ra ltrn c s nhng thng tin bit tm p ng ca h thng v nh gi cht lng cah. Bi ton ny lun gii c.

Thit k h thng: Bit cu trc v thng s ca i tng iu khin. Bi ton t ra lthit k b iu khin c h thng tha mn cc yu cu v cht lng. Bi tonni chung l gii c.

Nhn dng h thng: Cha bit cu trc v thng s ca h thng. Vn t ra l xcnh cu trc v thng s ca h thng. Bi ton ny khng phi lc no cng gii c.

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Cc nguyn tc iu khin

Cc nguyn tc iu khin

Cc nguyn tc iu khin c th xem l kim ch nam thit k h thng iu khint cht lng cao v c hiu qu kinh t nht.

Nguyn tc 1: Nguyn tc thng tin phn hi Mun qu trnh iu khin t cht lngcao, trong h thng phi tn ti hai dng thng tin: mt t b iu khin n i tngv mt t i tng ngc v b iu khin (dng thng tin ngc gi l hi tip). iukhin khng hi tip (iu khin vng h) khng th t cht lng cao, nht l khi cnhiu.

Cc s iu khin da trn nguyn tc thng tin phn hi l:

iu khin b nhiu (hnh 1.3): l s iu khin theo nguyn tc b nhiu t ura c(t) mong mun m khng cn quan st tn hiu ra c(t) . V nguyn tc, i vi hphc tp th iu khin b nhiu khng th cho cht lng tt.

S khi h thng iu khin b nhiu

iu khin san bng sai lch (hnh 1.4): B iu khin quan st tn hiu ra c(t) , so snhvi tn hiu vo mong mun r(t) tnh ton tn hiu iu khin u(t) . Nguyn tc iukhin ny iu chnh linh hot, loi sai lch, th nghim v sa sai. y l nguyn tcc bn trong iu khin.

S khi h thng iu khin san bng sai lch

iu khin phi hp: Cc h thng iu khin cht lng cao thng phi hp s iu khin b nhiu v iu khin san bng sai lch nh hnh 1.5.

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S khi h thng iu khin phi hp

Nguyn tc 2: Nguyn tc a dng tng xng Mun qu trnh iu khin c cht lngth s a dng ca b iu khin phi tng xng vi s a dng ca i tng. Tnh adng ca b iu khin th hin kh nng thu thp thng tin, lu tr thng tin, truyntin, phn tch x l, chn quyt nh,... ngha ca nguyn tc ny l cn thit k biu khin ph hp vi i tng. Hy so snh yu cu cht lng iu khin v b iukhin s dng trong cc h thng sau:

- iu khin nhit bn i (chp nhn sai s ln) vi iu khin nhit l sy (khngchp nhn sai s ln).

- iu khin mc nc trong bn cha ca khch sn (ch cn m bo lun c nctrong bn) vi iu khin mc cht lng trong cc dy chuyn sn xut (mc cht lngcn gi khng i).

Nguyn tc 3: Nguyn tc b sung ngoi Mt h thng lun tn ti v hot ng trongmi trng c th v c tc ng qua li cht ch vi mi trng . Nguyn tc bsung ngoi tha nhn c mt i tng cha bit (hp en) tc ng vo h thng vta phi iu khin c h thng ln hp en. ngha ca nguyn tc ny l khi thit kh thng t ng, mun h thng c cht lng cao th khng th b qua nhiu ca mitrng tc ng vo h thng.

Nguyn tc 4: Nguyn tc d tr V nguyn tc 3 lun coi thng tin cha y phi phng cc bt trc xy ra v khng c dng ton b lc lng trong iu kin bnhthng. Vn d tr khng s dng, nhng cn m bo cho h thng vn hnh anton.

Nguyn tc 5: Nguyn tc phn cp i vi mt h thng iu khin phc tp cn xydng nhiu lp iu khin b sung cho trung tm. Cu trc phn cp thng s dngl cu trc hnh cy, v d nh h thng iu khin giao thng th hin i, h thngiu khin dy chuyn sn xut.

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S iu khin phn cp

Nguyn tc 6: Nguyn tc cn bng ni Mi h thng cn xy dng c ch cn bngni c kh nng t gii quyt nhng bin ng xy ra.

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Phn loi iu khin

Phn loi iu khin

C nhiu cch phn loi h thng iu khin ty theo mc ch ca s phn loi. V dnu cn c vo phng php phn tch v thit k c th phn h thng iu khin thnhcc loi tuyn tnh v phi tuyn, bin i theo thi gian v bt bin theo thi gian; nucn c vo dng tn hiu trong h thng ta c h thng lin tc v h thng ri rc; nucn c vo mc ch iu khin ta c h thng iu khin n nh ha, iu khin theochng, iu khin theo di,...

Phn loi theo phng php phn tch v thit k

H thng tuyn tnh - H thng phi tuyn

H thng tuyn tnh khng tn ti trong thc t, v tt c cc h thng vt l u l phituyn. H thng iu khin tuyn tnh l m hnh l tng n gin ha qu trnhphn tch v thit k h thng. Khi gi tr ca tn hiu nhp vo h thng cn nm tronggii hn m cc phn t cn hot ng tuyn tnh (p dng c nguyn l xp chng),th h thng cn l tuyn tnh. Nhng khi gi tr ca tn hiu vo vt ra ngoi vng hotng tuyn tnh ca cc phn t v h thng, th khng th xem h thng l tuyn tnhc. Tt c cc h thng thc t u c c tnh phi tuyn, v d b khuch i thngc c tnh bo ha khi tn hiu vo tr nn qu ln, t trng ca ng c cng c ctnh bo ha. Trong truyn ng c kh c tnh phi tuyn thng gp phi l khe hv vng cht gia cc bnh rng, c tnh ma st, n hi phi tuyn... Cc c tnh phituyn thng c a vo HTK nhm ci thin cht lng hay tng hiu qu iukhin. V d nh t thi gian iu khin l ti thiu trong cc h thng tn la hayiu khin phi tuyn ngi ta s dng b iu khin on-off (bang-bang hay relay). Ccng phn lc c t cnh ng c to ra mmen phn lc iu khin. Cc ng nythng c iu khin theo kiu full on - full off, ngha l mt lng kh np vo mtng nh trc trong khong thi gian xc nh, iu khin t th ca phi tuyn.

H thng bt bin - h thng bin i theo thi gian

Khi cc thng s ca HTK khng i trong sut thi gian hot ng ca h thng, thh thng c gi l h thng bt bin theo thi gian. Thc t, hu ht cc h thng vtl u c cc phn t tri hay bin i theo thi gian. V d nh in tr dy qun ngc b thay i khi mi b kch hay nhit tng.

Mt v d khc v HTK bin i theo thi gian l h iu khin tn la, trong khilng ca tn la b gim trong qu trnh bay. Mc d h thng bin i theo thi gian

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khng c c tnh phi tuyn, vn c coi l h tuyn tnh, nhng vic phn tch v thitk loi h thng ny phc tp hn nhiu so vi h tuyn tnh bt bin theo thi gian.

Phn loi theo loi tn hiu trong h thng

H thng lin tc

H thng lin tc l h thng m tn hiu bt k phn no ca h cng l hm lin tctheo thi gian. Trong tt c cc HTK lin tc, tn hiu c phn thnh AC hay DC.Khi nim AC v DC khng ging trong k thut in m mang ngha chuyn mntrong thut ng HTK. HTK AC c ngha l tt c cc tn hiu trong h thng uc iu ch bng vi dng s iu ch. HTK DC c hiu n gin l h c cctn hiu khng c iu ch, nhng vn c tn hiu xoay chiu. Hnh 1.7 l s mtHTK DC kn v dng sng p ng qu ca h.

Cc thnh phn ca HTK DC l bin tr, khuch i DC, ng c DC, tachometerDC...

S HTK DC vng kn

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S HTK AC vng kn

Hnh 1.8 l s mt HTK AC c cng chc nng nh HTK hnh 1.7. Trongtrng hp ny, tn hiu trong h u c iu ch, ngha l thng tin c truyn inh mt sng mang AC. Ch rng bin iu khin u ra ca i tng vn ging nh HTK DC. HTK AC c s dng rng ri trong h thng iu khin my bay vtn la, nhiu v tn hiu l l vn phi quan tm. Vi tn s sng mang t 400Hz tr ln, HTK AC loi b c phn ln cc nhiu tn s thp. Cc thnh phn caHTK AC l thit b ng b, khuch i AC, ng c AC, con quay hi chuyn, myo gia tc... Thc t, mt h thng c th lin kt cc thnh phn AC v DC, s dngcc b iu ch v cc b gii iu ch thch ng vi tn hiu ti cc im khc nhautrong h thng.

H thng ri rc

Khc vi HTK lin tc, HTK ri rc c tn hiu mt hay nhiu im trong hthng l dng chui xung hay m s. Thng thng HTK ri rc c phn lm hailoi: HTK ly mu d liu v HTK s. HTK ly mu d liu dng d liu xung.HTK s lin quan n s dng my tnh s hay b iu khin s v vy tn hiu trongh c m s ha, m s nh phn chng hn.

Ni chung, mt HTK ly mu d liu ch nhn d liu hay thng tin trong mt khongthi gian xc nh. V d, tn hiu sai lch ca HTK ch c th c cung cp didng xung v trong khong thi gian gia hai xung lin tip HTK s khng nhn cthng tin v tn hiu sai lch. HTK ly mu d liu c th xem l mt HTK AC vtn hiu trong h thng c iu ch xung.

Hnh :minh ha hot ng ca mt h thng ly mu d liu. Tn hiu lin tc r(t) ca vo h thng, tn hiu sai lch e(t) c ly mu bi thit b ly mu, ng ra cathit b ly mu l chui xung tun t. Tc ly mu c th thng nht hoc khng.

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Mt trong nhng u im quan trng ca thao tc ly mu l cc thit b t tin trongh c th chia s thi gian dng chung trn nhiu knh iu khin. Mt li im khcl nhiu t hn.

Do my tnh cung cp nhiu tin ch v mm do, iu khin bng my tnh ngy cngph bin. Nhiu h thng vn ti hng khng s dng hng ngn cc linh kin ri rcch chim mt khong khng khng ln hn quyn sch ny. Hnh 1.10 trnh by ccthnh phn c bn ca b phn t li trong iu khin tn la.

S khi HTK ly mu d liu

S khi HTK tn la

Phn loi theo mc tiu iu khin

iu khin n nh ha

Mc tiu iu khin l kt qu tn hiu ra bng tn hiu vo chun r(t) vi sai lch chophp exl (sai s ch xc lp).

Khi tn hiu vo r(t) khng i theo thi gian ta c h thng iu khin n nh ha hayh thng iu chnh, v d nh h thng n nh nhit , in p, p sut, nng , tc,...

iu khin theo chng trnh

Nu r(t) l mt hm nh trc theo thi gian, yu cu p ng ra ca h thng sao chpli cc gi tr ca tn hiu vo r(t) th ta c h thng iu khin theo chng trnh.

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V d h thng iu khin my cng c CNC, iu khin t ng nh my xi mngHong Thch, h thng thu nhp v truyn s liu h thng in, qun l vt t nhmy...

iu khin theo di

Nu tn hiu tc ng vo h thng r(t) l mt hm khng bit trc theo thi gian, yucu iu khin p ng ra c(t) lun bm st c r(t), ta c h thng theo di. iu khintheo di c s dng rng ri trong cc HTK v kh, h thng li tu, my bay...

iu khin thch nghi

Tn hiu v(t) chnh nh li tham s iu khin sao cho h thch nghi vi mi bin ngca mi trng ngoi.

Nguyn tc t chnh nh

iu khin ti u - hm mc tiu t cc tr

V d cc bi ton qui hoch, vn tr trong kinh t, k thut u l cc phng php iukhin ti u.

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Lch s pht trin l thuyt iu khin

Lch s pht trin l thuyt iu khin

iu khin c in (classical control)

L thuyt iu khin c in (trc 1960) m t h thng trong min tn s (php bini Fourier) v mt phng s (php bin i Laplace). Do da trn cc php bin i ny,l thuyt iu khin c in ch yu p dng cho h tuyn tnh bt bin theo thi gian,mt d c mt vi m rng p dng cho h phi tuyn, th d phng php hm mt. L thuyt iu khin kinh in thch hp thit k h thng mt ng vo - mt ngra (SISO: single-input/single-output), rt kh p dng cho cc h thng nhiu ng vo -nhiu ng ra (MIMO: multi-input/multi-output) v cc h thng bin i theo thi gian.

Cc phng php phn tch v thit k h thng trong l thuyt iu khin c in gmc phng php Nyquist, Bode, v phng php qu o nghim s. thit k hthng dng phng php Nyquist v Bode cn m t h thng di dng p ng tn s(p ng bin v p ng pha), y l mt thun li v p ng tn s c th o cbng thc nghim. M t h thng cn thit k dng phng php qu o nghims l hm truyn, hm truyn cng c th tnh c t p ng tn s.

Hm truyn ca cc h thng phc tp c tnh bng cch s dng s khi hay s dng tn hiu. M t chnh xc c tnh ng hc bn trong h thng l khng cnthit i vi cc phng php thit k c in, ch c quan h gia ng vo v ng ra lquan trng.

Cc khu hiu chnh n gin nh hiu chnh vi tch phn t l PID (Proportional IntegralDerivative), hiu chnh sm tr pha,... thng c s dng trong cc h thng iukhin kinh in.

nh hng ca cc khu hiu chnh ny n biu Nyquist, biu Bode v qu onghim s c th thy c d dng, nh c th d dng la chn c khu hiuchnh thch hp.

iu khin hin i (modern control)

(t khong nm 1960 n nay)

K thut thit k h thng iu khin hin i da trn min thi gian. M t ton hcdng phn tch v thit k h thng l phng trnh trng thi. M hnh khng giantrng thi c u im l m t c c tnh ng hc bn trong h thng (cc bin trngthi) v c th d dng p dng cho h MIMO v h thng bin i theo thi gian. L

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thuyt iu khin hin i ban u c pht trin ch yu cho h tuyn tnh, sau c m rng cho h phi tuyn bng cch s dng l thuyt ca Lyapunov.

B iu khin c s dng ch yu trong thit k h thng iu khin hin i l biu khin hi tip trng thi. Ty theo cch tnh vector hi tip trng thi m ta cphng php phn b cc, iu khin ti u, iu khin bn vng...Vi s pht trin cal thuyt iu khin s v h thng ri rc, l thuyt iu khin hin i rt thch hp thit k cc b iu khin l cc chng trnh phn mm chy trn vi x l v mytnh s. iu ny cho php thc thi c cc b iu khin c c tnh ng phc tphn cng nh hiu qu hn so vi cc b iu khin n gin nh PID hay sm tr phatrong l thuyt c in.

iu khin thng minh (intelligent control)

iu khin kinh in v iu khin hin i, gi chung l iu khin thng thng(conventional control) c khuyt im l thit k c h thng iu khin cn phibit m hnh ton hc ca i tng. Trong khi thc t c nhng i tng iu khinrt phc tp, rt kh hoc khng th xc nh c m hnh ton. Cc phng php iukhin thng minh nh iu khin m, mng thn kinh nhn to, thut ton di truyn mphng/bt chc cc h thng thng minh sinh hc, v nguyn tc khng cn dng mhnh ton hc thit k h thng, do c kh nng ng dng thc t rt ln. Khuytim ca iu khin m l qu trnh thit k mang tnh th sai, da vo kinh nghimca chuyn gia. Nh kt hp logic m vi mng thn kinh nhn to hay thut ton ditruyn m thng s b iu khin m c th thay i thng qua qu trnh hc hay qutrnh tin ha, v vy khc phc c khuyt im th sai. Hin nay cc b iu khinthng thng kt hp vi cc k thut iu khin thng minh to nn cc b iu khinlai iu khin cc h thng phc tp vi cht lng rt tt.

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Mt s v d v cc phn t v h thng t ng

Mt s v d v cc phn t v h thng t ng

Cc phn t t ng

Nh cp , mt HTK gm cc phn t c bn sau:

* Phn t cm bin, thit b o lng

* i tng hay qu trnh iu khin

* Thit b iu khin, cc b iu khin th ng v tch cc

Cc loi cm bin, thit b o lng

Bin tr tuyn tnh, bin tr gc quay dng chuyn i s dch chuyn thnh inp. Ngoi ra cn c th chuyn i kiu in cm v in dung... Nguyn tc chung o cc i lng khng in nh nhit , quang thng, lc, ng sut, kch thc, dichuyn, tc ... bng phng php in l bin i chng thnh tn hiu in. Cu trcthit b o gm ba thnh phn: b phn chuyn i hay cm bin, c cu o in v ccs mch trung gian hay mch gia cng tn hiu v d nh mch khuch i, chnhlu, n nh. Cm bin xenxin lm phn t o lng trong cc h bm st gc quay,truyn ch th gc quay c ly xa m khng thc hin c bng c kh. Bin p xoayhay cn gi l bin p quay dng bin i in p ca cun s cp hoc gc quayca cun s cp thnh tn hiu ra tng ng vi chng. Bin p xoay sin, cos o gcquay ca rto, trn t cun s cp, thnh in p t l thun vi sin hay cos ca gcquay . Bin p xoay tuyn tnh bin i lch gc quay ca rto thnh in p t ltuyn tnh. Con quay 3 bc t do v con quay 2 bc t do c s dng lm cc b cmbin o sai lch gc v o tc gc tuyt i trong cc h thng n nh ng ngmca cc dng c quan st v ngm bn.

Cm bin tc - b m ha quang hc l a m trn c khc vch m nh sng cth i qua c. Pha sau a m t phototransistor chu tc dng ca mt ngun sng.ng c v a m c gn ng trc, khi quay nh sng chiu n phototransistor lcb ngn li, lc khng b ngn li lm cho tn hiu cc colecto l mt chui xung. Trna m c khc hai vng vch, ngoi A trong B c cng s vch, nhng lch 90o (vchA trc B l 90o) . Nu a m quay theo chiu kim ng h th chui xung B s nhanhhn chui xung A l 1/2 chu k v ngc li.

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Thit b o tc nh DC Tachometer, AC Tachometer, Optical Tachometer. Cm binnhit nh Pt 56?, Pt 100?, Thermocouple...

i tng iu khin

i tng iu khin c th l thit b k thut, dy chuyn sn xut, qui trnh cngngh... l mc tiu iu khin ca con ngi trong cc lnh vc khc nhau.

Cc phn t chp hnh thng dng trong KT l cc loi ng c bc, ng c DC,servomotor, ng c AC, ng c thy lc kh nn... ng c bc c dng nhv chnh xc do c cu trc rto v stato kh c bit. Rto thng thng l cc namchm vnh cu c cnh c x rnh rng ca sut chu vi ca rto, tp trung ngsc t ti cc mi rng. Tng t, stato c ch to thng dng c bn bi dy qunxen k theo cc t cc. Khi c dng in chy qua mt cun dy stato, rto s quay mtgc n v tr cn bng t thng l giao im ca hai rng stato v rto. Thay i tht cc cun dy 1, 2, 3, 4 rto s lch mt gc l 90o. C ba cch iu khin ng cbc: iu khin hnh trnh nng lng thp, iu khin thng, iu khin 1/2 bc.V cun dy stato c in tr thun rt nh khong 0,2? do vy thng iu khin bngcc ngun dng thng dng nht l transistor, Fet....

Mt loi o lng iu khin khc cng thng gp trong cng nghip l h thng nhit,v d nh l nung trong dy chuyn sn xut gch men, l sy trong dy chuyn chbin thc phm, h thng lm lnh trong cc dy chuyn ch bin thy sn. Yu cuiu khin i vi h thng nhit thng l iu khin n nh ha hoc iu khin theochng trnh. M hnh ton ca ng c DC v l nhit s c trnh by mc 2.2.2.

K thut giao tip my tnh

Thit b iu khin rt a dng, c th l mt mch RC, mch khuch i thut ton,mch x l hay my tnh PC. Trc y cc b iu khin nh PID, sm tr pha thngc thc hin bng cc mch ri (xem mc 2.2.2.2). Gn y do s pht trin ca lthuyt iu khin ri rc v k thut vi x l cc b iu khin trn c thc thibng cc chng trnh phn mm chy trn vi x l hay my tnh. Hin nay my tnh khng nh l thit b iu khin a nng v tin cy. Phn di y s trnh by mt svn lin quan n k thut giao tip my tnh.

B chuyn i ADC v DAC

Hnh : l s Card A/D v D/A 8 bit. Trong cc ng dng cn chnh xc cao hnc th s dng card A/D v D/A 12 bit.

Card giao tip vi my tnh

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V d Card giao tip s dng IC8255 gn trn slot m rng ca Main Board my tnh(H.1.13).

Cc loi giao thc truyn tin

RS232C serial Interface, chu ni 25 chn dng truyn d liu ni tip vi tc nhhn 20.000 bits/second (nm 1969). Khong 1975 n 1977 p dng RS-422, RS-423,RS-449. RS-449 chu ni 37 chn, tc truyn c th nhanh gp nm ln so viRS-232C.

Vo nm 1970-1975 pht trin Bus d liu song song vi IEEE-488.

Nm 1978 - IEEE - 583 c slots cho 25 moduls, ni trc tip vi Bus I/O ca my tnh,ni song song ti 7 CRATES.

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Card AD v DA 8 bit

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Card xut nhp

Cc ng dng ca h thng iu khin t ng

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minh ha mt h thng iu khin mc cht lng trong b.

Tc dng chy ng ra qua van V1 l bin i, h thng c th duy tr mc cht lng h= const vi sai s cho php kh chnh xc. Nu mc cht lng trong b khng ng, mtin p sai lch c to ra qua khuch i a vo b iu khin ng c iu chnhvan V2 khi phc li mc cht lng mong mun bng cch iu chnh tc dngchy ng vo.

Trong trng hp dng chy vo c tc hng s, phao c hai cp tip im thngng, thng m iu khin ng m ng c in AC. trnh ng c b ngngt khng dt khot, to hai mc tng ng vng tr Trigger Schmidt h.

H thng iu khin t ng mc cht lng trong b

minh ha mt h thng nh v dng cho b phng tn la.

H thng hi tip ny c thit k nh v b phng kh chnh xc da trn cc lnh tbin tr R1 l tn hiu vo c t xa h thng. Bin tr R2 cho tn hiu hi tip trv b khuch i vi sai, hot ng nh mt b pht hin sai lch. Nu c sai lch, ckhuch i a n ng c, iu chnh v tr trc ng ra tng ng vi v tr trc ngvo v sai lch bng 0.

Mt h thng t ng nh v tr dng cho

b phng tn la

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Robot l mt lnh vc rt quan trng trong ng dng cc HTK.

Vo thp nin 1960, ngi ta bt u nhn ra Robot l mt cng c quan trng trgip cng vic ch to, t cc ng dng ca chng trong nhiu h thng ch to khcnhau c pht trin nhanh chng. L thuyt iu khin t ng, nguyn tc iukhin thch nghi, cc hm Lyapunov c p dng c c Robot c ng theo mun hay lc cn thit. Lnh vc ca Robotics cng ty thuc vo cch s dng cccm bin quan st v cc my tnh lp trnh cho Robot hon thnh cng vic theo yucu.

Robot c sng to ra thc hin nhiu cng vic khc nhau, lm cu ni gia cclnh vc ch to, cc nhim v vn chuyn khng gian v chm sc y t. ng dng chyu ca Robot l t ng ha qu trnh sn xut. Robot c s dng trong dy chuynsn xut xe hi, l mt thnh phn trong tu con thoi khng gian ca NASA, l bn gipvic cho con ngi Robot tr gip trong cc bnh vin, thc hin cc cng vic ca yt chm sc bnh nhn. Cc Robot ny s dng cc cm bin quan st, siu m v hngngoi iu khin thang my, trnh cc vt cn dc theo ng i, mang cc khaythc n theo yu cu, ly thuc hay cc vt mu ca phng th nghim, ghi li tnh trngsc khe ca ngi bnh, bo co cng vic qun l

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Bi 2: M t ton hc h thng iu khinlin tcM t ton hc h thng iu khin lin tc

Khi nim

c c s cho phn tch, thit k cc h thng iu khin c bn cht vt l khc nhau,c s chnh l ton hc. Tng qut quan h gia tn hiu vo v tn hiu ra ca hthng tuyn tnh c th biu din bng phng trnh vi phn bc cao. Vic kho st hthng da vo phng trnh vi phn bc cao thng gp nhiu kh khn. C hai phngphp m t ton hc h thng t ng gip cho vic kho st h thng d dng hn, lphng php hm truyn t v phng php khng gian trng thi. Phng php hmtruyn t chuyn quan h phng trnh vi phn thnh quan h phn thc i s nhphp bin i Laplace, trong khi phng php khng gian trng thi bin i phngtrnh vi phn bc cao thnh h phng trnh vi phn bc nht bng cch t cc bin ph(bin trng thi). Mi phng php m t h thng u c nhng u im ring. Trongquyn sch ny chng ta s m t h thng bng c hai phng php.

Hm truyn t v i s s khi

Php bin i Laplace

1- nh ngha

Cho f(t) l hm xc nh vi mi t = 0, bin i Laplace ca f(t) l:

trong : s - l bin phc (bin Laplace) s j = s + ?

- l ton t bin i Laplace

F(s) - l nh ca hm f(t) qua php bin i Laplace.

Bin i Laplace tn ti khi tch phn biu thc nh ngha (2.1) hi t.

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2- Tnh cht ca php bin i Laplace

Tnh tuyn tnh

Nu hm f1(t) c bin i Laplace l

v hm f2(t) c bin i Laplace l

th:

nh ca o hm

Nu hm f(t) c bin i Laplace l

th:

trong f(0+) l iu kin u.

Nu iu kin u bng 0 th:

nh ca tch phn Nu hm f(t) c bin i Laplace l

th:

nh l chm tr

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Lm tr hm f(t) mt thi gian l T

Nu f(t) c lm tr mt khong thi gian T, ta c hm

f(t-T). Khi :

nh l gi tr cui

Nu hm f(t) c bin i Laplace l

th:

3- Bin i Laplace ca mt s hm c bn

Khi kho st h thng t ng ngi ta thng t tn hiu vo l cc tn hiu c bn. Vd nh kho st h thng iu khin n nh ha tn hiu vo c chn l hm nc, kho st h thng iu khin theo di tn hiu vo c chn l hm hm dc, nhiutc ng vo h thng c th m t bng hm dirac. Tn hiu ra ca h thng t ngcng c dng l t hp ca cc tn hiu c bn nh hm nc, hm m, hm sin, Do trong mc ny chng ta xt bin i Laplace ca cc hm c bn s dng trongvic phn tch v thit k h thng cc phn sau.

Hm xung n v (hm dirac) (H.2.2a)

Hm xung n v thng c s dng m t nhiu tc ng vo h thng.

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Cc hm c bn

a) Hm xung n v; b) Hm nc n v; c) Hm dc n v

d) Hm parabol; e) Hm m; f) Hm sin

Theo nh ngha:

Hm nc n v (H.2.2b)

Trong cc h thng iu khin n nh ha, tn hiu vo c dng hm nc n v.

Theo nh ngha php bin i Laplace ta c:

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Hm truyn t

1- nh ngha

Tn hiu vo v tn hiu ra ca h thng t ng

Quan h gia tn hiu vo v tn hiu ra ca mi h thng tuyn tnh bt bin lin tcu c th m t bi phng trnh vi phn h s hng:

trong cc h s

v

l thng s ca h thng

; n l bc ca h thng. H thng c gi l hp thc (proper) nu n = m, h thngc gi l khng hp thc nu n < m. Ch c cc h thng hp thc mi tn ti trongthc t.

Gi s iu kin u bng 0, bin i Laplace hai v phng trnh trn ta c:

G(s) gi l hm truyn ca h thng.

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nh ngha: Hm truyn ca mt h thng l t s gia bin i Laplace ca tn hiu rav bin i Laplace ca tn hiu vo khi iu kin u bng 0.

2- Hm truyn t ca cc khu hiu chnh

Trong h thng t ng cc khu hiu chnh chnh l cc b iu khin n gin cs dng bin i hm truyn t ca h thng nhm mc ch tng tnh n nh, cithin p ng v gim thiu nh hng ca nhiu ln cht lng ca h thng.

Thng khu hiu chnh l cc mch in. C hai dng mch hiu chnh l mch hiuchnh th ng v mch hiu chnh tch cc. Mch hiu chnh th ng khng c cc bkhuch i, li ca cc mch ny thng nh hn hay bng 1. Ngc li mch hiuchnh tch cc c cc khu khuch i, li ca cc mch ny thng ln hn 1. Phnny trnh by hm truyn mt s khu hiu chnh thng c s dng trong thit k hthng.

c tnh ca cc khu hiu chnh ny s c phn tch cc chng sau.

Khu hiu chnh th ng:

Cc khu hiu chnh th ng

a) Khu tch phn bc mt; b) Khu vi phn bc mt

c) Khu sm pha; d) Khu tr pha

Khu hiu chnh tch cc :

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Cc khu hiu chnh tch cc

a) Khu t l; b) Khu tch phn t l PI

c) Khu vi phn t l; d) Khu vi tch phn t l PID

S khi

mc 2.2.2 chng ta dn ra c hm truyn ca cc phn t c bn trong h thngiu khin. Trong thc t cc h thng thng gm nhiu phn t c bn kt ni vinhau. Mt cch n gin nhng rt hiu qu trong vic biu din cc h thng phc tpl dng s khi.

S khi ca mt h thng l hnh v m t chc nng ca cc phn t v s tc ngqua li gia cc phn t trong h thng. S khi gm c ba thnh phn l khi chcnng, b tng v im r nhnh.

- Khi chc nng: Tn hiu ra ca khi chc nng bng tch tn hiu vo v hm truyn

- im r nhnh: Ti im r nhnh mi tn hiu u bng nhau.

- B tng: Tn hiu ra ca b tng bng tng i s ca cc tn hiu vo.

Cc thnh phn c bn ca s khi

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a) Khi chc nng; b) im r nhnh; c) B tng

S dng tn hiu

S dng tn hiu v cng thc Mason

1- nh ngha

biu din h thng t ng, ngoi phng php s dng s khi, ta cn c th sdng phng php s dng tn hiu. Hy so snh hai hnh v di y, hnh 2.14b ls dng tn hiu ca h thng c s khi nh hnh 2.7a.

Biu din h thng bng s dng tn hiu

a) S khi; b) S dng tn hiu

nh ngha

S dng tn hiu l mt mng gm cc nt v nhnh.

- Nt: mt im biu din mt bin hay tn hiu trong h thng.

- Nhnh: ng ni trc tip hai nt, trn mi nhnh c mi tn ch chiu truyn ca tnhiu v c ghi hm truyn cho bit mi quan h gia tn hiu hai nt.

- Nt ngun: nt ch c cc nhnh hng ra.

- Nt ch: nt ch c cc nhnh hng vo.

- Nt hn hp: nt c c cc nhnh ra v cc nhnh vo.

Ti nt hn hp, tt c cc tn hiu ra u bng nhau v bng tng i s ca cc tn hiuvo.

- ng tin: ng gm cc nhnh lin tip c cng hng tn hiu i t nt ngunn nt ch v ch qua mi nt mt ln.

- li ca mt ng tin: tch ca cc hm truyn ca cc nhnh trn ng tin .

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- Vng kn: ng khp kn gm cc nhnh lin tip c cng hng tn hiu v ch quami nt mt ln.

- li ca mt vng kn: tch ca cc hm truyn ca cc nhnh trn vng kn .

2- Cng thc Mason

Hm truyn tng ng ca h thng t ng biu din bng s dng tn hiu cth tnh theo cng thc:

trong :

- li ca ng tin th k

- nh thc ca s dng tn hiu:

- tng li vng ca cc vng kn c trong s dng tn hiu.

- tng cc tch li vng ca hai vng khng dnh nhau.

- tng cc tch li vng ca ba vng khng dnh nhau.

- nh thc con ca s dng tn hiu.

c suy ra t bng cch b i cc vng kn c dnh ti ng tin Pk..

Ch : * khng dnh = khng c nt no chung.

* dnh = c t nht nt chung.

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Phng php khng gian trng thi

Khi nim

Nh trnh by u chng ny, quan h gia ng vo v ng ra ca h thng lintc bt k c th m t bng phng trnh vi phn bc n. Nghin cu h thng da trnphng trnh vi phn bc n rt kh khn, do cn m t ton hc khc gip cho vicnghin cu h thng d dng hn. Phng php hm truyn chuyn quan h phngtrnh vi phn cp n thnh phn thc i s nh php bin i Laplace. Nghin cu hthng m t bng hm truyn thun li hn bng phng trnh vi phn, tuy nhin hmtruyn c mt s khuyt im sau:

- Ch p dng c khi iu kin u bng 0.

- Ch p dng c cho h thng tuyn tnh bt bin, khng th p dng m t h phituyn hay h bin i theo thi gian.

- Nghin cu h thng trong min tn s.

Mt phng php khc c s dng kho st h thng t ng l phng phpkhng trng thi. Phng php khng gian trng thi chuyn phng trnh vi phn bcn thnh n phng trnh vi phn bc nht bng cch t n bin trng thi. Phng phpkhng gian trng thi khc phc c cc khuyt im ca phng php hm truyn.

Trng thi ca h thng, h phng trnh bin trng thi

Trng thi

Trng thi ca mt h thng l tp hp nh nht cc bin (gi l bin trng thi) m nubit gi tr ca cc bin ny ti thi im to v bit cc tn hiu vo thi im t = to, tahon ton c th xc nh c p ng ca h thng ti mi thi im t = to.

H thng bc n c n bin trng thi. Cc bin trng thi c th chn l bin vt l hockhng phi l bin vt l. V d ng c DC l h bc hai, c hai bin trng thi c thchn l tc ng c v dng in phn ng (bin vt l). Tuy nhin ta cng c thchn hai bin trng thi khc.

Phng php m t h thng bng cch s dng cc bin trng thi gi l phng phpkhng gian trng thi.

Vct trng thi

n bin trng thi hp thnh vct ct gi l vect trng thi, k hiu:

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Bng cch s dng cc bin trng thi, ta c th chuyn phng trnh vi phn bc n mt h thng thnh h n phng trnh vi phn bc nht vit di dng ma trn nh sau:

trong :

Phng trnh (2.17) c gi l phng trnh trng thi ca h thng. Nu A l ma trnthng, ta gi (2.172) l h phng trnh trng thi dng thng; nu A l ma trncho, ta gi (2.17) l h phng trnh trng thi dng chnh tc.

i vi cc h thng hp thc cht (bc t s hm truyn nh hn bc mu s) th D =0.

H thng m t bi h phng trnh trng thi (2.17) c th biu din di dng s trng thi nh sau:

S trng thi ca h thng

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Bi 3: c tnh ng hc ca h thngKhi nim v c tnh ng hc

Khi nim v c tnh ng hc

c tnh ng ca h thng m t s thay i tn hiu u ra ca h thng theo thigian khi c tc ng u vo. Trong thc t cc h thng iu khin rt a dng, tuynhin nhng h thng c m t bng m hnh ton hc c dng nh nhau s c ctnh ng hc nh nhau. kho st c tnh ng ca h thng tn hiu vo thngc chn l tn hiu c bn nh hm xung n v, hm nc n v hay hm iu ha.Ty theo dng ca tn hiu vo th m c tnh ng thu c l c tnh thi gian hayc tnh tn s.

c tnh thi gian

c tnh thi gian ca h thng m t s thay i tn hiu u ra ca h thng khi tnhiu vo l hm xung n v hay hm nc n v.

Tn hiu vo v tn hiu ra ca h thng

Nu tn hiu vo l hm xung n v

th p ng ca h thng l:

g(t) c gi l p ng p ng xung hay cn gi l hm trng lng ca h thng.

Vy p ng xung l p ng ca h thng khi tn hiu vo l hm xung n v. Theobiu thc p ng xung chnh l bin i Laplace ngc ca hm truyn.

Nu tn hiu vo l hm nc n v r(t) = 1(t) th p ng ca h thng l:

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Biu thc (3.2) c c do p dng tnh cht nh ca tch phn ca php bin iLaplace. t:

h(t) c gi l p ng nc hay cn gi l hm qu ca h thng.

Vy p ng nc l p ng ca h thng khi tn hiu vo l hm nc n v. Theo biuthc (3.3) p ng nc chnh l tch phn ca p ng xung.

V d : Cho h thng c hm truyn l:

Xc nh hm trng lng v hm qu ca h thng.

Gii. Hm trng lng:

Hm qu :

Thc hin php bin i Laplace ngc ta c kt qu nh

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trn. g

Nhn xt: bi trc ta bit c ba cch m t ton hc h thng tuyn tnh lin tc ldng phng trnh vi phn, hm truyn v h phng trnh trng thi. Do quan h giahm trng lng v hm qu vi hm truyn cho bi biu thc (3.1) v (3.3) ta thyrng c th dng hm trng lng hay hm qu m t ton hc h thng t ng.Khi bit hm trng lng hay hm qu th s suy ra c hm truyn d dngbng cc cng thc sau y:

V d: Cho h thng c p ng nc n v l:

Xc nh hm truyn ca h thng.

Gii. Theo bi, ta c:

c tnh tn s

c tnh tn s ca h thng tuyn tnh lin tc m t quan h gia tn hiu ra v tn hiuvo ca h thng trng thi xc lp khi thay i tn s ca tn hiu dao ng iu hatc ng u vo ca h thng.

Xt h tuyn tnh lin tc c hm truyn l G(s), gi s tn hiu vo l tn hiu hnh sin:

Tn hiu ra ca h thng l:

Gi s G(s) c n cc pi phn bit tha

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, ta c th phn tch C(s) di dng:

Bin i Laplace ngc biu thc trn, ta c:

Nu h thng n nh th tt c cc cc pi u c phn thc m. Khi :

Nu G(s) c cc bi th ta cng c th chng minh c p ng xc lp ca h thngc dng (3.6). Cc h s

v

xc nh bi cng thc:

rt gn biu thc ta c:

Biu thc cho thy trng thi xc lp tn hiu ra ca h thng l tn hiu hnh sin, cngtn s vi tn hiu vo, bin t l vi bin tn hiu vo (h s t l l

) v lch pha so vi tn hiu vo ( lch pha l

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).

nh ngha: c tnh tn s ca h thng l t s gia tn hiu ra trng thi xc lpv tn hiu vo hnh sin.

Ta rt ra:

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Cc khu ng hc in hnh

trn chng ta va cp n khi nim c tnh ng hc ca h thng t ng. Trongmc ny, chng ta s xt c tnh ng hc ca mt s khu c bn nh khu t l, viphn, tch phn, qun tnh bc mt, dao ng bc hai, Trn c s c tnh ng hcca cc khu c bn, mc s trnh by cch xy dng c tnh ng hc ca h thng tng.

Khu t l (khu khuch i)

Vy tn hiu ra ca khu t l bng tn hiu vo khuch i ln K ln. Hnh 3.2 m thm trng lng v hm qu ca khu t l.

c tnh thi gian ca khu t l

a) Hm trng lng; b) Hm qu

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c tnh tn s ca khu t l1. Biu Bode; b) Biu Nyquist

Cc biu thc trn cho thy c tnh tn s ca khu t l l hng s vi mi , do biu Bode v bin l mt ng song song vi trc honh, cch trc honh

; biu Bode v pha l mt ng nm ngang trng vi trc honh; biu Nyquistl mt im do vct

khng i vi mi . Xem hnh

Khu tch phn l tng

c tnh thi gian:

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Vy hm trng lng v hm qu ca khu tch phn l tng tng ng l hm ncn v v hm dc n v .Mt c im quan trng cn quan tm l hm qu cakhu tch phn l tng tng n v cng.

c tnh thi gian ca khu tch phn l tng a) Hm trng lng; b) Hm qu

Nu v L() trong h ta vung gc thng thng th th L() l ng cong.Tuy nhin do trc honh ca biu Bode c chia theo thang logarith c s 10 nnd dng thy rng biu Bode v bin ca khu tch phn l tng l ng thngc dc 20dB/dec. Biu Bode v pha ca khu tch phn l tng l ng nmngang do

vi mi . Biu Nyquist l na di ca trc tung do

c phn thc bng 0, phn o lun lun m

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c tnh tn s ca khu tch phn l tng a) Biu Bode; b) Biu Nyquist

Khu vi phn l tng

Hm qu :

Hm trng lng:

Hm qu ca khu vi phn l tng hm xung n v ,hm trng lng l o hmca hm qu , ch c th m t bng biu thc ton hc (hnh 3.7), khng biu dinbng th c.

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Hm qu ca khu vi phn l tng

c tnh tn s ca khu vi phn l tng hon ton tri ngc so vi khu tch phnl tng. Biu Bode v bin ca khu vi phn l tng l ng thng c dc+20dB/dec, biu Bode v pha l ng nm ngang

. Biu Nyquist l na trn ca trc tung do

c phn thc bng 0, phn o lun lun dng

c tnh tn s ca khu vi phn l tng a) Biu Bode; b) Biu Nyquist

Khu qun tnh bc nht

Hm truyn:

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c tnh thi gian:

Hm trng lng:

Hm qu :

Hm trng lng ca khu qun tnh bc nht l hm m suy gim v 0, hm qu tng theo qui lut hm m n gi tr xc lp bng 1. Tc bin thin ca hm trnglng v hm qu t l vi T nn T c gi l thi hng ca khu qun tnh bcnht. T cng nh th p ng cng nhanh, T cng ln th p ng cng chm. Hnh 3.8minh ha c tnh thi gian ca hai khu qun tnh bc nht c thi hng tng ng lT1 v T2, trong T1 < T2.

Thay t = T vo biu thc 3.42 ta c h(T) = 0,63 , do thi hng ca khu qun tnhbc nht chnh l thi gian cn thit hm qu tng ln bng 63% gi tr xc lp(gi tr xc lp ca h(t) = 1). Mt cch khc xc nh thi hng T l v tip tuyn vihm qu ti gc ta , khong cch t giao im ca tip tuyn ny vi ng nmngang c tung bng 1 chnh l T.

c tnh thi gian ca khu qun tnh bc nht a) Hm trng lng; b) Hm qu

c tnh tn s:

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Biu thc cho thy biu Bode bin l mt ng cong. C th v gn ng biu Bode bin bng cc ng tim cn nh sau:

- Nu

, do ta c th v gn ng bng ng thng nm trn trc honh ( dc bng 0).

- Nu

, do ta v gn ng bng ng thng c dc 20dB/dec.

Nh phn tch trn, ta thy ti tn s 1/T dc ca cc ng tim cn thay i, biu Bode l mt ng gp khc nn tn s 1/T gi l tn s gy ca khu qun tnh bcnht. Thay gi tr vo biu thc ta v c biu Bode v pha. mt s imc bit nh sau:

Hnh 3.9a minh ha biu Bode ca khu qun tnh bc nht. ng cong t nt biu Bode bin chnh l ng L() v chnh xc. Sai lch cc i gia ng

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cong v chnh xc v cc ng tim cn xut hin ti tn s gy, ti tn s ny gi trchnh xc ca L() l

, trong khi gi tr gn ng l 0dB, sai lch ny kh b c th b qua c. Do khiphn tch v thit k h thng t ng trong min tn s ta c th dng biu Bodebin v bng cc ng tim cn thay cho biu Bode bin v chnh xc.

v biu Nyquist ta c nhn xt sau:

iu ny chng t biu Nyquist ca khu qun tnh bc nht nm trn ng trntm

, bn knh

. Do pha ca G(j) lun m khi thay i t 0 n +8 nn biu Nyquist l na dica ng trn

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c tnh tn s ca khu qun tnh bc nht a) Biu Bode; b) Biu Nyquist

Khu vi phn bc nht

Hm truyn:

c tnh thi gian:

Hm qu :

Hm trng lng:

Hm qu ca khu vi phn bc nht l t hp tuyn tnh ca hm xung n v v hmnc n v (hnh 3.10). Ta thy rng khu vi phn l tng v vi phn bc nht c cim chung l gi tr hm qu v cng ln ti t = 0. Hm trng lng l o hmca hm qu , ch c th m t bng biu thc ton hc ,khng biu din bng thc.

Hm qu ca khu vi phn bc nht

c tnh tn s:

Phn thc:

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Phn o:

Bin :

Pha:

So snh biu thc (3.53) v (3.54) vi (3.45) v (3.46) ta rt ra c kt lun: biu Bode ca khu vi phn bc nht v khu qun tnh bc nht i xng nhau qua trchonh (hnh 3.11a).

Do G(j) c phn thc P() lun lun bng 1, phn o Q() c gi tr dng tng dn t0 n +8 khi thay i t 0 n +8 nn biu Nyquist ca khu vi phn bc nht l nang thng qua im c honh bng 1 v song song vi trc tung nh hnh 3.11b.

c tnh tn s ca khu vi phn bc nht a) Biu Bode; b) Biu Nyquist

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Khu dao ng bc hai

Hm truyn:

c tnh thi gian:

Hm trng lng:

Hm qu :

trong lch pha xc nh. Biu thc cho thy c tnh thi gian ca khu daong bc hai c dng dao ng suy gim, hm trng lng l dao ng suy gim v 0,hm qu l dao ng suy gim n gi tr xc lp l 1 (hnh 3.12).

- Nu =0:

, p ng ca h l dao ng khng suy gim vi tn s

, do

gi l tn s dao ng t nhin ca khu dao ng bc hai.

- Nu

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: p ng ca h l dao ng vi bin gim dn, cng ln dao ng suy gim cngnhanh, do gi l h s tt (hay h s suy gim).

c tnh thi gian ca khu dao ng bc hai a) Hm trng lng; b) Hm qu

c tnh tn s:

Bin :

Biu thc cho thy biu Bode bin ca khu dao ng bc hai l mt ng cong.Tng t nh lm i vi khu qun tnh bc nht, ta c th v gn ng biu Bode bin bng cc ng tim cn nh sau:

- Nu

th

, do ta c th v gn ng bng ng thng nm trn trc honh ( dc bng 0).

- Nu

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, do ta v gn ng bng ng thng c dc 40dB/dec.

Ta thy rng ti tn s 1/T dc ca cc ng tim cn thay i nn tn s 1/T gi ltn s gy ca khu dao ng bc hai.

Biu Bode v pha ca khu dao ng bc hai l mt ng cong, biu thc(3.62) ta thy biu Bode v pha c im c bit sau y:

Hnh 3.13a minh ha biu Bode ca khu dao ng bc hai. Cc ng cong biu Bode bin chnh l ng L() v chnh xc. Biu Bode bin chnh xc cnh cng hng

ti tn

, do d thy rng nu cng nh th nh cng hng cng cao. Khi =0 th tn scng hng tin n tn s dao ng t nhin

.

Biu Nyquist ca khu dao ng bc hai c dng ng cong nh minh ha hnh3.13b. Khi =0 th G(j) c bin bng 1, pha bng 0; khi

th G(j) c bin bng 0, pha bng 180o. Giao im ca ng cong Nyquist vitrc tung c

, do tng ng vi tn s

, thay

vo biu thc ta suy ra bin ti giao im vi trc tung l

.

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c tnh tn s ca khu dao ng bc hai a) Biu Bode; b) Biu Nyquist

Khu tr hon (khu tr)

Hm truyn:

c tnh thi gian:

Hm trng lng:

Hm qu :

c im ca khu tr l tn hiu ra tr hn tn hiu vo mt khong thi gian l T.

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c tnh thi gian ca khu tr a) Hm trng lng; b) Hm qu

c tnh tn s:

Bin :

Pha:

Biu Bode bin ca khu tr hon l ng thng nm ngang trng vi trc honhdo L() = 0 vi mi . rng biu thc (3.68) l phng trnh ca mt ng thngnu trc honh chia theo thang tuyn tnh. Tuy nhin do trc honh ca biu Bodeli chia theo thang logarith nn biu Bode v pha ca khu tr hon l ng congdng hm m, xem hnh 3.15a.

Do G(j) c bin bng 1 vi mi v c pha gim t 0 n

nn biu Nyquist ca khu tr l ng trn n v c mi tn ch chiu tng ca nh hnh 3.15b.

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c tnh tn s ca khu tr hon a) Biu Bode; b) Biu Nyquist

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c tnh ng hc ca h thng t ng

c tnh ng hc ca h thng t ng

c tnh thi gian ca h thng

Xt h thng c hm truyn:

Bin i Laplace ca hm qu l:

Ty theo c im ca h thng m c tnh thi gian ca h thng c th c cc dngkhc nhau. Tuy vy chng ta c th rt ra mt s kt lun quan trng sau y:

- Nu G(s) khng c khu tch phn, vi phn l tng th hm trng lng suy gim v0, hm qu c gi tr xc lp khc 0.

- Nu G(s) c khu tch phn l tng n = 0) th hm trng lng c gi tr xc lp khc0, hm qu tng n v cng.

- Nu G(s) c khu vi phn l tng ( m b = 0 ) th hm qu suy gim v 0.

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- Nu G(s) l h thng hp thc ( n m = ) th h(0)=0.

- Nu G(s) l h thng hp thc cht ( n m < ) th g(0)=0.

- Nu G(s) khng c khu tch phn, vi phn l tng v c n cc phn bit, H(s) c thphn tch di dng:

Bin i Laplace ngc biu thc (3.71) ta c hm qu ca h thng l:

Do hm qu l t hp tuyn tnh ca cc hm m c s t nhin. Nu tt c cccc pi u l cc thc th hm qu khng c dao ng; ngc li nu c t nht mtcp cc phc th hm qu c dao ng.

Trn y va trnh by mt vi nhn xt v c tnh thi gian ca h thng t ng.Thng qua c tnh thi gian chng ta c th bit c h thng c khu tch phn, viphn l tng hay khng? H thng ch gm ton cc thc hay c cc phc?

Nhng nhn xt ny gip chng ta c c hnh dung ban u v nhng c im cbn nht ca h thng, t chng ta c th chn c phng php phn tch, thit kh thng ph hp.

c tnh tn s ca h thng

Xt h thng t ng c hm truyn ) (s G . Gi s ) (s G c th phn tch thnh tch cacc hm truyn c bn nh sau:

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c tnh tn s ca h thng l:

Bin :

Biu thc cho thy biu Bode bin ca h thng bng tng cc biu Bode bin ca cc khu c bn thnh phn.

Pha:

Biu thc chng t biu Bode pha ca h thng bng tng cc biu Bode pha cacc khu c bn thnh phn.

T hai nhn xt trn ta thy rng v c biu Bode ca h thng, ta v biu Bode ca cc khu thnh phn, sau cng th li. Da trn nguyn tc cng th,ta c phng php v biu Bode bin gn ng ca h thng bng cc ng timcn nh sau:

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Phng php v biu Bode bin bng cc ng tim cn

Gi s hm truyn ca h thng c dng:

Bc 1: Xc nh tt c cc tn s gy

, v sp xp theo th t tng dn:

Bc 2: Nu tt c cc tn s

th biu Bode gn ng phi qua im A c ta :

Bc 3: Qua im A, v ng thng c dc:

* (- 20 dB/dec ) nu G(s) c a khu tch phn l tng

* (+ 20 dB/dec ) nu G(s) c a khu vi phn l tng

ng thng ny ko di n tn s gy k tip

Bc 4: Ti tn s gy

, dc ca ng tim cn

c cng thm:

* (- 20 dB/dec ) nu

l tn s gy ca khu qun tnh bc mt.

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* (+ 20 dB/dec ) nu

l tn s gy ca khu vi phn bc mt.

* (-40 dB/dec ) nu

l tn s gy ca khu dao ng bc hai.

* (+40 dB/dec ) nu

l tn s gy ca khu vi phn bc hai,

.

( l s nghim bi ti

)

ng thng ny ko di n tn s gy k tip.

Bc 5: Lp li bc 4 cho n khi v xong ng tim cn ti tn s gy cui cng.

V d :V biu Bode bin gn ng ca h thng c hm truyn:

Da vo biu Bode gn ng, hy xc nh tn s ct bin ca h thng.

Gii. Cc tn s gy:

Biu Bode qua im A c ta :

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Biu Bode bin gn ng c dng nh hnh 3.16. Theo hnh v, tn s ct binca h thng l 103rad/sec.

Biu Bode bin ca h thng

V d :. Hy xc nh hm truyn ca h thng, bit rng biu Bode bin gn ngca h thng c dng nh hnh 3.17.

Biu Bode bin ca h thng v d trn

Gii: H thng c bn tn s gy

. Da vo s thay i dc ca biu Bode, ta thy hm truyn ca h thng phic dng:

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Vn cn li l xc nh thng s ca h thng. Theo hnh v:

- dc on BC l 20dB/dec, m t im B n im C bin ca biu Bodegim 40dB (t 34dB gim xung 6dB), do t B n C tn s phi thay i l 2decade. Suy ra:

- dc on DE l +40dB/dec, m t im D n im E bin ca biu Bodetng 60dB (t 6dB tng ln +54dB), do t D n E tn s phi thay i l 1.5 decade.Suy ra:

Do hm truyn ca h thng l:

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Kho st c tnh ng hc ca h thng

Kho st c tnh ng hc ca h thng

S dng cng c h tr kho st c tnh ng hc ca h thng. V d cng cControl Toolbox 5.0 h tr y cc lnh kho st c tnh ng ca h thng, c phpcc lnh rt gi nh nn rt d s dng.

- V p ng xung: lnh impulse

- V p ng nc: lnh step

- V biu Bode: lnh bode

- V biu Nyquist: lnh nyquist

C th nhp chut vo mt im bt k trn c tnh ng hc m Matlab v c bit gi tr c th ca tung , honh ti im .

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Bi 4: Kho st tnh n nh ca h thngKhi nim v n nh

Khi nim v n nh

nh ngha

H thng c gi l trng thi n nh, nu vi tn hiu vo b chn th p ng cah cng b chn (Bounded Input Bounded Output = BIBO).

Yu cu u tin i vi mt h thng KT l h thng phi gi c trng thi nnh khi chu tc ng ca tn hiu vo v chu nh hng ca nhiu ln h thng.

H phi tuyn c th n nh trong phm v hp khi lch ban u l nh v khng nnh trong phm v rng nu lch ban u l ln.

i vi h tuyn tnh c tnh ca qu trnh qu khng ph thuc vo gi tr tc ngkch thch. Tnh n nh ca h tuyn tnh khng ph thuc vo th loi v gi tr ca tnhiu vo v trong h tuyn tnh ch tn ti mt trng thi cn bng.

Phn bit ba trng thi cn bng: Bin gii n nh, n nh v khng n nh. Trn hnh4.1 nu thay i nh trng thi cn bng ca qu cu, chng hn cho n mt vn tc banu b th qu cu s tin ti mt trng thi cn bng mi (Hnh 4.1a), hoc s daong quanh v tr cn bng (Hnh 4.1b v d), hoc s khng tr v trng thi ban u(Hnh 4.1c). Trong trng hp u, ta c v tr cn bng bin gii n nh, trng hpsau l n nh v trng hp th ba l khng n nh. Cng v tr b v d trn hnh 4.1,nu qu cu vi lch ban u l ln th cng s khng tr v trng thi cn bng banu c - Hai trng thi b v d ca qu cu ch n nh trong phm v hp m khngn nh trong phm vi rng.

Minh ha trng thi n nh

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Trong trng hp ny vic kho st tnh n nh c gii hn cho cc h tuyn tnh btbin theo thi gian. l nhng h thng c m t bng phng trnh vi phn tuyntnh h s hng v c th p dng c nguyn l xp chng.

n nh ca h tuyn tnh

Mt h thng KT c biu din bng mt phng trnh vi phn dng tng qut:

Phng trnh ng vi tn hiu vo h thng l r(t) v tn hiu ra c(t). Hm truyn t cah thng c m t bng (4.1) c dng:

Nghim ca (4.1) gm hai thnh phn:

trong :

co(t) - l nghim ring ca (4.1) c v phi, c trng cho qu trnh xc lp.

cq(t) - l nghim tng qut ca (4.1) khng c v phi, c trng cho qu trnh qu .

Dng nghim tng qut c trng cho qu trnh qu trong h thng:

trong pi l nghim ca phng trnh c tnh:

pi c th l nghim thc cng c th l nghim phc lin hp v c gi l nghim ccca h thng. a thc mu s hm truyn t l A(s) bc n do h thng c n nghimcc pi (Pole), i = 1, 2,..., n .

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Zero l nghim ca phng trnh B(s) = 0. T s hm truyn t G(s) l a thc bc m(m< n) nn h thng c m nghim zero - zj vi j = 1, 2,..., m

H thng n nh nu:

H thng khng n nh nu:

Trong phng trnh (4.4) h s i l hng s ph thuc vo thng s ca h v trng thiban u.

Phn bit ba trng hp phn b cc trn mt phng phc

s

1- Phn thc ca nghim cc dng ai > 0

2- Phn thc ca nghim cc bng khng ai = 0

3- Phn thc ca nghim cc m ai < 0

n nh ca h thng ch ph thuc vo nghim cc m khng ph thuc vo nghimzero, do mu s hm truyn t l A(s) = 0 c gi l phng trnh c tnh hayphng trnh c trng ca h thng.

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Phn b cc trn mt phng S

Kt lun:

1- H thng n nh nu tt c nghim ca phng trnh c tnh u c phn thc m:Re{pi} < 0, i < 0 cc nghim nm bn tri mt phng phc:

2- H thng khng n nh nu c d ch l mt nghim phng trnh c tnh (4.9) cphn thc dng (mt nghim phi) cn li l cc nghim u c phn thc m (nghimtri)

3- H thng bin gii n nh nu c d ch l mt nghim c phn thc bng khngcn li l cc nghim c phn thc m (mt nghim hoc mt cp nghim phc lin hpnm trn trc o). Vng n nh ca h thng l na tri mt phng phc s S.

p ng qu c th dao ng hoc khng dao ng tng ng vi nghim caphng trnh c tnh l nghim phc hay nghim thc.

Tt c cc phng php kho st n nh u xt n phng trnh c tnh (4.9) theomt cch no . Tng qut, ba cch nh gi sau y thng c dng xt n nh:

1- Tiu chun n nh i s Routh - Hurwitz

2- Tiu chun n nh tn s Mikhailov - Nyquist - Bode

3- Phng php chia min n nh v phng php qu o nghim s.

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Tiu chun n nh i s

Tiu chun n nh i s

iu kin cn

iu kin cn h thng n nh l tt c cc h s ca phng trnh c trng phikhc 0 v cng du.

V d: H thng c phng trnh c trng:

Tiu chun n nh Routh

Cho h thng c phng trnh c trng

Mun xt tnh n nh ca h thng theo tiu chun Routh, trc tin ta thnh lp bngRouth theo qui tc:

- Bng Routh c n+1 hng.

- Hng 1 ca bng Routh gm cc h s c ch s chn.

- Hng 2 ca bng Routh gm cc h s c ch s l.

- Phn t hng i ct j ca bng Routh (i = 3) c tnh theo cng thc:

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Pht biu tiu chun Routh

iu kin cn v tt c cc nghim ca phng trnh c trng nm bn tri mtphng phc l tt c cc phn t nm ct 1 ca bng Routh u dng. S ln i duca cc phn t ct 1 ca bng Routh bng s nghim nm bn phi mt phng phc.

V d : Hy xt tnh n nh ca h thng c phng trnh c trng l:

Gii

Bng Routh

V tt c cc phn t ct 1 bng Routh u dng nn tt c cc nghim ca phngtrnh c tnh u nm bn tri mt phng phc, do h thng n nh.

V d : Hy xt tnh n nh ca h thng t ng c s khi nh sau

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Hnh 4.3: S khi h thng t ng v d 4.2

Gii. Phng trnh c trng ca h thng l

V cc phn t ct 1 bng Routh i du hai ln nn phng trnh c tnh u c hainghim nm bn phi mt phng phc, do h thng khng n nh.

V d : Cho h thng c s khi nh sau

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Hnh 4.4: S khi h thng t ng v d 4.3

Xc nh iu kin ca K h thng n nh.

Gii. Phng trnh c tnh

iu kin h thng n nh

Cc trng hp c bit

- Trng hp 1: nu c h s ct 1 ca hng no bng 0, cc h s cn li ca hng khc 0 th ta thay h s bng 0 ct 1 bi s e dng, nh ty , sau qu trnhtnh ton c tip tc.

V d : Xt tnh n nh ca h thng c phng trnh c trng:

Gii

Bng Routh

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V cc h s ct 1 bng Routh i du hai ln nn phng trnh c tnh ca h thngc hai nghim nm bn phi mt phng phc, do h thng khng n nh.

- Trng hp 2: nu tt c cc h s ca hng no bng 0

- Thnh lp a thc ph t cc h s ca hng trc hng c tt c cc h s bng 0, gia thc l Ap(s).

- Thay hng c tt c cc h s bng 0 bi mt hng khc c cc h s chnh l cc hs ca

. Sau qu trnh tnh ton tip tc.

Ch : Nghim ca a thc ph Ap(s) cng chnh l nghim ca phng trnh c trng.

V d 4.5. Xt tnh n nh ca h thng c phng trnh c trng:

Xc nh s nghim ca phng trnh c tnh nm bn tri, bn phi hay trn trc oca mt phng phc.

Gii

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a thc ph

Nghim ca a thc ph (cng chnh l nghim ca phng trnh c trng)

Kt lun

- Cc h s ct 1 bng Routh khng i du nn phng trnh c trng khng cnghim nm bn phi mt phng phc.

- Phng trnh c tnh c hai nghim nm trn trc o.

- S nghim nm bn tri mt phng phc l 5 2 = 3.

=> H thng bin gii n nh.

Tiu chun n nh Hurwitz

Cho h thng c phng trnh c trng

Mun xt tnh n nh ca h thng theo tiu chun Hurwitz, trc tin ta thnh lp matrn Hurwitz theo qui tc:

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- Ma trn Hurwitz l ma trn vung cp nn.

- ng cho ca ma trn Hurwitz l cc h s t a1 n an.

- Hng l ca ma trn Hurwitz gm cc h s c ch s l theo th t tng dn nu bnphi ng cho v gim dn nu bn tri ng cho.

- Hng chn ca ma trn Hurwitz gm cc h s c ch s chn theo th t tng dn nu bn phi ng cho v gim dn nu bn tri ng cho.

Pht biu tiu chun Hurwitz

iu kin cn v h thng n nh l tt c cc nh thc con cha ng choca ma trn Hurwitz u dng,

V d 4.6. Cho h thng t ng c phng trnh c trng l

Hi h thng c n nh khng?

Gii. Ma trn Hurwitz

Cc nh thc

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V tt c cc nh thc con cha ng cho ca ma trn Hurwitz u dng nn hthng n nh.

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Phng php qu o nghim s

Phng php qu o nghim s

Khi nim

- Xt h thng c phng trnh c tnh

- Nghim ca phng trnh c tnh ng vi cc gi tr khc nhau ca K

Qu o nghim s

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V cc nghim ca phng trnh tng ng vi cc gi tr ca K ln mt phng phc.Nu cho K thay i lin tc t 0 n +8, tp hp tt c cc nghim ca phng trnh tothnh ng m nt nh trn hnh v. ng m nt trn hnh v c gi l qu onghim s.

nh ngha

Qu o nghim s l tp hp tt c cc nghim ca phng trnh c tnh ca h thngkhi c mt thng s no trong h thay i t

.

Qui tc v qu o nghim s

S h thng iu khin t ng

Xt h thng iu khin c s khi hnh 4.6. Phng trnh c tnh ca h

Mun p dng cc qui tc v qu o nghim s, trc tin ta phi bin i tng ngphng trnh c tnh v dng

trong K l thng s thay i.

t

Gi n l s cc ca G0(s), m l s zero ca Go(s)

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Sau y l 11 qui tc v qu o nghim s ca h thng c phng trnh c tnh cdng

Qui tc 1: S nhnh ca qu o nghim s = bc ca phng trnh c tnh = s ccca G0(s) = n.

Qui tc 2: Khi K = 0: cc nhnh ca qu o nghim s xut pht t cc cc ca Go(s).Khi K tin n

: m nhnh ca qu o nghim s tin n m zero ca Go(s), n-m nhnh cn li tinn 8 theo cc tim cn xc nh bi qui tc 5 v 6.

Qui tc 3: Qu o nghim s i xng qua trc thc.

Qui tc 4: Mt im trn trc thc thuc v qu o nghim s nu tng s cc v zeroca Go(s) bn phi n l mt s l.

Qui tc 5: Gc to bi cc ng tim cn ca qu o nghim s vi trc thc xcnh bi

Qui tc 6: Giao im gia cc tim cn vi trc thc l im A c ta xc nh bi

(pi v zi l cc cc v cc zero ca Go(s)).

Qui tc 7: im tch nhp (nu c) ca qu o nghim s nm trn trc thc v lnghim ca phng trnh:

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Qui tc 8: Giao im ca qu o nghim s vi trc o c th xc nh bng mt tronghai cch sau y

- p dng tiu chun Routh-Hurwitz.

- Thay

vo phng trnh c tnh (4.12), cn bng phn thc v phn o s tm c giao imvi trc o v gi tr K.

Qui tc 9: Gc xut pht ca qu o nghim s ti cc phc pj c xc nh bi

Dng hnh hc ca cng thc trn l

Qui tc 10: Tng cc nghim l hng s khi K thay i t

Qui tc 11: H s khuch i dc theo qu o nghim s c th xc nh t iu kinbin

V d 4.7. Cho h thng t ng c s khi nh sau

Hy v QNS ca h thng khi

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Gii. Phng trnh c tnh ca h thng

Cc cc: ba cc.

Cc zero: khng c.

=> QNS gm c ba nhnh xut pht t cc cc khi K = 0.

Khi

, ba nhnh ca QNS s tin n v cng theo cc tim cn xc nh bi:

- Gc gia cc tim cn v trc thc

- Giao im gia cc tim cn v trc thc

- im tch nhp l nghim ca phng trnh

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- Giao im ca QNS vi trc o c th xc nh bng mt trong hai cch sau y:

Cch 1

p dng tiu chun Routh

Bng Routh

iu kin h thng n nh

Vy h s khuch i gii hn l Kgh = 30.

Thay gi tr Kgh = 30 vo phng trnh (2), gii phng trnh ta c giao im caQNS vi trc o.

Cch 2

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Giao im (nu c) ca QNS v trc o phi c dng

Thay

vo phng trnh (1) ta c

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Tiu chun n nh tn s

Nguyn l gc quay

Xt h thng bc n c phng trnh c tnh h s hng:

a thc A(s) c vit di dng:

vi p1, p2,... pn l cc ca h thng, l nghim ca phng trnh c tnh.

Thay s = j vo ta c:

Gi s phng trnh c m nghim phi (c phn thc dng), cn (n - m) nghim tri(c phn thc m)

Gc quay ca vect a thc c tnh tn s A(j)

Khi tn s thay i t

n

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th s thay i gc quay ca vect a thc c tnh tn s A(j) s l:

K hiu ch s thay i gc quay Nu qui nh chiu quay dng l chiu ngc chiukim ng h th ta c biu thc sau i vi nghim tri v phi:

H c m nghim phi v (n - m) nghim tri:

Nguyn l gc quay

H thng bc n c m nghim phi v (n - m) nghim tri c vect a thc c tnh tns A(j) s quay mt gc l (n-2m)/2 vng kn theo chiu ngc chiu kim ng h khitn s ? bin thin t

n

Vct a thc c tnh tn s A(j) s quay mt gc bng hiu s nghim tri (n - m) vnghim phi (m) nhn vi khi bin thin t

n

.

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Tiu chun n nh tn s Mikhailov

Tiu chun n nh da vo nguyn l gc quay c A. V. Mikhailov pht biu vonm 1938: iu kin cn v h tuyn tnh n nh l biu vect a thc ctnh A(j) xut pht t na trc thc dng ti ? bng khng, phi quay n gc phn ttheo chiu ngc chiu kim ng h khi bin thin t 0 n

, vi n l bc ca phng trnh c tnh ca h thng

V d :. xt h bc ba n = 3

Cho bin thin t 0 n v cng bng phng php trn xy dng ton b biu vct a thc c tnh A(j).

- a thc c tnh (mu s hm truyn t ca h cn xt n nh trng thi h hoctrng thi kn) c phn tch thnh hai thnh phn:

Tiu chun n nh Nyquist

Cho h thng t ng c s khi

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Cho bit c tnh tn s ca h h G(s), bi ton t ra l xt tnh n nh ca h thngkn Gk(s).

Tiu chun Nyquist

H thng kn Gk(s) n nh nu ng cong Nyquist ca h h G(s) bao im (1, j0) 1/2 vng theo chiu dng (ngc chiu kim ng h) khi thay i t 0 n

, trong l l s cc ca h h G(s) nm bn phi mt phng phc.

V d :. Cho h thng hi tip m n v, trong h h G(s) c ng cong Nyquistnh hnh v. Bit rng G(s) n nh. Xt tnh n nh ca h thng kn

V G(s) n nh nn G(s) khng c cc nm bn phi mt phng phc. Do theo tiuchun Nyquist h kn n nh nu ng cong Nyquist G(j) ca h h khng bao im(1, j0). V vy:

Trng hp 1: G(j) khng bao im (-1, j0) ? h kn n nh.

Trng hp 2: G(j) qua im (-1, j0) h kn bin gii n nh;

Trng hp 3: G(j) bao im (-1, j0) ? h kn khng n nh.

Ch : i vi cc h thng c khu tch phn l tng, xc nh ng cong Nyquistc bao im (1, j0) hay khng, ta v thm cung -/2 bn knh v cng ln ( l s khutch phn l tng trong hm truyn h h).

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Tiu chun n nh Bode

Cho h thng t ng c s khi nh hnh Cho bit c tnh tn s ca h h G(s),bi ton t ra l xt tnh n nh ca h thng kn Gk(s).

Tiu chun Bode

H thng kn Gk(s) n nh nu h thng h G(s) c d tr bin v d tr phadng

V d : Cho h thng h c biu Bode nh hnh v. Hi h kn c n nh khng?

Gii. Trn biu Bode ta xc nh c:

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Do GM < 0 v FM < 0 nn h thng kn khng n nh.

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Bi 5: nh gi cht lng ca h thngiu khinCc tiu chun cht lng

n nh l iu kin cn i vi mt h KT, song cha phi l h thng cs dng trong thc t. Nhiu yu cu i hi h thng phi tha mn c cng mt lccc tiu chun cht lng khc nhau nh chnh xc, n nh, p ng qu , nhy, kh nng chng nhiu... Sau y l mt s tiu chun thng dng nh gicht lng h thng iu khin.

Sai s xc lp

Sai s l hiu s gia tn hiu vo v tn hiu hi tip. Mc ch mun tn hiu ra quavng hi tip lun lun bm c tn hiu vo mong mun. iu c ngha sai s xclp bng khng.

vt l ( qu iu chnh )

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Thi gian p ng

Thi gian ln nh l thi gian p ng ra t gi tr cc i (tp = tpeak).

Thi gian qu ts = tset xc nh bi thi im p ng ra t sau tr i khng vtra khi min gii hn sai s quanh gi tr xc lp. V d: c th l 2%, 5%...

d tr n nh

nh ngha: Khong cch t trc o n nghim cc gn nht (nghim thc hoc phc)c gi l d tr n nh ca h. K hiu khong cch ngn nht y l ?o, nu ?ocng ln th qu trnh qu cng nhanh v xc lp. p ng qu ca h bc n:

trong Re (pi + ?o) = 0

Tiu chun tch phn

Trong thc t mt h thng KT c thit k phi tha yu cu c hai ch xclp v qu . Qu trnh qu c th c nh gi thng qua gi tr tch phn ca sailch gia gi tr t v gi tr tc thi o c ca i lng cn iu chnh.

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Sai s xc lp

Xt h thng hi tip m c s khi nh hnh v:

H thng hi tip m

Sai s ca h thng l

Sai s xc lp

Sai s xc lp khng nhng ph thuc vo cu trc v thng s ca h thng m cnph thuc vo tn hiu vo.

Tn hiu vo l hm nc n v

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Tn hiu vo l hm dc n v

Tn hiu vo l hm parabol

Nhn xt

Ty theo s khu tch phn l tng c trong hm truyn h ( ) ( ) G s H s m Kp , Kv ,Ka c gi tr nh bng sau:

- Nu G(s)H(s) khng c khu tch phn l tng th h thng kn theo kp s thay ica tn hiu vo l hm nc vi sai s

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v khng theo kp s thay i ca tn hiu vo l hm dc v hm parabol.

- Nu G(s)H(s) c mt khu tch phn l tng th h thng kn theo kp s thay i catn hiu vo l hm nc vi sai s xle = 0, v theo kp s thay i ca tn hiu vo l hmdc vi sai s

v khng theo kp s thay i ca tn hiu vo l hm parabol

h thng c mt khu tch phn l tng gi l h v sai bc mt.

- Nu G(s)H(s) c hai khu tch phn l tng th h thng kn theo kp s thay i catn hiu vo l hm nc v hm dc vi sai s exl = 0, theo kp s thay i ca tn hiuvo l hm parabol vi sai s

h thng c hai khu tch phn l tng gi l h v sai bc hai.

- Nu G(s)H(s) c ba khu tch phn l tng th h thng kn theo kp s thay i catn hiu vo l hm nc, hm dc v hm parabol vi sai s

h thng c ba khu tch phn l tng gi l h v sai bc ba.

H thng c n khu tch phn l tng gi l h v sai bc n.

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p ng qu

p ng qu l p ng ca h thng khi tn hiu vo l hm nc n v.

H qun tnh bc mt

Hm truyn:

H thng kn ch c mt cc thc

Gin cc - zero ca h qun tnh bc nht

p ng qu ca h qun tnh bc nht

p ng ca h thng khi tn hiu vo l hm nc

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Nhn xt (xem hnh 5.4)

p ng qu ca khu qun tnh bc nht khng c vt l.

Thi hng T l thi im c(t) t 63.2% gi tr xc lp, T cng nh p ng cngnhanh.

Thi gian xc lp ts (settling time) l thi gian sai s gia c(t) v gi tr xc lp nhhn ( = 5% hay 2%).

Sai s xc lp bng 0.

H dao ng bc hai

Hm truyn

trong

H thng c cp cc phc lin hp

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Gin cc - zero ca h dao ng bc hai

p ng qu ca h dao ng bc hai

p ng ca h thng khi tn hiu vo l hm nc

trong lch pha xc nh bi

Nhn xt (xem hnh 5.6)

p ng qu ca khu dao ng bc hai c dng dao ng vi bin gim dn.

- Nu

, p ng ca h l dao ng khng suy gim vi tn

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gi l tn s dao ng t nhin.

- Nu

p ng ca h l dao ng vi bin gim dn

gi l h s tt (hay h s suy gim),

cng ln dao ng suy gim cng nhanh.

p ng ca khu dao ng bc hai c vt l.

Tng qut, vt l (POT Percent of Overshoot) c nh ngha l

(cmax - gi tr cc i ca c(t); cxl - gi tr xc lp ca c(t))

i vi h dao ng bc hai, vt l POT c tnh bi cng thc

Thi gian xc lp ts l thi gian sai s gia c(t) v gi tr xc lp nh hn e (e = 5%hay 2%).

i vi h bc hai

Thi gian ln tr: (rise time) l thi gian c(t) tng t 10% n 90% gi tr xc lp.

i vi h bc hai

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Ch : Nu

ta khng gi l h dao ng bc hai v trong trng hp ny p ng ca h khng cdao ng.

Nu

h thng kn c mt nghim kp (thc).

p ng ca h thng

Nu

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h thng kn c hai nghim thc phn bit

p ng ca h thng

H bc cao

Cp cc quyt nh ca h bc cao

H bc cao c nhiu hn hai cc. p ng tng ng vi cc cc nm cng xa trc osuy gim cng nhanh. Do c th xp x h bc cao v h bc hai vi cp cc l haicc nm gn trc o nht. Cp cc nm gn trc o nht ca h bc cao gi l cp ccquyt nh.

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Cc tiu chun ti u ha p ng qu

Tiu chun tch phn sai lch IE (Integrated Error)

i vi h c p ng qu khng dao ng (ng 1 hnh 5.3) th tiu chun IE chnhl din tch ca hm sai lch e(t) to vi trc thi gian t cn t gi tr cc tiu th chtlng t tt nht.

Tiu chun IE v IAE

Song i vi h c p ng qu dao ng n nh (ng 2) th tiu chun IE khngphn nh ng cht lng ca h thng do c min din tch m c tr bt i. Ktqu gi tr tch phn nh nhng qu trnh qu xu. V vy phi s dng tiu chuntch phn tr s tuyt i ca sai lch.

Tiu chun IAE (Integral of the Absolute Magnitude of the Error - tchphn tr tuyt i bin sai s)

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i vi h bc hai:

Tiu chun ISE (Integral of the Square of the Error - tch phn ca bnhphng sai s)

ISE xem nh nhng din tch b v bnh phng mt s nh hn 1 b hn tr s tuyti ca s y. Mt trong nhng l do khin tiu chun ISE thng c s dng l cngvic tnh ton v thc hin n gin. C th tnh c lng ISE theo bin i Fourierhoc theo cng thc.

i vi h bc hai:

Tiu chun ITAE (Integral of Time multiplied by the Absolute Value of theError- tch phn ca thi gian nhn vi tr tuyt i ca sai s)

i vi h bc hai:

Trong ba tiu chun ti u ha p ng qu va trnh by trn, tiu chun ITAEc s dng nhiu nht. p ng qu ca h thng bc n l ti u theo chunITAE th mu s hm truyn kn h bc n phi c dng

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Nu mu s hm truyn h kn c dng nh trn v t s hm truyn h kn ca h bcn l

th p ng qu ca h thng l ti u v sai s xc lp bng 0.

Tiu chun tch phn c tnh n nh hng ca tc thay i ca sai lche(t)

vi l hng s c chn thch hp cho tng trng hp.

V d: ln khng cho php dao ng ln. Ngc li, nh cho php qu dao ngln.

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Bi 6: Thit k h thng iu khin lin tcKhi nim

Thit k l ton b qu trnh b sung cc thit b phn cng cng nh thut ton phnmm vo h cho trc c h mi tha mn yu cu v tnh n nh, chnh xc,p ng qu , C nhiu cch b sung b iu khin vo h thng cho trc, trongkhun kh quyn sch ny chng ta ch yu xt hai cch sau:

Cch 1: thm b iu khin ni tip vi hm truyn ca h h, phng php ny gil hiu chnh ni tip (H.6.1). B iu khin c s dng c th l b hiu chnh smpha, tr pha, sm tr pha, P, PD, PI, PID, thit k h thng hiu chnh ni tipchng ta c th s dng phng php QNS hay phng php biu Bode. Ngoi ramt phng php cng thng c s dng l thit k theo c tnh qu chun.

H thng hiu chnh ni tip

Cch 2: iu khin hi tip trng thi, theo phng php ny tt c cc trng thi ca hthng c phn hi tr v ng vo v tn hiu iu khin c dng

(H.6.2). Ty theo cch tnh vct hi tip trng thi K m ta c phng php iukhin phn b cc, iu khin ti u LQR, .

H thng iu khin hi tip trng thi

Qu trnh thit k h thng l qu trnh i hi tnh sng to do trong khi thit k thngc nhiu thng s phi chn la.

Ngi thit k cn thit phi hiu c nh hng ca cc khu hiu chnh n chtlng ca h thng v bn cht ca tng phng php thit k th mi c th thit kc h thng c cht lng tt. Do cc phng php thit k trnh by trong bi ny

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ch mang tnh gi , l nhng cch thng c s dng ch khng phi l phngphp bt buc phi tun theo.

Vic p dng mt cch my mc thng khng t c kt qu mong mun trong thct. D thit k theo phng php no yu cu cui cng vn l tha mn cht lngmong mun, cch thit k, cch chn la thng s khng quan trng.

Trc khi xt n cc phng php thit k b iu khin, chng ta xt nh hng cacc b iu khin n cht lng ca h thng.

Phng php thay i thng s

Nguyn tc thit k h thng dng phng php hiu chnh thng s hay cn gi lQNS l da vo phng trnh c tnh ca h thng sau khi hiu chnh:

(6.1)

(6.2)

Ta cn chn thng s ca b iu khin GC(s) sao cho phng trnh (6.1) c nghim tiv tr mong mun.

Hiu chnh sm pha

thun li cho vic v QNS chng ta biu din hm truyn khu hiu chnh sm phadi dng sau :

(6.3)

Bi ton t ra l chn gi tr KC, a v T p ng ca h thng tha mn yu cu vcht lng qu ( vt l, thi gian xc lp, )

Ta bit cht lng qu ca h thng hon ton xc nh bi v tr ca cp cc quytnh. Do nguyn tc thit k khu hiu chnh sm pha dng phng php QNS lchn cc v zero ca khu hiu chnh sao cho QNS ca h thng sau khi hiu chnhphi i qua cp cc quyt nh mong mun. Sau bng cch chn h s khuch i

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KC thch hp ta s chn c cc ca h thng chnh l cp cc mong mun. Nguyntc trn c c th ha thnh trnh t thit k sau:

Trnh t thit k

Khu hiu chnh: Sm pha

Phng php thit k: QNS

Bc 1: Xc nh cp cc quyt nh t yu cu thit k v cht lng ca h thngtrong qu trnh qu :

Bc 2: Xc nh gc pha cn b cp cc quyt nh

nm trn QNS ca h thng sau khi hiu chnh bng cng thc:

(6.4)

trong pi v zi l cc cc ca h thng G(s) trc khi hiu chnh.

Dng hnh hc ca cng thc trn l:

(6.5)

Bc 3: Xc nh v tr cc v zero ca khu hiu chnh

V hai na ng thng bt k xut pht t cc quyt nh s* sao cho hai na ngthng ny to vi nhau mt gc bng *. Giao im ca hai na ng thng ny vitrc thc l v tr cc v zero ca khu hiu chnh.

C hai cch v thng dng:

- PP ng phn gic ( cc v zero ca khu hiu chnh gn nhau).

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- PP trit tiu nghim ( h bc ca h thng).

Bc 4: Tnh h s khuch i KC bng cch p dng cng thc:

Gii thch

Bc 1: Do cht lng qu ph thuc vo v tr cp cc quyt nh nn thit kh thng tha mn cht lng qu mong mun ta phi xc nh cp cc quyt nhtng ng. Gi cp cc quyt nh mong mun l

.

Bc 2: h thng c cht lng qu nh mong mun th cp cc quyt nh

phi l nghim ca phng trnh c tnh sau khi hiu chnh (6.1). Xt iu kin vpha:

(6.6)

trong zi v pi l cc zero v cc cc ca h thng h trc khi hiu chnh. t gcpha cn b

, t biu thc (6.6) ta suy ra:

Do s phc c th biu din di dng vct nn cng thc trn tng ng vi cngthc hnh hc sau:

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Bc 3: By gi ta phi chn cc v zero ca khu hiu chnh sao cho:

(6.7)

Do * v s* bit nn phng trnh (6.7) c hai n s cn tm l 1/T v 1/T. Chntrc gi tr 1/T bt k thay vo phng trnh (6.7) ta s tnh c 1/T v ngc li,ngha l bi ton thit k c v s nghim.

Thay v chn nghim bng phng php gii tch (gii phng trnh (6.7) nh va trnhby chng ta c th chn bng phng php hnh hc. Theo hnh 6.3 hai s phc

v

c biu din bi hai vct

v

, do

v

. Thay cc gc hnh hc vo phng trnh (6.7) ta c:

T phn tch trn ta thy cc v zero ca khu hiu chnh sm pha phi nm ti im Bv C sao cho BPC * = F . y chnh l c s ton hc ca cch chn cc v zero nh trnh by trong trnh t thit k.

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Quan h hnh hc gia v tr cc v zero ca khu hiu chnh sm pha vi gc pha cn b

Quan h hnh hc gia v tr cc v zero ca khu hiu chnh sm pha vi gc pha cnb

V d : Thit k khu hiu chnh sm pha dng phng php QNS.

Cho h thng iu khinnh hnh v. Hy thit k khu hiu chnh GC(s) p ngqu ca h thng sau khi hiu chnh tha: POT < 20%; tq < 0,5 sec (tiu chun 2%).

Gii: V yu cu thit k ci thin p ng qu nn s dng khu hiu chnh smpha:

Bc 1: Xc nh cp cc quyt nh

Theo yu cu thit k, ta c:

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Vy cp cc quyt nh l:

Bc 2: Xc nh gc pha cn b

Cch 1. Dng cng thc i s

Cch 2. Dng cng thc hnh hc

Bc 3: Xc nh cc v zero ca khu hiu chnh bng phng php ng phn gic.

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Bc 4: Tnh C K .

Vy hm truyn ca khu hiu chnh sm pha cn thit k l:

Nhn xt

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Qu o nghim s ca h thng trc khi hiu chnh khng qua im s* (H.6.4a) do h thng s khng bao gi t c cht lng p ng qu nh yu cu d c thayi h s khuch i ca h thng.

S thay i dng QNS khi hiu chnh sm pha a) QNS trc khi hiu chnh; b) QNS sau khihiu chnh

Bng cch s dng khu hiu chnh sm pha, qu o nghim s ca h thng b sadng v qua im s* (H.6.4b). Bng cch chn h s khuch i thch hp (nh thchin bc 4) h thng s c cp cc quyt nh nh mong mun, do p ng qu t yu cu thit k (H.6.5).

p ng nc ca h thng v d 6.4 trc v sau khi hiu chnh

Hiu chnh tr pha

Hm truyn khu hiu chnh tr pha cn thit k c dng:

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Bi ton t ra l chn gi tr KC, v T p ng ca h thng tha mn yu cu vsai s xc lp m khng lm nh hng n p ng qu (nh hng khng ngk).

Ta bit do khu hiu chnh tr pha c h s khuch i min tn s thp ln nnc tc dng lm gim sai s xc lp ca h thng. p ng qu ca h thng saukhi hiu chnh tr pha gn nh khng i th cp cc quyt nh ca h thng trc vsau khi hiu chnh phi nm rt gn nhau. t c iu ny ta phi t thm cc vzero ca khu hiu chnh tr pha sao cho dng QNS thay i khng ng k. y lnguyn tc cn tun theo khi thit k khu hiu chnh tr pha. Trnh t thit k di yc th ha nguyn tc trn:

Trnh t thit k

Khu hiu chnh: Tr pha

Phng php thit k: QNS

Bc 1: Xc nh t yu cu v sai s xc lp.

Nu yu cu v sai s xc lp cho di dng h s vn tc

th tnh bng cng thc:

trong KV v K*V l h s vn tc ca h thng trc v sau khi hiu chnh.

Bc 2: Chn zero ca khu hiu chnh sao cho:

trong

l cp cc quyt nh ca h thng sau khi hiu chnh.

Bc 3: Tnh cc ca khu hiu chnh:

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Bc 4: Tnh KC bng cch p dng cng thc:

trong

l cp cc quyt nh ca h thng sau khi hiu chnh. Do yu cu thit k khng lmnh hng ng k n p ng qu nn c th tnh gn ng:

Gii thch

Bc 1: Ta c h s vn tc ca h thng trc v sau khi hiu chnh l:

Do ta chn bng cng thc trn. Cc bc thit k tip theo m bo

.

Bc 2: Gi s1,2 l cp cc quyt nh ca h thng trc khi hiu chnh:

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Xt iu kin v pha. h thng c cht lng qu gn nh khng thay i th

. Suy ra:

(6.8)

Phn tch trn cho thy cc v zero ca khu hiu chnh tr pha phi tha mn biuthc (6.8). Khi thit k ta thng chn khu hiu chnh tr pha sao cho

, t c iu ny c th t cc v zero ca khu hiu chnh tr pha nm rt gngc ta so vi phn thc ca

. Do ta chn v tr zero sao cho:

Bc 3: Suy ra:

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rng bng cch chn nh trn 1/T cng nm rt gn gc ta do 1/.

Bc 4: bc 2 v 3 ta mi chn cc v zero ca khu hiu chnh tr pha tha mniu kin v pha. tha mn iu kin bin ta chn KC bng cng thc

C th d dng kim chng c rng do cch chn zero v cc ca khu hiu chnhnh bc 2 v bc 3 m bc 4 ta lun tnh c

. Nh vy KC tha mn gi thit ban u khi tnh h s bc 1.

Hiu chnh sm tr pha

Hm truyn khu hiu chnh sm tr pha cn thit k c dng:

trong :

l khu hiu chnh sm pha

l khu hiu chnh tr pha.

Bi ton t ra thit k

ci thin p ng qu v sai s xc lp ca h thng.

Trnh t thit k

Khu hiu chnh: Sm tr pha

Phng php thit k: QNS

Bc 1: Thit k khu sm pha

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tha mn yu cu v p ng qu (xem phng php thit k khu hiu chnhsm pha mc trc).

Bc 2: t

.

Thit k khu hiu chnh tr pha

mc ni tip vo

tha mn yu cu v sai s xc lp m khng thay i ng k p ng qu cah thng sau khi hiu chnh sm pha (xem phng php thit k khu hiu chnh trpha mc trc).

V d 6.6. Thit k khu hiu chnh sm tr pha dng phng php QNS.

Hy thit k khu hiu chnh GC(s) sao cho h thng sau khi hiu chnh c cp cc phcvi

; h s vn tc VK = 80.

Gii: H cha hiu chnh c

; VK = 8 .

V yu cu thit k b hiu chnh ci thin p ng qu v sai s xc lp nn GC(s)l khu hiu chnh sm tr pha.

Bc 1: Thit k khu hiu chnh sm pha GC1(s)

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- Cp cc quyt nh sau khi hiu chnh:

Gc pha cn b

- Gc pha cn b:

- Chn zero ca khu sm pha trng vi cc s = -0,5 ca G(s) h bc h thng saukhi hiu chnh.

T cc s*1 v hai na ng thng to vi nhau mt gc l * nh hnh 6.6. Cc cakhu sm pha ti im B.

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Bc 2: Thit k khu hiu chnh tr pha GC2(s)

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- Xc nh :

H s vn tc ca h sau khi hiu chnh sm pha:

H s vn tc mong mun:

- Xc nh zero ca khu tr pha:

- Xc nh cc ca khu tr pha:

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Tm li khu hiu chnh sm tr pha cn thit k l:

Phng php thay i cu trc

C nhng h thng iu khin d thay i thng s n mc no cng khng lm nn nh c. H thng nh vy c gi l h thng c cu trc khng n nh. Munlm cho h thng chuyn sang trng thi n nh ta phi thay i cu trc ca n.

Lm thay i cu trc tc l lm thay i cp ca phng trnh vi phn ca h thng thc tnh cht lng cng thay i.

Nguyn l bt bin v iu khin b

Mt h thng KT trong cc ta yi(t) v sai lch e(t) khng ph thuc vo cctc ng bn ngoi fi(t) c gi l h thng bt bin. gim nh hng ca nhiu vtng chnh xc ngi ta thng s dng nguyn tc b sai lch tc ng u vo vb nhiu.

Thit k h thng iu khin PID

B iu khin PID l trng hp c bit ca hiu chnh sm tr pha nn v nguyntc c th thit k b iu khin PID bng phng php dng QNS hoc dng biu Bode.

Mt phng php khc cng thng dng thit k b iu khin PID l phng phpgii tch. Sau y l mt v d:

V d 6.10. Cho h thng iu khin nh hnh v:

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Hy xc nh thng s ca b iu khin PID sao cho h thng tha mn yu cu:

- H c cp nghim phc vi = 0,5 , n = 8

- H s vn tc KV = 100.

Gii: Hm truyn b iu khin PID cn thit k:

H s vn tc ca h sau khi hiu chnh:

Theo yu cu bi KV = 100 nn suy ra:

Phng trnh c tnh ca h sau khi hiu chnh l:

(1)

h thng c cp cc phc vi

th phng trnh c tnh (1) phi c dng:

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Cn bng cc h s hai phng trnh (1) v (2), suy ra:

Vi KI = 100, gii h phng trnh trn ta c:

Vy hm truyn ca khu hiu chnh PID cn thit k l:

B iu khin PID c s dng rt rng ri trong thc t iu khin nhiu loi itng khc nhau nh nhit l nhit, tc ng c, mc cht lng trong bn cha...do n c kh nng lm trit tiu sai s xc lp, tng tc p ng qu o gim vt l nu cc thng s ca b iu khin c chn la thch hp. Do tnh thng dngca n nn nhiu hng sn xut thit b iu khin cho ra i cc b iu khin PIDthng mi rt tin dng. Trong thc t cc phng php thit k b iu khin PIDdng QNS, biu Bode hay phng php gi tch rt t c s dng do s khkhn trong vic xy dng hm truyn ca i tng. Phng php ph bin nht chn thng so cho cc b iu khin PID thng mi hin nay l phng php Zeigler-Nichols.

Phng php Zeigler-Nichols

Phng php Zeigler-Nichols l phng php thc nghim thit k b iu khin P,PI, hoc PID bng cch da vo p ng qu ca i tng iu khin. B iu khinPID cn thit k c hm truyn l:

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(6.9)

Zeigler v Nichols a ra hai cch chn thng s b iu khin PID ty theo c imca i tng.

Cch 1: Da vo p ng qu ca h h, p dng cho cc i tng c p ng ivi tn hiu vo l hm nc c dng ch S nh hnh 6.7, v d nh nhit l nhit, tc ng c,

p ng nc ca h h c dng S

Thng s b iu khin P, PI, PID c chn nh sau:

V d 6.11. Hy thit k b iu khin PID iu khin nhit ca l sy, bit c tnhqu ca l sy thu c t thc nghim c dng nh sau:

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Gii. Da vo p ng qu thc nghim ta c:

Chn thng s b iu khin PID theo phng php Zeigler- Nichols:

Cch 2: Da vo p ng qu ca h kn, p dng cho cc i tng c khu tchphn l tng, v d nh mc cht lng trong bn cha, v tr h truyn ng dng ngc,... p ng qu (h h) ca cc i tng c khu tch phn l tng khng cdng nh hnh 6.24 m tng n v cng. i vi cc i tng thuc loi ny ta chnthng s b iu khin PID da vo p ng qu ca h kn nh hnh 6.8. Tng dnh s khuch i K ca h kn hnh 6.8 n gi tr gii hn Kgh, khi p ng ra cah kn trng thi xc lp l dao ng n nh vi chu k Tgh.

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p ng nc ca h kn khi K = Kgh

Thng s b iu khin P, PI, PID c chn nh sau:

V d :Hy thit k b iu khin PID iu khin v tr gc quay ca ng c DC, bitrng nu s dng b iu khin t l th bng thc nghim ta xc nh c khi K = 20v tr gc quay ng c trng thi xc lp l dao ng vi chu k T = 1 sec.

Gii. Theo d kin ca bi ton, ta c:

Chn thng s b iu khin PID theo phng php Zeigler-Nichols:

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Tnh iu khin c v quan st c ca h thng tuyn tnh lin tc

Khi nim iu khin c, quan st c (Controllabbility and Observability) do R-Kalman, R.E. ra.

iu khin c ca mt h thng l vi mt tc ng u vo liu chuyn c trngthi ca h t thi im u vo t0 n thi im cui t1 trong khong thi gian hu hnh (t1-t0) hay khng?

Quan st c ca h thng l vi cc ta o c bin ra yi ca h, liu ta c thkhi phc c cc vector trng thi xi trong mt khong thi gian hu hn hay khng?

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Bi 7: M t ton hc h thng iu khinri rcH thng iu khin ri rc

Khi nim

Chng ny cp n mt loi h thng iu khin c hi tip, trong tn hiu timt hay nhiu im l mt chui xung, khng phi l hm lin tc theo thi gian. Tythuc vo phng php lng t ha tn hiu m ta c cc loi h thng x l tn hiukhc nhau. Phng php lng t ha theo thi gian cho tn hiu c bin lin tc,thi gian ri rc. H thng x l loi tn hiu ny c gi l h thng ri rc. Nu phplng t ha c tin hnh theo thi gian v c theo bin th kt qu nhn c ltn hiu s. H thng x l tn hiu s gi l h thng s. Trong h thng ri rc v hthng s, thng s iu khin - bin ca tn hiu ch xut hin ti cc thi im rirc cch u nhau ng bng mt chu k ly mu tn hiu. V c thi gian tr tt yu doly mu, vic n nh h thng tr nn phc tp hn so vi h lin tc, do i hinhng k thut phn tch v thit k c bit.

S pht trin mnh m ca k thut s, k thut vi x l v k thut my tnh lm chongy cng c nhiu h thng iu khin s c s dng iu khin cc i tng.H thng iu khin s c nhiu u im so vi h thng iu khin lien tc nh uynchuyn, linh hot, d dng i thut ton iu khin, d dng p dng cc thut ton iukhin phc tp bng cch lp trnh. My tnh s cn c th iu khin nhiu i tngcng mt lc. Ngoi ra, gi my tnh ngy cng h trong khi tc x l, tin cyngy cng tng ln cng gp phn lm cho vic s dng cc h thng iu khin s trnn ph bin. Hin nay cc h thng iu khin s c s dng rt rng ri, t cc biu khin n gin nh iu khin nhit , iu khin ng c DC, AC,... n cc hthng iu khin phc tp nh iu khin robot, my bay, tu v tr, cc h thng iukhin qu trnh cng ngh ha hc v cc h thng t ng cho nhng ng dng khcnhau.

S khi h thng iu khin s

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Hnh 7.1 trnh by s khi ca h thng iu khin s thng gp, trong h thng chai loi tn hiu: tn hiu lin tc c(t), uR(t) v tn hiu s r(kT), cht(kT), u(kT). Trungtm ca h thng l my tnh s, my tnh c chc nng x l thng tin phn hi t cmbin v xut ra tn hiu iu khin i tng. V cm bin v i tng l h thng lintc nn cn s dng b chuyn i A/D v D/A giao tip vi my tnh. Do phn tch v thit k h thng iu khin s trc tin ta phi m t ton hc c qutrnh chuyn i A/D v D/A. Tuy nhin, hin nay khng c phng php no cho phpm t chnh xc qu trnh chuyn i A/D v D/A do sai s lng t ha bin , v vythay v kho st h thng s hnh 7.1 ta kho st h ri rc hnh 7.2.

S khi h thng iu khin ri rc

Trong quyn sch ny, chng ta pht trin cc phng php phn tch v thit k hthng iu khin lin tc cho h thng iu khin ri rc. Nu phn gii ca phplng t ha bin nh c th b qua sai s th ta c th xem tn hiu s l tnhiu ri rc, iu c ngha l l thuyt iu khin ri rc trnh by trong quyn nyhon ton c th p dng phn tch v thit k cc h thng iu khin s.

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c im ly mu

Qu trnh ly mu d liu

Ly mu l bin i tn hiu lin tc theo thi gian thnh tn hiu ri rc theo thi gian.Xt b ly mu c u vo l tn hiu lin tc x(t) v u ra l tn hiu ri rc x*(t)(H.7.3). Qu trnh ly mu c th m t bi biu thc ton hc sau:

x*(t) = x(t).s(t)

trong s(t) l chui xung dirac:

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ng thi gi s rng x(t) = 0 khi t < 0, ta c:

Bin i Laplace hai v phng trnh (7.3) ta c:

Biu thc chnh l biu thc ton hc m t qu trnh ly mu.

nh l Shanon: c th phc hi d liu sau khi ly mu m khng b mo dng thtn s ly mu phi tha mn iu kin:

trong fc l tn s ct ca tn hiu cn ly mu.

Trong cc h thng iu khin thc t, nu c th b qua c sai s lng t ha thcc khu chuyn i A/D chnh l cc khu ly mu.

Khu gi d liu

Khu gi d liu l khu chuyn tn hiu ri rc theo thi gian thnh tn hiu lin tctheo thi gian.

Khu gi d liu c nhiu dng khc nhau, n gin nht v c s dng nhiu nhttrong cc h thng iu khin ri rc l khu gi bc 0 (Zero-Order Hold - ZOH)(H.7.4).

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Khu gi bc 0 (ZOH)

Ta tm hm truyn ca khu ZOH. rng nu tn hiu vo ca khu ZOH l xungdirac th tn hiu ra l xung vung c rng bng T (H.7.4b). Ta c:

R(s) = 1 (v r(t) l hm dirac)

Theo nh ngha:

Do :

Biu thc (7.6) chnh l hm truyn ca khu gi bc 0.

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Trong cc h thng iu khin thc t, nu c th b qua c sai s lng t ha thcc khu chuyn i D/A chnh l cc khu gi bc 0 (ZOH).

Nhn xt

Bng cch s dng php bin i Laplace ta c th m t qu trnh ly mu v gi dliu bng cc biu thc ton hc (7.4) v (7.6). Tuy nhin cc biu thc ton hc ny licha hm ex nn nu ta s dng m t h ri rc th khi phn tch, thit k h thng sgp nhiu kh khn. Ta cn m t ton hc khc gip kho st h thng ri rc d dnghn, nh php bin i Z trnh by di y chng ta s thc hin c iu ny.

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Php bin i Z

nh ngha

Cho x(k) l chui tn hiu ri rc. Bin i Z ca x(k) l:

(7.7)

trong : z = eTs (s l bin Laplace)

K hiu:

Nu x(k) = 0,

k < 0th biu thc nh ngha tr thnh:

Min hi t (Region of Convergence - ROC)

ROC l tp hp tt c cc gi tr z sao cho X(z) hu hn.

ngha ca php bin i Z

Gi s x(t) l tn hiu lin tc trong min thi gian, ly mu x(t) vi chu k ly mu T tac chui ri rc x(k) = x(kT).

Biu thc ly mu x(t):

Biu thc bin i Z:

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V z = eTs nn v phi ca hai biu thc (7.9) v (7.10) l nh nhau, do bn cht cavic bin i Z mt tn hiu chnh l ri rc ha tn hiu .

Php bin i Z ngc

Cho X(z) l hm theo bin phc z. Bin i Z ngc ca X(z) l:

vi C l ng cong kn bt k nm trong min hi t ROC ca X(z) v bao gc ta .

Tnh cht ca php bin i Z

Tnh tuyn tnh

Nu:

Th:

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Di trong min thi gian

Lm tr tn hiu Ko mu

Nu:

Th:

Nhn xt:

Nu trong min Z ta nhn X(z) vi

th tng ng vi trong min thi gian l tr tn hiu x(k) ko chu k ly mu.

V:

nn z1 c gi l ton t lm tr mt chu k ly mu.

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T l trong min Z

Nu:

Th:

o hm trong min Z

Nu:

Th:

nh l gi tr u

Nu:

Th:

nh l gi tr cui

Nu:

Th:

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Bin i Z ca cc hm c bn

Hm dirac

Theo nh ngha:

Vy:

(ROC: ton b mt phng Z)

Hm nc n v

Hm nc n v (lin tc trong min thi gian):

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Ly mu u(t) vi chu k ly mu l T, ta c:

Theo nh ngha:

Nu |z-1| < 1th biu thc trn l tng ca cp s nhn li v hn. p dng cng thctnh tng ca cp s nhn li v hn, ta d dng suy ra:

Vy:

Hm dc n v

Hm dc n v (lin tc trong min thi gian):

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Ly mu r(t) vi chu k ly mu l T, ta c:

Ta tm bin i Z ca r(k) bng cch p dng tnh cht t l trong min Z:

Ta c:

Vy:

Hm m

Hm m lin tc trong min thi gian:

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Ly mu r(t) vi chu k ly mu l T, ta c:

Theo nh ngha:

Nu

th biu thc trn l tng ca cp s nhn li v hn. p dng cng thc tnh tng cacp s nhn li v hn, ta suy ra:

Vy:

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Kt qu trn ta d dng suy ra:

Cc phng php tm bin i Z ngc

Cho hm ( ) X z , bi ton t ra l tm ( ) x k . Theo cng thc bin i Z ngc, ta c:

vi C l ng cong kn bt k nm trong ROC ca

v bao gc ta .

Tm x(k) bng cng thc trn rt phc tp, thc t ta thng p dng cc cch sau:

Cch 1: Phn tch X( z ) thnh tng cc hm c bn, sau tra bng bin i Z

V d 7.1. Cho:

Tm x(k).

Gii. Phn tch

, ta c:

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Tra bng bin i Z:

Suy ra: x(k) = (2k + 3k) u(k)

Cch 2: Phn tch ( ) X z thnh chui ly tha

Theo nh ngha bin i z:

Do nu phn tch X(z) thnh tng ca chui ly tha ta s c gi tr x(k) chnh lh s ca thnh phn zk.

V d :. Cho:

Tm x(k).

Gii:

Chia a thc ta c:

Suy ra: x(0) = 0; x(1) = 1; x(2) = 5; x(3) = 19; x(4) = 65,...

Cch 3: Tnh x(k) bng cng thc qui

V d :. Cho:

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Tm x(k).

Gii: Ta c:

Bin i Z ngc hai v phng trnh trn ( tnh cht di trong min thi gian), tac:

Vi iu kin u: x( k 1) = 0; x(k 2) = 0

Thay vo cng thc trn ta tm c:

x(0) = 0; x(1) = 1; x(2) = 5; x(3) = 19; x(4) = 65,...

Cch 4: p dng cng thc thng d

Nu

l cc bc mt th:

Nu

l cc bc p th:

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V d : Cho:

Tm x(k).

Gii. p dng cng thc thng d, ta c:

M:

Do : x(k) = 2k + 3k

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M t h thng ri rc bng hm truyn

Hm truyn ca h ri rc

Quan h gia tn hiu vo v tn hiu ra ca h thng ri rc c m t bng phngtrnh sai phn:

trong n m, n gi l bc ca h thng ri rc Bin i z hai v phng trnh ta c:

t:

G(z) c gi l hm truyn ca h thng ri rc.

Hm truyn (7.18) c th bin i tng ng v dng:

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Hai cch biu din trn hon ton tng ng nhau, trong thc t hm truyn dng thhai c s dng nhiu hn.

V d 7.5. Cho h thng ri rc m t bi phng trnh sai phn:

Tm hm truyn ca h thng.

Gii. Bin i Z hai v phng trnh sai phn m t h thng, ta c:

Tnh hm truyn h