Giáo trình lý thuyết điều khiển tự động 1

7/22/2019 Gio trnh l thuyt iu khin t ng 1

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I HC THI NGUYNKHOA CNG NGH THNG TIN

B mn Cng nghiu khin tng

DNG CHNH CNG

GIO TRNH

L THUYT IU KHINTNG 1

Thi Nguyn 11-20007


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Gio trnh L thuytiu khin tng 1

CHNG 1 : M T MT H THNG IU KHIN TNG 11.1 Cc khi nim cbn .......................................................................................................11.2. Cc phn t cbn ca h thng,iu khin tng......................................................21.3 Cc nguyn tc iu khin cbn....................................................................................2

1.4.2/ Phn loi theo tnh cht ca lng vo.....................................................................21.4.3/ Phn loi theo dng tn hiu s dng trong h thng. ..............................................3

1.4 Phn loi cc h thng iu khin tng.......................................................................41.4.1 Phn loi theo nguyn l xy dng. ..........................................................................41.4.4/ Phn loi theo dng phng trnh ton hc m t h thng.....................................41 4.5/ Phn loi theo tnh cht ca cc tc ng bn ngoi................................................51.4.6/ Phn loi theo s lng i lng cn iu khin....................................................5

1.5 Qu trnh thit lp mt h thng iu khin.....................................................................5Cu hi n tp chng 1.........................................................................................................6

CHNG 2 : M HNH TON HC CA H THNG IUKHIN..................................................................................................7

2.1 Cc khu cbn ...............................................................................................................7

2.1.1 Khu khuch i........................................................................................................72.1.2 Khu tch phn...........................................................................................................82.1.3 Khu vi phn..............................................................................................................82.1.4 Khu bc nht............................................................................................................82.1.5 Khu bc hai ..............................................................................................................82.1.6 Khu bc n .................................................................................................................8

2.2 M hnh trong min tn s................................................................................................82.2.1 Khi nim v php bin i Laplace v ng dng ....................................................82.2.2 Hm s truyn ca h thng KT. .......................................................................202.2.3 Hm truyn t ca mch in ................................................................................222.2.4 Hm truyn ca h thng ckh..............................................................................242.2.5 S tng ng gia h ckh vi mt mch in ................................................272.2.6 Hm truyn ca cc phn tin t ........................................................................29

2.3 M hnh ton hc trong min thi gian ..........................................................................302.3.1 Khi nim trng thi v bin trng thi ...................................................................302.3.2 H tuyn tnh h s hng. ........................................................................................322.3.3 ng dng biu din m hnh ton hc trn khng gian trng thi..........................32

2.4 Chuyn t hm truyn t sang khng gian trng thi v ngc li ..............................352.4.1 Chuyn t hm truyn t sang khng gian trng thi............................................352.4.2 Chuyn t khng gian trng thi sang hm truyn t............................................39

2.5 Tuyn tnh ha................................................................................................................40

Bi tp chng 2...................................................................................................................41CHNG 3 : P NG THI GIAN..............................................423.1 Cc c tnh ca h thng KT...................................................................................42

3.1.1 c tnh thi gian ....................................................................................................423.1.2 c tnh xung (Hm trng lng):..........................................................................423.1.3 Hm qu ..............................................................................................................423.1.4 c tnh tn s.........................................................................................................43

3.2 Cc khu ng hc in hnh .........................................................................................463.2.1 nh ngha cc khu ng hc in hnh ................................................................463.2.2.Cc khu nguyn hm. ............................................................................................463.2.3 Khu tch phn.........................................................................................................50

3.2.4.Khu vi phn............................................................................................................513.2.5 Khu tr ...................................................................................................................53

3.3 M hnh ZPK (Zero, Pole and Gain) ..............................................................................53


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3.4 H thng bc nht...........................................................................................................563.5 H thng bc 2................................................................................................................59

3.5.1 H thng p ng xung tt dn (Overdamped) .......................................................603.5.2 H thng p ng di tt dn (Underdamped)......................................................613.5.3 H thng p ng khng b nht (Undamped)........................................................623.5.4 H thng p ng tt dn ti hn (Critically Damped Response)...........................62

3.5.5 Tm p ng t do ...................................................................................................633.6 Mt s vn chung v h thng bc hai.......................................................................633.7 H thng bc hai di tt dn (Undierdamped) .............................................................65Bi tp chng 3...................................................................................................................70

CHNG 4 : CC PHNG PHP GIM THIU H THNGA CP..............................................................................................71

4.1 S khi ca mt h thng..........................................................................................714.1.1 H thng dng ni tip ............................................................................................714.1.2 H thng dng song song (Parallel Form)...............................................................724.1.3. H thng dng phn hi (Feedback Form).............................................................73

4.2 Phn tch v thit k h thng phn hi..........................................................................774.3 Grap tn hiu...................................................................................................................794.3.1 Cc khi nim cbn ..............................................................................................794.3.2 Cc dng biu din Graph tn hiu ..........................................................................804.3.3 Cc quy tc bin i Graph .....................................................................................804.3.4 Quy tc Masson .......................................................................................................81

Bi tp chng 4...................................................................................................................84

CHNG 5 : SN NH CA H THNG ...............................865.1 Khi nim vn nh h thng iu khin tng........................................................865.2 Nhn xt chung :.............................................................................................................875.3 Tiu chun n nh i s...............................................................................................87

5.3.1 Tiu chun Rao (Routh): .........................................................................................885.3.2 Tiu chun Hurwitz .................................................................................................895.3.3 Mt s trng hp ca tiu chun Routh - Hurwitz................................................905.3.4 S dng tiu chun Routh - Hurwitz thit k sn nh ...................................92

5.4 Xt n nh cho h c m t ton hc di dng m hnh trng thi.............................93Bi tp chng 5...................................................................................................................95

CHNG 6 : CHT LNG H THNG......................................966.1 M u...........................................................................................................................966.2 Sai strng thi xc lp (SSE) cho h thng phn hi n v ....................................97

6.2.1 SSE i vi T(s) ......................................................................................................97

6.2.2 SSE cho G(s) ...........................................................................................................986.3 Hng s sai s tnh v loi h thng.............................................................................1016.3.1 Hng s sai s tnh.................................................................................................1016.3.2 Loi h thng.........................................................................................................104

6.4 Cc tham s k thut rt ra t SSE...............................................................................1056.5 SSE cho nhiu...............................................................................................................1066.6 SSE cho h thng phn hi khng phi l n v.........................................................1086.7 nhy ........................................................................................................................110Bi tp chng 6.................................................................................................................113

CHNG 7: TNG HP H THNG IU KHIN..................1157.1 Khi nim .....................................................................................................................1157.2 Chn biu chnh.......................................................................................................115

7.2.1 Phn loi cc biu chnh...................................................................................1157.2.2 Phng php Ziegler- Nichols...............................................................................116


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7.2.3 Tiu chun phng ..................................................................................................1177.2.4 Phng php tng hng s thi gian (Kuhn).........................................................119

7.3 iu khin c v quan st c ...............................................................................120Cu hi n tp chng 7.....................................................................................................121

CHNG 8 : H THNG IU KHIN S .................................122 8.1 Mu..........................................................................................................................122

8.2 M hnh gi mu bc khng.........................................................................................1258.3 Bin i Z .....................................................................................................................1258.4 Hm truyn t .............................................................................................................1278.5 Sn nh.....................................................................................................................1298.6 Sai s xc lp ................................................................................................................130Ti liu tham kho..............................................................................................................133


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DANH MC HNH V

Hnh 1.1 . Siu khin ca l hi pht in .............................1Hnh 1.2: S tng qut h thng iu khin tng ......................2Hnh 1.3. S

nguyn tc

iu khi

n theo sai l

ch ............................2

Hnh 1.4: S nguyn tc iu khin b nhiu.................................3Hnh 1.5: S nguyn tc iu khin hn hp ...................................3Hnh 2.1: S phn chia h mt h thng iu khin thnh cc hthng .....................................................................................................7Hnh 2.2 : S mt h thng iu khin tng qut ...........................7Hnh 2.3: S khu khuych i tnh.................................................7Hnh 2. 4: S khu khuch i tng................................................8Hnh 2.5: in tr...............................................................................22 Hnh 2.6: in cm L..........................................................................22Hnh 2.7: Tin C ............................................................................23Hnh 2.8: S cc phn t mch in RLC mc ni tip................23Hnh 2.9: S cc phn t mch in RLC mc song song.............24Hnh 2.10: S biu din l xo .......................................................24Hnh 2.11: S biu din b gim chn du p .............................25Hnh 2.12: S biu din trng khi ...............................................25Hnh 2.13: S biu din thit b gim chn....................................26Hnh 2.14: S biu din lc tc ng ln trng khi ....................26Hnh 2.15: S biu din s tng ng gia mch c kh vmch in............................................................................................28Hnh 2.16: Biu din phn t khuch i thut ton..........................29Hnh 2.17 S h thng khuch i o ..........................................30Hnh 2.18: S khi biu din h thng iu khin trong khng giantrng thi .............................................................................................32Hnh 2.19: S mch RLC mc hn hp .........................................32Hnh 2.20: S mch RLC mc ni tip .........................................34Hnh 2.21: S mch RLC mc ni tip .........................................35Hnh 2.22: S biu din bng s khi trong gian trng thi.....39Hnh 3.1: c tnh tn s bin pha.................................................45Hnh 3.2 Biu din khu ng hc in hnh......................................46Hnh 3.3. c tnh thi gian ca khu khng qun tnh......................47Hnh 3.4: c tnh tn s ca khu khnng qun tnh ........................47Hnh 3.5: c tnh thi gian ca khu qun tnh bc nht..................47

Hnh 3.6: c tnh tn s ca khu qun tnh bc nht.......................48Hnh 3.7: c tnh thi gian ca khu bc hai....................................49


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Hnh 3.8: c tnh tn s ca khu bc hai.........................................50Hnh 3.9: c tnh thi gian ca tch phn .........................................51Hnh 3.10: c tnh tn s ca khu tch phn ...................................51Hnh 3.11: c tnh thi gian ca khu vi phn l tng...................52Hnh 3.12:

c tnh t

n s

ca khu vi phn l t

ng........................52

Hnh 3.13. c tnh qu v cc c tnh tn s ca khu tr..........53Hnh 3.14 : S b tr cc im cc v im khng.......................55Hnh 3.15: H thng i tng lm v d 3 ........................................55Hnh 3.16: H thng bc nht v phn bim cc ..........................56Hnh 3.17: p ng u ra ca h thng bc 1 vi tn hiu bc thangn v ..................................................................................................57 Hnh 3.18: ng c tnh p ng ca h thng bc nht................58Hnh 3.19: Cc h thng bc hai v p ng vi tn hiu bc thang nv .........................................................................................................60 Hnh 3.20: p ng bc hai to bi cc nghim phc........................62Hnh 3.21: p ng bc hai theo h s tt dn ...................................65Hnh 3.22: p ng bc hai ca h thng di tt dn.......................66Hnh 4.1: S khi ca h thng.....................................................71 Hnh 4.2: S khi ca h thng ni tip.........................................72Hnh 4.3: H thng ghp ni tip.......................................................72Hnh 4.4: S khi ca h thng mc song song.............................73Hnh 4.5: S khi ca h thng c phn hi.................................73Hnh 4.6: a) H thng phn hi m b) H thng phn hi dng c)Hm truyn ca h thng c phn hi.................................................73 Hnh 4.7: S khi h thng phn hi n v .................................74 Hnh 4.8 : Hnh bin i cc s khi cbn..................................76Hnh 4.9: Rt gn s p dng cc quy tc bin i .......................77Hnh 4.10: H thng c phn hi m ..................................................77Hnh 4.11: Skhi h thng phn hi bit trc h s khuch i ...78Hnh 4.12: S khi ca h thng phn hi khi h s khuch i Kcha bit..............................................................................................79Hnh 4.11 : Mt nt cbn .................................................................79Hnh 4.13 : Biu din mt nhnh cbn ............................................80Hnh 4.14: Graph biu din h thng ni tip.....................................80Hnh 4.15: Granph biu din h thng song song...............................80Hnh 4.16: Graph biu din c phn hi .............................................80Hnh 4.17: skhi minh ho quy tc Masson.....................................82

Hnh 5.1 : H thng c h s khch i K cha bit...........................92Hnh 6.1: Cc tn hiu th...................................................................96


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Hnh 6.2: Cc dng phn hi...............................................................97 Hnh 6.3: H thng c sai strng thi xc lp vi T(s) .................98Hnh 6.4: H thng khng c b tch phn .......................................100Hnh 6.5 : H thng c mt b tch phn..........................................100Hnh 6.6: H

th

ng c m

t b

tch phn...........................................102Hnh 6.7: H thng khng c b tch phn .......................................104Hnh 6.8: H thng khng c b tch phn .......................................106Hnh 6.9: H thng phn hi m c nhiu tc ng..........................106Hnh 6.10: H thng phn hi nhiu................................................107Hnh 6.11: H thng phn hi m c nhiu tc ng vi cc i tngthc ...................................................................................................108Hnh 6.12 : H thng phn hi khng phi l n v ........................108 Hnh 6.13: H thng phn hi khng phi l n v .........................109 Hnh 6.14: H thng phn hi m khng phi l n v c nhiu tcng ..................................................................................................110Hnh 6.15: nhy i vi h kn...................................................111 Hnh 6.16: nhy i vi SSE......................................................112Hnh 7.1: Cu trc cbn ca mt h thng iu khin...................115Hnh 7.2: c tnh qu ................................................................117 Hnh 7.3: S cu trc c h s khuych i K .............................117Hnh 7.4: Cu trc iu khin c phn hi n v ...........................118 Hnh 8.1 : Siu khin phn hi c s dng my tnh ..............122Hnh 8.2: Tn hiu c trch mu s dng trong my tnh s .........123Hnh 8.3: Tn hiu r(t) c trch mu..............................................123Hnh 8.4: Tch ca dng sng theo thi gian v tn hiu trch mu ..124Hnh 8.5: Tn hiu r(t) c trch mu..............................................125Hnh 8.6: H thng tn hiu trch mu ..............................................127Hnh 8.7: Mt phng phn b sn nh.........................................129Hnh 8.8: H thng iu khin phn hi c trch mu .............130Hnh 8.9: Sai s xc lp ca hiu khin s ...................................131


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1

CHNG 1 : M T MT H THNG IU KHIN TNG

1.1 Cc khi nim cbn

hiu c khi nim v h thng iu khin tng trc ht ta xem v dsau

Tn hiu cho

Hnh 1.1 . Siu khin ca l hi phtin

iu khin l tp hp tt c cc tc ng c mc ch nhm iu khin mt qu

trnh ny hay qu trnh kia theo mt quy lut hay mt chng trnh cho trc. iukhin hc l mt b mn khoa hc nghin cu nguyn tc xy dng cc hiu khin.

Qu trnh iu khin hoc iu chnh c thc hin m khng c s tham giatrc tip ca con ngi, th chng ta gi l qu trnh iu khin v iu chnh tng.

Tp hp tt c cc thit b m nh qu trnh iu khin c thc hin gi lh thng iu khin .

Tp hp tt c cc thit b k thut; m bo K hoc C tng mt qu trnh

no c gi l h thng K hoc C tng (i khi gi tt l h thng tng -HTT).


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2

1.2. Cc phn tcbn ca h thng,iu khin tng

i tng iu khin (Object), Thit b iu khin (Controller ), Thit b o

lng (Measuring device).- S tng qut

Hnh 1.2: S tng qut h thngiu khin tng

Mi h thng iu khin tng u bao gm 3 b phn cbn :- Thit biu khin C (Controller device).

- i tng iu khin (Object device).

- Thit bo lng (Measuring device).

u(t) tn hiu vo ; e(t) Si lch iu khin ; x(t) Tn hiu iu khin ; y(t) Tn hiura ; z(t) Tn hiu phn hi

1.3 Cc nguyn tc iu khin cbn

C 3 nguyn tc iu khin cbn :-Nguyn tc iu khin theo sai lch (Hnh l.3).

Hnh 1.3. S nguyn tc iu khin theo sai lch

Tn hiu ra y(t) c a vo so snh vi tn hiu vo u(t) nhm to nn tn hiutc ng ln u vo biu khin C nhm to tn hiu iu khin i tng O.

V hu ht cc h thng KT trong k thut l nhng h mch kn v qu trnhiu khin cc thit b k thut chung quy li l qu trnh iu chnh cc tham s can, nu di y chng ta s cp n s phn loi cc h thng KT mch kn vl thuyt v cc h.

1.4.2/ Phn loi theo tnh cht ca lng vo.

Tu theo tnh cht ca tc ng u vo, cc h thng KT c 3 loi :H thng n nh tng (iu chnh theo hng s) l h thng c lng vo

khng i. Nhim v ca h thng l duy tr mt hoc mt vi i lng vt l gi tr


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khng i. Th d nh h thng KT tc ng cnhit, h thng KTin p,tn s ca my pht, hn nh ng bay ca my bay khi gc li khng thay i ...

H thng iu chnh theo chng trnh l h thng c lng vo l cc hm bit trc, c th di dng chng trnh.Th d hiu khin ng bay nh trcca my bay khng ngi li, h thng iu khin cc my cng c: bo, phay vichng trnh nh trc trong b nhmy tnh...

H tng bm, gi tt l h bm l h thng c lng vo l cc hm thi giankhng bit trc, c th thay i theo quy lut bt k. Nhim v ca h l bo mlng ra phi 'bm" theo s thay i ca lng vo. Th d cc h nh l h bm ngb gc, cc h bm v tuyn in t ca cc i radar...

1.4.3/ Phn loi theo dng tn hiu sdng trong h thng.

Theo dng tn hiu s dng trong h thng, chng ta c cc tc ng lin tc v

cc h thng gin on (hay h ri rc).H tc ng lin tc (gi tt l h lin tc l h m tt c cc phn t ca h c

lng ra l cc hm lin tc theo thi gian.

Tn hiu di dng hm lin tc c th l tn hiu mt chiu (cha bin iu)hoc tn hiu xoay chiu ( c bin iu) tng ng chng ta c hKT mtchiu (DC) v h thng KT xoay chiu (AC) (th d h thng bm ng b cngsut nho dng ng cchp hnh 2 p ha).

-Nguyn tc iu khin theo phng php b nhiu (Hnh 1.4)

Hnh 1.4: S nguyn tc iu khin b nhiu

Nguyn tc b nhiu l s dng thit b b K gim nh hng ca nhiu l

nguyn nhn trc tip gy ra hu qu cho h thng (Hnh 1.4).-Nguyn tc iu khin theo sai lch v b nhiu (Hnh l.5)

Hnh 1.5: S nguyn tc iu khin hn hp


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Nguyn tc iu khin hn hp l phi hp c hai nguyn tc trn, va c hitip theo sai lch va dng cc thit b b nhiu.

1.4 Phn loi cc h thng iu khin tng.

1.4.1 Phn loi theo nguyn l xy dng.

Cc phn tc phn chia thnh cc loi: h thng K theo mch h, h thngK theo mch kn v h thng K hn hp .

Ngoi nhng nguyn l trn, t nhng nm 60 ca th k XX, trn csp dngiu khin hc trong cth sng vo k thut ra i mt loi hnh h thng tngm phng hot ng ca cth sng: l cc h t chnh, thch nghi. Nguyn l tchnh v thch nghi khng i hi phi bit y cc c tnh ca qu trnh iukhin v trong qu trnh lm vic, cc h thng ny t chnh v thch nghi vi cc iukin bn ngoi thay i.

L thuyt cc hK t chnh v thch nghi tr thnh mt nhnh pht trinquan trng ca l thuyt KT.

H tc ng gin on (gi tt l h gin on hay h ri rc) l cc h c cha tnht mt phn t gin on, tc l phn t c lng vo l mt hm lin tc v lngra l mt hm gin on theo thi gian.

Tu theo tnh cht gin on ca lng ra, cc h gin on c th phn chiathnh cc loi: h thng KT xung, h thng KT kiu r le v h thng KTs.

Nu s gin on ca tn hiu ra xy ra qua nhng thi gian xc nh (ta gi lgin on theo thi gian) khi tn hiu vo thay i, th ta c hKT xung.

Nu s gin on ca tn hiu xy ra khi tn hiu vo qua nhng gi tr ngngxc nh no (chng ta gi l gin on theo mc), th c thKT kiu rle. Hrle thc cht l h phi tuyn, v c tnh tnh ca n l hm phi tuyn. y l itng nghin cu ca mt phn quan trng trong l thuyt K.

Nu phn t gin on c tn hiu ra di dng m s (gin on c theo mc vc theo thi gian), th ta c hKT s. H thng KT s l h cha cc thit b s

(cc b bin i A/D, D/A, my tnh in t (PC), b vi x l.1.4.4/ Phn loi theo dng phng trnh ton hc m t h thng.

V mt ton hc, cc h thng KTu c th m t bng cc phng trnhton hc: phng trnh tnh v phng trnh ng. Da vo tnh cht ca cc phngtrnh, chng ta phn bit h thng KT tuyn tnh v hKT khng tuyn tnh(phi tuyn).

H thng KT tuyn tnh l h thng c m t bng phng trnh ton hctuyn tnh. Tnh cht tuyn tnh ca cc phn t v ca c h thng KT ch l tnh

cht l tng. V vy, cc phng trnh ton hc ca h thng l cc phng trnh c tuyn tnh ho, tc l thay cc s ph thuc gn ng tuyn tnh.


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5

H tuyn tnh c phng trnh ng hc vi cc tham s khng thay i th gi lhKT tuyn tnh c tham skhng thay i, hay hKT tuyn tnh dng, cnnu h thng c phng trnh vi tham s thay i th gi l hKT tuyn tnh ctham sbin thin, hay hKT tuyn tnh khng dng.

H thng KT phi tuyn l h thng c m t bng phng trnh ton hcphi tuyn. H phi tuyn l h c cha cc phn t phi tuyn in hnh, th d l hc cha cc phn t rle.

1 4.5/ Phn loi theo tnh cht ca cc tc ng bn ngoi.

Cc tc ng bn ngoi vo h tng c quy lut thay i bit trc hocmang tnh cht ngu nhin.

H thng tin nh l cc h c cc tc ng bn ngoi l tin nh, tc l bittrc cc quy lut thay i ca n ( th d xt h thng vi cc tc ng in hnh).

H thng khng tin nh (hay h ngu nhin) l cc hc xem xt nghin cukhi cc tc ng bn ngoi l cc tn hiu ngu nhin.

1.4.6/ Phn loi theo s lng i lng cn iu khin.

Tu theo s lng cn iu khin (lng ra ca h) chng ta c: h mt chiu vh nhiu chiu.

H thng KT mt chiu c cha mt i lng cn iu khin, cn hKTnhiu chiu l h c cha t hai i lng cn iu khin trln. Th d v h nhiuchiu c th l h thng KT mt my pht in, nu h thng KT cng mt lc

iu khin tng in p v tn s ca n.Ngoi cc cch phn loi chnh xt trn, tu thuc vo s tn ti sai s ca

htrng thi cn bng, chng ta phn bit hai loi h thng: h thng tnh (c sai stnh) v h phim tnh (khng c sai s tnh). Tu thuc vo quy lut (nh lut) iukhin (tc l dng ca tn hiu iu khin x(t) do ccu iu khin to ra), chng taphn bit cc biu khin t l (biu khin P), biu khin t l vi phn (biukhin PD), biu khin vi phn - tch phn (biu khin PID).

1.5 Qu trnh thit lp mt h thng iu khin

- Bc 1 : Chuyn i cc yu cu k thut thnh mt h thng vt l.

- Bc 2: V s khi chc nng. Chuyn i s miu tc tnh h thngthnh mt s khi chc nng. y l s miu t v cc phn chi tit ca h thngv mi quan h gia chng.

- Bc 3: Thit lp s nguyn l.

- Bc 4: S dng s nguyn l thit lp s khi hoc graph tn hiu hocbiu din khng gian trng thi.

- Bc 5 : Rt gn s khi.- Bc 6: Phn tch v thit k.


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6

Cu hi n tp chng 1

1 . H thng iu khin tng c th phn loi nh th no?

2. H thng iu khin c my phn t cbn?

3. Hy nu cc quy tc iu khin cbn iu khin mt h thng iu khin?4. Nu cc bc thit lp mt h thng iu khin?


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7

CHNG 2 : M HNH TON HC CA H THNG IU KHIN

Mi h thng c thchia lm nhiu phn sthun tin hn v mi phn scbiu din bng 1 hm ton hc gi l hm truyn t (transfer function)

Hnh 2.1: S phn chia h mt h thngiu khin thnh cc h thng

2.1 Cc khu cbn

Ta c mt h thng iu khin:

Hnh 2.2 : S mt h thngiu khin tng qut

a phn cc mch phn hi ca h thng iu khin l mch phn hi m.

Khi chng ta tin hnh phn tch h thng tt hay xu hay thit k biu khincho h thng u phi xut pht t m hnh ton hc ca h thng hay ni cch khc ta

phi tm c quan h gia u vo v u ra ca h thng.

2.1.1 Khu khuch i.

Hnh 2.3: S khu khuych i tnh

- Khu khuch i l tn hiu u ra l khuch i ca tn hiu u vo y = K.x (2.l)trong : K l h s khuch i

( Khuch i tnh l cc tn hiu u vo th tm c tn hiu u ra)

- Cng c h thng c khuch i nhiu tng


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Hnh 2.4: S khu khuch i tng

2.1.2 Khu tch phn

vi Ti l thi gian tch phn

2.1.3 Khu vi phn

TD l hng s thi gian vi phn2.1.4 Khu bc nht

Trong : K l h s truyn ca khu

T l hng s thi gian ca khu

Phn ng ca h thng tt hay xu ph thuc vo h sK, nhanh hay chm ph thuc

vo T.2.1.5 Khu bc hai

Trong : K l h s khuch i

T l hng s thi gian

suy gim tn hiu

y l m hnh ton hc ca mch RLC.2.1.6 Khu bc n

thng thng n>m.

2.2 M hnh trong min tn s

2.2.1 Khi nim v php bin i Laplace v ng dng

2.2.1.1 Khi nim v bn cht ca php bin i Laplaee :Khi s dng cc php bin i tn hiu h thng t min thi gian sang min


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khc thun tin trong vic x l tn hiu. Nh trong h thng lin tc ngi ta hays dng php bin i Lpalace bin i t min thi gian sang min tn s phc.Cc phng trnh vi tch phn s chuyn i thnh cc phng trnh i s thngthng.

Trong cc h thng ri rc ngi ta hay s dng php bin i Z chuyn tnhiu t min thi gian sang min tn s phc. Trong thc t ngi ta cn s dng cc

php bin i khc x l tn hiu nh gii tng quan, m ho c hiu qu, chngnhiu,

Thc hin cc php bin i c cng c ton hc nh my tnh s, cng c phbin v hiu qu l phn mm Matlab hay thc hin bin i bng tay.

a) Bin i Laplace thun

nh.ngha: Gi F(s) l bin i Laplace ca hm f(t), khi ta c:

trong :

-s = +j)

- e-st l ht nhn ca php bin i.

- F(s) l hm phc.

- f(t) l hm biu din trn min thi gian xc nh trn R.

thc hin c bin i Laplace hm f(t) phi l hm thc v tho mn mtsiu kin sau:

- f(t) l hm gc khi tho mn cc iu kin sau:

1. f(t) = 0 khi t < 0

2. f(t) lin tc khi t 0, trong khong hu hn bt k cho trc ch c hu hncc im cc tr.

3 . Hm f(t) gi l hm bc s m khi t nu tn ti mt s thc 0 v M >

0 th |f(t)| Met ,t> 0 , c gi l ch s tng ca hm f(t). Khi hm f(t) lhm bc s m nu hm f(t) tng khng nhanh hn hn hm et

- Nu f(t) l hm gc c ch s tng th tch phn I = +

0)( dttfe st Je s hi t

trong min Re(s) = > . Khi I = +

0)( dttfe st = F(s) S l mt hm phc.

V d l: Tm nh ca hm gc sau


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p dng cng thc bin i ta c

Tm bin i Laplace? Gii

V du 3: Tm nh Laplace ca hm f(t) = 4t2

T bng bin i Laplace ta cL{t

n} = 1

!+ns

n

p dng bin i tm nh Laplace ca hm f(t) = 4t22

L{4t2} =

312

8!24

ssx =+

b) Bin i Laplace ngc:

Bin i Laplace ngc l xc nh tn hiu f(t) tnh Laplace F(s) ca n.

Gi f(t) l gc ca nh F(s) Khi ta c:

nhng cng thc (2.8) ny t dng, ta hay p dng phng php bin i ngc hmF(s) c dng hm hu t.

Gi s f(t) c nh Laplace dng sau


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vi n > m.

Cc bc thc hin nhsau:

Bc 1 : Phn tch F(s) thnh tng cc hm phn thc ti gin

trong A, Aki, Bk, Ck l cc hng s. ak l im cc thc bi rk V k+ jk l im

cc phc ca F(s), ni cch khc chng l im m ti F(s) = .

Bc 2: Xc nh hm gc cho tng phn t.

V d 1 : Tm hm gc f(t) ca nh Laplace sau

Gii:

Bc 1 : Phn tch thnh tng cc phn thc ti gin

Bc 2 : Xc nh hm gc cho tng thnh phn

V d 2:

Ta thc hin chia t s cho mu s cho n khi s d cn li c bc ca t nhhn bc ca mu.

Thc hin bin i Laplace ngc c s dng bng bin i Laplace


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S dng phng php phn tch Xs=5

22 ++ss

thnh tng cc phn thc n

gin.Ta xt mt s trng hp sau:

Trng hp 1 : Nghim ca mu thc T(s) l thc v ring bit. Gi s nghim camu thc T(s) c hai nghim S1 = -1 v S2 = - 2.

Nghim ca mu thc l ring bit nn tng phn thc s c bc l 1 .

tm K1 ta nhn (2.) vi (s) tch Kl ring ra

Sau cho s - 1 , rt ra c Kl = 2. Lm tng t v cho s - 2 ta rt rac K2 = - 2.

Lc

Thc hin bin i Laplace ngc ca X(s) ta c

Mt cch tng qut khi mu s ca F(s) cos nghim thc v ring bit, ta thc hinnh sau:

Nu bc ca t nh hn bc ca mu ta thc hin tm cc h s Ki nh sau:

- Nhn hai v vi (s +pi) tm h s Ki.

- Cho s - p;, rt ra c Ki.

Trng hp 2: Mu s c nghim thc v lp li. Gi s nghim ca mu thc T(s)

c ba nghim si = -1 v s2,3 = - 2. Lc ta phn tch X(s) nh sau:


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Tm cc h s Kl, K2 V K3

tm K2 ta nhn hai v ca (2.) vi (s + 2)2

khi cho s - 2 ta tm c K2 = - 2

Tm K3 bng cch ly o hm (2.) theo bin s ta c

Cho s - 2 ta rt ra c K3 = - 2.

Thay K1 , K2 v K3 ta c

Thc hin bin i Laplace ngc ta c

x(t) = (2e-1 - 2te-2t - 2e-2/ )u(t)

Tng qut cho trng hp ny

thc hin c phi c iu kin bc ca t nh hn bc ca mu v c rnghim bi ti - pl tm Kl n Krcho phn thc c nghim bi, u tin ta nhn hai

v (2. l2) vi (s +pl)

r

. ta c

Ta c th tm ngay c Kl khi cho s - pl. tm K2 ta ly o hm (2.12)

theo bin s v cho s - pl. Ln lt ly o ta tm c K3n Kr Cng thc chung tm Kl n Kr l:


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Trng hp 3: Mu thc c nghim phc hay nghim o. Gi s mu s ca F(s) cnghim phc.

F(s) c th phn tch thnh cc phn thc nh sau

D dng tm c Kl = 3/5 khi cho s 0. tm K2 V K3 ta quy ng phn thcvi mu s chung nh nht l s(s2+ 2s + 5) bc cc phn thc

Thc hin ng nht thc hai v ta c

Thay cc h s ta c

T bng tra nh ca tch hm m v hm sin v cos

V

Cng hai cng thc trn ta c

Ta a cng thc (2.) v dng trn

Tra bng ta tm c hm gc nh sau


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Trong trng hp trn ta cng c th thc hin n gin bng cch phn tch thngthng

Kl d dng tnh c v bng 3/5.

Tng t ta tm c K3 l nghim phc lin hp ca K2.

Ta c

T ta tm c hm gc nh sau

P dng cng thc le ca hm sin v cos

Suy ra

Bin i Laplace mt shm n gin:


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2.2.1.2 Cc tnh cht ca php bin i Laplace :

1.Tnh cht tuyn tnh: L[a.f(t)]= a.L[f(t)] = a.F(s).2. Tnh cht xp chng: Nu fl(t) v f2(t) c nh bin i Laplace l F

1(s) v F

2(s) th

ta c:

L[f1(t) + f)(t)] = L[f1(t)] + L[f2(t)] = Fl(s) + F2(s)

V d : Tm nh ca hm hm f(t) cosat trong a l hng s.

Theo cng thc le ta c

Thc hin php bin i Laplace

3. Tnh cht tr (Chuyn dch thi gian -Translation in time):Nu f(t) c nh l F(s), a l mt s thc v f(t-a) = 0 khi 0


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p dng tnh cht tr ta c

4. Tnh cht vi phn phc (Complex difflrentiation): Nu f(t) c nh l F(s) th :

V d: L[t.e-as] - - dl[e-as]/ds = - d[l/(s+a)]/ds = 1/ (s+a)2

5. Tnh cht chuyn dch nh: Nu f(t) c nh l F(s) a l mt s thc bt k hay l mts phc khi :

6. Tnh cht vi phn thc: Nu f(t) c nh l F(s) th

7. Tnh cht tch phn thc Nu F(s) l nh ca f(t) th

8. Tnh cht gi tr cui:

Nu bin i Laplace ca f(t) v f (s) v nu gii hn f(t) tn ti khi t khi

:

9. Tnh cht gi tru: Nu tn ti )(lim0

tft

th

2.2.1.3 ng dng ca php bin i Laplace


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a) ng dng gii phng trnh vi phn tuyt tnh.

Khi chuyn phng trnh vi phn t min thi gian sang min nh phc trthnhphng trnh i s. Sau khi gii ra c nghim ta chuyn ngc v min thi gian.

V d 1 : Gii phng trnh vi phn sau vi cc skin u bng khng.

chuyn sang min nh Laplace vi y(0-) = 0 v (0-) = 0

Rt Y(s) ra ta c

Phn tch Y(s) thnh tng cc phn thc ti gin

Tm cc h s K1, K2 v K3.

Vy

Thc hin bin i Laplace ngc ta tm c

Trong cng thc trn c cha u(t) ni ln rng cc p ng s bng 0 cho n khi t =0.V vy cc p ng u ra cng bng 0 cho n kho t = 0. thun tin ta c th b khiu u(t) i, vy p ng u ra c th vit nh sau

V d 2: Gii phng trnh vi phn bng ton t Laplace sau

vi skin y(+0) = a v 0)0(

=+

dt

dy

Chuyn c hai v sang min nh phc nhton t Laplace


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Thc hin bin i Laplace ngc rt ra c y(t)

V d 3: Gii phng trnh vi phn sau

vi Skin : 0)0(

)0( =+

=+dt

dyy

Thc hin bin i Laplace

Suy ra

b) Gii mch inCho mch in sau

Gi s khi mch in ng ti thi im t - 0 th VC(0) = 1.0V. Tm dng ini(t) chy trong mch in. (trong V(t) = 5V, C = 1F, R = 1k.)

Gii :

Ta c phng trnh sau

Hay

thay cc thng su bi cho vo


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Theo u bi VC(0) = 1.0V nn ta c

Thay vo cng thc trn ta c

Thc hin tra bng bin i Laplace ta tm c i(t) nh sau

2.2.2 Hm s truyn ca h thng KT.

Nhm n gin ho cc phng php phn tch v tng hp h thng tngngi ta thng chuyn phng trnh ng hc ca hdng phng trnh vi phnvit vi cc nguyn hm x(t), y(t) thnh phng trnh vit di dng cc hm s X(s),Y(s) thng qua php bin i Laplace.

V d xt hm s x(t) - hm s ca bin s t (bin s thc, y t l thi gian) tagi l nguyn hm. Ta cho php bin i hm s x(t) thng qua tch phn:

trong : s = + j - bin s phc, bin i (2.l5) hm x(t) thnh hm bin s X(s)c gi l l bin Laplace, v X(s) c gi hm nh. Nh vy hm nh l mt hm

bin s phc s. Php bin i Laplace c k hiu sau:

Gi s nguyn hm x(t) c cc iu kin ban u khng, tc l vi t-o gi tr cua

hm x(t) v cc bc o hm dix(t)/dti; vi i = 1, 2, 3, . .., (n-l) u bng 0 tnh theotnh cht ca php bin i Laplace (nh l vnh o hm ca nguyn hm) chng


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ta c :

Nhn hai v ca phng trnh (2.6) vi e-st

, sau ly tch phn theo t t 0 n, tc l ly bin i Laplace ca hai v phng trnh, vi gi thit rng cc hm x(t),y(t) c cc iu kin ban u bng 0, da theo tnh cht tuyn tnh ca php bin iLaplace , phng trnh (2.6) s c dng:

y, Y(s), X(s) - l cc bin i Laplace ca hm lng ra v hm lng voca h.

Phng trnh (2.l7) c gi l phng trnh ng hc m t quan h vo ra cah vit di dng ton t Laplace.y l phng trnh i s, vi n v m l cc s mca bin s s gii phng trnh (2 .17) ng vi lng ra Y(s).

Chng ta k hiu:

v gi biu thc i s ny l hm s truyn (hoc hm truyn t) ca h thng tng (hay ca mt phn t ca n).

Khi

Hoc

Vy hm struyn (H S T) ca h thng (hay ca mt phn t) tng l t s

hm nh ca lng ra vi hm nh ca lng vo ca n (qua php bin i Laplace)vi gi thit tt c cc iu kin u bng khng.

Biu thc (2.19) cho chng ta thy, HST l mt hm phn s hu t ca bin s,

c bc cc a thc tho mn m n. Gi thit iu kin ban u ca cc hm lng vov lng ra u bng khng l ph hp vi iu kin thng gp trong cc h thngKT.

Phng trnh (2.20) cho php xc nh hm nh ca lng ra nu bit hm nhca lng vo v biu thc HST ca h. Nh vy HST hon ton xc nh cc tnh

cht ng hc ca h thng. xc nh nguyn hm ca lng ra, tc l xc nh y(t)khi bit x(t) c th bin i ngc Laplace, theo :


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l phng php ton t gii phng trnh vi phn. Nu Y(s) l hm ngin, chng ta c th s dng bng bin i Laplace ca cc hm n gin in hnh,c trong ph lc cc sch ni v bin i Laplace, tra cu nguyn hm y(t). Nuhm nh Y(s) l hm phc tp, cn phn tch chng thnh t hp tuyn tnh cc hmn gin, m chng ta bit nguyn hm ca n. Nguyn hm y(t) chnh l t hptuyn tnh ca cc nguyn hm thnh phn.

2.2.3 Hm truyn t ca mch in

Trong mch in c cc phn t cbn lin tr(R), in cm (L) v tin(C).

a) in trR

Hnh 2.5:in tr

in p ri t l thun vi cng dng in I chy qua in tr:

Thng qua php bin i Laplace ta c c hm truyn ca in trl

b) in cm L

Hnh 2.6:in cm L

in p ri trn in cm l

Thng qua bin i Laplace ta tnh c trkhng Z v hm truyn ca in cm L


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c) Tin C

Hnh 2.7: Tin C

in p ri trn in dung l

Trkhng v hm truyn t ca tin

d) Cc phn tR, L v C mc ni tip

Hnh 2.8: S cc phn tmch in RLC mc ni tip

Thc hin php bin i Laplace ta c

Rt ra c hm truyn l:

e) Cc phn tmc song song


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Hnh 2.9: S cc phn tmch in RLC mc song song

Dng in ca mch in l

Tng trca mch song song c tnh l

Hm truyn ca h thng l

2.2.4 Hm truyn ca h thng ckh2.2.4.1 Phn tchuyn ng thng

a) L xo

Hnh 2.10: S biu din l xo

trong : K l h sn hi ca l xo

Nu ta n l xo c chiu di L, di ng c mt lng X th cn mt lc tc ng lnl

Thng qua bin i Laplace to c hm truyn ca l xo nh sau:

b) B gim chn du p (khng kh)


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Hnh 2.13: S biu din thit b gim chn

tm c hm truyn ca h thng trc tin ta v biu din cc lc tc ngtrng khi

Hnh 2.14: S biu din lc tc ng ln trng khi

S dng nh lut Newton vit phng trnh chuyn ng


T ta rt ra hm truyn ca h thng l

2.2.4.2 Phn tchuyn ng quayTheo nh lut II Newton v chuyn ng quay th gia tc gc ca vt quay t lthun vi tng momen tc ng ln n, ta c phng trnh sau

trong :

J l mmen qun tnh tc ng ln vt.

l v tr gc quay ca vt th

M l m men tc ng ln vt

Cc mmen bn ngoi c to bi ng cdo ti trng tc ng ca l xo hoc


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vt gim chn. Hnh biu din s ca mt a quay trong cht lng lm cho trc lp

trn n b bin dng i mt gc .

Nu ta quay a vi mmen xon x, trc s quay i mt gc to nn m men ca lxo xon:

Mmen cn thit thng lc ma s ca cht lng:

trong C l h s ma st ca cht lng

Nh vy ta c phng trnh:

Thay vo ta c:

2.2.5 Stng ng gia h ckh vi mt mch in

S tng ng gia mch ckh v mch in

trng khi = M in cm = M

b giam chn = fv in tr= 1/fvl xo = K in dung = 1/K

lc tc ng = f(t) ngun p = f(t)

vn tc = v(t) dng vng = v(t)


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Hnh 2.15: S biu din stngng gia mch ckh v mch in

Khi so snh vi dng vng ta c mch tng ng ni tip, nu dng phngphp nt, th mch tng ng ng l mch song song.

Phng trnh chuyn ng l

i vi mch RLC ni tip l

hai cng thc trn khng tng thch vi nhau do khong cch v dng in khngtng thch vi nhau. Ta bin i s tng thch bng cch chuyn i t khong cchsang vn tc

Ta cng c th chuyn i sang h song song

trng khi = M in cm = M

b gim chn = fv in tr= l/fv

l xo = K in dung = 1/K

lc tc ng = f(t) ngun dng = f(t)

vn tc = v(t) in p nt = v(t)

Cng thc mch song song l


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2.2.6 Hm truyn ca cc phn tin t

Hnh 2.16: Biu din phn tkhuch i thut ton

- Sai lch in p u vo: V2(t) - V1(t).

- Trkhng u vo cao: Zl = (l tng).

- Trkhng u ra thp: Z0 - 0 (l tng).

- H s khuch i cao A = (l tng).

in p u ra c tnh l

Nu V2(t) c ni t th b khuch i c gi l khuch i o.

Lc V0(t) = -A V1(t).

Trong hnh 2.16 c, nu trkhng u vo cao th ta c Ia(s) = 0 suy ra I1(s) = -I2(s). Khi h s khuch i A ln, v1(t) = 0 th I1(s) = Vl(s)/Zl(s) v - I2(s) = -V0(s)/Z2(s).

Cho hai dng in ny bng nhau ta c

V d : Tm hm truyn ca mch khuch i o sau


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Hnh 2.17 S h thng khuch i o

Tng trZl(s) l

Tng trZ2(s) l

Thay Zl(s) v Z2(s) vo cng thc 2.

2.3 M hnh ton hc trong min thi gian

2.3.1 Khi nim trng thi v bin trng thi

2.3.1.1 Khi nim v trng thi

Khi nim trng thi c trong csca cch tip cn hin i trong m tnghc ca cc h thng c Turing ln u tin a ra nm 1936. Sau khi nimny c cc nh khoa hc Nga v Mng dng rng ri gii cc bi ton iukhin tng.

Trng thi ca h thngc c trng nh l lng thng tin ti thiu vh,cn thitxc inh hnh vi ca h trong tng lai khi bit tc ng vo. Ni mtcch khc, trng thi ca hc xc nh bi t hp cc to m rng c trngcho h.

Trng thi ca mt h thng l tp hp nh nht cc bin (gi l bin trng thi)m nu bit gi tr ca cc bin ny ti thi im to v bit cc tn hiu vo thi imt>t0 ta hon ton c th xc nh c p ng ca h thng ti mi thi im t>t0.

H thng bc n c n bin trng thi. Cc bin trng thi c th chn l bin vt l

hoc khng phi l bin vt l.

Theo quan im phn tch v tng hp h thng thng, ngi ta chia cc binc trng h thng hay c quan h nht nh vi n v cc nhm nh sau:

- Cc bin vo hay cc tc ng vo uic to ra bi cc h thng nm ngoicc hc xt.

- Cc bin ra yic trng cho p ng ca h theo cc bin vo nh.

- Cc bin trung gian Xic trng trng thi bn trong ca h.

2.3.1.2 Khi nim vc ttrng thi:n bin trng thi hp thnh vc tct


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gi l vc ttrng thi.

- Khng gian trng thi: khng gian n chiu l khng gian hp bi cc trc cacc bin trng thi.

V d ta c cc bin trng thi in p ca in trVR v in p ca tin vcc bin ny s hnh thnh 2 trc ca khng gian trng thi.

thun li trong thao tc vi cc i lng nhiu chiu, t hp cc bin vo cth trnh by di dng vc tcc tc ng vo:

T hp cc bin ra trnh by di dng vctra

Cc t hp cc to trung gian, c trng ni dung bn trong ca hc vitdng vc ttrng thi ca h.

Theo nh ngha trng thi ca h ti thi im bt k t > t0, trng thi ca h lmt hm ca trng thi ban u x(t0)v vc tvo r(t0,t), tc l:

Vc tra ti thi im t c quan hn tr vi x(t0) v u(t0, t)

Cc phng trnh (2.59) v (2.60) thng gi l phng trnh trng thi ca. Nuh thng c m t bi cc phng trnh vi phn tuyn tnh ,th phng trnh trngthi ca hc vit di dng sau : (Bng cch s dng cc bin trng thi, ta c thchuyn phng trnh vi phn bc n m t h thng thnh h gm n phng trnh vi

phn bc nht)

trong : x ( n x 1 ) vc tcc bin trng thi,

u (m x 1 ) vc tcc bin u vo

y (r x 1 ) vc tcc bin u ra.

A(t) - Ma trn h thng.

B(t) - Ma trn iu khin hay m trn u vo.

C(t) Ma trn ra.

D(t) - Ma trn vng.

Cc ma trn c cc phn t ph thuc vo bin t, ln lt c kch thc l: A (n x


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n), B (n x m), C (r x n ), D (r x m).

Hnh 2.18: S khi biu din h thngiu khin trong khng gian trng thi

Thc t cc h thng thc u c tnh qun tnh, do D l mt ma trn c cc phn tu bng khng.

2.3.2 H tuyn tnh h s hng.

H thng c m hnh trng thi l:

x =Ax + Buy = Cx + D (2.62)

Trong cc ma trn A, B, C v D l cc ma trn hng s.

A c gi l ma trn h thng. Nu s lm cho phng trnh det(sI - A) = 0 th sc gi l gi tr ring ca ma trn A (y chnh l im cc ca h thng). I l matrn n v, s l mt s phc, det l k hiu ca php tnh nh thc ma trn.

2.3.3 ng dng biu din m hnh ton hc trn khng gian trng thi

ng dng h phng trnh trng thi biu din cc h vt l phc tp. Bcu tin l chn vcttrng thi, vic la chn ny phi tun theo cc yu cu sau :

- Cc bin trng thi phi l ti thiu nhng vn phi m bo biu din y trng thi ca h thng.

- Cc bin trng thi phi c lp tuyn tnh.

V d 1 : Cho h thng vt l c s nh sau:

Hnh 2.19: S mch RLC mc hn hp

Xy dng m hnh trng thi cho i tng.

Gii:

Bc 1 :t tn cc dng in nhnh bao gm iR, iL v iC.

Bc 2: Chn cc bin trng thi bng cc vit phng trnh vi phn cho cc phn t


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cha nng lng bao gm tin C v in cm L

Ta chn iL v vC l cc bin trng thi, nhng do iC v vL khng phi l cc bin trng

thi nn ta phi vit di dng t hp tuyn tnh ca cc bin trng thi iL v vC , binu vo l v(t).

Bc 3: S dng l thuyt v mch in c th l vit phng trnh da vo nh lutKirchhoff. Ti nt 1 ta c

Mt khc ta c

Bc 4: Thay cng thc trn vi nhau ta thu c cng thc nh sau:

hoc

Bc 5: Rt ra cng thc ca tn hiu u ra iR(t)

kt qu cui cng l

tn hiu u ra

V d 2: Cho mch in gm ba phn t R, Lv C mc ni tip


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Hnh 2.20: S mch RLC mc ni tip

Ul l in p t vo mch. Tm m hnh trng thi.Gii:Ta c phn trnh in p ca mch l:

thay cc cng thc tnh in p ca cc phn t

trong

Trng thi ca mch c quyt nh bi in p ra u2 v dng in i. Ta gi u2V i l cc bin trng thi.

t:

u2 = x1i = x2

T cng thc (2.72 v (2.73) ta rt ra cng thc tnh dng in l

Dng chnh tc c vit nh sau:

Vit h trn di dng vctma trn

hay vit gn li


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gi l phng trnh trng thi ca h thng. Khng gian hai chiu gm trng thi dngin i = x2 v in p trn t l u2 = x1c gi l khng gian trng thi. V d 3:

Hnh 2.21: S mch RLC mc ni tip

Ta c

Thay i(t) =dt

dq vo cng thc trn ta c

Ta t i(t), q(t) l cc bin trng thi

vit di dng vctma trn

in p VL l bin trng thi u ra

2.4 Chuyn thm truyn t sang khng gian trng thi v ngc li

2.4.1 Chuyn thm truyn t sang khng gian trng thi

c th m phng c mt h thng trn my tnh th m hnh ton hc cai tng phi c biu din trn khng gian trng thi. V vy khi ta a m hnh cai tng biu din bng hm truyn t ta phi chuyn sang phng trnh trng thi.

- Chn cc bin trng thi, mi bin trng thi c xc nh bi o hm cabin trng thi trc .


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- Ta xt phng trnh vi phn sau:

Cch thun tin chn bin trng thi l chn bin u ra

Ly o hm hai v

Biu din trn khng gian trng thi

Biu din di dng vctma trn

Vit phng trnh trng thi u ra


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Cc bc thc hin bin i thm truyn sang h phng trnh trng thi:

- B1 : chuyn t hm truyn v phng trnh vi phn v thc hin php bin iLaplace ngc vi cc iu kin u bng khng.

- B2: Thc hin chn cc bin trng thi v biu din trong khng gian trngthi.

V d l: Mt i tng c hm truyn t l

Xy dng m hnh trng thi cho i tng. Xc nh cc gi tr ring.

Gii

Bc 1 : Tm phng trnh vi phn

Bc 2: La chn cc bin trng thi

Vit di dng vctma trn

Tm gi tr ring

Cc gi tr ring l sl = - 1 , s2 = -4V d 2 : Cho hm truyn sau :


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Chuyn i sang h phng trnh trng thi.

Gii:Bc 1 : Tm phng trnh vi phn

Thc hin php nhn cho

Chuyn i thnh phng trnh vi phn bng cch dng php bin i Laplace ngcvi iu kin u bng 0

Bc 2: La chn bin trng thi Chn cc bin trng thi nh sau:

Ly o hm c hai v phng trnh (2.89) ta s thu c h phng trnh trng thi

Vit di dng vctma trn

M hnh c biu din nh sau:


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Gii: Tu bi ta xc nh cc ma trn A, B, C v D

Ta tm (sI - A)-1

2.5 Tuyn tnh ha

- Cc h thng m ta xt vi gi thuyt l tuyn tnh. Trn thc t hu ht cc itng l phi tuyn.

- Trong h thng cng c th bao gm ci lng phi tuyn v tuyn tnh.

- Do thc t yu cu ngi thit k pha tuyn tnh ha mt si lng phi tuyn s dng.

- Cc bc thc hin tuyn tnh ha

+ Bc l: Vit phng trnh vi phn ca h thng. Vi gi thit tn hiu u vo nh

+ Bc 2: Tuyn tnh ha phng trnh vi phn, dng bin i Laplace vi iu kinu = 0.


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Bi tp chng 2

1. Tm hm truyn ca h thng sau

2. Gii phng trnh vi phn sau

3 . Tm hm truyn G(s) ca h thng khi bit c dng biu din trn khng giantrng thi


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khi iu kin ban u bng khng.

i vi h thng tng dng, c tnh qu khng ph thuc vo thi imbt u tc ng. c tnh hi c th c dng dao ng hay bin i mt cch niu. Hin nay, nh gi cht lng h thng tng ngi ta thng s dng ctnh qu h(t) bi v n th hin c cc ch tiu cht lng cbn ca h thng.

Gia c tnh qu xung v c tnh qu c lin h vi nhau, thay x(t) = l(t)trong tch phn sau ta c:

i bin t- = v v d = -dv, ta nhn c:

Vy, c tnh qu l tch phn ca c tnh qu xung. Ly o hm ca chai v ca (3.6), ta s nhn c:

Vy c tnh qu xung l o hm theo thi gian ca c tnh qu . Vit(3.7) di dng ton t, s c:

T (3.8) ta c mi lin h gia nh hm ca c tnh qu theo hm truyn cuah thng.

S dng bin i ngc Laplace, s xc nh c hm qu theo hm s

truyn.

3.1.4 c tnh tn s.

Cc c tnh tn s (TTS) c ngha quan trng trong m t h thng tngdng. C th nhn c cc c tnh ny khi nghin cu chuyn ng cng bc cah thng di tc ng ca tn hiu iu ho.

Tnh cht cbn ca h thng tuyn tnh l tnh cht xp chng. Gi s nhiulon g(t) tc ng ln h thng bng khng, ta c mi quan h gia lng ra y(t) vi


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lng vo x(t) c m t bng phng trnh vi phn sau:

Cc TTS xc nh mi lin h gia ph lng vo X(j) v ph lng Y(j).

Cc ph ny chnh l bin i Phuri ca cc hm thi gian tng ng:

y: F - k hiu bin i Phuri.

- bin thc, l tn s ca tn hiu iu ho.

Ly bin i Phuri phng trnh (3.13) vi iu kin ban u bng 0, s nhnc phng trnh biu din quan h gia ph lng ra vi ph lng vo ca hthng tng:

sau khi bin i, s nhn c:

T s ca cc a thc:

c gi l hm s truyn tn s ca h tng. . .

Ty ta thy c th nhn c biu thc hm truyn tn s t hm s truyn

bng cch thay bin phc s bng j.

Hm s truyn truyn tn s W(j) c th biu din di dng:


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Trn mt phng phc (hnh 3.1 ) ng vi tn s no th hm s truyn tn s

s xc nh vc t. di vc tbng A(), cn argument (gc hp thnh bi vc t

ny vi bn trc thc dng) l (). ng cong c v bi u mt ca vc t

ny, khi tn s bin thin t 0 n (i khi t -n ) gi l c tnh tn s(TTS) bin pha ca h thng tng.

Hnh 3.1:c tnh tn sbin pha

Hm truyn tn sc gi l hm tn sbin pha. Phn thc ca n P -

ReW(j) v phn o Q() = ImW(j) c gi l tng ng l hm tn s phn thc

v hm tn s phn o. th ca hm tn s phn thc P() gi l TTS phn thc.

Cn th ca hm tn s phn o - TTS phn o Q().

Mun A() = |W(j)| gi l hm tn s bin . th ca n gi l TTS

bin . cn () l argument ca W(j) c gi l hm tn s pha. th ca ngi l TTS pha.

Ngoi cc TTS ni trn, cn thng s dng cc TTS bin logaritL() v TTS pha logarit (). Gi:

l hm tn s bin logarit. th din t s ph thuc gia hm tn s bin

logarit L() vi hm logarrit tn s lg(), gi l TTS bin logarit.

Khi xy dng TTS bin logarit, trn trc honh ta t cc tn s theo t l

logarit, cc im chia tng ng vi gi tr lg(), nhng li ghi gi tr ca tin

cho vic c tn s, ch khng ghi gi tr lg(). Cn trc tung l L(). TTS phalogarit l th biu din s ph thuc gia hm tn s pha (theo) vi logarit lg().Trc honh ca TTS pha logarit ging trc honh ca TTS bin logarit.

n v ca L() tnh theo (3.20) l xibel [db]. n v c sca trc honhtrong TTS bin logarit la cc [de] hay octav [oc]. cc l rng khongtn s m trn tn s thay i 10 ln. Tng t, octavl khong m trn tn sthay i 2ln.

Khi xy dng TTS logarit th cn lu l trc tung v trc honh khng ct

nhau ti tn s= 0. Hai trc s ct nhau ti mt tn s thch hp no , bi v khi

0 th lg() - : ngha l = 0 th s tng ng vi im xa v cng. Trn thc


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t, ngi ta thng s dng TTS bin logarit tim cn. nghing ca c tnhtim cn thng dng l db/dc hay db/oc: Khi tc ng ln u vo h thng T lmt tn hiu iu ho th lng ra ca h thng cng thay i theo quy lut iu ho,nhng bin v pha ca lng ra v lng vo s khc nhau. Ni chung, bin v

pha lng ra s ph thuc vo tn s ca lng vo. Ngi ta gi t s gia bin

lng ra vi bin lng vo l m un, cn gc lch pha gia lng ra vi lngvo l argument ca hm s truyn tn s. V th, c tnh tn sbin din t sthay i ca t scc bin lng ra vi lng vo, cnTTS pha din t gc lch

pha gia lng ra vi lng vo. l ngha vt l i TTS.

3.2 Cc khu ng hc in hnh

3.2.1 nh ngha cc khu ng hc in hnh

Cc khu ng hc m phng trnh vi phn m t qu trnh ng hc cachng c bc nhhn hoc bng2, c gi l khu ng hc in hnh.

c im ca cc khu ng hc in hnh l ch c mt u vo v mt u ra,tn hiu u ra khng nh hng n tn hiu u vo.

Hnh 3.2 Biu din khu ng hc in hnh.

Cc khu ng hc in hnh bao gm: Khu nguyn hm, khu tch phn, khuvi phn, khu tr. Khu nguyn hm gm cc khu: Khu khuch i khu khng quntnh), khu qun tnh bc mt, khu qun tnh bc hai (Khu lao ng). Sau y takho st cc khu ng hc in hnh trn.

3.2.2.Cc khu nguyn hm.

a) Khu khuch i

- Khu khng qun tnh l khu m phng trnh ng hc c dng:

- Hm truyn t ca khu. W(s) = K- Cc c tnh thi gian.- Hm qu : h (t) =K.1(t) .Hm trng lng : g(t) = K.(t).


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Hnh 3.3.c tnh thi gian ca khu khng qun tnh

Cc c tnh tn s.

Hnh 3.4:c tnh tn sca khu khnng qun tnh

Hm truyn tn s: W(j) = K.c tnh BT : A() = K.c tnh PT : () = 0.c tnh BTL : L() = 20.lgk

b)Khu qun tnh bc nht:Phng trnh vi phn:

Trong: K l h struyn ca khu.T l hng s thi gian ca khu.

- Hm truyn t :

- Cc c tnh thi gian :Hm qu :

Hm trng lng :

Hnh 3.5:c tnh thi gian ca khu qun tnh bc nht


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Cc c tnh tn s:

Hnh 3.6:c tnh tn sca khu qun tnh bc nhtTTS bin pha :

TTS bin :

TTS pha:

TTS bin logarit (TBL

c)Khu qun tnh bc hai:

- Phng trnh:

Trong: T l hng s thi gian.K l h s huyn.l h s tt dn tng i ( 0).

- Hm truyn t:- Cc c tnh thi gian:

Trng hp thnht: 0


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dao ng.Trng hp th hai: = 0.

Hm qu : h(t) = K(l - cos1t)Hm trng lng : g(t) = K1sin1t

Vi = 1/T : Tn s ring dao ng. c tnh qu dao ng iu ho v cgi l khu dao ng iu ho.Trng hp thba: 1.

Hm qu :khi = 1 vi l nghim kp ca phng trnh c trng.h(t) = K(l - (l - t)e-t )Khi > 1 vi 1 , 2 l hai nghim thc ca phng trnh c trng.

Hm trng lng:khi = 1 vi l nghim kp ca phng trnh c trng.

Khi > l vi l, 2 l hai nghim thc ca phng trnh c trng.

Hnh 3.7:c tnh thi gian ca khu bc hai

- Cc c tnh tn s:


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Hnh 3.8:c tnh tn sca khu bc hai

TTS bin pha :

TTS bin :

TTS bin logatit (TBL L() = 201g A().3.2.3 Khu tch phn.

- Phng trnh vi phn ca khu tch phn.

Trong T = 1/Kc gi l hng s thi gian tch phn.- Hm truyn t ca khu.

Bin i phng trnh vi phn sang ton tLaplace ta c:

- Cc c tnh thi gian.Hm qu : h(t) =Kt.

Hm trng lng: k(t) = h'(t) = K.


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- Cc c tnh thi gian:

Khu vi phn l tng:Hm qu :

Hm trng lng:

Khu vi phn bc mt:Hm qu :

Hm trng lng:

Hnh 3.11:c tnh thi gian ca khu vi phn l tng- Cc c tnh tn s: Khu vi phn l tng :TTS bin pha :

TTS bin :

TTS pha 2/)( =

TTS bin logarit (TBL)

Hnh 3.12:c tnh tn sca khu vi phn l tng

Khu vi phn bc mt :TTS bin pha :

TTS bin :


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TTS pha

TTS bin logarit (TBL)

3.2.5 Khu trKhu chm sau l khu ng hc m sau mt khong thi gian xc nh th lng

ra lp li lng vo v tn hiu khng b mo.Phng trnh ng hc ca khu tr c dng :

Cc phn t thuc khu tr nh bng ti , ng ng dn nhit , ng ng dncht lng . . .

- Hm truyn t ca khu tr:

- c tnh thi gian:Hm qu :

Hm trng lng:

- Cc c tnh tn s:

TTS bin pha :

TTS bin :

TTS pha :TTS bin lgarit:

Hnh 3.13.c tnh qu v cc c tnh tn sca khu tr

3.3 M hnh ZPK (Zero, Pole and Gain)Ta xt hm truyn sau:


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A(s) l mu s ca hm truyn, B(s) l t s ca hm truyn.- im khng (Zeros) l l cc gi tr lm cho hm truyn G(s) bng 0 hay l nghim

ca phng trnh B(s) 0. Cc im khng c k hiu l zi (i:1.m).

- im cc (Poles) l cc gi tr lm cho hm truyn khng xc nh hay l nghim

ca phng trnh A(s) = 0. Cc im cc c k hiu l pi (i: 1 m).- H s khuch i tnh (Gain) k hiu l K.V d 1 : Tm cc im cc, im khng v h s khuch i ca h thng ca hmtruyn sau:

H s khuch i K = 5 .- im cc:- im khng:

V d 2:Ta c hm truyn sau

Phn tch thnh tng cc phn s bc nht

Quy ng mu s v ng nht hai v ta c

Gii h phng trnh:A + B = 1

5A = 2Suy ra: A = 2/5, B = 3/5


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Hnh 3.14 : S btr cc im cc v im khng

p ng u ra:

trong :5

2l thnh phn cng bc

5

3e

-5tl thnh phn t do.

Mt s kt lun:1. im cc ca hm truyn u vo ca h thng quyt nh dng ca p ng

cng bc.2. im cc ca hm truyn h thng quyt nh dng ca p ng t do.3. p ng u ra c dng hm me-tnu c im cc nm trn trc thc.4. im cc v im khng quyt nh bin ca cp ng cng bc v p

ng t do.V d 3 : Cho h thng c hm truyn nh sau:

Hnh 3.15: H thngi tng lm v d 3Tm hm p ng u ra c(t) bao gm hai thnh phn p ng t do v p ng

cng bc.


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Gii:- Kim tra xem cc im cc ca h thng to ra thnh phn p ng t do tun theoquy lut hm m.- im cc u vo to ra thnh phn p ng cng bc.

Thc hin bin i Laplace ngc ta c:

3.4 H thng bc nhtH thng bc 1 khng c im khng c biu din nh sau:

Hnh 3.16: H thng bc nht v phn bim cc

nu tn hiu u vo l bc thang n vR(s) = 1/s th p ng u ra C(s) l:

Thc hin bin i Laplace ngc ta c p ng u ra biu din trn min thi gian l

- im cc u vo ti thi im ban u to ra p ng cng bc cf(t) = 1

- im cc h thng ti - a tao ra p ng t do cn(t) = - e-at.


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Hnh 3.17: p ngu ra ca h thng bc 1 vi tn hiu bc thangn v

ti thi im t = 1/a ta c

T vic kho st c tnh ca i tng bc ta c cc khi nim sau:

- Hng s thi gian (constant time): gi 1/a l hng s thi gian ca p ng. Hng sthi gian c thc hiu nh l khong thi gian m e-at gim 37% gi tr ban uhay l khong thi gian p ng tn hiu bc thang n v tng ti 63% gi tr xc lp.

Nghch o ca hng s thi gian gi l tn s (1/s). V vy ta c th gi hng sa l tn s hm m. Hng s thi gian c xem nh l c tnh p ng thi gian cah thng bc 1 v vy n c quan h vi tc ca h thng tng ng vi tn hiu

bc thang n vu vo.- Thi gian tng Tr(rise time): thi gian tng c nh ngha l thi gian m c ctnh mp m i t 01. n 0.9 gi tr xc lp.

Thi gian tng c tnh bng s sai lch gia hai thi im c(t) = 0,9 v c(t) =0.1

Thi gian xc lp hay thi gian n nh Ts (settling time): thi gian xc lp lkhong thi gian m p ng t n v sai s trong khong 2% vi c(t) = 0.98 thayvo cng thc v rt ra c


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aTs

4= (3.39)

Hm truyn ca h thng bc 1 qa thc nghim:Trn thc t khng d dng tm c hm truyn ca h thng bi v cc thit b

trong h thng kh c th xc nh c. V vy hm truyn ca h thng c th xcnh c bng cch xc nh quan h gia u vo v u ra thng qua phn tchng c tnh ca i tng khi cho p ng u vo l tn hiu bc thang n v.Hm truyn c th xc nh ngay car khi ta khng bit c cu trc bn trong ca itng.

vi tn hiu vo l hm bc thang n v ta c th tnh c hng s thi gian vcc gi tr xc lp.

Xt v d sau:

Cc p ng u ra:

Nu ta xc nh c h s khuch i K v a t phng th nghim ta s xc nhc hm truyn ca i tng.

Gi s ta c p ng sau:

Hnh 3.18:ngc tnh p ng ca h thng bc nht

p ng ca h thng bc nht khng c qu iu chnh v sai lch im khng.Tng p ng ta xc nh hng s thi gian


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- Gi tr xc lp l gi tr m ng p ng t n bng 0.72.- Hng s thi gian l thi gian m ln bng 63% gi tr xc lp v bng 0.63

x 0.72 =0.45 hay bng 0.13 (s) suy ra a = 1/0.13 = 7.7

- p ng cng bc t n gi tr xc lp K/a = 0.72 suy ra K = 5.54

- Lc ta c hm truyn ca h thng l

Hm truyn ny cng rt gn vi hm truyn ca p ng trn

3.5 H thng bc 2


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Hnh 3.19: Cc h thng bc hai v p ng vi tn hiu bc thangn v

Ta c hm truyn tng qut ca h thng bc hai :

trong : K l h s khuch i.T l hng s thi gian.l suy gim.

3.5.1 H thng p ng xung tt dn (Overdamped)y l p ng khng eo ao ng trong khong gi trn nh nhng t ti

dao ng gii hn tt dn lu hn.Khi khu qun tnh bc hai c hai im cc thc th bao gm 2 khu qun tnh

bc mt ni tip nhau. Vi iu kin > 1 ta c

Hai im cc l:


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p ng u ra c mt im cc t gc to (do c p ng tn hiu bc thang

n v).- Hai im cc thc ca h thng.- im cc u vo s to ra thnh phn p ng cng c. Mi im cc ca h

thng s to ra p ng t do c dng hm m trong tn s hm m chnh bng vtr cc im cc.

p ng u ra s c dng:

ng c tnh ca h thng bc hai tt dn th hin hnh 3.19b3.5.2 H thng p ng di tt dn (Underdamped)

y l p ng c dao ng trong khong ng bao suy gim. H thng cngc nhiu ng bao th p ng t ti trng thi n nh cng lu.(Xem hnh 3 .19c)

Ta xt phng trnh c tnh:

Khi < 1 thi phng trnh (3.48) s c hai nghim phc lin hp hai nghim nyl hai im cc ca hm truyn.

Qu trnh qu xy ra trong khu bc hai l qu trnh dao ngl khu dao ng bc 2. p ng thi gian bao gm bin hm m gim to bi

phn thc ca im cc h thng v dng sng hnh sin to bi phn o ca im cch thng.


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Hnh 3.20:p ng bc hai to bi cc nghim phc

Hng s thi gian ca hm m bng phn thc ca im cc h thng. Gi tr caphn o l tn s thc ca dao ng hnh sin. Tn s dao ng hnh sin c gi l tns suy gim ca dao ng wd' p ng n nh c quyt nh bi im cc u vot gc to. Chng ta gi p ng ny l p ng di tt dn m tin ti gi trn nh qua p ng thi gian gi l dao ng suy gim.3.5.3 H thng p ng khng b nht (Undamped)

Nu im cc tin gn v khng cng b, ng bao gim cng lu, lc tac dao ng khng tt.

H thng bc hai ny s c: im cc nm gc to do p ng tn hiu bc

thang u vo v hai im cc ca h thng ch c phn o ( = 0).T hnh 3.19d

Hai im cc p1,2 = j3 to ra p ng dao ng hnh sin m tn s ca n bngv tr ca cc im cc nm trn trc o.

p ng u ra l:

Thay vo ta c

3.5.4 H thng p ng tt dn ti hn (Critically Damped Response)

y l p ng t ti gi trn nh nhanh nht. Gi tr gii hn lun lun bng1.


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Ta c p ng sau:

p ng ny c mt im cc nm ti gc to v hai im cc thc.

xem dng p ng hnh 3.19d3.5.5 Tm p ng tdo

p ng tt dn:Cc im cc : hai im thc - i , 2p ng:

p ng di tt dnCc im cc: 2 nghim phc

p ng:

p ng khng b nht Cc im cc: 2 im cc o j

p ng tt dn ti hnCc im cc: 2 im cc thc (kp) 1p ng:

3.6 Mt s vn chung v h thng bc haiTrong phn ny ta s xem xt hai khi nim ca hai thng s h thng bc 2 c

dng miu t ng c tnh p ng thi gian. l tn s t do (naturalfrequency) v h s tt dn (damping ratio).

- Tn stdo (Natural Frequency, n): l tn s ca dao ng trong h thng mkhng c s tt dn.

-H stt dn (Damping ratio ):


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Biu din h thng bc hai theo hai thng sn v

i vi h thng khng b nht ta c cc im cc nm trn trc o

Theo inh ngha tn s dao ng t do wn l tn s ca dao ng trong h thng.

V vy cc im cc nm trn trc o l bj . Suy ra

Vi gi thit h thng di tt dn im cc phc c phn thc l -a/2. lnca gi tr ny chnh l tn s gim hm m

suy ra a = 2n (3.64)vy hm truyn l

V d:Cho hm truyn sau:

So snh hai cng thc (3.66) v (3.65) ta c:

Ta tm cc im cc ca h thng:

Phng trnh c trng l: s2+ 4.2s + 36 = 0 C hai nghim phc:

ng c tnh p ng t gi tr ca


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Ta c cc p ng tng ng vi gi tr ca nh sau:

Hnh 3.21:p ng bc hai theo h stt dn3.7 H thng bc hai di tt dn (Undierdamped)

H thng di tt dn, m hnh vt l ph bin, c cc p ng n nht nnc xem xt c th hn. nh ngha cc thng sp ng ca h thng di tt dntheo thi gian v xem xt mi quan h vi v tr cc im cc.

Trc tin ta tm p ng ca h thng bc hai vi p ng tn hiu bc thangn v

gi thit < 1 v thc hin bin i ta c


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Thc hin php bin i Laplace ngc

trong :

Khi cng nh th p ng dao ng cng nhiu.

Hnh 3.22:p ng bc hai ca h thng di tt dn

Ngoi hai khi nim h s suy gim v tn sp ng t do n ta c thm cc khinim sau:

- Thi gian nh Tp (Peak Time): l thi gian m c(t) t max u tin.- Phn trm quiu chnh: %Os (Percent Overshoot): l khong m dng

sng vt qu gi trn nh C.- Thi gian tng Tr (rise time): thi gian tng c nh ngha l thi gian m

c c tnh mp m i t 0.1 n 0.9 gi tr xc lp.- Thi gian xc lp hay thi gian n nh Ts (settling time): thi gian xc lp l

khong thi gian m p ng t n v sai s trong khong 2%

a) Tnh Tp


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Bin i Laplace ngc ta c:

Cho c(t) = 0 suy ra

Hay

Khi n - 1 ng c tnh t gi tr max

b) Tnh phn trm qu iu chnh

T hnh v 3.22 ta c

cmax l gi tr khi ng c tnh t gi tr max ti thi im TP

i vi tn hiu bc thang n v

Thay vo cng thc (3.77) ta tm c phn trm qu iu chnh

Suy ra


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c) Tnh Ts

tm c Ts ta phi tm c thi gian m c(t) t n v gin nh trong

khong 2%

T cng thc (3.71 ) ta tnh bin ca c(t) t n 0.02

Vi gi thit :

Suy ra

Ly xp x cng thc (3.83)

d) Tnh TrTm nt bng cch cho c(t) = 0.9 v c(t) = 0.1 . Ly gn ng ta c thi gian tng

nTrV d: Cho hm truyn sau

Tnh Tp, %Os, Ts V Tr.

Gii:T hm truyn ta tnh c n = 10 , = 0.75Thay vo cng thc tnh Tp


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Ta c bng sau:

H s suy gim Thi gian tng thngthng

0.1 1.104

0.2 1.203

0.3 1.321

0.4 1.463

0.5 1.638

0.6 1.854

0.7 2.126

0.8 2.467

0.9 2.883

Da vo bng trn ta tnh c thi gian tng thng thng xp x 2.3 suy ra Tr= 0.23

v n = 10.


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Bi tp chng 31. Hy xc nh hm trng lng g(t) v hm qu h(t) ca nhng h tuyn tnh chm truyn t sau

2. Tm v tr cc im cc, im khng v v trn mt phng phc

3. Tm hm truyn v im cc ca h thng sau

4. Tm cc thng s ca h thng bc 2 , n, Tr, Ts, Tp % OS

5. Tm p ng u ra c(t) khi bit tn hiu tc ng l tn hiu bc thang n v


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Hnh 4.2: S khi ca h thng ni tipV d: Ta c m hnh nh sau:

Hnh 4.3: H thng ghp ni tip.Hnh 4.3a) hm truyn c tnh:

Hnh 4.3 b) hm truyn c tnh:

Hnh 4.3 c) ta tnh c hm truyn ca h thng bng mch vng hoc theo nt :

Nhng nu tnh theo cng thc ca s ni mc ni tip

Ta thy s khc nhau l do gia hai h thng tn ti mt h s t l. khcphc gia hai h thng ta mc thm mt khu khuch i nh hnh 4.3 d).4.1.2 H thng dng song song (Parallel Form)

H thng mc song song l h thng c tn hiu vo ca h thng l tn hiu vo


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ca cc phn t thnh phn, cn tn hiu ra ca h thng bng tng i s cua cc tnhiu thnh phn.

Hnh 4.4: S khi ca h thng mc song song

4.1.3. H thng dng phn hi (Feedback Form)H thng c mch mc phn hi gm hai mch: mch thun v mch phn hi.

tn hiu ra ca mch thun l tn hiu ra ca h thng v l tn hiu vo ca mch phnhi.H thng c hai dng phn hi:

- Phn hi m:

- Phn hi dng:

Hnh 4.5: S khi ca h thng c phn hi

Hnh 4.6: a) H thng phn hi m b) H thng phn hi dngc) Hm truyn ca h thng c phn hi

- Mch phn hi n v:


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Hnh 4.7: S khi h thng phn hi n v

Mt khc:

Hm truyn ca h thng c tnh l:

Cc k nng bin i s cbn:

- Chuyn tn hiu u vo:

T trc ra sau mt khi:

T sau mt khi ra trc mt khi:

Chuyn i tn hiu ra:T trc mt khi ra sau khi :


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Cc b cng lin nhau c thi ch cho nhau hoc cng xp chng li:

Cl(s) = C2(s) = R(s).G(s)- Cc b cng lin nhau c thi ch/cho nhau hoc cng xp chng li:

V d 1 : Rt gn h thng nh hnh sau


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Hnh 4.8 : Hnh bin i cc s khi cbn.

V d 2 : Rt gn s khi p dng cc quy tc di chuyn tn hiu


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Hnh 4.9: Rt gn s p dng cc quy tc bin i

4.2 Phn tch v thit k h thng phn hiMc ch : ng dng cc quy tc trn phn tch v thit k h thng bc 2.

phn trm qu iu chnh, thi gian nh, thi gian tng c thc tnh ton thm truyn ca h thng.

Xt h thng:

Hnh 4.10: H thng c phn hi mi tng c hm truyn l:


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Hm truyn ca h thng c tnh l:Trong :K: H s khuch i (t l gia in p u vo v u ra)

Khi h s K thay i, cc im cc thay i qua 3 ch hot ng ca h thngbc hai : dao ng tt dn, tt tt dn ti hn v di tt dn. V d K bin i trong rigia 0 v a2/4, cc im cc ca h thng l thc v c tnh ton l:

Khi h s K tng ln, cc im cc di chuyn dc theo trc thc v h thng vndao ng tt dn cho n khi K : a2/4. Ti h s khuch i ny, c hai nghim u lthc v bng nhau, h thng l tt dn ti hn.

i vi h s khuch i ln hn a2/4, h thng l di tt dn vi cc nghimPhc l:

Khi h s khuch i tng ln, phn thc l hng s v phn o tng ln. Do vy,thi gian nh gim xung ln v phn trm qu iu chnh tng ln, trong khisettling time vn l hng s.

- Tnh thi gian p ng nhanh.Bi ton: cho h thng

Hnh 4.11: Skhi h thng phn hi bit trc h skhuch i

Tm hng s thi gian nh Tp, phn trm qu iu chnh % v thi gian Ts.Gii:

Hm truyn ca h thng c phn hi l:T cng thc:

Mt khc:

Suy ra = 0.5


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Ta c cng thc tnh nh sau:

Tm h s khuch i ca h thng khi bit phn trm qu iu chnh

Hnh 4.12: S khi ca h thng phn hi khi h skhuch i K cha bit

Hm truyn ca h kn l:

T cng thc 5 = 2n v n = K tnh c:

K2

2= (4.21)

Nh ta bit phn trm qu iu chnh %Os l hm ca m y li lhm ph thuc h s K. Do vy khi ta thay cng thc (4.21) vo cng thc tnh

(4.18) , li l hm ph thuc h s khuch i K.Vi ch tiu cht Lng cho trc ta d dng tnh c h s = 0.591 .

Thay vo cng thc (4.21) ta tm c h s khuch i K = 17.892.Mc d ta thit k theo qu iu chnh nhng ta khng th la chn c thi

gian dao ng c bi v phn thc lun l -2.5 (khng cn tnh n h s khuch iK).4.3 Grap tn hiu4.3.1 Cc khi nim cbn

- Graph l mt hnh gm cc nhnh v cc nt.- Mi nt ca Graph c biu din bng mt im v ghi tn mt i lng no

trong h thng iu khin.

O

V(s)Hnh 4.11 : Mt nt cbn

- Nt gc l lng vo, nlng ra ca mt khu no .


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- Mt nhnh ni nt gc v nt ngn c mi tn, trn ghi gi tr ca hmtruyn t tng ng vi mto .

Hnh 4.13 : Biu din mt nhnh cbnHm truyn t ca mt nhnh bng t s ca gi tr nt ngn v gi tr nt gc:

S lin kt ca cc nhnh mt Graph tn hiu cho mt h thng iu khin.4.3.2 Cc dng biu din Graph tn hiu- Dng ni tip:

Hnh 4.14: Graph biu din h thng ni tip

- Dng song song:

Hnh 4.15: Granph biu din h thng song song

- Dng phn hi:

Hnh 4.16: Graph biu din c phn hi

4.3.3 Cc quy tc bin i Graph- Cc nhnh ni tip:

- Cc nhnh song song


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- Phn hi dng hoc m

- Kh nhnh to vng kn.

4.3.4 Quy tc MassonHm truyn t C(s)/R(s) trong mt h thng c biu din bng Graph c

tnh theo cng thc sau:

trong :- k l sng dng hng tu vo cho n u ra. ng dng l mt .

ng lin tc bao gm cc nhnh c cng mt hng khi u vo cho n u ram khi tn hiu truyn t qua mt nt ca n t gc n ngn chc mt ln.

- Tk l hm truyn t ca dng th k hng tu vo n u ra.- kl nh thc con ca Graph suy ra t bng cch bi cc vng kn.Li c dnh vi ng dng th k.

- .....1,,,

++= kji

kji

i ji

jii LLLLLL +

i

iL l tng s tt c cc hm truyn t ca cc vng kn c trong Graph.

Vng kn l mt ng khp kn bao gm cc nhnh lin tip c cng mt hng mtn hiu i qua mt nt ca mt nhnh no n chc mt ln.

+ jji

iLL,

l tng s cc tch hm truyn t ca hai vng kn khng dnh vo

nhau trong Graph.+

kji

kji LLL,,

l tng s cc tch hm truyn t ca ba vng kn khng dnh vo


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nhau trong Graph.V d:

Hnh 4.17: skhi minh ho quy tc Masson

Bc 1 : Xc nh ng tin trc

trong v d ny ch c mt ng duy nht l:

Bc 2: Xc nh hm truyn ca cc vng kn

Bc 3: Xc nh tch cc hm truyn ca hai vng kn khng dnh vo nhau

Bc 4: Xc nh tch cc hm truyn ca ba vng kn khng dnh vo nhau

Bc 5: Tnh

Bc 6: Xc nh ky ch c L3 l khng dnh n P


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T cc cng thc tnh ton trn ta thay vo cng thc tnh G(s) ta c:


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Bi tp chng 41 .Rt gn s h thng sau:

2. Cho h thng sau

Tm qu iu chnh OS%, thi gian qu v thi gian nh khi tn hiu u vo ltn hiu bc thang n v.3 . Tm tn hiu c(t) khi tn hiu u vo r(t) l tn hiu bc thang n v.


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4. Tm , N,OS%, thi gian nh v thi gian qu ca h thng sau

5 . S dng quy tc Masson tm hm truyn T(s) ca h thng cho di y


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CHNG 5 : SN NH CA H THNG

5.1 Khi nim vn nh h thng iu khin tngnh ngha:

n nh cu h thng l kh nng ca h thng ttrli trng thi xc lp sau

khi cc tc ng ph vtrng thi xc lp c mti. Thc cht khi ni ti n inhl ni ti mti lngc iu khin no n nh.

Mt h thng KT l mt h thng ng hc, thng c m t bng phngtrnh vi phn bc cao:

Nghim ca phng trnh vi phn ny gm hai thnh phn :

- yqd (t): l nghim tng qut ca (5.1) khi v phi bng 0, c trng cho qu trnhqu .

- y0 (t): l nghim ring ca (5.l) khi c v phi, n c trng cho qu trnh xclp .

Qu trnh xc lp l qu trnh n nh, v vy ch cn xt qu rnh qu . Nuqu trnh qu theo thi gian b trit tiu th hn nh, nu khng trit tiu th hkhng n nh. M nghim qu c biu din bng biu thc tng qut sau:

Trong si l nghim ca phng trnh c trng :

T nhng nhn xt trn ta c th kt lun nh sau: Mt h thng c gi l nnh nu qu trnh qu tt dn theo thi gian. H thng khng n nh nu qu trnh

qu tng dn theo thi gian. H thng bin gii n nh nu qu trnh qu khng i hoc dao ng khng tt dn.

Biu din bng biu thc ton hc nh ngha trn ta c h thng n nh khi:

v h khng n nh khi :

H thng c xt l h dng, ngha l cc h s ai khng bin i theo thi


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gian.

Nu a; < 0 Hn nh, nu i = 0 hbin gii n nh, nu vi i > 0 h khngn nh.

Khi si l cp nghim phc lin hp s; = i1

Nu i < 0 Hn nh , nu i = 0 hbin gii n nh , nu i >0 h khng nnh.

5.2 Nhn xt chung :

-H thng sn nh khi v chkhi tt c cc nghim ca phng trnh c tnh

c phn thc m (tt c cc nghim nm na bn tri mt phngphc).- H thng s bin gii n nh nu phng trnh c tnh c t nht mt

nghim thun o cn tt c cc nghim khc l nghim thc m hoc nghim phc cphn thc m (c t nht mt nghim nm trn trc o cn cc nghim cn li nm na tri mt phng phc).

- H thng s khng n nh nu phng trnh c tnh c t nht mt nghim cphn thc dng (c t nht mt nghim nm na phi mt phng phc).

Nh vy xt tnh n nh ca h thng ta cn phi tm nghim ca phng

trnh vi phn (5.1) ri ly gii hn. Vic ny rt kh khn, nn xt n nh ch cntm nghim ca phng trnh c trng (5.3). Trong thc t ngi ta tm mi quan hgia cc h s ca phng trnh c trng vt cc nghim c phn thc m nhgi tnh n nh ca h. l cc tiu chun n nh. C hai tiu chun n nh:

- Tiu chun n nh i s: Tm iu kin rng buc gia cc h s ca phngtrnh c tnh hn nh. l cc tiu chun Routh, Hurwitz.

- Tiu chun n nh tn s: Thng qua c tnh tn s ca h thng xt tnhn nh. l cc tiu chun n nh Mikhailp, Nyquist.

5.3 Tiu chun n nh i s.

iu kin n nh cn thit ca HTKT:

Gis h thng c phng trnh c tnh:

Nhvy phng trnh c tnh, c hai loi nghim :

C m nghim thc (si= - i) v (n - m) /2 nghim phc (si = k jk) .

Vi i, k, k u dng.

Phng trnh c tnh c chuyn sang dng:


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suy ra:

Nu ta khai trin phng trnh trn sc mt a thc c tt c cc h sudng.Nhvy iu kin cn thith thngn nh l tt c cc h sca phngtrnh c tnh phi dng (phi cng du).

5.3.1 Tiu chun Rao (Routh):

- Pht biu tiu chun: " iu kin cn v cho h thng tuyn tnh n nhl tt c cc s hng trong ct th nht ca bng Routh dng ".

- Cch thnh lp bng Routh :

Gi s cho phng trnh c tnh sau:

Hai hng u bng Routh c sp xp nh sau :

C0 C2 C4 C6

Cc s hng trong cc hng c tnh theo biu thc sau :

Nhn xt:

Mi mt s hng trong mt hng ca bng Routh l mt thng s c:

- T s : l mt nh thc hng hai mang du m vi ct th nht ca n cng lct th nht ca hai hng ng st trn hng c s hang ang tnh ,cn ct th hai ca


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nh thc chnh l ct ng st bn phi s hng ang tnh cng ca hai hng trn.

- Mu s : trong tt c cc s hng ca mt hng c chung mu s chnh l shng ng ct th nht v hng st ngay trn s hng ang tnh.

V d : Cho phng trnh c tnh ca h thng :

Lp bng Routh :

1 8 3

2 4 3

6 3

3 0

0

H thng n nh v tt c cc s hng trong ct th nht dng.

Mt s tnh cht ca bng Routh

- Khi lp bng Routh, gin n trong tnh ton, c th nhn hay chia cc h strong ct vi cng mt i lng, kt qu vn khng thay i.

- Trong trng hp h khng n nh, bao nhiu ln i du ct 1 th c bynhiu nghim na phi mt phng phc.

Nu tr s gn cui ct mt bng 0 (C1n = 0) c ngha l nghim kp thun o.

Tr s cui cng s khng tnh c v rn+1 = . Nu tr s cui cng bng 0 (Cn+l = 0)th phng trnh c trng c mt nghim bng 0 v an = 0.

- Nu cc h s ca mt hng bng 0, v c nghim phi hoc cp nghim nmtrn trc o.

5.3.2 Tiu chun Hurwitz

Pht biu tiu chun.

iu kin cn v h tuyn tnh n inh l h sa0 > 0 v cc nh thcHurwitz dng.

Thnh lp nh thc Hurwtiz.

nh thc Hurwitz lp t ma trn h s theo quy tc sau:

- Theo ng cho ca ma trn, vit cc h s t a1n an,

- Pha trn ng cho, cc h s tng dn, pha di gim dn.

- Cc h s nh hn a0 v ln hn anu bng 0.

Ma trn c dng nh sau:


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Cc nh thc Hurwitz dng tng ng vi :

Lu : Khi kho st tnh n nh vi a0 > 0, nu c h s bt k no m (ai < 0 )

th kt lun l h khng n nh.

Vi iu kin ai > 0 (i = 0,l,2...n) th ch cn xt i > 0 vi i = 2, ... n-l l c, v

Ch : Tiu chun n nh Hurwitz ch l mt dng biu din khc ca tiuchun Routh. N ch dng vi h thng c phng trnh c tnh bc thp (di bc4).

5.3.3 Mt s trng hp ca tiu chun Routh - Hurwitz

Hai trng hp c bit c th xy ra:- Xut hin s 0 ct th nht.

- Xut hin mt hng ton s 0.

a) Trng hp thnht: S 0 ct thnht

Nu c s 0 ct th nht th vic to ra hng tip theo s chia cho s 0.

trnh trng hp ny ta gn mt gi tr thay th s 0. Sau dng tnh ton

v Xt du cho ()

V d: Xc nh tnh n nh ca hm truyn h kn sau:


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Lp bng Routh v xt du

Nhn bng xt du c trong hai trng hp = th ct th nht i du hai lnc ngha l phng trnh c tnh c hai nghim nm bn phi trc o. Do vy hthng trn l khng n nh.

b) C mt hng ton s khng

Khi gp trng hp ny ta u tin ta quay li hng pha trn hng c ton s 0

v thnh lp mt a thc ph m s dng cc gi tr ca hng lm h s. a thcbt u vi lu tha ca s ct k hiu s v b bin tip theo v thc hin h bc athc ph.

V d: Xc nh s nghim nm bn phi trc o ca h kn sau:

Lp bng Routh

a thc ph : P(s) s4

+ 6s2

+ 8 (5.16)Ly vi phn a thc (5.16)


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S dng cc h s trong a thc 5.17 thay th hng c ton s 0. Sau khi thayv tnh ton ta thy ct u tin cc h su dng do vy khng c im cc no

nm bn phi trc o.5.3.4 Sdng tiu chun Routh - Hurwitz thit k sn nh

Cho h thng sau:

Hnh 5.1 : H thng c h skhch i K cha bitTm phm vi ca h s khuch i K h thng n nh, khng n nh hay bingii n

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