Lý Thuyết Điều Khiển Tuyến Tính - GS.TS. Nguyễn Doãn Phước - NXB KH&KT 2009

Nguyn Don Phc

l thuyt

iu khin tuyn tnh (In ln th t, c sa i v b sung)

Nh xut bn Khoa hc v K thut

H Ni 2009

2

Author: Nguyen Doan Phuoc

Assoc. Prof. of Department of Automatic Control, Hanoi University of Technology.

Title: Theory of Linear Control

This book aims to provide basic knowledges of linear control. It presents the conceptual steps to carry out a linear control problem such as modelling, analysis and controller design. Many examples are given in the book to illustrate the theory.

This book is the product of several courses given by the author at the Hanoi University of Technology (HUT). It is written for control engineering students and master students in Universities as a course and self study textbook.

Chu trch nhim xut bn: PGS. TS. T ng Hi Bin tp: Nguyn ng Trnh by v ch bn: Tc gi V ba: Trn Thng

In 1000 cun kh 1624 cm ti xng in NXB Vn ha dn tc. Giy php xut bn s 150604/2/2005. In xong v np lu chiu thng 7/2005.

3

Li ni u

Sau ln xut bn u tin nm 2002, tc gi nhn c rt nhiu ng gp t pha bn c c c ni dung vi cht lng tt hn cho nhng ln xut bn sau ny ny. Tc gi hy vng vi s sa i , cc bn sinh vin ang theo hc cc ngnh iu khin t ng, o lng v Tin hc cng nghip, T ng ha, hc vin cao hc, nghin cu sinh thuc cc ngnh lin quan, s c c mt ti liu vi cht lng tt hn h tr cho vic t hc, cng nh cho vic hiu k, hiu su bi ging.

L thuyt iu khin tuyn tnh l phn nn tng c bn v quan trng nht ca L thuyt iu khin ni chung. Rt nhiu cc pht trin mi v khi nim cng nh

phng php ca iu khin nng cao nh n nh u, n nh theo hm m, n nh ISS, iu khin tuyn tnh ha chnh xc, iu khin thch nghi khng nhiu ... u c c s gi v t tng t L thuyt iu khin tuyn tnh. Nm vng v lm ch L thuyt iu khin tuyn tnh s gip ta c c mt kin thc c bn chc chn t tin tin su hn vo cc lnh vc khc ca iu khin.

So vi ln xut bn th nht, ln xut bn th t ny, quyn sch c b cc li hon ton bng vic phn chia cc chng theo ch tng dng m hnh m t h thng c s dng. C th l:

Chng 1 c dnh cho phn nhp mn L thuyt iu khin tuyn tnh, cc bc c bn cn phi thc hin khi phi gii quyt mt bi ton iu khin.

Chng 2 trnh by cc bc thc hin bi ton iu khin khi m hnh ton hc ca i tng l m hnh trong min phc (min tn s).

Chng 3 l ni dung cc bc thc hin bi ton iu khin ng vi m hnh trng thi ca i tng (iu khin trong khng gian trng thi).

Chng 4 l ni dung tng bc thc hin bi ton iu khin khi i tng c m hnh khng lin tc, c xem nh phn nhp mn ca iu khin s.

trong , tng chng 2, 3 v 4 li c trnh by theo ng th t thc hin cc bc mt bi ton iu khin, nh: 1. Cng c ton hc cn thit, 2. Xy dng m hnh m t i tng, 3. Phn tch i tng v 4. Thit k b iu khin.

Cng so vi ln xut bn th nht, cc ln ti bn sau ny, tc gi a thm mt s ni dung c cho l cn thit ca iu khin nng cao, nhng c lin quan n m hnh tuyn tnh ca i tng. Cc phn c b sung thm bao gm:

Phn tnh tnh bn vng ca h tuyn tnh c m hnh ton hc ca i tng l hm truyn.

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Thut ton thit k b iu khin theo m hnh mu.

Phng php tham s ha Youla, phng php thit k b iu khin n nh mnh v n nh song hnh iu khin n nh bn vng i tng tuyn tnh (nguyn l iu khin a m hnh).

Thit k b iu khin tuyn tnh theo nguyn l bm tn hiu mu (tracking control).

Thit k b iu khin b bt nh cho i tng tuyn tnh.

Thit k b lc Kalman.

Cui cng, quyn sch c vit vi s gip , chia s rt to ln ca nhng thnh vin trong gia nh tc gi l v Ng Kim Th, con gi Nguyn Phc My v hai chu ngoi Bng, Bo. Khng c h chc chn quyn sch khng th hon thnh c. Quyn sch cn c hon thnh nh s c v, khuyn khch v to iu kin thun li ca cc ng nghip trong B mn iu khin T ng, Trng i hc Bch khoa, ni tc gi ang cng tc. Tc gi xin c gi ti gia nh v cc bn li cm n chn thnh.

Mc d rt n lc, song chc khng th khng c thiu st. Do tc gi rt mong nhn c nhng gp sa i, b sung thm ca bn c hon thin. Th gp xin gi v:

Trng i hc Bch khoa H Ni Khoa in, B mn iu khin T ng

[email protected]

H Ni, ngy 29 thng 10 nm 2009

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Mc lc

1 Nhp mn 11 1.1 Ni dung bi ton iu khin 11

1.1.1 Bi ton c tn hiu tin nh (iu khin tin nh)........................................................ 14 Khi nim tn hiu ......................................................................................................... 14 Phn loi tn hiu tin nh ........................................................................................... 15 Mt s tn hiu tin nh in hnh ................................................................................ 17 Chun ca tn hiu (hay hm s).................................................................................. 19

1.1.2 Bi ton c tn hiu ngu nhin (iu khin ngu nhin) ............................................... 21 Khi nim qu trnh ngu nhin.................................................................................... 21 Qu trnh ngu nhin dng v ngu nhin egodic........................................................ 22

1.2 Nhng cu trc c bn ca h thng iu khin 23

1.2.1 Phn loi h thng .......................................................................................................... 23

1.2.2 Xc nh tn hiu iu khin thch hp ............................................................................ 24

1.2.3 S dng b iu khin .................................................................................................... 25 iu khin h................................................................................................................ 25 iu khin phn hi trng thi...................................................................................... 26 iu khin phn hi tn hiu ra ..................................................................................... 26

Cu hi n tp v bi tp 27

2 iu khin lin tc trong min phc 29 2.1 Cc cng c ton hc 29

2.1.1 L thuyt hm bin phc................................................................................................. 29 nh ngha, khi nim hm lin tc, hm gii tch ........................................................ 29 Tch phn phc v nguyn l cc i modulus............................................................. 30 Hm bo gic (conform) ............................................................................................... 32

2.1.2 Chui Fourier v php bin i Fourier .......................................................................... 34 Chui Fourier (cho tn hiu tun hon)......................................................................... 34 Php bin i Fourier ................................................................................................... 38

2.1.3 Php bin i Laplace..................................................................................................... 46 Php bin i Laplace cho tn hiu lin tc .................................................................. 46 Php bin i Laplace cho tn hiu khng lin tc (bin i Z).................................... 48

2.1.4 Php bin i Laplace ngc.......................................................................................... 49 Bin i ngc hm hu t ........................................................................................... 49 Phng php residuence.............................................................................................. 52

2.1.5 Mt ng dng ca php bin i Laplace: Gii phng trnh vi phn ............................ 55

2.2 Xy dng m hnh ton hc 57

2.2.1 Phng trnh vi phn m t quan h vora................................................................... 60 2.2.2 Hm truyn, hm trng lng v hm qu ................................................................ 63 2.2.3 Php bin i s khi (i s s khi) .................................................................. 71

Hai khi song song ....................................................................................................... 71 Hai khi ni tip ............................................................................................................ 72

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H c hai khi ni hi tip............................................................................................. 72 Chuyn nt ni tn hiu t trc ra sau mt khi .......................................................... 73 Chuyn nt ni tn hiu t sau ti trc mt khi ......................................................... 73 Chuyn nt r nhnh tn hiu t trc ra sau mt khi................................................. 74 Chuyn nt r nhnh tn hiu t sau ti trc mt khi ................................................ 74 Chuyn nt r nhnh t trc ra sau mt nt ni ......................................................... 74 Chuyn nt r nhnh t sau ti trc mt nt ni ........................................................ 75

2.2.4 S tn hiu v cng thc Mason................................................................................. 77

2.2.5 th c tnh tn binpha ........................................................................................... 83 Khi nim hm c tnh tn .......................................................................................... 83 Xy dng hm c tnh tn bng thc nghim ............................................................. 85 th c tnh tn binpha ......................................................................................... 86

2.2.6 th c tnh tn logarith th Bode ......................................................................... 90

2.2.7 Quan h gia phn thc v o ca hm c tnh tnTon t Hilbert ........................... 96 Bi ton th nht: Xc nh hm truyn t phn thc hm c tnh tn....................... 97 Bi ton th hai: Xc nh hm truyn t phn o hm c tnh tn............................ 99 Ton t Hilbert: Trng hp tng qut........................................................................ 100

2.2.8 Xy dng m hnh ton hc ca cc khu ng hc c bn bng thc nghim ch ng........................................................................................................... 102

Khu qun tnh bc nht............................................................................................. 103 Khu tch phnqun tnh bc nht ............................................................................ 104 Khu tch phnqun tnh bc n ................................................................................. 105 Khu qun tnh bc hai ............................................................................................... 107 Khu qun tnh bc cao .............................................................................................. 109 Khu (b) Lead/Lag .................................................................................................... 111 Khu dao ng bc hai............................................................................................... 114 Khu chm tr (khu tr) ............................................................................................ 115

2.2.9 Ma trn hm truyn cho h MIMO ................................................................................ 117

2.3 Phn tch h thng 118

2.3.1 Nhng nhim v c bn ca cng vic phn tch ......................................................... 118

2.3.2 Xc nh tnh n nh t a thc c tnh ...................................................................... 120 Mi lin h gia v tr cc im cc v tnh n nh ca h thng .............................. 120 Tiu chun i s th nht: Tiu chun Routh........................................................... 122 Tiu chun i s th hai: Tiu chun Hurwitz .......................................................... 127 Tiu chun i s th ba: Tiu chun LienardChipart.............................................. 129 Tiu chun hnh hc: Tiu chun Michailov ............................................................... 131

2.3.3 Phn tch cht lng h kn t hm truyn ca h h................................................... 134 Xt tnh n nh: Tiu chun Nyquist.......................................................................... 134 Kim tra tnh n nh h kn nh biu Bode........................................................... 140 nh gi sai lch tnh ................................................................................................. 142 Thng s c trng ca qu trnh qu : qu iu chnh v thi gian qu .... 144 Thng s c trng ca qu trnh qu : Sai lch bm............................................ 147

2.3.4 Quan h gia cht lng h thng vi v tr im cc v im khng ca hm truyn............................................................................................................................. 150

Mt s kt lun chung................................................................................................. 150 iu kin tn ti qu iu chnh ............................................................................ 151 Khu thng tn v h pha cc tiu ............................................................................. 154

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Phn tch bng phng php qu o nghim s...................................................... 156 2.3.5 Phn tch tnh bn vng................................................................................................. 161

nh gi cht lng bn vng nh hm nhy............................................................ 162 nh gi tnh n nh bn vng vi sai lch m hnh khng c cu trc ................... 163 H va c tnh n nh bn vng va c nhy nh ............................................... 164 Tnh n nh bn vng ca h bt nh c cu trc: Tiu chun Kharitonov ............. 165 Bi ton m................................................................................................................. 169

2.4 Thit k b iu khin 170

2.4.1 Chn tham s cho b iu khin PID ........................................................................... 170 Hai phng php xc nh tham s PID ca ZieglerNichols .................................... 172 Phng php ChienHronesReswick....................................................................... 174 Phng php tng T ca Kuhn................................................................................... 176 Phng php ti u ln.......................................................................................... 177 Phng php ti u i xng...................................................................................... 183 Chn tham s PID ti u theo sai lch bm............................................................... 191

2.4.2 Phng php iu khin cn bng m hnh ................................................................. 193 Thit k b iu khin cn bng hm truyn ca h h (loop shaping) .................... 193 Thit k b iu khin cn bng hm truyn ca h kn............................................ 196 iu khin theo nguyn l m hnh ni (IMC) ............................................................ 199 Thit k b iu khin d bo Smith cho i tng c tr ......................................... 201

2.4.3 Thit k b iu khin theo m hnh mu..................................................................... 202 Thut ton tm nghim phng trnh Euclid................................................................ 204 Thut ton thit k hai b iu khin theo m hnh mu ........................................... 205

2.4.4 Tp cc b iu khin lm n nh i tng v khi nim n nh mnh, n nh song hnh.............................................................................................................. 207

Mt s khi nim c bn............................................................................................. 207 Ni dung phng php tham s ha Youla ................................................................ 208 Kh nng iu khin n nh mnh (strongly stable) ................................................. 212 B iu khin n nh song hnh (simultane stable).................................................. 213

2.4.5 iu khin tch knh..................................................................................................... 216 Tch knh trong ton b min thi gian ..................................................................... 216 Tch knh trong ch xc lp ................................................................................. 217


3 iu khin lin tc trong min thi gian 229 3.1 Cng c ton hc 229

3.1.1 Nhng cu trc i s c bn ....................................................................................... 229 Nhm .......................................................................................................................... 229 Vnh............................................................................................................................ 230 Trng......................................................................................................................... 230 Khng gian vector....................................................................................................... 231 Khng gian vector con................................................................................................ 232 a tp tuyn tnh......................................................................................................... 233 i s.......................................................................................................................... 233 Ideale .......................................................................................................................... 233

3.1.2 i s ma trn............................................................................................................... 234

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Cc php tnh vi ma trn........................................................................................... 235 nh thc ca ma trn................................................................................................. 236 Hng ca ma trn ....................................................................................................... 238 Ma trn nghch o ..................................................................................................... 238 Vt ca ma trn........................................................................................................... 239 Ma trn l mt nh x tuyn tnh ................................................................................ 240 Php bin i tng ng......................................................................................... 240 Khng gian nhn v khng gian nh ca ma trn...................................................... 241 Gi tr ring v vector ring......................................................................................... 242 Chun ca vector v ma trn...................................................................................... 244 Ma trn c cc phn t ph thuc thi gian................................................................ 245


3.2.1 Phng trnh trng thi .................................................................................................. 245 Cu trc chung............................................................................................................ 245 Quan h gia m hnh trng thi v hm truyn ........................................................ 249

3.2.2 Qu o trng thi......................................................................................................... 255 Ma trn hm m v cch xc nh.............................................................................. 256 Nghim ca phng trnh trng thi c tham s khng ph thuc thi gian .............. 262 Nghim ca phng trnh trng thi ph thuc thi gian............................................ 264 Qu trnh cng bc v qu trnh t do ...................................................................... 266

3.3 Phn tch h thng 267

3.3.1 Nhng nhim v c bn ca cng vic phn tch ......................................................... 267

3.3.2 Phn tch tnh n nh.................................................................................................... 268 Phn tch tnh n nh BIBO........................................................................................ 268 Tiu chun n nh Lyapunov Hm Lyapunov ......................................................... 271

3.3.3 Phn tch tnh iu khin c ...................................................................................... 276 Khi nim iu khin c v iu khin c hon ton ......................................... 276 Cc tiu chun xt tnh iu khin c cho h tham s hng.................................. 280 Tiu chun xt tnh iu khin c cho h tham s ph thuc thi gian.................. 284

3.3.4 Phn tch tnh quan st c......................................................................................... 289 Khi nim quan st c v quan st c hon ton .............................................. 289 Mt s kt lun chung v tnh quan st c ca h tuyn tnh ................................ 290 Tnh i ngu v cc tiu chun xt tnh quan st c ca h tham s hng ......... 293

3.3.5 Phn tch tnh ng hc khng...................................................................................... 295


3.4.1 B iu khin phn hi trng thi gn im cc........................................................... 297 t vn v pht biu bi ton................................................................................ 297 Phng php Ackermann ........................................................................................... 298 Phng php Roppenecker ........................................................................................ 304 Phng php modal phn hi trng thi..................................................................... 308

3.4.2 iu khin tch knh..................................................................................................... 317 B iu khin phn hi trng thi tch knh FalbWolovich...................................... 317 B iu khin tch knh SmithMcMillan................................................................... 321

3.4.3 iu khin phn hi trng thi ti u ............................................................................ 324 iu kin cn v cc bc tng hp b iu khin ti u .......................................... 324 Bn v tnh n nh ca h kn ti u v bi ton m................................................. 330

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Phng php tm nghim phng trnh Riccati........................................................... 332 3.4.4 iu khin bm (tracking control) bng phn hi trng thi ........................................ 334

3.4.5 iu khin phn hi trng thi thch nghi...................................................................... 337 Trng hp i tng c cht lng mong mun khi khng c nhiu .................. 338 Trng hp tng qut ................................................................................................. 340

3.4.6 iu khin phn hi tn hiu ra ..................................................................................... 341 t vn ................................................................................................................... 341 B quan st Luenberger ............................................................................................. 344 Gim bc b quan st Luenberger ............................................................................. 346 B quan st Kalman ................................................................................................... 347 Thit k b iu khin ti u phn hi u ra LQG.................................................... 350 Kt lun v cht lng h kn: Nguyn l tch ........................................................... 351 iu khin khng nhiu bng phn hi u ra........................................................... 355

3.4.7 Loi b sai lch tnh bng b tin x l ......................................................................... 356

3.4.8 Hin tng to nh (peak) v bi ton chn im cc................................................. 359


4 iu khin h khng lin tc 371 4.1 Tn hiu v cng c ton hc 371

4.1.1 Tn hiu khng lin tc u ........................................................................................... 371 M t qu trnh trch mu............................................................................................ 371 Dy s, tnh hi t v gi tr gii hn........................................................................... 372

4.1.2 Cng c ton hc .......................................................................................................... 374 Php bin i Fourier ri rc (DFT)............................................................................ 374 Php bin i Z thun................................................................................................. 377 Php bin i Z ngc................................................................................................ 380 Chui v tnh hi t ca chui .................................................................................... 383

4.1.3 Php bin i z .............................................................................................................. 384


4.2.1 Khi nim h khng lin tc.......................................................................................... 386

4.2.2 Phng trnh sai phn, hm trng lng v hm truyn............................................... 387 Phng trnh sai phn................................................................................................. 387 Dy gi tr hm trng lng (hm trng lng)........................................................... 390 Hm truyn ................................................................................................................. 390 Mt s kt lun chung................................................................................................. 393

4.2.3 M hnh trng thi ......................................................................................................... 394 Xc nh m hnh trng thi t phng trnh sai phn................................................ 394 Xc nh m hnh trng thi t hm truyn................................................................. 396 Xc nh m hnh trng thi h khng lin tc t m hnh trng thi h lin tc........ 396 Xc nh hm truyn t m hnh trng thi................................................................. 398 Xc nh hm trng lng t m hnh trng thi......................................................... 399

4.2.4 i s s khi h khng lin tc .............................................................................. 399 Hai khi ni tip: ......................................................................................................... 400 Hai khi song song: .................................................................................................... 400 H hi tip:.................................................................................................................. 400

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4.3 Phn tch h khng lin tc 404

4.3.1 Phn tch tnh n nh.................................................................................................... 404 Qu trnh t do, iu kin cn v h n nh.................................................... 404 Tiu chun SchurCohn-Jury ..................................................................................... 407 S dng cc tiu chun xt tnh n nh h lin tc ................................................... 410 Tiu chun Nyquist ..................................................................................................... 413

4.3.2 Tnh iu khin c v quan st c ........................................................................ 415 Phn tch tnh iu khin c.................................................................................... 415 Phn tch tnh quan st c ...................................................................................... 417

4.3.3 Chu k trch mu v cht lng h thng...................................................................... 421 Hin tng trng ph .................................................................................................. 421 Chn chu k trch mu ng nht im cc .......................................................... 422 Quan h gia chu k trch mu v tnh iu khin c, quan st c.................... 422 Quan h gia chu k trch mu v tnh n nh .......................................................... 423


4.4.1 Chn tham s cho b iu khin PID s....................................................................... 424 Cu trc b iu khin PID s.................................................................................... 424 Xc nh tham s cho PID s bng thc nghim ....................................................... 425

4.4.2 Cc phng php thit k trong min tn s ................................................................ 427 S dng nh x lng tuyn tnh thit k b iu khin........................................ 427 Thit k b iu khin khng lin tc theo m hnh mu ........................................... 430 Thit k b iu khin deadbeat............................................................................... 431

4.4.3 Cc phng php thit k trong min thi gian ............................................................ 435 iu khin phn hi trng thi gn im cc ............................................................. 435 B quan st trng thi tim cn v k thut gim bc b quan st............................ 435 Thit k b lc Kalman (quan st trng thi Kalman) ................................................ 437 iu khin phn hi u ra theo nguyn l tch ........................................................ 440 Thit k b iu khin deadbeat............................................................................... 441

4.4.4 Nhp mn iu khin d bo ........................................................................................ 443 Nguyn tc chung ca iu khin d bo (MPCmodel predictive control)............... 443 iu khin d bo h SISO trong min phc ............................................................. 443 iu khin d bo h MIMO trong khng gian trng thi........................................... 446


nh Laplace v nh Z ca mt s tn hiu c bn 451

Ti liu tham kho 452

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1 Nhp mn

1.1 Ni dung bi ton iu khin

iu khin h thng c hiu l bi ton can thip vo i tng iu khin hiu chnh, bin i sao cho n c c cht lng mong mun. Nh vy r rng khi thc hin mt bi ton iu khin, ta cn phi tin hnh cc bc sau y:

1) Xc nh kh nng can thip t bn ngoi vo i tng. V i tng giao tip vi mi trng bn ngoi bng tn hiu vora nn ch c th thng qua tn hiu vora ny mi c th can thip c vo n. Nh vy phi hiu r bn cht tn hiu i tng l tin nh, ngu nhin, lin tc hay khng lin tc.

2) Sau khi hiu r bn cht, phng tin can thip i tng th bc tip theo phi xy dng m hnh m t i tng. Hnh thc m t c dng nhiu trong iu khin l m hnh ton hc biu din mi quan h gia tn hiu votrng thitn hiu ra.

3) Vi m hnh ton hc c, tip theo ta phi xc nh xem i tng hin c nhng tnh cht g, cc c tnh no cn phi sa i v sa i nh th no h c c cht lng nh ta mong mun. Ni cch khc l phi phn tch h thng v phi ch r tng nhim v ca s can thip.

4) Khi xc nh c tng nhim v c th cho vic can thip ta s tin hnh thc hin vic can thip m c th l phi xc nh tn hiu kch thch u vo mt cch thch hp, hoc phi thit k b iu khin to ra c tn hiu u vo thch hp .

Tt

Khng tt

Xc nh loi tn hiu

Xy dng m hnh ton hc

Phn tch h thng

Xc nh tn hiu iu khin hoc thit

k b iu khin

nh gi cht lng

Kt thc

Bt u

Hnh 1.1: Trnh t cc bc thc hin mt bi ton iu khin

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5) Cui cng, do kt qu thu c hon ton c xy dng trn nn m hnh ton hc c ca i tng, song thc t li c p dng vi i tng thc, nn cn thit phi nh gi li cht lng ca kt qu can thip khi chng lm vic thc vi i tng. Nu iu cng mang li cht lng nh mong i th ta kt thc bi ton iu khin. Ngc li, ta phi quay li t u vi bc 1) hoc 2).

Hnh 1.1 cho ta mt ci nhn tng quan v cc bc phi thc hin trong mt bi ton iu khin. C th thy rng kt qu bi ton iu khin ph thuc rt nhiu vo bc xy dng m hnh ton hc m t i tng.

Vic xy dng m hnh cho i tng c gi l m hnh ha. Ngi ta thng phn chia cc phng php m hnh ha ra lm hai loi:

phng php l thuyt v

phng php thc nghim (nhn dng).

Phng php l thuyt l phng php thit lp m hnh da trn cc nh lut c sn v quan h vt l bn trong v quan h giao tip vi mi trng bn ngoi ca i tng. Cc quan h ny c m t theo quy lut lha, quy lut cn bng, di dng nhng phng trnh ton hc. iu kin c th xy dng c m hnh ton hc theo phng php l thuyt l phi bit c cu trc vt l bn trong h thng v cc phng trnh cn bng hal gia cc thnh phn bn trong .

V d 1.1: Xy dng m hnh bng phng php l thuyt

Chng hn ta phi xy dng m hnh cho i tng l mt chic xe chuyn hng. Tn hiu u vo tc ng y xe l lc u(t). Di tc ng ca lc u(t) xe s i c qung ng k hiu bi y(t). Hnh 1.2 m t cu trc vt l bn trong h.

Khi xe chuyn ng s c hai lc cn tr s chuyn ng ca xe (b qua ma st tnh). Th nht l lc ma st ng xc nh bi:

Fs = d dtdy

, d l h s ma st ng

v th hai l lc cn tr s thay i tc

Fgt = m 2

2

dt

yd, m l khi lng ca xe.

T hai phng trnh cn bng hal trn, cng nh theo nguyn tc bo ton nng lng chung, ta c c m hnh m t i tng, tc l m t quan h gia tn hiu vo u(t) v tn hiu ra y(t) nh sau (gi l m hnh vora):

udtdy

ddt

ydm =+

2

2 hm truyn ( )

(1 )k

G ss Ts

=

+ vi k =

d1

v T=dm

. S

y(t)

u(t)m

mydy

Hnh 1.2: H thng xe chuyn hng.

13

Trong cc trng hp m s hiu bit v nhng quy lut giao tip bn trong i tng cng v mi quan h gia i tng vi mi trng bn ngoi khng c y c th xy dng c mt m hnh hon chnh, nhng t nht t c th cho bit cc thng tin ban u v m hnh th tip theo ngi ta phi p dng phng php thc nghim hon thin nt vic xy dng m hnh i tng trn c s quan st tn hiu vo v ra ca i tng sao cho m hnh thu c tha mn cc yu cu ca phng php l thuyt ra. Phng php thc nghim c gi l nhn dng. Khi nim nhn dng (identification) c Zadeh nh ngha c th nh sau:

nh ngha 1.1 (Nhn dng): Nhn dng l phng php thc nghim xc nh mt m hnh c th trong lp cc m hnh thch hp, sao cho sai lch gia m hnh vi h thng l nh nht.

Nh vy c th thy bi ton nhn dng c ba c im nhn bit. l:

thc nghim, nhn bit qua vic o cc tn hiu vo v ra,

lp cc m hnh thch hp, c c t nhng thng tin ban u v h thng (gi chung li l thng tin Apriori),

sai lch gia m hnh c c v h thng l nh nht, c nhn bit t hm mc tiu m t sai lch v c thc hin bng phng php ti u.

Nhng phng php xc nh m hnh ton bng thc nghim, song khng c s nh gi sai lch gia m hnh v h thng v khng cn phi tm nghim ti u c c m hnh vi sai lch nh nht, c gi l phng php xp x m hnh (model estimation).

Tuy nhin, t nhiu l do, chng hn nh v b qua cc gi thit phi c cho cc nh lut cn bng c p dng, hay b qua s tc ng ca nhiu trong qu trnh o tn hiu vo v ra, ta khng th hy vng rng m hnh thu c, cho d bng l thuyt hay thc nghim, l m t tuyt i chnh xc h thng. Ni cch khc, gia m hnh v h thng thc lun tn ti sai lch nht nh v sai lch ny cng lun thay i theo thi gian lm vic, theo iu kin mi trng xung quanh . Bi vy, thng thng ngi ta cng rt tha mn, nu c c mt m hnh va c cu trc n gin, va m t chnh xc i tng vi mt s gi thit nht nh. Nhng iu ny cng dn n kh nng kt qu thu c (b iu khin) b ph thuc vo nhng gi thit ny v khi chng khng cn c tha mn, chng hn nh khi h thng thay i mi trng lm vic, hoc khi c nhng tc ng khng lng trc ca mi trng xung quanh vo h thng th chng s khng cn ng na v ta li phi thc hin li bi ton iu khin t u vi cc bc nu hnh 1.1.

Nhm hn ch vic phi thc hin li t u bi ton iu khin ch v khng lng trc c nhng sai lch c th c gia m hnh v i tng thc, ngi ta phi gi nh c s tn ti sai lch ny ngay khi phn tch v khi thit k b iu khin. cng chnh l ni dung ca hai chuyn ngnh ring c tn gi l:

14

iu khin bn vng: To ra c mt b iu khin mang li cht lng mong mun cho mt tp hp cc m hnh ca h thng (ch khng ch ring cho mt m hnh), hoc vi mt m hnh c cha sai lch bt nh b chn.

iu khin thch nghi: To ra c b iu khin c kh nng t chnh nh, t thay i theo s thay i ca sai lch (khng b chn) gia m hnh v i tng thc, sao cho cht lng ca h thng khng b thay i.

Quyn sch ny s trnh by chi tit tng bc khi thc hin mt bi ton iu khin tuyn tnh. Tuy nhin, do cc cng c ton hc c s dng phi ph hp vi kiu m hnh ton hc thu c cng nh chng loi tn hiu tc ng vo h thng, nn cc bc thc hin s c trnh by theo ba dng in hnh, c th l:

Chng 2 vi cc bc thc hin bi ton iu khin khi m hnh thu c l mt m hnh trong min phc (i tng iu khin c m t bng phng trnh i s trong min phc).

Chng 3 l ni dung cc bc thc hin bi ton iu khin ng vi lp cc m hnh trng thi (i tng iu khin c m t bng h cc phng trnh vi phn trong min thi gian).

Chng 4 l ni dung tng bc thc hin bi ton iu khin khi tn hiu vora tc ng ln i tng iu khin, hay h thng iu khin l loi tn hiu khng lin tc, hoc l tn hiu s.

1.1.1 Bi ton c tn hiu tin nh (iu khin tin nh)

Khi nim tn hiu

nh ngha 1.2 (Tn hiu): Tn hiu l mt hoc nhiu hm thi gian, mang thng tin vt l v c truyn ti bng mt i lng vt l (khc).

Nh vy tn hiu c ba c im nhn bit. l:

c m t bng mt (hoc nhiu) hm thi gian x(t),

hm thi gian phi mang mt thng tin vt l nht nh,

v hm phi truyn ti c cng bng mt i lng vt l.

V d 1.2: Minh ha khi nim tn hiu

iu khin mt bnh nc sao cho mc nc trong bnh lun l hng s khng i th cao ct nc trong bnh s l mt trong nhng thng s k thut c quan tm ca h thng. Gi tr v cao ct nc ti thi im t c o bi cm bin v c biu din thnh mt i lng in p di dng hm s ph thuc thi gian u(t) c n v l Volt. i lng vt l y l in p c s dng truyn ti hm thi gian u(t) mang thng tin v cao ct nc.

15

iu khin nhit th tt nhin nhit hin thi l mt thng s k thut ca h thng c quan tm. Gi tr nhit ti thi im t di dng gi tr ca hm s ph thuc thi gian i(t) c o bi cm bin v c biu din thnh mt i lng dng in c n v l Ampe. Nh vy tn hiu i(t) l mt hm thi gian mang thng tin v nhit trong phng ti thi im t v c truyn ti bi i lng vt l l dng in.

Ting ni l mt i lng vt l. Ting ni c bin i thnh dng in l mt i lng vt l khc truyn hu tuyn i xa. Dng in c m t bng mt hm thi gian i(t). Nh vy hm thi gian i(t) y l mt tn hiu, n mang thng tin ca ting ni v c truyn ti nh dng in. S

Nu trong i tng c nhiu tn hiu x1(t), x2(t), , xn(t) c quan tm cng mt lc th sau y ta s s dng k hiu vector:

x(t) = (x1(t), x2(t), , xn(t))T ch chng, trong ch s m T l k hiu ca php chuyn v vector (hay ma trn).

Phn loi tn hiu tin nh

Tn hiu tin nh l tn hiu nu nh ngha 1.2, nhng c m t ch bng mt hm thi gian x(t). Do c m t bng hm thi gian nn da vo tnh cht ca hm thi gian ngi ta phn loi tn hiu thnh tng cp phm tr nh sau:

1) lin tc v khng lin tc (phn loi thng qua min xc nh tR). Mt tn hiu c gi l lin tc, nu hm x(t) m t n lin tc tng on, ngc li n c gi l tn hiu khng lin tc. Khi nim hm x(t) lin tc trong mt on c hiu l

n lin tc ti mi im trong on , tc l vi mi t0 thuc on lun c:

0

0 0 0( ) lim ( ) ( 0) ( 0)t t

x t x t x t x t

= = = +

v gii hn ny khng ph thuc chiu tt0 t bn tri sang (lun c tt0), c k hiu l x(t0+0).

Tn hiu khng lin tc c m t bi dy cc ga tr {xk}, k=,1,0,1, ca n.

2) tng t v ri rc (phn loi thng qua min gi tr xR). Tn hiu tng t l tn hiu m hm x(t) m t n c min gi tr to thnh tng khong lin thng, ngc li n s c gi l tn hiu ri rc. Chng hn tn hiu c gi tr ch l nhng s hu t l tn hiu ri rc.

3) tun hon v khng tun hon. Tn hiu x(t) c gi l tun hon nu tn ti hng s T c x(t+T)=x(t), t. Hng s T c gi l chu k ca tn hiu tun hon.

4) nhn qu v phi nhn qu (causal v uncausal). Tn hiu nhn qu l hm x(t) tha mn x(t)=0 khi t

16

Vic phn chia chng thnh tng cp nh vy ni rng mt tn hiu khng th c cc tnh cht trong cng mt cp. Chng hn khng th c tn hiu va tng t, va ri rc, song li c tn hiu va khng lin tc v va ri rc. Tn hiu khng lin tc v ri rc c gi l tn hiu s.

Hnh 1.3 minh ha bn dng c bn ca tn hiu causal. Bn kiu tn hiu trn ch l s phn loi c bn theo min xc nh hoc theo min gi tr ca x(t). Trn c s bn kiu phn loi c bn m mt tn hiu x(t) khi c chung ng thi ti c min xc nh v min gi tr c th l:

dng tn hiu lin tctng t,

dng tn hiu khng lin tctng t,

dng tn hiu lin tcri rc,

dng tn hiu khng lin tcri rc,

V d 1.3: Khi nim tn hiu khng lin tcri rc (tn hiu s)

Gi s ta c tn hiu lin tctng t x(t). x l tn hiu x(t) bng nhng thut ton chy trn my tnh ngi ta cn phi trch mu tn hiu ti nhng im thi gian

cch u nhau Ta c gi l thi gian trch mu. Nu dy cc gi tr tn hiu {xk } ,

k=,1,0,1, thu c vi xk= x(kTa) c xem nh mt tn hiu th do min xc

nh ca {xk } l tp im m c

x(t)

t t

t

T 2T 3T

t

T 2T 3T

Lin tctng t

Khng lin tctng t

Lin tcri rc

Khng lin tcri rc (tn hiu s)

Hnh 1.3: Cc dng tn hiu c bn khc nhau.

2

33,74,14,5

2

3,84,2

x(t)

x(t) x(t)

17

{ t = kTa k Z} , Z l k hiu ch tp cc s nguyn khng lin thng, tc l khng to ra c mt khong bt k no nn dy {xk } lin

tc ti cc im trong , nn {xk } l tn hiu c dng khng lin tctng t.

Tn hiu khng lin tc tng t {xk } vn cha th x l c bng my tnh bi

my tnh ch lm vic c vi s hu t trong mt khong cho php, trong khi xk c th

l mt s thc bt k (v d nh s v t 2 , 3 , , ). Hn na, min gi tr cho php ca cc s hu t cn ph thuc my tnh, ngn ng lp trnh. Chng hn bin thc kiu double ca ngn ng lp trnh C ch lm vic c vi nhng s hu t trong khong

t 1,710308 n 1,710308 hoc vi bin kiu long double th khong cho php l t

1,1104932 n 1,1104932.

Bi vy bc tip theo cn phi lm l xp x cc gi tr xk thnh s hu t gn

nht, k hiu l kx

, nhng khng nm ngoi min cho php. Vic xp x {xk } thnh

{ kx } v hnh chung ri rc ha min gi tr ca x(t). Min gi tr ca { kx

} by gi l tp cc s hu t (cc im khng lin thng). V d

{ kx

Q 1,710308 kx 1,710308} , Q l tp cc s hu t v do dy { kx

} l tn hiu khng lin tc ri rc (tn hiu s). S

Mt s tn hiu tin nh in hnh

Trong v s cc cc tn hiu vi nhiu dng khc nhau, iu khin tuyn tnh c mt s quan tm c bit n mt s tn hiu in hnh thng gp trong ng dng (hnh 1.4). l cc tn hiu bc thang (Heaviside), tn hiu tng u, tn hiu xung vung v hm xung dirac. Tt c cc loi tn hiu ny u c mt im chung l causal (nhn qu), tc l x(t)=0 khi t

18

3) Tn hiu tng u c xc nh qua cng thc

r(t) = t1(t) =

k )

th dy hm {xk(t)} c gi l dy Cauchy.

Khc vi trng s thc R , m mi dy Cauchy u hi t (ti gi tr gii hn x no cng thuc R ) , th trong khng gian metric X ni chung l cha c m bo. Ni cch khc, khng phi mi dy Cauchy ca cc hm s ca mt khng gian metric X cng hi t ti mt hm s no thuc X.

Mt khng gian metric X c gi l khng gian (complete), nu mi dy Cauchy trong n u hi t (ti mt phn t cng thuc X).

Mt khng gian metric X c gi l khng gian compact, nu mi dy {xk(t)} trong n u cha mt dy con hi t.

Trong khng gian vector X xc nh trn trng s thc R, nu c thm nh x, khng nht thit phi tuyn tnh, : X R tha mn:

x0 v x=0 khi v ch khi x=0, ax = |a | x ng vi mi aR v xX , x+y x+y vi mi x ,yX .

th gi tr thc x c gi l chun ca phn t x v khng gian vector X c gi l khng gian chun. Do X l khng gian vector nn t chun x ta cng c c metric: d (x ,y ) = xy

Ngc li, mt khng gian metric cng s l khng gian chun vi x=d (x ,0) , nu metric ca n cn tha mn thm:

d (x+z , y+z ) = d (x ,y ) , tc l metric bt bin vi php dch chuyn vector.

d (ax ,ay ) = |a | d (x ,y ) , tc l n thun nht (homogen). Trong mt khng gian X c th c nhiu loi chun. Hai chun xa v xb ca n

c gi l tng ng nu tn ti hai s thc m v M lun c:

mxb xa Mxb

Cc khng gian chun thng gp l:

1) Khng gian Lp[a,b] gm cc tn hiu x ( t ) thc, xc nh trn khong kn [a ,b ] , c chun c nh ngha l:

xp := ( )b

pp

a

x t dt , trong 1 p

21

2) Khng gian L[a,b] l tp hp cc tn hiu x(t) thc, xc nh trn khong kn [a ,b ] , c chun c nh ngha l:

x := )(sup txbta

.

c bit, c hai loi chun trn vi p trong 1 p cn tha mn:

xy1 xpyq nu 1 1

1p q

+ = (nh l Hlder)

x+yp xp+yp (nh l Minkovski)

Chun bc 1 ca tn hiu cn c gi l cng sut P v chun bc 2 c gi l

nng lng E ca tn hiu. Vi Lp[,] ngi ta thng vit gn thnh Lp.

1.1.2 Bi ton c tn hiu ngu nhin (iu khin ngu nhin)

Khi nim qu trnh ngu nhin

Cc tn hiu m ta lm quen t trc n nay c chung mt c im l chng u c m t bng mt hm thi gian x(t) c th. Nhng tn hiu c gi l tn hiu tin nh. Vic chng m t c ch bng mt hm thi gian ni ln tnh tng minh rng trong cc hon cnh cng nh thi im ging nhau ta lun xc nh c cng mt gi tr nh nhau cho tn hiu.

Nhng tn hiu khng m t c tng minh bng mt hm thi gian c th m

thay vo l mt tp hp ca nhiu hm thi gian xi(t), c tn l tn hiu ngu nhin. Ty vo tng hon cnh, tng trng hp, m tn hiu ngu nhin s nhn mt trong

cc hm xi(t), iR, thuc mt tp hp X ( t ) no lm m hnh v ngay c hon cnh no, trng hp no n s c m hnh xi(t) ta cng khng bit c trc. Nhiu nht ta

ch c th bit c v xc sut n c m t bi xi(t).

Tp hp X ( t ) ca tt c cc m hnh xi(t) c th c ca tn hiu ngu

nhin c gi l qu trnh ngu nhin v m t tn hiu ngu nhin mt cch y ta phi m t tp hp

X ( t ) , bng cch xc nh cc tham s c trng v n.

C hai tham s thng c s dng m t qu trnh ngu nhin

X ( t ) . l:

x ( t )

tX ( t )

Hnh 1.7: Tn hiu ngu nhin c m t nh l phn t ca mt tp hp cc hm thi gian c cng tnh cht.

22

Gi tr trung bnh mx(t): Ti mt im thi gian t0 c th th cc hm xi(t0) u

l nhng s thc. Gi tr trung bnh ca tt c cc phn t xi(t0) l mx(t0). Cho t0

chy t n th mx(t0) s tr thnh hm mx(t) ph thuc thi gian. S dng

k hiu M { } ch php tnh ly gi tr trung bnh th mx(t)=M{X ( t ) } .

Hm tng quan rx(t,): Ti mt im thi gian t0 c th th hm tng quan

rx(t0,) l gi tr trung bnh ca tt c cc tch xi(t0)xj(t0+). Cho t0 bin thin

nh t th hm tng quan rx(t,) s l mt hm ca hai bin t v . Nh vy, hm

tng quan s l rx(t,)= M{X ( t )X ( t+ ) } .

Qu trnh ngu nhin dng v ngu nhin egodic

Nhng qu trnh ngu nhin X ( t ) thng gp trong thc t l cc qu trnh ngu nhin dng. l loi qu trnh ngu nhin m c hai tham s ngu nhin mx(t) v

rx(t,) m t n u khng ph thuc vo bin thi gian t. Nh vy, qu trnh ngu nhin dng c:

Gi tr trung bnh mx(t) ca n l mt hng s, k hiu l mxR.

Hm tng quan rx(t,) l hm ca mt bin , k hiu l rx().

Trong cc loi qu trnh ngu nhin dng, ta li quan tm nhiu ti qu trnh ngu

nhin egodic. y l loi qu trnh ngu nhin dng m , cc tham s mx v rx() ch cn c xc nh t mt phn t x(t) lm i din l . Nh vy th:

Gi tr trung bnh mx ca qu trnh ngu nhin egodic X ( t ) s l gi tr trung bnh ca mt phn t x(t):

mx =0

1lim ( )

T

Tx t dt

T (1.3) Hm t tng quan rx() ca qu trnh ngu nhin egodic X ( t ) l gi tr trung

bnh ca tch x ( t )x ( t+ ) vi x ( t ) l mt phn t ty ca tp hp X ( t ) :

rx() = 0

1lim ( ) ( )

T

Tx t x t dt

T

+ (1.4)

Trong iu khin, t khi ta ch lm vic vi mt tn hiu ngu nhin. Khi phi lm vic vi nhiu tn hiu ngu nhin th cn phi ti mi lin quan gia chng.

Cho hai qu trnh ngu nhin egodic X ( t ) v Y ( t ) . c trng cho s lin quan gia X ( t ) v Y ( t ) l hm h tng quan:

23

rxy() =

0

1lim ( ) ( )

T

Tx t y t dt

T

+ (1.5)

Hai qu trnh ngu nhin c k vng bng khng X ( t ) v Y ( t ) s c gi l khng tng quan nu rxy()=0. Chng hn nh hai thit b pht tn hiu ngu nhin khc

nhau, c cu trc khc nhau s pht ra hai qu trnh ngu nhin X ( t ) , Y ( t ) c lp vi nhau. Gia chng khng c mt s lin quan no v do phi c rxy()=0.

Hm tng quan rx() v rxy() ca cc qu trnh ngu nhin egodic X ( t ) , Y ( t ) lun tha mn:

rx() l hm chn v rx(0) |rx()| rxy() = ryx()

|rxy()| (0) (0)x yr r (0) (0)

2x yr r+

1.2 Nhng cu trc c bn ca h thng iu khin

nh ngha 1.3 (H thng): H thng c hiu l mt tp hp cc phn t (linh kin, thit b, thut ton ), c kt ni vi nhau thc hin mt nhim v c th. H thng lun c giao tip vi mi trng bn ngoi bng cc tn hiu vo v ra.

Nh vy c ba c im nhn bit mt h thng . l:

l tp hp gm nhiu phn t thc hin mt nhim v chung,

gia cc phn t c quan h qua li,

c giao tip vi mi trng xung quanh.

1.2.1 Phn loi h thng

Hnh 1.8 minh ha cu trc mt h thng gm 4 phn t vi cc c im nhn bit

trn. Cc tn hiu u vo ca h s c vit chung li thnh vector 1( , , )T

mu u u= . Tng t cc tn hiu u ra cng c vit chung li thnh 1( , , )

Tpy y y= .

Da theo cc c im nu trong nh ngha 1.3 m h thng c phn loi thnh:

1) H SISO (single inputsingle output) , nu h c mt tn hiu vo v mt tn hiu ra.

2) H MIMO (multi inputsmulti outputs), nu s tn hiu vo ra ca n nhiu vonhiu ra.

3) Theo nguyn l nh trn, mt h thng cn c th l MISO (nhiu vomt ra) hoc SIMO (mt vonhiu ra).

24

4) Lin tc, nu cc tn hiu vora ( ), ( )u t y t l lin tc, ngc li nu cc tn hiu vo

ra { },{ }k ku y , k=,1,0,1, l khng lin tc h s c gi l khng lin tc.

5) Tuyn tnh, nu nhim v chung ca n, m t bi m hnh ton:

:T u y6 hay ( )y T u= tha mn nguyn l xp chng, tc l nh x T tha:

1 1 1

( )n n n

i i i i ii i i

T a u a T u y= = =

= =

Ngc li, h s c gi l h phi tuyn.

6) Tham s hng, nu m hnh ton :T u y6 ca n khng thay i (theo thi gian v theo khng gian. Ngc li h s c gi l khng dng, nu m hnh ca n thay i theo thi gian (thng cn c gi l h nonautonom), hoc h phn b ri, nu m hnh ca n thay i theo khng gian.

7) H nhn qu (causal), nu m hnh ton ( )y T u= ca n tha mn:

( )( ) ( ) vi 0y t T u t = Nh vy, h nhn qu, tn hiu ra ( )y t thi im t ch ph thuc tn hiu vo

( )u t ng thi im t v qu kh ca n. Ngc li, nu tn hiu ra ( )y t thi

im t cn ph thuc tn hiu vo ( )u t c thi tng lai > t th n c gi l h phi nhn qu (uncausal).

8) H tnh (static), nu nu tn hiu ra ( )y t thi im t c xc nh chnh xc ch

cn qua tn hiu vo ( )u t ng thi im t . Ngc li n s c gi l h ng (dynamic), nu tn hiu ra ( )y t thi im t ch c th c xc nh chnh xc t

tn hiu vo ( )u t c thi im t v qu kh (hoc tng lai) ca n.

9) Hi tip (hay h kn), nu cc quan h bn trong gia cc phn t (c m t bng nhng ng ni trong hnh 1.8) to thnh t nht l mt vng kn. Ngc li n c gi l h h.

1.2.2 Xc nh tn hiu iu khin thch hp

i tng iu khin cng l mt h thng. H thng iu khin l mt h thng bao gm i tng iu khin v b iu khin. Kt qu ca bi ton iu khin cho mt i tng hay mt h thng, tm c theo trnh t cc bc nu trong hnh 1.1. Nhim v iu khin bao gm:

u1

u2

y1

y2

Hnh 1.8: H thng

2

3

4

1

25

1) xc nh tn hiu iu khin thch hp cho i tng (tn hiu u vo, hay tn hiu t trc), k hiu bng u(t),

2) thit k b iu khin to ra tn hiu iu khin thch hp cho i tng. Nh vy, nu xem b iu khin nh mt h thng th u ra ca n chnh l u(t) c a ti i tng iu khin, cn tn hiu u vo ca n c th l:

a) Mt tn hiu lnh w(t) t trc cho b iu khin (cu trc iu khin h).

b) Cc tn hiu trng thi x (t ) ca i tng (iu khin phn hi trng thi).

c) Tn hiu u ra y(t) ca i tng (iu khin phn hi u ra).

y l kiu bi ton iu khin m yu cu ch dng li vic xc nh tn hiu thch hp p t ti u vo ca i tng sao cho i tng c c cht lng bn trong v tn hiu u ra nh mong mun. Chng hn bi ton xc nh quy tc thay i in p u vo u(t) ca ng c (i tng iu khin) sao cho tc vng quay ca

ng c (tn hiu u ra) thay i t gi tr ban u y0 ti gi tr mong mun yT v nng lng tn hao cho qu trnh thay i tc vng quay l t nht (cht lng bn trong ca i tng).

c im ca hnh thc iu khin ny l iu khin mt chiu v trong qu trnh iu khin, h thng khng c kh nng thay i hoc hiu chnh li c. Nh vy, cht lng iu khin ph thuc hon ton vo chnh xc ca m hnh ton hc m t i tng cng nh phi c gi thit rng khng c tc ng nhiu khng mong mun vo h thng trong sut qu trnh iu khin.

1.2.3 S dng b iu khin

iu khin h

V bn cht, hnh thc iu khin ny cng ging nh bi ton tm tn hiu iu khin thch hp p t u vo ca i tng, nhng c b sung thm b iu khin to ra c tn hiu iu khin . V d iu khin tu thy i c theo mt qu o y(t) mong mun (tn hiu u ra), ngi ta phi tc ng bng lc w(t) vo tay li to ra c v tr u(t) ca bnh li mt cch thch hp. Trong v d ny, h thng tay libnh li c vai tr ca mt b iu khin.

Hnh thc iu khin h ny (hnh 1.10) l iu khin mt chiu v cht lng iu khin ph thuc vo chnh xc ca m hnh ton hc m t i tng cng nh phi

u(t)w(t) y(t) i tng iu khin

B iu khin

Hnh 1.10: Cu trc iu khin h.

u (t ) y ( t ) i tng iu khin

Hnh 1.9: Xc nh tn hiu iu khin

26

c gi thit rng khng c tc ng nhiu khng mong mun vo h thng trong sut qu trnh iu khin.

iu khin phn hi trng thi

i tng iu khin, cc tn hiu trng thi x1(t), x2(t), , xn(t), c vit

chung dng vector x (t )=(x1( t ) , x2 (t ) , , xn (t ))T , l thnh phn cha ng y nht cc thng tin cht lng ng hc h thng. N phn nh nhanh nht s nh hng ca nhng tc ng bn ngoi vo h thng, k c nhng tc ng nhiu khng mong mun. Bi vy, c th to ra c cho i tng mt cht lng mong mun, n nh vi cc tc ng nhiu, cn phi c c mt tn hiu p t u vo l u(t) phn ng kp theo nhng thay i trng thi ca i tng.

Hnh 1.11 biu din nguyn tc iu khin phn hi trng thi. B iu khin s dng tn hiu trng thi x(t) ca i tng to ra c tn hiu u vo u(t) cho i tng. V tr ca b iu khin c th l mch truyn thng (hnh 1.11a) hoc mch hi tip (hnh 1.11b).

H thng iu khin phn hi trng thi c kh nng gi c n nh cht lng mong mun cho i tng, mc d trong qu trnh iu khin lun c nhng tc ng nhiu. Xt phn ng ca ngi li xe lm v d, trong ngi li xe c xem nh l b iu khin v chic xe l i tng iu khin. Nhim v ca b iu khin l gi n nh tc xe v v tr ca xe phi lun nm trong phn ng bn phi vch phn cch. Nh vy ngi li xe (b iu khin) :

Da vo khong cch ca xe vi vch phn cch (trng thi ca i tng iu khin) a ra quyt nh phi nh tay li sang phi mnh hay nh.

Da vo tnh trng ca mt ng nh ln dc hay xung dc (tc ng ca tn hiu nhiu ti cht lng h thng) iu chnh s v bn p ga.

iu khin phn hi tn hiu ra

Tuy rng vector trng thi x ( t ) cung cp cho ta y nht cc thng tin v cht lng ng hc ca i tng, song khng phi mi trng thi ca i tng l o c

x

uw ye

uw y

x

Hnh 1.11: iu khin phn hi trng thi

a) b)

i tng iu khin

B iu khin

i tng iu khin

B iu khin

27

trc tip. V l , trong nhiu trng hp, ngi ta nh phi thay b iu khin phn hi trng thi x ( t ) bng b iu khin phn hi u ra y(t).

Hnh 1.12 m t nguyn tc iu khin phn hi u ra. B iu khin s dng tn hiu u ra y ( t ) ca i tng to ngc ra c tn hiu u vo u(t) cho n. V tr ca b iu khin c th l mch truyn thng (hnh 1.12a) hoc mch hi tip (hnh 1.12b).

Cho ti nay, bi ton iu khin phn hi tn hiu ra vn cn l mt bi ton m v cha c li gii tng qut cui cng, v tn hiu u ra y(t) thng khng mang c y thng tin ng hc v i tng. Song ring h tuyn tnh, mt iu may mn ln l do chng tha mn nguyn l tch c, nn bi ton iu khin phn hi tn hiu ra lun thay c bng hai bi ton: phn hi trng thi v quan st trng thi (hnh 1.13) v nh vy, n c gii quyt trit .

Cu hi n tp v bi tp

1) Chng minh rng mi tn hiu lin tc x(t) c min xc nh l tp compact u xp x c bng tng tuyn tnh ca tn hiu bc nhy n v hoc tn hiu tng u vi mt sai lch nh ty .

2) Chng minh rng khng gian L1 l ng vi tch chp, tc l nu c x(t)L1 v

y(t)L1 th cng c 1( ) ( ) * ( ) ( ) ( )z t x t y t x y t d L

= = 3) Cho tn hiu x(t) xc nh bi:

( ) 1( ) 1( ) , 0 1 , 0ax t t t t b a b= < < < <

uw ye

Hnh 1.12: iu khin phn hi u ra.

a) uw y

b) i tng iu khin

B iu khin

i tng iu khin

B iu khin

uw y

xHnh 1.13: iu khin phn hi u ra theo nguyn l tch.

i tng iu khin

B iu khin

B quan st trng thi

28

Hy ch rng n c cng sut ( )P x t dt

= l gi tr hu hn nhng li c nng

lng 2( )E x t dt

= l v hn.

4) Cho tn hiu x(t) xc nh bi:

1

( ) 1( ) , 1 , 02

ax t t t b a b= < <

Hy ch rng n c cng sut P v hn nhng li c nng lng E hu hn.

5) Cho x(t) tun hon vi chu k T. K hiu s(t) l hm trch mu c cng chu k trch

mu T v ( ) ( ) 1( ) 1( )x t x t t t T= l hm ly t x(t) ch trong mt chu k. Chng minh rng ( ) ( ) * ( )x t x t s t=

, trong k hiu * l ch php tnh tch chp:

( ) * ( ) ( ) ( ) ( ) ( ) ( ) * ( )x t s t x s t d x t s d s t x t

= = = 6) Xt h SISO vi tn hiu vo l u(t), tn hiu ra l y(t), m t bi m hnh vora c

dng phng trnh vi phn tuyn tnh h s hng:

( ) ( 1) ( ) ( 1)1 0 1 n n m m

n my a y a y b u b u b u + + + = + + +" " vi ( )

kk

kd x

xdt

=

a) Hy ch rng h l tuyn tnh dng.

b) K hiu y(t)=g(t) l p ng ca h khi u vo l hm xung dirac (t) vi trng thi u bng 0. Chng minh rng khi p ng ca h vi trng thi u bng 0 v tn hiu vo u(t) bt k cho trc s c dng tch chp:

y(t)=u(t)*g(t)

29

2 iu khin lin tc trong min phc

2.1 Cc cng c ton hc

2.1.1 L thuyt hm bin phc

nh ngha, khi nim hm lin tc, hm gii tch

Hm s f(s), bin i mt s phc s=+ j , vi , l hai s thc, 1j = thnh

mt s phc khc:

f(s) = u(,)+jw(,), (2.1)

trong cc k hiu u(,) ch phn thc v w(,) ch phn o ca n, c gi l hm bin phc hay gn hn l hm phc. Vi cc k hiu trn th r rng mt hm bin phc f(s) c biu din thnh hai hm thc hai bin u(,) v w(,).

Hnh 2.1 minh ha hm phc f(s) nh mt nh x t mt phng phc s vo mt

phng phc z=f(s). Mt hm phc f(s) c gi l lin tc ti s0 c z0=f(s0) nu vi mi

ln cn Z nh cho trc ca z0, chng hn nh mt mt trn c bn knh nh v tm l z0, lun tn ti mt ln cn S tng ng ca s0, sao cho min nh ca n l f(S ) nm trn trong Z , tc l (hnh 2.1):

f(S ) Z (2.2)

Khi ngi ta cng vit:

0

0 0lim ( ) ( )s s

f s f s z

= =

Hm phc f(s) lin tc ti

mi im s0 thuc min G c gi l lin tc trn G.

Xt mt hm f (s ) lin tc trn G. Nu ti sG tn ti gii hn:

jwj

s=+ j

u

z= f (s )=u+ jw

S Z

Hnh 2.1: Hm bin phc l nh x t mt phng phc vo mt phng phc.

30

s

sfssfs

+

)()(lim

0 < (2.3)

v gii hn ny khng ph thuc vo kiu ca s0, th hm f (s ) c gi l kh vi ti

s . Khi gi tr gii hn (2.3) c gi l o hm ca f (s ) ti s v k hiu bng ds

sdf )(.

Ch rng y phi c iu kin l gii hn (2.3) khng c ph thuc vo hnh thc tin v 0 ca s .

V d 2.1: Hm bin phc khng kh vi

Xt hm phc:

f (s ) = Re(s ) hm ly phn thc ca bin phc s .

Hm ny l khng kh vi, v nu cho s0 dc theo trc thc th gii hn (2.3) s c gi tr bng 1, nhng nu cho s0 dc theo trc o j th n li c gi tr bng 0. S

Nu hm f(s) kh vi ti mi im s thuc min G th n c gi l gii tch (hay holomorph) trn G. Theo Cauchy v Riemann th cn v f(s) gii tch trn G l:

( , ) ( , )u w

=

v

( , ) ( , )u w

=

(2.4)

tc l phn thc u(,) v phn o w(,) ca hm f (s ) phi tha mn phng trnh vi phn Laplace:

2 2

2 20

u u

+ =

v

2 2

2 20

w w

+ =

Php tnh ly o hm ca cc hm phc v c bn cng c thc hin ging nh hm thc. V d:

f (s ) = sn 1( ) ndf s nsds

=

f (s ) = sin(s ) ( ) cos( )df s sds

=

Tch phn phc v nguyn l cc i modulus

Xt mt hm phc z=f(s) lin tc ti mi im s=+ j thuc min S vi bin l C (hnh 2.2). Gi AB

JJJG l mt ng cong no nm trong S. Ta chia ng AB

JJJG thnh n

on bng cc im phc s1=A , s2 , , sn=B ty v gi: k = sksk + 1

Nu nh tn ti gi tr gii hn:

1

lim ( )n

k kn k

f s

=

31

v gi tr gii hn ny khng ph thuc vo cch chn cc im sk trn on ABJJJG

, th n

s c gi l gi tr tch phn ca hm z=f(s) tnh dc theo on ABJJJG

v k hiu bi:

1

( ) lim ( )B n

k kn kA

f s ds f s

=

= (2.5)

Theo cng thc nh ngha (2.5) v tch phn nh trn ta thy gi tr tch phn cn ph thuc vo dng ca ng cong AB

JJJGtrong min S.

V php tnh tch phn phc ta c nhng kt lun c bn sau ca Cauchy:

1) (nh l tch phn ca Cauchy) Nu hm z=f(s) khng nhng lin tc m cn gii tch trong S th vi k hiu C ch ng bin ca S, ta lun c:

( ) 0C

f s ds =v (2.6) Ni cch khc, gi tr tch phn ca hm z=f(s) tnh dc theo on ng cong khp kn C l bin ca min S m f (s ) gii tch trong , s c gi tr bng 0.

2) nh l tch phn ca Cauchy ch rng gi tr tch phn:

B

A

dssf )(

ca hm z= f (s ) tnh dc theo on ABJJJG

s khng ph thuc vo dng ng cong

ABJJJG

nu nh on ABJJJG

ny nm trong mt min S m f(s) gii tch trong .

3) (Cng thc tch phn Cauchy) Gi S l min m hm z=f(s) gii tch trong n v C l bin ca min S c chiu ngc kim ng h (min S lun nm pha bn tri nu i dc trn C theo chiu ny). Khi , ti mt im s bt k thuc S lun c:

1 ( )

( )2 C

ff s d

j s

= v (2.7)

1( ) 1 ( )

2 ( )

k

k nC

d f s fd

jds s

+= v (2.8)

Php tnh tch phn ca hm phc c thc hin ging nh hm thc. V d:

Cj

A=s1

s=+ j

s2 sn=Bsn1

SHnh 2.2: Gii thch khi nim tch

phn phc.

32

f (s ) = sin(s ) ( ) cos( )f s ds s k= + (k l hng s) f (s ) = es ( ) sf s ds e k= + (k l hng s) Ngoi ra, nu gi max sup ( )

Cf f

= th khi p dng tch phn Cauchy (2.7) v (2.8) vi chy dc bin ca S l C v im s nm c nh bn trong, ta s c vi mi n (hnh 2.2):

2

max max

0

( )1 ( ) 1( )

2 2 2 2

n n nnn

C C C

f f ff d df s d d

j s s s

= = v v v

1 1

max max max1 1

( ) , ( ) limn n

nf s f n f s f f

=

trong l khong cch t s ti ng bin C v sup l k hiu gi tr chn trn nh nht (ging nh gi tr ln nht, nu n tn ti). Vy:

nh l 2.1 (Nguyn l cc i modulus): Nu hm z=f(s) lin tc trong min kn S , gii tch bn trong min th |z| = |f(s)| s c gi tr cc i trn bin ca S .

Hm bo gic (conform)

Mt hm phc f (s ) gii tch trn G v c ( ) 0df sds

c gi l hm bo gic

(conform). ngha ca tn gi bo gic c gii thch nh sau:

Nu gi 1sl v 2sl l hai ng cong to vi nhau mt gc trong mt phng phc s,

cng nh 1zl v 2zl l hai ng nh ca n trong mt phng phc z= f (s ) , tc l:

1 1( )z sl f l= , 2 2( )z sl f l=

th khi hai ng nh 1zl ,2zl ny cng s to vi nhau mt gc ng bng trong mt

phng phc z= f (s )hnh 2.3. Xt mt im s c th v vector

dds

d

=

l tip tuyn ti vi

mt ng cong ls no . Khi , trong mt phng phc z=f(s) ca hm bo gic f(s), vector ds s c

bin i thnh du

dzdw

=

vi:

u u

du d d

= +

v w w

dw d d

= +

Nu kt hp thm cng thc (2.4) ca Cauchy v Riemann th:

jwj

s=+ j

u

z= f (s )=u+ jw

Hnh 2.3: Gii thch khi nim hm bo gic.

1sl

2sl

1zl

2zl

33

cos sinsin cos

u ud d

dz Zw w d d

= =

(2.9)

trong :

( ) jdf s Ze

ds

= hay ( )df s

Zds

= v ( )

arcdf s

ds =

Cng thc (2.9) cho thy hm bo gic z= f (s ) to ra du

dzdw

=

t d

dsd

=

bng cch xoay vector ds i mt gc bng ( )

arcdf s

ds = v ko di ln ca n thm

ra mt h s nhn l ( )df s

Zds

= . c bit, nu gi lz l nh ca ls trong mt phng

phc z= f(s), tc l lz = f(ls) th dz s l vector tip tuyn ca lz ging nh ds l vector tip tuyn ca ls.

V d 2.2: Mt s hm bo gic n gin

1) Hm tuyn tnh z=f(s)=as+b, vi a ,b l hai hng s phc. y l mt nh x tuyn tnh, bin i mt vector s bt k sang mt phng z bng cch xoay n i mt gc = arc(a ) , ko di n ra bng mt h s |a | v dch chuyn song song mt khong cch bng b.

Nh vy, hm ny s bo ton dng mt ng cong bt k ca mt phng cha s sang mt phng cha z (hnh 2.4a).

2) Hm nghch o 1zs

= . Hm ny bin i mt vector s thnh vector z bng cch ly

i xng qua ng trn n v v sau li ly i xng tip qua trc thc (hnh 2.4b). Nh vy, hm ny s bin i ton b phn bn trong ng trn n v ca mt phng s thnh phn pha ngoi ng trn n v ca mt phng z .

3) Hm bnh phng z=s2 , tc l nu c s=+ j th cng s c:

z = f(s) = u+jw = 22+2j

n bin i mi ng hyperbol vung gc vi nhau trong mt phng s=+ j l

22=hng s k1 v 2=hng s k2 , thnh nhng ng thng song song vi

hai trc ta trong mt phng z= f (s )=u+ jw l u=k1 v w=k2 , tc l chng cng vung gc vi nhau.

4) Hm ly cn bc hai z s= .

34

5) Hm phn thc as bzcs d

+=

+ vi a ,b,c ,d l nhng hng s phc tha mn adbc0.

Hm ny c to thnh t ba hm bo gic con l:

z1 = cs+d, 21

1z

z= v 2

a bc adz z

c c

= +

nn n cng l hm bo gic.

6) Tng qut ha tt c nhng trng hp trn, ta s d dng i n kt lun rng hm phc dng thchu t, l hm c cu trc dng t s ca hai a thc nguyn t cng nhau (hu t) vi h s ca cc a thc l nhng s thc:

2

0 1 22

0 1 2

( ) , v

mm

i jnn

b b s b s b sG s a b

a a s a s a s

+ + + +=

+ + + +R R""

l mt hm bo gic. S

2.1.2 Chui Fourier v php bin i Fourier

Chui Fourier (cho tn hiu tun hon)

Bn cnh vic kho st tn hiu trc tip t c tnh ca n trong min thi gian, chng hn nh tnh lin tc, khng lin tc, ri rc hay tng t , nhiu khi trong thc t li xut hin cu hi rng tn hiu c c tnh tn s nh th no v n c di tn s lm vic l bao nhiu? Cc cu hi dn ta n bi ton phi phn tch tn hiu lin tc x(t) thnh dng tng tuyn tnh ca cc hm iu ha c tn s lm vic xc nh. Xt tn hiu tun hon vi chu k T , tc l x(t+T)=x(t), t. Chui Fourier ca tn hiu tun hon x(t) c hiu l:

( )01

2( ) cos( ) sin( ) v jk tk k k

k kx t c e a a k t b k t

T

= =

= = + + = (2.10)

vi k kc c= (s phc lin hp) v cos sinje j = + (cng thc Euler). Gi tr 2

T

=

c gi l tn s c bn ca tn hiu. Khi li xut hin tip cc cu hi:

Chui v phi c tn ti khng?

Cc hng s ca chui c xc nh t x(t) nh th no?

v ton b cu tr li l ni dung phng php phn tch chui Fourier sau y.

s

b

z

s

z

Hnh 2.4: Minh ha v d 2.2

a) b)

35

1) Dirichlet: iu kin chui Fourier (2.10) v phi hi t l: Nu khong (0,T ) chia c thnh hu hn cc khong con sao cho hm x(t) l lin tc, n iu trong cc khong con .

Mt cch ni khc: Nu hm x(t) ch c hu hn cc im khng lin tc v cng ch c hu hn cc im cc tr.

Ch : iu kin Dirichlet ch l iu kin . Chng hn vn c th tn ti hm x(t) lin tc trong ton khong (0,T ) nhng khng kh vi ti mi im trong , nhng vn c chui Fourier (2.10) hi t. V d hm Weierstrass:

1

( ) cos( )k kk

x t a b t

=

= vi 0 + v b l s nguyn.

Hin nay vn cha c iu kin cn v .

2) Hm x(t) phi lin tc tng on v ti im khng lin tc t0 phi c:

0 0 01

( ) ( 0) ( 0)2

x t x t x t= + +

y l iu kin du bng trong (2.10) cng ng ti t0.

3) Gi s rng tn ti chui (2.10), khi phi c:

( ) 0 01 , 1,2, v 2k k kc a jb k c a= = = (2.11) Ta c th khng nh tnh ng n ca (2.11) qua php bin i n gin sau:

( )

1

0 01 1 1

01

( )

( )cos( ) ( )sin( )

jk t jk t jk t jk t jk tk k k k k

k k k k k

k k k kk

x t c e c c e c e c c e c e

c c c k t j c c k t

= = = = =

=

= = + + = + +

= + + +

0 0 , v ( )k k k k k ka c a c c b j c c= = + = ( )0 0 1 v 2k k kc a c a jb= =

4) 0

00 0 0

1( ) , , 1,0,1,

1 2 2( ) , ( )cos( ) , ( )sin( ) , 1,2,

Tjk t

k

T T T

k k

c x t e dt kT

a x t dt a x t k t dt b x t k t dt kT T T

= = = = = =

(2.12)

Tht vy, t (2.10) ta c:

( ) jm tmm

x t c e

=

= 2

( )

0 0 01

2 ( ) '

0

( )

' , ( ')

T T T j m k tjk t jm t jk t Tm m

m m

j m k tm k

m

x t e dt c e e dt c e dt

T c e dt Tc t Tt

= =

=

= =

= = =

0

1( ) , , 1,0,1,

Tjk t

kc x t e dt kT

= =

36

Cc cng thc cn li trong (2.12) c suy ra t (2.11).

Nh vy, vi (2.10) v (2.12) ta lun phn tch c mt tn hiu tun hon x(t) thnh tng tuyn tnh cc tn hiu iu ha c bn. iu c ngha ln trong ng dng, chng hn nh:

Thnh phn:

2 2 11 1 1 1 1 1 1 11

cos( ) sin( ) cos( ) vi v arctanb

a t b t A t A a ba

+ = = + =

c gi l n hi ca x(t). Cc thnh phn

2 2cos( ) sin( ) cos( ) vi , arctan , 1kk k k k k k k kk

ba k t b k t A k t A a b k

a + = = + = >

gi l a hi ca x(t). Phn tch n hi v a hi c s dng nhiu trong cc ngnh thuc lnh vc iu khin truyn ti in v in t cng sut, cng nh phn tch cc dao ng iu ha thnh phn ca tn hiu tun hon trong cc qu trnh vt l m hc, nhit hc, in, c

Tm nghim tun hon ca mt s phng trnh vi phn o hm ring m t qu trnh truyn sng, truyn nhit, nh:

2 2

2 20

u u

x y

+ =

, phng trnh Laplace vi nghim u(x,y)

2 2

2 20

u u

t x =

, >0 phng trnh truyn sng vi nghim u(x,t)

2

2 0u u

ct x

=

, c>0 phng trnh truyn nhit vi nghim u(x,t)

Lc nhiu vi tn s xc nh c trong tn hiu tun hon x(t).

Phn tch s giao thoa cc p ng xung trong h tuyn tnh.

Xp x mt tn hiu x(t) tun hon, lin tc tng on bng tng hu hn cc hm iu ha:

( )01

( ) cos( ) sin( )n

k kk

x t a a k t b k t =

= + + Ch rng khi , xung quanh im khng lin tc

t0 ca x(t), tng hu hn v phi vn l mt hm lin tc vi cc thnh phn dao ng c bin ln. Tng cc s hng n cng ln, bin dao ng ny cng ln. Hin tng c gi l hin tng Gibb (hnh 2.5).

Thit k tn hiu tun hon x(t) vi di tn s lm vic cho trc (bi ton ngc ca vic phn tch chui Fourier).

Ngoi ra, ta cn c th d dng kim chng c cc tnh cht sau ca php phn tch chui Fourier cho tn hiu tun hon x(t) theo (2.10) v (2.12):

Hnh 2.5: Hin tng Gibb

37

1) Nu x(t) l hm l th ak=0, k v nu x(t) l hm chn th bk=0, k=1,2,

2) Nu c ( ) ( ), 2T

x t x t t+ = th a0=a2=a4= =b2=b4= =0

3) Bnh phng sai lch:

( )20

( ) ( )T

Q x t y t dt= gia hm x(t) cho trc v hm y(t) xc nh bi:

( )01

( ) cos( ) sin( )n

k kk

y t a a k t b k t =

= + + s l nh nht, nu y(t) c cc h s ak , bk c tnh t x(t) theo (2.12).

4) Chui hm theo t v phi trong (2.10) s hi t u (uniformly) ti x(t) nu c:

( )1

k kk

a b

=

+ < tc l khi gii hn x(t) ca chui (2.10) cng l hm lin tc, kh vi, kh tch ging nh cc phn t ca chui.

Cui cng, chui Fourier cn p dng c cho c tn hiu tun hon khng lin

tc, c m hnh dng dy gi tr {xk}, k=,1,0,1,, trong xk=x(kTa) v Ta l chu

k trch mu t tn hiu lin tc x(t). V l tn hiu tun hon nn phi c xk+N=xk, k, trong N l chu k tun hon ca dy. Khi dy trn, hay hm m rng

( ) ( ) ( )x t x t s t=

cng c dng chui Fourier (2.10), nhng vi cc h s ck, ak, bk c

tnh theo:

1 1 1

00 0 0

1 2 2 2 2, cos( ) , sin( ) , 1,2,

N N N

i k i k ii i i

i ia x a x k b x k k

N N N N N

= = =

= = = =

21

0

1kiN j

Nk i

ic x e

N

=

=

Cc cng thc ny c suy ra t (2.12) bng cch thay du tch phn bng du tng , t bng a

iTiT

N= v dt bi a

TT

N= . Chng thng c gi l chui Fourier ri rc (DFS

Discret Fourier Series). Ni cch khc, bn cht ca DFS chnh l chui Fourier (2.10) c p dng cho tn hiu khng lin tc.

Ch : Tn gi chui Fourier ri rc y khng lin quan ti tnh cht min gi tr ca nh x nh ta phn loi chng trc. Ni cch khc, tn gi ri rc y khng hm rng min gi tr ca cc h s ca chui l tp im khng lin thng. Tn gi n gin ch mun ni rng chui Fourier (2.10) c p dng ring cho tn hiu

khng lin tc {xk}, k=,1,0,1,. Bi vy, cht ch v mt ngn t, ta nn gi n

l chui Fourier cho tn hiu khng lin tc thay v chui Fourier ri rc.

38

Php bin i Fourier

Chui Fourier (2.10) c ngha ng dng ln trong thc t, song li ch p dng c cho lp cc tn hiu tun hon. Nhm m rng kh nng ng dng ca chui Fourier cho c cc tn hiu khng tun hon x(t), v c gi t cng thc (2.12) tnh

h s ck, ngi ta a ra khi nim php bin i Fourier, nh ngha nh sau: Cho tn hiu x(t), khng phn bit l tun hon hay khng tun hon, cng nh l lin tc

hay khng lin tc. nh Fourier ca n, k hiu bi X(j), c hiu l:

( ) ( ) j tX j x t e dt

= vi nh x ngc 1( ) ( )2j tx t X j e d

= (2.13) tin cho vic trnh by, sau y ta s s dng k hiu:

: ( ) ( ) ( ) j tx t X j x t e dt

= 6F vi 1 1: ( ) ( ) ( )2j tX j x t X j e d

= 6F ch php bin i Fourier, tc l:

{ }( ) ( )x t X j=F cng nh { }1 ( ) ( )X j x t =F (2.14) Tng t nh chui Fourier, iu u tin m ta cn phi bn y l kh nng

hi t ca tch phn v hn trong (2.13).

iu kin tn ti nh Fourier: Hm x(t) phi c chun bc 1, tc l tch phn v hn th nht trong (2.13) phi hi t:

( )x t dt

< , hay x(t)L1 v khi cng s c:

( ) ( ) ( )j t j tx t e dt x t e dt x t dt

<

Nu x(t) khng lin tc ti t0 th nh ngc cng thc th hai trong (2.13)

cng ng ti t0, hm x(t) phi c gi tr ti t0 l:

0 0 01

( ) ( 0) ( 0)2

x t x t x t= + +

Tip theo, ta s kho st mt s tnh cht c bn ca php bin i Fourier (2.13) tin cho vic s dng sau ny.

1) Nu x(t) l hm chn th X(j) l hm thun thc (phn o bng 0) v nu x(t) l hm l th X(j) l hm thun o (phn thc bng 0).

Khng nh ny c suy ra mt cch n gin nh cng thc Euler:

( ) ( ) ( )cos( ) ( )sin( )j tX j x t e dt x t t dt j x t t dt

= =

39

v iu hin nhin rng tch phn c cn i xng ca hm l s bng 0.

2) Php bin i Fourier l tuyn tnh: 1 1

{ ( )} { ( )}n n

i i i ii i

a x t a x t= =

= F F , aiR 3) Php bin i Fourier l ni x (injective): x(t)y(t) F{x(t)}F{y(t)} 4) Nu c ( ) ( )x t x t= th cng c ( ) ( )X j X j = , trong a l k hiu ch s phc

lin hp ca a.

5) Nu x(t) c nh Fourier X(j) th:

6) y(t)=x(t) s c nh ( ) ( ) jY j X j e =

a) ( )( ) dx ty tdt

= s c nh ( ) ( )Y j j X j =

7) Parseval: Gia nng lng tn hiu x(t) v nh Fourier X(j) ca n c quan h:

2

2

1( ) ( ) ( ) ( ) ( )

2

1 1( ) ( ) ( ) ( )

2 2

1( )

2

j t

j t

x t dt x t x t dt x t X j e d dt

X j x t e dt d X j X j d

X j d

= = =

= = = =

(2.15) 8) RiemannLebesgue: nh Fourier X(j) l hm lin tc theo v lim ( 0X j

= .

Chng minh: Vi 0 th:

( )( ) ( ) ( ) ( )

( )

j tj t j j t

j t

X j x t e dt e x t e dt x t e dt

x t e dt

+

= = = =

=

Suy ra

2 ( ) ( ) ( ) ( ) ( )

( ) ( ) 0 khi

j t j t j tX j x t e dt x t e dt x t x t e dt

x t x t dt

=

Tnh lin tc ca X(j) c xc nhn bng cch vi mi >0 cho trc ta chn hai hng s dng a ,b sao cho:

( )4t a

x t dt

>

40

( ) ( )( )( ) ( ) ( ) ( )( ) ( 1) 2 ( ) sin

2

2 ( ) sin 2 ( ) sin2 2

2 ( ) 2 ( ) ( )2 2

2 2

j t j t

j t j t

t a t a

t a t a t a

X j X j x t e x t e dt

tx t e e dt x t dt

t tx t dt x t dt

tx t dt x t dt ab x t dt

+

> < 0. T mc 2.1.4 ta c bit rng G(j) trong (2.51) l nh Fourier ca g(t) th G(s) phi c bn knh hi t bng 0, tc l G(s) phi c tt c cc im cc nm bn tri trc o, hay hm truyn G(s) phi l mt hm bn.

Nh vy l ta nh ngha xong hm c tnh tn. Nhng hm c tnh tn i din cho tnh cht g ca h thng?. tr li cu hi ny, ta xt v d sau.

V d 2.25: ngha ca hm c tnh tn

Xt h c hm truyn:

2( 0,5)

( )1

sG s

s+

=

+

Kch thch h t trng thi 0 bng tn hiu iu ha u ( t ) = sin(t ) , vi nh Laplace

2 2( )U s s

=

+, ta c p ng:

Y(s) = 221

)5,0(2

+

+

+

sss

=

+

+

+

++ 2222

2

2 )21(

11

1

s

s

ss

Suy ra

y(t) = [ ]) cos() sin()21( 1

1 22

tte t

++

v khi t th do et 0 ta c:

y(t) = ) sin(1

4512

42

+

+

++t vi = arctan

221

+ (2.52)

Mt khc hm c tnh tn ca h l:

2 2 4

2 2 22( 0,5) 1 2 1 5 4

( )1 1 1 1

jjG j j ej

+ + + += = =

+ + + + (2.53)

vi 2arctan 1 2

=

+. Do nu so snh (2.52) vi (2.53) ta s i n c kt lun cho

trng hp t :

84

( ) ( ) sin( )y t G j t = + vi l gc pha ca G(j). S Tng qut ha kt lun ca v d 2.25 ta c nh l sau:

nh l 2.7: Nu kch thch mt h thng c hm truyn bn G(s) t trng thi 0, tc l ti thi im kch thch h c

1

1(0) (0)

(0) 0n

n

dy d yy

dt dt

= = = ="

bng tn hiu iu ha u(t)=ejt th khi t , tc l h ch xc lp, s c p ng y(t) c xc nh t hm c tnh tn G(j) nh sau:

( )( ) ( ) j ty t G j e += , vi gc pha = arcG(j) (2.54)

Trong cng thc (2.54) ta s dng k hiu arcG(j) ch gc pha ca G(j):

Im ( )

arc ( ) arctanRe ( )

G jG j

G j

=

v ReG(j) l phn thc, ImG(j) l phn o ca G(j). Ni cch khc:

G(j)= ReG(j)+jImG(j)

Chng minh:

Do trong nh l 2.7 c gi thit G(s) l hm bn nn G(j) l nh Fourier ca hm trng lng g(t). Khi , c hai tn hiu vo v ra u(t), y(t) u c nh Fourier U(j), Y(j) v hm c tnh tn nh ngha bi (2.51) cng l:

trng thi u 0

( )( )

( )Y j

G jU j

=

=

cng l m hnh m t h khi t theo ngha:

1( ) : ( ) ( ) { ( ) ( )}G j u t y t G j U j =6 F V ch xt h ch xc lp vi t nn ta c th xem cc tn hiu u(t), y(t) l xc nh vi t. Bi vy, nu kch thch h bng tn hiu tun hon u(t) chu k T u vo th cng vi nh Fourier U(j):

2

( ) 2 ( )kk

U j c kT

=

= vi 2

1( )

j tT

kc u t e dtT

= tn hiu y(t) u ra s c nh Y(j) l:

2

0

2 2 2( ) 2 ( ) ( ) 2 ( ) ( )

2 2 12 ( ) trong ( ) v ( )

n nn n

T jn tT

n n n nn

Y j c n G j c G jn nT T T

c n c c G jn c u t e dtT T T

= =

=

= =

= = =

85

Vy:

2

2

1 2( ) ( ) ( )

2

2 2 2( ) ( ) ( )

2 vi ( )

j t j tn

n

jn tj t Tn n

n n

jn tT

n n nn

y t Y j e d c n e dT

c G jn n e d c G jn eT T T

d e d c G jnT

=

= =

=

= =

= =

= =

iu khng nh tnh ng n ca nh l 2.7, rng tn hiu ra y(t) cng tun hon

vi chu k T v dn, n=,1,0,1, l cc h s chui Fourier ca n, tc l:

2

0

1( )

T jn tT

nd y t e dtT

= , n= ,1,0,1, S

Xy dng hm c tnh tn bng thc nghim

nh l 2.7, c bit l phn chng minh ca n, a ta n thut ton xc nh hm c tnh tn (cng l m hnh ch xc lp) cho h tuyn tnh, tham s hng, bng phng php thc nghim ch ng gm cc bc nh sau:

1) Thit k tn hiu u(t) tun hon vi chu k T chn trc c di tn s lm vic ln v xc nh cc h s chui Fourier cn, n=,1,0,1, ca n theo (2.12):

00

1( )

Tjn t

nc u t e dtT

= , n= ,1,0,1, , 0 2T

=

Ta cng c th chn trc hai dy gi tr thc ak, k=0,1, v bk, k=1,2, ri thit k u(t) tun hon vi chu k T theo:

0 0 01

( ) cos( ) sin( )n

k kk

u t a a k t b k t =

= + + Khi cc h s phc cn, n=,1,0,1, s l

( )0 0 1 , , 1,2, 2n n nc a c a jb n= = + = 2) Kch thch h u vo bng tn hiu u(t) va c thit k ri o tn hiu y(t) cng

tun hon vi chu k T u ra.

3) Tnh cc h s chui Fourier dn, n=,1,0,1, ca y(t). Nu kt qu o c

bc 2) ch l dy gm N gi tr y0, y1,, yN1 trong mt chu k, th cc h s dn, n=,1,0,1, s c tnh xp x nh cng thc DFS, tc l:

21

0

1niN j

Nn i

id y e

N

=

= , n=,1,0,1, Ch rng do dn trong cng thc trn l tun hon vi chu k N nn y ta ch

cn tnh N gi tr ca n trong mt chu k dn, n=0,1,,N1 l .

86

4) Tnh cc gi tr ca hm c tnh tn:

0( ) , , 1,0,1, n

n

dG jn n

c = =

ri biu din chng di dng th (binpha hoc Bode).

Cui cng, cng t cc kt qu phn tch trn, ta cn nhn thy l khi h c kch thch bng tn hiu iu ha:

u(t)=sin(t)

u vo, th vi ak=0, k v b1=1, b2=b3==0, tn hiu ra khi t s l:

( )( ) ( ) sin ( ) vi ( ) arc ( )y t G j t G j = + = th c tnh tn binpha

ng biu din hm c tnh tn G(j) di dng th theo tham s khi chy t 0 n trong h trc ta c trc tung ImG(j)v trc honh ReG(j) c gi l ng th c tnh tn binpha.

V d 2.26: Xy dng hm c tnh tn

Cho mt h thng c hm truyn

4

( )1

G ss

=

+

Hm c tnh tn ca h l

2 2

Re ( ) Im ( )

4 4 4( )

1 1 1G j G j

G j jj

= =

+ + +

Do c

[ReG(j)2]2 + [ImG(j)]2 = 4

nn khi chy t 0 n , th ca n s l na ng trn nm di trc honh (v khi hm G(j) lun c phn o nh hn 0) hnh 2.44.

Ngoi ra, v G (s ) cn l hm bn, nn vi th c tnh tn cho trong hnh 2.45 ta s xc nh c p ng y ( t ) ca h khi b kch thch t trng thi 0 bi tn hiu iu ha u ( t )=sin( t ) . Theo nh l 2.7, c bit t v d 2.25, th p ng l:

t : ( ) 2 2 sin( )4

y t t

= S


Xt h vi hm truyn

3

( )(1 2 )

G ss s

=

+

/4 =0 =

ReG ImG

21 3

10,61,52

Hnh 2.44: th c tnh tn ca v d 2.26.

22

87

H c hm c tnh tn

2 2

Re ( ) Im ( )

3 6 3( )

(1 ) 1 4 (1 4 )G j G j

G j jj j

= =

+ + +

v t ta c ng c tnh tn ca h cho trong hnh 2.45a). ng c tnh tn ny c mt ng tim cn l

ReG= 6 S


Cho h vi hm truyn

2

5( )

1 5 4G s

s s=

+ +

T G(s) ta c hm c tnh tn v cng l nh Fourier G(j) ca hm trng

lng g(t), v G(s) l hm bn (c hai im cc l 114

s = v s2=1):

2

2 2 4 2 4

5 5 20 25( )

1 5 4( ) 1 17 16 1 17 16G j j

j j

= =

+ + + + + +

th hm G(j) cho trong hnh 2.45b). Vi th ta thy khi s = 0,5 th

Re[G(j)] = 0 v |G(j)| = 2 Ni cch khc nu kch thch h t trng thi 0 bng tn hiu u(t)=sin(0,5t) th sau mt khong thi gian ln (t ), p ng ca h s l

( ) 2sin 0,52

y t t

= S

V d 2.29: S dng hm c tnh tn

Cho mt h thng c m t bi hm truyn 2

5( )

4 (1 0,5 )G s

s s=

+. Hy xc nh

im tn s 0 m ti c |G(j) |=1. H c hm c tnh tn:

ImG

=

s= 0,5

ReG

=0

5 3

0,10,3

1

2

6 ImG

ReG3

2

4

6

8

0,4

0,5

0,9

Hnh 2.45: th c tnh tn ca v d 2.27 v 2.28

a) b)

88

2

5( )

4 (1 0,5 )G j

j j

=

+

T hm c tnh tn ta c:

2

0 0

5( ) 1

4 (1 0,5 )G j

j j

= =

+

220

025

12 16

j j

+ =

6 4 20 0 08 16 25 0 + + = 0 = 1 S


Cho h vi hm truyn

2 3 4

1 2 3 4

( )1

bG s

a s a s a s a s=

+ + + +

v G(s) c gi thit l hm bn. Hy xc nh tn hiu u(t) sao cho sau mt khong thi gian t ln ( t ) p ng y(t) ca h cho trng hp c kch thch t trng thi 0 bng tn hiu s

c mt gc lch pha vi u(t) l = 900.

T G(s) ta c hm c tnh tn G(j) v cng l nh Fourier ca hm trng lng g(t), v G(s) l hm bn:

2 2 2 2

( ) ( )

Re ( ) Im ( )

s j

bA bBG j G s j

A B A BG j G j

=

= =

+ +

trong

A = 1a2 2 +a4

4 , B = a1 a3 3 .

Hnh 2.46 biu din ng c tnh tn ca h. Theo nh l 2.7 v cng thc (2.54),

nu c ( ) sj tu t e = u vo th p ng ca h khi t s l

( )arc ( )( ) ( ) s sj t G jsy t G j e +=

iu ni rng y(t) c mt gc lch pha vi u(t) l 2 = ta phi xc nh im

tn s s ca ( ) sj tu t e = sao cho:

Im ( )

arc ( )Re ( ) 2

ss

s

G jG j

G j

= = , ReG(js)=0

2 2

0bA

A B=

+ 1a2

2+a4 4 = 0

22 2 4

4

4

2sa a a

a

= S

ImG

=

s

ReG

=0

Hnh 2.46: th c tnh tn bin pha ca v d 2.30.

s

89

Trn y l nhng v d v vic xy dng ng c tnh tn binpha t hm truyn ca h thng. t ngc li vn l ta c bi ton phi tm hm truyn G(s) t ng th ca hm c tnh tn binpha G(j). Bi ton ngc ny ta thng gp khi phi thc hin cng vic m hnh ha h thng tuyn tnh bng phng php thc nghim, tc l sau khi quan st/o cc tn hiu vo/ra ca h thng ta c c th hm G(j) th bc tip theo l t th ca G(j)phi xc nh bc m , n cng nh

cc h s b0 , b1 , , bm v a0 , a1 , , an cho hm truyn (2.37).

Cha kha cho vic gii bi ton ngc l mi lin h gia dng ca th G(j) vi G (s ) ti nhng im tn s c bit m y ta quan tm hn c l hai im tn s =0 v = .

Trc ht ta xt mi lin h gia chng khi = . T G(s) cho trong cng thc (2.37) ta c

1

0 11

0 1

( ) ( ) ( ) ( )

( ) ( )

m mm n m

n nn

b j b j bG j j

a j a j a

+ + +=

+ + +

"" (2.55)

Suy ra:

nh l 2.8: Nu mt h thng tuyn tnh c hm truyn G(s) l hp thc th hm G(j) ti im tn s = s tha mn:

a) 0 nu

lim ( ) nu m

n

m nG j b

m na

90

nh l 2.9: th hm c tnh tn binpha G(j) dng (2.56) ti im tn s =0 v hm truyn G (s ) ca n c quan h:

a) Nu 0

lim ( )G j

l mt s thc hu hn khc 0, tc l th ca G(j) bt u

t mt im trn trc thc, th k= )(~

lim0

jG

v r=0. H c hm truyn vi

r=0 c gi l h c khu khuch i.

b) Nu 0

lim ( )G j

=0 th r0 c gi l h c khu tch phn.

2.2.6 th c tnh tn logarith - th Bode

Bn cnh vic biu din hm c tnh tn G(j) trong mt phng vi hai trc ta ReG(j) v ImG(j) nh mc 2.2.5 va trnh by ngi ta cn c mt phng php biu din khc l th c tnh logarith, hay cn gi biu Bode. y l cch biu din G(j)thnh hai th ring bit theo cho:

1) bin , hay gi tr logarith ca |G(j) | l L() = 20 lg |G(j) | , c n v l Dezibel (dB), 2) v pha, hay gi tr gc () = arcG(j) c n v l Grad.

C hai th ny u c trc honh l song khng c chia u theo gi tr ca m li theo lg(). L do cho vic chia ny l trong mt khong din tch v tng i nh, ta vn c c th minh ha y cho h thng thng qua G(j) cho mt di tn s rt ln, cng nh cng vic xy dng th ca

1 21 2

(1 )(1 ) (1 )( )

(1 )(1 ) (1 )m

n

T j T j T jG j k

T j T j T j

+ + +=

+ + +

/ / /"" (2.57)

c thc hin mt cch n gin l cng tr cc th thnh phn:

/ /1 1 1 1

( ) 20 lg 1 lg 1 20 lg ( ) ( )m n m n

k k k kk k k k

L T j T j k L L = = = =

= + + = +

trong

/ /( ) 20 lg 1k kL T j = + v ( ) 20 lg 1k kL T j = +

V d 2.31: Xy dng biu Bode ca khu khuch i

Khu khuch i c hiu l mt h ng hc c bn c hm truyn

G(s) = k

Khu ny c hm c tnh tn:

91

G(j)= k

Do (hnh 2.47):

L() = 20lg |k | v 0 nu 0( )180 nu 0

k

k =

92

V d 2.33: Xy dng biu Bode ca khu khuch ivi phn

Khu khuch ivi phn l mt h ng hc c bn c hm truyn

G(s) = 1+Ts

Khu ny c hm c tnh tn:

G(j)= 1 + Tj Suy ra

L() = 10lg(1+T22 ) v () = arctanT .

Nh vy, th L () c hai tim cn ng vi khi 0 v khi l

( )0 khi 0

( )20 lg lg khi

LT

=

+

Chng ct nhau ti im tn s 1

G T = c gi l tn s gy. ng tim cn th hai

ng vi trng hp l ng thng c dc 20 dB/dec. Hnh 2.49 biu din biu Bode ca khu cho. S

1 dec

L()

0

3dB

Hnh 2.48: Biu Bode ca khu qun tnh bc nht cho trong v d 2.32.

()

45

20dB

0

90

T= 0,1T=1T=10

T=0,1 T=1T=10

10G =1 100,1 1

L()

03dB

Hnh 2.49: Biu Bode ca khu cho trong v d 2.33.

()

45

20dB

0

90

T=0,1T=1T=10

T=0,1T=1T=10

1 dec 10 100,1 1G =1

93

V d 2.34: Xy dng biu Bode ca khu tch phn

Khu tch phn l h ng hc c bn c hm truyn 1

( )I

G sT s

= . Khu ny c

1

( )I I

jG j

T j T

= = 1( ) 20 lg lgI

LT

= v ( )

2 =

Suy ra biu Bode ca n c dng nh hnh 2.50. ng th L() l ng thng vi dc 20dB/dec. N s ct trc honh (l trc m ti L() c gi tr 0) ti

Lý Thuyết Điều Khiển Tuyến Tính - GS.TS. Nguyễn Doãn Phước - NXB KH&KT 2009

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Transcript of Lý Thuyết Điều Khiển Tuyến Tính - GS.TS. Nguyễn Doãn Phước - NXB KH&KT 2009