chuong 1_Điều khiển tối Ưu

Chng 1 : iu khin ti u

Trang 5

Chng 1

IU KHIN TI U

Vi nt lch s pht trin l thuyt iu khin . - Phng php bin phn c in Euler_Lagrange 1766 . - Tiu chun n nh Lyapunov 1892 . - Tr tu nhn to 1950 . - H thng iu khin my bay siu nh 1955 . - Nguyn l cc tiu Pontryagin 1956 . - Phng php quy hoch ng Belman 1957 . - iu khin ti u tuyn tnh dng ton

phng LQR ( LQR : Linear Quadratic Regulator ) .

- iu khin kp Feldbaum 1960 . - Thut ton di truyn 1960 . - Nhn dng h thng 1965 . - Logic m 1965 . - Lut iu khin h thng thch nghi m hnh tham chiu MRAS v b t

chnh nh STR 1970 ( MRAS : Model-Reference Adaptive System , STR : Self-Tuning Regulator ) .

- H t hc Tsypkin 1971 . - Sn phm cng nghip 1982 . - L thuyt bn vng 1985 . - Cng ngh tnh ton mm v iu khin tch hp 1985 .


Trang 6

1.1 CHT LNG TI U 1.1.1 c im ca bi ton ti u 1. Khi nim Mt h iu khin c thit k ch lm vic tt nht l h lun trng thi ti u theo mt tiu chun cht lng no ( t c gi tr cc tr ) . Trng thi ti u c t c hay khng ty thuc vo yu cu cht lng t ra , vo s hiu bit v i tng v cc tc ng ln i tng , vo iu kin lm vic ca h iu khin Mt s k hiu s dng trong chng 1 .

Hnh 1.1: S h thng iu khin .

H thng iu khin nh hnh trn bao gm cc phn t ch yu : i tng iu khin ( TK ) , c cu iu khin ( CCK ) v vng hi tip ( K ) . Vi cc k hiu : x0 : tn hiu u vo u : tn hiu iu khin x : tn hiu u ra = x0 x : tn hiu sai lch f : tn hiu nhiu Ch tiu cht lng J ca mt h thng c th c nh gi theo sai lch ca i lng c iu khin x so vi tr s mong mun x0 , lng qu iu khin ( tr s cc i xmax so vi tr s xc lp ( )x tnh theo phn trm ) , thi gian qu hay theo mt ch tiu hn hp trong iu kin lm vic nht nh nh hn ch v cng sut , tc , gia tc Do vic chn mt lut iu khin v c cu iu khin t c ch lm vic ti u cn ty thuc vo lng thng tin ban u m ta c c . y chng ta c th thy c s khc bit ca cht lng ti u khi lng thng tin ban u thay i ( Hnh 1.2 ) .


Trang 7

Hnh 1.2 : Ti u cc b v ti u ton cc .

Khi tn hiu iu khin u gii hn trong min [u1,u2] , ta c c gi tr ti u cc i 1J

ca ch tiu cht lng J ng vi tn hiu iu khin 1u

.

Khi tn hiu iu khin u khng b rng buc bi iu kin 1 2u u u , ta c c gi tr ti u 2 1J J

> ng vi 2u . Nh vy gi tr ti u thc s

by gi l 2J .

Tng qut hn , khi ta xt bi ton trong mt min [ ],m nu u no v tm c gi tr ti u iJ

th l gi tr ti u cc b . Nhng khi bi ton

khng c iu kin rng buc i vi u th gi tr ti u l ( )iJ extremum J = vi iJ l cc gi tr ti u cc b , gi tr J chnh l

gi tr ti u ton cc . iu kin tn ti cc tr : o hm bc mt ca J theo u phi bng 0 :

0=

u

J

Xt gi tr o hm bc hai ca J theo u ti im cc tr :

022

>

u

J : im cc tr l cc tiu

022

uuL (1.6)

Nu Luu l xc nh m th im cc tr chnh l im cc i ; cn nu Luu l khng xc nh th im cc tr chnh l im yn nga . Nu Luu l bn xc nh th chng ta s xt n thnh phn bc cao hn trong (1.1) xc nh c loi ca im cc tr . Nhc li : Luu l xc nh dng ( hoc m ) nu nh cc gi tr ring ca n l dng ( hoc m ) , khng xc nh nu cc gi tr ring ca n va c dng va c m nhng khc 0 , v s l bn xc nh nu tn ti gi tr ring bng 0 . V th nu 0=uuL , th thnh phn th hai s khng hon ton ch ra c loi ca im cc tr . 2. Ti u ha vi cc iu kin rng buc Cho hm ch tiu cht lng v hng ( )uxL , , vi vector iu khin

mRu v vector trng thi nRx . Bi ton a ra l chn u sao cho hm ch tiu cht lng L(x,u) t gi tr nh nht v tha mn ng thi cc phng trnh iu kin rng buc . ( ) 0, =uxf (1.7) Vector trng thi x c xc nh t mt gi tr u cho trc bng mi quan h (1.7) , v th f l mt h gm n phng trnh v hng , nRf . tm iu kin cn v ca gi tr cc tiu , ng thi tha mn

( ) 0, =uxf , ta cn lm chnh xc nh trong phn trc . u tin ta khai trin dL di dng chui Taylor , sau xc nh s hng th nht v th hai . Tha s Lagrange v hm Hamilton . Ti im cc tr , dL vi gi tr th nht bng 0 vi mi s bin thin ca du khi df bng 0 . Nh vy chng ta cn c: 0=+= dxLduLdL Tx

Tu (1.8)

v: 0=+= dxfdufdf xu (1.9)


Trang 13

T (1.7) ta xc nh c x t gi tr u c, bin thin dx c xc nh bi (1.9) t gi tr bin thin du c . Nh vy , ma trn Jacobi fx khng k d v : duffdx ux 1= (1.10) Thay dx vo (1.8) ta c : duffLLdL uxTxTu )( 1= (1.11) o hm ring ca L theo u cha hng s f c cho bi phng trnh :

( ) xTxTuuTuxTxTudf

LffLffLLu

L

=

==

1

0

(1.12)

vi ( )TxTx ff 1 = . Lu rng : u

dx

Lu

L=

=0

(1.13)

thnh phn th nht ca dL bng khng vi gi tr du ty khi 0=df , ta cn c :

0= xT

x

Tuu LffL (1.14)

y l iu kin cn c gi tr cc tiu . Trc khi i tm iu kin , chng ta hy xem xt thm mt vi phng php c c (1.14) . Vit (1.8) v (1.9) di dng:

0=

=

dudx

ffLL

dfdL

ux

Tu

Tx

(1.15)

H phng trnh tuyn tnh ny xc nh mt im dng , v phi c mt kt qu [ ]TTT dudx . iu ny ch xy ra nu ma trn h s ( ) ( )mnn ++1 c hng nh hn n+1 . C ngha l cc hng ca ma trn tuyn tnh vi nhau tn ti mt vector c n s hng nh sau:

[ ] 0.1 =

ux

Tu

TxT

ffLL (1.16)

Hay:

0=+ xTT

x fL (1.17) 0=+ u

TTu fL (1.18)


Trang 14

Gii (1.17) ta c :

1= x

Tx

T fL (1.19) v thay vo (1.18) c c (1.14) . Vector nR c gi l tha s Lagrange , v n s l cng c hu ch cho chng ta sau ny . hiu thm ngha ca tha s Lagrange ta xt du = 0 , t (1.8) v (1.9) ta kh dx c : dffLdL xTx 1= (1.20) V vy:

( ) ==

=

Tx

Tx

du

fLfL 1

0

(1.21)

Do - l o hm ring ca L vi bin iu khin u l hng s . iu ny ni ln tc dng ca hm ch tiu cht lng vi bin iu khin khng i khi iu kin thay i . Nh l mt cch th ba tm c (1.14) , ta pht trin thm s dng cho cc phn tch trong nhng phn sau . Kt hp iu kin v hm ch tiu cht lng tm ra hm Hamilton . ( ) ( ) ( )uxfuxLuxH T ,,,, += (1.22) Vi nR l tha s Lagrange cha xc nh . Mun chn x , u , c c im dng , ta tin hnh cc bc sau . bin thin ca H theo cc bin thin ca x , u , c vit nh sau : dHduHdxHdH TTuTx ++= (1.23)

Lu rng : ),( uxfHH =

= (1.24)

Gi s chng ta chn cc gi tr ca u tha mn : 0=H (1.25) Sau ta xc nh x vi gi tr ca u c bng phng trnh iu kin rng buc ( ) 0, =uxf . Trong trng hp ny hm Hamilton tng ng vi hm ch tiu cht lng:


Trang 15

LH f ==0 (1.26) Nhc li : nu f = 0 , ta s tm c dx theo du t (1.10) . Ta khng nn xt mi quan h gia du v dx thun tin trong vic chn sao cho : 0=xH (1.27) Sau , t (1.23) , bin thin dH khng cha thnh phn dx. iu ny mang li kt qu :

0=+= Txx fL

x

H (1.28)

hay 1= xTx

T fL . Nu gi nguyn (1.25) v (1.27) th: duHdHdL Tu== (1.29) V H = L, c c im dng ta phi p t iu kin: 0=uH (1.30) Tm li , iu kin cn c c im cc tiu ca L(x,u) tha mn iu kin rng buc f(x,u) = 0 gm c : 0==

fH (1.31a)

0=+= Txx fL

x

H (1.31b)

0=+= Tuu fL

u

H (1.31c)

Vi ( ),,uxH xc nh bi (1.22) . Cch thng dng l t 3 phng trnh cho xc nh x , , v u theo th t tng ng . So snh 2 phng trnh (1.31b) v (1.31c) ta thy chng tng ng vi 2 phng trnh (1.17) v (1.18) . Trong nhiu ng ng , chng ta khng quan tm n gi tr ca , tuy nhin ta vn phi i tm gi tr ca n v l mt bin trung gian cho php chng ta xc nh cc i lng cn tm l u , x v gi tr nh nht ca L . u im ca tha s Lagrange c th tm tt nh sau : trn thc t , hai i lng dx v du khng phi l hai i lng bin thin c lp vi nhau , theo (1.10) . Bng cch a ra mt tha s bt nh , chng ta chn sao cho dx v du c th c xem l hai i lng bin thin c lp vi nhau .


Trang 16

Ly o hm ring ca H ln lt theo cc bin nh trong (1.31) , nh th ta s c c im dng . Khi a ra tha s Lagrange , chng ta c th thay th bi ton tm gi tr nh nht ca L(x,u) vi iu kin rng buc f(x,u) = 0 , thnh bi ton tm gi tr nh nht ca hm Hamilton H(x,u,) khng c iu kin rng buc .

iu kin (1.31) xc nh mt im dng . Ta s tip tc chng minh y l im cc tiu nh thc hin trong phn trc . Vit chui Taylor m rng cho bin thin ca L v f nh sau :

[ ] [ ] )3(21 O

dudx

LLLL

dudxdudx

LLdLuuux

xuxxTTTu

Tx +

+

= (1.32)

[ ] [ ] )3(21 O

dudx

ffff

dudxdudxffdf

uuux

xuxxTTux +

+

= (1.33)

Vi:

xu

ff xu

=

2

a ra hm Hamilton , ta s dng cc phng trnh sau :

[ ] [ ] [ ] )3(211 O

dudx

HHHH

dudxdudx

HHdfdL

uuux

xuxxTTTu

Tx

T +

+

=

(1.34) By gi , c c im dng ta cn c 0=f , v ng thi thnh phn th nht ca dL bng 0 vi mi s bin thin ca dx v du . V 0=f nn 0=df , v iu ny i hi 0=xH v 0=uH nh trong (1.31) . tm iu kin cho im cc tiu , chng ta xt n thnh phn th hai . u tin , ta cn xem mi quan h gia dx v du trong (1.34) . Gi s rng chng ta ang im cc tr nn 0=xH , 0=uH v 0=df . Sau , t (1.33) ta c : )2(1 Oduffdx ux += (1.35)


Trang 17

Thay vo (1.34) ta c :

[ ] )3(21 1 Odu

Iff

HHHH

IffdudL uxuuux

xuxxTx

Tu

T +

=

(1.36)

m bo y l im cc tiu , dL trong (1.36) phi dng vi mi s bin thin ca du . iu ny c m bo nu nh ma trn un vi f lun bng 0 l xc nh dng .

[ ]uxxx

Tx

Tuuxuxxu

Tx

Tuuu

ux

uuux

xuxxTx

Tufuu

fuu

ffHffffHHffHI

ffHHHH

IffLL11

1

+=

==

(1.37)

Lu rng nu iu kin rng buc ( ) 0, =uxf vi mi x v u th (1.37) c rt li thnh Luu phng trnh (1.6) . Nu (1.37) l xc nh m ( hoc khng xc nh ) th im dng s l im cc i ( hoc im yn nga ) .

1.1.3 V d Ti u ha khng c iu kin rng buc V d 1.1 : Khng gian ton phng . Cho 2Ru v :

[ ]ussuqqqq

uuL T 212212

1211

21)( +

= (1)

uSQuu TT +=

21

(2)

im cc tr c xc nh bi : 0=+= SQuLu (3) SQu 1 = (4) vi u* dng ch bin iu khin ti u. Loi ca im cc tr c xc nh bng cch xt ma trn hessian QLuu = (5)


Trang 18

im u* l cc tiu nu Luu > 0 ( 011 >q v 02122211 > qqq ) . L im cc i nu Luu < 0 ( 011 qqq ) . Nu 0 0 . T (6) ta thy rng gi tr nh nht ca L l L* = -1/2 . Cc ng ng mc ca L(u) trong (7) c v trong Hnh 1.4 , vi u = [u1 u2]T . Cc mi tn l gradient .

++

+=+=

12 2121

uu

uuSQuLu (9)

Lu rng gradient lun lun vung gc vi cc ng ng mc v c hng l hng tng L(u) . Chng ta dng du * ch gi tr ti u ca u v L cn tm . Tuy nhin ta thng b qua du * .


Trang 19

Hnh 1.4 : Cc ng ng mc v vector gradient .

V d 1.2 : Ti u ha bng tnh ton v hng . Phn trn chng ta cp phng php gii bi ton ti u bng cch s dng cc vector v gradient . Sau y ta s tip cn bi ton vi mt cch nhn khc , xem chng nh l nhng i lng v hng . chng minh , ta xt :

22221

2121 2

1),( uuuuuuuL +++= (1)

Vi 21 ,uu l cc i lng v hng . im cc tr xut hin khi o hm ring ca L theo tt c cc i s phi bng 0 :

0211

=+=

uuu

L (2a)

012 212

=++=

uuu

L (2b)


Trang 20

Gii h phng trnh trn ta c : 1,1 21 == uu (3) Vy , im cc tr l (1 ,-1) . Biu thc (1) l mt dng m rng ca biu thc (7) trong v d 1.1 , nh vy chng ta va tm c mt kt qu tng t bng mt cch khc .

Ti u ha c iu kin rng buc V d 1.3 : Khng gian ton phng vi iu kin rng buc tuyn tnh . Gi s hm ch tiu cht lng c cho bi v d 1.1 vi cc i lng v hng 21 ,uu c thay th bng ux, :

[ ] [ ]

+

=

u

x

u

xuxuxL 10

2111

21),( (1)

Vi iu kin rng buc : ( ) 03, == xuxf (2) Hm Hamilton s l :

)3(21 22

++++=+= xuuxuxfLH T (3)

vi l mt i lng v hng . iu kin c im dng theo (1.31) l : 03 == xH (4) 0=++= uxH x (5) 012 =++= uxH u (6) Gii (4) , (5) , (6) ta c : x = 3 , u = -2 , = -1 . im dng l : ( ) ( )2,3, =ux (7) xc nh (7) l im cc tiu , tm ma trn un theo (1.37) : 2=fuuL (8)

0>=fuuL , v th ( ) ( )2,3, =ux l im cc tiu . Cc ng ng mc ca L(x,u) v iu kin rng buc (2) c v trong Hnh 1.5 . Grad ca f(x,u) trong h ta (x,u) c vit nh sau:


Trang 21

=

01

u

x

ff

(9)

c v trong Hnh 1.4 . V grad ca L(x,u) :

++

+=

12uxux

LL

u

x (10)

Ti im cc tiu (3,-2) , grad L(x,u) s c gi tr :

=

01

u

x

LL

(11)

Cn lu rng gradf v gradL tng ng vi nhau ti im dng . C ngha l im cc tiu xut hin khi iu kin rng buc (2) l ng tip tuyn ca cc ng ng mc ca L. Di chuyn hng dc theo ng thng f = 0 s lm tng gi tr ca L . Ta tm c gi tr ca L ti im cc tiu bng cch thay x = 3, u = -2 vo (1) , ta c L*=0,5 . V = -1 , gi nguyn gi tr u = -2 , thay i iu kin rng buc df ( dch chuyn ng thng trong Hnh 1.5 v pha phi ) s lm tng L(x,u) vi dL = -df = df . V d 1.4 : Hm ch tiu cht lng dng ton phng vi iu kin rng buc tuyn tnh - Trng hp v hng . Xt hm ch tiu cht lng dng ton phng :

+= 2

2

2

2

21),(

by

a

xuxL (1)

Vi iu kin rng buc tuyn tnh : ( ) cmuxuxf +=, (2) Cc ng ng mc ca L(x,u) l nhng ellip ; nu L(x,u) = F/2 , th bn trc chnh v bn trc ph l al v bl . iu kin rng buc f(x,u) l mt h cc ng thng cha thng s c . Xem Hnh 1.6 ( lu rng u l bin c lp , vi x c xc nh bi f(x,u) = 0 ) . Hm Hamilton l :

)(21

2

2

2

2

cmuxbu

a

xH ++

+= (3)


Trang 22

V iu kin c im dng : 0=+= cmuxH (4)

02 =+= axH x (5)

02 =+= mbuH u (6)

Hnh 1.5 : Cc ng ng mc ca L(x,u) v iu kin rng buc f(x,u) .

Hnh 1.6 : Cc ng ng mc ca L(x,u) v iu kin rng buc f(x,u).


Trang 23

gii h phng trnh ny , trc ht ta s dng phng trnh (6) a ra bin iu khin ti u theo tha s Lagrange . mbu 2= (7) By gi thay phng trnh (7) vo (4) kh u , kt hp vi (5) v c vit li :

=

0111

2

22cx

a

mb

(8)

Gii ra ta c gi tr ca im dng :

222

2

mbaca

x+

= (9)

222mba

c

+= (10)

Thay (9) , (10) vo (7) , ta c c gi tr u ti u :

222

2

mbamcb

u+

= (11)

xc nh im dng l cc tiu , dng (1.37) tm ra ma trn un :

2

2

21

a

m

bL fuu += (12)

0>fuuL v vy ta tm c mt im cc tiu .

Thay (9) v (11) vo (1) ta c gi tr ti u ca hm ch tiu cht lng :

222

2*

21

mbacL

+= (13)

kim chng (1.21) , lu rng:

=

=

=

c

LfL

du

*

0

*

(14)

Gradf trong min (u,x) l :

=

1m

ff

x

u (15)


Trang 24

c biu din trong Hnh 1.6 . GradL l :

=

2

2

a

xbu

LL

x

u (16)

v ti im dng (11) , (9) s c gi tr :

222

*

1 mbacm

LL

x

u

+

=

(17)

iu ny tng ng vi (15) , v vy im dng xut hin khi f(x,u) = 0 l ng tip tuyn vi mt ng ng mc ca L(x,u) . V d 1.5 : Hm ch tiu cht lng dng ton phng vi iu kin rng buc tuyn tnh . By gi ta tng qut ha v d 1.4 vi vector nRx , mRu , nRf ,

nR . Xt hm ch tiu cht lng dng ton phng:

RuuQxxL TT21

21

+= (1)

vi iu kin rng buc tuyn tnh : 0=++= cBuxf (2) vi Q , R v B l cc ma trn , c l vector n hng . Gi s Q 0 v R > 0 ( vi Q , R l ma trn i xng ) . Cc ng ng mc ca L(x,u) l cc ng ellip trong khng gian , v f(x,u)=0 l mt phng ct ngang qua chng . im dng xut hin khi gradf v gradL song song vi nhau . Hm Hamilton l :

)(21

21

cBuxRuuQxxH TTT ++++= (3)

v cc iu kin c im dng l : 0=++= cBuxH (4) 0=+= QxH x (5) 0=+= Tu BRuH (6)


Trang 25

gii cc phng trnh trn , u tin ta dng iu kin (6) tm u theo : TBRu 1= (7) T (5) ta c : Qx= (8) Kt hp vi (4) ta c : QcQBu += (9) dng kt qu ny thay vo (7) cho ta : )(1 QcQBuBRu T += (10) hay :

( ) QcBRuQBBRI TT 11 =+ ( ) QcBuQBBR TT =+ (11) V R > 0 v BTQB 0 , chng ta c th tm nghch o ca (R + BTQB) v v th gi tr u ti u l : QcBQBBRu TT 1)( += (12) So snh kt qu ny vi (11) trong v d 1.4 . Thay (12) vo (4) v (9) cho ta gi tr trng thi ti u v tha s Lagrange ti u :

( )( )1T Tx I B R B QB B Q c= + (13) ( )( )1T TQ QB R B QB B Q c = + (14) Bng b ca nghch o ma trn :

( ) cBBRQ T 111 += (15) nu 0Q . Cc kt qu trn s rt li thnh kt qu ca v d 1.4 trong trng hp v hng . xc nh bin iu khin (12) l mt cc tiu , ta s dng (1.37) xc nh ma trn un l xc nh dng vi gi tr ca R v Q c gii hn . QBBRL Tfuu += (16) S dng (12) v (13) th vo (1) ta c c gi tr ti u :


Trang 26

( )[ ]cQBQBBRQBQcL TTT 121

*

+= (17)

TcL21

* = (18)

V th :

=

c

L * (19)

V d 1.6 : Bi ton vi nhiu iu kin rng buc . Tm khong cch nh nht gia parabol : dbxaxy ++= 2 (1) vi ng thng : cxy += (2) Xem Hnh 1.7 . Trong bi ton ny s c hai iu kin rng buc : 0),( 1211111 == dbxaxyyxf (3) V : 0),( 22222 == cxyyxf (4) vi ( )11 , yx l 1 im trn parabol v ( )22 , yx l 1 im trn ng thng . Chng ta chn hm ch tiu cht lng l mt na ca bnh phng khong cch gia 2 im ny .

221

2212121 )(2

1)(21),,,( yyxxyyxxL += (5)

gii bi ton ny , ta x l bng cch t :

=

=

=

2

1

2

1

2

1,,

yy

ux

xxf

ff (6)

v s dng cch tip cn vector ; tuy nhin , s kt hp gia mt iu kin rng buc tuyn tnh v mt iu kin phi tuyn s lm phc tp thm bi ton . Thay vo ta s s dng cc i lng v hng .


Trang 27

Hnh 1.7 : Bi ton vi nhiu iu kin rng buc .

a ra mt tha s Lagrange cho mi iu kin rng buc , hm Hamilton l :

)()()(21)(

21

22212111

221

221 cxydbxaxyyyxxH +++=

(7) Khi , c im dng ta cn c : 02 111211 == bxaxxH x (8) 02212 =+= xxH x (9) 01211 =+= yyH y (10) 02212 =++= yyH y (11) 01

2111 == dbxaxyH (12)

0222 == cxyH (13)


Trang 28

Gii (12) c c 1y nh sau : dbxaxy ++= 1211 (14) T (9) v (11) , ta c : 21122 yyxx == (15) v s dng (14) vi cxy += 22 t (13) c c kt qu sau : cxdbxaxxx ++= 212112 (16) Khi :

( )cdxbaxx +++= 1212 )1(21

(17)

Theo (10) v (11) , 1 = -2 , vy t (15) v (17) ta c : 211 xx =

( )cdxbax ++= 1211 )1(21 (18)

Cui cng , ch rng (8) l : ( )( ) 012 11 =+ bax (19) hay : ( )( ) 0)1()1(2 1211 =+++ cdxbaxbax (20) Phng trnh bc 3 (20) c gii c gi tr ti u *1x t gi tr a, b, c, d cho trc . Nu ng thng ct ngang qua parabol th giao im s l kt qu ti u ( khi 1=2=0 ) ; ngc li , s c ch mt cp gn nhau nht (x1,x2) , (y1,y2) . Mt khi tm c x1 th ta s tm c x2 , y1 v y2 ln lt theo cc phng trnh (17) , (14) v (15) . Thay cc gi tr ti u ny vo (5) s cho chng ta khong cch ngn nht l *2L .


Trang 29

1.2 CC PHNG PHP IU KHIN TI U 1.2.1 Phng php bin phn c in Euler_Lagrange 1. Gii thiu Nhim v ca iu khin ti u l gii bi ton tm cc tr ca phim hm

[ ( ), ( )]L x t u t bng cch chn tn hiu iu khin u(t) vi nhng iu kin hn ch ca i lng iu khin v ta pha . Mt trong nhng cng c ton hc xc nh cc tr l phng php bin phn c in Euler_Lagrange . ng cc tr l nhng hm trn cn phim hm cng cc iu kin hn ch l nhng hm phi tuyn . Do phng php ny khng th p dng cho nhng trng hp m tn hiu iu khin c th l cc hm gin on . Trng hp khng c iu kin rng buc Cho u(t) l hm thuc lp hm c o hm bc nht lin tc . Trong mt phng (u,t) cho hai im (t0,u0) v (t1,u1) . Cn tm qu o ni hai im ny sao cho tch phn theo qu o )(tuu = cho bi :

=1

0

),,()(t

t

dttuuLuJ (1.38)

c cc tr . L l hm c o hm ring bc mt v bc hai lin tc vi mi bin ca n . thng nht , y ta ly t0 = 0 v t1 = T . Bin i ca J do u to nn l :

)()()( uJuuJuuJ +=+

++=T T

dttuuLdttuuuuL0 0

),,(),,(

dttuuLtuuuuLT

++=0

)],,(),,([ (1.39)

Phn tch (1.39) theo chui Taylor v ch kho st thnh phn bc mt ca J ta c :

dtuu

tuuLu

u

tuuLuuJ

T

])),,(()),,(([),(0

+

= (1.40)


Trang 30

v u v u lin h nhau bi :

)0()()(0

udttutuT

+=

Xem u l hm bin i c lp , biu thc (1.40) c th bin i ch cha u bng cch ly tch phn nhng thnh phn cha u :

...]),,(),,([),,(),(0

0 udtutuuL

dtd

u

tuuLu

u

tuuLuuJ

TT

+

= (1.41)

T iu kin cho u(0) = (T) = 0 , phn u ca v phi biu thc (1.41) bng 0 . Nu gia s J ca ch tiu cht lng J tn ti v nu J c cc tr i vi u* th : 0),( * = uuJ (1.42) l iu kin c bn ca php tnh bin phn . T cc biu thc (1.41) , (1.42) ta c :

0]),,(),,([),(****

0

*=

= udtutuuL

dtd

u

tuuLuuJ

T

(1.43)

T c th rt ra phng trnh Euler_Lagrange :

0),,(),,( =

u

tuuLdtd

u

tuuL

(1.44a)

Hoc c th vit n gin :

0=

u

Ldtd

u

L

(1.44b)

Trng hp c iu kin rng buc Nu ngoi ch tiu cht lng (1.38) cn c cc iu kin rng buc dng : 0),,( =tuui [0, ]t T , 1,i n= (1.45) th ch tiu cht lng J c dng :

=

+=T

i

n

iiia dttuuttuuLuJ

0 1)],,()(),,([),(

(1.46)


Trang 31

m i(t) vi i = 1,2,,n l hm Lagrange .V gii hn tha mn vi mi t nn hm Lagrange ph thuc thi gian . Tng t nh trn ta c phng trnh Euler_Lagrange tng qut :

0),,,(),,,( =

u

tuuLdtd

u

tuuL aa

(1.47)

m ),,()(),,(),,,(1

tuuttuuLtuuL in

iia

=

+= (1.48)

Khi iu kin rng buc c dng :

=T

ii qdttuu0

),,( (1.49)

th phng trnh Euler_Lagrange tng qut (1.47) c phim hm :

),,(),,(),,,(1

tuutuuLtuuLn

iia

=

+= (1.50)

Trong trng hp ny , i l cc h s khng ph thuc thi gian . Khi c iu kin rng buc dng (1.45) hoc (1.49) phi gii (n+1) phng trnh xc nh y*(t) v i*(t) vi i=1,2,,n . Phng trnh Euler_Lagrange vi tn hiu iu khin b hn ch Trong phn trn ta ch cp ti bi ton m trong tn hiu iu khin khng c gii hn no rng buc . Trong thc t , thng gptn hiu iu khin c rng buc dng 1u .

iu kin cn c cc tr : khi u(t) l ng cc tr th u+u v u-u l nhng hm cho php . By gi ta so snh tr s phim hm ng cc tr vi tr s ca n hm u+u v u-u . Nu min bin i ca u(t) l kn v u(t) ngoi bin th mt trong cc hm u+u hoc u-u s ra ngoi min cho php . Mt trong cc bin php khc phc kh khn trn l ng cc tr bin v : )(tu (1.51) V d , nu 1u , iu kin )(tu ngha l 1)( t . i bin ta c :

2z u = (1.52)


Trang 32

th bin mi z s khng c iu kin hn ch v bin gii ca bin u tng

ng vi z = 0 . By gi ch tiu cht lng =T

dttuuLuJ0

),,()( c bin

mi u = z2 + , t : 2u zz = +

v ch tiu cht lng J c dng :

++=T

dttzzzLJ0

2 ],2,[ (1.53)

V khng c iu kin hn ch nn phng trnh Euler_Lagrange c dng :

0=

z

Ldtd

z

L

(1.54)

y zu

Lz

u

Lz

u

u

Lz

u

u

Lz

L

22

+

=

+

=

zu

Lz

u

u

Lz

u

u

Lz

L 2

=

+

=

zu

Lu

Ldtd

zz

Ldtd

2)(2

+

=

v (1.54) s c dng :

02222 =

+

zu

Lu

Ldtd

zzu

Lz

u

L

hay : 02 =

u

Ldtd

u

Lz

(1.55)

Phng trnh trn tha mn vi z = 0 , ngha l ng cc tr c nhng gi tr bin v phng trnh Euler_Lagrange vn l phim hm xut pht :

0=

u

Ldtd

u

L

2. V d V d 1.7 :

Tm qu trnh ti u * 2x u= v * 12

duu

dt= cc tiu ha ch tiu cht lng

J :


Trang 33

22

0

( ) ( )T

J u u dt= (1)

vi iu kin u :

2 00

( )T

u t dt = (2)

v iu kin bin : 2 2(0) ( ) 0u u T= = (3) iu kin u c dng :

=T

ii qdttuu0

),,( (4)

Phng trnh Euler_Lagrange c dng tng qut :

2 2

0L d Lu dt u

=

(5)

vi phim hm :

22 2 1 2 1 2( , , )L u u u u = + (6)

T 2 phng trnh trn ta c : 02 21 = u (7) Do :

21

2

=u (8)

Ly tch phn , ta c :

11

2 2ctxu +==

2121

2 4)( ctcttu ++= (9)

xc nh 211 ,, cc ta dng cc iu kin bin : 00)0( 22 == cu

04

)( 1212 =+= TcTTu


Trang 34

v iu kin u :

=+=T

Tc

Tdttu0

02131

2 212)(

T 2 phng trnh trn ta xc nh :

30

124T

= (10)

20

16T

c

= (11)

T qu trnh ti u l :

0 02 2 3

6 12( ) ( )u t x t tT T

= = (12)

20 02 2 3

6 6( )u t t tT T

= (13)

tng ng vi Hnh 1.8(a) . iu khin ti u )(* tx bin i tuyn tnh cn 2u l hm parabol .

Ta th so snh tn hao nng lng ca trng hp ny vi trng hp bi ton ti u tc ng nhanh c c tnh thi gian nh Hnh 1.8(b) . C hai trng hp u c cng gi tr 0 , tng ng vi phn gch sc . Ta c th xc nh ua theo (2) :

/ 2 2

00

2 ( . )4

Ta

a

u Tu t dt = =

02

4au T

= (14)

Nh vy tn hao nng lng tng ng vi :

22 0

30

16Ta aJ u dt T

= = (15)

cn v d ta ang xt :

22 0

30

12( )T

J x dtT

= = (16)


Trang 35

Ngha l chng khc nhau 16 1.3312

aJJ

= = ln .

Hnh 1.8 : c tnh thi gian ca h tn hao nng lng ti thiu (a) v h tc ng nhanh (b) .

V d 1.8 : Xt bi ton ti u tc ng nhanh vi iu kin u :

2 00

T

u dt = (1)

22 0

0

( )T

u dt q= (2)

iu kin bin : 2 2(0) ( ) 0u u T= = (3) Vi bi ton tc ng nhanh , t (1.49) v (1.50) ta c th vit :

22 2 1 2 1 2 2 2( , , , ) 1 ( )L u u u u = + + (4)

Phng trnh Euler_Lagrange :

2 2

0L d Lu dt u

=

(5)

1 2 22 0u = (6)


Trang 36

1222

u=

(7)

Ly tch phn biu thc trn ta c :

12 1

2

( ) ( )2

u t x t t c= = +

(8)

212 1 2

2

( )4

u t t c t c= + + (9)

Kt hp (9) vi iu kin 2 (0) 0u = suy ra : 2c = 0 v 1124

c T= .

V iu kin 2 ( ) 0u T = ta c : 0 11 22

26

c TT

=

01 32

24T

= (10)

01 26

cT

= (11)

Th vo (8) , (9) c :

0 02 2 3

6 12( ) ( )u t x t tT T

= = (12)

20 02 2 3

6 6( )u t t tT T

= (13)

So snh vi v d trc , ta thy qu trnh ti u l hon ton ging nhau . V d 1.9 : Xt i tng c m hnh ton hc gn ng nh sau :

( )( ) ( ) *, uxgtxfx kiii += [ ]nkni ,1;,1 = (1) Trong ( )Tnxxxx ,...,, 21= vector trng thi ; ( )xgk - hm phi tuyn tng minh ; ( )( )txf ii , - hm phi tuyn khng tng minh ; ( )ti - cc nhiu ngu nhin ; u - tn hiu iu khin . Chn hm ch tiu cht lng c dng :

( ) ( )[ ] +=0

22 dtxxJ (2)


Trang 37

Trong l hm s kh vi hoc tuyn tnh tng on v ( ) 00 = . Hm c la chn da trn cc yu cu v ng hc ca h thng . Lut iu khin u m bo cc tiu ho ch tiu cht lng J c th c xc nh bng cch gii phng trnh Euler : 0=+ (3) o hm ca hm s c dng :

==

+

=

n

ii

i

n

ii

i dx

dxdtd

11

(4)

Kt hp (4) v (1) ta c :

( ) ( )( )

( ) ( )

=

=

==

+

+

=

++

=

n

ii

ik

iii

n

kii i

n

ii

i

n

ikii

i

duxg

dxxf

dx

duxgxf

dxdtd

11

11

,

,

(5)

Gii phng trnh (3) kt hp vi (5) , xc nh lut iu *u khin m bo cc tiu ho hm mc tiu J v nh hng ng hc h thng chuyn ng theo xu hng ( ) 0lim

x

t :

( ) ( )

++

= =

=

n

ii

i

n

kii

iiik

xfxxg

u11

1*

,

1

(6)

Lu rng lut iu khin *u ch c ngha khi ( ) 0xg k v 0

kx .

1.2.2 Phng php quy hoch ng Belman 1. Gii thiu Phng php quy hoch ng c da trn nguyn l ti u s khai ca Belman : Mt chin lc ti u c tnh cht khng ph thuc vo nhng quyt nh trc ( v d nh nhng lut iu khin ) song cc quyt nh cn li phi cu thnh nn chin lc ti u c lin quan vi kt qu ca nhng quyt nh truc .


Trang 38

Nguyn l ti u ca Belman : Bt k mt on cui cng no ca qu o ti u cng l mt qu o ti u . Nguyn l ny gii hn xem xt trn mt s cc ch tiu ti u . N ch ra rng phng n ti u phi c xc nh t trng thi cui i ngc v trc . iu kin p dng : nguyn l ti u l mt phng php s , ch p dng c khi h thng c phn cp iu khin v ta bit trc s mt li c xy dng bng thc nghim . V d n gin sau s ch ra nhng vn mu cht ca phng php ny . Bi ton ng bay ca my bay Mt my bay bay theo hng t tri sang phi nh Hnh 1.9 qua cc im a, b, c tng trng cho cc thnh ph , v mc nhin liu cn thit hon tt mi chng ng . Chng ta s dng nguyn l ti u ca Belman gii bi ton cc tiu ha nhin liu tiu hao . Lit k cc trng thi k t 0 n 4 trong qu trnh ra quyt nh nh Hnh 1.9 (u mi tn v con s trong khung bc u c th cha cn quan tm). Ti mi gi tr 1,....1,0 = Nk phi c mt quyt nh , v N l trng thi cui . Trng thi hin ti l nt m chng ta ang ra quyt nh . V th trng thi ban u l ax =0 . Ti trng thi 1 , cc kh nng c th l bx =1 hoc

dx =1 . Tng t vi cx =2 , e hoc g ; fx =3 hoc h v trng thi cui cng ixxn == 4 .

iu khin ku trng thi k n trng thi k+1 c hai gi tr 1=ku : i theo hng ln th 1=ku v 1=ku nu i theo hng xung .

n y chng ta c bi ton ti thiu ha nng lng tiu hao vi trng thi cui c nh , lut iu khin v cc gi tr trng thi . tm ra lut iu khin ng vi mc tiu hao nhin liu ti thiu , ta s dng nguyn l ti u ca Belman , c bt u 4== Nk . Khng c quyt nh no c yu cu y do ta gim 3=k . Nu fx =3 th lut iu khin ti u l 13 =u v chi ph l 4 . iu ny c th hin bng cch t (4) pha trn nt f v chiu mi tn theo chiu t f n i . Nu hx =3 th lut iu khin ti u l 13 =u v chi ph l 2 , c th hin nh trn hnh .


Trang 39

By gi gim k xung 2 . Nu cx =2 th 12 =u vi tng chi ph s l 4 + 3 = 7 . Nu ex =2 chng ta phi a ra mt quyt nh . Nu chn 12 =u n c f v sau n i , chi ph s l 4 + 3 = 7 .

Hnh 1.9 : Lut iu khin nng lng tiu hao ti thiu .

Mt cch khc , nu chng ta chn 12 =u ti e v i n h , chi ph s l 2 + 2 = 4 . V th , ti e cch la chn ti u l 12 =u vi chi ph l 4 .

Nu gx =2 th ch c mt s chn la duy nht l 12 =u vi chi ph di chuyn l 6 . Bng cch ln lt gim k v tip tc so snh cc phng n iu khin ti u c cho bi nguyn l ti u , chng ta c th in vo cc la chn cn li ( u mi tn ) v chi ph ti u c th hin trong Hnh 1.9 . D dng nhn ra rng chui iu khin c la chn l chui ti u . Ch rng khi k = 0 , lut iu khin c th l 10 =u hoc 10 =u cng cho chi ph l 8 ; lut iu khin khi k = 0 l duy nht . C nhiu im cn ch trong v d ny . Trc ht , ta c hai ng i t a n i vi cng mt chi ph l 8 : iheba ( ng nt m ) v

iheda ( ng nt t ) . Hin nhin gii php ti u trong quy hoch ng l khng duy nht . Th hai , gi nh chng ta c gng xc nh l trnh ti u i t a n i khi khng bit nguyn l ti u v i theo chiu thun . Ti a ta s so snh chi ph khi i n b hoc d , v chng ta quyt nh i n d . Tip tc nh vy ta s i n g . khng cn la


Trang 40

chn no khc l i n i qua h . Ton b chi ph cho phng n ny l 1 + 2 + 4 + 2 = 9 v khng phi l ti u . Cui cng chng ta ch ra rng nguyn l ti u ca Belman gip gim s lng php tnh ton cn thit bng cch gim s lng cc la chn c th thc hin . 2. H ri rc Phng php quy hoch ng cng c th d dng p dng cho h phi tuyn Ngoi ra , nu c cng nhiu iu kin rng buc i vi tn hiu iu khin v bin trng thi th ta c c li gii cng n gin . t :

( )1 ,kk k kx f x u+ = (1.56) vi s m k trn f th hin s thay i theo thi gian . Gi nh kt hp vi hm ch tiu cht lng :

( ) ( )1( ) , ,N ki i N k kk i

J x N x L x u

=

= + (1.57)

vi [ ],i N l thi gian ly mu . Chng ta cn ch ra s ph thuc ca J i vi trng thi v thi gian u . Gi s ta c c tn hao ti u ( )1 1k kJ x+ + t thi im 1k + n thi im cui N ng vi nhng phng n kh thi 1+kx , v chui cc phng n ti u t thi im 1+k n N cho mi 1+kx .

Ti thi im k , nu ta p dng mt lut iu khin ku bt k v s dng mt chui lut iu khin ti u k t v tr 1+k , lc tn hao s l : ( ) ( )1 1,kk k k k kJ L x u J x+ += + (1.58) vi kx l trng thi thi im k , v 1+kx c cho bi (1.56) . Theo nguyn l Belman th tn hao ti u t thi im k s l : ( ) ( ) ( )( )1 1min ,

k

kk k k k k k

uJ x L x u J x + += + (1.59)

v lut iu khin ti u *ku ti thi im k l ku lm cho tn hao t cc tiu . Phng trnh (1.59) chnh l nguyn l ti u cho h ri rc . Vai tr quan trng ca n l c th cho php chng ta ti u ha cng lc ti thi im a nhiu hn mt vector iu khin .


Trang 41

Trong thc t , ta c th nh r cc iu kin rng buc c thm vo chng hn nh yu cu lut iu khin ku thuc v mt b cc lut iu khin c chp nhn . V d 1.10 : Xt h : kkk uxx +=+1 (1) c hm ch tiu cht lng :

12 2

00

12

N

N kk

J x u

=

= + (2)

vi thi im cui cng N = 2 . Tn hiu iu khin b rng buc ly cc gi tr :

1, 0.5,0,0.5,1ku = (3) v bin trng thi b rng buc ly cc gi tr : 0,0.5,1,1.5kx = (4) iu kin rng buc i vi tn hiu iu khin khng phi l khng c l do , tn hiu iu khin ti u thi gian ti thiu ch ly cc gi tr 1 ( v d 1.12 ), trong khi tn hiu iu khin ti u nhin liu ti thiu nhn cc gi tr 0 , 1 . iu kin rng buc i vi bin trng thi trong bi ton ny cng hp l , v nu trng thi ban u ly mt trong cc gi tr chp nhn c (4) , th di nh hng ca cc tn hiu iu khin cho php (3) cc trng thi sau s ly cc gi tr nguyn v bn nguyn . iu kin rng buc (4) c th vit li l 0 0,0.5,1,1.5x = v 0 1.5kx (5) y l iu kin xc thc v rng buc bin v trng thi , thng l hp l trong cc tnh hung vt l . By gi , bi ton iu khin ti u l tm dy tn hiu iu khin chp nhn c 0u

, 1u

sao cho ch tiu cht lng 0J t gi tr cc tiu trong khi to

ra qu o trng thi chp nhn c 0 1 2, ,x x x

. Chng ta mun ku c xc

nh nh l lut iu khin hi tip trng thi . Theo (1.58) ta c :

21

12k k k

J u J += + (6)


Trang 42

( )mink

k ku

J J = (7)

tm ku v kJ

ng vi mi kx . Ta xut pht t trng thi cui cng .

k = N = 2 : 2 2J x

=

ng vi mi gi tr 0,0.5,1,1.5Nx = ta c cc gi tr 0,0.25,1, 2.25NJ

= .

k = 1 : 21 2 2/ 2J u J

= +

- 1 1.5x = : v 2 1 1x x u= + v 20 1.5x nn ta ch xt cc gi tr 1 0u

1 0u = 2 1.5 0 1.5x = + = 2 2.25J

=

2 21 2 2/ 2 0 / 2 2.25 2.25J u J

= + = + =

1 0.5u = ( )2 1.5 0.5 1x = + = 2 1J = ( )21 0.5 / 2 1 1.125J = + = 1 1u = ( )2 1.5 1 0.5x = + = 2 0.25J = ( )21 1 / 2 0.25 0.75J = + = Nh vy , tn hiu iu khin ti u vi 1 1.5x = l 1 1u

= v tn hao ti

u l 1 0.75J

= . Ta c c s nh sau vi mi tn ch ra trng thi ti u .

Tng t nh vy cho cc trng hp cn li ca 1x . Tip tc vi trng thi 0k = . Cui cng ta s c li kt qu nh Hnh 1.10 .


Trang 43

Hnh 1.10 : Li kt qu ca bi ton ti u gii bng phng php quy hoch ng .

3. Phng php iu khin s Chng ta c th ri rc ha , gii bi ton ti u cho h ri rc v sau dng khu gi bc 0 to ra tn hiu iu khin s . Cho h thng : ( , , )x f x u t= (1.60) Vi hm ch tiu cht lng :

( ) ( )( ) ( ) ( )( )0

0 , , ,T

J x T T L x t u t t dt= + (1.61)


Trang 44

ri rc h thng vi chu k ly mu giy, ta c th s dng hm xp x bc 1 : ( )1( ) /k kx k x x += (1.62) Vit (1.60) di dng : ( )1 , ,k k k kx x f x u k + = + (1.63) cho n gin ta nh ngha : ( )kx x k , ( )ku u k nh ngha hm ri rc : ( ) ( ), , ,k k k k k kf x u x f x u k + (1.64) Khi ta c th vit : ( )1 ,kk k kx f x u+ = (1.65) Phng trnh ny ng vi (1.56) . ri rc ho hm ch tiu , ta c th vit :

( ) ( )( ) ( ) ( )( )( )11

00 , , ,

kN

k k

J x T T L x t u t t dt

+

=

= + (1.66)

Trong :

TN = (1.67)

S dng hm xp x bc 1 cho mi i lng tch phn :

( ) ( )( ) ( )10

0 , , ,N

k kk

J x T T L x u k

=

= + (1.68)

nh ngha hm ri rc : ( )0 0J J ( ) ( )( ), ,S NN x x N N ( ) ( ), , ,k k k k kL x u L x u k (1.69) Khi ta c :

( ) ( ) ( )10

0 , ,N

S kN k k

kJ N x L x u

=

= + (1.70)

y l cng thc (1.57) .


Trang 45

Trong trng hp h thng tuyn tnh bt bin theo thi gian vi ch tiu cht lng dng ton phng : x Ax Bu= + (1.71)

( ) ( ) ( ) ( ) ( )0

1 102 2

TT T TJ x T S T x T x Qx u Ru dt= + + (1.72)

S dng hm xp x bc nht ri rc ho h thng tr thnh : ( )1k k kx I A x B u + = + + (1.73) ( ) ( )1

0

1 102 2

NT T S T SN N N k k k k

kJ x S x x Q x u R u

=

= + + (1.74)

Trong : ( )NS S N (1.75) QQ S = (1.76) RR S = (1.77) Tuy nhin trong trng hp ny ta c th lm tt hn xp x Euler (1.73) bng cch s dng chnh xc phng trnh trng thi (1.71) bao gm b ly mu v khu gi bc 1 : k

Sk

Sk uBxAx +=+1 (1.78)

Trong :

AS eA = (1.79)

( )0

S AB e B dt

= (1.80)

Khi h thng ny c ri rc ho , phng php quy hoch ng c th c p dng tnh *ku nh trong phn ri rc . iu khin s p dng trong thc t c th hin nh sau : ( ) ku t u= , ( )1k t k + (1.81) s dng phng php quy hoch ng , bin trng thi v gi tr iu khin trc ht phi c lng t ho , c gii hn theo mt s tp gi tr c th chp nhn . Mc lng t cng tt th tn hiu s cng chnh xc ; tuy nhin khi s lng c th chp nhn c ca xk v uk tng th


Trang 46

khi lng tnh ton tm *ku cng tng theo . Vn ny c th nhanh chng gy kh khn k c i vi cc my tnh ln .

1.2.3 Nguyn l cc tiu Pontryagin _ Hamilton 1. Nguyn l cc tiu ca Pontryagin. Cho h thng : ),,( tuxfx = (1.82) Kt hp hm ch tiu cht lng :

( )( ) +=T

t

dttuxLTTxtJ0

0 ),,(,)( (1.83)

Trng thi cui phi tha : ( )( ), 0x T T = (1.84) v x(t0) c cho trc . iu kin bi ton ti u l :

u

H

= 0 (1.85)

vi ( , , , ) ( , , ) ( , , )TH x u t L x u t f x u t = + (1.86) Gi s hm iu khin u(t) l rng buc trong mt vng gii hn cho php , c ngha l gi tr yu cu c ln nh hn gi tr cho . iu kin dng thay bng iu kin tng qut : ( , , , ) ( , , , )H x u t H x u u t + Tha tt c gi tr u Du * th hin ch s cht lng ti u . M bt k s bin thin no trong b iu khin ti u xy ra ti thi im t ( trong khi trng thi v bin trng thi nu c duy tr ) s tng n gi tr ca hm Hamilton . iu kin ny c vit nh sau :

( , , , ) ( , , , )H x u t H x u t Tha tt c gi tr u (1.87)

Yu cu ti u biu thc (1.87) c gi nguyn l cc tiu Pontryagin : Hm Hamilton phi c cc tiu ha tt c cc gi tr u cho gi tr ti u ca trng thi v bin trng thi .


Trang 47

Chng ta s thy nguyn l cc tiu hu dng nh th no . c bit ch khng th ni rng biu thc ( , , ) ( , , , )H x u H x u t chc chn phi ng . V d 1.11 : Ti u ha vi nhng rng buc Gi s chng ta mun ti u cc tiu hm :

L = 21

u2 2u + 1 (1)

Vi iu kin :

u 1 (2) Xem Hnh 1.11 . Nguyn l cc tiu : L(u*) L(u) tha u (3)

Hnh 1.11 : Ti u ho vi nhiu iu kin rng buc .

D dng thy c gi tr ti u ca u l : u

* = 1 (4)

Gi tr ti u ca L l :


Trang 48

L* = L(1) = - 21

(5)

Gi tr nh nht khng rng buc tm c bng cch gii :

u

L

= u -2 = 0 (6)

nhn c : u = 2 (7) v : L(2) = -1 (8) nh hn (5) ; nhng u=2 th khng nm trong khon 1u .

2. iu khin Bang-Bang Chng ta hy tho lun bi ton ti thiu thi gian tuyn tnh vi ng vo rng buc . Cho h thng : x = Ax + Bu (1.88) vi ch tiu cht lng :

J(t0) = T

tdt

0

1 (1.89)

Vi T t do . Gi s hm iu khin phi tha mn iu kin sau :

( ) 1u t [ ]0 ,t t T (1.90) Bi ton ti u t ra l tm tn hiu iu khin u(t) cc tiu ho J(t0) , tha mn iu kin (1.90) vi t , i t trng thi x(t0) n trng thi cui cng x(T) tha cng thc (1.84) ca hm . Hm Hamilton cho vn ny l :

1 ( )T TH L f Ax Bu = + = + + (1.91) iu kin dng c tm thy l :

0 = =

u

H BT (1.92)

N khng cha u bi v hm Hamilton tuyn tnh i vi u . R rng , H cc tiu chng ta nn chn u(t) sao cho T(t)Bu(t) cng nh cng tt ( c


Trang 49

ngha l gi tr cng xa v pha bn tri trn trc ta thc ; TBu = - l gi tr nh nht ) . Nu khng c s rng buc no trn u(t) , th iu ny s cho ra nhng gi tr v hn ( dng hoc m ) ca nhng bin iu khin . Vi kt qu ny , bi ton ti u t ra phi c nhng iu kin rng buc i vi tn hiu iu khin . Theo nguyn l cc tiu Pontryagin (1.87) , hm iu khin ti u u*(t) phi tha mn :

1 ( ) ( ) 1 ( ) ( )T TAx Bu Ax Bu + + + + ( ) ( )T TBu Bu (1.93) i vi tt c gi tr u(t) cho php . iu kin ny cho php chng ta biu din u*(t) di dng bin trng thi . thy iu ny , trc tin chng ta tho lun v trng hp mt ng vo . t u(t) l mt i lng v hng v t b tng trng cho vector ng vo . Trong trng hp ny d dng chn u*(t) ti thiu T(t) bu(t) . ( Ch : gi tr nh nht ngha l T(t)bu(t) nhn mt gi tr cng gn - cng tt ) . Nu T(t)b l gi tr dng , chng ta nn chn u(t) = -1 lm cho T(t)bu(t) c gi tr m nht . Mt khc , nu T(t)b l gi tr m , chng ta nn chn u(t) gi tr cc i l gi tr 1 gi tr T(t)bu(t) cng m cng tt . Nu gi tr T(t)bu(t) bng zero ti thi im t , khi u(t) c th nhn bt c gi tr no ti thi im ny . Quan h gia iu khin ti u v bin trng thi c th biu din bng hm sgn(w) :

( ) ( )1

sgn 1,11

w

=

000

w

w

w

>

=

=


Trang 58

( )( )

( )

11;00

0;11

dez w

=

11

1111

>

=


Trang 59

Ch tiu cht lng vi yu cu nhin liu ti thiu :

( ) ( )dttuJT

=0

0 (5)

Ta cha ch n trng thi thi gian cui T hoc t do hoc rng buc, mc d cui cng ta cng s xt n c 2 trng hp . Hm Hamilton : uvuH vy ++= (6) Trong [ ]Tvy = . Do phng trnh bin trng thi l : 0y = (7) v y = (8) iu kin tip tuyn yu cu :

( ) ( ) ( ) ( )TuTTuTH v+==0 (9) T (4) , (7) , (8) ta suy ra : ( )y yt const = (10) ( ) ( ) ( )v v yt T T t = + (11) Thnh phn bin trng thi v(t) l tuyn tnh . Ty thuc vo bin cha bit y v v(T) ( chng tu thuc vo gi tr ca trng thi u ) , v(t) c th l hng s (y = 0) , c th tng (y < 0) hoc gim (y > 0) . Xem Hnh 1.13 . Nguyn l cc tiu Pontryagin yu cu : ( ) ( )( )tdeztu v= (12) do iu khin ti u l :

( )[ ]

[ ]

=

1-1;0-01;0

1

tu

( )( )

( )( )( ) 1

111

11

>

=


Trang 60

By gi chng ta s xc nh lut iu khin ti u v i tm nhng thi im lc b iu khin chuyn i sang gi tr mi . B qua nhng khong thi gian ring bit khc , c 3 gi tr ca u(t) l : -1 , 0 , 1 . Hnh 1.14 cho ta qu o mt phng pha khi u = 1 v u = -1 . Nu u(t) = 0 t , khi trng thi xc nh bi : ( ) ( )0v t v= (14) ( ) ( ) ( )0 0y t y v t= + (15) Nhng ng nm ngang ca hng s v trong qu o mt phng pha c cho Hnh 1.15 . Qu o mt phng pha trong trng hp u = 0 l nhng ng m vic tiu th nhin liu l zero . nhin liu s dng l ti thiu , chng ta s cho h thng di chuyn theo ng u = 1 hoc 1 , dn trng thi n mt trong nhng ng nm ngang . Sau di chuyn dc theo ng nm ngang n v tr chuyn i qua ng u = -1 hoc 1 dn trng thi tin v zero . thy c lut iu khin Bang-off-bang , chng ta kt hp qu o ca hai Hnh 1.14 v 1.15 c Hnh 1.16 . Phn tip theo chng ta s tho lun ring nhng tnh hung cho hai vn thi gian cui t do v c nh .

Hnh 1.15 : Qu o mt phng pha trong trng hp u = 0 .


Trang 61

Hnh 1.16 : Lut iu khin Bang-Off-Bang .

Hnh 1.17 : Qu o trng thi nhin liu ti thiu .

Thi gian cui t do : Vi trng hp thi gian cui t do , khi lut iu khin ca bi ton nhin liu ti thiu s khng tn ti .


Trang 62

Thi gian cui c nh : Cho trng thi u nh m t Hnh 1.17 . i vi bi ton thi gian ti thiu th thi gian cui nh nht l :

( ) ( ) ( )2000

2

minvyvT ++= (16)

Gi nh rng thi gian T ca bi ton nhin liu ti thiu c c nh ti gi tr : minT T> (17) Khi lut iu khin ca bi ton nhin liu ti thiu l : -1 , 0 , 1 vi thi gian chuyn i t1 v t2 c xc nh . T 10 t t< < , u(t) = -1 , biu thc (15) v (16) trong v d Bang-bang tr thnh : ( )1 1 0 1v v t v t= = (18) ( ) ( ) ( )

21

1 10 0 2ty t y v t= + (19)

T 1 2t t t< < , u(t) = 0 , ta c phng trnh trng thi : ( )2 1v t v= (20) ( ) ( ) ( )2 1 1 2 1y t y t v t t= + (21) T 2t t T< < , u(t) = 1 , ta c : ( ) ( ) ( )2 20 v T v t T t= = + (22)

( ) ( ) ( )( ) ( )2

22 2 20 2

T ty T y t v t T t

= = + + (23)

trong ta c s dng iu kin bin (4) . Th (18) , (20) vo (22) , ta c : 102 tTvt += (24) Th (18) , (19) , (20) , (21) , (24) vo (23) v n gin ha , cho ra kt qu :

( ) 02

20

00102

1 =

++++

vTvytTvt (25)


Trang 63

vi nghim :

( ) ( ) ( )2

24 2002

00 vyTvTvt+++

= (26)

t (24) v thc t th t1 < t2 , ta c :

( ) ( ) ( )2

24 2002

001

vyTvTvt

+++= (27)

v :

( ) ( ) ( )2

24 2002

002

vyTvTvt

++++= (28)

v T > Tmin nn du ca biu thc trong cn l dng . Chng ta c th biu din bi ton nhin liu ti thiu ny thnh dng vng h nh sau :

( )

=

101

* tu

tt

ttt

tt

) (1)

t cc tiu . T hnh v ta tm c :

121

)()(

2 ++=

sssRsC

(2) hoc c dng : rccc =++ 2 (3) i vi tn hiu sai lch e , ta c : rreee 22 +=++ (4) Vi r(t) = 1(t) , ta c 0)0( =+r , 0)0( =+r . Do , vi + 0t ta s c : 02 =++ eee , 1)0( =+e , 0)0( =+e (5) By gi , chng ta t cc bin trng thi nh sau : ex =1 (6) exx == 12 (7) Khi phng trnh trng thi l : Axx = (8)

vi

= 2110

A

Ch tiu cht lng J c th vit li nh sau :

+

+

+=+=0 0

22

21

22 )()( dtxxdteeJ

[ ] dtx

xxx

=

+ 2

1

021 0

01

+

=

0

QxdtxT (9)

Nu ma trn A n nh th ch tiu cht lng J c th xc nh t (1.129) : )0()0( ++= SxxJ T (10)


Trang 71

vi S l nghim ca phng trnh Lyapunov : QSASAT =+ (11) Phng trnh c vit li nh sau :

=

+

001

2110

2110

2221

1211

2221

1211

ss

ss

ss

ss (12)

Phng trnh trn tng ng vi h phng trnh sau : 11221 = ss (13) 02 121122 =+ sss (14) 02 222111 = sss (15) =+ 22212212 22 ssss (16) Gii h phng trnh trn ta c :

+

++

=

41

21

21

41

S (17)

Ch tiu cht lng J c vit li : )0()0( ++= SxxJ T

)0(4

1)0()0()0(4

1 2221

21 +

++++++

++= xxxx

(18)

Th cc iu kin u 1)0(1 =+x , 0)0(2 =+x vo (18) ta tm c :

41+

+=J (19)

tm cc tr ca J ta cho o hm ca J theo bng 0 :

0411 2 =

+=

J

(20)

21 += (21)

Xt o hm bc hai ca J theo ti 2

1 += :


Trang 72

2

2 312

J

+=

31 2 0

1122

+= = >

+ +

(22)

Nh vy, ch tiu cht lng J s t cc tiu ti gi tr ti u 1 / 2 = + min 1J = + (23) V d 1.15 : Xc nh lut iu khin ti u ri rc bit h thng c i tng iu khin m t bi phng trnh trng thi sau :

( ) ( ) ( )0 1 00 0.1 0.01

x t x t u t

= +

(1)

Ch tiu cht lng :

( )11

2 2

00.001

N

k kk

J x u

=

= + (2)

Chu k ly mu T = 0.5 sec , N = 50 . Ta d dng xc nh c phng trnh trng thi h ri rc t phng trnh trng thi h lin tc :

1k d k d kx A x B u+ = +

k d ky C x=

vi : 1 0.4880 0.951d

A =

,

0.001230.00488d

B

=

, [ ]1 0dC = Nghim ca bi ton ti u c tnh theo (1.138) v (1.139) :

11 1( )T Tk k k k k k k kK B S B R B S A+ += + (3)

( ) 11 1 1 1T T Tk k k k k k k k k k k k kS A S S B B S B R B S A Q+ + + + = + + (4) vi : k dA A= , k dB B= ,

1 00 0k

Q =

, 0.001kR =


Trang 73

Ta tnh c K49 = 0 khi bit S50 = 0 . Tip theo ta tnh gi tr S49 :

49 491 00 0

S Q = =

(5)

Tip tc vi K48 v S48 :

[ ] [ ]1

481 0 0.00123

0.00123 0.00488 0.001 . 0.00123 0.00488 .0 0 0.00488

K

= +

[ ]1 0 1 0.488 1.228 0.5990 0 0 0.951

=

(6)

[ ]48 1 0 1 0 1 0 0.00123 1 00.00123 0.00488 .0.488 0.951 0 0 0 0 0.00488 0 0S

=

[ ]1

0.00123 1 0 1 0.488 1 00.001 0.00123 0.00488

0.00488 0 0 0 0.951 0 0

+ +

0.9985 0.48730.4873 0.2378

=

(7)

Tip tc tnh ton nh my tnh , ta s xc nh c vi k = 39 ma trn Kk s hi t v gi tr [25 63] .Vy iu khin ti u cui cng l : [ ]25 63k ku x= (8)

1.3.5 Nhn xt Phng trnh Riccati dng tng hp cc h tuyn tnh vi ch tiu cht lng dng ton phng . Vi cch gii quyt ny , ta va m bo c tnh n nh ca h thng ( do cch chn hm nng lng V(x) theo tiu chun n nh th hai ca Lyapunov ) , va cc tiu ho c ch tiu cht lng J theo yu cu bi ton t ra . Tuy nhin , c vi im ta cn ch : i vi bi ton p dng phng trnh Riccati th vic chn ma trn trng lng thch hp ch tiu cht lng rt quan trng v n nh hng rt nhiu n kt qu tnh ton . Bn cnh , khi xt h ri rc phi m bo s hi t ca Kk ; nu khng th cn phi tng s trng thi , khi khi lng tnh ton cng tng rt nhiu , ch ph hp khi gii bng my tnh .


Trang 74

1.4 NG DNG MATLAB GII BI TON TI U 1.4.1 Ti u ho tnh Bi ton ch tiu cht lng dng ton phng v iu kin rng buc tuyn tnh _ Trng hp v hng ( v d 1.4 ) Cho a = 3 , b = 2 , m = 1 , c = 1 . Vi : x(1) = x , x(2) = u . Khi bi ton tr thnh tm gi tr ti thiu ca :

8)2(

18)1()(

22 xxxf +=

vi iu kin rng buc : 01)2()1()( =+= xxxg

y ta s s dng hm lsqlin ( Optimization Toolbox ) vi kt qu l gi tr ti u ca x 2)( DCxxf = t gi tr nh nht ( 2DCx l norm ca ma trn vung [ ]DCx ) . Cng cc iu kin rng buc :

BeqxAeqBAx=

.

Cn lp cc thng s C , D , A , B , Aeq , Beq nhp vo theo c php : beq) Aeq, B, A, D, lsqlin(C, =x

Chng trnh : C = [1/(18^(1/2)) 0;0 1/(8^(1/2))]; D = [0;0]; Aeq = [1 1]; Beq = [1]; x = lsqlin(C, D, [], [], Aeq, Beq)

Chng ta s c kt qu : x =

0.6923 0.3077


Trang 75

1.4.2 iu khin ti u cnh tay my hi tip gc Xt m hnh cnh tay my hai on nh hnh :

V tr im cui ca cnh tay hai on c cho bi phng trnh sau :

( ) ( )1 1 2 1 2cos cosx L L = + + ( ) ( )1 1 2 1 2sin siny L L = + +

Phng trnh ng lc hc :

11

22

TA B ETC D F

= +

trong [ ]1 2 TT T T= l tn hiu iu khin . Vi cc trng thi :

1 1

2 1 1

3 2

4 3 2

x

x x

x

x x

=

= =

= = =


Trang 76

1 2

2 1 2

3 4

4 1 2

x x

x AT BT Ex x

x CT DT F

=

= +

= = +

Chn ch tiu cht lng J c dng :

( )2 2 2 21 1 2 20

J dt

= + + +

Vi phim hm dng :

1 1 1 1

2 2 2 2

e K ee K e

= + = +

vi 1 1 12 2 2

r

r

e

e

=

=

1 2,r r l gc t ca 1 2,

1 1 2

2 2 4

e x

e x

= =

= =

1 1 2

2 2 4

e x

e x

= =

= =

m bo cc tiu ho ch tiu cht lng J th 1 2,T T l nghim ca h phng trnh sau :

1 1

2 2

00

+ = + =

Gii h phng trnh trn ta c :

( )( )

11 1 2 11 1 1

2 2 4 22 2 2

11

e EK x KT K A K Be FK x KT K C K D

+ + = + +

Tn hiu iu khin T c tnh ton bng chng trnh Giai_PT.m


Trang 77

Chng trnh : Thng s u vo cho h thng (file thongso.m) : global m1 m2 L1 L2 a1 a2 I1 I2 m1 = 3.6745; m2= 1.0184; L1= 0.6519 ; L2= 0.6019; a1= 0.3365 ; a2= 0.2606; I1= 0.370 ; I2= 0.081;

Chng trnh tm tn hiu iu khin (file Giai_PT.m) : function [C]= Giai_PT (theta1, theta2, theta1_dot, theta2_dot, e1, e2) % Nhap thong so cho canh tay m1 = 3.6745; m2 = 1.0184; L1 = 0.6519; L2 = 0.6019; a1 = 0.3365; a2 = 0.2606; I1 = 0.370; I2 = 0.081; K1 = 0.5; K2 = 0.8; m11 = m1*a1*a1+m2*(L1*L1+2*L1*a2*cos(theta2)+a2*a2)+I1+I2; m12 = m2*a2*(a2+L1*cos(theta2))+I2; m22 = m2*a2*a2+I2; n1 = -m2*L1*a2*sin(theta2)*(2*theta1_dot*theta2_dot+theta2_dot*theta2_dot); n2 = m2*L1*a2*sin(theta2)*theta1_dot*theta1_dot; A = [m11 m12; m12 m22]; B = [n1; n2]; A = inv(A); B = A*B; A = [K1*A(1,1) K1*A(1,2); K2*A(2,1) K2*A(2,2)];


Trang 78

B = [e1+B(1,1)*K1-theta1_dot*(K1+1); e2+B(2,1)*K2-theta2_dot*(K2+1)]; C = inv(A)*B; u1 = C(1,1); u2 = C(2,1);

Kt qu m phng :

V tr t thay i theo hm xung vi 1

V tr t thay i theo hm xung vi 2


Trang 79


Trang 80

1.4.3 H thng tc ng nhanh Xt v d iu khin Bang-bang (v d 1.12) Vi iu kin u (0) 10y = , (0) 10v = chng ta s v qu o trng thi ti u bng chng trnh ex1.12 .

Chng trnh : function [x,u,t] = ex1.12 a = [0 1;0 0]; b = [0;1]; x0 = [10 10]; T = 0.025; N = 1200; x(:,1) = x0; eps = 1e-4; t=0:T:T*N; for k = 1:N sw = x(1,k) + 0.5 * x(2,k) * abs( x(2,k) ); if ( abs(sw) < eps ) if ( x(1,k) > 0 ) u(k) = 1; end if ( x(1,k) < 0 ) u(k) = -1; end else if ( sw > 0 ) u(k) = -1; end if ( sw < 0 ) u(k) = 1; end end if ( x(1,k)^2 + x(2,k)^2 < eps ) u(k) = 0; end y = lsim(a,b,eye(2),zeros(2,1),u(k)*ones(1,2),[(k-1)*T, k*T],x(:,k)); x(:,k+1)=y(2,:)'; end


Trang 81

Qu o trng thi ti u .

1.4.4 LQR lin tc v ri rc 1. H lin tc Xt h v hng :

x ax bu= + vi ch tiu cht lng :

( )0

2 2 21 1( ) ( )2 2

T

t

J s T x T qx ru dt= + +

Vi a = 0.05 , b = r =1 , x(0) = 10 , ta s dng chng trnh ex v fex v cc qu o ti u ng vi cc gi tr q = 0.01 , 0.1 , 1 , 10 , 100 .

Chng trnh : function [x,u,S,tf] = ex x0 = 10; a = .05; b = 1; r = 1;


Trang 82

[tb,S] = ode45('fex',-10,0,0); K = -b * flipud(S) / r; tf = flipud(-tb); x(1) = x0; u(1) = K(1) * x(1); for k = 1 : length(tf)-1 x(k+1) = expm( (a + b * K(k) ) * ( tf(k+1) - tf(k) ) ) * x(k); u(k+1) = K(k+1) * x(k+1); end

function sd = fex(t,s) q = 1; a = .05; b = 1; r = 1; sd = 2 * a * s(1) - ( b^2 * s(1)^2 ) / r + q;

Qu o trng thi x(t)


Trang 83

Tn hiu iu khin ti u u(t)

Li gii phng trnh Riccati s(t)


Trang 84

2. H ri rc Xt h v hng :

1k k kx ax bu+ = +

vi ch tiu cht lng :

( )12 2 21 12 2N

i N N k kk i

J s x qx ru

=

= + +

a = 1.05 , b = 0.01 , q = r = 1 , x0 = 10 , N = 100 . Chng ta s xt hai trng hp sN = 5 v sN = 500 bng chng trnh dex tm cc qu o ti u .

Chng trnh : function [x,u,K,S] = dex a = 1.05; b = 0.01; q = 1; r = 1; x0 = 10; s = 5; N = 100; S(N+1) = s; for k = N:-1:1 K(k) = ( a * b * s ) / ( r + s * b^2 ); s = q + ( r * s * a^2 ) / ( r + s * b^2 ); S(k) = s; end x(1) = x0; for k = 1:N u(k) = -K(k) * x(k); x(k+1) = a * x(k) + b * u(k); end


Trang 85

Gi tr tun t sk (sN = 5)

li hi tip ti u Kk (sN = 5)


Trang 86

Qu o trng thi ti u xk* (sN = 5)

Gi tr tun t sk (sN = 500)


Trang 87

li hi tip ti u Kk (sN = 500)

Qu o trng thi ti u xk* (sN = 500)


Trang 88

CU HI N TP V BI TP 1. Trnh by phng php bin phn c in Euler_Lagrange cho cc trng hp : khng c iu kin rng buc , c iu kin rng buc v khi tn hiu u vo b hn ch . 2. Ch tiu cht lng v d 1.9 c dng :

( )2 20

J dt

= +

Hy chng minh hm bin s ph c xc nh t iu kin cc tiu ca J nh sau :

0 + = 3. Pht biu nguyn l ti u ca Belman . Trnh by tng gii quyt bi ton ti u ca phng php quy hoch ng . 4. Trnh by nguyn l cc tiu ca Pontryagin 5. Pht biu tiu chun n nh th hai ca Lyapunov . 6. ng dng Lyapunov trong bi ton LQR lin tc . 7. Tm im (x,y) thuc parabol :

2 3 6y x x= + sao cho khong cch t (x , y) n im c to (2,2) l ngn nht . Tnh khong cch . 8. a. Tm hnh ch nht c din tch ln nht vi chu vi p nh nht . Ngha l tm x v y tho mn cc i ho hm :

( , )L x y xy= vi iu kin rng buc : ( , ) 2 2 0f x y x y p= + = b. Tm hnh ch nht c chu vi nh nht vi din tch cho trc l 2a . Ngha l cc tiu ho hm :

( , ) 2 2L x y x y= + vi iu kin rng buc : 2( , ) 0f x y xy a= = 9. Cho h thng :

1 2 23 1 0

x x u

= +

Tm cc gi tr ti u , ,x u L tho mn cc tiu ch tiu cht lng :


Trang 89

1 0 2 11 10 2 1 12 2

T TL x x u u

= +

10. Cho h thng : 2

2d y

udt

=

Tm tn hiu iu khin u tho mn cc tiu ch tiu cht lng : 1

2

1

12

J u u dt

= +

vi cc iu kin u : ( 1) (1) 0( 1) (1) 0

y yy y

= =

= =

11. Cho h thng : x x u= +

a. Tm phng trnh trng thi ca h thng . b. Tm iu khin ti u cc tiu ch tiu cht lng J :

12

0

J u dt=

vi x(0) = 0 v x(1) = 2 . c. Tm qu o trng thi ti u . 12. Cho h thng :

21k k k kx x u u+ = +

vi tn hao : 1

20

0

N

N k kk

J x x u

=

= +

Cho N = 2 . Tn hiu iu khin ch nhn cc gi tr : 1ku = hoc 1ku = . xk nhn cc gi tr -1, 0, 1, 2 . a. S dng phng php quy hoch ng tm lut iu khin hi tip trng thi ti u . b. Vi 0 2x = , hy tm tn hao ti u , trnh t iu khin v qu o trng thi .


Trang 90

13. Xt h tc ng nhanh c dng sau : 2

2d x

x udt

+ =

1u

Tm qu o pha ti u a h v gc to t mt im bt k . 14. Xt bi ton tc ng nhanh :

2

0 1 00 1

x x u

= +

( ) 1u t a. Gii phng trnh bin trng thi. Dng nguyn l cc tiu Pontryagin tm lut iu khin ti u . b. V qu o pha cho trng hp u = 1 v u = -1 . c. Tm phng trnh ng chuyn i . 15. Cho h thng :

1 2

2

x x

x u

=

=

( )2 2 21 1 2 20

1 22

J x vx x qx u dt

= + + +

vi ( )2 0q v > . a. Tm li gii cho phng trnh Riccati i s . b. Tm iu khin ti u v h thng vng kn ti u . c. V qu o nghim s ca h thng khi q thay i t 0 n . Vi gi tr no ca q th h thng n nh . 16. Cho h thng :

1 2

2 1 22x x

x ax x u

=

= +

v ch tiu cht lng :

( )2 2 21 20

1 22

J x x u dt

= + +


Trang 91

a. V qu o nghim s ca h h khi a thay i t 0 n . Vi gi tr no ca a th h thng n nh . b. Vi a = -8 tm li gii cho phng trnh i s Riccati v h s K . 17. Xt h ri rc :

1 2k k kx x u+ = +

a. Tm li gii xk vi k = 0 ; 5 nu x0 = 3 . b. Xc nh lut uk tn hao nng lng ti thiu a h thng t x0 = 3 v x5 = 0 . V qu o trng thi ti u . c. Tm lut hi tip trng thi Kk ti u sao cho ch tiu cht lng J t cc tiu :

( )42 2 250

5 0.5 k kk

J x x u=

= + +

Tnh hm tn tht J ti u vi k = 0 ; 5 . 18. Xt h ri rc :

1k k kx ax bu+ = +

( )13 3 30

1 12 3

N

N N k kk

J s x qx ru

=

= + +

vi xk , uk l v hng . a. Tm phng trnh trng thi , phng trnh bin trng thi v iu kin tnh . b. Khi no th ta c th tm c lut iu khin ti u uk . Vi iu kin , hy kh uk trong phng trnh trng thi . c. Tm li gii bi ton iu khin vng h ( trng thi cui xN c nh ,

0Ns = , q = 0 ) .

chuong 1_Điều khiển tối Ưu

Documents