sổ tay cdt chuong 30-tke toi uu_bang_h2

8/6/2019 s tay cdt Chuong 30-Tke Toi Uu_bang_H2

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30Thit k h thng iu khin bng ti u H2

Armando A. RodriguezArizona State University

30.1 Gii thiu ..................................................................1

30.2 Cu trc thit k h iu khin tng qut .................2

30.3 Bi ton phn hi u ra ........................................10

30.4 Bi ton phn hi trng thi ..................................36

30.5 Bi ton ni x u ra ............................................3730.6 Tm tt ......................................................... ......... .39

30.1 Gii thiu

Chng ny trnh by vic thit k h thng iu khin thng qua ti u (ton phng) H2. Mt khung thng nht datrn khi nim thit b ph bin v ti u ha theo trng s cho php nh thit k hng ti bi ton phn hi trng thi, tiu trng thi, phn hi u ra ng v nhiu cu trc tng qut tng t. Khung mu ny cho php ngi thit k kt hp ddng cc tham s thit k v/hoc cc hm trng s c th c s dng tc ng ti kt qu ti u, tha mn cc c tnhthit k mong mun v h thng ha qu trnh thit k. Cc li gii ti u nhn c thng qua cc phng trnh Riccati niting; nh phng trnh Riccati i s iu khin (CARE - Control Algebraic Riccati Equation) v phng trnh Riccati i

s lc (FARE- Filter Algebraic Riccati Equation). Trong khi cc hm trng s ng lm tng s chiu ca phng trnhRiccati ang c gii, th cc li gii thu c bng vic s dng cc phn mm thit k tr gip bi my tnh ngy nay(nh MATLAB, hp cng c iu khin bn vng, hp cng c tng hp, ).

Ni tm li, ti u 2H tng qut ha tt c cc h phng php thit k b lc v iu khin ton phng ni ting:

Phng php thit k iu chnh ton phng tuyn tnh (LQR-Linear Quadratic Regulator) [7,11].

Phng php thit k b lc Bucy - Kalman (KBF-Kalman-Bucy Filter) [5,6].

Phng php thit k Gaussian ton phng tuyn tnh (LQG - Linear Quadratic Gaussian) [4,10,11].

Ti u 2H c th c s dng thit k c h thng cc lut iu khin phn hi trng thi h s hng, c lngtrng thi, cc b iu khin ng hc u ra, v nhiu b iu khin khc.

HNH 30.1 H thng phn hi tng qut

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S tay C in t

30.2 Cu trc thit k h iu khin tng qut

Trong phn ny, chng ta trnh by mt m hnh tng qut thit k h thng iu khin (v c lng). Mc ch cuicng, chng ta nghin cu h thng phn hi c tng qut ha nh hnh 30.1. Trong hnh ny, G biu th thit b tng qut.G cha mt m hnh cho thit b thc P(h vt l) cn iu khin. N cng c th cha cc hm trng s (tn s ph thuc)b sung c s dng hng ti cc mc tiu thit k vng kn. Kmiu t mt b iu khin hoc b b cn thit k.

tng chnh: thit k thng qua ti u tng chnh y l cc vn quan trng ny sinh trong iu khin, c lng, lc, v cc phm vi khc ca k thut

c a ra di dng mt m hnh tng qut ha G v b iu khinKthit k bng cch ti thiu mt s chun (nh H2)trn ma trn hm truyn ca h kn wzT t tn hiu w n tn hiuz.

Tn hiu

nh gi s linh hot ca cu trc h phn hi tng qut , ch cn nghin cu c tnh ca cc tn hiu z, u, w, vytrong hnh trn. Cc tn hiu ny c miu t nh sau:

Tn hiu c iu chnh: Tn hiu znz R miu t tn hiu c iu chnh hoc tn hiu m chng ta mun gin nh ph thuc vo ng dng v mc ch mong mun. Cc tn hiu ny bao gm sai s bm, u vo c cuchp hnh hoc u vo iu khin, sai lch c lng tn hiu,

Tn hiu iu khin: tn hiu unu R miu t tn hiu iu khin hoc cc bin vn hnh l u ra ca b iukhin K. Cc tn hiu iu khin c th bao gm lung kh t vo my, in p cung cp cho mt ng c,Chng cng c th gm cc c lng trng thi c cung cp bi K. Mc ch l Ktc ng v phi hp tnhiu iu khin u gi cho tn hiu cn hiu chnh z nh. Trong thc t, thng chng ta c nhiu tn hiu cnhiu chnh hn l cc tn hiu iu khin ( z un n> ). Nn ch rng, mc d ni chung l nh vy, nhng nuchng ta mun iu khin c lp m bin, th chng ta cn t nht m bin iu khin c lp. Nguyn l c bn nyphi c duy tr trong thc t. Cng nhiu bin iu khin c lp u (v nguyn tc) cng d iu chnh tn hiuz.

Tn hiu ngoi sinh: Tn hiu wnw R biu din tn hiu ngoi sinh (hay tn hiu ngoi) tc ng vo h thng.Tn hiu ngoi sinh bao gm cc lnh tham chiu c a ra iu khin h thng, cc tc ng ca nhiu lnh, nhiu t cc u o,

Tn hiu o: Cc tn hiu yny R miu t cc gi tr o hoc cc tn hiu l cc bin trc tip ti b iu khin K.

Cc gi tr o bao gm mt phn hoc tt c cc bin trng thi, cc u ra c th o c, cc tn hiu iu khinc th o c, cc tn hiu ngoi sinh c th o c, Trong thc t, chng ta c nhiu tn hiu ngoi sinh hncc gi tr o ( w yn n> ). Ni chung, cng nhiu cc gi tr o c lp th cng nhiu thng tin hu dng.

Ch thch 30.1 (Dn ti nguyn l tch)

Vic kt hp tn hiu iu khin u vi tn hiu cn chnh z l rt t nhin. Ngi ta cho rng cp ny hon ton c thnh ngha mt bi ton iu khin hoc bi ton iu chnh. iu ny tng t vi tnh hung bn ti trong cc bi tonLQR truyn thng. Trong bi ton , ngi ta tha hip hnh ng (kch thc) iu khin thay cho tc iu chnh.

Tng t nh vy, vic kt hp tn hiu ngoi sinh w vi tn hiu oy cng rt t nhin. Ngi ta cho rng cp ny honton c th nh ngha mt bi ton c lng hoc bi ton phn tch tn hiu. iu ny tng t vi tnh hung bn titrong bi ton KBF truyn thng. Trong bi ton , ngi ta loi b nhiu thay cho tc to c lng.

Cc s kt hp nh l mt bi ton LQG truyn thng, cu trc tng qut a ra mt nguyn l tch t nhin. Thtvy, y s l mt trng hp ca bi ton phn hi u ra 2H m chng ta s nghin cu.

Bi ton ti u 2H tng qut

Bi ton ti uH2 tng qut c th c trnh by nh sau:

Tm b iu khin hu t thc (c chiu hu hn) ph hpKlm n nh G sao cho chun 2H ca ma trn hmtruyn h kn l nh nht:

2min ( )wzK

T KH (30.1)

Trong

2

1{ ( ) ( )}2

def

HF trace F j F j d

= H (30.2)

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Thit k h thng iu khin bng ti u H2

0{ ( ) ( )}Htrace f t f t dt

= (30.3)

2 ( )f

+=

L R (30.4)

vf l ma trn p ng xung c kt hp vi ma trn hm truyn F.

Ch thch 30.2 (S dng chun bnh phng: cc tn hiu ngoi sinh di rng)

Ch rng chun bnh phng o nng lng ca p ng xung v s bin i ca hm n v dirac ( )t l duy nht, ncho php chun bnh phng ph hp khi cc tn hiu ngoi sinh w l di rng trong t nhin. iu ny c th lun lun cl gii bng cch a ra cc b lc thng thp thch hp bn trong G. Cng nn ch rng cc tng cng c cc cchgii thch ngu nhin. Thay cc hm delta dirac bng n trng c cng mnh hn.

Ch thch 30.3 (bi ton iu khin v bi ton c lng)

Mc d chng ta ang tm mt b iu khin ti u 2H , nhng phi ch rng m hnh tng qut h thng s c th thitk ti u trng thi min l lut iu khin h s hng v ng.

Vi vn trnh by trn, rt thch hp nhc li kt qu c bn sau:

B 30.1 (Chun bnh phng ca mt h thng n nh)

Xt mt h hon ton n nh nhn qu LTI l [ ]A, B, ,=F C . Ta c

2 2 ( )

H H

c oF f CL C B L B+= = =H L R (30.5)

trong cL l tnh iu khin c ca h thng gramian v oL l tnh quan st c ca h thng gramian. Gramian iukhin c l:

0

Hdef

At H A t

c L e BB e dt

= (30.6)

l nghim bn xc nh dng i xng duy nht ca phng trnh i s Lyapunov

0H Hc cAL L A BB+ + = (30.7)

cL l nghim xc nh dng khi v ch khi ( , )A B iu khin c. Gramian quan st c l

0

Hdef

A t H At

o L e C Ce dt

= (30.8)

l nghim bn xc nh dng i xng duy nht ca phng trnh i s Lyapunov

0H Ho oA L L A C C + + = (30.9)

L0 l xc nh dng khi v ch khi (A, C) quan st c

Ch thch 30.4 (ChunH2 c th sai Chun 2L l quan trng)

Mt ch quan trng l chun 2 2/H L (hoc nng lng) ca mt hm c th rt nh, trong khi bn thn hm c thrt ln. Xt mt xung nh v cao, lm v d. S quan st ny l ti hn bi v c nhiu trng hp quan trng chng ta cpnhiu n cao ca hm hn l nng lng ca n. Mt v d in hnh ca trng hp ny l l thuyt n nh Nyquist cin [2,8]. Nyquist pht biu rng cng nh ca hm nhy S=1/(1+L) c kt hp vi vng phn hi m chun l rtquan trng di dng n nh bn vng ca vng phn hi. nhy ln ngha l th Nyquist gn vi im ti hn -1,ngha l mt s giao ng nh (hoc sai s m hnh khng bit) c th dn n h kn khng n nh. gii quyt vn cbn ny chng ta phi s dng hm trng s ph thuc tn s, nhng nhng g chng ta cn l mt chun m n tc ng trctip nh l cc c phn. iu ny dn n ci gi l chun H v chun L min l /H L vn ng vi l thuyt iukhin [4, 11].

Ch thch 30.5 (Tnh ton chun 2H trong MATLAB)

Chun 2H ca h [ ], , ,F A B C D= c th tnh c tnh s dng cc lnh MATLAB sau:

lc=lyap(a, b*b)

twonorn=sqrt(trace(c*lc*c))

hoc

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Vi cu trc cho ca m hnh tng qut G , chng ta c cc mi quan h ca h kn sau:

u Ky= (30.14)

21 22( )K G w G u= + (30.15)

1

22 21[ ]I KG KG w= (30.16)

1

22 21[ ]K I G K G w

= (30.17)1

22 21[ ]y I G K G w= (30.18)

11 12z G w G u= + (30.19)

11 12G w G Ky= + (30.20)

1

11 12 22 21[ [ ] ]G G K I G K G w= + (30.21)

T , ta c ma trn hm truyn ca h kn nh sau:

1

22 21[ ]wuT K I G K G= (30.22)

1

22 21[ ]wyT I G K G

= (30.23)1

11 12 22 21[ ]wzT G G K I G K G= + (30.24)

Chng ta gi mi phng trnh l mt bin i phn on tuyn tnh (LFT - linear fractional transformation) theo K.

Ch thch 30.8 (t yu cu h kn)

Trong php bin i trn, c gi s rng ma trn nghch o -122[I-G ]K tn ti. Yu cu ny c m bo bi gi

thit 22 0D = . Gi thit ny dn n 22 22( ) 0G j D = = v v vy tn ti ma trn nghch o.

V d sau y trnh by cch tnh bi ton nhy hn hp c trng lng 2H nh th no hng ti kt qu thit kh thng iu khin phn hi.

V d 30.1 (Bi ton nhy hn hp c trng lng2

H : nguyn l thit k)V d ny xt vic thit k mt b iu khin K cho m hnh p p p pA , B , C , DP = nh trong hnh 30.2. c K,

chng ta s tnh ti u 2H , s ti u (trc tip hoc gim tip) em li nhiu kt qu l iu quan trng trong thit kca mt vng phn hi tt.

HNH 30.2 Vng phn hi m chun

Cc kt qu thc hin h thng phn hi: ni chung, trong thit k mt b iu khin phn hi K nh hnh 30.2, ngithit k phi xt mi kt qu thc hin ca vng kn sau:

Tnh n nh ca h kn: h kn phi n nh. iu ny i hi tt c cc ma trn hm truyn h kn phi n nh.Mt ma trn hm truyn h kn hp thc cht l ma trn m chun 2H ca n l v hn, dn n ma trn hmtruyn l khng n nh (hoc bin gii n nh). Cc ma trn hm truyn hp thc cht n nh cn phi cchun 2H hn ch.

iu khin theo lnh. h kn phi chng t l iu khin c theo lnh tn s thp tt. Tc l, u ra y (khngb ln vi cc gi tr o ca m hnh chun ha) phi bm vo cc lnh tham chiu tn s thp rn phi c ara vi h thng phn hi. iu ny i hi ma trn hm truyn nhy.

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S tay C in t

1[ ]def

S I PK = + (30.25)

phi nh tn s thp

i

def

d yT SP= (30.26)

def

T I S= (30.27)

Gim nhiu : H kn phi t ra suy gim nhiu tn s thp tt. Vi nhiu u ra 0d c ly mu ti u ra ca i

tng, yu cu ma trn hm truyn nh tn s thp. Vi nhiu u vo id c ly mu ti u vo m hnh cai tng. iu ny yu cu

i

def

d yT SP= (30.26)

phi nh tn s thp.

Gim nhiu cm bin. H kn phi gim nhiu u o tn s cao n tt. iu ny ch yu cn ma trn hm truynnhy b

def

T I S= (30.27)

phi nh tn s cao.

Bn vng n nh: H kn phi t ra bn vng i vi ng hc phi m hnh tn s cao (v d, cc ch bin i,ng hc k sinh, thi gian tr,..); iu ny i hi nh ca ma trn hm truyn h kn phi nh tn s cao.

[ ]actP I P = + (30.28)

+ Sai s m hnh nhn: Vi mt m hnh cho

actP P= + (30.29)

trong actP tng ng vi i tng tht, P tng ng vi m hnh tnh c sai s, v tng ng vi sais nhn n nh ti u ra ca i tng, ma trn hm truyn h kn ph hp (khi c thm ) l T .

+ Sai s m hnh cng: trong actP tng ng vi i tng tht, P tng ng vi m hnh tnh thm c sai s, v tng ngvi sai s cng n nh ti u ra ca i tng, ma trn hm truyn h kn ph hp (khi c thm ) l KS .

Tc ng iu khin ph hp: h kn phi c tc ng iu khin c nh kch c ph hp vi cc lnh thamchiu in hnh v nhiu cm bin. iu ny ch yu i hi kch thc ca KS l iu khin c. Qu nhiuu voK(nh tc ng o hm) c th gip v mt n nh, t c mt gii hn di thng rng v mt giihn pha, nhng n a n kt qu trong iu khin l khng cn thit lm cc lnh tham chiu rv nhiu cmbin n.

Cc lit k trn c rt nhiu vn quan trng c gi gn trong qu trnh thit k h thng iu khin. Mt phn cngvic ca ngi thit k, c th l u tin v chn cc vn quan trng nht. on cui ca phn ny, chng ta s quay lihnh 30.2 v xt thay vo l h thng tng tng (ton hc) c m t nh hnh 30.3.

HNH 30.3 H phn hi m cho bi ton nhy hn hp c trng lng

Cc hm trng s v ma trn hm truyn h kn

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Hnh 30.3 gm c cc hm trng s xc nh s gip chng ta tnh ton ti u ha 2H (trc tip hoc gim tip) hngti cc vn cp trn. Trong hnh trn mt hm trng 1w trn tn hiu y (sai lnh bm), mt hm trng 2w trn tn

hiu iu khin u , v mt hm trng 3w trn u ra i tng 3z . Cc tn hiu cn iu chnh

1

2

3

z

z z

z

=

Lin quan ti cc tn hiu ngoi sinh nh sau:

1 1 1 1z W z W Sw= = (30.30)

2 2 2 2z W z W KSw= = (30.31)

3 3 3 3z W z W Tw= = (30.32)

T , ma trn hm truyn h kn t w ti z c cho bi cng thc

1

2

3

wz

W S

T W KS

W T

=

(30.33)

Trong wzT gm cc ma trn hm truyn nhy, chng ta ni rng, chng ta c bi ton nhy hn hp c trnglng.

Chn cc hm trng

Ni chung, cc hm trng 1 2 3, ,W W W c chn cc ma trn hm truyn n nh n phi c cu trc ng cho (tnht l ti trng thi ban u).

Trng s nhy: Ngi ta chn trng s nhy 1W trn S nh sau:1

1 y yn n

kW I

s = +

(30.34)

Trong 1k , 0 > . Tham s 1k thng c chn ln. Tham s thng c chn nh. Chn sao cho S bpht nng tn s thp Tht vy chng ta mun K lm cho S nh.

Trng s iu khin: Ngi ta chn trng s iu khin 2W trn KS nh sau:

2 2 u un nW k I = (30.35)

Trong 2 0k > dn n pht khng suy bin trn tn hiu iu khin u (tc l trn KS )

Trng s u ra: Ngi ta chn trng s u ra trn 3W trn T nh sau:

3 33

3

( )y yn n

k s zW I

s p +

= +

(30.36)

Vi 3 0k > v 3 3z p< . Sao cho hm trng pht nng T tn s cao.

Ni chung, phi thn trng khi la chn cu trc hm trng. Chn khng ph hp c th dn ti bi ton khng kh thi(ill-posed) v rt kh khn trong qu trnh thit k. V d, 1W phi l hp thc cht vi bi ton 2H - Nu khng chun 2H

ca 1W S l v ngha (no sense) (S tin ti nhn dng tn s cao). Trong khi cha c mt phng php thit k c h thngno trong vic la chn cc hm trng, cc cu trc trn l thun tin tnh ton trong cc ng dng.

Miu t vo/ ra cho i tng tng qut ho G

t c miu t vo/ra (ma trn hm truyn) cho i tng tng qut ho ca chng ta, chng ta phi biu din cctn hiu cn iu chnh 1 2 3, ,z z z v cc gi tr o y di dng cc tn hiu ngoi sinh w v tn hiu iu khin u .

1 1 1 1 1 1 ( )z W z W w Pu W w W Pu= = = (30.37)

2 2 2 2z W z W u= = (30.38)

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S tay C in t

3 3 3 3z W z W Pu= = (30.39)

3y w z w Pu= = (30.40)

T , ta c miu t vo/ra (ma trn hm truyn) cho i tng tng qut ha G:

11 12

21 22

G Gz wG Gy u

=

(30.41)

1 1 1

2 2

3 3

0

0

z W W P

z W wz W P uy I P

=

(30.42)

Miu t khng gian trng thi cho i tng tng qut ho G

Tip theo chng ta thu c khng gian trng thi 2 ca ca G. lm c nh vy, gi s vic miu t khng giantrng thi nh sau:

, , , p p P p pP A B C D x = v i trng th i (30.43)

[ ]1 1 1 1 1 1, , ,W A B C D x= v i trng th i (30.44)



thu c s miu t khng gian trng thi mong mun cho G , chng ta cn miu t cc tn hiu3 3

i i=1 i i=1({x } , ,{z } , )px y& & & theo cc tn hiu3

i i=1({x } , , , )px w u . y ch l vn tnh ton v c lm nh sau:

1 1 1 1 1 1 1 1 1 1 1( ) p p p p p px A x B y A x B w C x D u A x B C x B D u= + = + = & (30.47)

2 2 2 2x A x B u= +& (30.48)

3 3 3 3 3 3 3 3 3 3 3 3 ( ) p p p p p px A x B z A x B C x D u A x B C x B D u= + = + + = + +& (30.49)

p p p px A x B u= +& (30.50)

1 1 1 1 1 1 1 1 1 1 1 1( ) p p p p p pz C x D y C x D w C x D u C x D C x D w D D u= + = + + = + (30.51)

2 2 2 2z C x D u= + (30.52)

3 3 3 3 3 3 3 3 3 1 3 3 ( ) p p p p p pz C x D z C x D C x D u C x D C x D D u= + = + + = + + (30.53)

p p p p p py w C x D u C x w D u= = + (30.54)

Cc phng trnh trn c th c vit theo dng 2 ca chun:

11 12

11 11 12

21 21 22

x A B B x

z C D D w

y C D D u

=

&

(30.55)

Nh sau:

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1 1 1 1

2 2 2 1

3 3 3 3 2

3

1 1 1 1 1

2 2 2

3 3 3 3

p p

p p

p p p

p p p

p p

p p

x A B C B D

x A B x

x A B C B D x

x A B x

z C D C D D D x

z C D w

z C D C D D u

y C I D

=

&&&&

(30.56)

Kim tra gi thit

Trong vic chn hm trng, 1 2 3, ,W W W , chng ta phi m bo rng khng c gi thit no ca bi ton 2H chun b vi

phm. c nh vy, chng ta cn 11 0D = v 22 0D = . c 11 0D = ta cn

1 0D = (30.57)

c 22 0D = ta cn

0pD = (30.58)Kt qu ny dn n

1 1 1 2

2 2 1

3 3 3 2

3

1 1

2 2 2

3 3 3

p

p

p p p

p

p

p

x A B C B

x A x

x A B C x

x A B x

z C x

z C D w

z C C u

y C I

=

&&&&

(30.59)

Trong phn tip theo, s phi chp nhn thm cc gi thit na trong vic miu t khng gian trng thi 2 ca cho itng tng qut ho G . Cc gi thit thm ph thuc vo bi ton 2H c th c xt.

Bi ton nhy hn hp ti u 2H c trng lng

Nh ni trn, bi ton nhy hn hp ti u 2H c trng lng l i tm mt b iu khin n nh c chiu xc

nh sao cho 2wz HT nh nht; tc l:

2

1

2

3

min minwzK K

W S

T W KS

W T

=

H(30.60)

Chng ta s trnh by bi ton iu khin ti u ny v cc bi ton tng t nh vy - c th c gii quyt s dngphn mm thit k c tr gip bi my tnh (v d nh, MATLAB, robust control toolbox, tools ).

[ , , , ] mod(' ')a b c d lin filename=

Ch gii 30.9 (To i tng tng qut G)

i tng tng qut ha G rt d to vi SIMULINK. Cc khi u vo s dng xc nh cc tn hiu ngoi w v cctn hiu iu khin u . Cc khi u ra s dng xc nh cc tn hiu iu chnh z v cc gi tr o y . Lnh linmod sdng to biu khi (file SIMULINK) t c khng gian trng thi 2 ca cho G ( 1 2 1 2, [ ], [ ; ],A B B B C C C = =

11 12 21 22[ , , ]D D D D D= ). C php ca lnh nh sau:

[ , , , ] mod(' ')a b c d lin filename=Phng php ny cho php to m hnh i tng tng qut ha nhanh.

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S tay C in t

Tng quan v bi ton ti u 2H

3 bi ton c bn c xt trong chng ny:

1. Bi ton phn hi u ra 2H . Nghim ca bi ton ny l mt b b ng hc da trn m hnh ti u c cutrc

2 2

u

c f f

opt

c n n

A B G H C H K

G O

=

(30.61)

Trong cG l ma trn h s iu khin (phn hi trng thi) v fH l mt ma trn h s b lc (quan st). cG c tm bng cch s dng nghim ca phng trnh Riccati i s iu khin (CARE Control AlgebraicRiccati Equation) tng t nh bi ton iu chnh ton phng tuyn tnh (LQR Linear Quadratic Regulator).

fH c xc nh bng nghim ca phng trnh Riccati i s lc (FARE) tng t nh bi ton lc Kalman-

Bucy (KBF). Cu trc ca ,opt cK G v fH c coi nh l nghim ca bi ton iu khin Gaussian ton phng

tuyn tnh, bi ton em li nguyn l tch ni ting: Cc im cc ca h kn l gi tr ring ca 2 cA B G , v gitr ring ca 2fA H C .

2. Bi ton phi hi trng thi 2H . Nghim ca bi ton ny l mt b b h s hng ti u (phn hi trng thi) ccu trc

opt cK G= (30.62)

Trong cG l ma trn h s iu khin (phn hi trng thi) c tm bng vic s dng nghim ca CARE.

Tng t nh tm trong bi ton LQR. Cc im cc ca h kn l gi tr ring ca 2 cA B G . Ni tm li, bi tonny c coi nh l mt cch tnh cc ma trn h s iu khin cG c th c s dng trong ng dng phn hitrng thi hoc trong cc ng dng b da trn m hnh.

3. Bi tom ni suy u ra 2H . Nghim ca bi ton ny l mt b b h s hng ti u (tnh) c cu trc.

opt f K H= (30.63)

Trong fH l mt ma trn h s b lc (quan st) c tm bng vic s dng nghim ca phng trnh FARE

- tng t nh cc bi ton KBF. Cc im cc ca h kn l gi tr ring ca 2fA H C . Ni tm li, bi ton nyc coi nh l mt cch tnh cc ma trn h s lc fH c th c s dng trong ng dng c lng trng thihoc trong cc ng dng b da trn m hnh.

30.3 Bi ton phn hi u ra 2H

Trong phn ny, chng ta xt bi ton phn hi u ra 2H . Bi ton ny a n kt qu l b b (ng) da trn m

hnh, bao gm mt ma trn h s iu khin (phn hi trng thi) cG v mt ma trn h s b lc (quan st) fH . Bi tonny tng qut ha cc tng a ra trong l thuyt LQG c in.

Sau y l gi thit chun ca bi ton phn hi u ra 2H

Gi thit 30.1 (Bi ton phn hi u ra)Trong ton b phn ny, s c gi thit l

1. Gi s i tng 22G . 2 2( , , )A B C n nh v xc nh c

Gi thit ny l iu kin cn v tn ti mt b iu khin n nh hon ton K. Vi gi thit ny, b iukhin da trn m hnh lm n nh h kn trong hnh 30.1 l:

2 2 22( )

u

c f c f

c n n

A B G H C D G H K

G O

=

(30.64)

c 2( )cA B G v 2( )fA H C n nh, theo nguyn l tch c in ca l thuyt iu khin tuyn tnh.

Gi thit ny pht biu rng Tt c cc im cc ca h h xu (bn phi trc thc v c phn o) phi

iu khin c bng tn hiu iu khin u v quan st c bng cc tn hiu o y.Gi s l G tha mn gi thit. Xt tng ca b tch phn /I s (hm trng). tng ny c th lm tri

gi thit. S hp th mt b tch phn /I s (hm trng) ln tn hiu ngoi sinh w vo trong G, v d lm tri

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vi gi thit v tnh n nh dn n vic n s a im cc h h n trc o khi s khng iu khinc bng tn hiu u. S hp th mt b tch phn ln tnh hiu iu chnh z vo trong G tri vi gi thit xcnh c s a mt iu cc ca h h ln trc o dn n khng quan st c bng cc cc gi tr o y. Sdng /( )( 0)I s + > thay cho /I s c c G khng tri vi gi thit.

2. Gi thit trong s iu kin khng suy bin 21 12 0TR D D= >

Gi thit ny dn n 12u un XnD R c tt c cc hng ct (tc l, rank 12 uD n= ) v v vy tt c cc tn hiu u

(tn hiu iu khin) u nh hng ti tn hiu iu chnh z thng qua 12D (tc l, 12D khng c phn t nobng khng). Ma trn 12D phi nhiu hng v t ct:

(S tn hiu iu chnh) z un n> (s tn hiu iu khin) (30.65)

Ma trn 21 12u un XnTR D D R= cng c th c hiu nh l mt trng lng trn tn hiu iu khin u - nh ma

trn trng lng iu khin R trong bi ton LQR. Trong bi ton LQR ta ni trng lng iu khin R trn ul khng suy bin. R cng ln, tn hiu iu khin cng nh - b qua tc iu chnh. R ln rn n bngthng iu chnh nh. R cng nh, tn hiu iu khin cho php s cng ln theo th t tng iu chnh. Rnh dn n bng thng chnh nh ln.

3. Gi thit b chnh nh.2

1 12

j I A B

C D

c tt c cc hng ct ( )un n+ vi mi

Gi thit ny dn n ma tran hm truyn t tn hiu iu khin u ti tn hiu cn chnh z 2 1 12( , , , , )A B C D R

khng c im khng no bn phi trc o. Kt hp (1) v (2), s m bo rng Hamiltonian conH gm

2 1 12( , , , , )A B C D R c kt hp vi tn hiu iu khin v tn hiu cn chnh z s ph thuc vo dom(Ric).iu ny m bo li rng gii php ca phng trnh Ricati i s iu khin (CARE) em li ma trn h siu khin sao cho 2 cA B G n nh.

Gi thit ny dn ti G khng quan st c bng cc tn hiu iu chnh z , tc l tt c cc im cc cah h trn trc o phi quan st c bng cc tn hiu z. 1( , )A C , v vy khng th c ch o khng quanst c. y l mt iu kin cn nhng khng . Mt b tch phn a thm vo c cu o s tri vi iukin ny.

T iu kin 12D c tt c cc hng ct, gi thit ny tng ng vi cp

1 1

2 12 1 12 12 1( , ( ) )T TA B R D C I D R D C (30.66)

khng c cc ch o khng quan st c

Nu 12D l vung, th n c ma trn nghch o v gi thit tng ng vi1

2 12 1

TA B R D C khng c ccch o.

Nu 12 1 0TD C = (khng c quan h cho gia cc tn hiu iu khin v cc trng thi), do gi thit ny

tng ng vi 1( , )A C khng c cc ch o khng quan st c.

4. Gi thit trng s o khng suy bin. 21 12 0TD D = >

Gi thit ny dn n w21yn XnD R c tt c cc hng dng (tc l hng ca 21 yD n= ) v v vy cc gi tr o y

c lp tuyn tnh vi 21D (ngha l 21D khng c khng gian trng bn tri). Ma trn 21D v vy phi t hngv nhiu ct tc l

(S gi tr o) yn n< (s tn hiu ngoi sinh) (30.67)

Ma trn w21 12

yn XnTD D R = c th c hiu nh l cng ca cc gi tr o di tc ng ca nhiu y nh l ma trn cng nhiu cm bin trong bi ton KBF. Nh trong bi ton KBF, chng ta ni rng matrn cng lin kt vi cc gi tr o y l khng suy bin. cng ln, chng ta cng mun lc thngthp cc gi tr o y- b qua tc c lng. ln dn n rng bng thng nh c lng kt hp(quan st). cng nh, chng ta cng mun cc gi tr o ca b lc thng thp y cng nh - cn bng smin dch ca chng ta vi nhiu tng tc c lng. nh dn n rng di thng ln trong vic clng kt hp (quan st).

5. Gi thit v b lc. 12 12

j I A BC D

c tt c cc hng dng ( )yn n+ vi mi

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S tay C in t

Gi thit ny hm rng ma trn hm truyn t cc tn hiu ngoi sinh w ti cc gi tr o y khng c imkhng bn tri trc o. Kt hp (1) v (3), s m bo rng ma trn Hamiltonian filH gm 1 2 21( , , , , )A B C D -gm cc tn hiu ngoi sinh v gi tr o y- s ph thuc vo dom(Ric). iu ny m bo li rng nghim ca

phng trnh Ricati i s b lc (FARE) em li ma trn h s b lc ynXn

fH R sao cho 2fA H G n nh.

Gi thit ny hm rng G khng c cc ch o tc l khng iu khin c bng cc tn hiu ngoisinh w, hay tt c cc im cc h h trn trc o phi iu khin c bng cc tn hiu ngoi sinh w. 1( , )A B ,

v vy khng th c cc ch o khng iu khin c. y l iu kin cn, n cha . Mt b tch phna thm vo c cu o s tri vi iu kin ny.

T iu kin 21D c tt c cc hng dng, gi thit ny tng ng vi cp

1 1

1 21 2 1 21 21( , ( ))T TA B D C B I D D (30.68)

khng c cc ch o khng iu khin c

Nu 21D l vung, th n c ma trn nghch o v gi thit tng ng vi 1 21 2TA B D C khng c cc

ch o.

Nu 12 1 0TD C = (khng c quan h cho gia nhiu qu trnh v nhiu cm bin), th gi thit ny th tng

ng vi 1( , )A B khng c cc ch o iu khin c.

Ch thch 30.10 (quan h hai chiu)

Trong phn trnh by trn, ta cn ch cc quan h 2 chiu sau:

TA A (30.69)

2 2

TB C (30.70)

1 1

TC B (30.71)

12 21

TD D (30.72)

12 12 21 21

T TR D D D D= = (30.73)

Dn n:

Tn hiu iu khin u quan h 2 chiu vi gi tr o y

Tn hiu iu chnh z quan h 2 chiu vi tn hiu ngoi sinh w

Cc ma trn Hamiltonian

Kt hp vi bi ton iu khin ti u 2H l 2 ma trn Hamiltonian sau:

2 1

12 1 2

1 1 1 12

0T T

con T T T T

A BH R D C B

C C A C D =

(30.74)

1 1

2 12 1 2 2

1 1

1 12 12 1 2 12 1( ) ( )

T T

T T T T T

A B R D C B R B

C I D R D C A B R D C

= (30.75)

12

21 1 2

1 1 1 21

0T T Tfil T T

A CH D B C

B B A B D =

(30.76)

1 1

1 21 2 2 2

1 1

1 21 21 1 1 21 2

( )

( ) ( )

T T T

T T T

A B D C C C

B I D D B A B D C

=

(30.77)

Ma trn Hamiltonian th nht c k hp vi bi ton iu khin phn hi trng thi ti u hoc bi ton iu chnh. Matrn th 2 c kt hp vi bi ton lc ti u hoc bi ton c lng ti u.

Li gii ca bi ton phn hi u ra 2H c trnh by trong [11,pp.261-262].

nh l 30.1 (Nghim ca bi ton phn hi u ra 2H a ra cc gi thit chun)

Gi s G tha mn cc gi thit 30.1 nhc li cc gi thit bi ton phn hi u tra 2H . chng ta c:

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B iu khin ti u 2H nh nht n chiu (nh i tng tng qut ha G) v cho bi cng thc

2 2

u

c f f

optc n y

A B G H C H K

G O n

=

(30.78)

trong ma trn khuych i iu khin un XncG R c cho bi

1

2 12 1[ ]T T

cG R B X D C

= + (30.79)l nghim xc nh dng duy nht ca CARE:

1 1 1 1

2 12 1 2 12 1 1 12 12 1 2 2( ) ( ) (1 ) 0T T T T T T T A B R D C X X A B R D C C D R D C XB R B X + + = (30.80)

v ma trn khuych i b lc ynXn

fH R c cho bi

1

2 1 21[ ]T TfH YC B D

= + (30.81)

( ) 0filY Ric H = l nghim xc nh dng duy nht ca FARE:

1 1 1 1

1 12 2 1 21 2 1 21 21 1 2 2( ) ( ) ( ) 0T T T T T T A B D C Y Y A B D C B I D D B YC C Y + + = (30.82)

thm na, chun nh nht (min) c cho bi:

22 2

22 1/ 2

1( )wz opt c c f T K M B R G M = +L L

L(30.83)

1 1( ) ( )T T

c ctrace B XB trace RG YG= + (30.84)

trong

2 1 12[ , , ]c c n n cM A B G I C D G= (30.85)

2 1 21[ , , ] f f n n f M A H C I B H D= (30.86)

Cui cng, cc im cc ca h kn l gi tr ring ca 2 cA B G v 2fA H C .

Ch thch 30.11 (Tnh ton b iu khin ti u 2H trong MATLAB)

Sau y l chui cc lnh MATLAB c th c s dng tnh ton b iu khin ti u 2H optK v kt qu ma trn

hm truyn h kn wzT :

tss_g = mksys(a, [b1 b2], [c1; c2], [0ones(nz, nw) d12; d21 0ones(ny, nu),tss)

[ss_k ss_twz] = h21qg(tss_g, schur)

[a_k, b_k, d_k] = branch(ss_k, a,b,c,d)

Lnh mksys tp hp d liu khng gian trng thi 2 ca cho i tng chun ha G a vo cu trc d liu vc t ct(c gi l mt vc t cy) x l cch chn bin tss (khng gian trng thi 2 ca). Tt c cc thng tin v chiu c giim vo vc t ct. Lnh h21qg tnh ton b iu khin ti u 2H optK v h kn t tn hiu ngoi sinh w ti tn hiu iu

chnh z. Mt phng php vc t ring-gi tr ring l phng php mc nh gii 2 phng trnh Riccati i s lin quan.Phng php Schur da trn bin i n nguyn (unitary transformation) ma trn sang ma trn tam gic ca Schur - cth c s dng bng cch a thm la chn schur. Cc kt qu c lu trong vc t cy ss_k v ss_twz, theo tht. Lnh branh sau c s dng ly li s biu din ca khng gian trng thi cho optK t vc t cy ss_k.

Ch thch 30.12 (Mi quan h LQG, bin gii bn vng n nh)

nh l 30.1 gii thiu b iu khin phn hi u ra ti u 2H ging vi cu trc c tm thy trong cc bi tonLQG c in. Trong khi cc cng thc ca bi ton LQR, KBF, v LQG/LTR em li kt qu trong h kn c bin gii nnh bn vng, th b iu khin LQG khng cn c cc bin gii n nh [3].

Cng ging nh vy vi thit k phn hi u ra 2H . Chng ta s trnh by cc bc thao tc gii cc bi tonLQG/LTR sinh ra b iu khin da trn m hnh vi cc bin gii bn vng n nh mong mun so vi cc thit k phn

hi tm c vi cc bi ton LQR v KBF tnh ton ph hp (v d, gii hn h s tng ln v hn, gii hn ca hs gim xung t nht l 6dB, gii hn pha t nht l 060 ).

V d sau trnh by ti u ha nhy hn hp H2 c trng lng c th c s dng thit k mt b iu khin nhth no, cho mt h khng n nh vi thi gian tr.

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S tay C in t

V d 30.12 (Thit k nhy hn hp H2 c trng lng cho h khng n nh c thi gian tr)

Trong v d ny, chng ta xt mt h khng n nh c thi gian trn 0.05 = s (50ms). H c m hnh (xp x) nhsau:

1 2 / 1 40

1 2 / 1 40

s sP

s s s s

= + + (30.87)

c im thit k. Mc ch l thit k mt b iu khin K tha mn cc c im ca h kn nh sau: (1) h kn phin nh, (2) nhy nh hn -60 dB vi mi tn s di 0.1 rad/s, (3) h s nhy khng vt qu khong 2 v 3 rad/s, (4) nhy nh di 5 dB, (5) nhy b nh di 10 dB.

Bi ton nhy hn hp 2H c trng lng. thit k c cc c tnh trn, chng ta a ra mt bi ton nhyhn hp 2H c trng lng vi hm trng 1W trn nhy S v hm trng 2W trn KS; tc l

2

2

1

2

min minwzK K

W ST

W KS

=

H

H

(30.88)

Cc hm trng c s dng l:

11

1

10

0.01

kW

s p s= =

+ +(30.89)

2 22

2

( ) 0.1( 40)

2

k s z sW

s p s

+ += =

+ +(30.90)

1W pht nhy S nng tn s thp (v d di 0.001 rad/s). Trn 0.1 rad/s, 1W nh v 2W pht KS (vi bin ln

hn 1) cho ti khong 4 rad/s. nghim ti u ha 2H ph thuc mt cch rt phc tp vo cc tham s nh ngha 1W v 2W

, khng nn ngc nhin khi xc nh cc tham s ph hp.

Xy dng i tng tng qut ha. i tng tng qut ha G c xy dng s dng SIMULINK v lnh linmod.Kt qu trnh by khng gian trng thi 2 ca th nh sau:

1 1 2

2 1 11 12

2 21 22

0.01 0 40 1 1 0

0 2 0 0 0 1

0 0 0 1 0 00 -W0 0 40 39 0 10

10 0 0 0 0 010 3 0 0 0 0.1

0 0 40 1 1 0

P A B B

G W C D D

P C D D

= = =

(30.91)

Tnh b iu khin ti u 2H . Lnh mksys c s dng xy dng khng gian trng thi 2 ca trn vo mt cutrc d liu vc t cy. Sau s dng lnh h2lqg t c b iu khin ti u. Ch rng i tng chun ha l bc4 (P bc 2, trng s nhy 1W bc 1, trng s iu khin 2W bc 1). B iu khin ti u l:

2

191.0813( 40)( 2)( 0.526)

( 1.925)( 0.01)( 84.15 2133)opts s s

Ks s s s

+ + +=

+ + + + (30.92)

Cng bc 4 - bc ca i tng chun ha G. im cc ti 0.01s = l mt b tch phn xp x - mt h qu ca hmtrng nng 1W t trn nhy tn s thp.

Phn tch h kn: cc im cc thu c ca h kn (2 im cc ca 22P G= , 4 im cc ca b iu khin optK ) l:

1, 2.0786 0.8302, 40, 40, 39.9216s j= (30.93)

nhy thu c, KS v p ng tn s nhy b theo th t nh hnh 30.4-30.6. Cc hnh minh ha tt c cc ctnh thit k trn. nh nhy khong 4.855 dB. nh ca nhy b khong 8.71 dB. p ng KS nm gn trong phnd ca b b gia 0.1 v 10 rad/s.

Tnh chun 2H nh nht. Chun bnh phng nh nht c tnh s dng chui cc lnh MATLAB sau:

lc = lyap(acl, bclbcl)minnorm = sqrt( trace( cc1c*c))

Chun bnh phng nh nht c l 9.0648.

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V d n gin sau minh ha li gii ca bi ton phn hi u ra 2H c th c s dng gii bi ton LQG nh thno

HNH 30.4 p ng tn s nhy thit k2H

HNH 30.5 p ng tn s KS thit k2H

HNH 30.6 p ng tn s nhy b trong thit k2H

V d 30.3 (Thit k LQG/LTR cho m hnh tn la khng n nh bc nht)

Ta xt mt tn la khng n nh c miu t bng mt m hnh bc nht vi trng thi x (t th phng), u vo iukhin u ( lch bnh li thng bng), nhiu qu trnh (gi thi) 1w = , v nhiu cm bin 2w = .

Gi thit l trng tm ca tn la (c.g. - center of garavity) cui tu, tm p sut (c.p. center of pressure) trong sc

nng c tp trung. Gi thit ny dn n s khng n nh trong vic phng tn la. Cng gi thit rng m men qun tnhca tn la i vi trng lng l rt nh. Gi thit ny dn n mt m hnh bc nht n gin. Vn tc gc ca tn la c

gi thit l t l vi t th phng x

v nhiu qu trnh 1w = . Tn hiu cn chnhT

1 2[z ]z z= gm t th phng

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S tay C in t

1z x= v u vo iu khin trng lng 2z u= . y, 0 > l mt tham s thit k c chn

di y. T th phng o c. Gi tr o t th phng y c thm c nhiu cm bin 2w = .

M hnh tn la. M hnh tn la (c tng qut ha) nh sau:

x x u= + +& (30.94)

x

z u

= (30.95)

y x = + 30.96)

trong 0 > l tham s thit k c chn di y.

c tnh thit k. Mc ch l thit k mt b b ti u H2 da trn m hnh ph hp hu t thc, iu ny dn nmt h kn n nh vi mt im cc tri 5s = (setting time 1st s= ).

Xy dng i tng tng qut ha. m hnh trn c th c vit li nh sau:

[1 0]x x u

= + +

& (30.97)

2 2

010

0z x u

= + +

(30.98)

1 10 0y x u

= + + (30.99)

T , dn n

[ ]1 21, 1 0 , 1A B B= = = (30.100)

1 11 2 2 12

01, 0

0C D D

= = =

(30.101)

2 21 22 1 11, 0 , 0C D D = = = (30.102)

Cc gi thit bi ton H2. By gi chng ta kim tra tng gi thit ca bi ton phn hi trng thi, nh pht biutrong gi thit 30.2. T d liu trn, ta c 11 2 20 xD = , 22 1 10 xD = , v 2 2( , , )A B C th n nh v c th xc nh, v

12 12 0TR D D = = > (30.103)

21 21 0TD D = = > (30.104)

T

12 1 0TD C = (30.105)

1 21 0TB D = (30.106)

Cc iu kin v hng trc o gm 2 1 12( , , , )A B C D v 1 2 21( , , , )A B C D trong gi thit 30.2 tr nn tng ng vi

1( , )A C c cc ch khng quan st c tht v 1( , )A B c ch khng iu khin c tht. Khi , tt c cc gi thitbi ton phn hi u ra H2 c tha mn.

i tng. Cui cng, chng ta c hm truyn i tng (hay hm truyn tn la) 22P G= l

1

22 2 2( )P G C sI A B= = (30.107)

1

1s=

(30.108)

22G th khng n nh vi mt im cc bn na phi ca mt phng ti 1s = . 22G cng l pha cc tiu (tc l, khng cim khng no c Res>0).

Ma trn khuych i b lc fH . Phng trnh Riccati i s lc l

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1 2

1 1 2 2

11 0T T TAY YA B B YC C Y Y Y Y

+ + = + + = (30.109)

hay

2 2 0Y Y = (30.110)

p dng cng thc ton phng v chn nghim dng (n nh) ca phng trnh:

2Y = + + (30.111)

Dn n ma trn khuych i b lc:

1

2

11 1TfH YC

= = + + (30.112)

By gi chng ta chn c im cc tri:

2

11 1 1 5f A H C

= = + = (30.113)

Suy ra:

1

24= (30.114)

Hm truyn h h KBF l:

1

2( ) KF f G C sI A H

= (30.115)

6

1s

=

(30.116)

Chng ta s thy rng iu ny s xp x vi hm truyn h h optPK trong thit k cui cng. Trong trng hp ny,

KFG l kt qu ca hm truyn h h.

Ma trn h s iu khin cG : T phng trnh 1 21 0TB D = , kt hp vi phng trnh Riccati i s iu khin

1 2

1 1 2 2

11 0T TA X XA C C XB R B X X X X

+ + = + + = (30.117)

hay

2 2 0X X = (30.118)

p dng cng thc ton phng v chn nghim dng (n nh) ca phng trnh:

2X = + + (30.119)

Dn n ma trn khuych i iu khin:1

2

11 1TcG R B X

= = + + (30.120)

iu ny dn n im cc h kn (b iu chnh) ti

2

1 11 1 1 1cA B G

= + = + (30.121)

Ch rng vi ln (tham kho iu khin t trong cc bi LQR) chng ta c mt im cc h kn ti 1s = , vmt ti 1s = . Chng ta s chn tham s thit k nh (tham kho iu khin r trong cc bi ton iu khin LQR) v

vy im cc h kn (iu chnh) ny1s

th nhanh v im cc b lc h kn ti 5s = l im cc h kn tri.

B b da trn m hnh phn hi u ra ti u 2H : Kt qu b b da trn m hnh phn hi u ra ti u H2 l:

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18/39

S tay C in t

2 2

0u y

c f f

opt

c n n

A B G H C H K

G

=

(30.122)

Trong

2 2

1 11 1 1 1 1c fA B G H C

= + + (30.123)

11 1 1 1 1 24

= + + (30.124)

16 1

= + (30.125)

11 1fH

= + + (30.126)

1 1 24= + + (30.127)

6= (30.128)1

1 1cG

= + + (30.129)

T , hm truyn ca b b l:

1

2 2( )opt c c f f K G sI A B G H C H = + + 30.130)

6(1 1 1/ )

6 1 1/s

+ +=

+ + +(30.131)

Vi nh (iu khin r), iu ny dn n

6(1 )

1/optK

s

+

+(30.132)

Hm truyn h h: Hm truyn h h c kt hp l:

1 1

2 2 2 2( ) ( )opt c c f f PK C sI A B G sI A B G H C H = + + (30.133)

6(1 1 1/ )1

1 6 1 1/s s

+ +=

+ + + (30.134)

Vi nh (iu khin r), iu ny tr thnh

6(1/ )11 1/

optPK s s

+

(30.135)

Khi phc hm truyn vng h (LTR): T y, chng ta thy rng tham s trng iu khin tin ti 0 (iu khin r),hm truyn h h tin ti hm truyn h h KBF KPG ; tc l

22

0

6lim

1optG K

s +

=

(30.136)

KFG= (30.137)

iu ny cho thy rng tin ti 0 (iu khin r), hm truyn h h thc t optPK tin ti hm truyn h h mong

mun KFG . Th tc khi phc mt hm truyn h h mong mun nh trn (vi cc thuc tnh h kn mong mun) s dngmt b iu khin LQG c gi l LQG vi khi phc hm truyn h h hay LQG/LTR.

18


19/39


Chn im cc iu chnh h kn xa: vi , h kn n nh vi cc im cc l 5s = v 1s = . Chn

12500 = l tt. iu ny dn n mt im cc h kn p ng nhanh 50s v lm cho im cc b lc h kn 5s = l

im cc tri nh yu cu.

Gii hn bn vng n nh: Cc thit k LQG v 2H rt ni ting v khng cn gii hn bn vng n nh. Trong thc t, l iu khng tt [3]. Cc thit k LQG/LTR vi i tng pha cc tin (nh l 1/( 1)P s= c m bo gii hn

bn vng n nh. Cc thit k LQG/LTR em li gii hn gn vi cc gii hn ca thit k LQR v KBF; tc l gii hn hs tng ln v hn, gii hn ca h s gim xung t nht l 6dB, gii hn pha t nht l 060 . Thit k LQG/LTR cui cngca chng ta

6 50

1 50optPK

s s

= + (30.138)

cung cp mt gii hn h s tng ln v hn v gii hn h s gim xung l 1/ 6( 15.56 )dB . em li kt qu l tn s cth s khuych i l 5.92 /g rad s = v bin gii pha khong 99.590 l iu khng ti.

V d sau m rng tng LQG/LTR trnh by trong v d 30.3 vi h MIMO tng qut c th thit k vng phnhi (vi bin gii bn vng khng ng k) bng ti u H2.

19


20/39

S tay C in t

V d 30.4 (Thit k iu khin MIMO LQG v LQG/LTR bng ti u H2)

Chng ta xt mt i tng P (MIMO Multi Input Multi Output) c nh ngha bi khng gian trng thi

x Ax Bu= +& (30.139)

y Cx= (30.140)

Gi thit l i tng [A,B,C]P= l n nh v xc nh cMc nh l chng minh li gii ca bi ton phn hi u ra ti u 2H c trnh by c th s dng gii quyt

bi ton LQG MIMO. Chng ta mun trnh by mt phng php a n khi nim LTR nh chng ta s dng mt biu khin LQG da trn m hnh khi phc ma trn hm truyn mong mun vi cc thuc tnh h kn mong mun. ngc ca chng ta th khng phi l mt bi ton iu khin LQG ti u ngu nhin; m l thit k cc lut iu khin vi ccc tnh h kn mong mun.

Xy dng i tng tng qut G: Vi mc ch cui cng l mt b b da trn m hnh c nh ngha bi mt ma trnkhuych i iu khin cG v ma trn khuych i b lc fH , chng ta xt i tng tng qut ha sau:

x Ax L Bu= + +& (30.141)

u un n

Mxz

I

=

(30.142)

y Cx = + (30.143)

trong u l tn hiu iu khin, x l trng thi i tng (tng qut ha), 1w = miu t nhiu qu trnh trong phngtrnh trng thi, 2w = miu t nhiu cm bin trong phng trnh gi tr o, nxnA R , unxnL R , unxnB R , yn xnM R ,

yn xnC R , , 0, 0y un n = > > .

Cc gi thit tham s thit k: phi c 1 trong 2 gi thit sau:

( , )A L khng c cc ch khng iu khin c o v ( , )A M l xc nh c, hoc

( , )A L l n nh v ( , )A M khng c cc ch khng quan st c o.

y, L, M, , v c coi nh l cc tham s thit k c chn theo th t sau cho cc ma trn khuych i iukhin v ma trn khuych i b lc G , v fH sao cho kt qu b b da trn m hnh c c cc c tnh ca h knmong mun.

Khng gian trang thi 2 ca cho i tng tng qut ha G: M hnh trn c th c vit li di dng khng giantrng thi 2 ca nh sau:

0

00 0

0 0 0

0

y

y u y u y y

u u u u y u u

y u y y

n n

n nn n n n

n n n n n n n n

n n n n

A L Bx

xM

zI

yu

C I

=

&

(30.144)

Kim tra cc gi thit phn hi u ra 2H : By gi chng ta chc chn rng tt c cc gi thit ca bi ton phn hi ura 2H trong gi thit 30.2 c tha mn.

Gi thit 22P G= : t m hnh 22 [A,B,C]P G= = l n nh v c th xc nh c, sinh ra 2 2( , , )A B B C C = = l n inh v c th xc nh c.

Gi thit v b chnh nh: T

12

0y u

u u

n n

n n

DI

=

c tt c cc hng ct, sinh ra ma trn trng s iu khin 12 21 0u uT

n xnR D D I = = > khng suy bin.

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T 12 1 0TD C = , dn n iu kin hng (ct) trn trc o gm 2 1 12( , , , )A B C D trong gi thit 30.2 th tng

ng vi 1 12 12 1 12 12 1 1( , ( ) ( , )T TA B R D C I D R D C A C = khng c cc ch khng quan st c o. T ( , )A M

hoc c th xc nh c hoc khng c cc ch khng quan st c o, dn n

1, 0un n

MA C

=

Khng c cc ch khng quan st c o. Kt hp vi ma trn Hamiltonian conH do , hiu ca nghimphng trnh Riccati v ma trn khuych i iu khinABGCl n nh.

Gi thit v b lc: T y y21 n n[0 I ]u y Xn xnD = c tt c cc hng dng, dn n ma trn trng s o l khng suybin.

T 1 21 0TB D = , dn n iu kin hng (dng) trn trc o gm 1 2 21( , , , )A B C D trong gi thit 30.2 th tng

ng vi 11 21 2 1 21 21 1( , ( ) ( , )T TA B D C B I D D A B = khng c cc ch khng iu khin c o. T ( , )A L

hoc n nh hoc khng c cc ch khng iu c o, dn n 1( , 0 )ynxnA B L = khng c cc ch

o khng iu khin c o. Kt hp ma trn Hamiltonian conH nghim phng trnh Riccati v ma trn khuych

i b lc fH l fA H C n nh.

T nhng iu cho trn, dn n l tt c cc gi thit ca bi ton phn hi u ra 2H trong gi thit 30.2 c tha mn.

Ma trn khuych i iu khin: Ma trn khuych i iu khin cG c cho bi cng thc

1 T

cG R B X = (30.145)

trong 0X > l nghim dng duy nht ca CARE:1 0T TA X XA C C XBR BX + + = (30.146)

Thm na, A BG n nh.

Ma trn khuych i b lc: Ma trn khuych i b lc fH c cho bi cng thc

1T

fH YC

= (30.147)Trong 0Y > l nghim dng duy nht ca FARE:

1 0T T TAY YA LL YC CY + + = (30.148)

Thm na, A HC n nh.

HNH 30.7 Vng phn hi m vi b b da trn m hnh LQG v m hnh

B b (LQG) ti u 2H : B b ti u 2H ti thiu ho chun 2H ca ma trn hm truyn t cc tn hiu ngoi sinh

w

=

n cc tn hiu cn chnh

u u

x

n xn

Cz

I

=

c cho bi:

0u

c f f

optc n n

A BG H C H K G

=

30.149)

Ch rng du tr trn cG (ton b phn mu s bn tay tri ca optK ) c b i trong vic thc hin h thng phn hi

m nh hnh 30.7. Theo nguyn l tch, cc im cc ca h kn l cc gi tr ring ca cA BG v fA H C .

21


22/39

S tay C in t

Gii hn bn vng n nh: Phi nhn mnh rng kt qu b iu khin optK , mc d n nh nhng c cc gii hn bnvng n nh khng tt [3]. iu ny bt chp mt thc t l vng iu chnh c kt hp

1( ) LQ cG G sI A B= (30.150)

v vng lc

1( ) KF f G C sI A H = (30.151)

Khi coi nh cc ma trn hm truyn h h MIMO nm trong vng phn hi m, c cc gii hn bn vng n nh rt tt nhsau: gii hn khuych i tng ln v tn, gii hn khuych i gim xung t nht l 6 dB, v gii hn pha t nht l 060 .iu ny sy ra mt cu hi t nhin sau:

C cch no c th chn ma trn khuych i iu khin cG v ma trn khuych i b lc fH

sao cho kt qu b b da trn m hnh optK em li vng phn hi c gii hn tt nh trn?

Rt may, cu tr li l c! Hai phng php em li cc gii hn n nh c th so snh c ti u vo i tng v ura i tng (nhng khng ng thi) c trnh by ngay sau y.

Phng php khi phc hm truyn (LTR Lop Tranfer Recovery Methods) : L phng php thc hin t c mtthit k phn hi vi gii hn n nh tt nh sau. Qu trnh ny gm 2 bc:

1. Thit k h kn mong mun: Bc th nht l thit k mt ma trn hm truyn h h mong mun c cc c tnh ca

h kn mong mun. H h mong mun ny c th c kt hp vi u ra m hnh. Nu nh vy, chng ta k hiu nl oL . Trong trng hp ny, oL biu th optPK mong mun. Nu c kt hp vi u vo m hnh, chng ta k

hiu n l iL . Trng hp ny, iL biu th optK mong mun. (Ni chung l opt opt PK P K P ).

2. Khi phc h h mc tiu (mong mun) bng b b da trn m hnh : Bc th 2 l s dng b b da trn m hnh[ , , ]opt c f K A BG H C H G= khi phc h h mong mun ( oL hoc iL ).

Nu chng ta mun khi phc oL (tc l, cc c tnh tt ti u ra i tng), th chng ta mun opt oPK L . Cchny gi l khi phc hm truyn h h ti u ra i tng (LTRO -loop transfer recovery at the plant output).

Nu chng ta mun khi phc iL (tc l, cc c tnh tt ti u vo i tng), th chng ta mun opt iK L . Cchny gi l khi phc hm truyn h h ti u vo (LTRI -loop transfer recovery at the plant input).

Ch : Ni chung, Cc c tnh c kt hp vi ct vng lp u ra i tng (cc c tnh ca optPK ) th khc

(c l rt khc) vi cc c tnh kt hp ct vng lp ca ti u vo m hnh (cc c tnh ca optK P). N th thng

rt kh c optPK v optK P u c cc c tnh tt (v d gii hn,..). Ngi thit k phi cn bng cc c tnh ttti u ra i tng v cc c tnh tt ti u vo i tng, hoc ngc li.

Cc phng php da trn 2H cho LTRO v LTRI c trnh by nh sau:

Khi phc hm truyn h h ti u ra i tng (LTRO).

1. Thit k h h mong mun oL : Bc th nht l thit k mt h h mong mun1( )o fL C sI A H

= vi ccc tnh h kn mong mun (v d, tnh n nh, nhy, nhy b, gii hn bn vng n nh, ...). Bc ny cth c lm s dng nhiu phng php (tu thuc phng php no bn cm thy tha mi).

Mt th tc em li cc c tnh tt ti u ra i tng c da trn phng php KBF. tng ny l

chn ma trn thit k L sao cho cc gi tr n 1( ) FOL f G C sI A H = tt; v d, gi tr n nh nht ln ti cc tn

s thp, gi tr n ln nht nh ti cc tn s cao, cc gi tr n i qua 0 dB vi dc -20 dB/dec,...

Sau chng ta gii phng trnh Riccati i s b lc vi , , ,y yn xn

A L C I = - s dng 0 > iu chnh

di thng ca h h mong mun 1( )o KF f L G C sI A H = = . cng nh (hoc cng ln) th di thng cng ln

(hoc cng nh).

Nguyn tc thit k h h mong mun l o KFL G= .

Nhn li phng trnh trong min tn s Kalman (KFDE)1 1

[ ( )][ ( )] ( ) ( )

H

H

KF KF FOL FOLI G j I G j I G j G j

+ + = +

(30.152)

T , ta c

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21[ ( )] 1 [ ( )]i KF i FOL I G j G j

+ = + (30.153)

iu ny gi rng bng vic thit k FOLG , ta c th thit k h h mong mun o KFL G= . C th l, nu FOLG ln ti tn s thp, th ta cn

1( ) ( ) KF FOLG j G j

(30.154)

ti tn s thp. T y cho thy rng ma trn L nn c s dng thit k h h mong mun o KFL G= trong khi0 > s dng iu chnh di thng h h mong mun tng hay gim dn n gim hay tng di thng .

Kt qu o KFL G= m bo c cc c tnh h kn tt c miu t nh bn di.

S tng quan gi tr n trn dn n

min[ ( )] 1KFI G j + (30.155)

vi mi . iu ny dn n cc gi tr n nhy kt hp tha mn

max 1

min

1[ ( )] 1

[ ( )]KF KFS j

S j

= (30.156)

vi mi , trong

1( ) [ ( )] KF KF S j I G j = + (30.157)

T mi quan h gi tr n nhy trn, chng ta c c gii hn h KBF nh sau:

gii hn trn ca h s tng ln v hn

gii hn di h s gim t nht l1

(6 )2

dB

gii hn pha t nht 060

Cc bin gii ni trn thch hp vi s thay i h s khuych i ng thi v c lp khi ngt u ra. Ging nhvy vi bin gii pha. Cc bin gii trn khng m bo l thay i h s v pha ng thi. Nn ch rng cc

bin gii c th d dng thay i s dng tiu chun n nh Nyquist. T s quan h v gi tr nhy n trn, ta c mi quan h gi tr nhy n nh sau:

( )max max max[ ( )] [ ( )] 1 [ ( )] 2 6 KF KF KF T j I S j S j dB = + (30.158)

vi mi , trong

1[1 ] KF KF KF KF T I S G G= = + (30.159)

2. Khi phc h h mong mun oL s dng b b da trn m hnh: bc th 2 l s dng mt b b da trn m

hnh [ , , ]opt c f f cK A BG H C H G= trong cG c tm bng cch gii phng trnh CARE vi, , ,

u un xnA B M C R I = = vi l mt gi tr v hng dng rt nh. Khi nh, chng ta gi l bi ton iukhin r.

Nu m hnh [A,B,C]P= l pha cc tiu, th

0lim 0X

+= (30.160)

0lim cG WC

+= (30.161)

vi cc ma trn trc chun W (tc l T TW W WW I = = )

0lim opt oPK L

+= (30.162)

Trong trng hp , opt oPK L vi nh v v vy optPK s c gii hn bn vng gn vi gii hn bnvng ca oL (ti u ra m hnh) Bt k s dng phng php no thit k oL . u phi ch rng iu

kin pha cc tiu trn m hnh P l iu kin . N l khng cn thit. Hn na, cG khng cn tnh s dng

CARE. Trong thc t, cG tha mn iu kin gii hn cho vi ma trn c th nghch o v chc chn rng

23


24/39

S tay C in t

cA BG l n nh (vi nh), s dn n khi phc hm truyn ca h ti u ra i tng. iu ny emli kt qu ca cu trc cc b b da trn m hnh v khng cn g lm vi cc bi ton iu khin ti uv b lc.

Gi thit rng iu kin gii hn gia nguyn vi cc ma trn nghch o W , khi phc vng lp h mongmun ( )o fL C sI A H = c th c chng minh nh sau: Vi chng ta c:

c

WC

G (30.163)

T dn n

1( )opt c c f f PK PG sI A BG H C H = + + (30.164)

1

f

WC WC P sI A B H

+

(30.165)

1

1 1( ) ( ) fWC WC

P sI A I B sI A H

+

(30.166)

11 1( ) ( )

f

WC sI A WC sI AP I B H

+

(30.167)

1 1( ) fC sI A H WPP W

(30.169)

1( ) f oC sI A H L = (30.170)

tng chnh (di dng i s) l tin ti 0, ng phn hi C trong b b[ , , ]opt c f f cK A BG H C H G= c ct (hnh 30.7) v cc c tnh tt gi ti mt ci gi l lm mi v

(chng hn, ma trn hm truyn h h ti v l ( ) )o f L C sI A H = trong hnh 30.7 truyn ti tn hiu sai lch e (u vo b, hoc u ra m hnh) trong h phn hi.

Khi phc hm truyn h h ti u vo i tng (LTRI).

1. Thit k h h mong mun iL : Bc th nht l thit k mt h h mong mun1( )i c L G sI A B

= vi ccc tnh h kn mong mun (v d, tnh n nh, nhy, nhy b, gii hn bn vng n nh, ...). Bc nyc th c thc hin bng nhiu phng php (tu thuc phng php no bn cm thy tha mi).

Mt th tc em li cc c tnh tt ti u vo i tng c da trn phng php LQG. tng ny l chn ma trn thit k M sao cho cc gi tr n 1( )OLG M sI A B

= tt; v d, gi tr n nh nht ln ticc tn s thp, gi tr n ln nht nh ti cc tn s cao, cc gi tr n i qua 0 dB vi dc -20 dB/dec,...

Sau chng ta gii phng trnh Riccati i s iu khin (CARE) vi , , ,y yn xn

A B M I = , s dng

0 > iu chnh di thng ca h h mong mun 1( )i LQG cL G G sI A B= = . cng nh (hoc cng ln)

th di thng cng ln (hoc cng nh).

Nguyn tc thit k h h mong mun l o LQL G= .

Phng trnh trong min tn s LQ (LQDE)

1 1[ ( )] [ ( )] ( ) ( )

H

H

LQ LQ OL OLI G j I G j I G j G j

+ + = +

(30.171)

T , ta c

21[ ( )] [ ( )]i LQ i OL I G j I G j

+ = + (30.172)

iu ny tha nhn rng bng vic thit k OLG , ta c th thit k h h mong mun i LQL G= . C th l, nu

OLG ln ti tn s thp, th ta cn

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1( ) ( ) LQ OLG j G j

(30.173)

ti tn s thp. T y cho thy rng ma trn M nn c s dng thit k h h mong mun i LQL G= trongkhi 0 > s dng iu chnh di thng h h mong mun tng hay gim dn n gim hay tng di

thng . Kt qu i LQL G= m bo c cc c tnh h kn tt c miu t di y.

S quan h gi tr n trn dn nmin[ ( )] 1LQI G j + (30.174)

vi mi . iu ny dn n cc gi tr n nhy kt hp tha mn

( )max 1min

1[ ( )] 1 0

[ ( ) ]LQ LQS j dB

S j

= (30.175)_

vi mi , trong

1[ ] LQ LQS I G= + (30.176)

T mi quan h gi tr n nhy trn, chng ta c c gii hn h LQR nh sau:

Bin gii khuych i tng ln v hn

Bin gii khuych i gim t nht l1

(6 )2

dB

Bin gii pha t nht 060

Cc bin gii ni trn ph hp vi s thay i h s khuych i ng thi v c lp khi ngt u ra. Gingnh vy vi bin gii pha. Cc bin gii trn khng m bo l thay i h s v pha ng thi. Nn ch rng cc bin gii c th d dng thay i s dng tiu chun n nh Nyquist [2, 8].

T s quan h v gi tr nhy n trn, ta c mi quan h gi tr nhy n nh sau:

( )max max max[ ( )] [ ( )] 1 [ ( )] 2 6 LQ LQ LQT j I S j S j dB = + (30.177)

Vi mi , trong

1[1 ] LQ LQ LQ LQT I S G G= = + (30.178)

2. Khi phc h h mong mun iL s dng b b da trn m hnh: bc th 2 l s dng[ , , ]opt c f f cK A BG H C H G= trong fH c tm bng cc gii phng trnh FARE vi

, , ,u un xn

A L B C I = = vi l mt gi tr v hng dng rt nh. Khi nh, chng ta gi y l bi ton cmbin t.

Nu m hnh [A,B,C]P= l pha cc tiu, th

0lim 0Y

+= (30.179)

0lim fH BV + = (30.180)

for some orthonormal W(i.e., WTW= WWT= I )

Vi cc ma trn trc chun W (tc l T TW W WW I = = )

0lim opt iK P L

+= (30.181)

Trong trng hp , opt iK L vi nh v v vy optK P s c gii hn bn vng gn vi gii hn bn

vng ca iL (ti u vo m hnh) Bt k s dng phng php no thit k iL .

Phi ch rng iu kin pha cc tiu trn m hnh P l iu kin . N l khng cn thit. Hn na,

fH khng cn tnh s dng FARE. Trong thc t, (1) fH tha mn iu kin gii hn0

lim fH BV

+

= vi

mt s ma trn c th nghch o V v (2) chc chn rng fA H G l n nh (vi nh), s dn n khi

25


26/39

S tay C in t

phc hm truyn ca h ti u vo i tng. tc l lim opt iK P L= . iu ny em li kt qu ca cu trc cc

b b da trn m hnh v khng cn g lm vi cc bi ton iu khin ti u v b lc.

Gi thit rng iu kin gii hn0

lim fH BV

+

= gia nguyn vi cc ma trn nghch o V , khi phc hm

truyn ca h mong mun 1( )i c fL G sI A H = c th c chng minh nh sau: Vi chng ta c:

f BVH (30.182)

T dn n

1( )opt c c f f K P G sI A BG H C H P = + + (30.183)

1

c

BV BV G sI A BG C p

+ +

(30.184)

1

c

BV BV G sI A C p

+

(30.185)

1

1 1( ) ( )cBV BV

G sI A I C sI A p

+ +

(30.186)

1

1 1( ) ( )cBV BV

G sI A I C sI A p

+

(30.187)

1

1 1( ) ( )cBV BV

G sI A C sI A P

(30.188)

1

1( )c V VG sI A B P P

(30.189)

1( )c iG sI A B L = (30.190)

tng chnh (di dng i s) l tin ti 0, ng phn hi B trong b b[ , , ]opt c f c f K A BG H C G H = c ct (hnh 30.7) v cc c tnh tt gi ti mt ci gi l ci tin

^

u

(chng hn, ma trn hm truyn h h ti^

u l1( ) )iL G sI A B

= trong hnh 30.7 truyn ti u vo i tngu (u vo b) trong h phn hi.

Ch thch 30.13 ( Cc gii hn n nh v nh nhy)

nh trn th nhy rt quan trng trong thit k h thng phn hi. nh ln, chng hn, c th da vo im cc ca hkn nm gn trc o. iu ny r rng l khng mong mun. V vy chng ta mun nhy nh. C th c chng t rngnh nhy thit lp bin gii pha v h s khuych i cn thit.

Gi s rng nhy ti nh l c gii hn bi 1 ; tc l max ( )S j < vi mi . C th thy rng khi h knc c bin gii gii hn (pha v h s khuych i) bn vng n nh nhiu bin nh sau:

1GM

>

(30.191)

1GM

(30.193)

Cc gii hn trn c th d dnh t c s dng tin chun Nyquist SISO [2, 8] nh sau:

Nu

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( )S j < (30.194)

vi mi , th

11 ( )L j

< + (30.195)

vi mi . iu ny dn n th Nyquist kt hp vi L khng th i xuyn qua ng trn tm -1 v bn knh 1/ , im

xa nht bn tri l [( +1)/ ] , xa nht bn phi l [( 1)/ ] . Gii hn trn ca h s khuych i sau im xa nht bnphi ca ng trn. Gii hn di ca h s khuych i sau im xa nht bn tri ca ng trn. Gii hn pha c th tc vi hnh hc nh.

V d sau y xt ng dng ca l thuyt 2H cho mt tay my robot

V d 30.5 (Thit k iu khin TQG/LTR cho tay my r bt 560 PUMA)

Trong v d ny, chng ta trnh by ti u 2H c s dng thit k mt b iu khin LQG/LTR cho mt cnh tay robot560 nh th no. Cnh tay r bt c minh ha nh hnh 30.8.

M hnh tuyn tnh 2 bc t do [ , , ] p p pP A B C = c s dng bt u qu trnh thit k. Tuyn tnh ha m hnh phi

tuyn PUMA [9] im cn bng 0 01 290 0 = = (c 2 khp thng ng), dn n m hnh tuyn tnh sau:

p p p p px A x B = +& (30.196)

p p py C x= (30.197)

1 2[ ]T

pu = (30.198)

1 2 1 2[ ]Tpx = & & (30.199)

[ ]1 2T

py = (30.200)

HNH 30.8 Cnh tay r bt PUMA 560 hai bc t do

0.0000 0.0000 1.0000 0.0000

0.0000 0.0000 0.0000 1.0000

31.7613 33.0086 0.0000 0.0000

56.9381 187.7089 0.0000 0.0000

pA

=

(30.201)

0.0000 0.0000

0.0000 0.0000

1037.7259 3919.6674

3919.6674 2030.8306

pB

=

(30.202)

[ ]2 2 2 20pC I = (30.203)Cc im cc ca h thng l 14.1050, 4.5299s s= = . Phn tch gi tr ring thy rng tnh khng n nh nhanh (fast

instability) ti s=14.1050 ch yu c kt hp vi khp trn (khp ngn hn), tnh khng n nh chm hn (slower

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S tay C in t

instability) ch yu c kt hp vi khp thp hn (di hn). H thng khng c b tch phn no (tc l khng c gi trring no bng khng) v, ng nh mong i cc gi tr n [ ( )]i P j th khng thay i ti tn s thp (xem 30.9).

Mc tiu h kn

Tm mt b iu khin c thc hin vi mt vng phn hi m. H phn hi m th phi tha mn cc c tnh sau:(1) h kn n nh, (2) sai s gia trng thi n nh vi tn hiu step bng 0, (3) p ng tt ti tn s thp nh sau (p ngstep vi qu iu chnh nh 3 s), (4) s suy gim nhiu u vo (disturbance) tn s thp tt, (5) s suy gim nhiu o

(noise) tn s cao tt, (6) gii hn bn vng n nh tn s thp u ra m hnh tt,Mi bc ca qu trnh thit k h thng iu khin ny c m t ngay by gi. tng chnh l thnh lp mt m

hnh thit k dP t m hnh tht P . M hnh thit k dP l ci m chng ta phi thit k.

Bc 1: Thm vo m hnh P vi cc b tch phn c m hnh thit k [A, B, ]dP C=

t c sai s gia trng thi n nh vi tn hiu step bng 0, chng ta bt u bng vic tng thm m hnh

p p pA , B , CP = vi cc b tch phn- mi b trn mt knh iu khin- to thnh m hnh thit k [ ]A, B, CdP = ;

tc l 2 2( / )d XP P I s= . iu ny c lm nh sau:

2 2 2 40 0

p p

AB A

=

(30.204)

HNH 30.9 Cc gi tr n ca cnh tay r bt PUMA 560

2 2

4 2

1

0B

=

(30.205)

2 20 pC C = (30.206)

Trng thi ca h thng ny li

p

xx

x

=

trong ix l trng thi tch phn v px l trng thi m hnh. Cc gi tr n

ca m hnh tng dP

c dc -20 dB/dec ti tn s thp. Nh ta mong mun (hnh 30.10). Gi tr n nh nht i qua dB 0ti 1 rad/s. Gi tr n ln nht i qua dB ti 8 rad/s.

Bc 2: Thit k ma trn hm truyn h h mong mun 1( )o KF f L G C sI A H = =

Tip theo chng ta thit k mt ma trn hm truyn h h mong mun 1( )o KF f L G C sI A H = = c cc c tnh ca h

kn mong mun (v d, cc gi tr n nhy, v tr cc im cc, gii hn n nh, ...) ti u ra. lm c iu ny chngta s dng cc t tng ca b lc Kalman. Nh vng lp LQR c thit k khng cn pht quan h cho cc trng thi iukhin, vng lp lc Kalman c thit k tng t nh trnh by trong gii hn bn vng n nh mong mun (gii hnkhuych i trn v hn, gii hn khuych i di t nht l 6 dB, gii hn pha t nht l 060 ). Thit k vng lp l tngny c thc hin nh sau:

Xt h c ghp thm nh hnh 30.11S dng n thit k ma trn hm truyn mc tiu o KFL G= vi cc c tnh h kn mong mun ti u ra. l

c nh vy, ta bt u thit lp mt h c ghp thm 1( )FOLG C sI A L= vi

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L

H

LL

L

=

(30.207)

1 1[ ( ) ] L p p pL C A B = (30.208)

1( ) H p p LL A B L= (30.209)

HNH 30.10 Cc gi tr n ca m hnh thit k cnh tay r bt PUMA 560

HNH 30.11 H c ghp thm s dng thit k h h mc tiu

Ma trn LL kt hp vi cc gi tr n ca1( )FOLG C sI A L

= ti tn s thp. Ma trn HL kt hp vi cc gi tr

n ti tn s cao. Kt hp, LL v HL lm cho ph hp vi cc gi tr n ca1( )FOLG C sI A L

= trong tonmin tn s (xem hnh 30.12). Ti sao nh vy?

Vic la chn LL v HL ny dn n

1 1( ) ( ) FOL p p H p p p LI

G C sI A L C sI A B Ls

= +

(30.210)

1( ) p p H p LI

C sI A L B Ls

= + (30.211)

1 1( ) ( ) p p p p L p LI

C sI A A B L B Ls

= + (30.212)

1( ) p p p LI

C A B Ls

=

(30.214)

I

s= (30.215)

Kt qu tn s ct khuych i trong hnh 30.12 l 1 rad/s, nh mong mun

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S tay C in t

HNH 30.12 Cc gi tr n FOLG cnh tay r bt PUMA 560

Ti sao li kt hp cc gi tr n ca FOLG theo cch ny? T ng thc Kalman trong min tn s (FKDE)

2

1

1

[ ( )] 1 [ ( )]i KF FOL I G j G j + = + (30.216)

iu ny gi rng bng cch v hnh dng FOLG , chng ta c th nh hnh c mc tiu o KFL G= . c bit l,nu FOLG ln ti tn s thp, th chng ta cn (t ng thc mim tn s Kalman)

1( ) ( ) ( )o KF FOLL j G j G j

= (30.217)

ti tn s thp. iu ny cho thy rng ma trn L c s dng nh hnh thit k h h mc tiu o KFL G= trong 0> c s dng chnh di thng ca h h mc tiu gim/tng s tng/gim di thng ca h h mctiu.

Ch rng bng vic chn L , chng ta lm tt c cc ch khng n nh ca m hnh khng th iu khinc bng L . V vy, ( , )A L l khng iu khin c! Trong khi iu ny xut hin mt s bt li. Lm th no

ma trn Hamiltonian kt hp ph thuc vo dom(Ric) sao cho tn ti fH n nh. Mt iu kin cn v cho iu

ny l ( , )A C c th xc nh c (detectable) v ( , )A L c ch quan st c trn trc o. Tht vy mt trong 2iu kin c tha mn, chng ta c th s dng lnh are tm nghim n nh ca FARE.

Tip theo chng ta gii FARE vi 2 2 ( 0.1)xI = =

1 0T T TAY YA LL YC CY + + = (30.218)

HNH 30.13 Cc gi tr n ca h h mong mun KFG ca cnh tay r bt PUMA 560

vi 0Y . Lnh are c s dng lm iu ny, khi n mang li mt nghim n nh (tn ti duy nht mtnghim). Sau y chng ta thnh lp ma trn khuych i b lc.

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1T

fH YC = (30.219)

2.3635 0.0384

0.40585 0.3091

13.1371 4.2300

4.2300 30.4572

90.2377 83.4384100.9668 467.7679

=

(30.220)

Dn n cc im cc ca h kn mc tiu nh sau ( ( ))i fA H C :

3.1623, 3.1623, 4.5299, 4.5299, 14.1050, 14.1050s = (30.221)

Cc gi tr n i vi ma trn hm truyn h h mc tiu thu c 1( )o KF f L G C sI A H = = nh hnh 30.13.

Cc gi tr n h h mc tiu nh mong mun t KFDE c kt hp tn s thp vi dc =20 dB/dec.Chng ta gi li khong 1 rad/s, sau chng ta tch. iu ny nhn c t /FOLG I s= khng phi l mt vng lpc th thc hin c (khng c tnh n nh ca h kn). Ma trn khuych i b lc thu c em li di thng cnthit lm n nh cnh tay r bt khng n nh, vi cc tnh n nh ti 14.1050,4.5299s = . Mt gi tr n i qua

0 dB trn 10 rad/s, mt gi tr n khc khong 30 rad/s. c s dng iu chnh di thng.Cc gi tr n ca nhy mc tiu tng ng 1[ ] KF KF S I G

= + v cc gi tr n ca nhy b-1

KF[I+G ] KF KF T G= theo th t nh hnh 30.14 v 30.15. Kt hp gi tr n ca nhy v gi tr n ca nhy

th cn h h mc tiu s c:

Cc c tnh iu khin theo lnh tn s thp tt

c tnh gim thiu nhiu tn s thp tt

Gim thiu nhiu cm bin tn s cao, v

Gii hn n nh MIMO tt (gii hn khuych i tng gn v hn, gii hn khuych i gim t nht l 6 dB, v giihn pha t nht l 060 ) ti u ra.

HNH 30.14 cc gi tr n nhy mc tiu 1[ ] KF KF S I G= + ca cnh tay r bt PUMA 560

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S tay C in t

HNH 30.15 Cc gi tr n nhy b mc tiu 1[ ] KF KF KF T G I G= + ca cch tay r bt

Cc gi tr n ca nhy b cho thy rng mt b lc tin x l (a reference command pre-filter) W s lm gim qu iu chnh nh vo lnh tham chiu. Vic thit k b lc s c nghin cu di y.

Bc 3: Gii bi ton iu khin r khi phc h h mong mun ti u ra i tng

Tip theo chng ta gii bi ton iu khin LQR r c c ma trn khuych i iu khin cG sao cho b b da trn

m hnh ti u 2H c f f c[A-BG -H C, H , G ]dK = vi [A,B,C]dP = xp x (ti to) ma trn hm truyn h h mc tiu

o KFL G= ; tc l

d d o KF P K L G = (30.222)

iu ny c thc hin bng cch gii phng trnh Riccati i s iu khin sau (s dng lnh lqr) vi13

2 2 ( 10 )xR I = = :

1 0T T T XA A X C C XBR B X + + = (30.223)

Vi 0X > v dng ma trn khuych i iu khin1 T

cG R B X = (30.224)

987.9832 543.0034 3162945.2928 56.9921 13941.9005 2069.8324

543.0034 3657.5891 11.7867 3162634.3919 2069.7987 3765.7978

=

(30.225)

c c kt qu cc im cc iu chnh h kn nh sau ( ( ))i cA BG :

440.8808, 220.4404 381.7871, 1881.9053, 940.9527 1629.7168s j j= (30.226)

C tt c cc tha s suy gim ln hn hoc bng 0.5 = . Nh mt ch trong thc hnh b iu khin d rng thchin trong thi gian thc, ngi ta c th s dng cc cng ngh gim m hnh [10] b i cc im cc tn s rt cao

trong b b. lm c nh vy phi cho php s dng bc tch phn ln hn trong h nhng thi gian thc hoc trong hthc hin bng vi x l.

Bc 4: Xy dng b iu khin K cui cng

Tip theo chng ta thit lp b iu khin cui cng nh sau:

dKKs

= (30.227)

[ , , ]c f f cA BG H C H G

s

= (30.228)

[ ], ,k k kA B C = (30.229)

Biu din b iu khin ny di dng khng gian trng thi l:

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2 22 2

6 2

00,

0c

K K

c f f

GA B

A BG H C H

= =

(30.230)

[ ]2 2 2 60KC I = (30.231)

Vi cch chn ny, ta c

d

KPK P s= (30.232)

2 2 dI

P Ks

= (30.233)

d dP K= (30.234)

o KFL G = (30.235)

Trong sut phn xy dng b iu khin K, chng ta ti to ma trn hm truyn h h mong mun o GKL G= . Nh vy,K c cha P nghch o (t v phi) trong th t thit k o KFPK L G = . Kim tra gi tr n cho h thc PK thy rng gitr n ca h thc ph hp vi gi tr n mc tiu tng v ln hn 100 rad/s/

Khi phc hm truyn h h: Ti sao chng ta c th khi phc c h h mong mun? Vic khi phc c chophp bi cu trc da trn m hnh ca b b K, trc kia cc phng trnh Riccati s dng c c cc ma trn cG v

fH , v thc t l m hnh p p p[A , B ,C ]P= (v v m hnh thit k [A,B,C]=P(I/s)dP = ) l pha cc tiu. iu kin pha cc

tiu, ni mt cch c th, l mt iu kin cn m bo rng tn ti mt ma trn trc chun ( )T TU U U UU I = = tc l

0lim cG UC

+= (30.236)

Gii hn ny lin quan vi ma trn khuych i iu khin v ma trn C ca m hnh, tuy nhiu, c th c s chngt rng vic khi phc ma trn hm truyn h h c th thc hin c; tc l

0 0lim limd d o KF P K PK L G

+ + = = = (30.237)

Bc 5: lnh thit k b lc tin x l WLnh MATLAB

0 ( * * , , )t a b g h c h g = (30.238)

c s dng tm cc im khng ca b b. Cng l cc im khng ca ma trn hm truyn h kn t r n y. B bcui cng (nh h mc tiu KFG ) c cc im khng gn 1.2s . Dn n b lc tin x l

2 2

1.2

1.2W I

s =

+(30.239)

c thm vo ngoi vng lp lc u vo tham chiu. c c nh vy, chng ta chc chn rng u vo tham chiustep cho 1 v 2 l trng thi n nh (nh c b tch phn trong b iu khin) khng vt qu qu iu chnh trong

thi gian ngn.

p ng tn s nhy

Kt qu cc gi tr n nhy c v nh hnh 30.16. th cho thy rng lnh tham chiu tn s thp r s cbm st v cc nhiu u ra tn s thp od s b suy gim. Chnh xc hn, lnh tham chiu r vi ph tn di 0.3 rad/s

c bm trong khong 20 dB; tc l, vi mt sai lch trng thi n nh khong 10%. Tng t nh vy, nhiu u ra od vi ph tn di 0.3 rad/s s c suy gim trong khong 20 dB.

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S tay C in t

HNH 30.16 p ng tn s nhy PUMA 560 ca tn hiu sai lch

HNH 30.17 Tham chiu vi p ng tn s u ra PUMA 560

Tham chiu ti p ng tn s u ra

Ma trn hm truyn t lnh tham chiu r ti gc lin kt y l

1[ ]ryT I PK PKW = + (30.240)

Gi tr n ca n c v nh Hnh 30.17. th ny cho thy lnh tham chiu tn s thp s c bm trong trng thin nh v qu iu chnh nh s t c trong mt khong thi gian ngn.

HNH 30.18 u ra PUMA 560: p ng ca lnh tham chiu 1

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HNH 30.19 iu khin PUMA 560: p ng ca lnh tham chiu 2

p ng lnh tham chiu step 1

p ng lnh step vi 1 c v nh hnh 30.18. Nh mong i, 1 theo sau lnh step th tt, khng c qu iuchnh v thi gian qu khong 1.6s. p ng kt hp 2 nh, ch ra mt s mc ni cho trong h kn cui cng. Tn hiuiu khin tng ng c v nh hnh 30.19. Vi kch thc c chp nhn.

HNH 30.20 u ra PUMA 560: p ng ca lnh tham chiu 2

HNH 30.21 iu khin PUMA 560: p ng ca lnh tham chiu 2

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S tay C in t

p ng lnh tham chiu step 2

p ng lnh step vi 1 c v nh hnh 30.20. Nh mong i, 2 theo sau lnh step th tt, khng c qu iu chnh

v thi gian qu khong 3 s. p ng kt hp 1 nh, ch ra mt s mc ni cho trong h kn cui cng. Tn hiu iukhin tng ng c v nh hnh 30.21. Kch thc c chp nhn.

30.4 Bi ton phn hi trng thi 2H

Phn ny a ra cc phng php trnh by trong bi ton phn hi u ra c th sn sng tip tc cho php thit kcc lut iu khin phn hi trng thi h s hng tt u 2H (cc ma trn h s hng cG ) cng c.

Cu trc m hnh tng qut ca bi ton phn hi trng thi

Trong trng hp ny, m hnh tng qut G (gm P v cc hm trng) c dng sau:

1 211 12

1 1221 22

0

0 0z w

y w y u

n n

n n n n n n

A B BG G A B

G C DG G C D

I

= = =

(30.241)

T y dn n cc tn hiu o y l cc trng thi x ca m hnh tng qut G. c hiu l tt c cc ch ca A uquan st c bng 2 nxnC I= .

Cc gi thit ca bi ton phn hi trng thi

Cc gi thit ca bi ton phn hi trng thi l tp con ca cc gi thit trong bi ton phn hi u ra. Cc gi thit phnhi trng thi l nh sau:

Gi thit 30.2 (Bi ton phn hi trng thi 2H )

Trong sut mc ny, chng ta gi thit nh sau:

1. Gi thit m hnh 22G . 2( , )A B n nh

2. Gi thit hm trng iu khin khng n. 12 12 0TR D D= > ( 12D c tt c cc hng ct).

3. Gi thit b chnh nh.2

2 12

j I A B

C D

c tt c cc hng ct vi mi

Nn ch rng nu 12 1 0TD C = , th (3) tng ng vi 1( , )A C khng c cc ch o khng quan st c. Nu

1( , )A C l c th xc nh c, th iu ny c tha mn.

Lut iu khin phn hi trng thi ti u 2H

B iu khin ti u 2H nh sau

opt cK G= (30.242)

trong ma trn h s iu khin un XncG R= c cho bi

1

2 12 1[ ]T T

cG R B X D C = + (30.243)

trong 0X l nghim bn xc nh dng duy nht ca CARE:1 1 1 1

2 12 1 2 12 1 1 12 12 1 2 2( ) ( ) ( ) 0T T T T T T T A B R D C X X A B R D C C I D R D C XB R B X + + = (30.244)

Cc im cc h kn c c t lut iu khin phn hi trng thi h s hng trn l gi tr ring ca 2 cA B G . Chunh kn nh nht l

2 1 1min ( )T

wzK

T trace B XB=H

(30.245)

Thit k vng phn hi trng thi

Nu chn

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2B B= (30.246)

1 0un n

MC

=

(30.247)

12

0y

u u

n n

n n

D

I

=

(30.248)

u un nR I = (30.249)

th 12 1 0TD C = v chng ta c

2

1 TcG B X

= (30.250)

trong 0X l nghim bn xc nh dng duy nht ca CARE:

10T T TA X XA M M XB B X

+ + = (30.251)

LQFDE sau y c th nhn c t CARE:

1 1[ ( )] [ ( )] ( ) ( )

H

H

LQ LQ OL OLI G j I G j I G j G j

+ + = +

(30.252)

trong

1( )OLG M sI A B= (30.253)

1( ) LQ cG G sI A B= (30.254)

Cc tng thit k vng lp trnh by trc y c th p dng c. Ngi thit k c th s dng ma trn M v

i lng v hng 0 > thit k OLG c gng t c LQG mong mun. Ma trn M , c th l, c th c s dng lm cho ph hp cc gi tr ring cc tn s cao, tn s thp, tt c cc tn s,... Gi thit l ( , )A B l n nh v ( , )A M

khng c cc ch khng o khng quan st c, m bo tn ti mt nghim n nh. Thm na, kt qu vng lp LQG

s c nhy danh ngha v cc c tnh bn vng n nh h qu ca LQFDE. Kt qu ma trn h s iu khin cG c thc s dng trong mt vng phn hi trng thi, mt vng phn hi trng thi bin i, hoc trong mt b b da trn mhnh.

30.5 Bi ton ni x u ra 2H

Phn ny s cho thy lm th no cc phng php c trnh by trong bi ton phn hi u ra c th vn c tiptc cho php trong thit k c lng trng thi ti u 2H (cc ma trn h s lc fH ).

Cu trc i tng tng qut ca bi ton ni x u ra

Trong trng hp ny (i lp vi trng hp phn hi trng thi), i tng tng qut G (gm P v cc hm trng) cdng sau:

111 12

121 22

2 2

0 0

0 z w z u

y u

n n

n n n n

n n

A B I G G A B

G CG G C D

C D

= = =

(30.255)

T y dn n cc tn hiu iu khin u nm gn trong tt c cc trng thi x ca i tng tng qut G. c hiu ltt c cc ch ca A u iu khin c bng

2 nxn

B I= .

Cc gi thit ca bi ton ni x u ra

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S tay C in t

Cc gi thit ca bi ton ni x u ra l tp con ca cc gi thit trong bi ton phn hi u ra. Cc gi thit ni x ura nh sau:

Gi thit 30.3 (Bi ton ni x u ra 2H )

Trong sut mc ny, chng ta gi thit nh sau:

1. Gi thit i tng 22G . 2( , )A C c th xc nh c

2. Gi thit trng s o khng suy bin. 21 21 0TD D = > ( 21D c tt c cc hng dng).

3. Gi thit b lc.1 21

j I A B

C D

c tt c cc hng ct vi mi

Nn ch rng nu 1 21 0TB D = , th (3) tng ng vi 1( , )A B khng c cc ch o khng iu khin c. Nu

1( , )A B l n nh, th iu ny c tha mn.

Lut iu khin ni x u ra 2H

B iu khin ti u 2H nh sau

opt f K H= (30.256)

Trong ma trn h s b lc ynxn

fH R= c cho bi

1

2 1 21[ ]T T

fH YC B D= + (30.257)

Trong 0Y l nghim bn xc nh dng duy nht ca FARE:

1 1 1 1

1 21 2 1 21 2 1 21 21 1 2 2( ) ( ) ( ) 0T T T T T T A B D C Y Y A B D C B I D D B YC C Y + + = (30.258)

Cc im cc h kn c c t lut ni x u ra trn l gi tr ring ca 2fA H C . Chun h kn nh nht l

( )2 1 1minT

wzK

T trace CYC =H

(30.259)

Thit k vng c lng (lc)Nu chn

1 0 yn nB L = (30.260)

21 0 y u n ny yn nD I = (30.261)

2C C= (30.262)

y yn nI = (30.263)

th 1 21 0TB D = v chng ta c

2

1TfH YC

= (30.264)

trong 0Y l nghim bn xc nh dng duy nht ca FARE:1 0T T TYA AY LL YC CY + + = (30.265)

FFDE sau y c th nhn c t FARE:

1 1[ ( )][ ( )] ( ) ( )

H

H

KF KF FOL FOLI G j I G j I G j G j

+ + = +

(30.266)

trong 1( )FOLG C sI A L

= (30.267)

38


39/39


1( ) KF f G C sI A H = (30.268)

Cc tng thit k vng lp trnh by trc y c th p dng c. Ngi thit k c th s dng ma trn L v ilng v hng 0 > thit k FOLG c gng t c KFG mong mun. Ma trn L , c th l, c th c s dng

lm cho ph hp cc gi tr n cc tn s cao, tn s thp, tt c cc tn s,... Gi thit l ( , )A C l n nh v ( , )A L

khng c cc ch o khng iu khin c, th chc chn tn ti mt nghim n nh. Thm na, vng lp LQG thu cs c nhy danh ngha (nominal sensitivity) v cc c tnh bn vng n nh h qu ca KFDE. Ma trn h s b lcthu c fH (ni x u ra) c th c s dng trong mt vng c lng (phn hi), mt vng c lng (phn hi) bini, hoc trong mt b b da trn m hnh.

30.6 Tm tt

Chng ny trnh by mt m hnh tng qut thit k h thng iu khin bng ti u ha 2H . Trong khi trng tmvn l cc h lin tc LTI, nhng cc phng php ny rt linh hot v c ng dng rng ri. Chng c th c s dng thit k cc lut iu khin phn hi trng thi h s hng, cc c lng trng thi h s hng, cc b iu khin phn hiu ra ng, v hn th na. Cc hm trng c cung cp d dng trong cc m hnh tng qut trnh by. Chc nng nyc th c s dng thc hin cc mc tiu thit k cho h kn. Tt c cc t tng c trnh by c th c mrng vi s thay i kho lo (rt quan trng), em li h thng iu khin bng ti u 2H . bit thm chi tit bn cc th tm trong [8, 11].

Cc phng php trnh by trong chng ny c th c m rng sang h bt bin tuyn tnh ri rc. Cng c th mrng sang cc h thng d liu c ly mu.

Ti liu tham kho

[1] Chen, T. and Francis, B., Optimal Sampled-Data Control Systems, Springer, London, 1995.

[2] Dorf, R.C. and Bishop, R.H., Modern Control Systems, Addison Wesley, 8th edition, CA, 1998.

[3] Doyle, J.C., Guaranteed margins for LQG regulators, IEEE Transactions on Automatic Control, Vol. AC-23, No. 4, August 1978, pp. 756757.

[4] Doyle, J.C., Glover, K., Khargonekar, P.P., and Francis, B.A., State-space solutions to standard H 2 and H controlproblems, IEEE Transactions on Automatic Control, Vol. AC-34, No. 8, 1989, pp. 831847. Also seeProceedings of the 1988 American Control Conference, Atlanta, Georgia, June, 1988.

[5] Kalman, R.E., A new approach to linear filtering and prediction problems, ASME Journal of BasicEngineering, Vol. 85, 1960, pp. 3445.

[6] Kalman, R.E. and Bucy, R.S., New results in linear filtering and prediction problems,ASME Journal of BasicEngineering, 1960, pp. 95108.

[7] Kwakernaak, H. and Sivan, R., Linear Optimal Control Systems, Wiley-Interscience, New York, 1972.

[8] Rodriguez, A.A., A Practical Neo-Classical Approach to Feedback Control System Analysis and Design,Control3D,2000.

[9] Spong, M.W. and Vidyasagar, M., Robot Dynamics and Control, John Wiley and Sons, New York, 1989.

[10] Zhou, K., Doyle, J.C., and Glover, K., Robust and Optimal Control, Prentice-Hall, NJ, 1996.

[11] Zhou, K. and Doyle, J.C., Essentials of Robust Control, Prentice-Hall, NJ, 1998.

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