chuong 2_Điều khiển thích nghi

Chng 2 iu khin thch nghi

Trang 92

Chng 2

IU KHIN THCH NGHI

2.1 Khi nim 2.1.1 nh ngha

Thch nghi l qu trnh thay i thng s v cu trc hay tc ng iu khin trn c s lng thng tin c c trong qu trnh lm vic vi mc ch t c mt trng thi nht nh, thng l ti u khi thiu lng thng tin ban u cng nh khi iu kin lm vic thay i hay : iu khin thch nghi l tng hp cc k thut nhm t ng chnh nh cc b iu chnh trong mch iu khin nhm thc hin hay duy tr mt mc nht nh cht lng ca h khi thng s ca qu trnh c iu khin khng bit trc hay thay i theo thi gian.

H thng c m t trong hnh di y gm 2 vng: - Vng hi tip thng thng - Vng hi tip iu khin thch nghi

Kt lun

1. iu khin thch nghi lin quan n: - S khc nhau trong cc qu trnh ng hc - S khc nhau trong cc nhiu 2. Cc h thng thch nghi l phi tuyn

2.1.2 Nhn dng h thng

Lm th no c c m hnh?


Trang 93

- Vt l (hp trng) - Kinh nghim (hp en) - Kt hp ( hp xm)

K hoch ho thc nghim Chn la cu trc m hnh

- Cc hm chuyn i - p ng xung - Cc m hnh trng thi

Tham s thch nghi - Thng k - Cc vn nghch o

S hp l

2.1.3 c lng tham s thch nghi thi gian thc

1. Gii thiu 2. Bnh phng cc tiu v hi qui 3. H thng ng 4. Cc iu kin thc nghim 5. Cc v d 6. Cc kt lun

2.1.4 Phn loi

C th phn loi cc h thch nghi theo cc tiu chun sau : 1. H thch nghi m hnh tham chiu ( MRAS ) 2. B t chnh nh ( STR ) 3. Lch trnh li 4. H t hc 5. H t t chc


Trang 94

2.1.5 ng dng

T chnh nh Lch trnh li Thch nghi lin tc

Hnh 2.1 S cc ng dng

Qu trnh ng hc

Bin i Hng s

S dng b iu khin vi cc thng s bin i

S dng b bin i vi cc thng s hng

S bin thin khng bit trc

S bin thin bit trc

S dng b iu khin thch nghi

S dng lch trnh li


Trang 95

2.2 H thch nghi m hnh tham chiu MRAS (Model Reference Adaptive Systems)

2.2.1 S chc nng H thng thch nghi s dng m hnh chun l mt trong nhng phng php chnh ca iu khin thch nghi. Nguyn l c bn c trnh by hnh 2.2

Hnh 2.2 S khi ca mt h thng thch nghi m hnh tham chiu

M hnh chun s cho p ng ng ra mong mun i vi tn hiu t (yu cu). H thng c mt vng hi tip thng thng bao gm i tng v b iu khin. Sai s e l sai lch gia ng ra ca h thng v ca m hnh chun e = y - ym. B iu khin c thng s thay i da vo sai s ny. H thng c hai vng hi tip:hi tip trong l vng hi tip thng thng v vng hi tip bn ngoi hiu chnh tham s cho vng hi tip bn trong. Vng hi tip bn trong c gi s l nhanh hn vng hi tip bn ngoi. Hnh 2.2 l m hnh MRAS u tin c ngh bi Whitaker vo nm 1958 vi hai tng mi c a ra: Trc ht s thc hin ca h thng c xc nh bi mt m hnh, th hai l sai s ca b iu khin c chnh bi sai s gia m hnh chun v h thng. M hnh chun s dng

u

y uc

M hnh

C cu hiu chnh

B iu khin i tng

Tham s iu khin

ym


Trang 96

trong h thch nghi bt ngun t h lin tc sau c m rng sang h ri rc c nhiu ngu nhin. Chng ny tp trung vo tng c bn. vn c trnh by mt cch r rng, ta ch tp trung vo cu hnh trong hnh 2.2 c gi l h MRAS song song . y l mt trong nhiu cch c th xy dng m hnh chun. Chng ny cp chnh n h lin tc theo phng php trc tip c ngha l tham s c cp nht mt cch trc tip.

2.2.2 Lut MIT (Massachusetts Institude Technology) ( MIT = Massachusetts Institute Technology : Vin cng ngh Massachusetts)

Hnh 2.3 M hnh sai s H thng thch nghi m hnh tham chiu u tin c a ra gii quyt vn : cc c im ca mt m hnh tham chiu yu cu ng ra l qu trnh l tng cn c p ng i vi tn hiu iu khin nh th no. th minh ha trong hnh 2.2. Trong trng hp ny, m hnh tham chiu mang tnh song song hn l ni tip, ging nh cho SOAS (Self Oscillating Adaptive Systems). B iu khin c th c xem nh bao gm hai vng: mt vng pha trong gi l vng hi tip thng thng c qu trnh v b iu khin. Cc thng s ca b iu khin c chnh nh bi vng ngoi sao cho sai s e gia ng ra y v ng ra m hnh ym l nh nht. V vy vng ngoi cn c gi l vng chnh nh. Vn l xc nh c cu chnh nh cho h thng n nh, ngha l sai s bng zero. iu ny khng th thc hin c. C cu chnh nh vi thng s sau c gi l lut MIT, c s dng cho h MRAS u tin:

=

ee

dtd

s

pipipipi pipipipi

Khu tch phn

u

yuC e

e


Trang 97

Trong phng trnh ny e l sai s ca m hnh e = y ym. Cc thnh phn ca vector e/ l o hm nhy ca sai s i vi cc thng s chnh nh .Thng s xc nh tc thch nghi. Lut MIT c th c gii thch nh sau. Gi s rng cc thng s thay i chm hn nhiu so vi cc bin khc ca h thng. bnh phng sai s l b nht, cn thay i cc thng s theo hng gradient m ca bnh phng sai s e2. Gi s mun thay i thng s ca b iu khin sao cho sai s gia ng ra ca i tng v ca m hnh chun tin ti zero. t e l sai s v l thng s hiu chnh. Ch tiu cht lng :

J( ) = 21

e2 (2.1)

lm cho J() MIN th cn phi thay i cc thng s theo hng m ca gradient J, c ngha l :

=

=

e

eJ

t (2.2)

Gi s rng cc thng s cn thay i thay i chm hn nhiu so vi cc

bin khc ca h thng. V vy o hm

e c tnh vi gi thit l

hng s. Biu thc o hm

e gi l hm nhy ca h thng. Lut iu

chnh theo phng trnh (2.2) vi

e l nhy th c lin h ging nh

lut MIT. Cch chn hm tn tht theo phng trnh (2.1) c th l tu . Nu chn J( ) = e (2.3) Khi lut hiu chnh s l :

)(esignedtd

= (2.4)

Hoc

)(esignesigndtd

=

y gi l gii thut du - du. H ri rc s dng gii thut ny c ng dng trong vin thng ni i hi tnh ton nhanh v thc hin n gin.


Trang 98

Phng trnh (2.2) cn c p dng trong trng hp c nhiu thng s hiu chnh, khi tr thnh mt vector v

e

l gradient ca sai s i

vi cc thng s tng ng. ng dng ca lut MIT c biu din bng hai v d sau : V d 2.1 - Hiu chnh li nui tin Xt vn hiu chnh li nui tin vi m hnh v i tng u c hm truyn l G(S). Sai s l:

e = y ym = G(p) uc G(p) uc vi uc l tn hiu t, ym l ng ra m hnh, y l ng ra i tng, l thng s hiu chnh, v p = d/dt l ton t vi phn. nhy khi y bng :

e

= G(p)uc = ym /

Lut MIT c cho :

dtd

= - yme/

Nu du ca c bit, khi y a ra = / S thay i ca tham s t l vi tch sai s e v ng ra ca m hnh ym. V d trn khng dng vic xp x : Khi lut MIT c p dng vo nhng vn phc tp hn th cn phi c xp x tnh c nhy.

V d 2.2 MRAS cho h bc nht Xt h thng c m t bi phng trnh:

buaydtdy

+= (2.5)

vi u l bin iu khin, y l ng ra c o lng. Gi s mong mun c c h vng kn c m t bi:

dtdym

= - amym + bmuc

M hnh km theo hon ho c th t c vi b iu khin : u(t) = 0t uc(t) 0s y(t) (2.6) vi tham s t0 = bm / b ; s0 = (am a)/b


Trang 99

Ch hi tip s l dng nu am < a, ngha l m hnh mong mun th chm hn qu trnh. p dng lut MIT , s dng sai s e = y ym , vi y l ng ra h kn. Theo phng trnh (2.5) v (2.6) th:

y = 0

0

bsapbt

++uc

vi p l ton t vi phn. nhy c th tnh c bng cch ly o hm ring phn theo tham s ca b iu khin s0 v t0 :

0t

e

=

0bsapb++

uc

0s

e

= - 20

02

)( bsaptb++

uc = -0bsap

b++

y

Cc cng thc ny khng th dng v thng s i tng a v b cha bit. V vy cn phi lm xp x c c lut hiu chnh tham s thc t. thc hin iu ny, u tin cn quan st vi gi tr ti u ca tham s b iu khin, ta c :

p + a + bs0 = p + am Hn na cn ch l b c th c bao gm trong h s tc thch nghi . Bi v n xut hin trong tch b, iu ny i hi du ca b phi c bit. Sau khi xp x, lut cp nht cc tham s iu khin c c l:

(2.7)

V d trn ch cch s dng lut MIT to c lut hiu chnh thng s. Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng): M phng bng Matlab h MRAS trong v d 2.2 (V d 4.2 TLTK[1]) vi a = 1, b = 0.5, am = 2 v bm = 2. Tn hiu vo l sng vung vi bin bng 1 v = 2.

eyapdt

ds

euapdt

dt

m

c

m

+=

+=

1

1

0

0


Trang 100

Vi tnh cht sau cn ch : 1. Khng cn thit i hi mt m hnh km theo hon ho. Cc th tc c th c p dng cho h phi tuyn. Phng php ny cng c th c dng iu khin cho h bit trc mt phn.

2. Cu trc nh hnh 2.3 c mt php nhn gia e v

e.

Ly tch phn phng trnh (2.7) s cho ra cc tham s v c truyn n b iu khin s dng php nhn th hai. 3. S xp x l cn thit c c lut iu khin hiu chnh tham s thc t.

Lut MIT c th thc hin tt nu li thch nghi l nh. ln tu thuc vo bin ca tn hiu chun v li ca i tng. V vy khng th c mt gii hn c nh m bo an ton do lut MIT c th cho mt h vng kn khng an ton. Lut hiu chnh b sung c th c dng bng l thuyt n nh. Nhng lut ny tng t lut MIT nhng cc hm nhy th ng nhin l khc. ny c trnh by nhiu hn trong mc 2.2.4

2.2.3 Ni dung, phng php thit k MRAS C ba phng php c bn phn tch v thit k h MRAS :

Phng php tip cn Gradient Hm Lyapunov L thuyt b ng Phng php gradient c dng bi Whitaker u tin cho h MRAS. Phng php ny da vo gi s tham s ca b hiu chnh thay i chm hn cc bin khc ca h thng. Gi s ny tha nhn c s n nh gi cn thit cho vic tnh ton nhy v cho c cu hiu chnh thch nghi. Phng php tip cn gradient khng cho kt qu cn thit cho h thng kn n nh. B quan st c a ra p dng l thuyt n nh Lyapunov v l thuyt b ng c dng b sung cho c cu thch nghi. i vi h thng c tham s iu chnh c nh trong hnh 2.2, phng php thch nghi s dng m hnh chun cho mt cch hiu chnh tham s tng qut c c hm truyn h thng vng kn gn vi m hnh. y gi l vn m hnh km theo. Mt cu hi t ra l chng ta lm cho sai


Trang 101

lch nh nh th no, iu ny ph thuc bi m hnh, h thng v tn hiu t. Nu c th lm cho sai s bng 0 i vi mi tn hiu yu cu th gi l m hnh km theo hon ho.

M hnh km theo Vn m hnh km theo c th c gii quyt bng thit k phn s cc (miu t ngn gn v thit k phn cc c cho trong ph lc A (TLTK[1])). M hnh km theo l cch n gin thit lp hay gii mt vn iu khin tu ng. M hnh s dng c th l tuyn tnh hay phi tuyn. Cc tham s trong h thng c hiu chnh c c y cng gn vi ym cng tt i vi mt tp cc tn hiu vo. Phng php thch nghi l mt cng c thit k h MRAS, vn ny c trnh by trong mc 2.2.4. Mc d m hnh km theo hon ho ch c th t c trong iu kin l tng nhng phn tch trng hp ny s cho hiu bit su sc vo vn thit k. Xt h 1 u vo,1 u ra c th l lin tc hay ri rc c phng trnh:

y(t) = )(tuAB

(2.8)

vi u l tn hiu iu khin, y l ng ra. K hiu A, B l nhng a thc theo bin S hay Z. Gi s bc ca A bc ca B ngha l h thng l hp thc (i vi h lin tc) v nhn qu i vi h ri rc. Gi s h s bc cao nht ca A l 1.Tm b iu khin sao cho quan h gia tn hiu t uc v tn hiu ra mong mun ym c cho bi :

)(tuAB

y cm

m

m = (2.9)

vi Am, Bm cng l nhng a thc theo bin S hoc Z. Lut iu khin tng qut c cho bi :

(2.10) vi R, S, T l cc a thc. Lut iu khin ny c xem nh va c thnh phn hi tip m vi hm truyn S/R v thnh phn nui tin vi hm truyn T/R. Xem hnh 2.4

SyTuRu c =


Trang 102

Hnh 2.4 H vng kn vi b iu khin tuyn tnh tng qut

Kh u 2 phng trnh (2.8) v (2.10) c phng trnh sau cho h thng vng kn : cBTuyBSAR =+ )( (2.11) t c p ng vng kn mong mun, th AR + BS phi chia ht cho Am, cc zero ca i tng, khi cho B = 0, s l zero ca h kn nu khng b kh bi cc vng kn. Bi v cc im zero khng n nh khng th b kh nn c th phn tch thnh B = B+B-, trong B+ cha nhng thnh phn c th kh i, B- l thnh phn cn li. Theo phng trnh (2.11) AR + BS l a thc c trng ca h thng c phn tch thnh ba thnh phn : kh zero ca i tng:B+ ; cc mong mun ca m hnh c cho bi Am; cc cc ca b quan st A0. V th : AR + BS = B+A0Am (2.12) gi l phng trnh Diophantine ( hay l phng trnh nhn dng Benzout). V B+ c th kh nn :

(2.13) Chia phng trnh (2.12) cho B+ s c: A .R1 + B -.S = A0Am (2.14) V yu cu l phi ging p ng mong mun nn t s (2.11) phi chia ht cho Bm, nu khng th s khng c li gii cho bi ton thit k. V vy : Bm = B -.Bm (2.15) T = A0Bm iu kin m bo tn ti li gii l :

bc( A0) 2 bc(A) - bc( Am) - bc(B+) - 1

Cu y

B iu khin Qu trnh u

SyTuRu C = AB

1RBR+

=


Trang 103

bc( Am) - bc (Bm) bc( A) - bc(B) Nhng iu kin ny c cho trong ph lc A (TLTK[1]). Gi s tt c cc zero u b kh, khi c th vit (2.14) li nh sau :

A0Am = AR1 + b0S Nhn 2 v cho y v dng thm phng trnh (2.8) ta c :

A0.Am.y = BR1u + b0Sy = b0(Ru + Sy) (2.16) Cc thng s v tri bit, v phi cha bit. a thc T c c trc tip t phng trnh (2.15). Cc tham s m hnh ca phng trnh (2.16) by gi c th c dng c lng cc tham s cha bit ca b iu khin (chng 3 TLTK[1]). iu ny dn n h MRAS trc tip. Li gii tng qut c trnh by trong chng 4 TLTK[1].

H tuyn tnh tng qut H SISO c m t bi phng trnh sau:

Ay = Bu Vi c tnh h thng mong mun t c l:

Amym = Bmuc B iu khin: Ru = Tuc - Sy (*) H vng kn c m t:

CuBSARBTy+

=

Thay y vo (*) ta tnh c:

CuBSARAT

u+

=

Sai s l: e = y - ym By gi cn phi xc nh cc o hm ring ca sai s i vi tng tham s hiu chnh tm lut chnh nh thng s cc hm nhy. t ri , si , ti l cc h s ca a thc R, S, T. Cc hm nhy c cho bi:


Trang 104

m

CmC A

uBu

BSARBT

e +

=

=+

=

C

ik

i

uBSAR

BTApr

e2)( uBSAR

Bp ik

+

i = 1,. . , k

li ,,0 =

C

im

i

uBSAR

Bpt

e

+=

i = 0,,m

Trong k = bc(R), l = bc(S), m = bc(T). V phi cc phng trnh trn cn cha A, B l cc thng s cha bit nn khng tnh c cc hm nhy. Mt cch xp x c c lut cp nht c thc t l:

AR + BS A0AmB+

Suy ra cc hm nhy:

uAApB

r

e

m

ik

i 0

Tng t cho si v ti Tuy nhin v phi vn cn B- l cha bit. Nu tt c cc zero u c kh, khi ta c B- = b0. Nu du ca b0 bit c th c th thc hin c lut cp nht thng s. Thnh phn b0 c th c bao gm trong c . Nn c th suy ra lut cp nht hiu chnh cc thng s nh sau:

uAA

pe

dtdr

m

iki

0

= i = 1,, k = bc(R )

yAA

pe

dtds

m

ili

0

= li ,...,0= = bc(S)

Cm

imi u

AAp

edtdt

0

= mi ,...,0= = bc(T)

yBSAR

Bpu

BSARBTBp

s

eil

C

il

i +=

+=

2)(


Trang 105

Nhn xt:

- Cn phi xy dng 3 trng thi ca b lc mAA0

1cho lut hiu chnh trn.

- S thay i cc tham s ny t l vi tch sai s e v tn hiu b lc mAA0

1

- c c lut iu chnh cc tham s trn cn phi gi s cc zero phi n nh v du ca b0 phi c bit. - C th trnh c gi s ny bng cch s dng cc thut ton phc tp hn nh c lng trng thi

Tiu chun cc tiu ho - Lut MIT c th c s dng cho cc hm tn tht khc. - Lut hiu chnh cc thams s c th t c bng cch tnh gradient

hm tn tht i vi cc tham s v s thay i cc tham s phi ngc du vi gradient.

- Phng php ny cn bit cc tham s ca m hnh i tng tnh ton nhy. Tuy nhin iu ny l khng c thc v do c th s dng phng php xp x hay bng cc b c lng thng s.

Sai s v s hi t tham s H thng thch nghi s dng m hnh chun da vo tng l lm cho sai s e = y ym tin ti zero. iu ny khng c ngha l cc tham s iu khin tin ti gi tr ng ca n (v d nh trng hp tn hiu = 0).

V d 2.3 Hi t sai s Gi s h thng c s nh hnh 2.5: Ng ra: y = u Lut iu khin: u = uc M hnh: ym = 0uc Sai s: e = y ym = uc - 0uc = ( - 0)uc Lut hiu chnh tham s theo phng php gradient:


Trang 106

)( 02

=

= cue

edtd

Li gii cho phng trnh vi phn trn l:

tIet += ])0([)( 00 (*) Trong : duI

t

ct )(02

=

(0) l gi tr ban u ca . V v vy sai s e tr thnh:

e(t) = uc(t) tIe ])0([ 0 Do It >0 nn khi t th e(t) 0 ngay c khi tn hiu iu khin uc(t) 0.

Hnh 2.5 M hnh hi t sai s Gi tr gii hn ca ph thuc vo tnh cht ca uc() (hi t hoc phn k) ( do (t) tnh theo biu thc (*) ). V d trn cho bit c sai s e 0 tuy nhin tham s khng tin n gi tr ng ca n. y l tnh cht ca h thng thch nghi s dng m hnh chun. iu kin chnh xc hi t tham s l tn hiu kch thch phi lun tn ti.

0G(s)

-

s

pipipipi

pipipipi

G(s)

M hnh

y i tng

u

e

uc

+

-

ym


Trang 107

n nh ca vng iu khin thch nghi

v d trn bin thin tham s t l vi bnh phng tn hiu iu khin uc. iu ny hp l trong mt s trng hp l khi tn hiu iu khin uc cng ln th cng d pht hin gi tr b sai ca . Tuy nhin thay i ca tham s iu chnh ph thuc vo bin ca tn hiu iu khin c th dn n khng n nh. V d sau y cho lut iu khin khng ph thuc vo uc: V d 2.4 Gi s h thng c m hnh hnh 2.6:

Hnh 2.6 H thng thch nghi m hnh tham chiu cho vic chnh nh li nui tin

Vn l iu chnh 0. Gi s hm truyn c cho bi:

212

1)(asas

sG++

=

Sai s e = G(p)( - 0 ) uc

M hnh Gm

0 G

G pipipipi

pipipipi

-

s

cu

my

e

y +

-

C cu hiu chnh


Trang 108

Trong p biu th cho php ly o hm. V vy:

e

= G(p)uc = 0my

iu chnh tham s theo lut MIT:

m

m yey

ee

edtd

==

= 0 vi 0 =

H thng iu khin thch nghi v vy biu din c bi cc phng trnh vi phn sau:

cm

mm uyadt

dya

dtyd 0

212

2

=++ (I)

cuyadtdy

adt

yd =++ 2122

(II)

mmm yyyyedtd )( == (III)

Phng trnh (I) c th gii c nu cho sn hm uc , xem nh bin ym bit trc

o hm (II) ta c:

dtdu

tudtd

dtdy

adt

yda

dtyd c

c )(222

13

3

+=++

Thay (III) vo ta c:

dtdu

tutytytuty

dtdu

tuyyydtdy

adt

yda

dtyd

c

cmcm

c

cmm

)()()()()(

)()(

2

2

2

1

3

++=

+=++

Suy ra:

)()()()()()( 2222

13

3

tytudt

duttytytu

dtdy

adt

yda

dtyd

mc

c

mc +=+++

y l phng trnh vi phn tuyn tnh bin thin theo thi gian. hiu c h thng, ta thc hin cch th nh sau: - u tin gi s cu l hng s

0cu


Trang 109

- Ng ra m hnh khi s c gi tr cn bng l 0my . Gi s c cu hiu chnh thch nghi c ni vo khi t n im cn bng (trng thi cn bng). Khi phng trnh (II) trn s c cc h s hng v c li gii trng thi cn bng l: 2

000 /)( auyty cm ==

n nh nu 21aa > 20

2

00 )( cmc ua

yu =

Lut hiu chnh b sung Nhng hiu bit c c t vic tnh ton trong v d 2.3 ch ra rng cn phi b sung cho lut MIT. Lut MIT l phng php gradient c bn. gim c c bng lut MIT c quyt nh bi tham s , s ny l do ngi dng chn. C th t c phng php gradient b sung m t l hiu chnh khng ph thuc vo bin ca tn hiu (t) yu cu. Mt kh nng l lm chun ho v thay th lut MIT bi:

+

=

ee

ee

dtd

T

Tham s > 0 c a vo trnh trng hp chia cho 0. C th nhn thy rng t l hiu chnh tham s ph thuc vo bin ca tn hiu yu cu mt lng nh bi v do nhiu o lng.

2.2.4 Thit k MRAS dng l thuyt n nh ca Lyapunov Vi lut hiu chnh tham s c c t phng php Gradient c trnh by trong mc 2.2.3 ly gn ng l c c lut hiu chnh tham s da vo kinh nghim c v hp l ri chng ta th ch ra rng sai s m hnh s tin n 0. Mt kh nng khc c c vng ngoi ca h thng thch nghi s dng m hnh chun l tm ra lut hiu chnh m m bo sai s tin v 0. Nhng nghin cu cho lut hiu chnh nh vy c thc hin trong mt khong thi gian di. tng c bn thit k lut hiu chnh da vo l thuyt n nh c trnh by trong mc ny v c th hin theo lch s pht trin.


Trang 110

tp trung vo vn chnh trnh nhng chi tit khng cn thit, t hiu chnh li nui tin ca h thng c bit trc c dng trong mc ny. H thng dng y ging nh hnh 2.6 nhng c cu thch nghi th khc. Vn l tm lut hi tip bo m sai s e = y ym trong hnh 2.6 tin n 0, cn bit rng vn iu khin h thng vi c tnh ng hc bit trc v h s li cha bit th khng qu kh. Vn ring bit c chn trnh by tng hn l trnh by mt vn thc t. Mt khi tng c bn c pht trin, s m rng n nhng cu hnh tng qut th tng i d hiu hn, chi tit c trnh by trong TLTK[1].

Phng php th hai ca Lyapunov Minh ha bng th phng php Lyapunov Hnh 2.7 (a), (b) v (c) biu din cc trng thi cn bng v nhng ng cong tiu biu tng ng i vi h thng n nh, n nh tim cn v khng n nh. Trong hnh 2.7 (a), (b) hoc (c), vng S() gii hn cho trng thi ban u x0, v vng S() tng ng vi gii hn cho qi o xut pht ti x0. Ch rng nhng nh ngha c cp trc y khng ch ra chnh xc vng ca iu kin cho php ban u. V vy cc nh ngha p dng cho vng ln cn ca trng thi cn bng (l trng thi ti mi o hm u trit tiu), tr khi S() tng ng vi trng thi ban u ca i tng. Ch l trong hnh 2.7 (c), ng cong ri vng S() v dn n trng thi cn bng khng n nh. Tuy nhin, chng ta khng th ni rng ng cong s i n v tn bi v n c th n gn mt vng trn gii hn pha ngoi vng S(). (Nu mt h thng tuyn tnh bt bin theo thi gian l khng n nh, cc ng cong bt u gn vi trng thi cn bng khng n nh i n v cc. Nhng trong trng hp ca h thng phi tuyn, iu ny tht s khng cn thit). S hiu bit v cc nh ngha ni trn l yu cu ti thiu hiu vic phn tch n nh ca cc h thng tuyn tnh v phi tuyn c mt trong phn ny. Ch rng nhng nh ngha ny khng ch hn ch cc khi nim v s n nh ca mt trng thi cn bng. Thc ra, nhng cch nh ngha khc cng c s dng.Chng hn, trong cc l thuyt iu khin thng thng hoc kinh in, ch c cc h thng n nh tim cn mi c gi l h thng n nh, cn cc h thng khc n nh theo Lyapunov, nhng khng n nh tim cn, c gi l khng n nh.


Trang 111

(a) (b) (c) Hnh 2.7 (a) Trng thi cn bng n nh

(b)Trng thi cn bng tim cn (c)Trng thi cn bng khng n nh V d 2.5 Xt h thng c m t bi phng trnh trng thi sau:

1x = x2 - x1( 21x + 22x )

2x = - x1 - x2( 21x + 22x ) Trng thi cn bng (o hm = 0) ti gc ta (x1 = 0, x2 = 0). Nu chng ta nh ngha mt hm v hng V(x) nh sau:

V(x) = 21x + 22x l hm xc nh dng, sao cho o hm theo thi gian hm V(x) theo mt ng cong bt k

V (x) = 2 1x 1x + 2 2x 2x = -2( 21x + 22x )2

l hm xc nh m. iu ny cho thy rng V(x) tng lin tc theo ng cong bt k; v vy V(x) l hm Lyapunov. Hm V(x) tr thnh v hn vi lch v hn t trng thi cn bng, trng thi cn bng gc ca h thng l n nh tim cn trong vng rng. Ch rng nu chng ta V(x) nhn gi tr hng s 0, C1, C2,. . . (0 < C1 < C2


Trang 112

Khi vng trn V(x) = Ck nm hon ton trong vng trn V(x) = Ck+1, mt ng cong i din i qua vng bin gii ca cc ng vin V t ngoi vo trong. T y, biu din hnh hc ca hm Lyapunov c th c pht biu nh sau: V(x) l thc o khong cch ca bin trng thi x t gc to ca trng thi trung gian. Nu khong cch gia gc v bin trng thi tc thi x(t) tng lin tc khi t tng {V[x(t)] < 0 } th x(t) 0. Qu o (1) trn hnh 2.8 l chuyn ng n nh tim cn v gc ta , song khng tho tiu chun n nh th 2 ca Lyapunov: hm )(xV khng phi l hm xc nh m vi mi bin trng thi x. Tiu chun n nh th 2 ca Lyapunov l iu kin , khng phi l iu kin cn nh gi tnh n nh ca nghim phng trnh vi phn phi tuyn. Nu tho tiu chun th h n nh. Nu khng tho, vn kt lun v tnh n nh cn b ng, ph thuc vo: 1.Chn hm V(x) 2.Chn bin trng thi x

Hnh 2.8 Cc vng trn hng s V v hai qu o n nh

V=C1

V=C2

V=C3

V tng

x1

x2 (1) (2)


Trang 113

V d thit k MRAS dng Lyapunov Gi s tt c cc bin trng thi ca h thng u o lng c, nh l v n nh ca Lyapunov c th dng thit k lut iu khin thch nghi m bo s n nh cho h thng vng kn, v d sau trnh by tng ny.

V d 2.6 H MRAS bc nht da vo l thuyt n nh.

Xt bi ton nh trong v d 2.2. Khi tham s ca i tng c bit lut iu khin theo phng trnh 2.6 cho kt qu mong mun. Mt h thch nghi s dng m hnh chun m c th tm ra cc h s t0 v s0 khi tham s a, b khng c bit c th t c nh sau :

Sai s : e = y - ym

Ly o hm v s dng phng trnh 2.5, 2.14 v m hnh mong mun kh o hm y v ym , ta c :

dtde

= -ame + (am a b 0s )y + (b 0t - bm)uc

Ch rng sai s e s tin n 0 nu cc tham s ny bng vi gi tr mong mun. By gi ta cn c gng xy dng mt c cu hiu chnh tham s sao cho cc thng s t0 v s0 tin n gi tr mong mun. S dng cho mc ch ny, hm Lyapunov c dng :

V(e, 0t , 0s ) = 21 [e2 +

b1 (b 0s + a - am)2 + b

1 (b 0t bm)2]

Hm ny s bng 0 khi e = 0 v cc tham s b iu khin bng vi gi tr ti u. o hm ca V l :

dtdV

= edtde

+ 1 (b 0s + a am) dt

ds0 +

1 ( b 0t - bm) dt

dt0

= -ame2 +

1 ( 0bs + a am )( dt

ds0 - ye ) +

1 ( 0bt bm )( dt

dt0 + uce )

Nu cc tham s c cp nhp bi:

dtdt0

= -uce (2.17)


Trang 114

dtds0

= ye

ta c

2eadtdV

m=

Nh vy: Hm V s gim khi e khc 0. V vy c th kt lun l sai s e s tin v 0. Tuy nhin cn ch l cc tham s t0 v s0 s hi t n gi tr cn bng nu khng c cc iu kin khc tc ng vo. V vy lut ny tng t nh lut MIT nhng nhy c thay i bi tn hiu khc. Lut hiu chnh cc thng s lm n nh cho h thng m cc bin trng thi c th o lng c xy dng bng s tng qut ho trc tip ca k thut dng trong v d sau . Lut hiu chnh theo phng trnh 2.17 t c bng cch p dng l thuyt n nh tng t nh bng lut MIT ( so snh vi v d 2.2) trong c hai trng hp, lut hiu chnh c th vit nh sau :

dtd

= e

vi l vector cc tham s , = [-uc y]T khi s dng lut theo Lyapunov v

= [-uc y]T/(p + am) nu s dng lut MIT vector c th c gii thch nh l gi tr m ca gradient hm tn tht. Phng php Lyapunov by gi c p dng cho vic hiu chnh h li nui tin. V d 2.7 y ch xt vn hiu chnh li nui tin. Sai s c cho bi

e = G(p)( - 0 )uc gii thiu mt khng gian trng thi biu th cho hm truyn G. Quan h gia tham s v sai s e c vit bi:

cuBAxdtdx )( 0 += (2.18)

e = Cx


Trang 115

Nu h ng nht x = Ax l n nh tim cn v c tn ti 2 ma trn P v Q xc nh dng sao cho: QPAPAT =+ (2.19) Chn hm Lyapunov nh sau :

V = 21 [xTPx + ( - 0)2]

o hm V v s dng phng trnh sai phn 2.18 c :

dtdV

=

2 (

dtd

dtdxPxPx

dtdx TT )() 0++

S dng phng trnh 2.18 ta c :

dtdV

=

2 {[Ax + B cu ( - 0)]TPx + xTP[Ax + B cu ( - 0)]}

+( - 0)dtd

= -

2

xTQx + ( - 0)(

dtd

+ cu BTPx)

Nu lut hiu chnh tham s c chn l :

PxBudtd T

c

= (2.20)

th o hm ca hm Lyapunov s m khi x 0. Vi lut hiu chnh theo phng trnh 2.20 vector trng thi x v c sai s e = Cx v vy s tin n khng.Tuy nhin ch l sai s tham s - 0 khng cn thit l phi tin n khng.

V d h bc hai MRAS

V d 2.8 Xt G(s) = )( assK+

v m hnh l Gm(s) = AmBm

= 22

2

2

++ ss


Trang 116

a thc A0, R, S v T c chn bi : A0(s) = s + 0a R(s) = s + 1r

S(s) = 0s s + 1s T(s) = 0t s + 1t

Phng trnh Diophantine 2.7 cho li gii sau : 1r = 2 + 0a - a

0s = (2 0a + 2 - a 1r )/K 1s = a0

2/ K

0t = 2/ K

1t = Ka /2

0

n gin ha, ta chn : Q(s) = A0(s).Am(s) P1(s) = Am(s) P2(s) = A0(s) Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng): M phng bng Matlab h bc hai MRAS trong v d 2.8 (V d 4.8) vi = 1, = 0.7, = 1, a0 = 2, a =1 v K = 2.Gi s rng 00 bb = . H thng MRAS ri rc H MRAS c thc hin cho h lin tc khng c nhiu, nhng c th thc hin c MRAS cho h ri rc. Thut gii trn c th c dng cho trng hp h ri rc. B c lng c th da vo chun bnh phng ti thiu. Phn ny dnh trnh by trong b iu khin t chnh nh trong phn 2.3

MRAS cho h thng ch bit c tng phn Trong phn trc ta gi s tt c m hnh ca i tng l cha bit.Trong mt s trng hp c tnh ng hc ca h thng c bit mt phn, cn li l khng bit. S bit trc ny c th c kt hp vo h MRAS. iu ny c th thc hin tu thuc ch yu vo tham s v cu trc ca m hnh i tng. Phng php ny c minh ha bng v d .


Trang 117

iu khin thch nghi cho tay my Gi s cc bin trng thi c o lng y , c th tm c mt bin sai s tuyn tnh i vi cc tham s, iu ny lm d dng trong vic xy dng h thch nghi s dng m hnh chun n nh. iu ny c minh ha bng vic iu khin tay my khi m c tnh ng hc l phi tuyn. Mt thao tc trc tip c m t bi m hnh : H(q) q + C(q, q ) q + G(q) = T (2.21) vi q l vector ta tng qut. H l ma trn qun tnh, C l ma trn tt, G l vector trng trng. Bin iu khin l moment t vo c cu chp hnh.Phng trnh m t tay my c tnh cht :

21

dtd ( q TH q ) = q TH( q ) q + q TC( q, q ) q = q T( T G ) (2.21a)

iu ny c gii thch l o hm ca ng nng q TH q bng vi cng sut c cung cp bi c cu chp hnh v moment trng lc. V d:Tay my hai khp ni Xt tay my hai khp ni vi ti cha bit trong hnh di y. Khp ni th hai vi ti cha bit c xem nh l c thm mt khp ni vi 4 tham s cha bit: khi lng me, moment qun tnh Ie, khong cch t trng tm n khp ni th hai cel , gc e so vi khu lin kt th hai. H thng c m t bi phng trnh (2.21) vi

++

++++=

224232

2423224231

sincossincossin2cos2

qqqqqq

H

+

+++=

4433

121212413 cos)(YY

qeeYYG

vi:

12

2111111

2122124

2122123

222122

2222121

/

)sin()cos()cos()sin(

)cos()cos(2)sin()sin(2

lgelmllme

qqeqqY

qqeqqY

qqqqY

qqqqqY

cc

=

=

++=

++=

+=

=


Trang 118

Tay my hai khp ni vi ti cha bit vi g l gia tc thng v bn tham s cha bit 41 ,, l nhng hm c cc tham s vt l cha bit.

ecee

ecee

cee

eceeec

llmllm

lmIlmlmIlmI

sincos

14

13

212

21

221111

=

=

+=

++++=

Bn tham s cha bit ceee lIm ,, v e c xc nh duy nht bi 41 ,, . H thng c th c vit li:

TqqqT = ),,( vi T c cho bi:

e

l1

me

l c1 q1

m 1

l ce


Trang 119

( )

=

=

+++

+++++

2

1211

4321

41231221

222121221221

)cos(,,,

)sin()cos(0)sin()sin(2)cos()cos(2

qee

T

YqqYqqqqYqqqqYqqqqeqeq

T

vi )cos( 12 qee = v 21 , l cc moment tc dng vo. c tnh ng hc c th c vit di dng tuyn tnh theo cc tham s vi gi s l tt c cc trng thi v gia tc c th o lng c. V d c th c tng qut ho v phng trnh (2.21) c th c vit thnh:

0),,()(),()( qqqqGqqqCqqHT T= vi GCH ,, v l bit trc hay c th o lng c. D l m hnh khng tuyn tnh, n vn tuyn tnh theo cc tham s c th thay i. Mt iu quan trng l kin thc bit trc c dng v h thng khng xem nh l m hnh hp en vi tham s thay i theo thi gian. M hnh th vn cn cha tho mn bi v gia tc phi c o cng vi v tr v vn tc.

t qu o tham kho cho v tr v vn tc l qm v mq . a ra phng trnh Lyapunov nh sau:

( ) ~~~~~)(~21 ++= Tp

TT qKqqqHqV

Trong : 0~,~,~ === mm qqqqqq ; pK v l nhng ma trn xc nh dng. Ly vi phn V , s dng phng trnh (2.21a) cho ta:

~~)~(~

~~)~~(~

~~~~~~

21

~~

++=

+++=

+++=

Tpmm

T

Tpm

T

Tp

TTT

qKqCqHGTq

qKqCqHqHq

qKqqHqqHqV

a ra lut iu khin:

qKqKGqCqHT dpmm ~~ ++= (2.21b)


Trang 120

Lut iu khin bao gm thnh phn nui tin t thnh phn bit ca m hnh v thnh phn t l v hi tip vn tc, ngha l:

~~)~~~~(~ ++= TdmmT qKGqCqHqV Trong :

)()()(~),(),(),(~

)()()(~

qGqGqGqqCqqCqqC

qHqHqH

=

=

=

t:

~~~~ Tmmm GqCqH =+ t c nh vy l do m hnh tuyn tnh i vi cc tham s. Hn na

),,,( mmmm qqqq = c ngha l ch c gia tc ca m hnh phi c bit, khng phi l gia tc thc. Dn n:

)~~(~~~ qqKqV TmTdT ++= ngh cp nht thng s:

)(~~ 11 mTmTm qqq === (2.21c) Hm V tho tnh cht ca hm Lyapunov l xc nh dng v o hm:

qKqV d ~~=

l bn xc nh m. iu ny c ngha l h vng kn n nh v vn tc khi xc lp bng khng. B iu khin cng c th c b sung m bo l sai s v tr bng 0. Lut iu khin theo phng trnh (2.21b) v tham s c cp nht theo phng trnh (2.21c) l cc hm ca bin mm qqqq ,,, v mq , nhng gia tc ca khp ni khng cn thit phi o c. rng lut iu khin l trng hp c bit ca h MRA tng qut vi mqqe = .

2.2.5 Kt lun

Cc tng c bn da trn MRAS c trnh by trong phn ny bao gm :


Trang 121

1. Phng php gradient 2. Thit k theo Lyapunov v siu n nh 3. S gia sai s Trong mi trng hp lut cp nht tham s cho di dng :

dtd

= (2.22)

hay di dng chun ho :

dtd

=

T+

(2.23)

Trong phng php gradient, vector l gi tr m ca gradient sai s theo cc tham s. c lng thng s hay xp x c th c dng trong phng php gradient. Trong nhng trng hp khc l vector li c c bng cch lc ng vo, ra v tn hiu t. S hng l s gia sai s c gii thch l sai s d bo ca vn c lng.Thng dng s gia sai s tuyn tnh theo cc thng s. Phng php gradient linh hot v n gin p dng vo mi cu trc h thng. Cch tnh ton i hi phi xc nh c hm nhy bi v lut hiu chnh da vo vic tnh gradient, c th khng nh l phng php s hi t, c cho bi li thch nghi c chn l nh. Hn na, gi tr ban u ca tham s phi chn h thng vng kn l n nh. Phng php ny s gy khng n nh nu h s li thch nghi ln. Vn l kh tm c gii hn n nh trc. H MRAS tng qut c a ra da vo vic thit k m hnh km theo. Thut gii ny bao gm nhng trng hp c bit ca vic thit k MRAS c trnh by trong cc phn trn. Vic c lng tham s c th c thc hin vi nhiu cch khc so vi phng trnh 2.22 v 2.23.

2.3 B t chnh nh (STR Self Tuning Regulator) 2.3.1 t vn S tng ng chc chn Thng s c lng Phng php gradient Bnh phng cc tiu Cc phng php thit k b iu khin


Trang 122

PID V tr cc LQG (Linear Quadratic Gaussian) B t chnh nh (STR) da trn quan im phn tch, nh gi cc thng s cha bit. tng c bn c minh ho trong hnh 2.9 . Cc thng s cha bit c nh gi trc tuyn (on-line) bng cch dng phng php c lng qui. Cc thng s c lng c xem nh l thng s thc, khng tin cy ca cc c lng l b qua. y gi l qui tc tng ng nht nh (certainty equivalence principle).

Hnh 2.9 M hnh t chnh nh

Nhiu phng php c lng khc nhau c th c vn dng nh xp x c on, bnh phng ti thiu..... Khi design hnh 2.9 tng trng

Thit k b iu khin

S thch nghi

Qu trnh B iu khin

Ng vo

Tham chiu

Cc tham s b iu khin

Cc tham s qu trnh c tnh

Ng ra

B t chnh nh


Trang 123

cho bi gii trc tuyn cc bi ton thit k h thng vi cc thng s cha bit trc. y l bi ton thit k c bn. in hnh cho phng php ny l phng php khc bit cc tiu, bnh phng tuyn tnh, t cc, model following. Phng php thit k c la chn ph thuc vo c tnh ca h thng vng kn. Mc tiu ca mc ny l a ra quan im c bn v tnh cht ca cc b t chnh nh. B t chnh nh ban u ch p dng cho cc h thng ly mu d liu, nhng cc thut ton lin tc v hn hp (hybrid) cng c pht trin. Trong mc ny, gi s h thng l SISO : A(q)y(t) = B(q)u(t) + C(q)e(t) (2.24)

y : u ra u : u vo {e(t)} : chui phn b Gausse A, B, C : cc a thc theo q (ton t sai phn ti). Gi thit bcA = bcC = n v bcA - bcC = d0. Qu trnh iu khin thng c m t dng ton t q-1. a thc c tnh c dng:

)()( 1* = zAzzA n n = bcA. Khi m hnh (2.24) c m t nh sau:

)()()()()()( 1*01*1* teqCdtuqBtyqA += B t chnh nh da trn quan im c lng cc thng s ca qu trnh. Phng php d hiu l c lng cc thng s ca hm truyn ca qu trnh v nhiu (thut ton thch nghi gin tip). Cc thng s ca b chnh nh s khng c cp nht trc tip m l gin tip thng qua c lng m hnh ca h thng. B iu khin thch nghi loi ny da trn phng php bnh phng ti thiu v iu khin bm theo (Kalman 1958). Phng php ny khng da vo c tnh vng kn ca h thng. Cc thng s ca b chnh nh cng c th c lng trc tip gi l thut ton thch nghi trc tip. C 2 phng php trc tip v gin tip u gi l iu khin t chnh nh.

2.3.2 B t chnh nh gin tip Trong phn ny, gi s m hnh ca h thng c phng trnh 2.24. Cch d dng nht l to b t chnh nh theo nh phn 2.3.1 c lng cc thng s ca a thc A, B, C.


Trang 124

Xt trng hp xc nh (e(t) = 0). Nhiu phng php qui cp c th c s dng c lng cc thng s ca A, B.

T = [b0 b1 ... bm a1 ... an ] T(t 1) = [u( t d0) ... u(t d0 m ) y(t 1) ... y(t n)]

trong 0dmn = . Khi b c lng bnh phng cc tiu c cho bi:

[ ][ ] )28.2(/)1()1()()(

)27.2()1()1()1()1()1()()26.2()1()1()()()25.2()()()1()(

1

=

+=

=

+=

tPttKItPttPtttPtK

tttyt

ttKtt

T

T

T

Trong trng hp nhiu l ngu nhin, phng php bnh phng ti thiu cho ra cc c lng sai lch nu C(q) qn. Lc ny, chng ta phi dng cc phng php nh cc i qui, bnh phng cc tiu tng qut.

Tnh hi t Nu tn hiu u vo c kch thch y v cu trc ca m hnh cn c lng thch hp th cc c lng s hi t n mt gi tr thc nu h thng vng kn n nh. iu kin hi t cho cc phng php khc nhau l khc nhau. Trong c 2 trng hp nhiu xc nh (e(t) = 0) v nhiu ngu nhin (e(t) khng ) th iu kin hi t ph thuc tn hiu u vo, qu trnh v nhiu ca h thng. Tn hiu iu khin u(t) c pht i qua khu hi tip. iu ny lm phc tp vic phn tch nhng n cn thit yu cu h thng vng kn phi n nh. Trong MRAS vic phn bit tnh hi t s c cp r hn chng 6 (TLTK[1]).

Bi ton thit k nn tng cho nhng h thng bit trc Nhiu phng php thit k c s dng trong cc b t chnh nh ph thuc vo c tnh ca h thng vng kn. Phng php thit k thng s dng l t cc (pole placement). Phng php da theo m hnh mu (mode following) v phng php t cc c cp phn 2.2 v ph lc A (TLTK[1]).


Trang 125

Xt m hnh ca h thng c phng trnh 2.24 v p ng ca h thng vng kn mong mun l : Am(q).y(t) = Bm(q).uc(t) (2.29) B iu khin l: )()()()()()( tyqStuqTtuqR c = (2.30) R1 v S l gii php cho phng trnh Diophantine mAASBAR 01 =+

(2.31) trong

)35.2()34.2()33.2()32.2(

1

0

RBR

BAT

BBBBBB

m

mm

+

+

=

=

=

=

Mt vi iu kin phi tho mn chc rng b iu khin l nhn qu (causal) (xem ph lc A TLTK[1] ). Cc phng trnh trn l c bn cho nhiu bi ton thit k khc nhau.

Mt kiu mu cho mt b t chnh nh gin tip B t chnh nh gin tip da trn thit k t cc c th biu din trong thut ton sau:

Thut ton 2.1 - B t chnh nh gin tip D liu : Hm truyn p ng xung vng kn mong mun Bm/Am v a thc quan st mong mun A0 c cho trc. Bc 1: c lng cc h s ca a thc A, B, C trong phng trnh (2.24)

dng phng php bnh phng ti thiu t cc phng trnh (2.25) (2.28) Bc 2: Thay A, B, C bng cc c lng t c bc 1 v gii phng trnh (2.31) tm R1, S. Tnh R bng phng trnh (2.35) v T bng phng trnh (2.34). Bc 3 : Tnh tn hiu iu khin t phng trnh (2.30) Lp li bc 1, 2, 3 mi chu k ly mu.


Trang 126

Mt s vn cn ch vi thut ton ny : + Bc ca cc a thc phng trnh 2.24 hoc gii hn bc cao nht phi bit trc. + Tha s chung ca cc c lng A, B c kh nng gii c phng trnh 2.31 + Phi m bo h thng vng kn l n nh. + Cc tn hiu nn kch thch lin tc m bo s hi t ca cc thng s.

.

V d 2.9 B t chnh nh gin tip vi nhiu xc nh Xt h thng c hm truyn :

G(s) = )1(1+ss

Hm truyn ny c xem nh l m hnh c bn ca ng c. Hm truyn p ng xung vi chu k ly mu h = 0.5 l :

H(q) = AB

=

61.061.1090.0107.0

2 +

+

qqq

= )61.0)(1()84.0(107.0

+

qqq

H thng c ly mu c 1 zero = -0.84 bn trong vng trn n v vi h s tt nh. Gi s h thng vng kn mong mun l :

m

m

AB

=

50.032.118.0

2 + qq

iu ny tng ng vi h thng c tn s dao ng t nhin 1 rad/sec v h s tt = 0.7 Gi s a thc quan st l :

A0 = (q 0.5)2 Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng) ng dng Matlab m phng h thng trong v d 2.9 (V d 5.1 (TLTK[1]).Kt qu nhn c c m t hnh (5.2), (5.3) v (5.4) trong TLTK[1]. Hnh 5.2 biu din tn hiu u ra v tn hiu iu khin ca h thng thc khi mt b t chnh nh gin tip c s dng vi phng php bnh phng cc tiu v zero z = - 0.84 ca h thng thc b kh.


Trang 127

Hnh 5.3 ch ra vic c lng cc thng s ca h thng hi t nhanh n cc thng s ca m hnh thc.C s dao ng ln ca tn hiu iu khin do vic kh zero. Dao ng ny l kt qu ca s chn la km trong bi ton thit k c bn ch khng phi ph thuc vo b t chnh nh. Dao ng ny c th trnh c bng cch thay i thit k m khng kh zero ca h thng thc ( chng hn Bm = B). Hnh 5.4 ch ra kt qu khi thay i thit k khng c zero no b kh. p ng ca h thng vng kn by gi c tho mn.

V d 2.10 B t chnh nh vi nhiu ngu nhin : Xt h thng c m t nh sau :

y(t) + ay(t 1) = bu(t 1) + e(t) + c e(t 1) vi a = - 0.9, b = 3, c = -0.3. Bi ton thit k c bn c s dng l iu khin sai lch cc tiu. B iu khin sai lch cc tiu c cho nh sau :

u(t) = - b

ac y(t) = - 0.2y(t)

iu ny dn n h thng vng kn : y(t) = e(t) Phng php cc i qui c s dng c lng cc thng s cha bit a, b v c. Cc c lng t c t phng trnh 2.25 2.28 vi :

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

][

=

=

=

tttyt

ttytut

cab

T

T

T

B iu khin l:

)()()()(

)()()(

0

0

tbtatc

ts

tytstu

=

=

Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng): ng dng Matlab m phng b t chnh nh trong v d 2.10 (V d 5.2 TLTK[1]). Xem kt qu m phng trong hnh (5.5), (5.6) v (5.7) ca TLTK[1]. Hnh 5.5 ch ra kt qu ca m phng thut ton ny. Hnh 5.6 biu din hm chi ph :


Trang 128

V(t) = =

t

iiy

1

2 )(

Khi s dng b iu khin sai lch cc tiu ti u v b t chnh nh gin tip. ng cong cho tn hao tch lu ca STR gn vi ng cong ti u. iu ny c ngha b t chnh nh gn nh ti u ngoi tr khong t qu khi khi ng. Hnh 5.7 biu din thng s ca b iu khin )(0 ts . Tm tt Thut ton t chnh nh gin tip l nhng ng dng n gin ca tng t chnh nh. Chng c th c p dng ti nhiu phng php thit k b iu khin v c lng thng s. C 3 kh khn chnh vi phng php ny. Phn tch tnh n nh l phc tp bi v cc thng s chnh nh ph thuc vo cc thng s c lng. Thng th cn phi gii cc phng trnh tuyn tnh trong cc thng s b iu khin. L trnh t cc thng s qu trnh n cc thng s t chnh c th c cc im k d. iu ny xy ra trong cc phng php thit k da vo phng php t cc, chng hn, nu m hnh c lng c chung cc v zero. Cc cc v zero chung cn phi loi b trc khi tin hnh phng php t cc. Do vic phn tch tnh n nh ch thc hin trong mt s t trng hp. m bo cc thng s hi t n cc gi tr chnh xc th cu trc ca m hnh phi chnh xc v tn hiu u vo phi kch thch lin tc.

2.2.3 B t chnh nh trc tip Khi lng tnh ton cho cc thut ton phn trc tn nhiu thi gian v tnh n nh rt kh phn tch. Nhiu thut ton khc c xut vic tnh ton thit k n gin hn. tng l dng cc c tnh, cc cc v zero mong mun vit li m hnh h thng sao cho cc bc thit k l khng ng k. iu ny dn ti vic thng s ho li m hnh. Nhn phng trnh Diophantine (2.31) vi y(t) v dng m hnh c phng trnh 2.24 th :

[ ] )()()()36.2()()()(

)()()(

1

11

10

tCeRtSytRuBtCeRtSyBtBuR

tSyBtAyRtyAA m

++=

++=

+=

Ch rng phng trnh 2.36 c th c xem nh l mt m hnh ca h thng c thng s ho trong B-, R v S. Vic c lng cc thng s ny to ra cc a thc R v S ca b chnh nh mt cch trc tip. Kt hp


Trang 129

phng trnh 2.34 , tn hiu iu khin c tnh t phng trnh 2.30 . Lu m hnh phng trnh 2.36 l phi tuyn tr phi B- l hng s. Cch khc thng s ho l vit m hnh phng trnh 2.36 nh:

)37.2(10 CeRySuRyAA m ++=

trong RBR = v SBS = Ch a thc R phng trnh (2.36) l monic (a thc c h s bc cao nht bng 1) nhng R phng trnh (2.37) th khng phi monic. Cc a thc R v S c mt tha s chung tng trng cho cc zero tt km. Tha s chung ny nn kh b trc khi tnh ton lut iu khin.

Thut ton 2.2 - B t chnh nh trc tip : Bc 1: c lng cc h s ca a thc R v S m hnh phng trnh (2.37). Bc 2: Kh cc tha s chung trong R v S t c R v S. Bc 3: Tnh tn hiu iu khin t phng trnh 2.30 m R v S c c bc 2. Lp li bc 1, 2, 3 mi chu k ly mu. Thut ton ny trnh vic c lng phi tuyn nhng cn phi c lng nhiu thng s hn khi dng phng trnh 2.36 v cc thng s ca a thc B- c c lng 2 ln. Bc 2 do rt kh thc hin. V vic c lng cc thng s phng trnh 2.36 tng i kh nn ta xt trng hp c bit B- l hng s. Gi s tt c cc zero c th b kh ( 0bB = )

[ ] )38.2()()()()( 100 tCeRtSytRubtyAA m ++=p ng mong mun nh sau:

)()( 0 tTubtyA cmm = Trong : bc(A) = n v 0A chia ht cho T. Sai s (t) = y(t) - ym c cho bi:


Trang 130

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

1

0

0 teAACR

tTutSytRuAA

bt

m

c

m

++=

By gi ta xem xt cc trng hp khc nhau. u tin gi s e = 0. a thc quan st c th c chn t do, khi dng m hnh lin tc theo thi gian th iu cn thit phi gi s b0/(A0Am) l SPR (Strictly Positive Real = Thc dng cht) t c mt MRAS n nh. Ta cng cn lu rng hm truyn c cc h s l s thc dng tho iu kin cn n nh c gi l PR (Positive Real). Hm l SPR nu n n nh vi d tr dng nh tu . Mt iu kin tng t cng l cn thit cho cc m hnh ri rc theo thi gian. Vit li m hnh nh sau:

trong

)()()(1)(

)()()(1)(

)()()(1)(

1*1*0

1*1*0

1*1*0

tuqAqA

tu

tyqAqA

ty

tuqAqA

tu

c

m

cf

m

f

m

f

=

=

=

iu ny tng ng vi trng hp P = Q = A0Am phn 2.2 . Tnh hi t by gi s ph thuc vo du ca b0. iu ny ch ra mi lin h gia MRAS v STR.

Thut ton 2.3 - B t chnh trc tip vi nhiu xc nh D liu : Cho trc gii hn thp nht ca thi gian tr d0 v du ca b0, p ng xung hm truyn vng kn mong mun b0/A*m v a thc quan st mong mun A0. Bc 1 : c lng cc h s ca a thc R*, S*, v T* phng trnh 2.38 dng phng php c lng qui. Bc 2 : Tnh tn hiu iu khin t :

R*u(t) = - S*y(t) + T*uc(t)

)]()()([

])()()([)(

0*

0*

0*

0

0000

dtuTdtySdtuRbAAtu

TAAtyS

AAtuRbt

cfff

m

c

mm

+=

+=


Trang 131

Lp li cc bc 1, 2 mi chu k ly mu. Thut ton ny tng ng vi b iu khin thch nghi dng m hnh chun phn 2.2 . Ch thut ton yu cu b0 phi bit trc. Nu khng bit trc b0 th cng c th c lng c bng cch thay phng trnh 2.38 bng :

A0Amy(t) = Ru(t) + Sy(t) +R1C.e(t) m R by gi khng phi l monic.

Cc b iu khin thay i cc tiu v mc trung bnh di chuyn (Minimum Variance and Moving average)

Cc thut ton iu khin trong trng hp nhiu ngu nhin cho h thng c m t bi phng trnh 2.24 s c xem xt. u tin gi s m hnh bit trc, e l mt nhiu ngu nhin v uc = 0. a thc ca b quan st ti u cho m hnh phng trnh 2.24 l A0 = C. Tiu chun thit k l thay i cc tiu hoc trung bnh di chuyn. Nu qu trnh l cc tiu pha, b chnh nh thay i cc tiu c cho bi: R*(q -1)u(t) = - S*(q -1)y(t) (2.39) Trong R* v S* l nghim c bc cc tiu ca phng trnh Diophantine A* (q -1)R* (q -1) + q 0d B* (q -1)S*(q -1) = B* (q-1)C* (q -1) (2.40) vi d0 = Bc (A) - Bc (B). B iu khin thay i cc tiu tng ng vi m hnh mong mun vi mt khong tr d0 bc, A*m = 1. T phng trnh 2.40 th R* phi chia ht cho B* :

R* = R*1.B*

Trong : Bc(R1) = d0 1. Phng trnh 2.40 c vit li : A*R*1 + q 0d S* = C*

C*y(t) = A*R*1y(t) + S*y(t d0) = B*R*1u(t d0) + S*y(t d0) + R*1C*e(t)

= R*u(t d0) + S*y(t d0) + R*1C*e(t) phng trnh ny c th c vit li :

y(t + d0) = *1

C[R*u(t) + S*y(t)] + R*1e(t + d0) (2.41)


Trang 132

vi b iu khin phng trnh 2.39 th u ra ca h thng vng kn tr thnh :

y(t) = R*1(q-1).e(t) Ng ra v vy l mt trung bnh di chuyn vi bc (d0 -1). Trong strm (1970) ch ra rng b chnh nh s cc tiu s thay i ng ra. Mt c im quan trng l ng ra tr thnh mt trung bnh di chuyn bc (d0 1). Ch s t nhin d0 c din t nh l s mu tri qua u ra thay i khi u vo thay i. B iu khin thay i cc tiu c hn ch l tt c cc zero ca qu trnh u b kh. iu ny c ngha s l kh khn nu B c cc zero bn ngoi vng trn n v. Cc kh khn ny s trnh c b iu khin trung bnh di chuyn. B iu khin ny lm cho ng ra c bc ln hn (d0 1). B iu khin c xut nh sau: tha s B+ v B- trong B vi B+ c cc zero tt nhanh ( zero well damped). Xc nh R* v S* t :

A*R* + q- 0d B*S* = B+ *C*

Phng trnh 2.41 cho ta:

y(t + d) = *

1C

[R*u(t) + S*y(t)] + R*1e(t + d) (2.42)

Trong : +

= BRR *1*

V ng ra c iu khin l mt qu trnh trung bnh di chuyn vi bc (d 1) nn chng ta gi l iu khin trung bnh di chuyn. Ch khng c zero no b kh nu B+ * = 1, c ngha d = bc (A) = n. C 2 lut iu khin thay i cc tiu v trung bnh di chuyn dn n m hnh tng ng ca phng trnh 2.41 v 2.42 . S khc nhau duy nht l gi tr ca d m s iu khin s zero ca qu trnh b kh. Vi d = d0 = Bc(A) - Bc(B) : tt c zero b kh. Vi d = Bc(A) : khng c zero no b kh. Lc vi A*0 trong phng trnh 2.38 cng c th to ra m hnh ca phng trnh 2.42 :

y(t + d) = *

*

0

CA [R*uf(t) + S*yf(t)] + R*1e(t + d) (2.43)

Nu B+ cha tt c cc zero n nh ca h thng th n s tng ng nh b iu khin thay i cc tiu cn ti u trong strm (1970)


Trang 133

B t chnh nh thay i cc tiu v trung bnh di chuyn Thut ton 2.4 - Thut ton t chnh nh trc tip c bn

D liu : Cho trc khong d bo d. Gi k v l tng ng l s thng s trong R* v S*. Bc 1: c lng cc h s ca a thc R* v S*

y(t + d) = R*(q -1)uf(t) + S*(q -1)yf(t) + (t + d) (2.44) trong : R*(q -1) = r0 + r1q -1 +. . . + rkq k S*(q -1) = s0 + s1q -1 + . . . + slq l V

uf (t) = )(1

1*0

qA u(t)

yf (t) = )(1

1*0

qA y(t)

s dng cc phng trnh 2.25 2.28 vi

(t) = y(t) - R*uf (t d) S*yf (t d) = y(t) - T(t d) (t 1)

T = )(1

1*0

qA[u(t) . . . u(t k) y(t) . . . y(t l)]

T = [r0. . . rk s 0 . . .sl] Bc 2: Tnh lut iu khin )()()()( 1*1* tyqStuqR = (2.45) Vi R* v S* c thay bng cc c lng tng ng trong bc 1. Lp li cc bc 1 v 2 mi chu k ly mu. Ch : Thng s r

0 c th c lng hoc gi s bit trc. cc trng hp sau thun li ta vit R* nh sau:

R*(q -1) = r 0 (1 + '1r '1 ... krq ++ kq )

V s dng

(t) = y(t) 0r uf(t d) - T(t d) (t 1)


Trang 134

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

001*0

ltytykturturqA

T=

]......[ 0''1 lkT ssrr= Tnh cht tim cn M hnh phng trnh 2.41 v 2.42 c din t nh l vic thng s ho li m hnh phng trnh 2.24 . Chng tng ng vi m hnh phng trnh 2.44 trong thut ton 2.4 nu A0 c chn bng C. Vector hi qui khng tng quan vi sai s v phng php c lng bnh phng ti thiu s hi t ti thng s tht. Mt kt qu ng kinh ngc l cng t chnh nh chnh xc khi A0 C. Kt qu sau ch ra cc thng s t chnh nh chnh xc c ga tr tng ng vi thut ton 2.4 khi A0 C.

nh l 2.1 Tnh cht tim cn Xt thut ton 2.4 vi A*0 = 1 dng phng php c lng bnh phng cc tiu. Thng s b0 = 0r c th c nh hoc c c lng. Gi s vector hi qui c gii hn, v cc c lng l hi t. H thng vng kn t c trong iu kin gii hn c c im

kdddtutyldddtyty

++==+

++==+

,...,1,0)()(,...,1,0)()(

(2.46)

trong du gch ch gi tr trung bnh theo thi gian; k, l l s cc thng s c lng trong R* v S*.

Chng minh: M hnh ca phng trnh 2.44 c th c vit li: y(k + d) = T(k) + (k + d)

v lut iu khin tr thnh:

0)()( =+ dkkT Ti mt trng thi cn bng, cc thng s c lng l nhng hng s. Hn na, chng tho mn cc phng trnh chun, trong trng hp ny c vit li nh sau:

==

+=+t

k

Tt

kdtkk

tdkyk

t 11)()()(1)()(1

S dng lut iu khin


Trang 135

==

++=+

t

k

Tt

k ttdkdtkk

tdkyk

t 11)]()()[()(1lim)()(1lim

Nu thng s c lng )( t hi t khi t , v cc vector hi qui b gii hn th v phi s tin ti zero. Phng trnh 2.46 by gi ko theo

1*0 =A v xc nh v s hi qui vector trong thut ton 2.4

nh l 2.2 Tnh cht tim cn 2 Gi s thut ton 2.4 vi phng php c lng bnh phng cc tiu c p dng cho phng trnh 2.24 v: min(k, l) n 1 (2.47) C ngha tn hiu ra l qu trnh c mc trung bnh di chuyn bc (d -1). Nu cc c lng tim cn ca R v S lin quan vi nhau, nghim trng thi cn bng l:

)()( tyty + = 0 = d, d + 1,..... (2.48)

Chng minh: H thng vng kn c m t nh sau: )()(* tSytuR =

A* y(t) = B*u(t d0) + C*e(t) V vy

(A*R* + 0dq B*S*)y = R*C*e (A*R* + 0dq B*S*)u = eCS **

Tn hiu c nh ngha (A*R* + 0dq B*S*) = C*e (2.49) V vy:

*Ry = v *Su = iu kin ca phng trnh 2.46 a n

)()( tytR + = 0 = d, d + 1, ..., ld + )()( tytS + = 0 = d, d + 1, ..., kd +

t


Trang 136

)()()( tytC y += cc phng trnh trn c th c vit li:

l

l

l

k

k

k

ssss

ssss

ssss

rrrr

rrrr

rrrr

210

210

210

210

210

210

00

000

00

000

++

)(

)(

dC

lkdC

y

y

= 0

Cy( ) = 0 = d, d + 1, . . . , d + k + l Hm tng quan tho mn phng trnh:

F*(q -1)Cy( ) = 0 0 H thng phng trnh 2.49 c bc

n + k = n + max(k,l) Nu

k + l + 1 n + max(k, l) hoc tng ng vi

min(k, l) n 1 dn n

Cy( ) = 0 = d, d + 1,... l iu cn chng minh.

2.3.4 Kt ni gia MRAS v STR Cc h thng thch nghi dng m hnh chun trc tip c cp trong phn 2.2. Trong ph lc A (TLTK[1]) cng ch ra m hnh km theo v t cc l lin quan vi nhau. By gi chng ta s chng t b chnh nh trc tip dng phng php t cc thut ton 2.2 l tng ng vi mt MRAS. Trong trng hp nhiu xc nh, khi B- l hng s, m hnh ca qu trnh c vit li nh sau:


Trang 137

y(t) = )( 0dtTf Trong thut ton gin tip, cc thng s c c lng bng cc thng s ca b chnh nh. Phng php bnh phng cc tiu c s dng cho vic c lng v (t) c vit li: (t) = y(t) - )( ty = )()( 0dtty Tf (2.50) Thng s cp nht c th c vit li:

)()()()1()( 0 tdttPtt Tf += (2.51) Ch rng theo phng trnh 2.50 th

)()( 0 tgraddtTf = Vector )( 0dtTf din t nh l o hm ca nhy. Vic cp nht thng s phng trnh 2.51 l mt phin bn ri rc theo thi gian ca lut MIT. S khc bit chnh l sai s m hnh e(t) = y(t) - ym(t) c thay bng gi tr thng d (t) v li MRAS c thay bng ma trn P(t) cho phng trnh 2.28. P lm thay i hng ca gradient v to ra mt chiu di bc thch hp. Ngc li, lut MIT cng c th xem nh l mt thut ton gradient cc tiu e2, phng trnh 2.51 dc xem nh l mt phng php Newton cc tiu 2(t). Gi tr thng d c xem nh s gia sai s. Ch rng trong cc k thut nhn dng nh cc b t chnh nh chng ta thng c gng t c mt kiu mu tng t vi

Tfty =)( Vi phng php m hnh chun th thng xuyn ch c th t mt m hnh kiu ))(()( TfpGty = Vi G(p) l SPR.

V d 2.11 - B t chnh nh trc tip vi thay i cc tiu M hnh ca qu trnh phng trnh 2.44 l :

y(t + 1) = )1()()( 00 +++ ttystur Gi s 0r c nh ti gi tr 0r = 1. Ch iu ny khc vi gi tr tht l bng 3. Thng s s0 c c lng dng phng php bnh phng cc tiu. Lut iu khin tr thnh:


Trang 138

)(

)(0

0 tyr

stu =

Bi tp v nh (dng lm bi tp trong phn Cu hi n tp v bi tp cui chng): Dng Matlab m phng cho v d 2.11 (V d 5.3 TLTK[1]). Xem kt qu m phng hnh 5.8 v 5.9 trong TLTK[1]. Hnh 5.8 biu din t s 00 / rs , n nhanh chng hi t n mt gi tr ca b iu khin thay i cc tiu ti u thm ch 0r khng bng gi tr tht ca n. Hnh 5.9 biu din hm tn hao khi dng b t chnh nh v b iu khin thay i cc tiu ti u.

2.3.5 iu khin d bo thch nghi Thut ton 2.4 l cch thc hin mt b iu khin vi tm d bo thay i. Bi ton iu khin c bn l b iu khin trung bnh di chuyn. B iu khin trung bnh di chuyn cng c th p dng c cho cc h thng khng cc tiu pha nh c minh ha phn B chnh nh trc tip. Nhiu cch khc c iu khin d bo s c cp trong ti liu, mt vi trong s ny s c tho lun v phn tch. Cng nh i vi cc thut ton trc, xc nh bi ton iu khin c bn l rt quan trng hiu r cc tnh cht tim cn ca thut ton. Trc tin ta s phn tch trng hp cc thng s l bit trc. Thut ton iu khin d bo da trn mt m hnh ca qu trnh gi thuyt v cc tn hiu iu khin tng lai. iu ny to ra mt chui cc tn hiu iu khin. Ch c mt tn hiu u tin l c p dng cho qu trnh v mt chui cc tn hiu iu khin mi c tnh ton khi thc hin php o c mi. y gi l b iu khin li tm (receding horizon controller).

D bo ng ra tng c bn trong cc thut ton iu khin d bo l vit li m hnh qu trnh c c mt biu thc r rng cho ng ra mt thi im tng lai. Xt m hnh : A* (q -1) y(t) = B* (q -1) u(t d0) (2.52) 1 = A* (q -1)F*(q -1) + q d G*d(q -1) (2.53) trong

bc( F*d ) = d 1 bc( G*d) = n 1


Trang 139

Ch s d l tm d bo vi d bc. Gi s d d0. Vic ng nht a thc phng trnh 2.52 c s dng d bo ng ra d bc pha trc. V vy : y(t + d) = A*F*d y(t + d) + G*d y(t) = B*F*d u(t + d d0) + G*d y(t) B* (q -1)F*d (q -1) = R*d (q -1) + q (d - d 0 + 1) R *d (q -1) Bc(R*d) = d d0 Bc( R *d) = n 2 Cc h s ca R*d l nhng gii hn d d0 + 1 u tin ca p ng xung ca h thng vng h. iu ny c th thy nh sau:

q - 0d B*/A* = q - 0d B*(F*d + **

AG

q dd )

= )( 1*0 qRq dd + q ( d + 1) *dR (q -1) + )()()(

1*

1*1*

qAqGqB d q ( d + 0d ) (2.54)

y( t + d) = )( 1* qRd u(t + d d0) + *dR (q -1) u(t 1) + G*d (q 1) y(t) = )( 1* qRd u(t + d d0) + dy (t) (2.55)

)( 1* qRd u(t + d d0) ph thuc vo u(t), . . . , u(t + d d0), dy (t) l hm ca u(t 1), u(t 2),... v y(t), y(t -1)....Bin dy (t) c hiu nh l iu kin d bo ca y(t + d) vi gi s u(t) v cc tn hiu iu khin tng lai l zero. Ng ra thi im (t + d) v vy ph thuc vo cc tn hiu iu khin tng lai ( nu d > d0), tn hiu iu khin, cc ng vo v ng ra thi im trc. Cng c th gi s tn hiu iu khin duy tr hng s: u(t) = u(t + d) = .... = u(t + d d0) (2.56) Cch khc xc nh lut iu khin l mang y(t + d) n mt gi tr mong mun trong khi cc tiu mc tiu iu khin theo tm d bo:

+

=

dt

tkku 2)( (2.57)

iu khin khng thay i theo thi gian: Chn ng ra c d bo bng vi ng ra mong mun ym v gi s vn gi phng trnh 2.56 :

[R*d(1) + q -1 *dR (q -1)]u(t) + G*d (q 1) y(t) = ym(t + d)


Trang 140

Lut iu khin l:

u(t) = 11**1*

)()1()()()(

+

+

qqRRtyqGdty

dd

dm (2.58)

Tn hiu iu khin ny s c s dng cho qu trnh. ln ly mu k tip, mt php o mi t c v lut iu khin 2.58 dc s dng tip. Ch gi tr ca tn hiu iu khin thay i theo thi gian ch khng phi c nh. y ta s dng qui tc iu khin li tm. Ch lut iu khin l khng i ngc vi b iu khin LQ c nh tm. By gi chng ta s phn tch h thng vng kn khi s dng phng trnh 2.58 iu khin qu trnh 2.52.Vic thc hin cc php tnh ton t sai phn ti l cn thit c th quan st cc cc ban u. Phng trnh 2.53 dc vit li theo ton t sai phn ti nh sau: q n + d - 1 = A(q)Fd(q) + Gd(q) (2.59) a thc c tnh ca h thng vng kn l:

P(q) = A(q) [q n 1Rd(1) + dR (q) ] + Gd (q) B(q) Bc(P) = Bc(A) + n - 1 = 2n 1 Phng trnh thit k 2.59 c th c s dng vit li hm P(q): B(q)q n + d - 1 = A(q) B(q)Fd(q) + Gd (q) B(q)

= A(q)[q n -1 Rd (q) + dR (q)] + Gd (q) B(q) V vy :

A(q) dR (q) + Gd (q) B(q) = B(q) q n + d -1 - A(q)q n 1Rd(q) Cho ta :

P(q) = q n 1A(q)Rd(1) + q n 1[qd B(q) - A(q)Rd(q)] Nu h thng n nh th cc s hng pha sau ca 2.54 s bin mt khi d . Do :

dlim P(q) = q n -1 A(q)Rd(1) nu A(z) l mt a thc n nh.

V d 2.12 - iu khin d bo Xt qu trnh :

y(t + 1) = ay(t) + bu(t) Phng trnh 2.59 cho ta :


Trang 141

qd = (q a)(q d 1 + f1q d 2 + . . . + fd 1) + g0 V vy:

F(q) = q d 1 + aq d 2 + a2q d 3 +. . . + a d 1 G(q) = a d Rd(q) = bF(q) dR (q) = 0 v khi ym = 0, lut iu khin tr thnh:

u(t) = - )...1( 1+++ dd

aaba y(t) = - )1(

)1(

d

d

abaa y(t)

Phng trnh c tnh ca h thng vng kn l:

P(q) = q a + 1

)1(

d

d

a

aa

c cc:

pd = 1

d

d

a

aa

V tr ca cc c cho bi: 0 pd < a a 1 (h thng n nh)

0 pd < 1 a > 1 ( h thng khng n nh)

Cc vng kn vi cc gi tr khc nhau ca a v b c ch hnh 5.16 (TLTK[1]). V d cng cho thy vic quan st l y th tm d bo phi t 5 10 mu. Cng c th tng qut ho kt qu v d 2.12 cho cc h thng bc cao hn. i vi cc h thng thay i chm hoc khng n nh th p ng vng kn ca n s rt chm khi tng tm d bo. V vy gii hn 2.56 khi s l khng hu ch.

N lc iu khin cc tiu Thut ton iu khin l s mang y(t + d) ti ym(t + d) trong khi cc tiu phng trnh 2.57 . Phng trnh 2.55 c vit li:


Trang 142

y(t + d) = R*d(q -1)u(t + d d0) + )(tyd = rd 0 u(t + ) + . . .+ rdu(t) + )(tyd

= d d0. Gii thiu hm Lagrange: 2J = u(t)2 + . . .+ u(t + )2 + 2[ym(t + d) - )(tyd - R*d(q -1) u(t + )]

Cho o hm ring i vi cc bin u(t), . . . ,u(t + ) v bng 0 ta c: u(t) = rd

.

.

.

u(t + ) = 0dr ym(t + d) - )(tyd = 0dr u(t + ) + . . . + rd u(t)

Cc phng trnh ny cho ta:

u(t) =

)()( tydty dm +

trong :

=

d

idi

r

r=0

2

S dng nh ngha )(tyd cho ta: u(t) = ym(t + d) - *dR u(t 1) - *dG y(t)

hoc

u(t) = *1

* )()(d

dm

RqtyGdty

+

+

= )(

)()()1(1 qRq

tyqGndtyd

n

dm

+

++

(2.60)

S dng phng trnh 2.60 v m hnh ca phng trnh 2.52 cho a thc c tnh vng kn:

P(q) = A(q) [q n - 1 + dR (q)] + Gd(q)B(q) Phng trnh ny c dng nh 2.58 vi Rd(1) c thay bng . iu ny c ngha cc cc vng kn tin gn ti zero ca q n 1A(q) khi A(q) l n nh


Trang 143

v khi d . iu g s xy ra khi h thng khng n nh Hy xt v d sau y:

V d 2.13 - iu khin n lc cc tiu Xt h thng tng t nh v d 2.12 . B iu khin n lc cc tiu trong trng hp ny c cho bi:

= b 1)1(22

...1

+++d

d

a

aa = )1(

)1(21

2

aa

abd

d

cho ta ( khi ym = 0)

u(t) = -

da y(t) = - )1()1(

2

212

d

d

abaa

y(t)

Cc ca h thng vng kn l:

pd = a - 1)1(

2

212

d

d

a

aa =

1212

d

d

a

aa

cho ta:

dlim pd = a | a | 1 (h thng n nh)

dlim pd = 1/a | a | > 1 (h thng khng n nh)

v d ny, b iu khin n lc cc tiu s to ra mt h thng vng kn tt hn nu iu khin tng lai c gi s l hng s.

iu khin d bo tng qut: Cc b iu khin d bo cp t trc ch xem xt gi tr ng ra ch mt thi im tng lai. Nhiu tng qut ho khc nhau ca iu khin d bo c xut m trong hm tn hao l cc tiu:

J(N1, N2, Nu) = E{ = =

++++2

1 1

22 )1()]()([N

Nk

N

km

u

ktuktykty } (2.61)

Trong = 1 q -1 l ton t vi phn. S la chn cc gi tr khc nhau ca N1, N2, Nu s a ra cc phng php khc nhau. Phng php iu khin d bo tng qut c minh ho bng cch dng hm tn hao 2.60 v m hnh qu trnh: A*(q)y(t) = B*(q -1)u(t d0) + e(t) / (2.62)


Trang 144

M hnh ny c gi l CARIMA ( Controlled AutoRegressive Intergrating Moving Average). N c thun li l b iu khin bn thn s cha mt khu tch phn. Ging nh phng trnh 2.53 ta c ng nht: 1 = A*(q)F*d(q - 1)(1 q -1) + q d G*d (q 1) (2.63) Cng thc ny c s dng xc nh ng ra d bc k tip:

y(t + d) = F*dB* u(t + d d0) + G*dy(t) + F*de(t + d) F*d c bc d -1. B d bo vi sai s qun phng ti u vi ng ra c o c n thi im t v chui ng vo bt k l: )( dty + = F*dB* u(t + d d0) + G*dy(t) (2.64) Gi s u ra mong mun ym(t + k), k = 1, 2, ...l c sn. Hm tn hao 2.61 s c cc tiu cho ra mt chui cc tn hiu iu khin tng lai. Ch gi tr mong i 2.61 s c c tng ng vi d liu c c ti thi im t vi gi s cc o c tng lai khng c sn. iu ny c ngha ch c tha s u tin ca chui iu khin l c s dng. Cc php ton s lp li khi c c mt o c mi. B iu khin vi kt qu nh th gi l iu khin hi tip ti u vng h. Nh tn ca n, gi s s dng hi tip nhng n ch c tnh ton ch da vo thng tin c sn thi im hin ti. Dng phng trnh 2.55 :

y(t + 1) = R*1(q 1) u(t + 1 d0) + 1y (t) + F1*e(t + 1) y(t + 2) = R*2(q 1) u(t + 2 d0) + 2y (t) + F2*e(t + 2)

.

.

.

y(t + N) = R*N(q 1) u(t + N d0) + Ny (t) + *NF e(t + N) Mi gi tr ng ra bao gm cc tn hiu iu khin tng lai ( nu d > d0), ng vo o c v tn hiu nhiu tng lai. Cc phng trnh trn c th c vit li:

y = Ru + y + e trong :

y = [y(t + 1) . . . y(t + N)]T u = [u(t + 1 d0) . . . u(t + N d0)]T

y = [ y 1(t) . . . y N(t)]T


Trang 145

e = [F1*e(t + 1) . . . *NF e(t + N)]T

T phng trnh 2.54 ta thy cc h s ca R*d chnh l (d d0 + 1) s hng u ca p ng xung q d 0 B*/ (A*) v cng ging nh (d d0 +1) s hng u ca p ng bc q d 0 B*/ A*. Do ma trn R l ma trn tam gic di:

R =

021

01

0

000

rrr

rr

r

NN

Nu h thng c thi gian tr (d0 > 1) th (d0 1) hng u ca R s l zero. Gi:

ym = [ym(t + 1) . . . ym(t + N)]T Gi tr mong i ca hm tn hao c vit li: J(1, N, N) = E{( y ym)T(y ym) + uTu}

= (Ru + y - ym)T(Ru + y - ym) + uTu Cc tiu ho biu thc ny theo u ta c: u = (RTR + I ) 1RT(ym - y ) (2.65) Thnh phn u trong u l u(t) l tn hiu iu khin ng dng cho h thng. Ch b iu khin t ng c mt khu tch phn. iu ny l cn thit b cho s hng nhiu sai lch phng trnh 2.62 Vic tnh ton phng trnh 2.65 lin quan ti ma trn nghch o NxN, m N l tm d bo ca hm tn hao. gim khi lng tnh ton th ta c th gii hn cc tn hiu iu khin tng lai. Chng hn, ta gi s vic tng tn hiu iu khin l bng zero sau Nu bc (Nu < N):

u (t + k 1) = 0 vi k > Nu iu ny c ngha tn hiu iu khin sau Nu bc s l hng s. So snh vi iu kin khng ch phng trnh 2.57 . Lut iu khin ( phng trnh 2.65) s thay i: u = (R1TR1 + I ) 1R1T(ym - y ) (2.66) R1 l ma trn NxNu


Trang 146

R1 =

uNNNNrrr

r

rr

r

21

0

01

0

000

Ma trn ly nghch o by gi c bc NuxNu. Ng ra v cc tm iu khin c chn nh sau: N1: Nu thi gian tr bit trc th N1 = d0, ngc li chn N1 = 1. N2: Tm ng ra cc i N2 c chn sao cho N2h c gi tr bng vi thi gian ln ca h thng, trong h l thi gian ly mu ca b iu khin. Nu: Thng Nu = 1 s c c kt qu tt i vi nhng h thng n gin. i vi cc h thng phc tp, Nu t nht phi bng vi s cc khng n nh hoc s cc gy dao ng tt yu. b iu khin d bo tng qut c kh nng thch nghi th iu cn thit l phi c lng A* v B* mi bc thi gian. Cc gi tr d bo ng vi cc tm d bo khc nhau s c tnh ton v tnh tn hiu iu khin phng trnh 2.66 . B iu khin d bo thch nghi v vy s l mt thut ton iu khin gin tip. Phng trnh 2.64 c tnh bng cch qui n gin khi lng tnh ton.Cui cng, Nu thng c gi tr nh ma trn nghch o c bc thp. Tn hiu iu khin u(t) t phng trnh 2.66 l:

u = [ 1 0 . . . 0] [R1TR1 + I ] 1R1T[ym - y ] = [1 . . . N] [ym - y ] Hn na, t phng trnh 2.62 , s dng phng trnh 2.54

y =

+

+

)()1(

)()1(

**

*

1*

1

tyGtuR

tyGtuR

NN

=

+

+

*1

*

**

*

11

*

**

1

0

0

NdN

d

GqBAR

GqBAR

y(t)

H thng vng kn c phng trnh c tnh:


Trang 147

A* + [1 . . . N]

+

+

**1**

*

1*1**

1

0

0

Nd

N

d

GBqAR

GBqAR

ng nht phng trnh 2.63 cho ta: B* = A* B*F*d + q dGd* B*

= A*[ *dR + q ( d - 0d + 1) *dR ] + dq ** BGd iu ny cho ta phng trnh c tnh:

A* + [1 . . . N]

NN qRAB

qRAB

)(

)(

***

*

1**

= A* + =

N

ii

ii RABq

1

*** )( (2.67)

Phng trnh 2.67 cho ra mt biu thc ca phng trnh c tnh vng kn nhng vn cn kh khn a ra mt kt lun tng qut v tnh cht ca h thng vng kn ngay c khi qu trnh bit trc. Nu Nu = 1 th:

i =

=

+N

jj

i

r

r

1

2

Nu ln, h thng vng kn s khng n nh khi h thng vng h khng n nh. Tuy nhin nu c 2 tm iu khin v tm d bo u tng th bi ton s tng t nh bi ton iu khin LQ vi tm c nh v do n s c c tnh n nh tt hn.

2.3.6 Kt lun Trong phn ny chng ta xem xt nhiu b t chnh nh khc nhau. tng c bn l c lng cc thng s cha bit ca h thng v thit k b iu khin. Cc thng s c lng gi s bng vi thng s thc khi thit k b iu khin. Thnh thong cng bao gm cc c lng cha chc chn vo trong thit k. Bng cch kt hp cc phng php c lng khc


Trang 148

nhau v cc phng php thit k khc nhau ta s c c cc b t chnh vi cc tnh cht khc nhau. Trong phn ny ta ch cp tng c bn v cc tnh cht tim cn. Tnh hi t ca c lng v tnh n nh ca h thng s c tho lun trong chng 6 (TLTK[1]). Kha cnh quan trng nht ca cc b t chnh nh l a ra cc thng s ho. Mt thng s ho li c th t c bng cch s dng m hnh h thng v p ng vng kn mong mun. Mc tiu ca vic thng s ho li l thc hin c lng trc tip cc thng s ca b iu khin sao cho m hnh mi tuyn tnh vi cc thng s. Ch c vi thut ton t chnh nh c cp v gii quyt trong phn ny. Vic kt hp cc phng php c lng khc nhau v vn thit k c bn s to ra cc thut ton vi cc tnh cht khc nhau. Mc tiu ca phn ny l a ra mt cm nhn cch pht trin v phn tch cc thut ton. Khi thc hin mt b t chnh th vic la chn bi ton thit k c bn l rt quan trng. Mt phng php thit k m khng ph hp cho h thng bit trc th cng s khng tt hn khi h thng cha bit trc. B t chnh nh cng c kh nng p dng cho cc h thng MIMO. Trng hp MIMO l rt kh phn tch. Kh khn chnh l xc nh c kin thc u tin cn thit trong h MIMO l g. Cng tng i n gin khi a ra mt thut ton t chnh tng ng vi b t chnh nh trc tip tng qut cc trng hp hn ch khi cc ma trn tng tc ca h thng bit trc. 2.4 Chnh nh t ng v lch trnh li

2.4.1 Gii thiu 1. Chnh nh v thch nghi 2. Kin thc u tin 3. Gi tr ban u ca b iu khin thch nghi 4. iu khin PID 5. Cc vn vn hnh 6. Giao tip iu khin Mt loi c bit ca thch nghi vng h hay s thay i cc tham s b iu chnh c cp trong phn ny. Trong nhiu trng hp, c th bit c s thay i ng hc ca qu trnh theo cc iu kin vn hnh. Ngun gc ca s thay i ng hc c th l tnh phi tuyn. C th thay i tham s ca b iu khin bng cch gim st cc iu kin vn hnh


Trang 149

ca qu trnh. Khi nim ny gi l lch trnh li, v m hnh u tin c s dng ch iu chnh li ca qu trnh.

2.4.2 K thut chnh nh 1. Phng php Zeigler Nichols Lut iu khin PID:

++=

t

di

c dtdeTdsse

TteKtu

0

)(1)()(

2. Phng php p ng qu M hnh 3 thng s:

sLesT

ksG

+=

1)(

Phng php p ng nc:

Phng php Zeigler Nichols:

Thi gian

L T

a

k 0.63k


Trang 150

B iu khin aKc Ti / L Td / L Tp / L P PI

PID

1 0.9 1.2

3 2

0.5

4 5.7 3.4

Nhng kh khn i vi phng php Zeigler Nichols: - Kh xc nh cc thng s - Tt qu chm - Hai thng s th khng Phng php din tch:

keAT

kA

LT

1

0

=

=+

3. Phng php p ng tn s tng: Cho chy b iu khin t l, tng li cho n khi h thng bt u dao ng. Quan st li Ku gii hn v Chu k gii hn Tu . Lp li: Xc nh c tnh p ng tn s.

A0

A1

k

L + T


Trang 151

Cc thng s b iu khin:

B iu khin Kc / Ku Ti / Tu Td / Tu Tp / Tu P PI

PID

0.5 0.4 0.6

0.8 0.5

0.12

1 1.4

0.85 Thc nghim:

G(j)

)(1N

PID

Relay

-1

A

T

y u Qu trnh


Trang 152

Kt qu thc t - Thng tin bit trc? - Bt u thc nghim nh th no? - Hi tip n bin gii hn ca dao ng. - Hiu chnh lut Zeigler Nichols: Thay i cc gi tr trong bng. S dng 3 thng s: Ku, Tu v Kp. - Lm sao ng u vi nhiu c Nhiu ti Nhiu o T tr S lp li trc tuyn tng: Tm cc nt c trng ca p ng trc tuyn i vi im t hoc cc nhiu ti. Hiu chnh b iu khin da trn cc c tnh quan st c.

c tnh: h s tt d v vt l o

21

23

ee

eed

=

1

2

e

eo =

B iu khin hiu chnh da trn lut th v sai. D dng i vi PI v kh khn hn i vi PID.

Tp

e1

e2

e3


Trang 153

Thng tin bit trc Tin chnh nh

2.4.3 Lch trnh li V d cc bin lch trnh Tc sn xut Tc my S t l v p lc ng

Thnh thong c th tm thy nhng bin i ph c tng quan tt vi nhng thay i ca qu trnh ng hc. V th c th lm gim nh hng ca tham s bin ng ch n gin bng vic thay i tham s ca b iu chnh nh cc hm ca cc bin ph (xem hnh 2.10)

Hnh 2.10 M hnh lch trnh li

Lch trnh li c th c xem nh h thng iu khin hi tip m li hi tip c chnh bi b b c cung cp trc.

Qu trnh

Lch trnh li

B iu khin

Tn hiu iu khin

Ng ra

iu kin vn hnh

Cc thng s b iu khin

Tn hiu vo


Trang 154

u, khuyt im ca lch trnh li Mt hn ch ca lch trnh li l b vng h. Khng c hi tip b cho sai s lch trnh. Hn ch khc ca lch trnh li l vic thit k tn nhiu thi gian. Tham s b iu chnh phi c chn cho nhiu iu kin vn hnh v c tnh k thut phi c kim tra bng nhiu qu trnh m phng. Nhng kh khn ny trnh c nu lch trnh da vo cc php chuyn i phi tuyn. Lch trnh li c u im l cc tham s b iu chnh c th p ng rt nhanh vi s thay i ca qu trnh. Khi khng c c lng tham s, nhn t gii hn ph thuc vo tc p ng cc php o ph vi s thay i ca qu trnh.

2.4.4 Xy dng lch trnh La chn cc bin lch trnh Hon thin vic thit k iu khin cho nhng iu kin vn hnh khc nhau. S dng vic chnh nh t ng. S bin i. Tht kh tm lut chung cho vic thit k b iu chnh theo lch trnh li. Vn chnh l vic quyt nh cc bin s dng lm bin lch trnh. R rng cc tn hiu ph phi phn nh iu kin vn hnh ca i tng. S c nhng trnh by l tng n gin cho cc tham s b iu chnh lin quan n cc bin lch trnh. V th cn c kin thc tt v h ng hc ca qu trnh nu lch trnh li c s dng. Cc khi nim tng qut sau c th phc v cho mc ch ny. - Tuyn tnh ho c cu dn ng phi tuyn. - Lp trnh li da vo o c cc bin ph - Vn hnh da vo hiu sut - Cc php bin i phi tuyn. Cc khi nim ny c minh ho trong cc v d sau.

V d 2.14 Xem h thng vi 1 valse phi tuyn.Tnh phi tuyn c gi s l:

v = f(u) = u4 , u 0


Trang 155

Hnh (a)

Hnh (b) t 1 f l hm ngc xp x ca c tnh valse. b cho tnh phi tuyn, ng ra ca b iu chnh c cung cp thng qua hm ny trc khi n c p vo valse (xem hnh (a)). ta c quan h : v = f(u) = f [ 1 f (c)] Vi c l ng ra ca b iu chnh PI. Hm f [ 1 f (c)] c li t thay i hn hm f.

Nu 1 f chnh xc l hm ngc ca f th : v = c. Gi s f(u) = u4 c xp x bi 2 ng thng: mt ng ni t im (0 , 0) n im (1.3 , 3) v ng thng th hai ni gia 2 im (1.3 , 3) v (2 , 16), c v trong hnh (b) . Khi :

-1

c u y v

Qu trnh

f PI 1 f

0 0.5 1 1.5 2

u

v

f

f 5 10

15

20

yr


Trang 156

1 f (c) =

+

163,139.10538.0

30,433.0

cc

cc

Hnh (c) Hnh (c) cho thy s thay i trong tn hiu chun ti 3 iu kin vn hnh khc nhau khi s dng hm 1 f nh hnh (a) . So snh vi h thng trong hnh 2.2 (TLTK[1]) . Ta thy c s ci thin trong c tuyn ca h thng vng kn. Dng hm ngc 1 f trong h thng s cho p ng bng phng hn trong cc bi ton iu khin valse phi tuyn. V d trn cho thy tnh n gin v tin dng trong vic b cho h thng phi tuyn tnh bit trc. Trong thc t thng xp x h phi tuyn

y

20 40 60 80 100 0

yr

0.2

0.3

20 40 60 80 100 0

yr

1.1

1

20 40 60 80 100 0

yr

y

5.1

5


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bng mt vi on thng (nhiu hn 2). C nhiu b iu khin vng n thng mi s dng phng php b ny. Trong v d trn khng c s o c no ca iu kin vn hnh ngoi tr vic iu chnh ng ra. Trong cc trng hp khc, tnh phi tuyn c xc nh t s o c mt vi bin s.

2.4.5 ng dng Lch trnh li l phng php rt hu dng. N yu cu phi c kin thc tt v qu trnh v cc bin ph c th c o c. Mt thun li ln ca phng php ny l b iu chnh thch nghi (p ng) nhanh khi cc iu kin thay i. Mt s ng dng nh: nh hng cho tu, kim sot nng pH, kim sot kh t, iu khin ng c v iu khin bay. c im ca van

c tnh van ph thuc vo vic ci t.

tuyn tnh

Dng chy

V tr

M tnh theo %

M nhanh

A

B C


Trang 158

Lch trnh cho ng ra b iu khin

Lch trnh cho bin qu trnh

FT

FIC

LT

LIC


Trang 159

Lch trnh cho bin ngoi

2.4.6 Kt lun Lch trnh li l cch tt b cho c tnh phi tuyn bit trc. B iu chnh c th phn ng nhanh vi s thay i ca cc iu kin. Mt hn ch ca k thut ny l thit k tn nhiu thi gian nu khng dng php chuyn i phi tuyn v t ng chnh nh. Mt hn ch khc l cc tham s iu khin c thay i trong vng h, khng c hi tip t c tnh lm vic ca h thng. Phng php ny khng th dng c nu c tnh ng hc ca qu trnh hoc nhiu khng c bit trc y , chnh xc.

2.5 Bi tp ng dng Matlab

1.M hnh: H thng ga t ng trn t

ng lc hc ca t trn ng: t vn hnh trn ng nh moment sinh ra t ng c, thng qua h thng truyn ng, chuyn thnh lc ko tip tuyn ti cc bnh xe ch ng y t dch chuyn ln pha trc. Lc ko tip tuyn ny lun cn bng vi cc lc cn tc ng vo t theo nh lut I Newton:

Fko = Fcn ln + Fcn khi ng + Fcn leo dc + Fcn qun tnh

TIC

TT FT


Trang 160

Tng cc lc cn i vi t khng ph thuc tuyn tnh vo vn tc ca t v cc thnh phn lc cn ny c nhng h s ph thuc vo iu kin lm vic ca t nh loi ng, mp m, nghing ca mt ng, loi lp xe, nhit mi trng, gi, ti trng ca xe, tnh trng ca ng c, ca h thng truyn ng, mn ca lpCc iu kin lm vic ny khng c nh m thay i mi khi t vn hnh v trong lc t vn hnh.

2.Phng trnh trng thi: i tng vn hnh trn ng l mt i tng phi tuyn ch bao gm mt tn hiu iu khin vo l m cnh bm ga ca ng c ( hay

chuong 2_Điều khiển thích nghi

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