Chuong 3_dieu Khien Ben Vung_1

7/21/2019 Chuong 3_dieu Khien Ben Vung_1

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Chng 3

IU KHIN BN VNG

3.1 Gii thiu

3.1.1 Khi nim iu khin bn vng

H thng iu khin bn vng lm cho cht lng ca sn phm n nh, khng phthuc vo sthay i ca i tng cng nhca nhiu tc ng ln hthng.Mc chca iu khin bn vng l cht lng vng kn c duy tr mc d c nhng sthayi trong i tng.

P0:M hnh chun (m hnh danh

nh)

P :M hnh thc tvi sai lch

so vi m hnh chun

Hnh 3.1 : M hnh iu khin bn vng

Cho tp m hnh c sai s P v mt tp cc chtiu cht lng, gis

P0 P l m hnh danh nh dng thit kbiu khin K.Hthng

hi tip vng kn c gi l c tnh :

- n nh danh nh: nu K n nh ni vi m hnh danh nh P0- n nh bn vng: nu K n nh ni vi mi m hnh thuc P

- Cht lng danh nh: nu cc mc tiu cht lng c tha i vi m hnh danhnh P0

- Cht lng bn vng: nu cc mc tiu cht lng c tha i vi mi m hnh thuc

P

Mc tiu bi ton n nh bn vng l tm biu khin khng chn nh m hnh danh

nh P0m cn n nh mt tp cc m hnh c sai s P


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3.1.2 Chun ca tn hiu

3.1.2.1 Khi nim chun

Trong iu khin ni ring cng nh trong cc cng vic c lin quan n tn hiu nichung,thng thng ta khng lm vic ch ring vi mt tn hiu hoc mt vi tn hiuin hnh m ngc li phi lm vic vi mt tp gm rt nhiu cc tn hiu khc nhau.Khi phi lm vic vi nhiu tn hiu khc nhau nhvy chc chn ta s!gp bi ton sosnh cc tn hiu chn lc ra c nhng tn hiu ph hp cho cng vic.

Cc khi nim nhtn hiu x1(t) tt h"n tn hiu x2(t) chthc sc ngh#a nu nhchngcng c chiu theo mt tiu chun so snh no . Cng nhvy nu ta kh$ng nhr%ng x1(t) ln h"n x2(t) th phi chr php so snh ln h"n c hiu theo ngh#a no,x1(t) c gi trcc i ln h"n , c n&ng lng ln h"n hay x1(t) ch'a nhiu thng tin h"n

x2(t)..Ni mt cch khc ,trc khi so snh x1(t) vi x2(t) chng ta phi gn cho m(imt tn hiu mt gi trnh gi tn hiu theo tiu chun so snh c la chn .

nh ngha: Cho mt tn hiu x(t) v mt nh xx(t) )||x(t)|| R+ chuyn x(t) thnhmt sthc d"ng ||x(t)||.Sthc d"ng ny s!c gi l chun ca x(t) nu n thamn:

a. ||x(t)|| *0 v ||x(t)|| = 0 khi v chkhi x(t) =0 (3.1)

b. ||x(t)+y(t)|| +||x(t)|| + ||y(t)|| x(t), y(t) (3.2)

c. ||ax(t)|| = |a|.||x(t)|| x(t) v Ra . (3.3)

3.1.2.2 Mt schun thng dng trong iu khin cho mt tn hiu x(t):

- Chun bc 1: dttxtx

= |)(|||)(|| 1 (3.4)

- Chun bc 2:

= dttxtx 22 |)(|||)(|| . (3.5)

Bnh ph"ng chun bc hai chnh l gi tro n&ng lng ca tn hiu x(t).

-Chun bc p: p pp dttxtx

= |)(|||)(|| vi p N (3.6)

- Chun v cng: |)(|sup||)(|| txtxt

= (3.7)

y l bin hay nh ca tn hiu

Khi nim chun trong nh ngh#a trn khng bgii hn l chcho mt tn hiu x(t) mcn c p dng c cho cvector tn hiu gm nhiu ph,n tv m(i ph,n t li l

mt tn hiu.

Xt mt vector tn hiu:


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x(t) =

)(

)(1

tx

tx

n

- Chun 1 ca vectorx:

=

=n

i

ixx1

1 (3.8)

- Chun 2 ca vectorx:

=

=n

i

ixx1

2

2 (3.9)

- Chun v cng ca vectorx:

ni

ixx,...,2,1

max=

= (3.10)

3.1.2.3 Quan hca chun vi nh Fourier v nh Laplace:

-phc vmc ch sdng khi nim chun vo iu khin ,ta c,n quan tm ti mi

lin quan gia chun tn hiu x(t) l ||x(t)|| vi nh Fourier X(j ) cng nhnh Laplace

X(s) ca n.

nh l 3.1: (Parseval) Chun bc hai ca mt tn hiu x(t) v nh Fourier X(j) ca nc quan h:

djXdttxtx222 |)(|

2

1|)(|||)(||

2

== (3.11)

Cho tn hiu nhn qucausal x(t). Gi X(s) l nh Laplace ca n .Gisr%ng X(s) cdng thc -hu t.vi bc ca a th'c tskhng ln h"n bc a th'c m/u s,t'c l:

n

n

m

m

sasaa

sbsbb

sA

sBsX

+++

+++==

.....

.....

)(

)()(

10

10 vi m < n (3.12)

nh l 3.2: Xt tn hiu nhn qucausal x(t) c X(s) dng (3.12) .-chun bc 1 cax(t) l mt shu hn ||x(t)||1= K < th iu kin c,n v l tt ccc im cc caX(s) phi n%m bn tri trc o (c ph,n thc m) .

3.1.3 i sma trn

3.1.3.1 Mt sma trn thng gp:


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- Mt ma trn A=(aij) c shng b%ng sct c gi l ma trn vung. -ng cho nicc ph,n taii trong ma trn vung c gi l ng cho chnh .-ng cho cn li

c gi l ng cho ph.

A =

nnnn

n

n

aaa

aaa

aaa

21

22221

11211

(3.13)

- Mt ma trn vung A=(aij) c aij= 0 khi i 0j ,t'c l cc ph,n tkhng n%m trn ngcho chnh u b%ng 0, c gi l ma trn ng cho. Ma trn ng cho c k

hiu b1i:

A =

nna

aa

00

0000

22

11

= diag(aij) (3.14)

- Ma trn ng cho I = diag(1) =

100

010

001

gi l ma trn "n v.

- Ma trn vung A=(aij) c aij= 0 khi i > j (hoc i < j) c gi l ma trn tam gic+ Ma trn tam gic di

A=

nnnn aaa

aa

a

21

2221

11

0

00

(3.15)

+ Ma trn tam gic trn

A=

nn

n

n

a

aa

aaa

00

0 22211211

(3.16)

3.1.3.2 Cc php tnh vma trn:

- Php cng / tr2: Cho hai ma trn A=(aij) v B=(bij) cng c m hng v n ct .Tng hayhiu A B = C =(cij) ca chng c nh ngh#a l mt ma trn cng c m hng v n ctvi cc ph,n t


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cij= aij+ bij i=1,2,..,m v j=1,2,..,n.

- Php nhn vi s thc: Cho ma trn A=(aij) c m hng v n ct v mt sv hng

thc(ph'c) x ty .Tch B = xA = Ax = (b ij) c hiu l ma trn cng c m hng v nct vi cc ph,n t

Bij= x.aij i=1,2,.m v j=1,2,..,n

- Php chuyn v: Ma trn chuyn vca ma trn A=(aij) vi m hng v n ct l ma trnAT = (aji) c n hng v m ct c to t2ma trn A qua vic hon chuyn hng thnh ct

v ngc li ct thnh hng.

- Php nhn ma trn: Cho ma trn A=(aik) c m hng v p ct v ma trn B=(bkj) c phng v n ct ,t'c l :

+ A=(aik) i=1,2,....,m v k=1,2,.,p+ B=(bkj) k=1,2,.,p v j=1,2,..,n

Tch AB = C =(cij) ca chng l mt ma trn c m hng v n ct vi cc ph,n t

Cij= =

p

k

kjikba1

Mt ma trn vung A nnR c gi l ma trn trc giao nu ATA=AAT=I

3.1.3.3 Hng ca ma trn:

Cho n vector vi i=1,2,,n Chng s! c gi l c lp tuyn tnh nu $ng th'c

a1v1+a2v2+.+anvn=0 trong ail nhng sthc (hoc ph'c) s!ng khi v chkhia1= a2= ..=an= 0

Xt mt ma trn A=(aij) bt k c m hng v n ct .Nu trong sm vector hng c nhiunht p +m vector c lp tuyn tnh v trong sn vector ct c nhiu nht q +n vectorc lp tuyn tnh th hng ma trn "c hiu l:

Rank(A) = min{p,q}

Mt ma trn vung A kiu (n n) s!c gi l khng suy bin nu Rank(A)=n .Ngcli nu Rank(A)


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AB = BA =I (ma trn "n v) (3.21)

Th ma trn B c gi l ma trn nghch o ca A v k hiu l B = A-1.

Do phi tn ti chai php nhn AA-1v A-1A cho ra kt quc cng kiu nn ma trn Aphi l mt ma trn vung,t'c l phi c m = n.H"n na do det(I) = 1 0 nn:

det(A)det(A-1) 0 => det(A) 0 v det(A-1) 0. (3.22)

Vy A phi l ma trn khng suy bin.

Ma trn nghch o A-1ca A c tnh cht sau:

- Ma trn nghch o A-1ca A l duy nht (3.23)

- Tp hp tt ccc ma trn vung cng kiu v khng suy bin cng vi php nhn ma

trn to thnh mt nhm (khng giao hon). (3.24)

- Nghch o ma trn kiu (22):

=

=

ac

bd

Adc

baA

)det(

11 (3.25)

- (AB)-1= B-1A-1 (3.26)

- (A-1)T= (AT)-1 (3.27)

- Nu A = diag(ai) v khng suy bin th A-1= diag

ia

1 (3.28)

- A-1=)det(A

Aadj (3.29)

trong Aadjl ma trn c cc ph,n ta 3ij= (-1)i+jdet(Aij) vi Aijl ma trn thu c t2

A b%ng cch bi hng th'j v nhct th'i.

- Cho ma trn A Rn n khng suy bin . Nu U Rn mv V Rn ml hai ma trn

lm cho (I+VTA-1U) cng khng suy bin th

(A+UVT)-1= A-1 A-1U(I+VTA-1U)-1VTA-1 (3.30)

- Cho ma trn vung A =

43

21

AA

AAkhng suy bin,trong A1,A2,A3,A4 cng l cc

ma trn.

Nu A1khng suy bin v B = A4 A3A1-1A2cng khng suy bin th

+=

=

1113

1

12

11

113

121

111

1

43

211

BAAB

BAAAABAAA

AA

AAA (3.31)

Nu A4khng suy bin v C = A1 A2A4-1A3cng khng suy bin th


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+

=

=

132

13

14

14

13

14

142

111

43

211

AACAAAACAA

AACC

AA

AAA (3.32)

3.1.3.5 Vt ca ma trn:

Cho ma trn vung A=(aij) ,i,j=1,2,,n kiu (nxn).Vt ca A c hiu l tng gi trcc ph,n ttrn ng cho chnh ca A v c k hiu b%ng trace(A):

trace= =

m

i

iia1

(3.33)

Vt ca ma trn c cc tnh cht:

a. trace(AB) = trace(BA) (3.34)

b. trace(S

-1

AS) = trace(A) vi S l ma trn khng suy bin bt k (3.35)3.1.3.6 Gi trring v vector ring:

Sthc c gi l gi trring v vector x c gi l vector ring bn phi 'ng vi

gi trring ca A tha mn:

Ax = x x (3.36)

(A - I)x = 0 x (3.37)

Gi trring v vector ring ca ma trn A c nhng tnh cht sau:

a. Hai ma trn t"ng "ng A v S-1AS lun cng gi trring, ni cch khc gi trring

ca ma trn bt bin vi php bin i t"ng "ng:det(A- I)=det(S-1AS- I) (3.38)

b. Cc gi trring ca ma trn bt bin vi php chuyn v, t'c l:

det(A- I)=det(AT- I) (3.39)

c. Nu A khng suy bin th AB v BA c cng cc gi trring ,t'c l:

det(AB- I)=det(BA- I) (3.40)

d. Nu A l ma trn i x'ng (AT=A) th cc vector ring 'ng vi nhng gi trringkhc nhau s!trc giao vi nhau

Trong Matlab ,sdng hm eig(A) tm ma trn ring v vector ring.3.1.3.7 Tnh ton ma trn:

Cho ma trn X = (xij) Cm nl mt ma trn thc (hoc ph'c) v F(X) C l mt v

hng thc hoc ph'c ca X .-o hm ca F(X) i vi X c nh ngh#a

=

)()( XF

xXF

X ij (3.41)


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Cho A v B l nhng ma trn ph'c vi khng gian t"ng thch .Mt scng th'c o

hm :

( )

( )

( )

1

(3.42)

( ) (3.43)

2 ( ) (3.44)

( ) (3.45)

( ) (3.46)

T T

k k T

T T

T T

T

Trace AXB A BX

Trace X k X X

Trace XBX XB B BX

X AX AX A XX

Trace AX B BAX

=

=

= =

= +

=

3.1.3.8 Chun ca ma trn:Ngi ta c,n n chun ca ma trn l nh%m phc vvic kho st tnh gii tch ca

n.C nhiu chun khc nhau cho mt ma trn A=(aij) ,i=1,2,,m;j=1,2,,n.

Nhng chun thng thng c sdng:

- Chun 1 ca ma trnA

=

=

m

i

ijnj

aA1

11max (3.47)

- Chun 2 ca ma trnA

)(max *12

AAA ini

= (3.48)

- Chun v cng ca ma trnA

=

=

n

j

ijmi

aA1

1max (3.49)

- Chun Euclide ca ma trnA(chun Frobenius)

)(2

AAtraceaA T

i j

ijF== (3.50)

vi *A l ma trn chuyn vv ly lin hip. )( *AAi l trring ca ma trn AA* lmt sthc khng m.

3.1.4 Trsuy bin ca ma trn li chnh(Principal gain)

Trsuy bin ca ma trnA(mx l) c k hiu l )(Ai c nh ngh#a nhsau:

kiAAA ii ,...2,1)()(* == (3.51)

vi },min{ lmk= .


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Nu chng ta biu din ma trnAdi dngA(s) v t js= )0(


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Hnh 3.2 : Shthng dng phn tch n nh ni

nh ngha :

Hhi tip hnh 3.2 c gi l n nh ni nu tt ccc hm truyn t t2w1, w2ne1, e2u n nh.

-iu kin n nh ni cht h"n iu kin n nh da trn hm truyn vo-ra thngthng, v n trnh vic khcc cc v zero khng n nh gia cc khu lin tip nhau.Khi thnh lp hm truyn vo-ra, c thxy ra hin tng khcc v zero khng n nhca cc khu lin tip nhau. Nhvy, iu kin n nh ni bo m cc tn hiu bntrong hthng u hu hn khi tn hiu vo l hu hn.

V d, ta kho st iu kin n nh ni ca hthng hnh 3.2:

21

11

2

212122

21

11

1

121211

)()(

)()(

wGKIGwGKIe

GKeGwwGewe

KwKGIwKGIeKGeKwwKewe

+=

++=+=

+=++=+=

Suy ra:

1 11 1

1 12 2

( ) ( )

( ) ( )

e w

e w

=

I KG I KG K

I GK G I GK

-iu kin n nh ni ca hl cc hm truyn 1( )KG , 1( )KG K, 1( )GK G ,1( )GK u n nh.

3.1.6 nh l li nh(Small Gain Theorem)

Cho hthng c biu din nhhnh 3.3:Gi il trring ca G

G

K

w1e1

e2w2

+

++

+


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Hnh 3.3 : Hthng hi tip vng kn

-nh l li nhc pht biu nhsau:

Githit r%ng G(s) n nh, (G(j)) l bn knh phca G(j). Hthng vng kn n

nh nu ( ) 1max))((


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b. )1(,1))((


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3.1.7.3 -iu kin n nh bn vng i vi sai snhn 1,u ra

Hnh 3.6 : Sai snhn u ra

Vi = 1))((),()()( jsss OO , (3.62)

Ta c:

( ) ( ) ( )[ ( ) ( ) ( )]Ov s G s K s s w s v s= + (3.63)

hay

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

v s I G s K s G s K s s w s= + (3.64)

vy

)()(

)()()(

sKsGI

ssKsGM O

+=

(3.65)

Kt lun:Hthng vng kn hnh 3.6 n nh bn vng khi v chkhi:

1)()(

)()()(0 khi nhiu o s! lm sai lch tt c tnhiu o. Nu c mt vi tn hiu nhiu tdo v blc ph'c tp c bit n nhblc

Deyst c sdng gii quyt vn ny.H"n na gisr%ng (A, )W tm c

c ngh#a l nhiu qu trnh kch thch tt ccc trng thiTrong matlab sdng lnh Kalmantnh khu lc kalman lin tc t2m hnh sys ca itng:

[kest,L,P] = kalman(sys,W,V[,N,sensor,known])

W,V l cc ma trn hip ph"ng sai m tc im nhiu hthng v nhiu o lng.

N :mc nh b%ng 0

Hai vector sensor,known :ch'a chsca cc bin ra o c v cc ,u vo ta bit.

Kt qutnh kest :chnh l m hnh trng thi ca khu lc Kalman

Ma trn L :ma trn blc Kalman phn hi sai lch quan st.


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P :l ma trn hip ph"ng sai ca sai lch t#nh

3.2.3 Gii thut thit kLQG

Biu chnh ton ph"ng tuyn tnh (LQR) v blc Kalman c sdng vi nhau thit k b iu chnh ng. Th tc ny c gi l thit k b tuyn tnh tonph"ng Gaussian (LQG). -iu thun li quan trng ca vic thit kLQG l cu trc cabiu khin c cho b1i thtc. -iu ny lm cho cc bLQG c thit krt cch cho vic iu khin cc hthng hin i (v dnhiu khin khng gian v hng

khng ) khi cu trc biu khin khng bit trc c.

Gisph"ng trnh o lng ng ra c cho b1iwBuAxx ++= (3.138)

y=Cx+v (3.139)

vi x(t)R , u(t) l biu khin ng vo, w(t) l nhiu qu trnh, v v(t) l nhiu o.

Gisph"ng trnh hi tip trng thi ,y

u=-Kx+r (3.140)

c thit k, vi r(t) l ng vo .-li trng thi hi tip l K c chn b1i mt s

k8thut ch$ng hn nhk8thut LQR .Nu ph"ng trnh iu khin (3.140) c thay

vo (3.138) th hthng iu khin vng kn c tm thy nhsau:

wBrxBKAx ++= )( (3.141)

Thit khi tip trng thi ,y rt c quan tm nu cc iu kin c gith hthng vng kn m bo n nh.H"n na,sdng hi tip trng thi tt ccc nghimcc ca ph"ng trnh (A-BK) c tht tu6 nhmong mun .Kt qucc ph"ng trnhthit kca hi tip trng thi "n gin h"n ph"ng trnh cho hi tip ng ra . Tuy nhin

lut iu khin (3.138) khng ththc hin khi tt ccc trng thi khng tho c.

By gibquan st hoc blc Kalman

LyBuxLCAx ++= )( (3.142)

c thit k. - l li L ca blc c tm ra b%ng nhng k#thut tho luncung cp c lng trng thi.Khi tt c cc trng thi khng tho v iu khin(3.138) khng th thc hin trong thc t, gi s r%ng c lng hi tip )(tx thay thcc trng thi thc x(t) lut iu khin hi tip l

u = -Kx +r (3.143)

Nu K c chn s dng ph"ng trnh Riccati LQR v L c chn b1i s dngphong trnh Ricati ca blc Kalman.-iu ny c gi l thit kLQG


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-iu quan trng ca cc kt quny l trng thi hi tip ca K v li ca bquan stL c thc thit kring r!.

3.2.4 V d!:M hnh con l"c ngc:

Xt hthng con lc ngc nhhnh sau.Con lc ngc c gn vo xe ko b1i ngc" in.Chng ta ch xt bi ton hai chiu,ngh#a l con lc ch di chuyn trong mtph$ng.Con lc ngc khng n nh v n lun ng xung tr2khi c lc tc ng thchhp.Gi s khi lng con lc tp trung 1 ,u thanh nh hnh v! (khi lng thanhkhng ng k).Lc iu khin u tc ng vo xe.Yu c,u ca bi ton l iu khin vtr ca xe v gi cho con lc ngc lun th$ng 'ng. Bi ton iu khin h con lcngc chnh l m hnh ca bi ton iu khin nh hng tu v trkhi c phng

vo khng gian.

lsinx

u

mg

m

lc

os

l

M

y

x

Hnh 3.11: M hnh con l"c ngc

Ch thch :

M: trng lng xe (Kg)l: chiu di con lc ngc (m)

g: Gia tc trng trng (m/s2)

: Gc gia con lc ngc v ph"ng th$ng 'ng (rad)

m: Trng lng con lc ngc(Kg)

u: lc tc ng vo xe (N)

x: vtr xe (m)

Trc tin ta hy xy dng m hnh ton hc ca hcon lc ngc

Gi (xG,yG) l toca vt nng 1,u con lc,ta c:


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.sin

.cosG

G

x x l

y l

= +

=

p dng nh lut II Newton cho chuyn ng theo ph"ng x,ta c:

udt

xdm

dt

xdM G =+

2

2

2

2

Thay xG1biu th'c trn ta c:

ulxdt

dm

dt

xdM =++ )sin(

2

2

2

2

Khai trin cc o hm,rt gn ta c:

umlmlxmM =++

)(cos)(sin)(2

Mt khc, p dng nh lut II Newton cho chuyn ng quay ca con lc quanh trc tac:

sin)sin(cos.2

2

2

2

mglldt

ydml

dt

xdm GG =

Thay vo ta c:

sinsincoscos)sin(2

2

2

2

mgllldt

dmllx

dt

dm =

+

Khia trin cc o hm 1biu th'c trn v rt gn ta c:

lmMml

mlgmMu

mmM

mgmlux

mglmlxm

)()(cos

)sin(cos)(sin)(cos

)(cos

sincos)(sin

sincos

2

2

2

+

++=

+=

=+

Chng ta s!vit ch"ng trnh m phng c tnh ng ca i tng

Chng ta thy r%ng hcon lc ngc l hphi tuyn , c thiu khin hcon lcngc b%ng ph"ng php LQG chng ta c,n m hnh tuyn tnh.Gi s gc nh

chng ta c thxp xsinb%ng 0,cosb%ng 1 v cng gisnh 02

.Vicc iu kin trn,chng ta c thtuyn tnh ho cc ph"ng trnh phi tuyn:

mgmlxm

umlxmM

=+

=++

)(


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-t cc bin trng thi:

=

=

=

=

xx

xx

x

x

4

3

2

1

(3.144)

Kt hp vi hai ph"ng trnh trn ta suy ra hph"ng trnh bin trng thi nhsau:

+=

=

+

=

=

uM

gxMl

mx

xx

uMl

gxMl

mMx

xx

1

1

14

43

12

21

(3.145)

Vit li di dng ma trn

+

+

=

M

Ml

x

x

x

x

gM

m

gMl

mM

x

x

x

x

1

0

1

0

000

1000

000

0010

4

3

2

1

4

3

2

1

u (3.146)

Ph"ng trnh 1ng ra,chng ta gishai trng hp:

=

4

3

2

1

1000

0100

0010

0001

x

x

x

x

y (3.147)

Nu cho c hai bin trng thi(vtr x v gc lch )th :

=

4

3

2

1

0100

0001

x

x

x

x

y (3.148)

Chng ta s! kho st h con lc ngc c cc thng s nhsau:M=1kg,l=1mLy gi trgia tc trng trng g=9.8m/s2

Ph"ng trnh (3.146) tr1thnh:


30/33

u

x

xx

x

x

x

x

x

+

=

1

01

0

00098.0

100000078.10

0010

4

3

2

1

4

3

2

1

(3.149)

Thit kLQG:

Lc Kalman:

- Bquan st:

Nu khng c ng phn hi qua L th x~ khng tim cn vx c v

vy L c chn sao cho x~

xBquan st c thit ktheo githuyt

Gista c hthng:

+=

++=

Cxy

BuAxx , : nhiu (3.150)

,v l cc nhiu trng c phn bgaussian vi{ }{ } 0.

0.

>=

>=

WE

VE

T

T

{ } ,0. =TE c lp nhau

Bquan st c ph"ng trnh trng thi:

=

++=

xCy

yyLBuxAx

)( (3.151)

Hnh 3.12 Blc Kalman

B

vDuCxy

Buxx

++=

++=

C

A

L

++-

++

y

u y


31/33

LxLCAx

xCLxAx

xCCxLxAxxxx

+=

+=

++==

~)(~)~(~~

)(~~~

(3.152)

Lc Kalman c xy dng trn c"s1:xc nh L sao cho k6vng ton { }xxE T cc tiu.1= VPCL T (3.153)

Trong P l nghim ca ph"ng trnh Riccati

}.{

}.{

01

T

T

TTT

vvEV

wwEW

CPVPCWAPPA

=

=

=++

(3.154)

Biu khin LQG (Linear Quard Gaussian):

Trong biu khin LQ ta hi tip trng thi tuy nhin trong thc tnhiu khi ta phiquan st ly c bin trng thi c lng (do khng o c) v hi tip trng thic lng => LQG

Hnh 3.13: Biu khin LQG

-iu khin LQG l kt hp iu khin LQR vi lc Kalman .

Bc 1:Thit kiu khin LQR=>KC

-

A

L

C

u y

+

++

+

-

B

Kc

Cxy

BuAxx

=

++=

-


32/33

Bc 2:Thit kblc Kalman =>L

3.3 iu khin bn vng H3.3.1 Biu Bode a Bin (Multivariable Bode Plot)

Bin ca ma trn hm truyn ton ph"ng )H(j: ti bt k6mt t,n s j no, phthuc vo hng tn hiu kch thch ,u vo.Bin ca ma trn hm truyn H( j )

c bao pha trn b1i gi trsuy bin cc i, k hiu ))(( jH , pha di b1i gi tr

suy bin cc tiu ca n, k hiu ))(( jH . Chnh v vy,chng ta c,n tnh ton hai gitrrng buc ny.

V d!: Biu -Bode Bin -HMIMO:

Gishthng a bin:

(3.155)

Cxxy =

=

0000

0001,

c hm truyn hMIMO 22 l:

)()()()(11

sDsNBAsICsH

== Vi: 165176908)( 234 ++++= sssssD

+

+

+

=

150

073

150

079

50

07

10

01)( 23 ssssN .

Hm H(s) l ma trn 22 , n c hai gi trsuy bin. Ch r%ng gi trsuy bin l lintc, ngai tr2ga trsuy bin cc i v cc tiu. Nhng gi trsuy bin c thgiao nhau ,c minh ch'ng b%ng hnh hc.

buAxuxx +=

+

=

00

10

00

01

3800

8300

0012

0021


33/33

Hnh 3.14 Biu Bode Bin hMIMO ca gi trsuy bin trong min tn s

Trong Malab dng hm sigma(H)

-minh ha skhc bit gia thtrsuy bin hMIMO v gin Bode hSISOring bit, xt h thng sau. Hm truyn ca h thng ny l ma trn vung c hng 2.

Hm truyn hSISO ring bit trong hthng vng h12 ng vo/2 ng ra l:

)2.20)(615.3)(0163.0(

8.14)(11

+++

=

ssss

sH

]063.3)4225.0)[(2.20)(615.3)(1)(0163.0(]49.2)55.0)[(237.2(9.36

)(22

22

12++++++

+++=

sssss

ssssH

]063.3)4225.0)[(2.20)(615.3)(0163.0()283.2)(573.2(65.2

)(2221 +++++

+=

sssss

ssssH

]063.3)4225.0)[(2.20)(615.3)(1)(0163.0(]446.0)139.0[(79.0

)(22

22

22++++++

++=

sssss

ssH

--contiep---

Chuong 3_dieu Khien Ben Vung_1

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