chuong 3 dieu khien thich nghi

8/13/2019 Chuong 3 Dieu Khien Thich Nghi

1/12

Ly thuyet ieu khien hien aiHuynh Minh Ngoc http://www.ebook.edu.vn

122

Chng 3: ieu khien thch nghi3.1.ieu khien thch nghi mo hnh tham chieu MRAS.

Hnh 3.1 la he thch nghi mo hnh tham chieu goc, c e ngh biWhitaker vao nam 1958 trong o gii thieu hai y tng mi. Th nhat, ch tieu cua

he c ac ta bi mo hnh; th hai tham so cua bo hieu chnh c chnh da trensai so gia mo hnh tham chieu va he thc. He thch nghi mo hnh tham chieu phatsinh ban au t bai toan t chnh trong he lien tuc thi gian tien nh. Y tng valy thuyet a c m rong bao gom he ri rac thi gian va he co nhieu ngau nhien.

ym

Tham so hieu chnh

Uc u y

Hnh 3.1 : S o khoi cua he thch nghi mo hnh tham chieuLuat MIT:C cau chnh nh vi thong so sau c goi la luat MIT, c s dung cho heMRAS au tien (hnh 3.1-1):

=

e

edt

d

e uc-ykhau tch phan

e u

Hnh 3.1-1:Mo hnh sai so.

oi tngBo ieukhien

C che ieuchnh

Mo hnh thamchieu

s


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123

Gia smuon thay oi thong so cua bo ieu khien sao cho sai so gia ngo ra cuaoi tng va cua mo hnh chuan tien ti zero. at e la sai so va la thong sohieu chnh. Ch tieu chat lng :

2

2

1eJ = (3.1-1)

e lam cho J() min (toi thieu) th can phai thay oi cac thong so theo hng amcua graien J,ngha la:

=

=

ee

J

t (3.1-2)

Gia s rang cac thong so can thay oi thay oi cham hn nhieu so vi cac bien

khac cua he thong.V vay ao ham

ec tnh vi gia thiet la hang so.Bieu

thc ao ham

ec goi la ham o nhay cua he thong.Luat ieu chnh theo

phng trnh (3.1-2) vi

ela o nhay th co lien he giong nh luat MIT. Cach

chon ham ton that theo phng trnh (3.1-1) co the la tuy y..Neu choneJ =)( (3.1-3)

Khi o luat hieu chnh se la:

)(esigne

dt

d

= (3.1-4)

Hay )(.. esigne

sign

dt

d

=

Nay goi la giai thuat dau-dau. He ri rac s dung giai thuat nay c ng dungtrong vien thong ni oi hoi tnh toan nhanh va thc hien n gian.Phng trnh (3.1-2) con c ap dung trong trng hp co nhieu thong so hieu

chnh, khi o tr thanh moat vect va

e la graien cua sai so oi vi cac

thong so tng ng. ng dung luat MIT c bieu dien bang hai v du sau:V du 3.1-1: Hieu chnh o li nuoi tienV du 3.1-2: MRAS cho he bac nhat (Nguyen Th Phng Ha, 2005[1])

Xet he thong c mota bi phng trnh:buay

dt

dy+== (1)

Vi u la bien ieu khien, y la ngo ra c o lng.Gia s muon co c he vong kn c mo ta bi

cmmmm ubya

dt

dy+= (2)

Mo hnh kem theo hoan hao co the at c vi bo ieu khien:)()()( 00 tystuttu c = (3)

Vi tham so baasbbt mm /)(;/ 00 ==


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124

Chu y hoi tiep se la dng neu am


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125

Hnh 3.1-2: ap ng ngo ra y, ngo ra tham chieu ym, va tn hieu ieu khien.Nhan xet:-He thong vong kn a at en ap ng mong muon ch sau moat thi gian ngan.-Toc o hoi tu phu thuoc vao hai thong so la va b.

*Thiet ke MRAS dung ly thuyet on nh LyapunovTh du 3.1-3: He MRAS bac nhat da vao ly thuyet on nh.

Xet he thong c mota bi phng trnh:

buaydt

dy+== (1)

Vi u la bien ieu khien, y la ngo ra c o lng.Gia s muon co c he vong kn c mo ta bi

cmmmm ubya

dt

dy+= (2)

Mo hnh kem theo hoan hao co the at c vi bo ieu khien:)()()( 00 tystuttu c = (3)

Khi tham so cua oi tng c biet luat ieu khien theo phng trnh (3) cho ket quamong muon. Mot he thch nghi s dung mo hnh chuan ma co the tm ra cac he so t0 vas0 khi tham so a va b khong biet co the at c nh sau:Sai so :e=y-ym,Ham Lyapunov:

])(1

)(1

[2

1),,( 20

20

200 mm bbt

baabs

besteV +++=

Neu cac tham so c cap nhat:

yedt

ds

eudt

dtc

=

=

0

0

Ta c:2ea

dt

dVm=

3.2.ieu khien thch nghi t chnh STR.


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Bo ieu khien t chnh

ac ta Tham so oi tng

Tham sobo ieu khien

Tham chieu u yNgo vao

Hnh 3.2 : S o khoi cua bo ieu khien t chnh

3.2.1.Biu khin tchnh STR :

3.2.1.1 tng cbn:

Trong hthng thch nghi, ngi ta gisrng tham siu khin c iuchnh lin tc. iu ny c ngha l tham siu khin thay i theo qu trnh. Tuynhin, vic phn tch cc tnh cht hi tv n nh i vi cc hthng nhvy lkh. n gin bi ton, ngi ta gisrng qu trnh c cc tham sl hng snhng khng bit. Khi bit c qu trnh, thtc thit kxc nh mt tp hp cctham siu khin mong mun. Biu khin thch nghi phi hi tvcc gi trtham sny ngay ckhi qu trnh khng bit. Biu khin vi tnh cht ny cgi l tchnh, bi v n tinh chnh biu khin n cht lng mong mun.

Biu khin tchnh (STR) da vo tng tch c lng tham schabit ra khi thit kbiu khin. tng ny c minh ha Hnh 3.2.

Tham scha bit c c lng trc tuyn, sdng cc phng php clng quy. Cc tham sc lng c coi nhng, ngha l khng xt n khng chc chn. iu ny c gi l Nguyn l Tng ng Chc chn. Nhiusc lng khc nhau c thc sdng, chng hn nhbnh phng cctiu, bnh phng cc tiu m rng, Cc phng php thit kc thc sdng l thay i cc tiu, bnh phng tuyn tnh, t cc v m hnh tham

oi tngBo ieukhien

c lngThiet ke boieu khien


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127

chiu. Phng php thit kc chn phthuc vo yu cu cht lng vng knv iu khin thc t. Di y l biu khin tchnh nh t cc.

Khi c lng tng trng cho c lng trc tuyn tham squ trnhsdng cc thut ton bnh phng cc tiu. Khi Thit k tng trng cho ligii trc tuyn i vi bi ton thit kcho hthng vi tham s bit. y cgi l bi ton thit kcbn. Khi Biu khin tnh ton tn hiu iu khintcc tham sc tnh ton khi trc. i khi thut ton STR c n gin

bng cch tham sha li v c lng trc tip cc tham sbiu khin, khngcn phi c lng cc tham squ trnh.

3.2.1.2 Biu khin hai bc tdo:

Hnh 3.3: Biu khin hai bc tdo

Xt qu trnh c m tbi hthng mt ng vo, mt ng ra (SISO) :

( ) ( ) ( ) ( ) ( ) ( )1 1 1q y k B q u k C q e k = +

V biu khin hai bc tdo :

( ) ( ) ( ) ( ) ( ) ( )1 1 1P q u k R q w k Q q y k =

Trong :( )

( )

( )

1 1 2 ...0 1 2

1 1 2 ...0 1 2

1 1 2 ...0 1 2

npP q p p q p q p qnp

nqQ q q q q q q q qnq

nrR q r r q r q r qnr

= + + + +

= + + + +

= + + + +

np, nq, nr l bc ca cc a thc P, Q, R tng ng.


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128

Biu khin ny gm c thnh phn nui tin (feedforword) vi hm truyn

( )( )

1

1

R q

P q

v thnh phn hi tip (feedback) vi hm truyn

( )( )

1

1

Q q

P q

. Do , n c

hai bc tdo. Skhi ca hthng vng kn c minh ha Hnh 3.3.Khu(k) ta c p ng vng kn :

( ) ( ) ( )BR CP

y k w k e kAP BQ AP BQ

= ++ +

Trong AP + BQ l a thc c trng vng kn.Mong mun c p ng vng kn:

( ) ( )B

my k w km Am=

T, ta c (gise(k)=0):

BBR mP BQ Am

=+

Mt cch tng qut, ( ) ( )deg degP BQ Am+ > . Suy ra, BR v AP + BQ

c tha schung A0, gi l a thc quan st.Theo nh mong mun, ch kh b cc zero n nh. Do , phn tch

B thnh B+B-.

Trong : B+: cha cc zero n nh v bkhB- : cha cc zero khng n nh v khng bkh

T, ta c:

0

0

BB mBR B B Rm

P BQ A A AAP B B Qm m

+ = =

+ + +

Suy ra:

1P P B+

=

Do :

1 0

P B Q A Am

+ =


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129

iu kin biu khin nhn qu:

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

deg deg

deg deg

deg deg deg deg

deg 2deg deg deg 10

Q P

R P

A B A Bm m

A A B Am

+

3.2.1.3 Cc thut ton t cc:

Thut ton 3:Thit kt cc vi tt czero khng bkh

Cc dliu cho: a thc A, BCc chtiu cht lng: a thc A

m, B

m, A

0Cc iu kin tng thch:B chia ht Bmdeg(Am)-deg(Bm) deg(A)-deg(B)deg(A0) 2deg(A) - deg(Am) 1

Bc 1. Gii phng trnh: AP + BQ = A0Am

Bc 2. Tnh 0BmR A

B=

Bc 3. Tnh lut iu khin

( ) ( ) ( ) ( ) ( ) ( )1 1 1P q u k R q w k Q q y k = Trong a thc R c thc tnh theo 2 cch sau:

Cch 1. li DC

t '0 0R r A=

( ) ( ) ( ) ( )

''0 00

0

Br ABR By k w k w k r w k

AP BQ A A Am m= = =

+

ng ra bm theo ng vo trng thi xc lp th:( )

( )

1'0 1

A qmrB q

==

=

Cch 2. tng nguyn l m hnh mi:


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130

( ) ( ) ( )

( ) ( )

( )1

e k w k y k

BRw k w k

AP BQ

BR

w kD

=

= +

=

Vi D = ama0Gi F l a thc pht sinh ca w(k) , tc l Fw(k)=[0]V dnu w(k) l hm nc th: 11F q= Cn nu w(k) l hm sine th: 1 20 )1 2cos( q qF T

+=

sai sxc lp bng 0, th D -BR = FS vi S l a thc bt k.V

( ) ( )

( ) ( )

[ ]0

D BRe k w k

D

FS Sw k Fw k

D D

=

= =

=

Gii phng trnh ny sc R.

3.2.2. Biu khin PID:

3.2.2.1 Gii thiu:

Biu khin PID c sdng rt rng ri trong thc tiu khin nhiuloi i tng khc nhau nhnhit l nhit, tc ng c, mc cht lng trong

bn cha. Theo cc bo co th hn 98% cc qu trnh trong cng nghip sdngbiu khin PID. L do biu khin ny c sdng rng ri nhvy l n ckhnng trit tiu sai sxc lp, tng p ng qu , gim vt lnu cc thamsbiu khin c chn la thch hp. Do sthng dng ca n nn nhiu hngsn xut thit biu khin cho ra i cc biu khin thng mi rt thngdng. Thc tcc phng php thit kbiu khin PID dng quo nghim s,

gin Bode hay phng php gii tch rt t c sdng do skh khn trongvic xy dng hm truyn i tng. Phng php phbin nht chn tham s

b iu khin PID thng mi hin nay l phng php Ziegler-Nichols.

Biu khin PID c hm truyn nhsau:Trong min lin tc:

( ) ip dK

K ss

G s K += +

Trong min ri rc:


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11 10

10

1) (1 )

2 1( i dp

K T KqK q

q TG q

+= + +

Di y l sbiu khin PID.

Hnh 3.4: Skhi biu khin PID

Cng thc tnh tn hiu iu khin u(k) tsai se(k):

1( ) ( ) ( ) ( )

0 d

k du k K e k e s ds T e k p T dt

i

= + +

Trong :u(k): tn hiu iu khine(k): l sai sc nh ngha e(k) = w(k) y(k)w(k): l tn hiu ty(k): l ng ra

3.2.2.2 Phng php p ng qu :Nhiu m hnh cng nghip c p ng nc nhHnh 3.5


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Hnh 3.5: p ng nc ca mt qu trnh cng nghip tiu biu

p ng ny c thxp xbng hm truyn sau :

( ) 1k Ls

G s eZN Ts

= +

Trong :k l li tnhL l thi gian trT l hng sthi gian

Ch :

L

Ta k=

Mt cch n gin xc nh cc tham sca biu khin PID l da vop ng nc c Ziegler-Nichols pht trin. Phng php ny chsdng hai thamsnhBng 3.1 sau.

Biu khin Kp Ti TdP 1/a

PI 0.9/a L/0,3PID 1.2/a 2L L/2Bng 3.1

Quy lut ny c pht trin bi cc m phng thc nghim nhiu hthngkhc nhau. N c nhc im l h thng vng kn vt lnhiu. C thchnh sacc tham sc kt qutt hn.Bai tap:1. ieu khien thch nghi mo hnh tham chieu MRAS.2.ieu khien thch nghi t chnh STR

3. Luat MIT.4. Giai thuat ieu khien t chnh dung phng phap at cc.5. ieu khien PID so.6.Thiet ke MRAS dung ly thuyet on nh Lyapunov7.ng dung Matlab mo phong v du 3.1-3.


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Chng 4: ieu khien m4.1. He thong vi s khong chac chan4.1.1.c lng trang thai vi nhieu ngau nhienTrong phan trc chung ta gia s thong tin o lng trong bo c lng la hoan hao.

Tuy nhien nay la trng hp kho v moi cam bien co moat so khiem khuyet. Th dunhieu ien cua cam bien co the c mo ta nh la bien ngau nhien them vao phngtrnh ngo ra:

)()(.)( kTekTxCkTy += (1)

Moi thanh phan cua vect nhieu e(kT) c mo hnh nh la chuoi cac bien ngau nhien, dang t nh la so ngau nhien.Khi nhieu o lng hien dien, phng trnh (1) c dung co tnh toan nhieu. Cau trucc lng c thay oi:

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

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

kTxCkTekTCxKkTukTx

kTxCkTyKkTukTxTkTx

+++=

++=+(2)

4.1.2. He m

4.1.2.1.LOGIC MHe m la he thong da tren tri thc hay da tren luat. Trai tim cua

he m la c s tri thc gom co cac luat Neu-Th. Chang han luat neu thsau :

Neu toc o xe la cao th ap dung t lc .trong o t cao va t c ac trng bi ham thanh vien tng ng.

He m c ng dung trong may giat, on nh anh so, trong xe hi, che

tao xi mang, ieu khien m ng xe la,vv...

4.1.2.1.1.Tap m

nh ngha tap hp kinh ien

Tat ca nhng oi tng xac nh nao o hp lai tao thanh mot tap hp, moioi tng la mot phan t cua tap hp.Th du : Tat ca nhng ngi Viet nam tren the gii tao thanh tap hp ngi Vietnam. Moi ngi Viet nam la mot phan t cua tap hp o.

Neu a la phan t cua tap hp E ta noi a thuoc E va viet Ea . Neu akhong la phan t cua E ta noi a khong thuoc E va viet Ea .Th du : 4 tap hp cac so chan.

Tap rong la tap hp khong co phan t nao.Neu moi phan t cua A cung la phan t cua B th ta noi A bao ham trong

B hay A la tap con cua B. Ky hieu : BA .Con co mot phng phap khac nh ngha tap ro A - phng phap thanh

vien, gii thieu ham thanh vien zero- mot (con goi la ham ac trng) oi vi Ava c ky hieu )(xA e ma :

chuong 3 dieu khien thich nghi

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