chuong 1 ht dieu khien phi tuyen

7/22/2019 Chuong 1 HT Dieu Khien Phi Tuyen

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Trng ai hoc Cong nghiep Tp. Ho Ch MinhKhoa Cong nghe ien t

Bo mon ieu khien t ong

Bai giang Ly thuyet ieu khien hien aiBien soan: Huynh Minh Ngoc

Lu hanh noi boTp. Ho Ch Minh, thang 1 nam 2012


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Huynh Minh Ngoc http://www.ebook.edu.vn2

MUC LUCLi noi au 3Chng 1: He thong ieu khien phi tuyen 4Chng 2: ieu khien toi u 28Chng 3: ieu khien thch nghi 119Chng 4: ieu khien m va mang nron 130Chng 5: Cac ng dung 151Tai lieu tham khao 209


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Ch ng trnh chi ti tMn h c: L thuy t iu khi n hi n i

H : i hc in t t ng

Ch ng 1: H th ng iu khi n phi tuy n1.1.Khi ni m1.2.Ph ng phpm t phng pha.1.3. X p x m hnh ton h c phi tuy n.1.4.Ph ng phphm m t .1.5. L thuy t n nh Lyapunov1.5.1.Ph ng php th nht.1.5.2.Ph ng php th hai.1.6. Ph ng php n nh tuy t i Popov.

Ch ng 2: iu khi n ti u2.1. T i u ha t nh.2.1.1. T i u ha khng c rang bu c.2.1.2. T i u ha v i rang bu c ng thc

B nhn Lagrange v hm Hamilton2.1.3.Ph ng php nghi ms.2.2.LQR cho h th i gian h u hn.2.3. iu khi n ti u ca h thng r i r c th i gian.2.4. iu khi n ti u ca hlin tc th i gian.2.5. LQR bm theo.2.6.Nguyn l c c tiu Pontryagin2.6.1. Bi ton tr ng thi cu i t do: thi t k LQR2.6.2. Nguyn l c c tiu Pontryagin

H c rang bu c Nguyn l c c tiu Pontryagin

2.7.Ph ng php quy h ach ng ca Bellman.2.8. LQR v i hi ti png ra.2.9.B lc Kalman.Lc Kalman.LQG2.10. Thi t k iu khin H .

Ch ng 3: iu khi n thch nghi

3.1. Khi ni m3.2. iu khi n thch nghim hnh tham chi u MRAS.3.2.1. MRAS3.2.2. Lu t MIT.3.2.3. iu khin thch nghi s dngl thuy t n nh Lyapunov.3.3. iu khi n thch nghi t ch nh STR. So snh v i b iu khi n PID s .Ch ng 4: iu khi n m vm ng n ron.4.1. H thng v i skhng ch c chn.4.2. H m Logic m : t p m ,lut h pthnh m ,hm ,b iu khin m .4.3.M ng n ron

Ch ng 5: Cc ng d ng


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5.1. iu khi n thch nghi t ch nh dung ph ng php t c c.5.2. iu khi n m nhi t l in.5.3. iu khi n ti u cho i t ng robot (tay my).


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LI NOI AULy thuyet ieu khien hien ai la mon hoc chuyen nganh cua nganh

ong. Sinh vien phai hoc qua mon ly thuyet ieu khien t ong trc e chn mon ly thuyet ieu khien hien ai. Noi dung bai giang gom co cac cChng 1: He thong ieu khien phi tuyen.Chng 2: ieu khien toi u.Chng 3: ieu khien thch nghi.Chng 4: ieu khien m va mang nron.Chng 5: Cac ng dung.

Noi dung bai giang bam theo e cng chi tiet mon hoc cua khoa Coien t. oi tng giang day la sinh vien ai hoc nganh ien t- t ongtai lieu tham khao tot cho sinh vien cac nganh Vien thong, may tnh va nghiep.

V bai giang c bien soan lan au nen khong tranh khoi sai sotthanh cam n cac gop y cua ong nghiep, va ban oc e bai giang hn. Th gop y xin gi ve a ch: Bo mon ien t t ong, khoa Cong trng ai hoc Cong nghiep Tp. HCM, so 12 Nguyen Van Bao, P.4, Q. GoHCM. T: 38940390, email:[email protected] [email protected].

Thang 1 nam 2012Tac gia

Huynh Minh Ngoc


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Chng 1: He thong ieu khien phi tuyen1.1. Khai niem ve oi tng phi tuyen :Th du : Xet he sau

X M( ) N( ) y-

Hnh 1.1G1(s), G2(s) la khau tuyen tnh.N la khau phi tuyen.Ham truyen he kn :

1)()(

)( 00

+= sG sG

sG k , vi G0(s)=G1(s).G2(s)Phng trnh ac trng: 1+ G0(s)=0Thng ngo vao he phi tuyen M( )=M.sin tNgo ra he phi tuyen N( )=N1.sin( t+1)+ N2.sin(2 t+2)+ N3.sin(3 t+3)+..Tnh chat va ac iem rieng cua he phi tuyen:-Nguyen l xep chong khong ap dung cho he phi tuyen.-S on nh cua he phi tuyen lai phu thuoc ieu kien va ban chat cua tncac thong so he.-oi vi he tuyen tnh hoan chuyen hai phan t trong mot tang khong an

hoat ong. ieu nay khong ung neu mot phan t la phi tuyen.Cac ac iem rieng cua he phi tuyen:-Cac chu trnh gii han.-Dao ong t kch.-Nhay cong hng.-Phat sinh hai phu.Cac phng phap nghien cu he phi tuyen :-Phng phap mat phang pha.-Phng phap can bang ieu hoa.-Tieu chuan on nh Liapunov.-Tieu chuan on nh tuyet oi Popov.1.2 Cac phng phap nghien cu he phi tuyen1.2.1. Phng phap mat phang phaMat phang pha va tnh chat cua no

Xet he phi tuyen bac hai (n=2) c mo ta dang hai phng trnh vi pnhat vi cac bien trang thai x1 va x2 :

),(

),(

212

22

.

2111

1

.

x x f dt

dx x

x x f dt

dx x

==

== (1.1)

Hoac c mo ta di dang mot phng trnh

G1(s) N G2(s)


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),(),(

211

212

1

2

x x f x x f

dxdx

= (1.2)

vi cac ieu kien au x1(0) va x2(0).iem can bang xc thoa:

0),(

0),(

2122

2

.

2111

1

.

===

===

x x f dt

dx x

x x f dt

dx x

Phng phap mat phang pha la phng phap o hoa e tm ap ng qua onhat hay bac 2 ti ieu kien au hay ngo vao n gian. Bat chap cac ranghu ch bi v tnh trc quan phng phap nay cung cap va bi v nhieu he thap ng bac 2. Xet phng trnh phi tuyen

0),(),( =++ x x xh x x x g x &&&&& Thay the vao

dxdy

y xdxdy

y x y x ==== &&&&& ,

Sau o phng trnh giam thanh phng trnh bac nhat:0),(),()( =++ x y xh y y x g

dxdy

y

Sap xep lai ta co phng trnh mat phang pha

y x y xh y y x g

dxdy ),(),( =

Mat phang pha la o th cua y theo x nh hnh ve. Tai moi iem (x,y), dy/dx latuyen cua quy ao pha thong qua iem o.ng Isocline la ng thang quy ao hang so. Phng trnh isocline cho dy/d

m y x g x y xh

y+

=),(

),(

Th du: Xet phng trnh02 2 =++ x x x nn &&&

Phng trnh mat phang pha: y

x ydxdy nn )2(

2 +

=

Phng trnh isocline: m x

y nn

+

=

2

2

ng isocline la cac ng thang qua goc toa o, minh hoa hnh ve vicua m.

1.2.2. Phng phap tuyen tnh hoa ieu hoa:1.2.2.1. Khai niem

Phng phap tuyen tnh hoa ieu hoa hay con c goi la phng phapta a xuat hien ong thi trong vong mot thang cua nam 1948 nhieu nAnh,..

Viec dung ham mo ta la mot co gang e m rong gan ung hamac lc cua he tuyen tnh sang he phi tuyen. Phng phap tuyen tnh hoa


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phng phap khao sat trong mien tan so a c ng dung cho cac he phcao (n>2) do de thc hien va tng oi giong tieu chan Nyquist.Y tng c ban : Ham mo ta hay con goi la he so khuech ai phc cuac nh ngha la t so cua thanh phan c ban cua ap ng au ra mo

tuyen oi vi bien o tn hieu sin cua tn hieu vao. Noi chung ham mo ta bien o va tan so cua tn hieu vao va phc tap bi v dch pha co the xvao vi thanh phan c ban au ra. Ta se nghien cu phng phap phan tcva so sanh no vi khai niem ham truyen oi vi he tuyen tnh.

Neu au vao phan t phi tuyen la tn hieu hnh sin, phep phan tch hgia s la au ra cung la tn hieu tuan hoan co cung chu k c ban nh cunhap. V vay viec phan tch ch lien quan en cac thanh phan c ban cua dara. Tat ca cac hai phu va thanh phan mot chieu eu bo qua. S tha nhabi v cac thanh phan hai thng rat nho so vi thanh phan chnh. Hn nthng lam suy giam cac thanh phan hai do tac dung loc von co cua no.phi tuyen khong tao ra cac thanh phan mot chieu do tnh oi xng va cung bat k hai phu nao. V vay trong nhieu trng hp(khong phai la tat ca) thban la nhng thanh phan co y ngha au ra dung cu phi tuyen.

Neu mot he cha nhieu hn mot phi tuyen, ta phai gop tat ca lai vc mot ham mo ta to hp.

Can kiem tra s chnh xac cua phng phap ham mo ta v o la mot pgan ung. Tuy nhien phng phap nay a ra cac ket qua hp l va co li lacho cac he thong bac bat k nao va ap dung kha n gian. Ket qua nhan tra bang cac ky thuat khac hay mo phong tren may tnh.

Mac du co nhc iem, ky thuat ham mo ta la cong cu co ch ethiet ke cac he phi tuyen. Ham mo ta nh mot dang ham truyen tong quaphi tuyen.

e rut ra bieu thc toan hoc cho ham mo ta ta hay xet mot he phquat mo ta hnh 8.2 . Theo nh ngha ve ham mo ta, ta gia s au vao phi tuyen N(M,) c cho bi:m(t)=Msin t (1)Tong quat , au ra trang thai xac lap cua dung cu phi tuyen c bieu dien

...)3sin()2sin()sin()( 332211 ++++++= t N t N t N t n (2)

r(j) m( t) n( t) c(j)

-

Hnh 1.2

Bang nh ngha, ham mo ta co dang11),( = je

M

N M N (3)

G1(j) N G2(j)


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Chu y la ham mo ta phu thuoc vao bien o va tan so cua tn hieu vao.tuyen c e cap co o li va dch pha thay oi theo bien o va tan so tn1.2.2.2.Ham mo ta cua cac phi tuyen thong dung:Trong phan nay , dan ra cac ham mo ta oi vi cac phi tuyen thong dung.

s dung pho bien nhat la chuoi Fourier cua dang song ngo ra dung cu phi xet thanh phan c ban. Ta hay xet thanh phan phi tuyen N(M,) trong mot he hoi tiepto hp trnh bay hnh 1.2. Gia s au vao m(,t) c cung cap bi tn hieu sinm(t)=Msin t (4)Ta bieu dien dang song ngo ra bang chuoi Fourier cho bi bieu thc

=

=

++=11

0 sincos2

)(k

k k

k t k Bt k A A

t n (5)

trong o

)(cos)(2 2/

2/

t d t k t nT

AT

T

k

= ; k=0,1,2, (6)

)(sin)(2 2/

2/

t d t k t nT

BT

T k

= ; k=0,1,2, (7)

Tong quat, neu n(t)=-n(-t), la ham leAk=0. Neu n(t)=n(-t), la ham chanBk=0.

V chung ta ch lien he vi thanh phan ra c ban ng vi tan so nen ch can xac nhA1 va B1. Vi m(t)=Msin t ham mo ta co the co c t bieu thc

1

112/1

212111 tan)()(),( B A

M A

M B

M A

j M B

M N

+=+= (8)

a)Ham mo ta cua vung chet (Dead zone)Hnh 1.3 mo ta ac tnh cua vung cheat, moi lien he gia vao va ra c biecac phng trnh:

0)( =t n khi D


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T (14) v ham mo ta la t so cua bien o thanh phan c ban cua au vaco the bieu dien nh sau:

)cos

2

(2

)( 1111 t t

M

D K

M

B M N dz

== (15)

hay )sin)cos(sin2

(2

)( 1111 M D

M D

M D K

M B

M N dz ==

n(t)

-D 0 +D m(t)

K1

Hnh 1.3. ac tnh phi tuyen cua vung chet.

Hnh 1.4. Dang song vao va ra t dung cu phi tuyen co ac tnh vung cheat b)Khau bao hoa(Saturation)Hnh 1.5 minh hoa ac iem phi tuyen bao hoa. S lien he gia vao va ra cu

nay co the bieu dien bang cac phng trnh sau:t M k t n sin)( 1= khi S


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S k t n 1)( = khi m(,t)>S (17)S k t n 1)( = khi m(,t)


12/29


)(sin)(4 2/

01 t d t t n B

= (18)Thay phng trnh (16) va (17) vao (18) ta c

)(sin)sin()(sin)sin([4 2/

110

111

1

t d t t M K t d t t M K B t

t

+= (19)trong o )(sin 11 M

S t = hay

M S

t =1sin .

Giai (19) dan ra bieu thc

)cos(2

111

1 t t M S M K

B

+= (20)

T (20) v ham mo ta la t so cua bien o thanh phan c ban cua au vaco the bieu dien nh sau:

)cos(2)( 1111 t t

M S K

M B M N sat

+== (21)

hay )sin)cos(sin(2)( 1111 M S

M S

M S K

M B

M N sat +==

c)ham mo ta cua khau khe h (Blacklash)

)12

(sin 11 =

M D

t hay 12sin 1 = M D

t .

M

M

D

M

D A )2

2(

21 =

M t M D

t B )cos)12

(2

(1

111

=

)/(tan1

)( 1112

12

1 B A B A M M N backlash

+=

d)Ham mo ta cua phan t on/off co t tre.

e)Khau Rle 3 v tr co tre

)sin(sin)(

2)cos(cos

)(2

2121

++

+=

h D A K

jh D A

K N N N

h D M A

M D

A +=== ;sin;1sin 21

x(t)=Msin(t), M>D+h. f)Khau so sanh co tre.Trigger Schmitt khong ao.

)sin(cos4 max0

j AV V

N H

=

H V M

D M

A A === ;1sin


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1.3 Che o t dao ongHam mo ta cua mot phan t phi tuyen co the dung xac nh s ton tai chan trong he ieu khien hoi tiep phi tuyen bang phng phap xap x. Chung phi tuyen hnh 1.2. Neu ta gia s ham phi tuyen la N(M,) va R(j)=0, ta hay xac

nh cac ieu kien ton tai dao ong trong he.Vi m(t)=Mcos tThanh phan c ban cua n(t) c cho bi:

)],(cos[),()( 111 M t M M N t n += Co the viet lai

0]1)(),([ )(1 1 =++ t j j Mee jG M N (*)

vi G(j)=G1((j)G2 (j) la dch pha do G(j).oi vi trng hp duy tr dao ong, thanh phan trong dau ngoac vuong cu

bang 0 v M0. Do o

),(1

)(

0)(),(1

M N jG

jG N

=

=+

Neu to hp bien o va tan so co the tm c ma thoa man phng trnh ieu khien hoi tiep ton tai che o dao ong.1.4. Xap x cua mo hnh toan hoc phi tuyen:*He thong phi tuyen:Mot he la phi tuyen neu nguyen l xep chong khong ap dung c.*Tuyen tnh hoa he phi tuyen:* Xap x tuyen tnh cua mo hnh toan hoc phi tuyen:Xet he co ngo vao x(t) va ngo ra y(t). Quan he gia y va x la:

Y=f(x) (2-218)Neu ieu kien lam viec bnh thng tai, , th phng trnh (1) c khai trien chuoiTaylo:

...)(!2

1)()(

)(

22

2

+++=

=

x xdx

f d x x

dxdf

x f y

x f y (2-219)

trong o ao ham 22

, dx f d

dxdf

,.. c tnh tai x = . Neu x u nho, ta bo qua thanhphan bac cao trong . Phng trnh (2-219) c viet

)( x xk y y += (2-220)trong o

)( x f y =

dxdf

k = tai x = .

Phng trnh (2-220) viet lai:)( x xk y y = (2-221)


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ch ra y t le vi x . Phng trnh (4)(2-228) cho mo hnh toan hoc tuyen tnhoi vi he phi tuyen (2-219) gan iem lam viec.

oi vi ham hai bien),( 21 x x f y =

Khai trien Taylo:

...])())((2)([!2

1

)]()([),(

2222

2

2

221121

22

1121

2

222

111

21

++

+

+

+

+=

x x x

f x x x x

x x f

x x x

f

x x x f

x x x f

x x f y

trong o vi phan tng phan c tnh tai1 x x = , 2 x x = .Mo hnh toan hoc tuyen tnh cua he phi tuyen trong lan can iem lam viebi:

)()( 222111 x xk x xk y y +=

trong o

221

221

,22

,11

21 ),(

x x x x

x x x x

x f

k

x f

k

x x f y

==

==

=

=

=

Th du: Cho z=xy (2-222), tuyen tnh hoa (2-222) phng trnh phi tuyen trong 5x7, 10y 12. Tm sai so cua phng trnh tuyen tnh hoa c dung e tnh

x=5, y=10.Giai:V 5x7, 10y 12 nen chon 6611.611,6 ==== z y x M rong phng trnh (2-222) vao chuoi Taylo quanh iem y x == , va bo qua thanhphan bac cao:

)()( y yb x xa z z += trong o

6),(

11),(

,

,

==

=

==

=

==

==

x y

y x f b

y x

y x f a

y y x x

y y x x

V vay phng trnh tuyen tnh la:)11(6)6(1166 += y x z

hay 66611 += y x z Khi x=5, y=10, gia tr cua z cho bi phng trnh tuyen tnh la:Z=11x+6y-66=55+60-66=49Gia tr chnh xac z=xy=50. Sai so la 50-49=1 hay 2%.1.5 Tieu chuan on nh Liapunov


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16/29


hay x A xi = . vi

.........

...

...

2

2

1

2

2

1

1

1

x f

x f

x f

x f

A

=

Thanh lap phng trnh ac trng tng ng phng trnh xap x tuyen tnh:Det(sI-A)=0 (1)

Vi I la ma tran n v co rank la n (bac cua phng trnh).Liapunov chng minh rang neu nghiem cua phng trnh ac trng (1) co phkhac khong th cac phng trnh xap x tuyen tnh luon cho ap so ung v

nh cua he phi tuyen.* Neu tat ca cac nghiem cua phng trnh ac trng eu co phan thc amtuyen se on nh trong pham vi hep.

=< niS i ,1;0Re * neu ch co mot trong so cac nghiem cua phng trnh ac trng co phan thhe phi tuyen khong on nh.* Neu co du ch la mot nghiem cua phng trnh ac trng co phan thc batat ca nghiem con lai eu co phan thc am th khong the ket luan ve tnh phi tuyen theo anh gia nghiem cua phng trnh tuyen tnh gan ung c.

Th du: Xet mot he tuy ong n gian co s o nh hnh

UM -

v=0 UN u

-

Hnh 1.7: S o he tuy ong n gian.UN=sin (ac tnh phi tuyen).Ham truyen cua ong c G(s):

)1()(

+=

Ts s K

sG

UM-ien ap tng ng vi moment tai at vao ong c.Xet on nh cua he trang thai can bang theo phng phap th nhat cua LiGiai:

Thanh lap he phng trnh bien trang thai cho heat x1= ta co:

N G(s)


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21 x

dt dx

=

M U T K

xT

xT K

dt dx

+= 212 1sin

hay

=

==

2

1

2

1 )(;);( f

f x f

x

x x x f

dt dx

Phng trnh cha thanh phan sinx, do o la phng trnh phi tuyen. Trang thai c nh ngha la

0=dt dx do vay

==

0

0

2

1

f

f

Phng trnh trang thai can bang la02 = x

1sin01

sin 1212 ==+= M M M U U xU T

K x

T x

T K

dt dx

S dung phng phap th nhat e khao sat oi vi phi tuyen nho

T x

T K

x f

x f

x f

x f

A 1cos

10

1

2

2

1

2

2

1

1

1

=

=

Phng trnh ac trng det(sI-A)=0

0cos)1

(1cos1

1cos

10

0

01

11=++=+

=

xT K

T s s

T s x

T K

s

T x

T K

s

s (2)

Xet cac trng hp cu the:1.UM=0 khong co tac ong nhieuT phng trnh cua trang thai can bang ta co* Sinx1=UM=0

,..)4,2,0(21 = m x

cosx1=1Phng trnh ac trng (2) co dang0)

1( =++

T K

T s s

vi K>0, Res1,2


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Mot nghiem co phan thc dng va mot nghiem co phan thc am, ap dungth nhat ket luan he khong on nh trong pham vi hep va iem can bang ktrong pham vi hep.

2.0cos

1sin

1

1

1

===

=

xU x

U

m

M

Phng trnh ac trng co dang0)

1( =+

T s s

Mot nghiem s=0 va mot nghiem s=-1/T


19/29


hien tam trong lng con lac. Toa o theo truc hoanh va truc tlac c cho bi :

sinsin2

l X L

X x +=+= (1)

coscos2

l L y == (2)

Fy

G Fx f x

mg Fx Fy Mg

Fy /2 Fy /2Hnh 2: S o than t do cua he con lac ngc tren xe.

S o than t do cua xe va con lac c minh hoa hnh 2x va Fy the hien lc phan ng tai iem truc. Xem xet con lac nTong lc ta co cac phng trnh sau :

sincos 2&&&&& ml ml X m F x

+= (3) cossin 2&&& ml ml mg F y = (4)

&& J l F l F x y = cossin (5)Xem xet lc phng ngang tac ong len xe, ta co la :

x x F f X M =&& (6)

Thay (6) vao (3), ta c : x f ml ml X m M =++

2)(sin)(cos)( &&&&& (7)S dung (3), (4), (5) va tnh toan rut gon , ta at c :

0sin34cos =+ mg ml X m &&&& (8)

Phng trnh (7) va (8)mo ta ong lc hoc he con lac ngc .nh ngha bien trang thai =1 x , va

.

2 = x Ta suy ra :

2

2

)(cos34

.coscossinsin

aml l

f aaml g x

=

&&&


20/29


hay :2

2

)(cos3

)(4.coscossinsin)(

ml m M l

f ml g m M x

++

=&

&& (9)

trong o chung ta a thay

12

2

mL J = , va m M a +=1


21/29

V tr xe :

m M ml f

X x+

+=

])cos()[sin( 2 &&&&&

hay : 2

2

)(cos)(34

)cos()sin()sin(3

4

3

4

mm M

mg ml f X

x

+

+=

&

&& (10)

trong o , thong so con lac la :Khoi lng xe M=1 kg, khoi lng con lac m=0,1 kg, na chieul=0,5 m (chieu dai con lac L=2l=1m), gia toc trong trng g=9,8 m2 .Phng trnh (9) va (10) c dung mo ta ac tnh ong cua oi tng.Phng trnh phi tuyen (7)va (8) mo ta he.e ieu khien PID ta phai tuyen tnh hoa he tai vi tr can bang0= . Gia s goc u

nho e chung ta xap x sin = , cos =1 va 2

&

=0. Khi o thay cac xap x nay vao (7va (8) ta co x f ml X m M =++

&&&&)( (11)

.34

mg ml X m =+ &&&& (12)

Trc tien chung ta tm ham truyen cua he con lac ngc. Lay bien hai ve (11)va (12) vi ieu khien au zero ta c:

u smls s X sm M =++ )()(.)( 22 (13))()(

3

4)( 22 smg smls s X ms =+ (14)

V )( s la goc lech so vi phng thang ng, nh la ham cua trang thaiphng trnh (14) ta c

)(.3.4

)( 2 sl

s g

s X

= (15)

Thay phng trnh (2-74) (15)vao (2-72) (13), ta c:

)()()(.3.4

)( 222 sU smls s sl

s g

m M =+

+ (16)

Sap xep lai ta c ham truyen sau:

g m M sl m M

ml sU s

)().3

)(4(

1)()(

2 +++= (17)

Bieu dien he dang khong gian trang thai, ta at

====

x x

x x

x

x

&

&

4

3

2

1

(18)


22/29

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

22

)4(3134

m M

mg u X

+

=

&& (19)

ml m M l

u g m M

+

+=

3)(4

)( && (20)

Phng trnh tuyen tnh hoa co the c bieu dien nh sau:

++

+=

=+

+

+=

=

um M

xm M

mg x

x x

uml m M l

xml m M l

g m M x

x x

.)4(

4)4)(3/1(

.)().3/4(

1)().3/4(

)(

14

43

12

21

&

&

&

(21)

Viet lai (21) di dang ma tran:

u

m M

ml m M l

x

x

x

x

m M

mg

ml m M l

g m M

x

x

x

x

.

)4(40

)(34

10

000)4(

31

1000

000)(

34

)(0010

4

3

2

1

4

3

2

1

+

+

+

+

+

+

=

&&&&

(22)

V ta xet v tr xe va v tr goc cua con lac, ngo ra nh sau:

u

x

x

x x

t y

t y.

0

0

0100

0001

)(

)(

4

3

2

1

2

1

+

=

(23)

Thay gia tr M=1 kg; m=0,1 kg;l=0,5m;g=9,8m/s2 vao (22), ta c:

u

x x

x

x

x x

x

x

.

975,00

463,1

0

000717,01000

00078,15

0010

4

3

2

1

4

3

2

1

+

=

&&&&

(24)

1.5.3.Phng phap th hai cua LiapunovMot trong nhng phng phap co hieu lc nhat e khao sat bai toa

chuyen ong la phng phap th hai hay con goi la phng phap trcLiapunov. Theo phng phap nay tieu chuan on nh chuyen ong co the atiep vao he phng trnh vi phan cua chuyen ong b nhieu ma khong thotch phan he phng trnh.

ong hoc he phai c mo ta bi khong gian trang thai.


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23

Th du, he phi tuyen co the c mo ta bi tap hp n phng trnh vtuyen bac nhat.

),,...,,( 21 t x x x f x ni =& i=1,2,...,nDang mo hnh khong gian trang thai la :

),( t x f x =&

trong o

=

=

=

),,...,(

.

.

.

),,...,(

),(;

.

.

.

;

.

.

.

1

1111

t x x f

t x x f

t x f

x

x

x

x

x

x

nn

n

nn &

&

&

trong o x la vect trang thai, va tat ca phan t la bien trang tha(x1=...=xn=0) la khong gian trang thai va gia s la iem can bang, ngha lai=0,i=1,2,...,n.nh l Liapunov ve on nh tiem can :Neu tm c mot ham V(x)=V(x1,x2,..,xn) xac nh dau , sao cho ao ham cutheo phng trnh vi phan cua chuyen ong b nhieu

),...,,( 21 n x x x f x =& (3)cung la ham xac nh dau, song trai dau vi dau cua ham V(x) th chuyen nhieu se on nh tiem can.nh l on nh Liapunov : Bay gi co the tom tat cho khong gian trang thaMot he ong lc bac n la on nh tiem can neu ham xac nh dng V(t)

co ao ham theo thi gian la am doc theo quy ao cua he thong.Trong thc te de tm mot ham la xac nh dng, nhng them vao o ham VdV/dt

>=>=

n

qq

qqq (5)


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24

Neu ham V(x) xac nh am th ieu kien (5) c thay the bang ieu kien

0

;...0;02221

12112111


25/29


25

(ii) Vi hien tng tre : tnh phi tuyen vi bo nh, trong o u phu thuocs thi gian cua e ; th du : khau rle v tre, khau khe h trong liekh.

(iii) Tnh phi tuyen tong quat : thay oi theo thi gian va co le bao go

tng tre.

r(t) + e u c(t)

-

(a)

u

K

f(e)

e(b)

Hnh 1.8 : He thong va tnh phi tuyen

Gia s la r(t) va ngo vao nhieu bat k la b chan va tch phan vuo


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26

S dung c thc hien bang ham ap ng tan so hieu chnh G*(j) c nh nghabiRe G*=Re G Im G*= Im G 0 (2)Trong dien at o hoa sau cua ieu kien Popov cho on nh tiem can toan cu

nh l Popov : Vi bat k ieu kien au, ngo ra he thong la b chan va co tien ve zero khi t neu o th cua G*(j) nam hoan toan ve ben phai cua Popov, ma cat truc thc tai -1/K o doc 1/q. ay gii han ve q va K tnh phi tuyen :

(i) -


27/29


27

(b)

q=0

-1/K

G*

(c)Hnh 1.9: Phng phap Popov

Trong hnh 1.9 (a) o th cc cua )0(*

G nam ben phai cua ng thang Popov nhminh hoa, v vay he thong la on nh tiem can. Trong hnh 1.9 (b), ng dang bi -1/K cho gia tr K cc ai ma nh l am bao on nh, va the thphi tuyen gia tr n, bat bien theo thi gian (i), ma trong o q>0 la cho phetnh phi tuyen (iii) rang buoc ve q th chat che hn. Nh c minh hoa hnh 1q b rang buoc la 0 (zero), gia tr cc ai cua K bay gi ma tng ng vng cua G*. Neu trong (c) tnh phi tuyen la loai (i), gia tr cc ai K se tphan giao cua G* vi truc thc am bi v tan tai iem nay la ng thangnhan c. That la quan trong quan sat rang phan giao cua G va G* vi trucgiong nhau, va rang phan giao nay cung cho o li on nh cc ai cua hbang tieu chuan Nyquist. Nh vay, vi tnh phi tuyen (i) phan on nh Popov 1.9 (c), va cung nh (a), la giong nh phan Hurwitz (ngha la pham vi o li ohe la tuyen tnh).Bai tapChng 1 : He thong ieu khien t ong phi tuyen

1. ac iem va tnh chat cua he phi tuyen. Cho th du ve he phi tuyen.2. Trnh bay phng phap mat phang pha.3. Trnh bay phng phap tuyen tnh hoa ieu hoa (ham mo ta).4. Tieu chuan on nh Liapunov

5. Mot he KT gom mot khau phi tuyen bao hoa va mot khau tuyentruyen G(s)


28/29


28

r(t)=0 e(t) u(t) c(t)

Hnh 1B.5

u

-D +D e

E(t)=Msin tG(s)= K T1=0,1 sec va T2=10sec

S(T1s+1)(T2s+1)a.Chng minh rang ham mo ta cua khau bao hoa co dang

+= ))(cos(arcsin)arcsin(2

M D M D

M D K

N N

, ieu kien M >= D.

Vi M D

t =1sin

b.Cho D=1, KN=5. Tm ieu kien cua K e he phi tuyen on nh trang thbang.6. Cho he thong co s o khoi tren hnh 8B.6

2)( s K

sG =

r(t) x u F(x) c(t)

F(x)Za

x

N G(s)

N G(s)


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29

-Za Hnh 1B.6

Xet on nh tai trang thai can bang (r(t)=0) theo phng phap quy ao pha vphap Liapunov.

7.

Tieu chuan on nh Popov.

chuong 1 ht dieu khien phi tuyen

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