Chuong9_CSTD

Mon hoc Mon hoc

C S T ONGC S T ONG

Bin son: TS. Hunh Thi Hong B mn iu khin t ng

Khoa in in Ti hc Bch Khoa TPHCM

Email: [email protected]: www4.hcmut.edu.vn/~hthoang/

Ging vin: HTHong, NVHo, NHong, BTHuyn, HHPhng, HMTr

9 September 2011 H. T. Hong - HBK TPHCM 1

Chng 9Chng 9

THIET KETHIET KE HE THONG IEU KHIEN RI RACHE THONG IEU KHIEN RI RAC

9 September 2011 H. T. Hong - www4.hcmut.edu.vn/~hthoang/ 2

Khi nim

Noi dung chng 9Noi dung chng 9

Khi nim B iu khin sm-tr pha & PID ri rc Thit k h thng ri rc min Z g Tnh iu khin c v quan st c ca h ri rc Thit k h thng ri rc dng k thut phn b cc c lng trng thi h ri rc


Cac bo ieu khien ri racCac bo ieu khien ri rac


Cac s o ieu khien thng dungCac s o ieu khien thng dung

ieu khien noi tiepC(s)+ T G(s)ZOH

R(s) ieu khien noi tiep

GC(z)T

H(s)H(s)

ieu khien hoi tiep trang thai

+r(k) c(k)u(k) Cd)()()1( kukk dd BxAx

x(t)

K

)()()( dd


Ham truyen cua cac khau c ban ri racHam truyen cua cac khau c ban ri rac

Kh i h Khau vi phan

Vi phanu(t)e(t)

Vi phan

tde )()(Kh i h li dttdetu )()( Khau vi phan lien tuc:

Khau vi phan ri rac: TkekTekT ])1[()()( Khau vi phan ri rac:T

kTu ])[()()( zEzzEU )()()(

1

zG 11)( Ham truyen khau vi phan ri rac:T

zU )()()(


zTzGD )( Ham truyen khau vi phan ri rac:

Ham truyen cua cac khau c ban ri racHam truyen cua cac khau c ban ri rac

Kh t h h h hu(t)e(t)

Khau tch phan Tch phan

dt

)()( Kh t h h li t detu 0

)()( Khau tch phan lien tuc:

Kh h h ikT

kTTk

)1(

Khau tch phan ri rac: dekTu 0

)()( dedeTk

)1(0

)()(

kkTk ()()(kT

dTkkT )(])1[()( kTeTkeTTku ()]1[(2

])1[(

Tk

dtteTkukTu)1(

)(])1[()(

)()()()( 11 EETUU )()(2

)()( 11 zEzEzTzUzzU

Ham truyen khau tch phan ri rac: 1)( zTzGI9 September 2011 H. T. Hong - www4.hcmut.edu.vn/~hthoang/ 7

y p 12

)( zzGI

Ham truyen cua bo ieu khien PID ri racHam truyen cua bo ieu khien PID ri rac

Bo ieu khien PID lien tuc: Bo ieu khien PID lien tuc:

sKsKKsG DPPID )(

Bo ieu khien PID ri rac:

s

zz

TK

zzTKKzG DIPPID

111

2)(

P I D

zKz 1z

zTK

zzTKKzG DIPPID

11

)( hoac

P I D


P I D

Ham truyen cua bo ieu khien ri racHam truyen cua bo ieu khien ri rac

Bo ieu khien sm pha tre pha lien tuc:

ba tre phasm phaba

Bo ieu khien sm pha, tre pha lien tuc:

bsasKsGC

)( Ri rac hoa, s dung phng phap tch phan hnh thang:

)2()2()( aTzaTKG

Bo ieu khien sm pha, tre pha

)2()2()()()( bTzbTKzGC

p , p

C

CCC pz

zzKzG )(

)2()2(

aTaTzC )2(

)2(

bTbTpC

Cpz )1,1( CC pz

)2( aT )2( bT

tre phasm phaCC pz


CC pz tre pha

Phng phap thiet ke he thong ieu khien ri racPhng phap thiet ke he thong ieu khien ri rac

C h 1 hi k i i h h i khi li Cach 1: Thiet ke gian tiep he thong ieu khien lien tuc, sau ori rac hoa ta c he thong ieu khien ri rac. Chat lng cuah i h t l h li t h k l T he ri rac xap x chat lng he lien tuc neu chu ky lay mau T unho.

Cach 2: Thiet ke trc tiep he thong ieu khien ri rac.Phng phap thiet ke: QNS, phng phap phan bo cc, phngg p p Q , p g p p p , p gphap giai tch,


Thiet ke bo ieu khien ri rac trong mien ZThiet ke bo ieu khien ri rac trong mien Z


Trnh t thiet ke khau sm phaTrnh t thiet ke khau sm pha ri racri rac dung QNSdung QNS

)()( C pzzzKzG Kh hi h h thi t k )( )( CCC

CCC pzpz

KzG Khau hieu chnh can thiet ke Bc 1: Xac nh cap cc quyet nh t yeu cau thiet ke ve chat

l h th t t h

oquagianThiPOT lovot o

2*2,1 1 nn jslng cua he thong trong qua trnh qua o:

**2,1

Tsez

Bc 2: Xac nh goc pha can bu e cap cc quyet nh nam*z

o,... qua gian Thi nnTezr * 2* 1 nTz

Bc 2: Xac nh goc pha can bu e cap cc quyet nh namtren QNS cua he thong sau khi hieu chnh bang cong thc:

2,1z

m in i zzpz *1*10* )arg()arg(180

i

ii

i zzpz1

11

1 )arg()arg(180trong o pi va zi la cac cc va zero cua G(z) trc khi hieu chnh. *0* )(180 zzG ccencuacccactgoc


1)(180 zzG ccencuacccac tgoc *1 )( zzG cc encua zero cac t goc

Trnh t thiet ke khau hieu chnh sm pha dung QNS (tt)Trnh t thiet ke khau hieu chnh sm pha dung QNS (tt)

Bc 3: Xac nh v tr cc va zero cua khau hieu chnh Bc 3: Xac nh v tr cc va zero cua khau hieu chnhVe 2 na ng thang bat ky xuat phat t cc quyet nh saocho 2 na ng thang nay tao vi nhau mot goc bang * . Giao

*1z

g g y g giem cua hai na ng thang nay vi truc thc la v tr cc vazero cua khau hieu chnh.Co hai cach ve thng dung:Co hai cach ve thng dung: PP ng phan giac (e cc va zero cua khau H/C gan nhau) PP triet tieu nghiem (e ha bac cua he thong)PP triet tieu nghiem (e ha bac cua he thong)

Bc 4: Tnh he so khuech ai KC bang cach ap dung cong thc:

1)()( *1zzC zGzG


Th du thiet ke bo ieu khien sm pha ri rac dung QNSTh du thiet ke bo ieu khien sm pha ri rac dung QNS

C(s)R(s) C(s)+ T G(s)ZOHR(s)

GC(z)

50)(G 10T

TK bo ieu khien sm pha GC(z) sao cho he thong sau khi hieu)5(

)( sssG sec1.0T

C

chnh co cap cc quyet nh vi , (rad/sec)707.0 10n



Giai:

G )( )5(50)( sssG

Phng trnh ac trng: 0)(1 zG Giai:

ssGzzG )()1()( 1 Z

50

)5( ss

)5(50)1( 2

1

ssz Z

)]501()150[( 5.05.05.0 eezez

)()1(5

)]5.01()15.0[()1(10 5.021

ezzeezezz

180210 z

)1()1( aTaTaT aTeezeaTza Z

)607.0)(1(18.021.0)(

zzzzG


)()1()( 22 aTezzaass Z


Cap cc phc mong muon: Cap cc phc mong muon:jrez *2,1

493.010707.01.0 eer nTtrong o:

707.0* 21 493.0 jez 707.0707.01101.01 22 nT

2,1 493.0 ez320.0375.0* 2,1 jz



G h b Im z

321* )(180

Goc pha can bu: Im z+j 0.375+j0.320

01 9.152

02 9.125 P

03 6.14

0* 84Re z

01 +1123

A B

*084 pc zc

j


j


Ch kh hi h h b h h t i t Chon cc va zero cua khau hieu chnh bang phng phap triettieu nghiem:

607.0 Cz607.0Cz C

ABOBOApC 607.0OB578.0AB

029.0Cp



T h K Tnh KC: 1)()( * zzC zGzG1)18.021.0()607.0( zzK 1

)607.0)(1()029.0( 320.0375.0 jzC zzz

K

]180)32003750(210[ j 1)1320.0375.0)(029.0320.0375.0(

]18.0)320.0375.0(21.0[

jjjKC

2670 1702.0471.0

267.0 CK 24.1CK

K t l H t b i khi thi t k l

607.024.1)( zzGC

Ket luan: Ham truyen cua bo ieu khien can thiet ke la:


029.0)( zC


Q hi h th t khi hi h hQuy ao nghiem so cua he thong trc va sau khi hieu chnh


Trnh t thiet ke khau tre phaTrnh t thiet ke khau tre pha ri racri rac dung QNSdung QNS

zz )( )( CCC

CCC pzpz

zzKsG Khau hieu chnh can thiet ke

Bc 1: at Xac nh t yeu cau ve sai so xac lapCp1 Bc 1: at . Xac nh t yeu cau ve sai so xac lap.PK VK aKhoac hoac

C

C

z 1

*PK

*VK

*aK

hoac hoac Bc 2: Chon zero cua khau hieu chnh rat gan iem +1:

1Cz Bc 3: Tnh cc cua khau hieu chnh:

)1(1 CC zp Bc 4: Tnh KC thoa man ieu kien bien o:


1)()( * zzC zGHzG

Th du thiet ke bo ieu khien tre pha ri rac dung QNSTh du thiet ke bo ieu khien tre pha ri rac dung QNS

C(s)R(s) C(s)+ T G(s)ZOHR(s)

GC(z)

50)(G 10T

TK bo ieu khien tre pha GC(z) sao cho he thong sau khi hieu)5(

)( sssG sec1.0T

C

chnh co he so van toc 100* VK



Giai: Phng trnh ac trng trc khi hieu chnh:

0)(1G

Giai:

sGzzG )()1()( 1 Z )5(

50)( sssG0)(1 zG

sG )()(

)5(50)1( 2

1z Z )5(ss

)()1(5)]5.01()15.0[()1(10 5.02

5.05.05.01

ezzeezezz

)1()1( aTaTaT aTeezeaTza Z )()1(5 ezz

)6070)(1(

18.021.0)( zzzzG


)()1()( 22 aTezzaass Z )607.0)(1( zz


PTT trc khi hie chnh PTT trc khi hieu chnh0

)6070)(1(18.021.01

zzz

)607.0)(1( zz

Cc cua he thong trc khi hieu chnh547.0699.02,1 jz



B 1 X h Bc 1: Xac nh He so van toc trc khi hieu chnh:

1 )()1(lim1 11

zGzT

KzV

18.021.0)1(li1 1 zK

100*KHe so van toc mong muon:

)607.0)(1()1(lim

1.01

1 zzzK zV 9.9VK

100VKHe so van toc mong muon:

9.9VKD 100*

V

V

KDo o:

099,0


099,0


B 2 Ch kh t h t +1 Bc 2: Chon zero cua khau tre pha rat gan +1Chon: 99.0 Cz 99.0Cz

Bc 3: Tnh cc cua khau tre pha)1(1 CC zp 999.0Cp)99.01(099.01

999,099,0)(

szKzG CC

Bc 4: Xac nh he so khuech ai1)()( * zzC zGzG

1)607.0)(1(

)18.021.0()999.0()99.0(

547.0699.0

jzC zz

zzzK

9 September 2011 H. T. Hong - HBK TPHCM 26

j

1007.1 CK


QNS trc va sau khi hieu chnh


Th du thiet ke bo ieu khien PID dung PP giai tchTh du thiet ke bo ieu khien PID dung PP giai tch

C( )R( ) C(s)+ T G(s)ZOHR(s)

GC(z)

H(s)

11010)( ssG 05.0)( sH sec2T

Thiet ke khau hieu chnh GC(z) sao cho he thong kn co cap ccphc vi 0 707 2 rad/sec va sai so xac lap oi vi tnphc vi =0.707, n=2 rad/sec va sai so xac lap oi vi tnhieu vao la ham nac n v bang 0.



Khau hieu chnh can thiet ke la khau PI (v yeu cau sai so xac Khau hieu chnh can thiet ke la khau PI (v yeu cau sai so xaclap bang 0)

11

2)( zTKKzG IPC 12)( zPC

Phng trnh ac trng cua he thong sau khi hieu chnh la:0)()(1 zGHzGC

sHsGGH )()()1()( 1 Ztrong o:

05.010)1( 1 Z

11

szzGH

)()()1()( 1 Z

)110()1( 1

ssz Z

)1(05.0)1(2.0

1 ez

zz

TK

zzTKKzG DIPPID

111

2)(

))(1(1.0)()1( 2.0

1 ezzz

091.0)(GH


P I D)819.0(

)( zzGH


Do o phng trnh ac trng cua he thong la: Do o phng trnh ac trng cua he thong la:

08190

091.011

21

zzzTKK IP 819.012 zz

0)819009100910()819109100910(2 KKzKKz 0)819.0091.0091.0()819.1091.0091.0( IPIP KKzKKz(do T=2)




trong o:

059.02707.02 eer nT828270701221 22 T

828.2*2,1 059.0

jez 828.2707.01221 nT

018.0056.0*2,1 jz Phng trnh ac trng mong muon: Phng trnh ac trng mong muon:

0)018.0056.0)(018.0056.0( jzjz2


00035.0112.02 zz


Can bang cac he so phng trnh ac trng cua he thong va Can bang cac he so phng trnh ac trng cua he thong vaphng trnh ac trng mong muon, ta c:

1120819109100910 KK

0035.0819.0091.0091.0112.0819.1091.0091.0

IP

IP

KKKK

13.609.15

I

P

KK

1113.609.15)(

zzzGCKet luan:

0)819.0091.0091.0()819.1091.0091.0(2 IPIP KKzKKz


00035.0112.02 zz

Thiet ke bo ieu khien ri racThiet ke bo ieu khien ri racet e bo eu e acet e bo eu e actrong khong gian trang thaitrong khong gian trang thai


Tnh ieu khien cTnh ieu khien c

)()()1( kukk dd BxAx Ch h th

)()()()()1(

kkykukk

d

dd

xCBxAx Cho he thong:

HT i l i khi h i l k HT c goi la ieu khien c hoan toan neu ton tai luat k u(k) co kha nang chuyen he t trang thai au x(k0) en trang thai cuoi x(kf) bat ky trong khoang thi gian hu han k0 k kf . f y g g g 0 f

Mot cach nh tnh, he thong ieu khien c neu moi bien trang thai cua he eu co the b anh hng bi tn hieu ieutrang thai cua he eu co the b anh hng bi tn hieu ieu khien.


ieu kien can va u e he thong ieu khien cieu kien can va u e he thong ieu khien c

)()()( kkk

)()()()()1(

kkykukk

d

dd

xCBxAx oi tng:

Ma tran ieu khien c (Controlability matrix)

][ 12 ndddddd BABABABC ][ ddddddd BABABAB C

ieu kien can va u e he thong ieu khien c la:

nrank )(C


PP phan bo cc thiet ke bo ieu khien hoi tiep trang thaiPP phan bo cc thiet ke bo ieu khien hoi tiep trang thai

r(k) y(k)u(k) x(t)+r(k) y(k)u(k) Cd)()()1( kukk dd BxAx

x(t)

Bc 1: Viet phng trnh ac trng cua he thong kn

K

Bc 1: Viet phng trnh ac trng cua he thong kn0]det[ KBAI ddz (1)

Bc 2: Viet phng trnh ac trng mong muon Bc 2: Viet phng trnh ac trng mong muon

0)(1

n

iipz (2)

Bc 3: Can bang cac he so cua hai phng trnh ac trng (1) va

1i),1( , nipi la cac cc mong muon


(2) se tm c vector hoi tiep trang thai K.

PP phan bo cc. Th du 1PP phan bo cc. Th du 1

Ch h th i khi

+r(k) c(k)u(k) Cd)()()1( kukk BxAx x(k)

Cho he thong ieu khien

+

K

Cd)()()1( kukk dd BxAx

K

31601 0920

368.00316.01

dA

316.0092.0

dB 010dCHay xac nh vector hoi tiep trang thai K sao cho he thong kn cocap nghiem phc vi =0.707, n=10 rad/sec



Ph t h t h th k Phng trnh ac trng cua he thong kn0]det[ KBAI ddz

0316.0092.0

368.00316.01

1001

det 21

kkz

0316.0368.0316.0

092.0316.0092.01det

21

21

kzkkkz

316.01A

0)092.0316.0(316.0)316.0368.0)(092.01( 2121 kkkzkz0)368.0316.0066.0()368.1316.0092.0( 2121

2 kkzkkz

092.0

368.00d

B

A


316.0dB



trong o:

493.010707.01.0 eer nT70707070110101 22 T 707.0707.01101.01 nT

707.0* 493.02,1 jez 320.0375.0* 2,1 jz Phng trnh ac trng mong muon: Phng trnh ac trng mong muon:

0)320.0375.0)(320.0375.0( jzjz 024307502


0243.075.02 zz


Can bang cac he so phng trnh ac trng cua he thong va Can bang cac he so phng trnh ac trng cua he thong vaphng trnh ac trng mong muon, ta c:

750)368131600920( kk

243.0)368.0316.0066.0(75.0)368.1316.0092.0(

21

21

kkkk

047.112.3

2

1

kk

Ket luan: 047.112.3K0)368.0316.0066.0()368.1316.0092.0( 2121

2 kkzkkz


0243.075.02 zz


Cho he thong ieu khien: Cho he thong ieu khien:

r(k) c(k)+ ZOHT 0 1u(k) uR(t)

10x1x21

1 1T=0.1

++ k2

1s s

k1

1. Viet phng trnh trang thai mo ta he h2. Hay xac nh vector hoi tiep trang thai K = [k1 k2] sao cho hethong kn co cap nghiem phc vi =0.5, n=8 rad/sec.3. Tnh ap ng cua he thong vi gia tr K va tm c khi tn


hieu vao la ham nac n v. Tnh o vot lo, thi gian qua o.


Giai: Giai:1. Viet phng trnh trang thai mo ta he h:

c(t)uR(t) xx1 1 h li c(t)uR(t) 10x1x21

1s s

1B1: PTTT mo ta he lien tuc:

sXsX )()( 2 )()( sXssX )()( txtx s

sX )( 21 )()(2 sUsX R

)()( 21 sXssX )()( 21 txtx

)()()1( 2 sUsXs R )()()( 22 tutxtx R1)(2 s )()()( 2 R 22 R

)(0)(10)( 11 tu

txtxR

)(

1)(10)( 22tu

txtx R

)(010)(10)( 1 txtxtc9 September 2011 H. T. Hong - www4.hcmut.edu.vn/~hthoang/ 42

)(010)(10)( 21 txtxtc


B2: Ma tran qua o:B2: Ma tran qua o:

1)( -ss AI 1

101

s

s1

1010

1001

s

)1(11

)( sss

1010

1

10

)1()(

s

ssss

)]([)( 1 st L

sss

1)1(

111L

1)1(

11 11sss

LL

as10

110 1

sL

tet )1(1)(


tet 0)(


)()()1( kkk BAB3: PTTT mo ta he ri rac h:

)()()()()1(

kkckukk

d

dd

xCBxAx

10 )(Td A

1.0

1.0

0)1(1

ee

905.00095.01

dA

T

d d0

)( BB

1.0

0 10

0)1(1

d

ee

1.0

0

)1(

dee

0

1.0 e 0 0

11.0 10 1.0e 0950005.0

dB0 e

010 CC 0)1(1)(

eet t

t 11.0e 095.0


010 CCd1.0

T


2 Tnh o li hoi tiep trang thai K:2. Tnh o li hoi tiep trang thai K:

Phng trnh ac trng cua he kn:

0]d [ KBAI 0]det[ KBAI ddz

0005.0095.0101det 21 kkz 095.0905.0010 21 0

005.0095.0005.01det 21

kkz 0

095.0905.0095.0det

21

kzk

0)005.0095.0(905.0)095.0905.0)(005.01( 2121 kkkzkz 0)005.0095.0(905.0)095.0905.0)(005.01( 2121 kkkzkz0)905.0095.00045.0()905.1095.0005.0( 2121

2 kkzkkz



C t h Cap cc quyet nh mong muon:jrez * 2,1

67.085.01.0 eer nT693.05.0181.01 22 nT n

693.0* 67.02,1 jez

Ph h

428.0516.0* 2,1 jz

0)428.0516.0)(428.0516.0( jzjzPhng trnh ac trng mong muon:


0448.003.12 zz


C b h PTTT h k PTTT Can bang cac he so PTTT cua he kn va PTTT mong muon:

03.1)905.1095.0005.0( 21 kk 448.0)905.0095.00045.0( 21 kk

0.441k

8956044KVay

895.62k 895.60.44KVay

0)905.0095.00045.0()905.1095.0005.0( 21212 kkzkkz


0448.003.12 zz


3 Tnh ap ng va chat lng cua he thong :3. Tnh ap ng va chat lng cua he thong :

Phng trnh trang thai mo ta he kn:

)()()()()1(

kkckrkk

d

ddd

xCBxKBAx


Thi k b l hi iThi k b l hi iThit k b c lng trng thi ri rcThit k b c lng trng thi ri rc


Khi nim c lng trng thiKhi nim c lng trng thi

thc thi c h thng iu khin hi tip trng thi: cn phi o thc thi c h thng iu khin hi tip trng thi: cn phi o c tt c cc trng thi ca h thng.

Trong mt s ng dng, ch o c cc tn hiu ra m khng th o tt c cc trng thi ca h thng.

Vn t ra l c lng trng thi ca h thng t tn hiu ra o lng clng c

Cn thit k b c lng trng thi (hoc quan st trng thi)


Tnh quan st cTnh quan st c

)()()1( kukk dd BxAxCh h h

)()()()()1(

kkykukk

d

dd

xCBxAx Cho h thng

He thong tren c goi la quan sat c hoan toan neu cho tn He thong tren c goi la quan sat c hoan toan neu cho tn hieu ieu khien u(k) va tn hieu ra y(k) trong khoang k0 k kf ta co the xac nh c trang thai au x(k0).

Mot cach nh tnh, he thong la quan sat c neu moi bien trang thai cua he eu anh hng en au ra y(k).g y


iu kin cn v h thng quan st ciu kin cn v h thng quan st c

)()()1( kukk dd BxAxi

)()()()()1(

kkykukk

d

dd

xCBxAx i tng

Cn c lng trng thi t thng tin bit trc v m hnh)( kx

Ma trn quan st c (Observability matrix)

Cn c lng trng thi t thng tin bit trc v m hnh ton hc ca i tng v d liu vo ra ca i tng.

)(kx

dd

d

ACC

2ddAC

O

1nddAC iu kin cn v h thng quan st c l:


nrank )(O

Th d kho st tnh quan st cTh d kho st tnh quan st c

)()()1( kukk dd BxAxCh i

)()()()()1(

kkykukk

d

dd

xCBxAx Cho i tng

14809670 2310trong :

522.0297.0

148.0967.0dA

264.0231.0

dB 31dCHy nh gi tnh quan st c ca h thng

Giai: Ma tran quan sat c:

Hy nh gi tnh quan st c ca h thng.

dd

d

ACC

O

714.13

077.01

O

Do 484.1)det( O 2)( Orank He thong quan sat c


He thong quan sat c

B quan st trng thiB quan st trng thi

u(k) y(k)x(k)u(k) y(k))()()1( kukk dd BxAx x(k) Cd

+

CdBd 1z

L

)( kx

++++

)1( kxCd )( ky

Bd

Ad

z+

)()())()(()()()1(

kkykykykukk

d

dd

xCLBxAx B quan st trng thi:


Tnlll ][ 21 Ltrong :

Thit k b quan st trng thiThit k b quan st trng thi

Yu cu: Yu cu: B quan st trng thi phi n nh, sai s c lng trng thi

tim cn tin v 0. t h h b t h h i t h c tnh ng hc ca b quan st nhanh so vi c tnh

ng hc ca h thng iu khin.

Cn chn L tha mn: Tt c cc nghim ca phng trnh u

nm trong vng trn n v.

Cn chn L tha mn:0)det( ddzI LCA

Cc nghim ca phng trnh nm xa vng trn n v hn so vi cc cc ca phng trnh

0)det( ddzI LCA

0)det( KBAzI 0)det( KBA ddzI Ty theo cch thit k L ta c cc b quan st trng thi khc nhau:

B quan st trng thi Luenberger


q g g B lc Kalman ( L thuyt iu khin nng cao)

Trnh t thit k b quan st LuenbergerTrnh t thit k b quan st Luenberger

B 1 Vi t h t h t b t t th i Bc 1: Viet phng trnh ac trng cua bo quan sat trang thai0]det[ ddz LCAI (1)

B 2 Vi t h t h t t Bc 2: Viet phng trnh ac trng quan sat mong muon

0)(1

n

iipz (2)

Bc 3: Can bang cac he so cua hai phng trnh ac trng (1) va

1i),1( , nipi la cac cc mong muon cua bo quan sat

Bc 3: Can bang cac he so cua hai phng trnh ac trng (1) va (2) se tm c vector L.


Th d thit k b quan st trng thiTh d thit k b quan st trng thi

Th du: Cho oi tng mo ta bi phng trnh trang thai: Th du: Cho oi tng mo ta bi phng trnh trang thai:

)()()()()1(

kkykukk dd

xCBxAx

)()( kky d xC

148.0967.0

A 231.0

B 31dC

Gi kh th t thi h th H thit k

522.0297.0dA 264.0dB 31dC

Gi s khng th o c cc trng thi ca h thng. Hy thit k b quan st trng thi Luenberger, sao cho cc cc ca b quan st trng thi nm ti 0 13 v 0 36trng thi nm ti 0.13 v 0.36 .


Th d thit k b quan st trng thi (tt)Th d thit k b quan st trng thi (tt)

Gii: Phng trnh ac trng cua b quan sat Luenberger

0]det[ ddz LCAI

Gii:

0]det[ ddz LCAI 031

52202970148.0967.0

1001

det 1

ll

z522.0297.010 2 l

0352202970

3148.0967.0det 11

lzlllz

(1)0)549.0753.2413.1()489.13( 21212 llzllz

3522.0297.0 22 lzl

Phng trnh ac trng cua bo quan sat mong muon:0)36.0)(13.0( zz


(2)00468.049.02 zz

Th d thit k b quan st trng thi (tt)Th d thit k b quan st trng thi (tt)

C b h h i h t h (1) (2) Can bang cac he so cua hai phng trnh (1) va (2), suy ra:

0468054907532413149.0489.13 21

llll

0468.0549.0753.2413.1 21 ll Giai he phng trnh tren, ta c:

544.1653.2

2

1

ll

T544.1653.2L Ket luan


M phng b quan st trng thi ri rc M phng b quan st trng thi ri rc


Kt qu m phng c lng trng thiKt qu m phng c lng trng thi


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