sổ tay cdt chuong 24-kg trang thai

8/6/2019 s tay cdt Chuong 24-KG Trang Thai

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24Phn tch khng gian trng thi

v cc thuc tnh h thng

Mario E. SalgadoUniversidad Tcnica FedericoSanta Mara

Juan I. YuzUniversidad Tcnica FedericoSanta Mara

24.1 M hnh: Nhng khi nim c bn .........................1

24.2Bin trng thi: Nhng khi nim c bn ................2

24.3Miu t khng gian trng thi cho cc h thng thigian lin tc ......................................................... .....5

24.4Miu t khng gian trng thi cho h thng thi gianri rc v d liu ly mu .......................................15

24.5Cc m hnh khng gian trng thi cho cc h thng kt ni .....................................................................25

24.6Cc thuc tnh ca h thng ............................. .......27

24.7B quan st trng thi ..............................................42

24.8Phn hi trng thi ...................................................47

24.9Phn hi trng thi quan st c ..........................49

24.1 M hnh: Nhng khi nim c bn

Mi lin h thit yu gia mt k s/nh khoa hc v mt h thng l da vo kh nng m th thng ca h theo cch thun tin hiu v xc nh ng x ca h.

Bt c s m t no h tr mi lin h u c th gi l mt m hnh. Trong l thuyt hthng, m hnh ng vai tr c bn v chng cn thit cho vic phn tch, tng hp v thit k ttc cc loi h thng c th tng tng c.

Khng ch c mt m hnh duy nht cho mt h thng cho trc. Th nht, v m hnh c thtun theo nhiu mc ch khc nhau. Th d, khi x l vi ng c in, chng ta c th quan tmn qu trnh bin i nng lng c-in, hoc c th m hnh ho ng c ging nh mt h

thng nhit hay h thng c nghin cu s rung ng hoc sc bn vt liu vv Th hai, vtrn thc t cc m hnh thng khng chnh xc v cc h thng thc thng v cng phctp. Mt trong nhng quyt nh then cht i vi cc k s khi thc hin m hnh ho h thng lphi quyt nh c tnh thit yu no m m hnh phi th hin v quyt nh cng lin quanmt thit vi mc ch ca m hnh.

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S tay C in t

L thuyt ny h tr vic m hnh ho l bi bn thn n l mt lnh vc rng ln, nhngnguyn tc u tin, l thuyt tn hiu, ton hc v cng c s c kt hp theo nhiu cch khcnhau to ra cc phng php lun phong ph. Mt m hnh him khi c xy dng ch theomt hng, qu trnh xy dng m hnh thng xuyn tng tc qua li vi nhau v cc tin trnhca n ph thuc vo cht lng kt qu thu c khi s dng m hnh trong mt ng dng ringr. Cc tng tc cng c th bao gm vic thay i trong phng php lun ca vic m hnhho.

Trong chng ny, chng ta s gii quyt vi mt lp cc m hnh c bit miu t cc hthng ng. H thng ng l cc h thng m cc bin h thng u ph thuc ln nhaukhng ch theo ton hc m cn theo cch m chng ta quan st s can thip ca cc hiu ngtch lu v tc thay i. Cc m hnh cho h thng ng c th c xy dng trong min thigian lin tc, ri rc hoc trong cu trc thi gian lin tc-ri rc(i vi cc h thng hn hp,cc h thng ly mu).Chng ta s cp n c ba trng hp .

y chng ta s nhn mnh vo cc khi nim, nhng thuc tnh c bn, cc gii thch vt lv cc v d. Chng ta cng s khng i su vo vic chng minh v cng khng pht trin cc lthuyt phc tp cho ngi c d hiu. Bn c c th tm hiu su v l thuyt thng qua cc tiliu [6, 8, 10-14].

24.2 Bin trng thi: Nhng khi nim c bn

Gii thiu

Mt trong nhng lp m hnh c s dng thng xuyn nht l nhng m hnh c xcnh bi cc phng trnh da trn mt nhm cc bin ni ti ca h thng. Nhng bin ni ti c hiu l cc bin trng thi. Gi tr ca chng ti nhng khong thi gian tc thi nht nhc hiu nh trng thi h thng, mc d chng ta thng din t bin h thngv trng thih thngtheo ngha ging nhau.

nh ngha trn tr nn khng r rng khi n c t vo mt tp bin h thng c th. cim ni bt ca mt tp bin trng thi c lm r trong nh ngha sau:

Mt tp bin trng thi ca h thng cho trc l mt tp bin ni ti, do bt k mt binh thng no cng c th c tnh ton nh mt hm ca trng thi hin ti v cc u vo hinti v tng lai ca h thng.

Trong nh ngha ny, chng ti mun nhn mnh ngha vt l ca cc bin trng thi. Tuynhin, cng c th c thm nhng nh ngha tru tng khc. nh ngha ny cng ng rngnu chng ta bit trng thi ti thi gian t chng ta c th tnh ton c nng lng cha trong hthng ti thi im . Nng lng cha trong h thng ph thuc vo mt s bin h thng (tc, in p, dng in, v tr, nhit , p sut) v tt c chng, theo nh ngha, u c th ctnh ton t trng thi h thng.

nh ngha trn gi rng c th ngh v trng thi theo mt cch tng qut hn: cc bintrng thi c th c chn nh mt hm (vd: mt t hp tuyn tnh) ca cc bin ni ti ca hthng. S tng qut ho ny s to nn mt s khong cch gia trng thi v gii thch vt l can. Tuy nhin, cng c thun li khi to ra cu trc tng qut hn . N cng lm r thm: s lachn cc bin trng thi l khng duy nht.

Mt quan st quan trng khc l thi gian tin trin ca trng thi, bn thn biu trng thi

c th c tnh ton t gi tr hin ti ca trng thi v cc u vo hin ti v tng lai. V vycc m hnh lin quan u l phng trnh vi phn bc mt (thi gian lin tc) hoc phng trnhsai phn mt bc (thi gian ri rc).

Cc m hnh khng gian trng thi c bn

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Phn tch khng gian trng thi vi cc thuc tnh h thng

Nu chng ta biu din x l vector tng ng vi mt bin trng thi ring, dng tng qut cam hnh bin trng thi l nh sau:

i vi h thng thi gian lin tc:

F(x( ), u( ), )dx

t t tdt

= (24.1)

y( ) G (x( ),u( ), )t t t t = (24.2)trong u(t) l vector u vo v y(t) l vector u ra ca h thng.

i vi cc h thng thi gian ri rc:

d[ + 1] = F (x[ ], u[ ], )x t t t t (24.3)

dy[ ] = G (x[ ], u[ ], )t t t t (24.4)

tng t nh trng hp thi gian lin tc, u[t] l vector u vo v y[t] l vector u ra ca hthng.

Ch rng, trong sut chng ny chng ta s s dng k hiu t biu din thi gian lin tcv ri rc nhng s ch s dng [ v ] i vi thi gian ri rc.

c c ci nhn s b ban u v khi nim khng gian trng thi, ta xem v d sau:

V d 24.1

Trong hnh 24.1, mt ngoi lcf(t) c t vo h khi lng-l xo. V tr d(t) c o tngng vi v tr khi lng khi l xo trng thi t do khng c ngoi lc tc ng. S dchchuyn ca khi lng b cn li bi lc ma st trt t l vi tc ca vt nng v(t).

Chng ta bit rng tnh ton v tr v tc vt nng chng ta phi bit tc ban uca vt nng v lc ko ban u ca l xo. V vy vect trng thi phi c hai thnh phn, v d

x(t) = [x1(t)x2(t)]T, v mt la chn trng thi ban u l:

1( ) ( )x t d t = (24.5)

12 ( ) ( ) ( )x t v t x t = =g

(24.6)

Vi la chn ny c th p dng nh lut Newton c c:

2 1 2

( )( ) ( ) ( ) ( ) ( ) ( )

dv tf t M Kd t Dv t M x t Kx t Dx t

dt= + + = + +

g

(24.7)

trong D l hng s t l ma st nht. Chng ta c phng trnh trng thi nh sau:

1 2( ) ( )x t x t =g

(24.8)

2 1 2

1( ) ( ) ( ) ( )

K Dx t x t x t f t

M M M= +

g

(24.9)

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S tay C in t

HNH 24.1 H c

Ta cng thy nng lng w(t) c lu gi trong h thng c a ra bi:

2 21 1

w(t) ( ( )) ( ( )) ( ) ( )2 2

TK d t M v t x t Lx t = + = (24.10)

trong l ma trn ng cho: =diag ,2 2

K M

Cui cng, tnh khng duy nht ca vector trng thi c th c nh gi nu thay v cc la

chn c to ra theo (24.8) ta chn trng thi mi ( )x t lin quan vix(t) bi mt ma trn khng

suy bin:

( ) ( )x t x t = (24.11)

Vn ny s c cp chi tit phn Bin i trng thi tng ng.

Tn hiu v miu t khng gian trng thi

Khng gian trng thi cng c th c s dng miu t nhiu loi tn hiu bng cch sdng m hnh c dng:

( )A ( ), y(t) =C ( )

dx tx t x t

dt= i vi cc tn hiu thi gian lin tc (24.12)

[ ] [ ] [ ]q q[t +1]= A , Cx x t y t x t = cho nhng tn hiu thi gian ri rc (24.13)

minh ho tng ny, chng ta xt tn hiu thi gian lin tc c dng:

f(t) = 2 + 4cos(5t) sin(5t) (24.14)

tn hiu ny c th c gii thch nh l nghim ca phng trnh vi phn thun nht:

2

3

( ) ( )25 0

d f t df t

dtdt+ = , gi thit: f(0) = 6,

.

(0) 5f = v (0) 100f

= (24.15)

nu ta chn nh bin trng thi,1 2( ) ( ), ( ) ( )x t f t x t f t

= = v..

3( ) ( )x t f t = , th m hnh khng

gian trng thi cho tn hiu ny l:

[ ]

0 1 0( )

0 0 1 ( ), ( ) 1 0 0 x(t)

0 25 0

dx tx t y t

dy

= = =

(24.16)

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Theo cch dng ny ca m hnh khng gian trng thi, cc bin trng thi khng c nghavt l c th. Tuy nhin, miu t ny c bit c ch trong l thuyt khi phc tn hiu v khi x lnhiu khi tng hp h thng iu khin.

24.3 Miu t khng gian trng thi cho cc h thng thi gian lin tc

Phn ny gii thiu m t khng gian trng thi cho h thng thi gian lin tc. Phn tch tp

trung vo lp cc h thng tuyn tnh v bt bin. Trc tin ta s xem xt lm th no xydng mt m hnh tuyn tnh t cc phng trnh phi tuyn (24.1) v (24.2).

giai on ny, c mt gii hn na l cc h thng nghin cu khng c thi gian tr thunty. c tnh ny s sinh ra mt vect trng thi v hn chiu. Tuy nhin chng ta s thy trongmc 24.4 rng lp h thng ny c th c x l thnh cng bng m hnh d liu trch mu.

Tuyn tnh ho

V chng ta s tp trung vo cc h thng thi gian bt bin, nn (24.1) v (24.2) c th vit linh sau:

F( ( ), ( ))dx

x t u t dt

= (24.17)

( ) G( ( ), ( )y t x t u t = (24.17)

Gi thit rng m hnh (24.17) v (24.18) c t nht mt im cn bng cho bi {xQ, uQ, yQ).Ba vect ny tho mn:

0 = F(xQ, uQ) (24.19)

yQ = G(xQ, uQ) (24.20)

Ch rng im cn bng ny c xc nh bi o hm trng thi bng khng.

Nu by gi ta xem xt vng ln cn xung quanh im cn bng, sau ta c th tnh xp xm hnh (24.17) v (24.18) bng cch ct cc s Taylor trong m hnh nh sau:

Q Q

Q Q

Q Q Q Q

x=x x=xu=u u=u

F Fx F(x , u ) (x( ) x ) (u( ) u )

x u

t t

= + +

&

(24.21)

Q Q

Q Q

Q Q Q Qx=x x=x

u=u u=u

G Gy( ) G(x ,u ) (x( ) x ) (u( ) u )

x ut t t

= + +

& (24.22)

Phng trnh (24.21) v (24.22) c th vit nh sau:

( )( ) ( )

d x tA x t B u t

dt

= + (24.23)

( ) ( ) ( )y t C x t D u t = + (24.24)

trong :

( ) ( ) ( ) ( ) ( ) ( )Q Q Qx t x t x u t u t u y t y t y = = = (24.25)

v

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S tay C in t

, , ,= = = == = = =

= = = =

Q Q Q QQ Q Q Q

x x x x x x x xu u u u u u u u

F F G GA B C D

x u x u (24.26)

V d sau y s minh ho cho vic tuyn tnh ho.

V d 24.2

Xt h thng trong hnh 24.2

Qu cu st chu tc ng bi hai lc: sc nng ca bn thn vt, mg, v lc ht in t, f(t).in t c iu khin bi ngun in p, e(t) > 0, t

Hnh 24.2 H thng ht t tnh

Lc hp dn tc ng ln khi cu, f(t), ph thuc vo khong cch h(t) v dng in i(t). Miquan h ny c th m t gn ng nh sau:

1

2

( ) ( )( )

Kf t i t

h t K=

+(24.27)

trong K1 vK2 l hng s dng.

S dng quy tc th nht ta c:

( )( ) ( ) Ldi t e t Ri t dt

= + (24.28)

( )( )

dh tv t

dt= (24.29)

1

2

( )( ) ( )

( )

K dv t f t i t mg m

h t K dt = = +

+(24.30)

Tip theo ta chn cc i lng sau nh cc bin trng thi: dng in i(t), v tr khi cu h(t),v tc khi cu v(t)

[ ] T1 2 3( ) ( ) ( ) ( ) [ i(t) h(t) v(t) ]T

x t x t x t x t = = (24.31)

sau t (24.28) (24.30) ta c th t miu t h thng nh trong (24.1)

1 1( )( ) 1

( ) ( )dx tdi t R

x t e t dt dt L L

= = + (24.32)

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2

3

( )( )( )

dx tdh tx t

dt dt = = (24.33)

3 1

1

2 2

( )( )( )

( ( ) )

dx t Kdv tx t g

dt dt m x t K = =

+(24.34)

Trc khi xy dng m hnh tuyn tnh, im cn bng cn c tnh ton. u vo iu khin

cah thng ny l ngun pe(t)

. im cn bng nhn c khie(t) = E

Q, v vy trng thi cnbng c th tnh nh (24.32) - (24.34), bng cch t tt c cc o hm bng khng, c ngha l:

1 1

10 QQ Q Q

ERx E x

L L R + = = (24.35)

3 30 0Q Qx x = = (24.36)

11 1

1 2 1 2 2

2 2

0( )

Q

Q Q Q

Q

K EK Kx g x x K K

m x K mg mgR = = =

+(24.37)

Cc thit lp ny xy dng m hnh tuyn tnh theo s gia tng ca u vo e(t) v giatng trng thi x(t) = [x1(t) x2(t) x3(t)]T. Kt qu l:

1 1( ) 1( ) ( )d x t R x t e t dt L L

= + (24.38)

2

3

( )( )

d x tx t

dt

= (24.39)

2

31 2

1

( )( ) ( )

Q Q

d x t Rg Rmg x t x t

dt E K E

= (24.40)

Nu ta nh ngha v tr qu cu h(t) l u ra ca h thng, ta c th so snh cc phng trnhtrn vi (24.23) v (24.24) c:

2

1

1

0 0 00 0 1 , 0 , 1 , 0

0 00g

Q Q

R

L LA B C D

R Rmg

E K E

= = = =

(24.41)

Trong kt qu trn chng ti b qua k hiu ng trc, nhng ngi c cn nh rng mhnh trn y l tuyn tnh theo s gia tng ca trng thi, cc u vo v u ra xung quanh imcn bng chn.

Cc m hnh khng gian trng thi tuyn tnh

im khi u ca chng ta by gi l m hnh khng gian trng thi tuyn tnh bt bin.

( )

A ( )+B ( )dx t

x t u t dt

= (24.42)

y(t) = C ( ) + D ( )x t u t (24.43)

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S tay C in t

Vi gi thitx(t0) = x0, nghim ca phng trnh (24.42) l:

0

0

A(t-t ) ( )

0 0x(t) = e ( )

t

A t

t

x e Bu d t t + (24.44)

trong ma trn chuyn tipeAttha mn:

1

1

!

At k k

ke I A t k

== + (24.45)Bn c c th kim tra (24.44) l tha mn (24.43) bng cch dng quy tc Leibniz cho o

hm ca tch phn.

Vi kt qu trn, li gii cho (24.43) c cho bi:

0

0

( ) ( )

0( ) ( ) ( )

tA t t A t

t

y t Ce x C e Bu d Du t = + + (24.46)

ng lc hc h thng

Trng thi ca h thng gm hai thnh phn: thnh phn khng cng bc, xu(t), v thnh phncng bc x(t), trong :

0( )

0( ) A t tux t e x

= (24.47)

0

0

( )( ) ( )

tA t t

f

t

x t e Bu d = (24.48)

hiu r m hnh khng gian trng thi v li gii ca n, ta xt trng hp to = 0 v u(t) = 00t , ngha l trng thi ch c thnh phn khng cng bc.

0( ) Atx t e x= (24.49)

Gi thit rng nA v n gin, n c cc tr ring 1 2, ..... n vi n vector ring(c lptuyn tnh) v1, v2vn. Khi lun tn ti mt tp cc hng s 1 2, ..... n sao cho:

0

1

,n

x v =

= l l ll

(24.50)

kt qu bit t i s tuyn tnh cho thy rng cc tr ring ca Akl 1 2, ......k k k

n vi cc vectorring tng ng v1, v2vn.V vy:

01 1 1

1( )

!k

n nt At k k

kv

x t e x I A v t e vk

= = =

= = + = ll l

l l l ll l

1 2 3 (24.51)

Phng trnh ny cho thy rng thnh phn khng cng bc ca trng thi l t hp tuyntnh ca cc mode t nhin {e }tl , m mi mode trong s chng lin quan ti mt tr ring ca A.

V vy, ma trn A xc nh: cu trc ca p ng khng cng bc s n nh (hoc nhng mt khc) ca h thng

tc p ng

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Khi ma trn A khng c mt tp n vector ring c lp, cc dng Jordan c th c s dng(xem [9,10]).

Cu trc ca p ng khng cng bc

Khi thiu u vo, trng thi nh l t hp ca cc mode t nhin ph thuc vo cc lp hm c nh ngha: tt c c iu khin bi cc s m vi c phn thc hoc phc. V vynhng m hnh bao gm cc hng s, s m thc, cc sng sin thun ty, cc sng sin b iu

bin theo hm m, v mt s cc hm c bit khc xut hin t cc tr ring lp. minh ha lp lun trn v mi quan h vt l ca chng, xt h thng trong v d 24.1. Vi

h thng :

0 1A K D

M M

=

(24.52)

V vy, cc tr ring ca h thng l cc nghim ca phng trnh:

2det( ) 0D K

I AM M

= + + = (24.53)

c ngha l:

2

1,2 22 4

D D K

M MM = (24.54)

Do , khi suy gim bng khng (D = 0), cc tr ring ca h thng u l mt cp ca cc sphc lin hp, v t hp hai mode t nhin (phc) sinh ra mt dao ng c duy tr vi tn s

gc0

/K M = . iu ny ph hp vi cm nhn vt l ca chng ta, v chng ta cho rng daong c duy tr khi h thng c iu kin u khc khng ngay c khi ngoi lc f(t) bng khng.

Khi h thng b suy gim nh (D2


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S tay C in t

HNH 24.3 p ng khng cng bc ca h khi lng - l xo

Cu trc ca p ng cng bc

Khi trng thi ban u bng khng, trng thi ny s ch th hin thnh phn cng bc. Thnhphn cng bc ca trng thi s bao gm cc mode t nhin v mt s mode cng bc hoc cbit thm vo, m cc mod ny ph thuc vo bn cht ca u vo u( t) ca h thng. Ni chungcc mode cng bc ca u vo cng s xut hin trong trng thi. Tuy nhin, mt vi trnghp c bit xut hin khi mt s mod cng bc ca u(t) trng vi mt s mod t nhin ca hthng .

Tnh n nh ca h thng

n nh trong cc h thng tuyn tnh, bt bin cng c th c phn tch bng cch s dngma trn trng thi A.

Tt c cc bin h thng c th c biu din nh cc hm tuyn tnh ca trng thi v uvo h thng. Khi u vo h thng u(t) l mt vect ca cc hm thi gian b gii hn, th tnhkhng gii hn ca cc bin h thng ph thuc vo trng thi b gii hn.

V vy ta c kt qu sau:

nh l 24.1.Xt h thng vi miu t trng thi (24.42) v (24.43) trong B, CvD c ccphn t b gii hn. Khi trng thi ca h (v v vy u ra ca h) b gii hn vi tt c ccu vo b gii hn nu v ch nu cc tr ring ca A c cc phn thc m.

minh ha nh l ny, chng ta xt li h trong v d 24.2. Ma trn A (trong m hnh c tuyn tnh ho) c cho bi:

2

1

0 0

0 0 1

0g

Q Q

R

LA

R Rmg

E K E

=

(24.56)

v tr ring ca n l nh thc ca det(I A) = 0 trong :

2 2

1 1

det( )Q Q

R Rmg Rmg I A

L K E K E

= + + (24.57)

C th thy rng tp cc tr ring ca ma trn bao gm mt phn l thc v ln hn khng.iu ny cho thy h thng khng n nh. Kt qu ny cng ph hp vi suy lun vt l. Thtvy, t nht v mt l thuyt, chng ta c th nh v c qu cu trong qu trnh cn bng (iu

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ny c miu t bi x2Qtrong (24.37)). Tuy nhin, y l im cn bng khng n nh, v ngaykhi ta tc ng nh vo qu cu, n tng tc c v pha mt t hoc v pha nam chm.

Tc p ng v cng hng

Ngay c nu h thng n nh vn cn mt s vn lin quan n cc thuc tnh c bn khc.

u tin, trong h thng n nh, phn thc ca tr ring xc nh tc m ti mode linquan hi t v khng. Cc mod thp nht, cc mode tri xc nh tc m ti u ra h thng

t c gi tr trng thi tnh ca n, c ngha l xc nh tc p ng ca h thng. V d, nucc tr ring tri ca h thng l 12 0 , 0j = > , th cc mode t nhin c t hp s to ra

mt sng sin suy gim theo hm m 0( ) sin( )tt Ae t = + . Ta thy rng tn hiu ny gim

nhanh hn khi ln hn.

Vn th hai, c bit quan trng cho cc cu trc linh hot, l s c mt ca cng hng,lin quan n cc tr ring phc. Trong cc h thng vt l, s tn ti ca cc tr ring phc c linh mt thit ti s hin din ca hai dng nng lng. Cng hng miu t dao ng (suy gimchm) gia hai dng nng lng ny. Trong cc mch in, cc dng nng lng l nng lngtnh in trong cc t in v nng lng in t trong cc cun cm. Trong cc h c, ta c ngnng ca cc khi lng chuyn ng v th nng trong cc l xo. Cc cu trc linh hot c th cnhiu mode cng hng. Mt trong nhng vn chnh vi cng hng xy ra khi u vo channg lng tn s gn vi tn s cng hng. V d, nu h thng c tr ring 12 = -0.05 j, c

ngha l tn s cng hng l 1 rad/s v ngoi ra, mt trong cc thnh phn u vo l mt sngsin vi tn s 0.9 rad/s, th u ra ca h thng c dao ng (cng bc) rt ln vi bin banu tng gn nh tuyn tnh v sau n nh mt gi tr khng i. Trong thc t, hin tngny c th ph hu h thng (trng hp cu Tacoma).

Bin i tng ng trng thi

Chng ta ni rng vic chn cc bin trng thi l khng duy nht. Ta c h thng vi uvo u(t), u ra y(t) v hai la chn khc nhau ca cc vect trng thi: ( ) nx t vi 4 ma trn

trng thi (A, B, C, D), v ( ) nx t vi (A,B,C,D) . Tn ti ma trn khng suy bin n nT saocho:

1

( ) ( ) ( ) ( )x t Tx t x t T x t

= = (24.58)t ta c cc tng ng sau:

1 1, ,A TAT B TB C CT = = = (24.59)

Cc la chn bin trng thi khc nhau c th hoc khng th p ng cc phng php phntch h thng khc nhau. i khi ch l mt vn n gin ton hc, nh ta thy phn 24.6.Trong cc trng hp khc, quyt nh c a ra bng cch xem xt kh nng o lng cc binh thng no . Tuy nhin, iu quan trng l cc c tnh h thng c bn khng thay i, khngph thuc vo m t trng thi c chn. Chng uc lin h vi thc t cc tr ring ca hthng l bt bin tng ng vi cc bin i tng ng, v:

1 1 1det( ) det( ) det( )det( )det( )I A TT TAT T I A T = = (24.60)

det( )I A= (24.61)

V vy, s n nh, bn cht ca p ng khng cng bc v tc p ng u bt bin vicc bin i tng ng.

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S tay C in t

V d 24.3

Xt mch in nh hnh 24.4.

Chn vect trng thi x(t) = [x1(t) x2(t)]T = [iL(t) vc(t)]T t u(t) = vf(t). S dng quy tc thnht ta c:

1 21

1 2

110

( )

( ) ( )11Ldx t

x t u t R RdtR C

C R R C

= + +

(24.62)

HNH 24.4 Mch in

Vect trng thi la chn l [ ] [ ]1 2 2x( ) ( ) ( ) ( ) ( )TTt x t x t i t i t = = . R rng ta c:

2

2

11( ) ( )

0 1T

Rx t t

R

=

1 44 2 4 43(24.63)

Khng gian trng thi v cc hm truyn t

Miu t khng gian trng thi ca cc h tuyn tnh bt bin l mt cch miu t khc tt hncch c a ra bi cc hm truyn t. Ni ng ra, miu t khng gian trng thi c phm vi

rng hn nh s thy trong phn ny.Cho h tuyn tnh bt bin vi u vo ( ) mu t v u ra ( ) py t , hm truyn

( ) p mH s , c nh ngha bi phng trnh sau:

( ) ( ) ( ),Y s H s U s= trong [ ]( )

( )( )

i

ijj

Y sH s

Y s= (24.64)

c ngha l thnh phn (i,j) trong ma trn H(s) l bin i Laplace ca p ng theo u ra ith khixung n v c t u vojth, vi iu kin u bng khng v vi cc u vo cn li bngkhng vi tt c t 0.

Mt khc, nu ta bin i Laplace (24.42) v (24.43) vi cc iu kin u bng khng, ta c:

1

( ) ( ) ( )X s sI A BU s

= (24.65)

1

( )

( ) ( ) ( ) ( ( ) ) ( )H s

Y s CX s DU s C sI A B D U s= + = +1 4 442 4 4 43 (24.66)

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n gin v c th i su vo phn tch hn, trong phn cn li ca mc ny ta tptrung vo lp h thng scalar, c ngha l h thng vi mt u vo v mt u ra (SISO). iuny c ngha l m = p = 1, B tr thnh vect ct, C l vect hng, v ( )D H= (trong cc hthng thc thng gi sao cho ( ) 0D H= = ). Vi h thng SISO, H(s) l thng ca cc athc theos, c ngha l:

( ) det( )

( )

det( )

CAdj sI A B D sI AH s

sI A

+ =

(24.67)

trong k hiu Adj(o) l ma trn lin hp ca (o).

Vn quan trng l cc im cc ca hm truyn t l cc tr ring ca ma trn A. Tuynhin, iu ny khng ng, ni chung, tp cc im cc ca hm truyn l ng nht vi tp cctr ring ca ma trn A. iu ny c th c hiu r qua v d sau.

V d 24.4

t:

[ ]2 1 1

, , 0 1 , 00 3 0.5

A B C D

= = = = (24.68)

Khi :

13 1 11

( ) ( ) [0 1]0 2 0.5( 2)( 3)

sH s C sI A B

ss s + = = ++ +

(24.69)

0.5( 2) 0.5

( 2)( 3) ( 3)

s

s s s

+= =

+ + +(24.70)

V vy, hm truyn t ch c mt im cc, mc d ma trn A c hai tr ring. Chng ta thyrng c mt im cc - khng b loi H(s). Hin tng ny c gn lin vi vn thuc tnhca h thng, l ch trng tm ca 24.6.

Xt h thng cm ng t trong v d 24.2. Nu ta nh ngha dng in i(t) l u ra ca hthng, ta c th ngay lp tc thy rng hm truyn t t u vo e(t) ti u ra ny ch c mt

im cc. iu ny tri vi thc t rng chiu ca trng thi bng ba. gii thch cho iu ny,trong m hnh vt l c n gin ha, dng in i(t) khng b tc ng bi v tr v tc ca qu cu kim loi (ch rng, ta b qua s thay i ca t cm do cc thay i v tr caqu cu).

Kt qu quan trng c c l hm truyn khng th cung cp cng mt lng thng tin nhm hnh khng gian trng thi cho cng mt h thng.

Vn th v l lm th no nhn c mt miu t khng gian trng thi t mt hmtruyn cho trc. Ngi c phi bit rng m hnh khng gian trng thi kt qu khng th hinc cc s trit tiu im cc - khng. V l do , miu t nhn c c bit nh mt sthc hin ti thiu.

C nhiu phng php i t hm truyn ti mt m hnh khng gian trng thi. Sau y,chng ta s trnh by mt trong nhng phng php .

Xt hm truyn t cho bi:

1 2

0 1 2 1 0

1

0 1 1 0

( ) .....( ) ( ) ( )

( ) .....

n n

n nT Tn n

n

B s b s b s b s bH t H H

A s s a s a s a

+ + + += + = +

+ + + +(24.71)

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S tay C in t

Trc tin chng ta nh li rng ( )TD H= . Do chng ta tp trung vo hm truyn( ) ( ) ( )T TH s H s H = l hm truyn thch hp.

Xt bin tip theo ( )v t l c bin i Laplace Vl(s) tho mn:

{ }1

0

( ) ( ), 1,2,......( )

sV t U s n

A s

= l

l l (24.72)

iu ny c ngha rng:

1( )

( ) , {2,....,n}dv t

v tdt= ll l (24.73)

11

( ) ( )n

Y s b V s=

= l ll

(24.74)

1

0

1

10 0 0

( ) ( )

( )( ) ( ) ( ) ( )

( ) ( ) ( )

n

n n

sV s V s

A s s sU s U s U s a U s

A s A s A s

=

= = + l

l

ll1 4 2 43 1 4 2 43

(24.75)

By gi chn cc bin trng thi:

( ) ( )x t v t =l l (24.76)

T cc phng trnh trn, ta c:

0 1 2 2 1

00 1 0 0 0

00 0 1 0 0

,

0

1n n

A B

a a a a a

= =

K

LM

M M M K M M

L

(24.77)

0 1 2 1, ( )

nb b b b T C D H

= = L (24.78)

V d 24.5

Hm truyn t ca mt h thng c cho bi:

2 3 2

4 10 4 10( )

( 2) ( 1) 3 4

s sH s

s s s s

= =

+ + (24.79)

Khi thc hin ti thiu cho h thng ny l:

0 1 0 0

0 0 1 , 0

4 0 3 1

A B

= =

(24.80)

[ ]10 4 0 , 0C D= = (24.81)

Kt qu quan trng c rt ra l hm truyn t ca h thng l bt bin vi cc bin itrng thi tng ng.

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24.4 Miu t khng gian trng thi cho h thng thi gian ri rc vd liu ly mu

Phn ny s gii thiu khi qut v miu t khng gian trng thi cho cc h thng thi gian rirc, ch yu da trn cc kt qu c trnh by cho trng hp thi gian lin tc.

Cc m hnh thi gian ri rc c th c t hai ngun khc nhau:

T h thng thi gian ri rc thun ty, thng l khng tuyn tnh, c cc bin cnh ngha ch thi im xc nh tk. Cc h thng nh vy c th thy trong cc hthng kinh t, l thuyt x l stochastic

T vic ri rc ha h thng thi gian lin tc. Trong trng hp ny chng ta ch quantm vi gi tr ca mt vi bin h thng ti nhng thi im xc nh. Cc m hnh nythng c s dng khi cc h thng s, nh cc vi iu khin, cc my tnh, PLCstng tc vi cc h thng thc vi min thi gian lin tc, nh cc cu trc c kh,van, bn cha, mch in tng t hoc c mt qu trnh x l cng nghip 1. Chngc gi l cc h thng d liu trch mu.

Trong c hai trng hp trn, phn tch ca chng ta s tp trung vo lp cc m hnh tuyntnh v bt bin.

Tuyn tnh ha cc h thng thi gian ri rcThi gian ri rc tng ng vi (24.3) v (24.4) c cho bi cc phng trnh phi tuyn

sau:

[ ]1 ( [t], [t])dx t F x u+ = (24.82)

[t]= ( [t], [t])dy G x u (24.83)

Tuyn tnh ha ca cc m hnh ca cc h thng thi gian ri rc s cho ra ng ging ving ca h thi gian lin tc. Xt im cn bng c cho bi {xQ, uQ, yQ}:

( , )Q d Q Qx F x u= (24.84)

( , )Q d Q Qy G x u= (24.85)Lu rng im cn bng c xc nh bi mt tp cc gi tr khng i ca trng thi v cc

gi tr khng i ca u vo tha mn (24.82) v (24.83). iu ny to ra mt u ra ca hthng l khng i. V vy, m hnh ri rc c th c tuyn tnh ha xung quanh im cn bngny. nh ngha:

[t] = x[t] - , [t] = [t] - , [t] = [t] -Q Q Qx x u u u y y y (24.86)

Chng ta c m hnh khng gian trng thi:

d d[t+1] =A [t]+ [t]x x B u (24.87)

d d[t] = [t]+ [t]y C x D u (24.88)

Trong :

1 Qua cc b bin i s - tng th v tng t - s (DAC v ADC tng ng)15


16/51

S tay C in t

, , ,Q Q Q Q

Q Q Q Q

d d d d d d d d

x x x x x x x x

u u u u u u u u

F F G GA B C D

x u x u= = = == = = =

= = = =

(24.89)

Cc h thng d liu ly mu

Nh chng ta ni, cc m hnh thi gian ri rc thng t c nh ly mu cc u vov cc u ra ca h thng thi gian lin tc. Khi thit b s c s dng hot ng da trnmin thi gian lin tc, cc tn hiu lnh ch cn c nh ngha nhng thi im xc nh chkhng phi ton b thi gian. Tuy nhin, c th hot ng trn h thng thi gian lin tc,chng ta cn mt tn hiu vi thi gian lin tc. iu ny thng c xy dng bng vic gimu bc khng, s to ra mt tn hiu bc thang. Ngoi ra, khi ta mun o theo kiu s mt bin hthng, chng ta o ti nhng thi im xc nh. C ngha l chng ta phi ly mu cc tn hiuu ra. Hnh (24.5) m t khi nim ny. Nu ta gi thit vic ly mu l tun hon, vi chu k ,th ta ch quan tm ti cc tn hiu ti k. Tip theo ta sb khi i s,bng cch dng u(k) =u[t] cho u vo, y(k) = y[t] cho u ra, v x(k) = x[t] cho trng thi h thng.

HNH 24.5 H thng d liu ly mu

Nu ta xem xt m hnh khng gian trng thi tuyn tnh bt bin, lin tc, c nh ngha bi(24.42) v (24.43), vi trng thi ban u x(k0) = x0, chng ta c th s dng phng trnh(24.44) tnh ton gi tr tip theo ca trng thi:

0

0 0 0

0

( ) ( )

0 0( ) ( ) ( )

k

A k k A k

k

x k e x k e Bu d +

+ +

+ = + (24.90)

Hn na, nu gi mu bc khng c s dng, c ngha l u() = u(k0) cho k0 < k0 +, chng ta s nhn c:

0 0 0

0

( ) ( ) ( )A Ax k e x k e d B u k

+ = + (24.91)

V, nu ta bit trng thi v u vo ti thi gian k0, th u ra c xc nh bi phngtrnh (24.43):

0 0 0( ) ( ) ( )y k Cx k D u k = + (24.92)

By gi chng ta c th kt lun rng mt m hnh thi gian lin tc vi cc ma trn khnggian trng thi {A, B, C, D}, v chng ta ly mu cc u vo, cc u ra, mi giy th h thngd liu ly mu tng ng s c miu t bi m hnh khng gian trng thi ri rc:

( ) ( ) ( )d dx k A x k B u k + = + (24.93)

( ) ( ) ( )d dy k C x k D u k = + (24.94)

Trong :

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0

, , ,A Ad d d d A e B e d B C C D C

= = = = (24.95)

C nhiu phng php khc nhau nhn c Ad c nh ngha bi (24.95), nhng cchn gin tnh ma trn ny l dng bin i Laplace. iu ny s dn ra:

1 1{( ) }Ad tA e sI A

== = L (24.96)

V d 24.6

Xt h thng c trong v d 24.1 c miu t bi m hnh khng gian trng thi:

1 1

22

0 1 0( ) ( )

( )1( )

( )

x t x t f tK M

x tx t M D M

= +

g

g(24.97)

Trong f(t)l ngoi lc, ta c th chn v tr khi lng, x1(t), hoc tc vt, x2(t), nh lcc u ra h thng.

Vi mc ch minh ha s, chng ta t M= 1kg,D = 1.2N s/m, vK= 0.32 N/m.

Ma trn Ad nhn c t (24.96), bng cch p dng bin i ngc Laplace:

1 0.4 0.8 0.4 0.8

1

0.4 0.8 0.4 0.8

1 2 2.5( )

0.32 1.2 0.8( ) 2d t

s e e e eA

s e e e e

=

= = + + L (24.98)

Ma trn B c c t (24.95):

0.4 0.8 0.4 0.8

0.4 0.8 0.4 0.8

0

02 2.5( )

10.8( ) 2d

e e e eB d

e e e e

= +

(24.99)

0.4 0.8

0.4 0.8

6.25 3.125 3.125

2.5( )d

e eB

e e

+ + =

Ch rng, c Ad v Bd u l hm ca . V vy, chu k ly mu c vai tr quan trng trongng x ng ca h thng trch mu nh chng ta s thy trong cc mc con sau:.

Cc m hnh khng gian trng thi tuyn tnh

Chng ta s phn tch m hnh khng gian trng thi bt bin tuyn tnh:

[ 1] [ ] [ ]d dx t A x t B u t + = + (24.100)

[ ] [ ] [ ]d dy t C x t D u t = + (24.101)

y c th l m hnh thi gian ri rc c tuyn tnh ho ging nh (24.87) v (24.88), hocmt h thng d liu ly mu ging nh (24.93) v (24.94) trong c b khi i s thigian.

Li gii ca phng trnh (24.100) v (24.101), x[t0] = x0, c cho bi:0

0 0

( ) 1( ) ( ) 1

0 0 0

0

[ ] [ ]t t

t t t t i

d d d

i

x t A x A B u i t t t

=

= + + (24.102)

Trong 0( )t tdA l ma trn chuyn tip.

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S tay C in t

Bn c c th d dng kim tra li (24.102) tha mn (24.100). Vi kt qu trn, li gii cho(24.101) c cho bi:

0

0 0

( ) 1( ) ( ) 1

0 0

0

[ ] ( [ ]) [ ]t t

t t t t i

d d d d d d

i

y t C A x C A B u i t D u t

=

= + + + (24.103)

H thng ng lc

Trng thi ca h thng gm hai thnh phn: thnh phn khng cng bc, xu[t], v thnh phncng bc xf[t], trong :

0( )

0[ ] t tu dx t A x

= (24.104)

0

0

( ) 1( ) 1

0

0

[ ] [ ]t t

t t i

f d d x t A B u i t

=

= + (24.105)

hiu r hn m hnh khng gian trng thi v li gii ca n, xt trng hp khi t0 = 0 vu[t] = 0, 0t , c ngha l trng thi ch c thnh phn khng cng bc, khi :

0[ ] tdx t A x= (24.106)

Hn na, gi thuyt rng dAn n v n gin hn, n cha n tr ring khc nhau l , vi

n vect tr ring c lp tuyn tnh vl . Khi lun tn ti n hng s l tha mn:

0

1

,n

x v =

= l l ll

(24.107)

Mt kt qu ni ting t ton i s cho chng ta thy rng cc tr ring ca kdA lkl , vi

k tng ng vi cc vc t ring vl .

{01 1

[ ]t

n nt t t

d d d

v

x t A x A v A V

= =

= = = l l

l l l ll l

(24.108)

1[ ]

nt

x t v == l l ll (24.109)Cng thc ny cho thy thnh phn khng cng bc ca trng thi l t hp tuyn tnh ca

cc mode t nhin, t{ }l , v mi phn gn vi mt tr ring ca Ad, m n cng c bit nh lcc tn s c bn ca m hnh. V vy, chng ta li bit rng ma trn A d xc nh:


S n nh ca h thng

Tc p ng


Khi khng c u vo, trng thi nh l t hp ca cc mod t nhin thuc mt lp hm c nh ngha: nng lng ca cc tr ring ca mode, gm c thc hoc phc. Cc mode lcc hm ri rc lin quan n cc hng s, cc hm m thc, sng sin c bn, cc sng sin biu ch theo hm m, v mt s hm c bit khc sinh ra t cc tr ring lp li.

18


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minh ha cc vn trn v lm sng t ngha vt l ca chng, xt h thng c lymu trong v d 24.6. Nu = 1, ma trn khng gian trng thi l:

0.8913 0.5525 0.3397,

0.1786 0.2283 0.5525d dA B

= =

(24.110)

V vy, cc tr ring ca h thng l cc nghim ca phng trnh

0.8913 0.5525det( ) det0.1786 0.2283d

I A

= (24.111)

( 0.6703)( 0.4493) 0 = = (24.112)

Tc l 1 = 0.6703, 2 = 0.4493 v p ng khng cng bc l:

1 2[ ] (0.6702) (0.4493)t tux t C C = + (24.113)

Trong C1 v C2 ch ph thuc vo cc iu kin u. Chng ta c th thy rng, khi t tin tiv cng, xu[t] suy gim v khng, do | 1,2 | < 1. Ngoi ra, cc tr ring l cc s thc dng, nnkhng c dao ng cc mode c bn. iu ny ph hp vi vic chn tham s trong v d 1.6lm cho h khi lng - l xo trnh c rung qu mc.

Cu trc ca p ng cng bc

Xt phng trnh (24.102). Khi trng thi u l khng, trng thi s ch th hin thnh phncng bc. Tuy nhin, thnh phn cng bc ny s vn bao gm cc mode t nhin cng vi mts mod cng bc hoc ring bit ph thuc vo bn cht ca u vo u[t] ca h thng. Nichung, cc mod cng bc ca u vo cng s xut hin trong trng thi. Tuy nhin cc trnghp c bit s xut hin khi mt mod cng bc ca u[t] trng vi mod h thng c bn.

Tnh n nh h thng

n nh trong h thng tuyn tnh bt bin cng c th c phn tch bng cch s dng matrn trng thi Ad. Nh chng ta ni, tt c cc bin h thng u c th c biu din nh cchm tuyn tnh ca trng thi v u vo h thng. Khi u vo h thng u[t] l mt vect ca hmthi gian b gii hn, th gii hn bin ca cc bin h thng ph thuc vo trng thi b gii hn,chng ta c kt qu sau:

nh l 24.2 Xt h thng vi miu t trng thi (24.100) v (24.101) trong Bd, Cd v Ddc cc phn t b gii hn. Khi trng thi h thng c gii hn theo tt c cc u vo giihn nu v ch nu cc tr ring ca Ad nm trong ng trn n v, v d | l | < 1, l .

Tc p ng v cc cng hng

Ta nh li rng cc mode c bn ca h thi gian ri rc l nng lng ca cc tr ring l . Vcc tr ring c th lun c miu t nh cc i lng phc, nn chng ta c th vit ccmode c bn nh sau:

( ) ( ) ,

t j j t t

e e

= = = l ll

l l l l l (24.114)V vy chng ta c

0 | |< < l xc nh tc , ti mode suy gim ti khng vi h thng n nh ( |l| < 1), hoc ln ti v cng vi cc h thng khng n nh ( |l | > 1).

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S tay C in t

< l xc nh tn s ca mode t nhin, c o theo radians

Mc d cc mode t nhin ca cc h thng n nh suy gim ti khng, nhng cc tnh cht tnhin ca chng xc nh p ng qu ca h thng

minh ha nhng vn ny, p ng bc vi iu kin u bng khng, thng c sdng

V d 24.7

Xt h thng thi gian ri rc mt u vo-mt u ra (SISO)

[ 1] [ ] [ ]x t x t u t + = +l (24.115)

[ ] (1 ) [ ]y t x t = l (24.116)

nhn c p ng bc ta s dng phng trnh (24.103), trong x 0 = 0, u(t) = 1, 0

11

0

[ ]t

t i

d d d

i

y t C A B

=

= (24.117)

11 1

0

1(1 ) (1 )

1

ttt i t

ti

=

= = ll l l l

l

(24.118)

1 t= l (24.119)

HNH 24.6 p ng bc ca h thng vi cc tr ring khc nhau

Tn hiu u ra, y[t] = yh[t] + yp[t], c biu din trn hnh (24.6), vi cc gi tr khc nhauca tr ring l .

Qu nhn c bi yh[t] = -tl v p ng trng thi n nh bi yp[t] = 1.

Chng ta thy trong phng trnh (24.114), tr ring h thng khng nhng xc nh s ttdn ca p ng qu m cn xc nh tn s dao ng (khi tr ring c thnh phn o khckhng). Vn kh nng khi cc mode cng hng tn ti cng l vn chng ta thy trong nidung ca h thng thi gian lin tc, c ngha l, u vo h thng cha mt sng sin hoc mtloi tn hiu khc, vi nng lng mt tn s gn vi mt trong cc tn s t nhin ca h thng.u ra h thng vn cn b gii hn, mc d n tng ti bin khng mong mun.

V d 24.8Xt h thng thi gian ri rc c miu t bi m hnh khng gian trng thi:

1.2796 0.81873 1[ 1] [ ] [ ]

1 0 0x t x t u t

+ = +

(24.120)

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[ ] [0 0.5391] [ ] y t x t = (24.121)

Cc tr ring ca h thng nhn c t Ad:

/ 4

1,20.6398 0.6398 0.9048 ( )jj e = = (24.122)

V cc mode t nhin lin quan, c trong p ng qu l:

41,2 0.9048 0.9048 cos sin4 4

j tt t t

e t j t

= = (24.123)

Cc mode t nhin u suy gim nh, bi v | 1,2 | gn vi 1, v chng s a ra mt dao ngvi tn s /4.

Trong cc th hnh 24.7, ta thy r rng u ra cng hng mnh. th trn tng ng

vi u vo [t]=sin( )4

u t

, c ngha l, tn s u vo trng vi tn s ca cc mod t nhin.

Trong th di, u vo l mt sng vung ca tn s /12. Trong trng hp ny, hi bc baca u vo c tn s bng vi tn s ca mode t nhin.

nh hng ca cc chu k ly mu khc nhau.

Phng trnh (24.95) cho thy rng A d v Bdph thuc vo vic chn chu k ly mu . Vicla chn ny cng quyt nh v tr ca cc tr ring trong h thng. Nu ta xem phng trnh(24.96), gi thit rng A c ng cho ha, ta c:

1 1{ . .... } { ,..., }n ndiagdA e diag e e = = (24.124)

HNH 24.7 nh hng cacng hng ti u ra h thng

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S tay C in t

HNH 24.8 nh hng ca vic ly mu ti cc mode t nhin

Trong 1 n{ ,..., } u l cc tr ring ca cc h thng thi gian lin tc c bn. V vynhng tr ring c gn vi tr ring ca h thng d liu ly mu bng cng thc:

n e = ll (24.125)

Trong hnh 24.8 ta thy p ng ca h thng c ly mu trong v d 24.6, chn 1x [t] l u

ra ca h thng, khi iu kin u l T0x [1 0]= , vi cc gi tr khc nhau ca . Quan st thyrng trc honh tng ng vi t, do cc thi im tc thi thc l t.

HNH 24.9 H thng nhit vi tr thi gian

Mt vn c bn lin quan n vic ly mu tn hiu trong min thi gian lin tc l chu kly mu c chn nh gi c tnh t nhin thit yu ca tn hiu c ly mu. minh

ha mt la chn ti , gi thit rng tn hiu 0( ) sin( )f t A t = c ly mu mi giy, vi =0

2 / , l l . Khi tn hiu thi gian ri rc kt qu l ( ) 0,f t t = .

H thng d liu ly mu v cc tr thi gian

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Nh chng ta cp 24.3, khng th s dng m hnh khng gian trng thi trong minthi gian lin tc miu t h thng vi cc tr thi gian, v chng l cc h thng c th nguynv tn. Cng c th ni rng ta c th thay i vn ny bng cch s dng cc tn hiu ly mu.iu ny c thc hin nh trong v d sau.

V d 24.9

Xt h thng nhit trong hnh 24.9

Nhit c o, y(t), ca dng chy ph thuc vo nng lng to ra bi ngun nhit. Ngunny c iu khin bi mt tn hiu iu khin u(t). Cc thay i ca u(t) s lm thay i nhit y(t) vi mt tr thi gian ng k. V vy h thng c tuyn tnh ha c th c biu din bihm truyn t:

( )( )

( )

sY s e K H s

U s s

= =+

(24.126)

Trong U(s) v Y(s) l bin i Laplace ca u(t) v y(t)

Tip theo, chng ta gi thit rng u vo v u ra c ly mu mi [s]. Thi gian tr theo giy, l hm ca vn tc dng chy v n gin ha ta gi thit rng l bi s ca khongthi gian ly mu , c ngha l = m, m + . Cc tr ny c th hin h s mz trong mus ca bin i Z ca hm truyn t. Ni cch khc, tr lm xut hin mt tp m im cc ti gcta .Hn na, tr ring ca h thi gian lin tc ti s = - tr thnh tr ring ca h n thi gianri rc ti z e = (xem phng trnh (24.125)). Kt qu cho hm truyn t c dng:

( ) 1( )

( ) ( )mY z K e

H zU z z z e

= =

(24.127)

V hm truyn t ny c th biu din nh m hnh khng gian trng thi ri rc

1 2[ 1] [ ]x t x t + = (24.128)

2 3[ 1] [ ]x t x t + = (24.129)

...

1[ 1] [ ]m mx t x t ++ = (24.130)

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

Kx t e x t e u t

+ ++ = + (24.131)

1[ ] [ ]y t x t = (24.132)

V vy chng ta c th xem cc bin trng thi 1 1[t],...,x [t]mx + nh nhit ti cc im cchu nhau, gia ngun nhit v cm bin nhit.

Khi thi gian tr khng l bi s ca chu k ly mu , th mt im cc b sung hmtruyn ban u v mt im khng na xut hin hm truyn t ri rc.

Bin i tng ng trng thi tng bin i trng thi thng qua mt php bin i tng ng cng c p dng vi

cc h ri rc. Cc thuc tnh ca h thng vn khng thay i.

Khng gian trng thi v cc hm truyn t

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S tay C in t

Vi cc h thng ri rc, mi quan h gia cc m hnh khng gian trng thi v m hnh hmtruyn t v c bn l ging nhau nh trong trng hp thi gian lin tc (xem mc "Cc hmkhng gian trng thi v hm truyn"). Nh chng ta ni, miu t khng gian trng thi ca hthng tuyn tnh bt bin l mt cch m t khc tt hn cch c a ra bi cc hm truyn t,mc d trong mt s trng hp n cung cp nhiu thng tin hn v h thng.

Cho h thng bt bin ri rc tuyn tnhvi u vo m[t]u v u ra py[t] , hm truyn

t, H[z] p m , c nh ngha bi cng thc:

,

[ ][ ] [ ] [ ], [ [ ]]

[ ]i

i j

j

Y zY z H z U z H z

U z= = (24.133)

V d, thnh phn (i,j) trong ma trn H(z) l bin i Z ca p ng u ra th ikhi mt hmdelta Kronecker n v c a vo u vo th j, vi iu kin u bng khng v cc u vocn li bng khng vi 0t .

Mt khc, nu ta dng bin i Z vi m hnh khng gian trng thi ri rc (24.100) v(24.101) vi iu kin ban u bng khng, chng ta c:

1[ ] ( ) [ ]d dX z zI A B U z= (24.134)

[ ] [ ] [ ]d dY z C X z D U z= + (24.135)

dn ti:

1( ) [ ]d d d d C zI A B D H z + = (24.136)

Trong phn tch sau y, chng ta s tp trung vo lp cc h scalar, c ngha l1, , Td dm p B C = = l cc vc t ct v [ ]dD H= . Khi chng ta c th thyH[z] l thng

ca cc a thc theoz, c ngha l:

( ) det( )[ ]

det( )d d d d d

d

C Adj zI A B D zI AH z

zI A

+ =

(24.137)

Trong Adj(o) l ma trn lin hp ca (o)

Tng t trng hp thi gian lin tc, ta li c cc im cc ca hm truyn t l cc trring ca Ad. Tuy nhin, ni chung l khng ng tp cc im cc ca hm truyn l ngnht vi tp cc tr ring ca ma trn. N l quan trng thc hin cc m hnh hm truyn c thgiu i cc s trit tiu gia cc im cc v im khng vi cc h qu c m t trong ccmc con "Tnh iu khin c, t c v n nh" v "Tnh quan st, khi phc c v phthin c".

Kt qu quan trng cho h thng ri rc cng ging nh h thng lin tc l: hm truyn tc th khng cung cp cng mt lng thng tin nh m hnh khng gian trng thi chocng mt h thng.

Mt cch c c m hnh khng gian trng thi l s dng phng php c cp trongphn Khng gian trng thi v cc hm truyn t, bng cch s dng bin i Z thay cho bini Laplace v s dng:

[ ] { [ ]} [ ] { [ 1]}F z Z f t zF z Z f t = = + (24.138)

V d 24.10

Cho hm truyn t ca h thng:

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2

2

2 1 1.8 0.04[ ] 2

( 0.8)( 0.6) 1.4 0.48

z z zH z

z z z z

+ += = +

+(24.139)

Khi thc hin ti thiu cho h thng ny l:

0 1 0,

048 1.4 1d dA B

= =

(24.140)

[0.04 1.8], 2d dC D= = (24.141)

Trong cc m hnh ri rc, cng xy ra hm truyn t ca h thng l bt bin vi bin itng ng trng thi

24.5 Cc m hnh khng gian trng thi cho cc h thng kt ni

xy dng cc m hnh khng gian trng thi cho cc h thng phc tp, i khi c th biudin chng nh l s kt ni gia cc h thng n gin. S kt ni ny thng l s kt hp caba cu trc kt ni c bn: ni tip, song song, v phn hi. Trong ba trng hp c bn , mcch ca chng ta l lm sao nhn c m hnh khng gian trng thi cho h thng hn hp.

Trong phn tch di y, chng ta s s dng hai h thng, c cho bi:

H thng 1: 1 1 1 1 1( ) ( ) ( )dx t A x t B u t

dt= + (24.142)

1 1 1 1 1( ) ( ) ( )y t C x t D u t = + (24.143)

H thng 2: 2 2 2 2 2( )

( ) ( )dx t

A x t B u t dt

= + (24.144)

2 2 2 2 2( ) ( ) ( )y t C x t D u t = + (24.145)

Kt ni ni tip

H thng kt ni nh biu din trong hnh 24.10 c bit nh l mt s kt ni ni tip.

xy dng c m hnh khng gian trng thi mong mun, trc tin chng ta quan st thy y 2(t)= u1(t). Cng nh vy, u vo hn hp ca h thng l u( t) = u2(t), v u ra hn hp ca hthng l y(t) = y1(t). Nh vy ta c:

1 1 1 21 1 2

2 2 22

( ) ( )( )

0 ( )( )

x t x t B D A B C u t

A x t Bx t

= +

g

g(24.146)

1

1 1 2 1 2

2

( )( ) [ ] [ ] ( )

( )

x ty t C D C D D u t

x t

= +

(24.147)

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S tay C in t

HNH 24.10 Kt ni ni tip

Kt ni song song

Kt ni h thng c ch ra trn hnh 24.11 c bit nh l kt ni song song. nhn cm hnh khng gian trng thi mong mun, ta thy rng u vo l u( t) = u1(t) = u2(t) v u ra choton b h thng l y(t) = y1(t) + y2(0). Ta c:

1 1 11

2 2 22

( ) ( )0 ( )0 ( )

( )

x t x t BA u tA x t B

x t

= +

g

g(24.148)

1

1 2 1 2

2

( )( ) [ ] [ ] ( )

( )

x ty t C C D D u t

x t

= + +

(24.149)

HNH 24.11 Kt ni song song

Kt ni phn hi

Kt ni h thng c biu din trn hnh 24.12 l kt ni phn hi (vi phn hi m), tngng vi cu trc c bn ca mt vng iu khin, trong S1 l i tng iu khin v S2 l biu khin. xy dng m hnh khng gian trng thi hn hp chng ta thy rng u vo ca

ton h thng tha mn phng trnh u(t) = u(t) + y1(t), v u ra ca ton h thng l y( t) = y1(t).Hn na, chng ta gi thit rng S1 (i tng iu khin) l hon ton chnh xc, c ngha l, D 1= 0. T ta c:

HNH 24.12 Kt ni phn hi

1 1 1 21 1 2 1 1 2

2 1 2 2 22

( ) ( )( )

( )( )

x t x t B DA B D C B C u t

B C A x t Bx t

= +

g

g(24.150)

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1

1

2

( )( ) [ 0]

( )

x ty t C

x t

=

(24.151)

p dng cng cc kt qu, vi nhng sa i thch hp, ti cc h thi gian ri rc kt ni. Ccchi tit hn c th tm thy ch khc nh trong [15].

24.6 Cc thuc tnh ca h thng

Tnh iu khin c, t c v n nh

Mt vn rt quan trng m chng ta phi quan tm lin quan n h thng iu khin sdng cc m hnh khng gian trng thi l chng ta c th li h thng thng bi u vo iukhin ti mt v tr nht nh trong khng gian trng thi c hay khng. Ta phi nh rng cctrng thi ca h thng thng l cc bin ni ti nh nhit , p sut, mc ca thng cha ikhi, nhng bin l cc bin ti hn m chng ta mun gi gia cc gi tr xc nh.

Tnh iu khin c

Tnh iu khin c lin quan ti trng thi u x0 c th c li ti im gc trong thigian hu hn bng cch s dng u vo u(t) c hay khng.

V d 24.11

Nu ta xt m hnh c xc nh trong (24.152), ta ch rng u vo u(t) khng c tcng ti trng thi x2(t).

1 1

22

( ) ( )0 1 1( )

0 0 ( ) 0( )

x t x t u t

x tx t

= +

g

g(24.152)

Cho trng thi ban u [x1(0),x2(0)]T, u vo u(t) c th c chn lix1(t) v khng, trongkhix2(t) khng thay i.

Chng ta c nh ngha chnh thc sau:

nh ngha 24.1 Trng thi x0 c gi l c th iu khin c nu tn ti mt khong huhn [0, T] v mt u vo {u(t), t [0, T]} sao cho x(T) = 0. Nu tt c cc trng thi u c thiu khin c, th h thng c gi l c th iu khin c hon ton.

Tnh t c

Mt khi nim lin quan l tnh t c, i khi c s dng trong cc h thng ri rc.N c nh ngha mt cch chnh thc nh sau:

nh ngha 24.2 Mt trng thi 0x c gi l c th t c, t im gc, nu cho x(0) = 0,

tn ti mt khong thi gian hu hn [0, T] v mt u vo {u(t), t [0, T]} sao cho ( )x T x= . Nutt c cc trng thi u c th t c th h thng c gi l c th t c hon ton.

Vi cc h tuyn tnh bt bin lin tc, khng c s khc bit gia kh nng iu khin chon ton v kh nng t c. Tuy nhin, v d sau y minh ha rng c mt cht khc nhau

trong cc trng hp thi gian ri rc. Xt h thng v u ra:

0.5 1 0.5 1[ 1] [ ] [ ] [0]

0.25 0.5 0.25 0.5

d

t

A

x t x t x t x

+ = = 1 4 42 4 43(24.153)

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S tay C in t

Chng ta c th thy rng h thng ny hon ton iu khin c khi x[ t] = 0, 2t v2[0]x . iu ny ch ra rng tt c cc trng thi ban u u iu khin c. Tuy nhin,

khng c trng thi khc khng no l c th t c.

Thng thng, trong cc h tuyn tnh bt bin, tnh iu khin c v tnh t c c thc s dng thay th cho nhau.

Kim tra tnh iu khin c

By gi chng ta s trnh by mt cch phn loi xc nh tnh iu khin c hon tonca mt h thng.

nh l 24.3 Xt m hnh khng gian trng thi tuyn tnh bt bin trong n nA :

( ) ( ) ( )x t Ax t Bu t = +& (24.154)

( ) ( ) ( )y t Cx t Du t = + (24.155)

Tp hp tt c cc trng thi iu khin c l khng gian hng ca ma trn iu khin c[A,B]c trong :

2 1[ , ] [ ]nc A B B AB A B A B

= L (24.156)

M hnh c th iu khin c hon ton nu v ch nu [A,B]c c hng dng y

V d 24.12

Xt m hnh khng gian trng thi a ra trong (24.152), vi cc ma trn khng gian trngthi:

0 1 1,

0 0 0A B

= =

(24.157)

Ma trn c th iu khin c cho h thng ny c cho bi:

1 0[ , ] [ ]

0 0c A B B AB

= =

(24.158)

R rng, hng ca [A,B]c = 1, do h thng l khng th iu khin c hon ton.

Kt qu trn y p dng cho m hnh lin tc, v n cng nh vy vi tnh t c ca ccm hnh ri rc.

Chng ta cng c th thy rng tnh iu khin c ca h thng khng ph thuc vo vicchn cc bin trng thi. thy iu ny, xt mt php bin i tng ng c nh nghatrong mc nh "Bin i tng ng trng thi". Khi , thy rng 1A

iiT A T= , chng ta c:

1[ , ] [ , ]c cA B T A B = (24.159)

iu ch ra rng [A, B]c v [A,B]c c cng hng.

Bn c c th kim tra thy m hnh khng gian trng thi c s dng miu t cc tnhiu trong phn Tn hiu v m t khng gian trng thi l khng th iu khin c. Thc vy,lun ng rng bt k m hnh khng gian trng thi no c B = 0 u khng th iu khinchon ton.

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S thiu tnh iu khin c i khi l mt c tnh ca cu trc. Tuy nhin, trong mt strng hp, n ph thuc vo gi tr ca cc thng s nht nh. Ta s minh ha iu ny trong vd sau.

V d 24.13

Xt mch in t trong hnh 24.13

HNH 24.13 Mch in t

Trc tin ta xy dng m hnh khng gian trng thi cho mch. Chn 1 1( ) ( )Rx t i t = v2 3( ) ( )Cx t v t = nh l cc bin trng thi. Bng cch dng nguyn tc u tin trn na tri ca

mch, ta c:

3 1 1 2 1 2 1

1 2

( )( ), , ,iC i R R C R R

v v vdi C v v i i i i i

dt R R+ +

+

= = = = (24.160)

Suy ra:

1 1 21

1 1 2 1 1 2

( ) ( ) 1( ) ( )

+= +R

R i

di t R Ri t v t

dt C R R C R R(24.161)

1 1( ) ( ) ( )R iv t R i t v t + = + (24.162)

Tng t, t na phi ca mch ta c:

3

3

3 3 3 3

( ) 1 1( ) ( )C C

dv tv t v t

dt R C R C = + (24.163)

0 3( ) ( )Cv t v t = (24.164)

B khuych i thut ton (l tng) m bo rng ( ) ( )v t v t + = , v vy chng ta c th t hpcc m hnh khng gian trng thi cho trong (24.161) (24.164) nhn c:

1 21

1 1 2 1 1 1 2

3 1 3

3 33 3 3 3

( ) 1( ) 0( )

( )( ) ( ) 11

R

R

iC C

R Rdi t

C R R i t C R Rdt

v tdv t R v t

R C R C R C dt

+ = +

(24.165)

1

0

3

( )( ) [0 1]

( )R

C

i tv t

v t

=

(24.166)

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S tay C in t

V vy ta c ma trn iu khin c:

1 2

2

1 2 1 1 2 1

2 1 3 3

2

3 3 3 3 2 1

( )1

( )[ , ] [ ]

( )1

( )

c

R R

R R C R R C A B B AB

R C R C

R C R C R C

+ = = +

+

(24.167)

v:

21 1 3 32

1 2 3 1 2

det( [ , ]) ( ) (24.168)( )c

RA B R C R C

R R R C C = + (24.168)

y chng ta c th thy rng h thng hon ton iu khin c nu v ch nu

1 1 3 3R C R C .

iu ny c mt gii thch rt quan trng nu ta phn tch trn phng din hm truyn t.Vic p dng bin i Laplace cho cc phng trnh (24.161) (24.164), hm truyn t t ( )iv t

ti 0 ( )v t (cn nh l ( ) ( )v t v t + = ) c cho bi:

0 0 3 3 1 1

1 2

3 3 1 2 1

1 1

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

+

+= = +

+ +i i

sv s v s v s R C R C

R Rv s v s v s s sR C R R C

(24.169)

y, ta c th thy rng s mt tnh iu khin c hon ton, khi 1 1 3 3R C R C , c c t(24.168) ngha l c mt s trit tiu im khng - cc trong hm truyn t, c ngha l, imkhng t na tri ca mch trong hnh (24.13) b trit tiu bi im cc t phn kia ca mch. Vn ny s c phn tch c th trong mc "Khai trin chnh tc".

Gramian ca tnh iu khin c

Kim tra tnh iu khin c cho ta cu tr li c hay khngv tnh iu khin c ca m

hnh h thng. Tuy nhin, kt lun mt h thng iu khin c hon ton khng cho bit v mc ca tnh iu khin c. Vi h thng n nh, chng ta c th xc nh mc iu khintrng thi h thng qua nng lng ca tn hiu u vo u( t) gy ra t t= tin ti trngthi 0(0)x x= ti 0t= :

0 02

( ) ( ) ( ) ( )TJ u u t dt u t u t dt

= = (24.170)

C th thy nng lng iu khin ti thiu l:

1

0 0( ) ToptJ u x P x

= (24.171)

Trong :

0

T At T A t P e BB e dt

= g

(24.172)

Ma trn P c gi l gramian ca tnh iu khin c, v n o lng kh nng iukhin c ca vect trng thi x(0). Nu ma trn ny nh, ngha l ta cn nhiu nng lng u

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vo iu khin u(t) li vect trng thi n x0. Thc vy, ta c th c lng c tc ng cnthit cho mi mt trong cc bin trng thi, bng cch to ra, v d 0x [0, , 0,1,0, ,0]

T= K K .

Cn nhn mnh rng s tn ti ca tch phn c nh ngha (24.172) ch c m bo nutr ring ca A c phn thc m, tc l h thng phi c n nh.

Cng nh vy, gramian P ca tnh iu khin c c nh ngha (24.172) tha mnphng trnh Lyapunov:

0T T AP PA BB+ + = (24.173)Vi cc h thng ri rc chng ta c cc phng trnh sau cho gramian ca tnh iu khin

c:

0

( )k k T k d d d d d k

P A B B A

=

= (24.174)

n tha mn:

0T Td d d d d d A B A P B B + = (24.175)

Tng c nh ngha trong (24.174) b chn nu v ch nu h thng ri rc l n nh, v d,tr ring ca n nm bn trong ng n v.

V d 24.14

Ta c th phn tch m hnh ca v d 24.13, mch in t c miu t bi m hnh khnggian trng thi (24.165) v (2.166). Nu chng ta mun nh gi thng tin c c t gramian catnh iu khin c, c nh ngha trong (24.172), khi m hnh gn nh mt kh nng iukhin c hon ton, th chng ta c th chn cc gi tr ph hp cho cc thng s m bo

1 1 3 3R C R C .

Nu ta chn:

3 3

1 2 3 110 0.9 10R R R C F = = = = (24.176)

M hnh s c miu t bi:

1 1

33

3

20 0.01( )( ) 0( )9 9

( )( ) 10 1 1

R R

i

CC

i ti tv t

v tv t

= +

g

g(24.177)

1

3

0

( )( ) [0 1]

( )

R

C

i tv t

v t

=

(24.178)

Nu ta xt ln tng i ca cc phn t ca B, ta c th c quyn ni rng tc ng cau vo ( )u t ln trng thi 1( )Ri t s yu hn nhiu so vi tc ng ln trng thi 3 ( )Cv t . kimtra iu ny, ta c th tnh gramian ca tnh iu khin c c nh ngha (24.172), bngcch gii:

0

T T

AP PA BB= + + (24.179)3

11 12 11 12

3 21 22 21 22

20 20 0.010 10 0.01

0 19 9 99

10 1 0 1 1

p p p p

p p p p

= + +

(24.180)

Ta c:31


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S tay C in t

6

14736624.0 1224.90.28 10 0.000258620

,1224.9 1.30.000258620 0.99999948

P

= = (24.181)

V vy ta c th nhn c nng lng iu khin ti thiu li trng thi ( )x t , t 0

t= n 0x 0t= , t phng trnh (24.171).

0[1, 0] ( ) 4736624.0T opt x J u= = (24.182)

0[1,0] ( ) 1.3T optx J u= = (24.183)

Do ta c th xc minh rng, nng lng iu khin t c 1(0) 1Ri = l bc su ca ln ln hn so vi nng lng cn thit t c 3 (0) 1Cv = .

Ngoi ra, nu ta thay cc gi tr ca tham s trong phng trnh (26.169), chng ta c hmtruyn t cho bi:

0

11( ) 1 9

20( ) 195

i

sV s

V s s s

+ +=

+ +(24.184)

T chng ta thy c mt s trit tiu im khng - cc giGramian c trin khai cho c trng hp khng n nh, xem [16]

Khai trin chnh tc v tnh n nh

Nu mt h thng khng th iu khin c hon ton, n c th khai trin thnh mt hthng con c th iu khin c v mt h thng con khng th iu khin hon ton c theocch sau:

B 24.1Xt h thng c hng{ [A, B]} = < nc k . Khi tn ti mt bin i tng ng

Tsao cho 1x T x= ,

_ _1 1A T AT B T B = = (24.185)

A v B c dng:

_ __

_ _12

_,

00

cc

nc

A A BA B

A

= =

(24.186)

Trong Ac ckchiu v (A , B )c c l c th iu khin c hon ton.

Kt qu trn y cho thy trng thi no ta c th hoc khng th li ti khng. nh giiu ny, ta biu din cc phng trnh trng thi v u ra di dng:

_ _

12

_ 00

cc c c

ncncnc

x A A x BuxAx

= +

g

g (24.187)

c

c nc

nc

xy C C Du

x

= +

(24.188)

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Khng gian con c th iu khin c ca m hnh khng gian trng thi bao gm tt ccc trng thi c to ra thng qua mi t hp tuyn tnh c th ca cc trng thi trong xc . S

n nh ca khng gian con ny c xc nh bi v tr ca cc tr ring ca Ac .

Mt khc, khng gian conkhng th iu khin c bao gm tt c cc trng thi c tora thng qua mi t hp tuyn tnh c th ca cc trng thi trong xnc . S n nh ca khng gian

con ny c xc nh bi v tr ca tr ring ca Anc .

V vy, u vo s khng c tc ng no trn khng gian con khng iu khin c, mongmun ca chng ta l khng gian con khng iu khin c ny n nh, trng thi ca n s tinti im ban u. Trong trng hp ny m hnh khng gian trng thi c gi l c th n nhc.

Mt c tnh quan trng ca cc miu t (24.187) v (24.188) pht sinh t thc t l hmtruyn t c cho bi:

1( ) ( )c ccH s C sI A B D= + (24.189)

Phng trnh (24.189) ni rng cc tr ring ca khng gian con khng iu khin c khngph thuc vo tp cc im cc ca hm truyn t. iu ny ch ra rng c mt s trit tiu tt ccc im cc tng ng vi cc nghim ca (sI - Anc ).

Dng chnh tc ca tnh iu khin cB 24.2 Xt m hnh khng gian trng thi c th t c hon ton cho mt h SISO. Khi

tn ti mt bin i tng ng, bin i m hnh khng gian trng thi sang dng chnh tctnh iu khin c nh sau:

0

1

'

2

1

0 0 0 1

1 0 0 0

0 1 0 , 0

0 0 1 0n

A B

= =

K

K

K

M M K M M M

K

(24.190)

Trong 1

1 1 0 det( )n n

n I A

+ + + + = L l a thc ring ca A.B 24.3 Xt m hnh khng gian trng thi c th iu khin c hon ton cho mt h

SISO. Khi tn ti mt bin i tng ng, bin i m hnh khng gian trng thi thnhdng chnh tc ca b iu khinnh sau:

1 2 1 0

'' ''

1

01 0 0 0

, 00 1 0 0

00 0 1 0

n n

A B

= =

K

K

K

MM M O M M

K

(24.191)

Trong 11 1 0 det( )n n

n I A

+ + + + = L l a thc ring ca A.

Tnh quan st c, tnh khi phc c, v tnh xc nh c

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S tay C in t

Nu ta xt m hnh khng gian trng thi ca mt h thng, c th phng on rng nu quanst u ra trong mt s qung thi gian th c th bit c mt s thng tin v trng thi. Thuctnh ny ca m hnh c gi l tnh quan st c (hoc tnh khi phc c).

Tnh quan st c

Tnh quan st c lin quan n iu c th ni v trng thi nu ta o u ra i tng iukhin.

V d 24.15

Nu ta xt h thng nh ngha bi m hnh khng gian trng thi:

1 1 1

2 22

( ) ( )( ) 1 0, ( ) [1 0]

1 1 ( ) ( )( )

x t x t x ty t

x t x t x t

= =

g

g(24.192)

Ta c th thy u ra ( )y t ch c xc nh bi 1( )x t , v bin trng thi 2 ( )x t khng c nhhng n u ra. V vy h thng khng l khng th quan st c hon ton.

Ta c nh ngha sau:

nh ngha 24.3 Trng thi 0 0x c gi l khng th quan st c nu cho 0(0)x x= v

( ) 0u t = vi 0t th c ( ) 0y t = vi 0t , c ngha l, ta khng th thy bt k tc ng no ca

0x trn u ra h thng.

H thng c gi l cth quan st c hon ton nu khng tn ti trng thi u khckhng no m khng quan st c.

Tnh khi phc c

C mt khi nim khc, lin quan mt thit vi quan st c, c gi l tnh khi phcc. Tnh khi phc c lin quan n iu c th ni v ( )x T , nu quan st c cc gi trqu kh ca u ra y, vi 0 t T . Vi cc h thng lin tc, tuyn tnh bt bin, s khc bitgia kh nng quan st c v kh nng khi phc c l khng cn thit. Tuy nhin, v dminh ha sau y cho thy trong thi gian ri rc, hai khi nim l khc nhau. Xt:

0[ 1] 0, [0]x t x x+ = = (24.193)

[ ] 0y t = (24.194)

H thng ny r rng c th khi phc c cho mi 1T , v ta bit chc chn rng [T]=0x vi 1T . Tuy nhin, n khng quan st c hon ton v ( ) 0y t = , k m khng ph thuc vo

0x .

Kim ta tnh quan st c

Vic kim tra tnh quan st c ca mt h thng tun theo nh l sau:

nh l 24.4Xtm hnh khng gian trng thi tuyn tnh, lin tc, bt bin vi n nA

( ) ( ) ( )x t Ax t Bu t = +& (24.195)

( ) ( ) ( )y t Cx t Du t = + (24.196)

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Tp hp tt c cc trng thi khng quan st c l khng gian rng ca ma trn quan stc 0[A,C] trong :

0

1

[ , ]

n

C

CAA C

CA

=

M(24.197)

H thng hon ton quan st c nu v ch nu 0[A,C] c hng ct n y .

V d 24.16

Xt m hnh khng gian trng thi sau:

3 2 1, , [1 1]

1 0 0A B C

= = =

(24.198)

Ma trn quan st c c cho bi:

0

1 1[ , ]

4 2

CA C

CA

= =

(24.199)

V vy hng 0[A,C] = 2, cho thy h thng l c th quan st c hon ton.

V d 24.17

Xt m hnh c nh ngha trong (24.192), ta c:

1 0, [1 0]

1 1A C

= =

(24.200)

Ma trn ca kh nng quan st c l :

0

1 0[ , ]

1 0A C

=

(24.201)

V vy hng 0[A,C] = 1 < 2 v h thng l khng th quan st c hon ton.

Kt qu trn y cng p dng cho cc m hnh ri rc.

Tnh quan st c ca h thng khng ph thuc vo s la chn cc bin trng thi. C thchng minh rng hng ca ma trn c nh ngha trong phng trnh (24.197) khng thay ikhi bin i tng ng T c s dng (xem mc "Bin i tng ng trng thi").

Mt tnh quan st c

S mt tnh quan st c c th xut pht t cc cc c tnh cu trc ca h thng. Tuynhin, n cng c th xy ra khi cc thng s h thng no nhn mt vi gi tr c bit. Cngging nh tnh iu khin c c phn tch trong phn Tnh iu khin c, quan st

c v tnh n nh.

V d 24.18

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S tay C in t

Xt mch in t trong hnh 24.14. Ta c th thy ging nh trong hnh 24.13, trong phntri v phi hon i cho nhau, ta c th s dng cng thc tng t c m hnh khng giantrng thi. Cc bin trng thi c chn l 1 3x ( ) ( )Ct v t= v 2 1x ( ) ( )Rt i t= .

Vi na tri ca mch, ta c:

3

3

3 3 3 3

( ) 1 1( ) ( )C C i

dv tv t v t

dt R C R C = + (24.202)

3( ) ( )Cv t v t + = (24.203)

HNH 24.14 Mch in t

V vi na phi, ta c:

1 1 1

1

1 1 2 1 1 2

( ) 1( ) ( )R R

di t R Ri t v t

dt C R R C R R +

= +

(24.204)

0 1 1( ) ( ) ( )Rv t R i t v t = + (24.205)

B khuych i thut ton theo cch ni trn m bo ( ) ( )v t v t + = , v vy ta c th kt hpcc m hnh trng thi a ra trong cc phng trnh (24.202) (24.205):

3

3 3 3

3 3

1 21 1

1 2 1 1 1 2

1( ) 0 1( )

( )

( ) ( )1 0

C

C

i

R R

dv t

R C v tdtR C v t

R Rdi t i t R R C C R Rdt

= + +

(24.206)

3

0 1

1

( )( ) [1 ]

( )C

R

v tv t R

i t

=

(24.207)

Ma trn ca kh nng quan st c c cho bi:

1

1 2

3 3 2 1 2 1

1

[ , ] 1 1c

RC

C A R RCA

R C R C R C

= = +

(24.208)

xc nh kh nng quan st c hon ton, ni cch khc, ta cn tnh nh thc ca matrn

1 1 3 3

3 3 1

1det( [ , ]) ( )c C A R C R C

R C C = + (24.209)

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T ta kt lun rng, h thng ca m hnh ny l c th quan st c hon ton nu v chnu, 1 1 3 3R C R C , ging vi iu kin ta thu c v d 24.13.

p dng bin i Laplace vo cc phng trnh (24.204) (24.203) ta c hm truyn t t( )iV s ti 0 ( )V s :

0 0 3 31 1

1 2

3 31 2 1

11( ) ( )( )

1( ) ( ) ( )i i

sV s V s R C V s R C

R RV s V s V s ssR C R R C

+

+

= = + ++ (24.210)

iu kin 1 1 3 3R C R C = to ra s mt kh nng quan st c, dn ti mt s trit tiu imcc - khng trong m hnh hm truyn t, c ngha l, im cc na tri ca mch hnh 24.14b trit tiu bi im khng na phi. C mt cht khc bit gia cc hm truyn t trong(24.210) v (24.169). Kt qu cui cng l ging nhau, nhng th t s trit tiu l khc nhau vimi trng hp. S trit tiu cc im khng - cc lin quan n s mt tnh quan st c honton v s trit tiu cc im cc - khng lin quan n s mt tnh iu khin c hon ton.Nhng vn ny s c bn mt cch chi tit hn trong mc nh Khai trin chnh tc.

Gramian ca tnh quan st c

Kim tra tnh quan st c theo nh l 24.4 s tr li c hay khng v tnh quan st chon ton ca mt m hnh. Tuy nhin, i khi ta quan tm ti mc ca tnh quan st c camt m hnh c th. T ta c th c lng nng lng ca tn hiu u ra ( )y t , khi khng c

u vo ( ( ) 0u t = ) v trng thi l 0(0)x x= ti 0t=

2

0

0 0

( ) ( ) ( ) ( )

= = T

TE x y t dt y t y t dt (24.211)

C th chng minh rng nng lng u ra l:

2

0 0 0

0

( ) ( ) TE x y t dt x Qx

= = (24.212)

Trong :

0

T A t T At Q e C Ce dt

= (24.213)

Ma trn Q c gi l gramian ca tnh quan st c, v n o lng tnh quan st cca vect trng thi (0)x . Nu ma trn ny nh, c ngha rng chng ta c mt s ng gp yu

ca trng thi u 0x n nng lng u ra ( )y t . Thc vy, ta c th nh gi tc ng ca mi

bin trng thi bng cch ly, v d T0 [0, ,0,1,0, ,0]x = K K .

Ch rng, s tn ti ca tch phn c nh ngha trong (24.213) c m bo nu v chnu h thng n nh, c ngha l, nu v ch nu cc tr ring ca A c phn thc m.

Ngoi ra, gramian ca tnh quan st c Q cnh ngha trong (24.213) tha mn phngtrnh Lyapunov:

0T TA Q QA C C + + = (24.214)

Vi h thng ri rc n nh, gramian ca tnh iu khin c c nh ngha bi:

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S tay C in t

0

( )T k T k d d d d d k

Q A C C A

=

= (24.215)

N tha mn:

0T T

d d d d d d A Q A Q C C + = (24.216)

V d 24.19

Ta s s dng m hnh trong v d 24.18, c miu t bi m hnh khng gian trng thi(24.206) v (24.207), nh gi li ch ca gramian ca tnh quan st c (24.213), c bit lkhi m hnh gn mt tnh quan st c hon ton, c ngha l, khi 1 1 3 3R C R C

Gi thit 1 2 3 1, , , ,R R R C v 3C c cng cc gi tr nh trong v d 24.14, chng ta c:

3 3

3

11

1 0( )( ) 1

( )200( )10

( ) 9

C C

i

RR

v tv tv t

i ti t

= +

g

g(24.217)

33

0

1

( )( ) [1 10 ]

( )C

R

v tv t

i t =

(24.218)

Nu ta xt ln tng i ca cc thnh phn ca ma trn C, chng ta c th on c ura 0 ( )v t s ch yu c quyt nh bi trng thi 1( )Ri t . xc minh iu ny, chng ta tnhgramian ca tnh quan st c c nh ngha trong (24.172) bng cch gii:

0 T TA Q QA C C = + + (24.219)

3

11 12 11 12 3

33

21 22 21 22

1 01 10 10 1 102020 10100

99

q q q q

q q q q

= + +

(24.220)

Chng ta c:

0.57 69.8369.83 225000Q = (24.221)

T ta c th tnh s ng gp ca mi trng thi vi nng lng tng u ra. Bng cchlm nh vy, chng ta xc minh c rng bin trng thi 1( )Ri t c mt tc ng trn u ra ln

hn tc ng ca 3 ( )Cv t , nh c nh ngha phng trnh (24.212)

0 0[1,0] ( ) 0.57Tx E x= = (24.222)

0 0[1,0] ( ) 225000Tx E x= = (24.223)

Hm truyn t l:

1

1( ) 1920( ) 19

o

i

sV sV s ss

+ += ++

(24.224)

Ta quan st thy rng c mt s trit tiu im cc - khng gi.

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Nguyn tc i ngu

Chng ta thy mt s ging nhau gia cc kt qu nh l 24.3 v 24.4, v gia cc nhngha v gramian ca (24.172) v (24.213). l nguyn tc i ngu, nguyn tc ny c phtbiu nh sau:

nh l 24.5 (i ngu)Xt m hnh khng gian trng thi miu t bi mt h thng c s d liu(4-tuple (A, B, C, D)). Khi h thng l c th iu khin c hon ton nu v ch nu hthng i ngu (AT, BT,CT,DT) l c th quan st c hon ton.

Khai trin chnh tc v Tnh xc nh c

nh l trn thng c s dng i t kt qu ca tnh iu khin c sang kt qu catnh quan st c v ngc li. i ngu ca b 24.1 l:

B 24.4 Nu hng 0{ [A, C]}= k < n , th tn ti mt bin i tng ng T m1 1x , ,T x A T AT C CT = = = , t C v A c dng:

0

0

21

0, 0

no

AA C C

A A

= =

(24.225)

Trong 0

A c k chiu v cp (0 0,C A ) l quan st c hon ton.

Kt qu ny c mt tng ng thch hp vi thuc tnh iu khin c v s khai trin linquan . nh gi iu ny, ta s dng b 24.1 biu din cc phng trnh trng thi (c bin i) v phng trnh u ra theo dng c phn chia nh sau:

0 00 0

21

( ) 0 ( )( )

( )( ) no nonono

x t A x t Bu t

A A x t Bx t

= +

g

g(24.226)

0

0

( )( ) 0 ( )

( )no

x t y y C Du t

x t

= +

(24.227)

Miu t trn y ni ln ti sao c th gp kh khn khi c iu khin h thng ch bng u ra.u ra khng c thng tin trn trng thi xno .

Khng gian con c th quan st c ca mt m hnh l khng gian bao gm tt c cctrng thi c sinh ra qua mi t hp tuyn tnh c th ca cc trng thi ca xo . Kh nng n

nh ca khng gian con ny c quyt nh bi v tr ca cc tr ring cao

A .

Khng gian con khng th quan st c ca mt m hnh l khng gian bao gm tt c cctrng thi c sinh ra qua mi t hp tuyn tnh c th ca cc trng thi ca xno . Kh nng n

nh ca khng gian con ny c quyt nh bi v tr ca cc tr ring cano

A .Nu khng gian

con khng th quan st c l c th n nh th chng ta ni rng h thng l c th xc nhc.

Mt c tnh quan trng ca cc m t (24.226) v (24.227) xut pht t thc t l hm truyn

t c cho bi:1

0 00( ) ( )H s C sI A B D= + (24.228)

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S tay C in t

Phng trnh (24.228) ni rng cc tr ring ca khng gian con khng th quan st ckhng ph thuc vo tp cc im cc ca hm truyn h thng. iu ny ch ra rng c mt s

trit tiu ca tt c cc im cc tng ng vi cc nghim ca (no

sI A ).

Dng chnh tc ca tnh quan st c

C cc cp i ngu ca cc dng chnh tc c a ra trong b 24.2 v 24.3. V d, cpi ngu trong b 24.3 l:

B 24.5 Xt mt h thng SISO quan st c hon ton. Khi tn ti mt bin i tngng bin i m hnh v dng chnh tc ca b quan st:

11

0

0 0

1

( ) ( ) ( )1

0 0

nnb

x t x t u t

b

= +

M O M&

M M(24.229)

( ) [1 0 0] ( ) ( ) y t x t Du t = +L (24.230)

Khai trin chnh tc

S hiu bit hn na v cu trc ca cc h thng tuyn tnh ng nhn c bng cch chxem xt nhng h thng ch quan st hoc iu khin c tng phn. Nhng h thng ny c thc phn tch thnh nhng h thng quan st hon ton v iu khin c hon ton.

Hai kt qu ca b 24.1 v 24.4 c th c kt hp cho c nhng h thng m khng phil quan st c hon ton v cng nh iu khin c hon ton. Chng ta c th thy nh sau:

nh l 24.6 (nh l khai trin chnh tc ) Xt h thng c miu t bi dng khng giantrng thi. Khi lun tn ti mt bin i tng ng T sao cho m hnh c bin i cho

1x T x= c dng:

13 1

21 22 23 24 21 2

33

34 44

0 0

, , 0 0

0 0 0 000 0

coA A B

A A A A BA B C C C

AA A

= = =

(24.231)

Trong :

i. H thng con [1 1

, ,co

A B C ] l va c th iu khin c hon ton v va c th quan

st c hon ton v c cng hm truyn nh h thng ban u (xem b 24.6)

ii. H thng con:

1

1

21 23 2

0, , 0

coA BC

A A B

(24.232)

L c th iu khin c hon ton.

iii. H thng con:

13 11 2

33

, ,00

coA A BC C

A

(24.233)

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L c th quan st c hon ton.

Khai trin chnh tc c miu t trong nh l 24.6 dn ra mt h qu quan trng cho hmtruyn t ca m hnh ch c khng gian ph c th iu khin c hon ton v c th quan stc hon ton.

B 24.6Xt ma trn hm truyn H(s) cho bi:

( ) ( ) ( )Y s H s U s= (24.234)

Khi :

1 111

( ) ( )coH C sI A B D C sI A B D = + = + (24.235)

Trong 1,

coC A , v

1B u ging nh trong phng trnh (24.231). Miu t trng thi ny

l mt s thc hin ti thiu ca hm truyn t.

Nu M l ma trn vung bt k v k hiu {M} l tp cc tr ring ca M th:

22 33 44{ } { } { } { } { }coA A A A A = (24.236)

Trong :

{ }A = cc tr ring ca h thng,

{ }co

A = cc tr ring ca h thng con c th iu khin c v c th quan st c,

22{ }A = cc tr ring ca h thng con c th iu khin c nhng khng th quan st

c,

33{ }A = cc tr ring ca h thng con c th quan st c nhng khng th iu khin

c,

44{ }A = cc tr ring ca h thng con khng th quan st vkhng th iu khin c.

Ta thy rng tnh iu khin c ca mt h thng cho trc ph thuc vo cu trc ca cccng vo, c ngha l, ch m cc u vo c th iu chnh c c t vo. Do , cc trngthi ca mt h thng con cho trc c th khng iu khin c vi mt u vo cho trc,

nhng c th iu khin c hon ton vi u vo khc. S khc bit ny l quan trng khi thitk h thng iu khin v khng phi tt c cc u vo no ca i tng iu khin cng iukhin c (v d cc nhiu) v v vy, khng th c s dng li i tng iu khin tinhng trng thi nht nh.

Tng t nh vy, tnh quan st c ph thuc vo u ra no ang c xem xt. Nhngtrng thi no c th khng quan st c t u ra cho trc, nhng c th hon ton quan stc t mt s u ra khc. iu ny cng c tc ng ng k n cc h thng iu khin phnhi u ra, v mt s trng thi c th khng xut hin u ra ang c o v phn hi ca itng iu khin Tuy nhin, chng c th xut hin cc bin ni ti ct yu v v vy l quantrng vi vn iu khin.

Kim tra PBH

Mt cch kim tra khc cho tnh iu khin c v quan st c cung cp bi b cbit n nh l php kim tra PBH sau:

B 24.7Xt mt m hnh khng gian trng thi (A, B, C). Khi :

i. H thng l khng th quan st c hon ton nu v ch nu tn ti mt vect khckhng nx v mt i lng v hng n sao cho:

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S tay C in t

, 0 Ax x Cx= = (24.237)

ii. H thng khng th iu khin c hon ton nu v ch nu tn ti mt vect khc khngnx v mt i lng v hng sao cho:

, 0T T Tx A x x B= = (24.238)

24.7 B quan st trng thi

Cc khi nim c bn

Khi cc bin trng thi c o lng gim st, thc hin cc h thng iu khin , hoccho cc mc ch khc m phi i mt vi nhng vn kh khn v cng ngh v kinh t. Bquan st l cch c lng cc bin trng thi da trn m hnh h thng, o lng u ra y(t),v u vo u(t). y l s khi qut ha ca vic o lng gin tip mt bin h thng bng cchs dng m hnh h thng v o mt vi bin khc d o hn.

ng hc b quan st

Gi thit rng h thng c m hnh khng gian trng thi cho bi (24.42) v (24.43) vi D = 0(mt h thng chnh xc nh gi thit). Khi , cu trc chung ca mt b quan st trng thi h

thng kinh in c ch ra trn hnh 24.15, trong ma trn J l h s khuych i ca b quanst

HNH 24.15 B quan st trng thi kinh in

Phng trnh ca b quan st l:

$$ $( ) ( ) ( ( ) ( ))

d xAx t Bu t J y t C x t

dt= + + (24.239)

Mt cu hi t ra l: nu chng ta bit m hnh chnh xc ca h thng v u vo h thng,th ti sao ta cn phi cung cp u ra h thng? Cu tr li l cn phi o u ra v chng ta khngbit trng thi ban u ca h thng. iu ny c th c nh gi bng phng trnh sai s clng trng thi, ( ) ( ) ( )x t x t x t = % . Phng trnh ny c th nhn c bng cch tr (24.42) cho(24.239). T ta c:

%%( )

( ) ( )

d x t

A JC x t dt = (24.240)

T (24.240) ta thy sai s c lng s hi t v khng vi mt sai s ban u khc khng nuv ch nu tt c cc tr ring ca ma trn A JC c cc phn thc m, c ngha l a thc

( ) det( ) E s sI A JC = + ca b quan st l Hurwitz hon ton.

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Nhn xt

Phng trnh (24.240) ch hp l nu m hnh m t ng h thng. Cc sai s ca vicm hnh ha s tc ng n b quan st. iu ny thng dn n cc sai s c lngtrng thi khc khng.

Nu cp (A, C) l c th quan st c hon ton, th tr ring ca A-JC c th cnh v ty (trong min n nh). V vy, tc hi t c lng l do s la chnca ngi thit k. Cc tr ring ny l cc im cc ca b quan st.

Nu cp (A, C) l c th xc nh, th b quan st s dn ra mt cch tim cn sai strng thi tnh khng, mc d khng phi tt c cc tr ring ca A-JC u c th ct theo mun.

Nu h thng l khng th quan st c hon ton, v khng gian con khng th quanst c cha cc m hnh khng n nh, th b quan st s khng bao gi hi t.

minh ha cc phng php ca b quan st, chng ta tham kho v d 24.5

V d 24.20

Gi thit rng chng ta mun cc cc ca b quan st vi m hnh trng thi trong v d 24.5c t s = -4,s = -6, vs = -8. Ta c th tnh h s khuych i J ca b quan st, bng cchs dng phn mm nh MATLAB chng hn. Khi , ta c c:

[ ]4.5247 7.5617 4.1543T

J= (24.241)

nh gi ng hc b quan st, gi thit rng trng thi ban u ca h thng lTx(0) = [-1 2 1] v u vo h thng l mt sng vung ca bin 1 v tn s 1 rad/s. B

quan st c khi to vi x(0) 0= . Khi chun ca sai s c lng, x( )t , s c dng nhch ra trong hnh 24.16.

HNH 24.16 Sai s c lng trng thi

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S tay C in t

HNH 24.17 H thng quay

iu quan trng c ch ra trong v d ny l i tng iu khin khng n nh. C ngharng trng thi v c lng trng thi tng khng gii hn. Tuy nhin, vi gi thit rng m hnhl hon ton ng, th sai s c lng s hi t v khng.

V d 24.21

Hnh 24.17 biu din s ca mt h thng quay iu khin bi m men ( )t . Nng lngh thng c truyn qua h thng bnh rng vi hai bnh xe c bn knh r1 v r2 v qun tnhtng ngI1 vI2. Chuyn ng ca hai trc quay b gim bi ma st vi h s D1 vD2, n hixon ng k ca trc 2 cng c m hnh ha. Ti ca h thng c m hnh ha l qun tnh

I3. Ta mun c lng tc ti 3 da trn vic o tc 1 ca trc 1.

Trc tin ta cn xy dng m hnh khng gian trng thi. lm vic , chng ta chn mttp ti thiu cc bin h thng xc nh nng lng c lu gi trong h thng. H thng c bnthnh phn c th cha nng lng: 3 qun tnh v mt n hi. Nng lng trong I1 vI2 c thc tnh t hoc 1 hoc 2 , c ngha l chng ta ch cn mt trong cc tc , v chng thamn:

1 2

1 1 2 2

2 1

( )( ) ( ) ( ) ( )

( )

t rand t t t t

t r

= = (24.242)

V vy, cc bin trng thi c chn theo nh hng vt l l:

1 1( ) ( )x t t = (24.243)

2 2 3( ) ( ) ( )x t t t = (24.244)

3 3( ) ( )x t t = (24.245)

T cc nguyn tc u tin ta c:

1

1 1 1 1

( )( ) ( ) ( )

d tt D t I t

dt

= + + (24.246)

2 2

1 2 2 2 2 2 3

1

( )( ) ( ) ( ) ( ( ) ( ) )

r d tt t D t I K t t

r dt

= = + + (24.247)

3

2 3 2 3

( )0 ( ( ) ( ) )

d tK t t I

dt

= + (24.248)

V chng ta chn 1( )t nh l bin h thng c th o c, nn ta nhn c

2 2

1 2 2 1 1 2 2 2

22 2 2 2

1 2 2 1 1 2 2 1 2 2

1 2 2 1

1

2

2

3

0

( )0 1 ( ) ( )0

0 0 0B

A

r D r D r r K r

r I r I r I r I r I r I

rdx tx t t

dt r K

I

+ + + +

= +

1 44 2 4 431 4 4 4 4 442 4 4 4 4 4 43

(24.249)

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S tay C in t

V d 24.22

Mt h thng c m hnh khng gian trng thi c cho bi:

[ ]2 1 1

, , 1 1 , 01 3 0.5

A B C D

= = = = (24.259)

Gi thit rng, chng ta mun c tnh bin h thng ( ) ( )Tz t y x t = trong Ty =[1 1]. c

lng da trn c s quan st ph hp l ( )z t$ c cho bi:

$( )Tz x t =$ (24.260)

Khi , nhiu khi c tnh ( )z t l ( )vz t , c bin i Laplace tha mn:

1( ) ( ) ( ), ti ( ) ( )Tv v vZ s H s V s H s sI A JC J = = + (24.261)

HNH 24.18 c tuyn lc ca b quan st

Tip theo ta xt hai la chn khc nhau cho a thc E(s) ca b quan st . l:

1( ) ( 0.5)( 0.75)E s s s= + + v 2 ( ) ( 10)( 20)E s s s= + + (24.262)

Ngi c c th thy rng cc b quan st c dn ra s c cc tc rt khc nhau. Bquan st u chm hn nhiu b th hai.

Vi cc la chn ny chng ta tnh cc h s khuch i1

J v2

J , v cc hm lc tng ngca b quan st:

1

1 1 1 2

1.875 5.625( ) ( )

1.25 0.375T sH s sI A J C J

s s

+= + =

+ +(24.263)

1

2 2 2 2

144 432( ) ( )

30 200T sH s sI A J C J

s s

+= + =

+ +(24.264)

so snh c hai trng hp, ta tnh v v p ng tn s ca mi b lc. Kt qu c ch ratrn hnh (24.18).

T hnh 24.28 ta thy rng, vi tn s cao, b lc chm khng nhiu tt hn b lc nhanh.

V d trn y minh ha s cn bng gia tc ca b quan st v tnh khng nhiu. Mt

cch h thng gii quyt vn kh x ny l s dng l thuyt lc ti u, nh b lcKalman_Bucy. Ngi c quan tm c th tham kho [2].

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24.8 Phn hi trng thi

Cc khi nim c bn

Khi tt c cc trng thi h thng u c th o c v h thng c th t c honton(theo ngha c gii thch trong mc "Kh nng iu khin, kh nng t c v khnng n nh"), th chng ta c th iu khin h thng bng cch s dng phn hi trng thi

t c lnh y ca ng hc vng kn. tng ny c th hin trong hnh (24.19)Hnh 24.19 biu din dng c bn nht ca phn hi trng thi: u vo ca h thng l mt

thnh phn t l vi trng thi (thnh phn cn li l tn hiu ngoi ( )r t )

Phn hi trng thi l mt tng rt n gin v ngy th. Suy xt k cho thy tng ny cmt s thiu st v nhng c tnh tim n mi nguy him nh:

i hi nhiu bin nh cc bin trng thi. iu ny khng ch l rt t m trong mts trng hp, khng th thc hin c.

Mi php o trng thi l mt ngun gc ca sai s v chnh xc ca n l hu hn

Mi php os a vo nhiu vi tc ng xu n vic thc hin h thng iu khin.

Thc hin ton b mt cch chnh xc da trn mt hm chnh xc ca mt tp hpphc tp cc thit b. iu ny s t ra mt s vn lin quan n s thc hin v kh

nng tch hp h thng

HNH 24.19 Phn hi trng thi

Mc d c cc im yu, nhng phn hi trng thi t bn thn n l mt khi nim mnh, vnl c scho nhiu cu trc iu khin mnh v tinh vi hn. L do ch yu cho iu ny l bt kb iu khin tuyn tnh no cng c th c gii thch nh mt s kt hp ca b quan st trngthi v phn hi trng thi.

ng hc ca phn hi

Gi thit rng h thng c iu khin c hm truyn t H(s) v khng gian trng thi cbiu din bi (24.42) v (24.43) vi D = 0. Nu u vo ca i tng iu khin c iu khintheo:

( ) ( ) ( )u t Kx t r t = + (24.265)

Khi biu din khng gian trng thi cho ton b vng iu khin c cho bi:

( )( ) ( ( ) ( ))

dx tAx t B Kx t r t

dt= + + (24.266)

( ) ( )Y t Cx t = (24.267)

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S tay C in t

C th thy rng mi quan h gia ( )R s v Y(s) c cho bi

1 1 1

( )

( ) ( ) ( ( ) ) ( )H s

Y s C sI A B I K sI A B R s = + 1 44 2 4 43 (24.268)

iu ny cho thy vng phn hi trng thi s bo ton cc im khng ca h thng v dchcc im cc ti cc nghim ca nh thc (sI - A +BK)

Phn hi trng thi ti u. iu khin ti u

Xt h thng tuyn tnh bt bin c khng gian trng thi c biu din bi (24.42) v(23.43) vi D = 0, vi gi thit trng thi ban u x(0) = x o

Gi thit rng mc ch iu khin l li h thng t trng thi ban u x o ti gi tr nh nhtc th cng sm cng tt trong khong [0, tf]. Chng ta cn yu cu thm rng, qu trnh li khngyu cu qu nhiu n lc iu khin. Khi vn iu khin ti u c xc nh l vn tmiu khin ti u u(t) trong khong thi gian [0, tf] sao cho mt hm gi bc hai c ti thiu.Hm gi ny c chn l:

0

0

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

T T T

u f f f J x x t Qx t u t Ru t dt x t Q x t = + + (24.269)

Trong ,n n n nfQ Q u l cc ma trn xc nh, i xng, khng m v m mR

l ma trn xc nh, i xng, dng. Cc yu cu i vi cc ma trn trng l c t sao chohm gi c ngha. Nu Q c php l m, th gi ti u c th thm ch l m trong khi trngtrng thi c th tng khng gii hn v ln. Ngoi ra, nu ta cho php R c cc tr ring gc(c ngha l R c php l mt ma trn xc nh khng m, thay v yu cu n hon ton dng)th tn hiu iu khin u(t) cng c th tng khng gii hn (theo cc hng ca cc vect tr ringlin quan).

Mt lut iu khin tuyn tnh bt bin nhn c bng cch tim cn khi ft . Vi iukin ny, lut iu khin ti u c cho bi:

0 0( ) ( )u t K x t = (24.270)

Vi0 1 TK R B P = (24.271)

trong P l li gii duy nht khng m ca phng trnh i s Riccacti:

10 T TQ P BR B P P A A P = + + (24.272)

li gii ny tn ti, cn tha mn nhng iu kin k thut nht nh (vn ny cbn lun chi tit, v d trong [5]).

Nhn xt

Li gii cho vn LQR s ti thiu ha hm gi tr (24.269) v, khi ft , h thnglun n nh.

Vn quan trng l chn cc ma trn trng lng Q v R nh th no. La chn thngthng cho Q l TQ C C= . Vi la chn ny, u ra ca h thng c t trc tip vohm gi.

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Vi Q cho trc, kch thc ca R nh hng mnh n v tr cc im cc ca vng lpkn. R cng ln th vng lp iu khin cng chm.

bit thm v b iu khin ti u bc hai, c th c cc ti liu v d [1, 3, 4, 8, 9]

sổ tay cdt chuong 24-kg trang thai

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