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1/20

26Phng php qu o nghim

Hitay zbayThe Ohio State University

26.1 Gii thiu ........................................................... .1

26.2 Cc v tr im cc mong mun ........... ......... .....3

26.3 Xy dng qu o nghim ..................................5

26.4 Qu o nghim b sung ............................... ...13

26.5 Qu o nghim cho h thng c tr thi gian .14

26.1 Gii thiu

Qu o nghim l phng php ha c s dng trong vic phn tch v thit k h thng phn hi. Phng phpny c cng b chnh thc bi W.R.Evans[3,4], ngi nhn gii thng Richard E.Bellman v iu khin t Hing iu khin t ng M vo nm 1988 cho ng gp quan trng ny.

tho lun v phng php qu o nghim, trc tin ta phi xem li nh ngha c bn v tnh n nh ca u vohu hn v u ra hu hn (BIBO) ca h thng phn hi tuyn tnh bt bin c ch ra trong hnh 26.1, trong itng iu khin v b iu khin c biu din tng ng bi cc hm truyn t ( )P s v ( )C s 1. i tng iu khin,

( )P s , bao gm qu trnh vt l c iu khin, cng nh c cu chp hnh v ng lc hc cm bin.

H thng phn hi c gi l n nh nu khng c cc hm truyn kn t u vo ngoi vi r v v ti cc tn hiu niti e v u , c im cc no trong na mt phng ng bn phi, : { : Re( ) 0}s s+ = . iu kin cn thit cho n nhh thng c phn hi l cc im khng ca P(s) (tng ng vi C(s)) na mt phng ng bn phi u khc bit so vicc im cc ca C(s) (tng ng viP(s)). Khi c iu kin ny, ta ni rng khng c s mt n nh do trit tiu im cc-khng trong vic thit lp ( ) ( ) : ( )P s C s G s= , khi kim tra tnh n nh ca h thng c phn hi tr thnh tng ngvi vic kim tra tt c cc nghim ca:

1 ( ) 0G s+ = (26.1)

u nm trong na mt phng h bn tri : { : Re( ) 0}s s- =


2/20

S tay C in t

1

( )( ) vi ( ) ( )

( )

m

j

j

N sF s N s s z n m

D s == = - (26.2)

1

( ) ( )n

i

i

D s s p=

= -

Vi 1, , mz zK v 1, , np pK l im cc v im khng ca h kn. Trong trng hp ny, cc im cc l nghim caphng trnh c tnh:

( ) : ( ) ( ) 0s D s KN sc = + = (26.3)

Qu o nghim thng thngnhn c bng cch v cc nghim 1( ), , ( )nr K r K K ca a thc c tnh ( )sc trn mtphng phc, viKthay i t 0 n + . Tng t i vi gi tr m ca K cho qu o nghim b sung. Vi s tr gipca th qu o nghim, ngi thit k xc nh cc gi tr c th chp nhn c ca thng s K dn n mt tp ccim cc ca h kn trong "vng mong mun" trn mt phng phc. C mt vi yu t phi c xem xt theo khi nimvng mong mun ca mt phng phc m 1( ), , ( )nr K r K K cn nm . iu ny s c cp n trong phn sau.Phn 26.3 a ra cc th tc xy dng qu o nghim, v cc v d thit k c a ra trong phn 26.4.

Ngoi h s K ra, qu o nghim cng c th c v vi cc thng s h thng khc. V d phng trnh c tnh choh thng ( ) ( )G s G sl= cho bi:

(1 ) 1( ) ( ) ( ), ( ) , ( ) 1(1 ) c I

sG s P sC s P s C s K s s T s

lll

- = = = ++

cng c th c chuyn thnh dng cho trong (26.3). y cK v 1T l cc thng s ca b iu khin PI (t l - tch phn),v 0l > l thng s thay i ca i tng iu khin. Ch rng pha ca i tng l:

1( ) 2tan ( )2

P jp

w l w- =- -

Do thng s l c th coi l h s tr pha khng chc chn (v d, mt tr thi gian nh khng chc chn ca i tngiu khin c th c m hnh ha theo cch ny, xem [9]). D thy rng phng trnh c tnh l:

2 1( 1) (1 ) 0cI

s s K s s

T

l l + + - + =

HNH 26.2 Qu o nghim tng ng vi 1/K l=

V bng cch sp xp li cc h s bng vic nhn l , phng trnh ny c th bin i thnh:

21

21

( / )11 0

( / )c c

c c

s K s K T

s s K s K T l

+ ++ =

- -

T nh ngha1 2

1, ( ) ( / )c cK N s s K s K T l-= = + + v 2 1( ) ( / )c cD s s s K s K T = - - , ta thy phng trnh c tnh c

th vit dng (26.3). By gi, biu qu o nghim c th nhn c t d liu ( )N s v ( )D s c nh ngha

trn; iu ch ra cch m cc cc ca h kn di chuyn khi 1l - thay i t 0 ti + , vi tp c nh cc thng s b iukhin cho cK v 1T . V d vi cK =1 v 1T =2.5, qu o nghim c biu din hnh 26.2.

2


3/20

Phng php qu o nghim

Th tc xy dng qu o nghim s c a ra phn 26.3. Hu ht cc tnh ton lin quan trong mi bc ca th tcny u c th thc hin bng tay. V vy mt th xp x biu din qu o nghim c th v d dng. C mt s phnmm c th t ng to ra qu o nghim t cc s liu 1, , mz zK v 1, , np pK .

Nu c sn mt chng trnh s tnh nghim ca mt a thc, th ta cng c th nhn c qu o nghim tng ngvi mt thng s s c nhp vo trong phng trnh c tnh mt cch khng tuyn tnh. minh ha, ta xt v d sau:

0( ) ( )G s G sw= trong :

0 2 20 0

( 0.1) ( 0.2)( ) ( ) ( ), ( ) , ( )( 1.2 )( 0.1) ( 2)

s sG s P sC s P s C ss s s s

ww w

- -= = =+ + + +

y, 0 0w l thng s c th thay i ca i tng. Ch rng phng trnh c tnh:

( ) ( )

0 02

(1.2 )( 0.1)( 2)1 0

( 0.1)( 0.2) 0.2 0.1

s s s

s s s s s

w w+ + ++ =

+ + + - -(26.4)

khng th biu din c di dng ( ) ( ) 0D s K N s+ = vi ch mt thng sK. Tuy nhin, vi mi 0w ta c th tnhnghim ca (26.4) theo phng php s v v chng trn mt phng phc khi 0w thay i trong phm vi cn thit. Hnh 26.3minh ha c bn nhnh 1 4( ), , ( )r K r K K ca qu o nghim cho h thng ny khi 0w tng t khng ti v cng. Hnh nhnc bng cch tnh nghim ca (26.4) vi mt tp cc gi tr ca 0w , nh s dng MATLAB

HNH 26.3 Qu o nghim tng ng vi 0w

26.2 Cc v tr im cc mong mun

Biu din h thng phn hi ph thuc nhiu vo v tr cc im cc ca h thng iu khin kn ( ) 1, ,ir K n= K . Trc

ht, n nh ta mun ( )ir K - vi mi 1, ,i n= K . R rng, c mt im cc "gn" trc o th s nguy him, c nghal nhng s bt n nh ca i tng cng c th dn n h thng phn hi khng n nh. Do v tr im cc mongmun phi sao cho tnh n nh c duy tr trc nhng bt n nh vy (hoc vi s tn ti ca nhng thng s d thay i)ca i tng. Vi cc h bc hai, ta c th xc nh cc php o th ca tnh n nh chc chn theo cc v tr im cc cth c gn vi c tnh ca p ng bc. Vi cc h bc cao hn, cch tng t ch c th s dng vi cc cc tri.

Vi h thng iu khin phn hi c bn trong hnh 26.1, gi thit rng hm truyn t vng kn t( )r t

ti( )y t

c dngsau:

20

02 20 0

( ) , 0 1,2

T ss s

wz w

zw w= < <

+ +

V ( )r t l hm bc n v. Khi , u ra l:

0

2( ) 1 sin( ), 0

1

t

d

e y t t t

zw

w qz

-

= - + -

Trong 20: 1dw w z= - v-1: os ( )cq z= . Vi mt s gi tr thng thng ca z, p ng step ( )y t c dng ging

nh trong hnh 26.4. qu iu chnh c nh ngha l:

: 100% p ss

ss

y yPO

y

-=

3


4/20

S tay C in t

Trong py l gi tr nh. Bng tnh ton n gin c th thy gi tr nh ca ( )y t xut hin ti thi gian /p dt p w= ,v:

2/ 1: 100%PO e pz z- -=

HNH 26.4 p ng step ca h bc hai

HNH 26.5 PO v z

Hnh 26.5 biu din PO theo z. Thi gian xc lp c xc nh l khong thi gian nh nht st , sau khi p ng ( )y t nm trong khong 2% gi tr mong mun, c ngha l:

( ) : min{ : ( ) 0.02 }ss sst s t y t y y t t = - "

i khi 1% hoc 5% c s dng xc nh thi gian xc lp thay cho 2%, v c bn khng c g khc bit. Vi png ca h bc hai, ta c:

0

4( )t s

zw

V vy, c mt p ng xc lp nhanh th 0zw phi ln.

Cc cc ca h kn l:

21,2 0 0 1r j zw w z = - -

V vy, khi thi gian xc lp ln nht cho php v PO c xc nh ta c th xc nh vng cc v tr cc mong munbng cch xc nhz v 0zw ti thiu cho php. V d, cho PO v st c gii hn bi:

10% 8sPO and t s

4


5/20

Phng php qu o nghim

HNH 26.6 Vng ca cc im cc mong mun ca h kn

Yu cu v PO ng rng 6z , tng ng vi 053q ( nh rng os( )c q z= ). Yu cu v thi gian xc lp c

tha mn nu v ch nu 1,2Re( ) 0.5r - . V vy vng cc cc ca h kn l vng m mu trong hnh 26.6. Hnh ny cng

minh ha vng cc cc mong mun ca h kn vi cc yu cu thit k tng t trong trng hp thi gian ri rc.

Nu bc ca hm truyn t vng kn ( )T s ln hn hai, th bc s ph thuc vo v tr ca cc im cc v imkhng ca n, c th xp x p ng bc ca h kn bng p ng ca mt h bc hai. V d, xt h bc ba

20

2 20 0

( )( 2 )(1 / )

T ss s s r

w

zw w=

+ + + trong 0r zw?

p ng qu cha s hng ne- . So snh vi hnh bao 0te zw- ca s hng sin, ne- gim rt nhanh, v p ng tng thtng t nh p ng ca h bc hai. V vy, tc ng ca cc th ba 3r r= - l khng ng k.

Xt v d khc:

20

2 20 0

[1 / ( )]( )

( 2 )(1 / )

s rT s

s s s r

w e

zw w

+ +=

+ + +trong 0 re< =

Trong trng hp ny, mc d rkhng cn ln hn nhiu so vi 0zw , im khng ti ( )r e- + trit tiu tc ng ca

im cc ti r- . thy c iu ny, xt khai trin tng phn ( ) ( ) ( )Y s T s R s= vi ( ) 1/R s s=

0 1 2 3

1 2

( )A A A A

Y ss s r s r s r

= + + +- - +

Trong 0 1A = v

20

3 2 20 0

lim( ) ( )2 ( )s r

A s r Y sr r r

w e

zw w e-

= + = - + +

V 3 0A khi 0e nn s hng 3nA e- l khng ng k trong ( )y t .

Tm li, nu c s trit tiu im cc-khng xp x na mt phng tri, th cp im cc-khng ny c th c lp rat hm truyn t ( )T s xc nh PO v st . Cng nh vy, cc im cc gn nht vi trc o c nh hng ln ti p ng

qu ca ( )y t . khi qut nhn xt ny, cho 1, , nr rK l cc cc ca ( )T s , nh vy 2 1Re( ) Re( ) Re( ) 0kr r r=


6/20

S tay C in t

( ) ( ) ( )s D s KN sc = + (26.5)

Trong ( )D s v ( )N s l cc a thc monic (c ngha l h s ca nng lng cao nht c chun ha v 1). Nu N v/hoc D khng monic, h s cao nht c th b nhp voK.

Nguyn tc qu o nghim

Cn nhc li rng qu o nghim thng thng ch ra v tr ca cc cc ca h kn khi Kthay i t 0 n + . Nghimca ( )D s , 1, , np p l cc cc, v nghim ca 1( ), , , mN s z z K l cc im khng, ca h h, ( ) ( )G s K F s= . V ( )P s v

( )C s l chun, ( )G s l chun, v v vy n m . Do bc ca a thc ( )sc l n v c chnh xc n nghim.

Cho cc cc ca h kn, c nngha nghim ca ( )sc c k hiu l 1( ), , ( )nr K r K K . Ch rng l cc hm ca K;khi s ph thuc voKl r rng, c th vit n gin l 1, , nr rK . Cc im trong tha mn (26.5) vi mt sK>0 nmtrn qu o nghim. R rng l, mt im r nm trn qu o nghim nu v ch nu:

1( )

KF r

= - (26.6)

iu kin (26.6) c th c tch ra lm hai phn:

1

( )K

F r=-

(26.7)

0 (2 1) 180 ( ), 0, 1, 2,...K l F r l = = - + - = o o (26.8)

Quy tc pha (26.8) xc nh cc im trong nm trn qu o nghim. Quy tc v ln (26.7) xc nh h skhuych iK>0 m qu o nghim cho n l ti im cho trc r. Bng cch s dng nh ngha ca ( )F s , (26.8) c thc vit li nh sau:

1 1

(2 1) 180 ( ) ( )n m

i j

i j

l r p r z = =

+ = - - - o (26.9)

Tng t, (26.7) tng ng vi:

1

1

n

iim

jj

r pKr z

=

=

-=-

(26.10)

Xy dng qu o nghim

C mt s gi phn mm cho php to qu o nghim t ng vi /F N D= cho trc. C th l lnh rlocus vrlocfind trong MATLAB. Trong nhiu trng hp, qu o nghim gn ng c v bng tay s dng quy tc diy. Nhng quy tc ny c rt ra t cc nh ngha c bn (26.5),(26.7) v (26.8)

1. Qu o nghim c n nhnh: 1( ), , ( )nr K r K K

2. Mi nhnh bt u ( 0K @ ) ti im cc ip v kt thc (khi K ) ti mt im khng jz , hoc hi t vng tim cn, jMe al , trong M v:

2 1180 , 0,.....,( 1)l

ll n m

n ma

+= = - -

-o

3. C (n m- ) ng tim cn vi cc gc al . Trung tm ca ng tim cn (c ngha l im giao nhau trn trcthc ca chng) l:

1 1

n m

i ji j

a

p z

n ms

= =-

==

4. Mt imx nm trn qu o nghim nu v ch nu tng s im cc 'ip s v im khng 'jz s bn phi

ca x (c ngha l tng s im cc ca 'ip s vi Re( )ip x> cng vi tng s im khng 'jz s viRe( )jz x> ) l l. V ( )F s l hm hu t vi h s thc nn cc im cc v im khng xut hin trong lin hp

phc, do khi tnh s im cc v im khng bn phi ca mt im x chng ta ch cn xt cc imcc v im khng trn trc thc.

6


7/20

Phng php qu o nghim

5. Cc gi tr ca K qu o nghim i ngang qua trc o c th xc nh t vic kim tra tnh n nh Routh-Hurwitz. Ta c th t s jw= trong (26.5) v gii vi w v K thc tha mn:

( ) ( ) 0D j KN jw w+ =

Ch rng y c hai phng trnh, mt cho phn thc v mt cho phn o.

6. Cc im t gy (giao nhau ca hai nhnh trn trc thc) l cc li gii kh thi (tha mn quy tc 4) ca:

( ) 0dF sds

= (26.11)

7. Gc ca im xut pht ( 0K @ ) t mt im cc phc, hoc im n t mt im khng phc ( K + ), cth xc nh t quy tc pha. Xem v d di y.

Chng ta tng bc xy dng qu o nghim theo quy tc trn cho:

( 3)( )

( 1)( 5)( 4 2)( 4 2)

sF s

s s s j s j

+=

- + + + + -

Trc tin, lit k cc im cc v im khng:

1 2 3 4 14 2, 4 2, 5, 1, 3 p j p j p p z = - + = - - = - = = -

Do , 4n = v 1m = .1. Qu o nghim c bn nhnh

2. Ba nhnh hi t v ng tim cn c gc l 600, 1800, v -600, v mt nhnh hi t v 1 3z = -

HNH 26.7 Gc ca im xut pht t 4 2j- +

3. Trung tm ca ng tim cn l ( 12 3)/ 3 3s = - + = -

4. Cc khong ( , 5]- - v [-3,1] u nm trn qu o nghim.

5. Trc o i qua cc nghim kh thi ca:

4 3 2( 12 47 40 100) ( 3) 0 j j K jw w w w w- - + - + + = (26.12)

vi w v K thc. Cc phn thc v o ca (26.12) l:

4 247 100 3 0Kw w- - + =

2( 12 40 ) 0j Kw w- + + =

suy ra hai cp li gii kh thi l ( 100/ 3, 0)K w= = v

( 215.83, 4.62)K w= = .

6. Cc im t gy u l li gii ca:

4 3 23 36 155 282 220 0s s s s+ + + + =

V cc nghim ca phng trnh l -4.55 1.11j v 1.45 1.11j- nn khng c li gii no trn trc thc, do

khng c im t gy.7. xc nh gc ca im xut pht t im cc phc 1 4 2p j= - + , cho D biu din mt im trn qu onghim gn im cc phc 1p , v t iv, 1, ,5i = K , l cc vect c v t ip cho 1, ,4i = K v t 1z cho

5i = , nh biu din trn hnh 26.7. Cho 1 5, ,q qK l gc ca 1 5, ,v v . Quy tc pha ch ra rng:7


8/20

S tay C in t

1 2 3 4 5( ) 180q q q q q+ + + - = o

(26.13)

khi D tin ti 1p , 1q tr thnh gc ca im xut pht v cc iq 's cn li c xp x bi gc ca cc vect v tcc im cc khc, v t im khng ti im cc 1p . Do 1q c th c gii t (26.13), trong

0 1 0 1 22 3 4 5

90, tan (2), 180 tanq q q - - - , v0 1 1

5 290 tanq - + . T

01 15q - .

Qu o nghim chnh xc cho v d ny c biu din trong hnh 26.8. T kt qu ca mc 5 nu trn, v dng ca qu

o nghim, c th kt lun rng h thng phn hi n nh nu:33.33 215.83K< , th qu o nghim c ba nhnh, khong [-1, 0] nm trn qu o nghim, ba

ng tim cn c cc gc 0 0 0{60 ,180 , 60 }- vi trung tm ti a 1s = - ; v ch c mt im t gy ti1

31- + , xem hnh

26.9. T v tr ca cc im t gy, tm, v gc ca cc ng tim cn, c th suy ra l hai nhnh (mt nhnh bt u ti1p = -1, v nhnh cn li bt u ti 3 0p = ) lun nm bn phi ca 1p . Mt khc, iu kin thi gian xc lp ch ra rng

phn thc ca cc im cc ca h kn tri phi t hn hoc bng -1. Do , mt thnh phn tch phn n gin khng th lmc iu ny. By gi ta th b iu khin PI c dng:

( ) , 0cc cs z

C s K K s

- = >

Trong trng hp ny, ta c th chn 1cz = - trit tiu im cc ti 1p = -1, v h thng tr thnh h thng bc hai.

Qu o nghim cho ( ) 1/ ( 2)F s s s= + c hai nhnh v hai ng tim cn, vi tm a 1s = - v gc0 0{90 , 90 }- ; im

gy cng ti -1.Cc nhnh -2 v 0, hng v nhau, gp nhau ti -1, v c xu hng ko di v tn dc theo Re( ) 1s =- . Cccc ca h kn l:8


9/20

Phng php qu o nghim

1,2 1 1 , tai / 0.72cr K K K = - - =

HNH 26.9 Qu o nghim ca v d 1

Sai s trng thi n nh, khi ( )r t l hm dc n v, 2/K.Kcn phi tht ln tha mn iu kin thit k th 3. R

rng, 1,2Re( ) 1r = - vi mi 1K , tha mn yu cu v thi gian xc lp. Phn trm qu iu chnh nh hn 10% nu

zca nghim 1,2r ln hn 0.6. Mt php tnh n gin cho thy 1/ Kz= , v vy iu kin thit k c tha mn nu K=

1/0.36, v d 2cK = . Do b iu khin PI tha mn vn thit k s l:

1( ) 2

sC s

s

+ =

B iu khin s hy mt cc n nh ca i tng iu khin (ti s = -1). Nu c mt s khng chc chn trong v trca im cc ny, s trit tiu hon ton s khng xy ra v h thng s l bc ba vi im cc th ba ti 3 1r @- . V imkhng ti 0 1z = - s gn trit tiu hiu ng ca im cc ny nn p ng ca h thng s gn vi p ng ca h bc hai.Tuy nhin, ta phi cn thn nu cc s trit tiu im cc khng gn trc o, v trong trng hp ny, cc nhiu nh trong

v tr im cc c th dn n s thay i ln trong p ng h thng phn hi, nh minh ha trong v d sau.V d 2

Mt cu trc linh hot vi cc im cc suy gim nh c hm truyn t nh sau:

21

2 2 21 1

( )( 2 )

P ss s s

w

zw w=

+ +

S dng qu o nghim, ta c th thy b iu khin:

2 21 1

2

( 2 )( 0.4)( )

( ) ( 4)cs s s

C s Ks r s

zw w+ + +=

+ +

n nh h thng phn hi vi r ln v mt la chn cK

ph hp. V d, cho1 2, 0.1w z= =

v 10r= . Th qu onghim ca ( ) ( ) ( )/F s P s C s K = trong 21cK K w= , c biu din trong hnh 26.10. ChoK= 600, cc im cc ca hkn s l:

{ 10.78 2.57, 0.94 1.61, 0.2 1.99, 0.56} j j j- - - -

9


10/20

S tay C in t

HNH 26.10 Qu o nghim cho v d 2(a)

HNH 26.11 Qu o nghim cho v d 2(b)

V cc im cc 0.2 1.99j- b trit tiu bi cp im khng ti cng mt im trong hm truyn 1( 1 )T G G G -= + ca h kn, nn cc cc tri ti -0.56 v 0.94 1.61j- (c cc phn thc m tng i ln v t l suy gim khong 0.5).

By gi, gi thit rng b iu khin ny c c nh v cc im cc phc ca i tng iu khin c thay i mtcht bng cch t 0.09z= v 1 2.2w = . Qu o nghim tng ng vi h thng c biu din trong hnh 26.11. V ccim cc phc suy gim nh khng c trit tiu hon ton, nn c hn hai nhnh gn trc o. Hn na, vi cng gi tr K=600, cc im cc ca h kn l:

{ 10.78 2.57, 1.21 1.86,0.05 1.93, 0.51} j j j- - -

Trong trng hp ny, h thng phn hi l khng n nh.

V d 3

Mt trong nhng v d quan trng ca h thng c in t l ng c mt chiu. Hm truyn gn ng ca ng c mtchiu c dng:

( ) , 0( 1/ )

mm m

m

KP s

s st

t= >

+

Nhn thy rng nu mt ln, th ( ) ( )m bP s P s , trong 2( ) /b bP s K s= l hm truyn ca mt rm cng. Trong v d

ny, lp tng qut cc i tng ( )mP s s c xt. Gi thit rng 1/m mp t= - v mK c cho trc, b iu khin bcnht:

( ) ccc

s zC s K

s p

-= - (26.14)

s c thit k. Mc ch l t cc im cc ca h kn ra xa trc o. V bc ca ( ) ( ) ( )/ m cF s P s C s K K = bng ba, quo nghim c ba nhnh. Gi thit cc im cc mong mun c cho trc l 1 2,p p v 3p . Th, vn t im cc rtcuc l tm { , , }c c cK z p sao cho cc phng trnh c tnh l:

10


11/20

Phng php qu o nghim

1 2 3

3 21 2 3 1 2 1 3 2 3 1 2 3

( ) ( )( )( )

( ) ( )

s s p s p s p

s p p p s p p p p p p s p p p

c = - - -

= - + + + + + -

Nhng phng trnh c tnh thc, theo cc thng s b iu khin cha bit, l:

3 2

( ) ( )( ) ( )

( ) ( )

m c c

m c m c c

s s s p s p k s z

s p p s p p K s Kz

c = - - + -

= - + + + -

Trong K := m cK K . t cc h s ca ( )sc mong mun bng vi cc h s ca ( )sc thcs nhn c ba phngtrnh theo ba tham s cha bit:

1 2 3m c p p p p p+ = + +

1 2 1 3 2 3( )m c p p K p p p p p p+ = + +

1 2 3cKz p p p=

T phng trnh th nht xc nh c cp ,Kt c t phng trnh th hai, v cz t phng trnh th ba.

Vi cc gi tr s khc ca 1 2, ,mp p p v 3p , dng qu o nghim l khc nhau. Di y l mt s v d, vi cc qu

o nghim tng ng trong hnh 26.12 - 26.14(a) 1 2 30.05, 2m p p p p= - = = = -

11.70, 5.95, 0.68c cK p z= =- = -

HNH 26.12 Qu o nghim cho v d 3(a)

HNH 26.13 Qu o nghim cho v d 3 (b)

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S tay C in t

HNH 26.14 Qu o nghim cho v d 3(c)

(b) 1 2 30.5, 1, 2, 3m p p p p= - = - = - = -

8.25, 5.50, 0.73c cK p z= = - = -

(c) 1 2 30.5, 11, 4 1, 4 1m p p p j p j= - = - = - + = - -

35, 14, 5.343c cK p z= = - = -

V d 4

Xt hm truyn ca h h:

2

2

( 3 3)( )( ) ( )

( 3 3)( )c

c

c

s s s z P sC s K

s s s s p

- + -=

+ + -

Trong cK l h s khuych i ca b iu khin, cz v cp l im khng v im cc tng ng. Thy rng qu o

nghim c bn nhnh tr trng hp c cz p= . Cho cc im cc tri mong mun ca h kn l 1,2 0.4r = - . Sai s trng thi

tnh cho u vo tham chiu dc n v l:c

ss

c c

pe

K z=

Ty theo chng ta mun to t l /c c cK z p ln nh c th.

Phng trnh c tnh l:

2 2( ) ( 3 3)( ) ( 3 3)( )c c cs ss s s p K s s s z c = + + - + - + -

v c dng mong mun l:

23 4( ) ( 0.4) ( )( )s s s r s r c = + - -

cho cc gi tr 3,4r

vi 3,4Re( ) 0r < , c ngha l:

0.4 0.4( ) 0, ( ) 0s sd

s sds

c c=- =-= = (26.15)

Cc iu kin (26.15) a ra hai phng trnh:

0.784(0.4 ) 4.36 (0.4 ) 0c c c p K z + - + =

4.36 0.784 1.08(0.4 ) 3.8 (0.4 ) 0c c c cK p K z - - + + + =

T zc vpc c gii theoKc. sau bng cc php thay th n gin, t s c dt ln nht, Kczc/pc c th gim ti:

3.4776 0.784

24.2469 3.4776c c c

c c

K z K

p K

-=

-

Gi tr ln nht ca cK l 0.1297; t = - 0.9508cp v = - 1.1637cz . Cho b iu khin ny, cc im cc ca hhn hi l:

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Phng php qu o nghim

{ 1.64 0.37, 1.64 0.37, 0.40, 0.40}j j- - - - -

Qu o nghim c biu din trong hnh 26.15.

HNH 26.15 Qu o nghim cho v d 4

26.4 Qu o nghim b sung phn trc, thng s Kca qu o nghim c gi thit l dng, pha v quy tc ln c thit lp da trn gi

thit ny. C mt s trng hp m h s khuych i ca b iu khin cng c th m. V vy, hnh nh hon chnh nhnc bng cch v qu o nghim thng thng (vi K>0) v qu o nghim b sung (vi K


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S tay C in t

HNH 26.16 Qu o nghim b sung cho v d 3

HNH 26.17 Qu o nghim thng thng v qu o nhim b sung cho v d 4

V d 4

Trong v d ny, nuKtng t - n + , cc cc ca h kn s di chuyn dc theo qu o nghim b sung, v sau l qu o nghim thng thng, nh biu din trong hnh 26.17.

26.5 Qu o nghim cho h thng c tr thi gian

H thng iu khin phn hi chun c xt trong phn ny c ch ra trong hnh 26.18, trong b iu khin Cvi tngPc dng:

( )( )

( )c

c

N sC s

D s=

v

0 0

( )( ) ( ) ( )

( )phs

P

N sP s e P s where P s

D s

-= =

vi ( ,c cN D ) v ( ,p pN D ) l cc cp a thc cng nguyn t vi cc h s thc.2S hng hse- l hm truyn t ca thnhphn tr (trong hnh 26.18 u vo ca i tng b tr h giy). Ni chung, i tng c tr thi gian khi c:

Tr trong qu trnh x l ca cm bin Tr ca phn mm ca b iu khin

Tr truyn ti trong qu trnh x l

Trong trng hp ny, hm truyn t h l:

0( ) ( ) ( )hs

hG s G s e P s-= =

2 Mt cp a thc c gi l cp cng nguyn t nu chng khng c chung nghim.14


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Phng php qu o nghim

Trong 0 0( ) ( ) ( )G s P sC s= tng ng vi trng hp khng tr 0h =

Ch rng bin v pha ca ( )G jw c xc nh t cc ng nht thc:

0( ) ( )G j G jw w= (26.18)

0( ) ( )G j h G jw w w = - + (26.19)

HNH 26.18 H thng phn hi vi mt tr thi gian

Tnh n nh ca cc h thng tr

Tnh n nh ca h thng phn hi c ch ra trong hnh 26.18 tng ng vi vic c tt c c cc nghim ca:

( ) ( ) ( )hss D s e N sc -= + (26.20)

na tri mt phng hi, - , trong ( ) ( ) ( )c pD s D s D s= v ( ) ( ) ( )c pN s N s N s= . Ta gi thit rng khng c s trit tiu

cc im cc_khng khng n nh khi thc hin 0( ) ( )P sC s , v rng deg( ) deg( )D N> ( yNvD khng cn l a thc

monic). Thc ra ( )sc khng phi l mt a thc v n l mt hm siu vit cas. Cc hm c dng (26.20) thuc mt lphm c bit gi l cc gia thc (quasi-polynomials). Cc im cc ca h kn l nghim ca (26.20).

bit (xem 1, 10):

Nu kr l nghim ca (20), th kr cng th (c ngha l cc nghim xut hin theo dng cp lin hp phc nhthng thng)

C nhiu v hn im cc , 1,2, ,kr k = K tha mn ( ) 0krc =

V cckr

c th c lit k theo cch mRe( 1) Re( )

k k

r r+ ; hn na

Re( )k

r - khi k .

V d

Nu ( ) /hshG s e s-= , th cc im cc ca h thng vng h kr , vi 1,2, ,k= K l cc nghim ca:

21 0k kh jh

j k

k k

e ee

j

s wp

s w

- -+ =

+(26.21)

Trong k k kr js w= + vi mt s ,k ks w . Ch rng 2 1j ke p = vi mi 1,2, ,k= K Phng trnh 26.1 tngng vi tp cc phng trnh sau:

khk ke j

s s w- = + (26.22)

(2 1) ( ),k k kk h j p w s w - = + + (26.23)

iu ny kh l th v v vi 0h = ch c mt nghim 1r=- , nhng thm ch vi 0h > rt t cng c v hn nghim.T iu kin v ln (26.22), c th biu din:

0 1k ks w (26.24)

Cng nh vy, vi 0ks , pha ( )k kjs w + nm gia / 2p- v / 2p+ , v vy (26.23) a ra:

02k k

hp

s w (26.25)

Bng cch kt hp (26.24) v (26.25), c th chng minh rng h thng phn hi khng c nghim trong na mt phngng bn phi khi / 2h p< . Hn na, h thng khng n nh nu / 2h p . C th l, vi / 2h p= c hai nghim trn

trc o, ti 1j . Cng c th d dng thy rng, vi bt k 0h > , khi k , cc nghim hi t v:

15


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S tay C in t

1 2ln 2k

kr j k

h h

pp

-

Khi 0h , ln ca cc nghim tin ti .

Nh biu din v d trn, (iii) chng t rng vi s thc s cho bt k, ch c hu hn nghim kr trong min ca mtphng phc.

: { : Re( ) }s ss s=

C th vi 0s = , iu ny c ngha l gi a thc ( )sc c th c hu hn nghim trong na mt phng phi. V nhhng ca cc cc ca h kn c cc phn thc m rt ln b b qua (n mc ng x vo ra ca h kn b lin quan), nnch c hu hn nghim tri kr vi 1, ,k m= K phi c tnh ton cho mi mc ch thc t.

Cc nghim tri ca mt gi a thc

Ta xt vn sau: cho ( ), ( )N s D s v 0h , tm cc nghim tri ca gi a thc:

( ) ( ) ( )hss D s e N sc -= +

Vi mi gi tr c nh 0h > , c th thy rng tn ti axms sao cho ( )sc khng c nghim no trong vng axms , xem

[11] vi mt thut ton n gin c lng axms

, da trn tiu chun Nyquist. Cho 0h > v mt vng mt phng phcc nh ngha bi min axRe( ) mss s , vn l tm nghim ca ( )sc trong min ny.

R rng, mt im r js w= + trong l nghim ca ( )sc nu v ch nu:

( ) ( )h jhD j e e N js ws w s w- -+ =- +

Ly bnh phng ln ca c hai v ca cng thc trn, ( ) 0rc = suy ra

2( ) : ( ) ( ) ( ) ( ) 0hA x D x D x e N x N x ss s s s s-= + - - + - =

trong x jw= . S hng ( )D xs + c mt vi hm ( )D s c c tnh xs + . Cc s hng khc ca ( )A xs cng c

tnh tng t. Vi mi gi tr s c nh, hm ( )A xs l mt a thc ca bin x. Nu x l mt im khng ca ()As , th ( )x- cng l mt im khng.

Nu ( )A xs c mt nghim xl c phn thc bng 0, t r xs= +l l . Tip theo, c lng ln ca ( )rc l ; nu bng 0,

th nl l nghim ca ( )sc . Ngc li, nu ( )A xs khng c nghim trn trc o, th ( )sc khng th c nghim vi phn thc

l gi tr c nh ca s m t ()As c xy dng.

Thut ton

Cho min( ), ( ), ,N s D s h s v axms :

Bc 1. Ly cc gi tr 1, , ns s nm trong khong mins v axms sao cho min 1, 1i is s s s= = + , v axM ms s= . Vimi gi tr is thc hin cc th sau.

Bc 2. Xy dng a thc ( )iA x theo:

2( ) : ( ) ( ) ( ) ( )ihi i i iA x D x D i x e N x N x ss s s s-= + - - + -

Bc 3. Vi mi nghim trc o xl ca iA , thc hin php kim tra sau:

Kim tra nu c ( ) 0i xc s + =l , th ir xs= + l l nghim ca ( )sc ; nu khng th loi bxl .

Bc 4. Nu i M= th dng, nu khng th tng i thm mt v nhy ti bc hai.

V d

Ta s tm cc nghim tri ca:

1 0hse

s

-

+ =

Vi mt tp cc gi tr gii hn ca h . Nh rng (26.26) c cp nghim 1j khi / 2 1.57h p= = . Hn na, ccnghim tri ca (26.26) u na mt phng phi nu 1.57h > , v chng bn na mt phng tri nu 1.57h < . Do ,

16


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Phng php qu o nghim

vi, (1.2,2.0)h ta cn cc nghim tri gn trc o. Ly min 0.5s = - v ax 0.5ms = , vi 400M = khong cch tuyntnh is gia chng. Trong trng hp ny:

2 2 2( ) ihi i A x e x ss -= - -

Ngay c khi 2 2ih ies s- , ( )iA x c hai nghim:

2 2

, 1,2

ih

l i x j e

s

s

-

= - =lVi mi gi tr c nh is tha mn iu kin ny, t ir xs= +l l (cn nh rngxl l hm ca is , do rl l mt hm

ca is ) v c lng:

( ) : 1lhr

i

l

ef

rs

-

= +

Nu ( ) 0if s = , th rl l nghim ca (26.26). Cho 10 gi tr khc nhau ca (1.2,2.0)h , hm ( )f s c v trn hnh26.19. Hnh ny ch ra cc gi tr kh thi ca is vi rl (c nh ngha t is ) l mt nghim ca (26.26).

HNH 26.19. Tm cc nghim tri

HNH 26.20 Cc nghim tri khi h thay i t 1.2 n 2.0

Cc nghim tri ca (26.26), khi h thay i t 1.2 n 2.0, c ch ra trong hnh 26.20. Vi 1.57h < , tt c cc nghimu nm trong - . Cho 1.57h > , cc nghim tri u trong + , v vi 1.57h = th cc nghim tri 1j .

Qu o nghim s dng xp x Pad

Trong phn ny chng ta gi thit rng 0h > c c nh v ta c gng nhn c qu o nghim, tng ng vi hs khuych i K khng chc chn/iu chnh c, tng ng vi cc cc tri. Vn ny c th c gii quyt bng vic

tnh theo phng php s cc nghim tri ca a thc quasi:( ) ( ) ( ) hss D s KN s ec -= + (26.27)

17


18/20

S tay C in t

viKthay i, bng cch s dng phng php c gii thiu trong phn trc. Trong phn ny mt phng php khc sdng xp x Pad ca s hng hse- ca tr thi gian s c a ra. C th hn l tm a thc ( )hN s v ( )D s tha mn:

( )

( )hs h

h

N se

D s

- = (26.28)

do cc nghim tri:

( ) ( ) ( ) ( ) 0h hD s D s KN s N s+ = (26.29)

gn ng vi cc nghim tri ca ( )sc , (26.27). Chng ta phi thc hin xp x (26.28) nh th no?bng cch dng ccphp o tnh n nh v bn vng t tiu chun n nh Nyquist, ta c th ch ra rng ta c th xt hm gi sau xc nh ln ca sai s xp x:

max ( ) ( ): sup( ) ( )

jh hh

h

K N j N je

D j N j

w

w

w w

w w

-D = -

Trong axmK l gi tr ti a ca thng sKkhng chc chn/iu chnh c.

Xp x Pad bc l c nh ngha nh sau:

0( ) ( 1)

k k k

h kkN s c h s== -

l

0

( ) k kh kk

D s c h s=

= l

Trong h s kc c tnh t:

(2 )! !, 0,1,...,

2 ! !( )!kk

c kk k

-= =

-

l ll

l l

Xp x bc mt v bc hai dng:

2

2

1 / 2, 1

1 / 2( )( ) 1 / 2 ( ) / 12, 2

1 / 2 ( ) / 12

h

h

hs

hsN sD s hs hs

hs hs

- =

+= - + = + +

l

l

Vi cc s liu cho { ax, , ( ), ( )mh K N s D s }, ta tm bc l nh nht ca xp x Pad, sao cho h dD (hoc'

ax/h mK dD ) vi sai s d cho, hoc sai s tng i 'd cho nh th no? Cu tr li nm kt qu sau y [7]:vi mt bc xp x l cho trc, ta c:

2 1 42 ,

( ) 4( ) 4

2,

jh h

h

ehN j eheN j

eh

w

ww

w

ww

+

-

-

l ll

l

T ta c th gii quyt vn la chn bc xp x bng cch dng th tc sau y:

1. Xc nh tn s xw sao cho:

max ( ) ,( ) 2

K N j

D j

w d

w vi mi xw w

v gi tr khi to 1=l .

2. Vi mi 1l xc nh:

{ }4max ,x ehw w=ll

v v hm

18


19/20

Phng php qu o nghim

2 1max

max

( ) 42 ,

( ) 4( ) :

( ) 42 ,

( ) t

K N j eh

D j eh

K N j

D j eh

w ww

ww

ww w

w

+ F =

l

l

ll

l

HNH 26.21 Nghim tri vi 1=l

3. Kim tra nu:

[0, ]max ( )

xl

w ww d

F (26.30)

Nu ng th dng li, gi tr ny ca l tha mn gii hn sai s yu cu l h dD . Nu khng th tng l ln 1,

v nhy n bc 2. Ch rng v tri ca bt ng thc (26.30) l gii hn trn ca hD .

V ta gi thit ( ) ( )Deg D Deg N> , cui cng gii thut s qua bc 3 vi mt s hu hn 1l . Ti mi ln lp li, ta

phi v hm sai s ( )wF l v kim tra xem gi tr nh ca n c nh hn dhay khng. Thng thng, khi dgim, xw tng,

v ko theo l tng ln. Mt khc, vi cc gi tr l rt ln, ln tng i 0 /c cl ca cc h s tr nn rt ln, khi s cnhng kh khn trong phn tch v m phng. Cng nh vy, khi thi gian tr h tng ln, l phi tng ln gi mc sai sxp x dc nh. y l kh khn c bn lin quan n cc h c tr thi gian.

V d

t 2( ) 1, ( ) 2 2N s s D s s s= + = + + , 0.1h = , v ax 20mK = . Khi , vi ' 0.5d= , p dng th tc trn ta tnh

c 2=l l xp x nh nht tha mn 'ax/h mK dD < . V vy, mt xp x bc hai ca thi gian tr l d on cc

cc tri vi [0,20]K . Hnh 26.21 biu din cc qu o nghim xp x nhn c t xp x Pad ca bc =1,2,3l . C

mt s khc bit ng k gia cc qu o nghim ca =1l v =2l . Trong vng Re(s) -12 , cc nghim tri d on u

xp x ging nhau vi =2,3l , vi [0,20]K . Do ta c ni rng s dng xp x bc cao s khng to ra bt k sai khc

ng k no khi d on ng x ca cc cc tri vi di h s K cho trc.

References

[1] Bellman, R. E., and Cooke, K. L., Differential Difference Equations, Academic Press, New York, 1963.

[2] Dorf, R. C., and Bishop, R. H., Modern Control Systems, 9th ed., Prentice-Hall, Upper Saddle River, NJ, 2001.

[3] Evans, W. R., Graphical analysis of control systems, Transac. Amer. Inst. Electrical Engineers, vol. 67(1948), pp. 547551.

[4] Evans, W. R., Control system synthesis by root locus method, Transac. Amer. Inst. Electrical Engineers,vol. 69 (1950), pp. 6669.

[5] Franklin, G. F., Powell, J. D., and Emami-Naeini, A., Feedback Control of Dynamic Systems, 3rd ed.,Addison Wesley, Reading, MA, 1994.

[6] Kuo, B. C.,Automatic Control Systems, 7th ed., Prentice-Hall, Upper Saddle River, NJ, 1995.[7] Lam, J., Convergence of a class of Pad approximations for delay systems, Int. J. Control, vol. 52 (1990), pp.

9891008.

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S tay C in t

[8] Ogata, K., Modern Control Engineering, 3rd ed., Prentice-Hall, Upper Saddle River, NJ, 1997.

[9] zbay, H., Introduction to Feedback Control Theory, CRC Press LLC, Boca Raton, FL, 2000.

[10] Stepan, G., Retarded Dynamical Systems: Stability and Characteristic Functions, Longman Scientific& Technical, New York, 1989.

[11] Ulus, C., Numerical computation of inner-outer factors for a class of retarded delay systems, Int. J. SystemsSci., vol. 28 (1997), pp. 897904.

20

sổ tay cdt chuong 26-pp qdao nghiem

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