In cuon 1.pdf

Gio trnh L thuyt iu khin t ng 1

CHNG 1 : M T MT H THNG IU KHIN T NG

1.1 Cc khi nim c bn hiu c khi nim v h thng iu khin t ng trc ht ta xem v d

sau

Tuc

bi

My pht in o

thng s v in U, IO2 T P

My tnh

Khng ch tc

Va

Va

Va

L HI

Tn hiu ch oHnh 1.1: S iu khin ca l hi pht in

iu khin l tp hp tt c cc tc ng c mc ch nhm iu khin mt qu

trnh ny hay qu trnh kia theo mt quy lut hay mt chng trnh cho trc.

iu khin hc l mt b mn khoa hc nghin cu nguyn tc xy dng cc

h iu khin.

Qu trnh iu khin hoc iu chnh c thc hin m khng c s tham

gia trc tip ca con ngi, th chng ta gi l qu trnh iu khin v iu

chnh t ng.

Tp hp tt c cc thit b m nh qu trnh iu khin c thc hin gi

l h thng iu khin . 1

Gio trnh L thuyt iu khin t ng 1 Tp hp tt c cc thit b k thut, m bo K hoc C t ng mt qu

trnh no c gi l h thng K hoc C t ng (i khi gi tt l h

thng t ng HTT).

1.2 Cc phn t c bn ca h thng iu khin t ng i tng iu khin (Object), Thit b iu khin (Controller ), Thit b o

lng (Measuring device).

- S tng qut

2

OC

M

-

z(t)

u(t) e(t) x(t) y(t)

Hnh 1.2: S tng qut h thng iu khin t ng Mi h thng iu khin t ng u bao gm 3 b phn c bn :

- Thit b iu khin C (Controller device).

- i tng iu khin (Object device).

- Thit b o lng (Measuring device).

u(t) tn hiu vo ; e(t) Si lch iu khin ; x(t) Tn hiu iu khin ; y(t) Tn

hiu ra ; z(t) Tn hiu phn hi

1.3 Cc nguyn tc iu khin c bn C 3 nguyn tc iu khin c bn :

-Nguyn tc iu khin theo sai lch (Hnh 1.3).

OC

M

-

z(t)

u(t) e(t) x(t) y(t)

Hnh 1.3: S nguyn tc iu khintheo sai lch Tn hiu ra y(t) c a vo so snh vi tn hiu vo u(t) nhm to nn tn

hiu tc ng ln u vo b iu khin C nhm to tn hiu iu khin i

tng O.

Gio trnh L thuyt iu khin t ng 1 -Nguyn tc iu khin theo phng php b nhiu (Hnh 1.4)

3

OC

K

u(t) x(t) e(t)

y1(t)

y(t)

Hnh 1.4: S nguyn tc iu khin b nhiu Nguyn tc b nhiu l s dng thit b b K gim nh hng ca nhiu l

nguyn nhn trc tip gy ra hu qu cho h thng (hnh 1.4).

-Nguyn tc iu khin theo sai lch v b nhiu (Hnh 1.5)

O C

K

u(t) y(t) x(t)e(t)

y1(t)

M

-

z(t)

Hnh 1.5: S nguyn tc iu khin hn hp Nguyn tc iu khin hn hp l phi hp c hai nguyn tc trn, va c hi

tip theo sai lch va dng cc thit b b nhiu.

1.4 Phn loi cc h thng iu khin t ng. 1.4.1 Phn loi theo nguyn l xy dng.

Cc phn t c phn chia thnh cc loi: h thng K theo mch h, h

thng K theo mch kn v h thng K hn hp .

Ngoi nhng nguyn l trn, t nhng nm 60 ca th k XX, trn c s p

dng iu khin hc trong c th sng vo k thut ra i mt loi hnh h

thng t ng m phng hot ng ca c th sng: l cc h t chnh, thch

nghi. Nguyn l t chnh v thch nghi khng i hi phi bit y cc c

tnh ca qu trnh iu khin v trong qu trnh lm vic, cc h thng ny t

chnh v thch nghi vi cc iu kin bn ngoi thay i.

L thuyt cc h K t chnh v thch nghi tr thnh mt nhnh pht trin

quan trng ca l thuyt KT.

Gio trnh L thuyt iu khin t ng 1 V hu ht cc h thng KT trong k thut l nhng h mch kn v qu

trnh iu khin cc thit b k thut chung quy li l qu trnh iu chnh cc

tham s ca n, nu di y chng ta s cp n s phn loi cc h thng

KT mch kn v l thuyt v cc h .

1.4.2/ Phn loi theo tnh cht ca lng vo.

Tu theo tnh cht ca tc ng u vo, cc h thng KT c 3 loi:

H thng n nh t ng (iu chnh theo hng s) l h thng c lng vo

khng i. Nhim v ca h thng l duy tr mt hoc mt vi i lng vt l

gi tr khng i. Th d nh h thng KT tc ng c nhit, h thng

KT in p, tn s ca my pht, h n nh ng bay ca my bay khi gc

li khng thay i ...

H thng iu chnh theo chng trnh l h thng c lng vo l cc hm

bit trc, c th di dng chng trnh.Th d h iu khin ng bay

nh trc ca my bay khng ngi li, h thng iu khin cc my cng c:

bo, phay vi chng trnh nh trc trong b nh my tnh...

H t ng bm, gi tt l h bm l h thng c lng vo l cc hm thi

gian khng bit trc, c th thay i theo quy lut bt k. Nhim v ca h l

bo m lng ra phi "bm" theo s thay i ca lng vo. Th d cc h nh

l h bm ng b gc, cc h bm v tuyn in t ca cc i radar...

1.4.3/ Phn loi theo dng tn hiu s dng trong h thng.

Theo dng tn hiu s dng trong h thng, chng ta c cc tc ng lin tc

v cc h thng gin on (hay h ri rc).

H tc ng lin tc (gi tt l h lin tc) l h m tt c cc phn t ca h

c lng ra l cc hm lin tc theo thi gian.

Tn hiu di dng hm lin tc c th l tn hiu mt chiu (cha bin iu)

hoc tn hiu xoay chiu ( c bin iu) tng ng chng ta c h KT

mt chiu (DC) v h thng KT xoay chiu (AC) (th d h thng bm ng

b cng sut nh dng ng c chp hnh 2 p ha).

4

Gio trnh L thuyt iu khin t ng 1 H tc ng gin on (gi tt l h gin on hay h ri rc) l cc h c

cha t nht mt phn t gin on, tc l phn t c lng vo l mt hm lin

tc v lng ra l mt hm gin on theo thi gian.

Tu theo tnh cht gin on ca lng ra, cc h gin on c th phn chia

thnh cc loi: h thng KT xung, h thng KT kiu r le v h thng

KT s.

Nu s gin on ca tn hiu ra xy ra qua nhng thi gian xc nh (ta gi l

gin on theo thi gian) khi tn hiu vo thay i, th ta c h KT xung.

Nu s gin on ca tn hiu xy ra khi tn hiu vo qua nhng gi tr ngng

xc nh no (chng ta gi l gin on theo mc), th c th KT kiu rle.

H rle thc cht l h phi tuyn, v c tnh tnh ca n l hm phi tuyn. y l

i tng nghin cu ca mt phn quan trng trong l thuyt K .

Nu phn t gin on c tn hiu ra di dng m s (gin on c theo mc

v c theo thi gian), th ta c h KT s. H thng KT s l h cha cc

thit b s (cc b bin i A/D, D/A, my tnh in t (PC), b vi x l.

1.4.4/ Phn loi theo dng phng trnh ton hc m t h thng.

V mt ton hc, cc h thng KT u c th m t bng cc phng trnh

ton hc: phng trnh tnh v phng trnh ng. Da vo tnh cht ca cc

phng trnh, chng ta phn bit h thng KT tuyn tnh v h KT khng

tuyn tnh (phi tuyn).

H thng KT tuyn tnh l h thng c m t bng phng trnh ton

hc tuyn tnh. Tnh cht tuyn tnh ca cc phn t v ca c h thng KT

ch l tnh cht l tng. V vy, cc phng trnh ton hc ca h thng l cc

phng trnh c tuyn tnh ho, tc l thay cc s ph thuc gn ng

tuyn tnh.

H tuyn tnh c phng trnh ng hc vi cc tham s khng thay i th

gi l h KT tuyn tnh c tham s khng thay i, hay h KT tuyn tnh

dng, cn nu h thng c phng trnh vi tham s thay i th gi l h

KT tuyn tnh c tham s bin thin, hay h KT tuyn tnh khng dng.

5

Gio trnh L thuyt iu khin t ng 1 H thng KT phi tuyn l h thng c m t bng phng trnh ton

hc phi tuyn. H phi tuyn l h c cha cc phn t phi tuyn in hnh, th d

l h c cha cc phn t rle.

1.4.5/ Phn loi theo tnh cht ca cc tc ng bn ngoi.

Cc tc ng bn ngoi vo h t ng c quy lut thay i bit trc hoc

mang tnh cht ngu nhin.

H thng tin nh l cc h c cc tc ng bn ngoi l tin nh, tc l

bit trc cc quy lut thay i ca n (th d xt h thng vi cc tc ng in

hnh).

H thng khng tin nh (hay h ngu nhin) l cc h c xem xt nghin

cu khi cc tc ng bn ngoi l cc tn hiu ngu nhin.

1.4.6/ Phn loi theo s lng i lng cn iu khin.

Tu theo s lng cn iu khin (lng ra ca h) chng ta c: h mt chiu

v h nhiu chiu.

H thng KT mt chiu c cha mt i lng cn iu khin, cn h

KT nhiu chiu l h c cha t hai i lng cn iu khin tr ln. Th d

v h nhiu chiu c th l h thng KT mt my pht in, nu h thng

KT cng mt lc iu khin t ng in p v tn s ca n.

Ngoi cc cch phn loi chnh xt trn, tu thuc vo s tn ti sai s

ca h trng thi cn bng, chng ta phn bit hai loi h thng: h thng tnh

(c sai s tnh) v h phim tnh (khng c sai s tnh). Tu thuc vo quy lut

(nh lut) iu khin (tc l dng ca tn hiu iu khin x(t) do c cu iu

khin to ra), chng ta phn bit cc b iu khin t l (b iu khin P), b

iu khin t l vi phn (b iu khin PD), b iu khin vi phn - tch phn

(b iu khin PID).

1.5 Qu trnh thit lp mt h thng iu khin - Bc 1: Chuyn i cc yu cu k thut thnh mt h thng vt l.

- Bc 2: V s khi chc nng. Chuyn i s miu t c tnh h

thng thnh mt s khi chc nng. y l s miu t v cc phn

chi tit ca h thng v mi quan h gia chng.

6

Gio trnh L thuyt iu khin t ng 1 - Bc 3: Thit lp s nguyn l.

- Bc 4: S dng s nguyn l thit lp s khi hoc graph tn

hiu hoc biu din khng gian trng thi.

- Bc 5: Rt gn s khi.

- Bc 6: Phn tch v thit k.

7

Gio trnh L thuyt iu khin t ng 1 Cu hi n tp chng 1 1. H thng iu khin t ng c th phn loi nh th no?2. H thng iu khin c my phn t c bn? 3. Hy nu cc quy tc iu khin c bn iu khin mt h thng iu khin? 4. Nu cc bc thit lp mt h thng iu khin?

8

Gio trnh L thuyt iu khin t ng 1 CHNG 2: M HNH TON HC CA H THNG IU KHIN

Mi h thng c th chia lm nhiu phn s thun tin hn v mi phn s c biu din bng 1 hm ton hc gi l hm truyn t (transfer function)

H thng (System)

u ra u vo

Hnh 2.1 : S phn chia h mt h thng iu khin thnh cc h thng

H thng con (subsystem)

H thng con(subsystem)

H thng con (subsystem)

u ra u vo

2.1 Cc khu c bn Ta c mt h thng iu khin:

9

Hnh 2.2 : S mt h thng iu khin tng qut

B iu khin

C1

E

o lng

i tng

Chp hnh

CR

a phn cc mch phn hi ca h thng iu khin l mch phn hi m. Khi chng ta tin hnh phn tch h thng tt hay xu hay thit k b iu khin cho h thng u phi xut pht t m hnh ton hc ca h thng hay ni cch khc ta phi tm c quan h gia u vo v u ra ca h thng. 2.1.1 Khu khuch i

x y

K

Hnh 2.3 : S khu khuch i tnh - Khu khuch i l tn hiu u ra l khuch i ca tn hiu u vo

y = K.x (2.1) trong : K l h s khuch i ( Khuch i tnh l c c tn hiu u vo th tm c tn hiu u ra)

Gio trnh L thuyt iu khin t ng 1 - Cng c h thng c khuch i nhiu tng

10

x y

K1 K2 K3

Hnh 2.4: S khu khuch i tng 2.1.2 Khu tch phn

)()(1)(0

0 += tti

ydttxT

ty (2.2)

Vi Ti l thi gian tch phn 2.1.3 Khu vi phn

dtdxTy D= (2.3)

TD l hng s thi gian vi phn 2.1.4 Khu bc nht

xKydtdyT .=+ (2.4)

trong : K l h s truyn ca khu T l hng s thi gian ca khu Phn ng ca h thng tt hay xu ph thuc vo h s K, nhanh hay chm ph thuc vo T. 2.1.5 Khu bc hai

)()(22 tKxtydtdyT

dtdyT =++ (2.5)

Trong : K l h s khuch i T l hng s thi gian suy gim tn hiu y l m hnh ton hc ca mch RLC. 2.1.6 Khu bc n

)(...)(... 111

1011

1

10 txbdtxdb

dtxdb

dtxdbtya

dtyda

dtyda

dtyda mmm

m

m

m

nnn

n

n

n

++++=++++

(2.6)

thng thng nm. 2.2 M hnh trong min tn s 2.2.1 Khi nim v php bin i Laplace v ng dng

Gio trnh L thuyt iu khin t ng 1 2.2.1.1 Khi nim v bn cht ca php bin i Laplace :

Khi s dng cc php bin i tn hiu h thng t min thi gian sang min khc thun tin trong vic x l tn hiu. Nh trong h thng lin tc ngi ta hay s dng php bin i Lpalace bin i t min thi gian sang min tn s phc. Cc phng trnh vi tch phn s chuyn i thnh cc phng trnh i s thng thng.

Trong cc h thng ri rc ngi ta hay s dng php bin i Z chuyn tn hiu t min thi gian sang min tn s phc. Trong thc t ngi ta cn s dng cc php bin i khc x l tn hiu nh gii tng quan, m ho c hiu qu, chng nhiu,.

Thc hin cc php bin i c cng c ton hc nh my tnh s, cng c ph bin v hiu qu l phn mm Matlab hay thc hin bin i bng tay. a) Bin i Laplace thun nh ngha: Gi F(s) l bin i Laplace ca hm f(t), khi ta c:

==0

)()]([)( dtetftfsF stL (2.7)

trong : - js += - l ht nhn ca php bin i. ste

- F(s) l hm phc. - f(t) l hm biu din trn min thi gian xc nh trn R.

thc hin c bin i Laplace hm f(t) phi l hm thc v tho mn mt s iu kin sau: - f(t) l hm gc khi tho mn cc iu kin sau: 1. f(t) = 0 khi t < 0 2. f(t) lin tc khi t0, trong khong hu hn bt k cho trc ch c hu hn cc m cc tr. 3. Hm f(t) gi l hm bc s m khi t nu tn ti mt s thc 0 v M >0 th 0,)( > tMetf t , c gi l ch s tng ca hm f(t). Khi hm f(t) l hm bc s m nu hm f(t) tng khng nhanh hn hn hm et.

- Nu f(t) l hm gc c ch s tng th tch phn s hi t

trong min Re(s) = > . Khi s l mt hm phc.

+ =0

)( dttfeI st

)()(0

sFdttfeI st == + 11

Gio trnh L thuyt iu khin t ng 1 V d 1: Tm nh ca hm gc sau

f(t)

Gio trnh L thuyt iu khin t ng 1 Gi s f(t) c nh Laplace dng sau

nn

mm

sasaasbsbb

sAsBsF +++

+++== LL

10

10

)()()( (2.9)

vi n m. Cc bc thc hin nh sau: Bc 1: Phn tch F(s) thnh tng cc hm phn thc ti gin

= = +

+++=l

k kk

kkkkr

ii

k

ki

sCsB

asAAsF

k

122

1 )()(

)()(

(2.10) trong A, Aki, Bk, Ck l cc hng s. ak l im cc thc bi rk v kk j + l im cc phc ca F(s), ni cch khc chng l im m ti F(s) = . Bc 2: Xc nh hm gc cho tng phn t.

- L -1{ } )(tAA = - L -1 )(1

)!1()(

1

ti

etAas

A taikii

k

kik

=

- L -1 )(1)cos()(

)(22

tteBs

sBk

tk

kk

kk k =

+

- L -1 )(1)sin()( 22

tteCs

Ck

tk

kk

kk k =

+

V d 1: Tm hm gc f(t) ca nh Laplace sau

)1(1)( 2 += sssF

Gii: Bc 1: Phn tch thnh tng cc phn thc ti gin

2

111

1)(sss

sF ++= Bc 2: Xc nh hm gc cho tng thnh phn

f(t) = (e t 1 + t)1(t) V d 2:

2762)( 2

23

+++++=

ssssssF

Ta thc hin chia t s cho mu s cho n khi s d cn li c bc ca t nh hn bc ca mu. 13


221)( 2 ++++= ssssF

Thc hin bin i Laplace ngc c s dng bng bin i Laplace

++++=

52)()()( 2

1

sst

dttdtf L

S dng phng php phn tch 5

2)( 2 ++= sssX thnh tng cc phn thc n gin. Ta xt mt s trng hp sau: Trng hp 1: Nghim ca mu thc T(s) l thc v ring bit. Gi s nghim ca mu thc T(s) c hai nghim s1 = -1 v s2 = - 2.

)2)(1(2)( ++= sssX

Nghim ca mu thc l ring bit nn tng phn thc s c bc l 1.

21)2)(1(2)( 21 +++=++= s

KsK

sssX

tm K1 ta nhn (2.) vi (s+1) tch K1 ring ra

)2()1(

)2(2 2

1 +++=+ s

KsKs

Sau cho s - 1, rt ra c K1 = 2. Lm tng t v cho s - 2 ta rt ra c K2 = - 2. Lc

22

12

)2)(1(2)( ++=++= sssssX

Thc hin bin i Laplace ngc ca X(s) ta c )()22()( 2 tueetx tt =

Mt cch tng qut khi mu s ca F(s) cos nghim thc v ring bit, ta thc hin nh sau:

)()()()(

)()())(()(

)()()(

2

2

1

1

21

n

n

m

m

nm

psK

psK

psK

psK

pspspspssB

sAsBsF

+++++++++=++++==

LL

LL (2.11)

Nu bc ca t nh hn bc ca mu ta thc hin tm cc h s Ki nh sau: - Nhn hai v vi (s + pi) tm h s Ki. - Cho s - pi, rt ra c Ki.

14

Gio trnh L thuyt iu khin t ng 1 Trng hp 2: Mu s c nghim thc v lp li. Gi s nghim ca mu thc T(s) c ba nghim s1 = -1 v s2,3 = - 2. Lc ta phn tch X(s) nh sau:

)2()2(1)2)(1(2)( 32

212 +++++=++= s

Ks

KsK

sssX

Tm cc h s K1, K2 v K3

2)2(

2

121 =+= ss

K

tm K2 ta nhn hai v ca (2.) vi (s + 2)2

321

2

)2(1

)2()1(

2 KsKs

Kss

+++++=+

Khi cho s - 2 ta tm c K2 = - 2 Tm K3 bng cch ly o hm (2.) theo bin s ta c

3122 )1()2(

)1(2 KK

sss

s++

+=+

Cho s - 2 ta rt ra c K3 = - 2. Thay K1, K2 v K3 ta c

)2(2

)2(2

12

)2)(1(2)( 22 +++=++= ssssssX

Thc hin bin i Laplace ngc ta c )()222()( 22 tueteetx ttt =

Tng qut cho trng hp ny

)()()()()(

)()()()(

)()()(

211

1

2

1

1

21

n

nrrrr

nr

psK

psK

psK

psK

psK

pspspssB

sAsBsF

+++++++++++=+++==

LL

L (2.12)

thc hin c phi c iu kin bc ca t nh hn bc ca mu v c r nghim bi ti - p1. tm K1 n Kr cho phn thc c nghim bi, u tin ta nhn hai v (2. 12) vi (s + p1)r ta c

)()(

)()(

)()()(

)()()()()(

)()()(

1

2

11

113

21211

21

1111

n

nr

rr

rr

nr

rr

psKps

psKps

KpsKpsKpsKpspsps

sBpssFpssF

+++++

++++++++++=

++++=+=

+

L

LL

(2.13)

15

Gio trnh L thuyt iu khin t ng 1 Ta c th tm ngay c K1 khi cho s - p1. tm K2 ta ly o hm (2.12) theo bin s v cho s - p1. Ln lt ly o ta tm c K3 n Kr. Cng thc chung tm K1 n Kr l:

1!0,1)()!1(

1

1

11

1

===

rids

sFdi

Kps

i

i

i (2.14)

Trng hp 3: Mu thc c nghim phc hay nghim o. Gi s mu s ca F(s) c nghim phc.

)52(3)( 2 ++= ssssF

F(s) c th phn tch thnh cc phn thc nh sau

52)52(3

2321

2 ++++=++ ss

KsKs

Ksss

D dng tm c K1 = 3/5 khi cho s 0. tm K2 v K3 ta quy ng phn thc vi mu s chung nh nht l b c cc phn thc )52( 2 ++ sss

356

533 3

22 +

++

+= sKsK

Thc hin ng nht thc hai v ta c

560

56

530

53

33

22

==

+

==

+

KK

KK

Thay cc h s ta c

522

535

3

)52(3)( 22 ++

+=++= sss

sssssF

T bng tra nh ca tch hm m v hm sin v cos

{ } 22)( )(cos ++ += as asAtAe atL V

{ } 22)(sin ++= as BtBe atL Cng hai cng thc trn ta c

{ } 22)( )(sincos ++ ++=+ as BasAtBetAe atatL Ta a cng thc (2.) v dng trn

16

Gio trnh L thuyt iu khin t ng 1 ( ) ( )( )( ) 222 21

2211

535

3

)52(3)( ++

++=++= ss

sssssF

Tra bng ta tm c hm gc nh sau

+= ttetf t 2sin212cos

53

53)(

Trong trng hp trn ta cng c th thc hin n gin bng cch phn tch thng thng

2121

)21)(21(3

)52(3)(

321

2

jsK

jsK

sK

jsjssssssF

+++++=+++=++=

K1 d dng tnh c v bng 3/5.

)2(203

)21(3

212 jjss

Kjs

+=+=

Tng t ta tm c K3 l nghim phc lin hp ca K2. Ta c

++++

++=21

221

22035

3)(

jsj

jsj

ssF

T ta tm c hm gc nh sau

( ) ( ) ( ) ( )[ ]

+

+=

++=

+

jeeeee

ejejtf

tjtjtjtjt

tjj

22

24

203

53

22203

53)(

2222

2121

p dng cng thc le ca hm sin v cos

jee

ee

tjtj

tjtj

2sin

2cos

22

22

=

+=

Suy ra

+= ttetf t 2sin212cos

53

53)(

17

Gio trnh L thuyt iu khin t ng 1 Bin i Laplace mt s hm n gin:

x(t) X(s) X(t) X(s)

(t) 1 )!1n(

et t1n

n)s(

1

+

1(t) s

1 sint 22s +

tu(t) 2s

1 cost 22ss

+

tnu(t) 1ns

!n+ sin(t)e-t 22)s( ++

e-t +s1 cos(t)e-t 22)s(

s

+++

btat ee ))(( bsasab++

)()(1

abbe

abae

ab

btat

))((

1bsass ++

2.2.1.2 Cc tnh cht ca php bin i Laplace :

1. Tnh cht tuyn tnh: L[a.f(t)]= a.L[f(t)] = a.F(s). 2. Tnh cht xp chng: Nu f1(t) v f2(t) c nh bin i Laplace l F1(s) v

F2(s) th ta c: L[f1(t) f2(t)] = L[f1(t)] L[f2(t)] = F1(s) F2(s)

V d : Tm nh ca hm hm f(t) = cosat trong a l hng s. Theo cng thc le ta c

jatjatjatjat

eeeeat

+=+=21

21

2cos

Thc hin php bin i Laplace

{ } 2222211

211

21

21

21cos

ass

asjasjas

jasjaseeat jatjat +=+

++=++=

+= LL

3. Tnh cht tr (Chuyn dch thi gian -Translation in time): Nu f(t) c nh l F(s), a l mt s thc v f(t-a) =0 khi 0

Gio trnh L thuyt iu khin t ng 1 V d: Tm nh Laplace ca hm gc c th nh sau

f(t)

1 2

1

0

-1

2

Gio trnh L thuyt iu khin t ng 1 2.2.1.3 ng dng ca php bin i Laplace a) ng dng gii phng trnh vi phn tuyt tnh. Khi chuyn phng trnh vi phn t min thi gian sang min nh phc tr thnh phng trnh i s. Sau khi gii ra c nghim ta chuyn ngc v min thi gian. V d 1: Gii phng trnh vi phn sau vi cc s kin u bng khng.

uydtdy

dtyd 3232122

2

=++ chuyn sang min nh Laplace vi y(0-) = 0 v 0)0( =y&

ssYssYsYs 32)(32)(12)(2 =++

Rt Y(s) ra ta c

)8)(4(32

)3212(32)( 2 ++=++= sssssssY

Phn tch Y(s) thnh tng cc phn thc ti gin

84)8)(4(32)( 321 ++++=++= s

KsK

sK

ssssY

Tm cc h s K1, K2 v K3.

1)4(

32

2)8(

32

1)8)(4(

32

81

41

01

=+=

=+=

=++=

s

s

s

ssK

ssK

ssK

Vy

81

421)( +++= ssssY

Thc hin bin i Laplace ngc ta tm c )()21()( 84 tueety tt +=

Trong cng thc trn c cha u(t) ni ln rng cc p ng s bng 0 cho n khi t = 0. V vy cc p ng u ra cng bng 0 cho n kho t = 0. thun tin ta c th b k hiu u(t) i, vy p ng u ra c th vit nh sau

tt eety 8421)( += V d 2: Gii phng trnh vi phn bng ton t Laplace sau

02322

=++ ydtdy

dtyd

vi s kin y(+0) = a v bdt

dy =+ )0( Chuyn c hai v sang min nh phc nh ton t Laplace

20


[ ]

212

)2)(1()3(

)23()3()(

)3()()23(

0)(2)0()(3)0()0()(

2

2

2

+++

+=++++=++

++=++=++

=+++

++

sba

sba

ssbaas

ssbaassY

baassYss

sYyssYdt

dysysYs

Thc hin bin i Laplace ngc rt ra c y(t) tt ebaebaty 2)()2()( ++= vi t 0.

V d 3: Gii phng trnh vi phn sau

35222

=++ ydtdy

dtyd

vi s kin 0)0()0( =+=+dt

dyy

Thc hin bin i Laplace

[ ] [ ]222222

2)1(5)1(3

2)1(1023

53

523)(

3)()52(

+++++

=++=

=++

ss

sssssY

ssYss

Suy ra )2cos(

53)2sin(

103

53)( tetety tt = vi t 0.

b) Gii mch in Cho mch in sau

Gi s khi mch in ng ti thi im t 0 th vC(0) = 1.0V. Tm dng in i(t) chy trong mch in. (trong V(t) = 5V, C = 1F, R = 1k) Gii: Ta c phng trnh sau

+= idtCRitv 1)( hay

+= idtRCitCv )( thay cc thng s u bi cho vo

+=+=

idti

idti36

636

1010.5

10.1010.5


21

Gio trnh L thuyt iu khin t ng 1 [ ]

++= =

s

idt

sII

st 03

6

1010.5

Theo u bi vC(0) = 1.0V nn ta c [ ] [ ][ ] 6

0

060

10

110

11)0(

=

==

====

t

ttC

idt

idtidtC

V

Thay vo cng thc trn ta c

( )1000110.4

10110.4

10.4101

10.41010.5101

101010.5

33

6

63

6663

63

6

+=+==+

==

+

++=

ssI

Issss

Is

ssII

s

Thc hin tra bng bin i Laplace ta tm c i(t) nh sau teti 1000310.4)( =

2.2.2 Hm s truyn ca h thng KT.

Nhm n gin ho cc phng php phn tch v tng hp h thng t ng ngi ta thng chuyn phng trnh ng hc ca h dng phng trnh vi phn vit vi cc nguyn hm x(t), y(t) thnh phng trnh vit di dng cc hm s X(s), Y(s) thng qua php bin i Laplace.

V d xt hm s x(t) hm s ca bin s t (bin s thc, y t l thi gian) ta gi l nguyn hm. Ta cho php bin i hm s x(t) thng qua tch phn:

=0

.).()( dtetxsX st (2.15)

trong : s = + j - bin s phc, bin i (2.15) hm x(t) thnh hm bin s X(s) c gi l l bin Laplace, v X(s) c gi hm nh. Nh vy hm nh l mt hm bin s phc s. Php bin i Laplace c k hiu sau:

L{x(t)}=X(s) hoc x(t) X(s) Gi s nguyn hm x(t) c cc iu kin ban u khng, tc l vi t=0 gi tr

ca hm x(t) v cc bc o hm dix(t) / dti vi i = 1, 2, 3, , (n-1) u bng 0, tnh theo tnh cht ca php bin i Laplace (nh l v nh o hm ca nguyn hm) chng ta c:

ni

sXsadt

txdaL iiii

i

,,3,2,1

)(..)(

L==

(2.16)

22

Gio trnh L thuyt iu khin t ng 1 Nhn hai v ca phng trnh (2.6) vi e-st , sau ly tch phn theo t t 0

n , tc l ly bin i Laplace ca hai v phng trnh, vi gi thit rng cc hm x(t), y(t) c cc iu kin ban u bng 0, da theo tnh cht tuyn tnh ca php bin i Laplace , phng trnh (2.6) s c dng:

)()()()(

)()()()(

11

10

11

10

sXbsXbsXsbsXsbsYassYasYsasYsa

mmmn

nnnn

++++==++++

LL (2.17)

y, Y(s), X(s) l cc bin i Laplace ca hm lng ra v hm lng vo ca h.

Phng trnh (2.17) c gi l phng trnh ng hc m t quan h vo ra ca h vit di dng ton t Laplace.y l phng trnh i s, vi n v m l cc s m ca bin s s gii phng trnh (2.17) ng vi lng ra Y(s).

)()(1

110

11

10 sXasasasabsbsbsbsY

nnnn

mmmm

++++++++=

LL (2.18)

Chng ta k hiu:

nnnn

mmmm

asasasabsbsbsbsW ++++

++++=

11

10

11

10)(LL (2.19)

v gi biu thc i s ny l hm s truyn (hoc hm truyn t) ca h thng t ng (hay ca mt phn t ca n).

Khi Y(s) = W(s)X(s) (2.20) Hoc W(s) = Y(s) / X(s) (2.21)

Vy hm s truyn (H S T) ca h thng (hay ca mt phn t ) t ng l t s hm nh ca lng ra vi hm nh ca lng vo ca n (qua php bin i Laplace) vi gi thit tt c cc iu kin u bng khng.

Biu thc (2.19) cho chng ta thy, HST l mt hm phn s hu t ca bin s, c bc cc a thc tho mn m n. Gi thit iu kin ban u ca cc hm lng vo v lng ra u bng khng l ph hp vi iu kin thng gp trong cc h thng KT.

Phng trnh (2.20) cho php xc nh hm nh ca lng ra nu bit hm nh ca lng vo v biu thc HST ca h. Nh vy HST hon ton xc nh cc tnh cht ng hc ca h thng. xc nh nguyn hm ca lng ra, tc l xc nh y(t) khi bit x(t) c th bin i ngc Laplace, theo :

[ ] +

==

j

j

st dsesYj

sYLty ).(2

1)()( 1 (2.22)

23

Gio trnh L thuyt iu khin t ng 1 l phng php ton t gii phng trnh vi phn. Nu Y(s) l hm

n gin,chng ta c th s dng bng bin i Laplace ca cc hm n gin in hnh, c trong ph lc cc sch ni v bin i Laplace, tra cu nguyn hm y(t). Nu hm nh Y(s) l hm phc tp, cn phn tch chng thnh t hp tuyn tnh cc hm n gin, m chng ta bit nguyn hm ca n. Nguyn hm y(t) chnh l t hp tuyn tnh ca cc nguyn hm thnh phn.

2.2.3 Hm truyn t ca mch in

Trong mch in c cc phn t c bn l in tr (R), in cm (L) v t in (C).

a) in tr R

Hnh 2.5: in tr

in p ri t l thun vi cng dng in I chy qua in tr:

RZtvR

tiRitv === )(1)()(

Thng qua php bin i Laplace ta c c hm truyn ca in tr l

RUI 1G R == (2.23)

b) in cm L

Hnh 2.6 : in cm L

24


in p ri trn in cm l

== 0

)(1)()()( dttvL

tidt

tdiLtv (2.24)

Thng qua bin i Laplace ta tnh c tr khng Z v hm truyn ca in cm L

LsUIGLsZ

LL

1=== (2.25)

c) T in C

Hnh 2.7 : T in C

in p ri trn in dung l

dttdvCtidtti

Ctv )()()(1)(

0

== (2.26)

Tr khng v hm truyn t ca t in

CsU

IGC

ZC

C === 1 (2.27)

d) Cc phn t R, L v C mc ni tip

Hnh 2.8 : S cc phn t mch in RLC mc ni tip

25


Vrrr

rrr

rV

UUdt

dURCdt

UdLC

dtUdC

dtdi

dtdUCiidt

CU

UdtdiLRiU

=++

===

++=

2

2

2

2

0

1 (2.28)

Thc hin php bin i Laplace ta c

(LCs2 + RCs + 1) Ur = Uv (2.29)

Rt ra c hm truyn l:

LCs

LRs

LCUUsG

V

r

11)(

2 ++== (2.30)

e) Cc phn t mc song song

UV UrR L C

I

Hnh 2.9: S cc phn t mch in RLC mc song song

Dng in ca mch in l

ZUI = (2.31)

Tng tr ca mch song song c tnh l

RLsRLsRLCs

CsLsRZ++=++=

2

/11111 (2.32)

Hm truyn ca h thng l

26


sC

LCs

RCs

ZUIsG

1

111)(

2 ++=== (2.33)

2.2.4 Hm truyn ca h thng c kh

2.2.4.1 Phn t chuyn ng thng

a) L xo

Hnh 2.10: S biu din l xo

trong : K l h s n hi ca l xo

Nu ta n l xo c chiu di L, di ng c mt lng X th cn mt lc tc ng ln l

F(t) = Kx(t) (2.34)

Thng qua bin i Laplace to c hm truyn ca l xo nh sau:

KsXsFGloxo == )()( (2.35)

b) B gim chn du p (khng kh)

Hnh 2.11: S biu din b gim chn du p

di ng pt tng vi vn tc v, ta cn tc ng ln mt lc l f

dttdxftvftf vv)()()( == (2.36)

27

Gio trnh L thuyt iu khin t ng 1 trong fv l h s gim chn

Thc hin bin i Laplace

sfsXsFG vVD == )()( (2.37)

c) Trng khi

Hnh 2.12: S biu din trng khi

Theo nh lut II Newton tng cc lc bn ngoi tc ng vo mt trng khi s bng tch ca trng khi v gia tc ta c

=== 22 )()( dt txdMdttdvMMaf (2.38) Thc hin php bin i Laplace ta c hm truyn ca trng khi l

2

)()( Ms

sXsFGM == (2.39)

d) Thit b gim chn

Thit b gim chn bao gm trng khi l x0 - b gim chn

Hnh 2.13: S biu din thit b gim chn

28

Gio trnh L thuyt iu khin t ng 1 tm c hm truyn ca h thng trc tin ta v biu din cc lc tc

ng trng khi

Hnh 2.14: S biu din lc tc ngln trng khi

S dng nh lut Newton vit phng trnh chuyn ng

)()()()(22

tftKxdt

tdxfdt

txdM v =++ (2.40)


( ) )()(2 sFsXKsfMs v =++ (2.41) T ta rt ra hm truyn ca h thng l

KsfMssFsXsG

v ++== 2 1)(

)()( (2.42)

2.2.4.2 Phn t chuyn ng quay

Theo nh lut II Newton v chuyn ng quay th gia tc gc ca vt quay t l thun vi tng momen tc ng ln n, ta c phng trnh sau

2

2

dtdJM = (2.43)

trong :

J l mmen qun tnh tc ng ln vt.

l v tr gc quay ca vt th

M l m men tc ng ln vt

29

Gio trnh L thuyt iu khin t ng 1 Cc mmen bn ngoi c to bi ng c do ti trng tc ng ca l xo

hoc vt gim chn. Hnh biu din s ca mt a quay trong cht lng lm cho trc lp trn n b bin dng i mt gc .

Nu ta quay a vi mmen xon x, trc s quay i mt gc to nn mmen ca l xo xon:

M1 = k (2.44)

Mmen cn thit thng lc ma st ca cht lng:

dtdCM =2 (2.45)

trong C l h s ma st ca cht lng

Nh vy ta c phng trnh:

2

2

21 dtdJMMxM == (2.46)

Thay vo ta c:

kdtdC

dtdJx ++= 2

2

(2.47)

2.2.5 S tng ng gia h c kh vi mt mch in

S tng ng gia mch c kh v mch in trng khi = M in cm = M b gim chn = fv in tr = 1/fvl xo = K in dung = 1/ K lc tc ng = f(t) ngun p = f(t) vn tc = v(t) dng vng = v(t)

30


Hnh 2.15: S biu din s tng ng gia mch c kh v mch in

Khi so snh vi dng vng ta c mch tng ng ni tip, nu dng

phng php nt, th mch tng ng ng l mch song song. Phng trnh chuyn ng l

)()()( 2 sFsXKsfMs v =++ (2.48) i vi mch RLC ni tip l

)()(1 sEsICs

RLs =

++ (2.49) hai cng thc trn khng tng thch vi nhau do khong cch v dng in khng tng thch vi nhau. Ta bin i s tng thch bng cch chuyn i t khong cch sang vn tc

)()()()(2

sFsVsKfMsssX

sKsfMs

vv =++=++ (2.50)

Ta cng c th chuyn i sang h song song trng khi = M in cm = M b gim chn = fv in tr = 1/fvl xo = K in dung = 1/ K lc tc ng = f(t) ngun dng = f(t) vn tc = v(t) in p nt = v(t)

Cng thc mch song song l )()()11( sIsE

LsRCs =++ (2.51)

31

Gio trnh L thuyt iu khin t ng 1 2.2.6 Hm truyn ca cc phn t in t

Hnh 2.16 : Biu din phn t khuch i thut ton

- Sai lch in p u vo: v2(t) v1(t). - Tr khng u vo cao: Z1 = (l tng). - Tr khng u ra thp: Z0 = 0 (l tng). - H s khuch i cao A = (l tng). in p u ra c tnh l

v0(t) = A(v2(t) v1(t)) (2.52) Nu v2(t) c ni t th b khuch i c gi l khuch i o. Lc v0(t) = A v1(t). Trong hnh 2.16 c, nu tr khng u vo cao th ta c Ia(s) = 0 suy ra I1(s)=-I2(s). Khi h s khuch i A ln, v1(t) = 0 th I1(s) = V1(s)/Z1(s) v - I2(s) = - V0(s)/Z2(s). Cho hai dng in ny bng nhau ta c

)()(

)()(

1

1

2

0

sZsV

sZsV = hay l

)()(

)()(

1

2

1

0

sZsZ

sVsV = (2.53)

V d : Tm hm truyn ca mch khuch i o sau

Hnh 2.17 S h thng khuch i o

Tng tr Z1(s) l

1016.210360

103601106.5

11

1)(3

36

11

1 +=+=

+=

sx

xsx

RsC

sZ

Tng tr Z2(s) l

32


sx

sCRsZ

73

222

10102201)( +=+= Thay Z1(s) v Z2(s) vo cng thc 2.

sss

sZsZ

sVsV 547.22951.45232.1

)()(

)()( 2

1

2

1

0 ++== 2.3 M hnh ton hc trong min thi gian 2.3.1 Khi nim trng thi v bin trng thi

2.3.1.1 Khi nim v trng thi Khi nim trng thi c trong c s ca cch tip cn hin i trong m t ng hc ca cc h thng c Turing ln u tin a ra nm 1936. Sau khi nim ny c cc nh khoa hc Nga v M ng dng rng ri gii cc bi ton iu khin t ng. Trng thi ca h thng c c trng nh l lng thng tin ti thiu v h, cn thit xc nh hnh vi ca h trong tng lai khi bit tc ng vo. Ni mt cch khc, trng thi ca h c xc nh bi t hp cc to m rng c trng cho h. Trng thi ca mt h thng l tp hp nh nht cc bin (gi l bin trng thi) m nu bit gi tr ca cc bin ny ti thi im t0 v bit cc tn hiu vo thi im t>t0 ta hon ton c th xc nh c p ng ca h thng ti mi thi im t>t0. H thng bc n c n bin trng thi. Cc bin trng thi c th chn l bin vt l hoc khng phi l bin vt l. Theo quan im phn tch v tng hp h thng thng, ngi ta chia cc bin c trng h thng hay c quan h nht nh vi n v cc nhm nh sau: - Cc bin vo hay cc tc ng vo ui c to ra bi cc h thng nm

ngoi cc h c xt. - Cc bin ra yi c trng cho p ng ca h theo cc bin vo nh. - Cc bin trung gian xi c trng trng thi bn trong ca h.

2.3.1.2 Khi nim vc t trng thi: n bin trng thi hp thnh vc t ct

[ ]Tnxxxx ...21= (2.54) gi l vc t trng thi. - Khng gian trng thi: khng gian n chiu l khng gian hp bi cc trc ca cc bin trng thi.

33

Gio trnh L thuyt iu khin t ng 1 V d ta c cc bin trng thi in p ca in tr vR v in p ca t in vC cc bin ny s hnh thnh 2 trc ca khng gian trng thi. thun li trong thao tc vi cc i lng nhiu chiu, t hp cc bin vo c th trnh by di dng vc t cc tc ng vo:

[ ]Tn tutututu )(...)()()( 21= (2.56 ) T hp cc bin ra trnh by di dng vct ra

[ ]Tn tytytyty )(....()()( 21= (2.57 ) Cc t hp cc to trung gian, c trng ni dung bn trong ca h c vit dng vc t trng thi ca h .

[ Tnxxxx ...21= ] (2.58) Theo nh ngha trng thi ca h ti thi im bt k t > t0, trng thi ca h l mt hm ca trng thi ban u x(t0)v vc t vo r(t0,t), tc l:

x(t) = F[x(t0),u(t0,t) ] (2.59) Vc t ra ti thi im t c quan h n tr vi x(t0) v u(t0 ,t)

y(t) = [x(t0),u(t0,t)] (2.60) Cc phng trnh (2.59) v (2.60) thng gi l phng trnh trng thi ca h. Nu h thng c m t bi cc phng trnh vi phn tuyn tnh ,th phng trnh trng thi ca h c vit di dng sau : (Bng cch s dng cc bin trng thi, ta c th chuyn phng trnh vi phn bc n m t h thng thnh h gm n phng trnh vi phn bc nht)

( )

+=+=

)()()().()()().()().(tutDtxtCtytutBtxtAtx&

(2.61)

trong : x (n x1) vc t cc bin trng thi,

u (m x 1) vc t cc bin u vo

y (r x 1) vc t cc bin u ra.

A(t) - Ma trn h thng. B(t) - Ma trn iu khin hay m trn u vo. C(t) - Ma trn ra.

D(t) - Ma trn vng.

Cc ma trn c cc phn t ph thuc vo bin t, ln lt c kch thc

l: A(n x n), B(n x m), C(r x n ), D(r x m).

34


Hnh 2.18: S khi biu din h thng iu khin trong khng gian trng thi Thc t cc h thng thc u c tnh qun tnh, do D l mt ma trn c cc phn t u bng khng. 2.3.2 H tuyn tnh h s hng.

H thng c m hnh trng thi l:

uDxCyuBxAx

+=+=&

(2.62)

Trong cc ma trn A, B, C v D l cc ma trn hng s.

A c gi l ma trn h thng. Nu s lm cho phng trnh det(sI - A) = 0

th s c gi l gi tr ring ca ma trn A (y chnh l im cc ca h

thng). I l ma trn n v, s l mt s phc, det l k hiu ca php tnh nh

thc ma trn.

2.3.3 ng dng biu din m hnh ton hc trn khng gian trng thi

ng dng h phng trnh trng thi biu din cc h vt l phc tp.

Bc u tin l chn vct trng thi, vic la chn ny phi tun theo cc yu

cu sau:

- Cc bin trng thi phi l ti thiu nhng vn phi m bo biu din

y trng thi ca h thng.

- Cc bin trng thi phi c lp tuyn tnh.

35

Gio trnh L thuyt iu khin t ng 1 V d 1: Cho h thng vt l c s nh sau:

Hnh 2.19: S mch RLC mc hn hp

Xy dng m hnh trng thi cho i tng.

Gii:

Bc 1: t tn cc dng in nhnh bao gm iR, iL v iC. Bc 2: Chn cc bin trng thi bng cc vit phng trnh vi phn cho cc

phn t cha nng lng bao gm t in C v in cm L

LCC v

dtdiLi

dtdv

C == (2.63) Ta chn iL v vC l cc bin trng thi, nhng do iC v vL khng phi l cc bin trng thi nn ta phi vit di dng t hp tuyn tnh ca cc bin trng thi iL v vC , bin u vo l v(t). Bc 3: S dng l thuyt v mch in c th l vit phng trnh da vo nh lut Kirchhoff. Ti nt 1 ta c

LC

LRC

ivR

iii

+=+=

1 (2.64)

Mt khc ta c vL = - vC + v(t) (2.65)

Bc 4: Thay cng thc trn vi nhau ta thu c cng thc nh sau:

)(

1

tvvdtdiL

ivRdt

dvC

C

LCC

+=

+= (2.66)

hoc

36


)(11

11

tvL

vLdt

di

iC

vRCdt

dv

C

LCC

+=

+= (2.67)

Bc 5: Rt ra cng thc ca tn hiu u ra iR(t)

CR vRi 1= (2.68)

kt qu cui cng l

)(10

01

11

.

.

tvLi

v

L

CRCiv

L

C

L

C

+

=

(2.69)

tn hiu u ra

=L

cR i

vR

i 01 (2.70)

V d 2: Cho mch in gm ba phn t R, Lv C mc ni tip

Hnh 2.20: S mch RLC mc ni tip

U1 l in p t vo mch. Tm m hnh trng thi. Gii: Ta c phng trnh in p ca mch l:

u1 = uR + uL + uC (2.71) thay cc cng thc tnh in p ca cc phn t

21 dtdiLR uiu ++= (2.72)

trong == idtCuu C 12 (2.73) Trng thi ca mch c quyt nh bi in p ra u2 v dng in i. Ta gi u2 v i l cc bin trng thi. t:

u2 = x1 37

Gio trnh L thuyt iu khin t ng 1 i = x2

T cng thc (2.72 v (2.73) ta rt ra cng thc tnh dng in l

12

2

11 uL

uL

iLR

dtdi

dtduCi

+=

=

1212

21

11

1

uL

xLRx

Lx

xC

x

+=

= (2.74)

Dng chnh tc c vit nh sau:

12

1211

11

.01.0

uL

xLRx

Lx

uxC

xx

+=

++=

&

& (2.75)

Vit h trn di dng vct ma trn

12

1

2

1 10

1

10u

Lxx

LR

L

Cxx

+

=

&&

(2.76)

hay vit gn li uBxAx +=& (2.77)

gi l phng trnh trng thi ca h thng. Khng gian hai chiu gm trng thi dng in i = x2 v in p trn t l u2 = x1 c gi l khng gian trng thi. V d 3:

Hnh 2.21: S mch RLC mc ni tip

Ta c

)(1dtdiLR tvidt

Ci =++ (2.78)

Thay dtdqti =)( vo cng thc trn ta c

)(122

tvqCdt

dqRdt

qdL =++ (2.79)

38

Gio trnh L thuyt iu khin t ng 1 Ta t i(t), q(t) l cc bin trng thi

)(11 tvL

iLRq

LCdtdi

idtdq

+=

= (2.80)

vit di dng vct ma trn

vLq

i

LR

LCiq

+

=

101 10&&

(2.81)

in p vL l bin trng thi u ra

vRiqC

vL += 1 (2.82)

hay uiq

RC

vL +

= 1 (2.83)

2.4 Chuyn t hm truyn t sang khng gian trng thi v ngc li 2.4.1 Chuyn t hm truyn t sang khng gian trng thi

c th m phng c mt h thng trn my tnh th m hnh ton hc ca i tng phi c biu din trn khng gian trng thi. V vy khi ta a m hnh ca i tng biu din bng hm truyn t ta phi chuyn sang phng trnh trng thi. - Chn cc bin trng thi, mi bin trng thi c xc nh bi o hm ca bin trng thi trc . - Ta xt phng trnh vi phn sau:

ubyadtdya

dtyda

dtyd

ni

n

nn

n

0011

1

1 ... =++++

(2.84)

Cch thun tin chn bin trng thi l chn bin u ra

1

1

1

2

2

3

2

1

...

=

=

==

n

n

n dtydx

dtydx

dtdyx

yx

(2.85)

39

Gio trnh L thuyt iu khin t ng 1 Ly o hm hai v

n

n

n dtydx

dtydx

dtydx

dtdyx

=

=

=

=

1

3

3

3

2

2

2

1

...

&

&

&

&

(2.86)

Biu din trn khng gian trng thi

ubxaxaxaxxx

xxxxxx

nnn

nn

012110

1

43

32

21

...

...

+==

===

&&

&&&

(2.87)

Biu din di dng vct ma trn

u

bxx

xxx

aaaaaaaxx

xxx

n

n

nn

n

+

=

0

1

3

2

1

1543210

1

3

2

1

0

000

1000000

000100000001000000010

MM

KK

MKMMMMMMKKK

&&M&&&

(2.88)

Vit phng trnh trng thi u ra

[

=

n

n

xx

xxx

y

1

3

2

1

00001 MK ] (2.89)

Cc bc thc hin bin i t hm truyn sang h phng trnh trng thi: - B1: chuyn t hm truyn v phng trnh vi phn v thc hin php

bin i Laplace ngc vi cc iu kin u bng khng. - B2: Thc hin chn cc bin trng thi v biu din trong khng gian

trng thi.

40


V d 1: Mt i tng c hm truyn t l 45

25)()()( 2 ++== sssR

sCsW .

Xy dng m hnh trng thi cho i tng. Xc nh cc gi tr ring.

Gii

Bc 1: Tm phng trnh vi phn

rcdtdc

dtcdsRsssC 2545)(.5)45).(( 2

22 =++=++ (2.90)

Bc 2: La chn cc bin trng thi

==

===

212

21

12

1

5425 xxrxxx

xdtdcx

cx

&&

& (2.91)

Vit i dng vct ma trn

[ ]

=

+

=

xy

uxx

0 1250

.5- 41 0&

(2.92)

Tm gi tr ring

sI - A = (2.93)

+=

54

154

100

0s

sS

S

det(sI - A) = s(s + 5) + 4 = s2 + 5s + 4 = 0 (2.94)

91625 == (2.95) Cc gi tr ring l s1 = -1, s2 = -4

V d 2: Cho hm truyn sau:

2426924

)()()( 23 +++== ssssR

sCsG

Chuyn i sang h phng trnh trng thi. Gii: Bc 1: Tm phng trnh vi phn Thc hin php nhn cho

( ) )(24)(24269 23 sRsCsss =+++ (2.96) Chuyn i thnh phng trnh vi phn bng cch dng php bin i Laplace ngc vi iu kin u bng 0

41

Gio trnh L thuyt iu khin t ng 1 rcccc 2424269 =+++ &&&&&& (2.97)

Bc 2: La chn bin trng thi Chn cc bin trng thi nh sau:

cxcxcx

&&&

===

3

2

1

(2.98)

Ly o hm c hai v phng trnh (2.89) ta s thu c h phng trnh trng thi

1

3213

32

21

2492624xcy

rxxxxxx

xx

==+=

==

&&&

(2.99)

Vit di dng vct ma trn

[ ]

=

+

=

3

2

1

3

2

1

3

2

1

001

2400

92624100010

xxx

y

rxxx

xxx

&&&

(2.100)

M hnh c biu din nh sau:

Hnh 2.22: S biu din bng s khi trong gian trng thi

42

Gio trnh L thuyt iu khin t ng 1 2.4.2 Chuyn t khng gian trng thi sang hm truyn t

M hnh ton hc trong gian trng thi c biu din nh sau:

uDxCyuBxAx

+=+=&

(2.101)

Thc hin chuyn i Laplace vi iu kin u bng 0

sX(s) = AX(s) + BU(s) (2.102) Y(s) = CX(s) + DU(s) (2.103)

T rt X(s) ra: (sI A)X(s) = BU(s) (2.104) X(s) = (sI A)-1BU(s) (2.105)

Trong I l ma trn n v Thay X(s) vo (2.94) rt ra c

Y(s) = C(sI A)-1BU(s) + DU(s) (2.106) Ta gi [C(sI A)-1BU(s) + DU(s)] l ma trn hm truyn bi v n quan h vi vct bin ra Y(s) v vct bin vo U(s). Nu U(s) = U(s) v Y(s) = Y(s) l cc i lng v hng ta c th tm hm truyn nh sau:

DBAsICsUsYsT +== 1)_()()()(

(2.107)

Vi d: Cho phng trnh trng thi bit u ra l Y(s) v u vo l U(s)

[ ]xyuxx

00100

10

321100010

=

+

=& (2.108)

Gii: T u bi ta xc nh cc ma trn A, B, C v D

[ ] 000100

10

321100010

==

=

=

DC

BA (2.109)

Ta tm (sI - A)-1

43


123

)12()3(1

1323

)det()()(

32110

01

321100010

000000

)(

23

2

2

1

+++

+++++

==

+

=

=

sss

ssssss

sss

AsIAsIadjAsI

ss

s

ss

sAsI

(2.110)

Thay (sI - A)-1, B, C, D vo ta c hm truyn

123)23(10)( 23

2

+++++=sss

sssT (2.111)

2.5 Tuyn tnh ha - Cc h thng m ta xt vi gi thuyt l tuyn tnh. Trn thc t hu ht cc i tng l phi tuyn. - Trong h thng cng c th bao gm c i lng phi tuyn v tuyn tnh. - Do thc t yu cu ngi thit k pha tuyn tnh ha mt s i lng phi tuyn s dng. - Cc bc thc hin tuyn tnh ha + Bc 1: Vit phng trnh vi phn ca h thng. Vi gi thit tn hiu u vo nh + Bc 2: Tuyn tnh ha phng trnh vi phn, dng bin i Laplace vi iu kin u = 0.

44

Gio trnh L thuyt iu khin t ng 1 Bi tp chng 2 1. Tm hm truyn ca h thng sau a) G(s) = V0(s)/Vi(s) b) G(s) = V0(s)/Vi(s)

c) G(s) = VL(s)/V(s) d) G(s) = X1(s)/F(s)

e) G(s) = V0(s)/Vi(s)

2. Gii phng trnh vi phn sau

063 22

3

3

=++dtdy

dtyd

dtyd vi 14)0(,2)0(,5)0( 2

2

=+=+=+dtyd

dtdyy

3. Tm hm truyn G(s) ca h thng khi bit c dng biu din trn khng gian trng thi

a) b)

[ ]xyrxx

0011000

523100010

=

+

=&

[ ]xyrxx

631641

423350832

=

+

=&

45


CHNG 3: P NG THI GIAN

3.1 Cc c tnh ca h thng KT 3.1.1 c tnh thi gian

Mt trong cc c tnh quan trng ca h thng t ng l c tnh thi gian.

Ngi ta thng s dng c tnh thi gian m t h thng t ng tuyn tnh

dng v khng dng. c tnh thi gian l phn ng ca h thng i vi tc

ng no khi iu kin ban u bng 0. Cc tc ng thng c s dng l

tc ng xung n v (t) v tc ng bc thang n v 1(t).V vy ngi ta thng nh ngha c tnh thi gian l s thay i ca tn hiu ra theo thi gian

khi u vo tc ng l cc hm chun. Trong c tnh thi gian ngi ta quan

tm n hai c tnh c bn: c tnh qu v c tnh xung (Hm trng

lng).

3.1.2 c tnh xung (Hm trng lng):

Phn ng ca h thng i vi tc ng xung n v khi iu kin ban u

bng 0 c gi l hm qu xung g(t) (hm trng lng). Hm xung n v

(t-) c dng tng t nh (t), nhng sai lch theo thi gian mt khong l . Tng t, phn ng ca h thng t ng tuyn tnh i vi tc ng a(t-) s l ag(t-); vi a l hng s no . Ti trc thi im c tc ng, gi tr ca c tnh qu xung bng 0

ngha l:

g(t-) = 0, t < ng thc ny c gi l iu kin vt l thc t.

Dng khi nim c tnh qu xung, ta c th tm c phn ng ca h

thng i vi tc ng bt k cho trc.

Gia c tnh qu xung v hm s truyn ca h thng c quan h vi

nhau. nh hm trng lng s bng:

46


==0 0

)( )()()()()( dvevgsXdtetgdtexsY svtsst (3.1)

M Y(s) = X(s) . W(s) (3.2)

So snh (3.1) v (3.2) s c:

=0

)()( dtetgsW st (3.3)

Nh vy hm s truyn l bin i Laplace ca hm qu xung. Tt nhin,

hm qu xung c th nhn c theo hm s truyn bng cch bin i

ngc Laplace:

{ })()( 1 sWLtg = (3.4) 3.1.3 Hm qu

c tnh qu h(t) l phn ng ca h thng i vi tc ng bc thang

n v khi iu kin ban u bng khng.

i vi h thng t ng dng, c tnh qu khng ph thuc vo thi

im bt u tc ng. c tnh h(t) c th c dng dao ng hay bin i mt

cch n iu. Hin nay, nh gi cht lng h thng t ng ngi ta

thng s dng c tnh qu h(t) bi v n th hin c cc ch tiu cht

lng c bn ca h thng.

Gia c tnh qu xung v c tnh qu c lin h vi nhau, thay x(t)

= l(t) trong tch phn sau ta c:

==00

d)t(gd)t(g)t(l)t(h (3.5)

i bin t- = v v d = -dv, ta nhn c:

===

t

0

t

0

dv)v(gdv)v(gdv)v(g)t(h (3.6)

Nh vy, c tnh qu l tch phn ca c tnh qu xung. Ly o hm

ca c hai v ca (3.6), ta s nhn c:

dt

)t(dh)t(g = (3.7)

47

Gio trnh L thuyt iu khin t ng 1 Vy c tnh qu xung l o hm theo thi gian ca c tnh qu .

Vit (3.7) di dng ton t, s c:

W(s) = s.H(s) (3.8)

Trong H(s) = L[h(t)] v W(s)=L[g(t)] (3.9)

T (3.8) ta c mi lin h gia nh hm ca c tnh qu theo hm truyn

ca h thng.

)(1)( sWs

sH = (3.10)

S dng bin i ngc Laplace, s xc nh c hm qu theo hm s

truyn.

= )(1)( 1 sW

sLth (3.11)

3.1.4 c tnh tn s.

Cc c tnh tn s (TTS) c ngha quan trng trong m t h thng t

ng dng. C th nhn c cc c tnh ny khi nghin cu chuyn ng

cng bc ca h thng di tc ng ca tn hiu iu ho.

Tnh cht c bn ca h thng tuyn tnh l tnh cht xp chng. Gi s nhiu

lon g(t) tc ng ln h thng bng khng, ta c mi quan h gia lng ra y(t)

vi lng vo x(t) c m t bng phng trnh vi phn sau:

xbdt

dxb

dt

dxba

dt

dya

dt

dya m1mm

m

0n1nn

n

0 +++=+++ LL (3.12)

Cc TTS xc nh mi lin h gia ph lng vo X(j) v ph lng raY(j). Cc ph ny chnh l bin i Phuri ca cc hm thi gian tng ng:

{ }{ }

==

==

)t(yFdte)t(y)j(Y

)t(xFdte)t(x)j(X

0

tj

0

tj

(3.13)

y: F - k hiu bin i Phuri.

- bin thc, l tn s ca tn hiu iu ho.

48

Gio trnh L thuyt iu khin t ng 1 Ly bin i Phuri phng trnh (3.13) vi iu kin ban u bng 0, s

nhn c phng trnh biu din quan h gia ph lng ra vi ph lng vo

ca h thng t ng:

)()()()()()()()( 01

10 jXbjXjbjYajYjajYja mmnnn ++=+++ LL (3.14) sau khi bin i, s nhn c:

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

10

110

jWjXajajabjbjbjXjY

nnn

mmm

=++++++=

LL (3.15)

T s ca cc a thc:

)()(

)()()()()( 1

10

110

jDjE

ajajabjbjbjW

nnn

mmm

=++++++=

LL (3.16 )

c gi l hm s truyn tn s ca h t ng.

T y ta thy c th nhn c biu thc hm truyn tn s t hm s

truyn bng cch thay bin phc s bng j. Hm s truyn truyn tn s W(j) c th biu din di dng:

)()()()()( jeAjQPjW =+= (3.17) y

=+=

)()()()()( 22

jarctgWQPA (3.18)

Nu 2

)( jarctgW

Th )()()(

PQarctg= (3.19)

Trn mt phng phc (hnh 3.1) ng vi tn s no th hm s truyn tn

s s xc nh vc t. di vc t bng A(), cn argument (gc hp thnh bi vc t ny vi bn trc thc dng) l (). ng cong c v bi u mt ca vc t ny, khi tn s bin thin t 0 n (i khi t - n ) gi l c tnh tn s (TTS) bin pha ca h thng t ng.

49


50

Hnh 3.1: c tnh tn s bin pha

W(j)

jQ()

P()

Hm truyn tn s c gi l hm tn s bin pha. Phn thc ca n

P() =ReW(j) v phn o Q() = ImW(j) c gi l tng ng l hm tn s phn thc v hm tn s phn o. th ca hm tn s phn thc P() gi l TTS phn thc. Cn th ca hm tn s phn o - TTS phn o Q().

Mun A() =W(j) gi l hm tn s bin . th ca n gi l TTS bin . Cn () l argument ca W(j) c gi l hm tn s pha. th ca n gi l TTS pha.

Ngoi cc TTS ni trn, cn thng s dng cc TTS bin logarit

L() v TTS pha logarit (). Gi: )(lg20)(lg20)( jWAL == (3.20)

l hm tn s bin logarit. th din t s ph thuc gia hm tn s bin

logarit L() vi hm logarrit tn s lg(), gi l TTS bin logarit. Khi xy dng TTS bin logarit, trn trc honh ta t cc tn s theo t

l logarit, cc im chia tng ng vi gi tr lg(), nhng li ghi gi tr ca tin cho vic c tn s, ch khng ghi gi tr lg(). Cn trc tung l L(). TTS pha logarit l th biu din s ph thuc gia hm tn s pha () vi logarit lg(). Trc honh ca TTS pha logarit ging trc honh ca TTS bin logarit.

n v ca L() tnh theo (3.20) l xibel [db]. n v c s ca trc honh trong TTS bin logarit la cc [dc] hay octav [oc]. cc l rng

khong tn s m trn tn s thay i 10 ln. Tng t, octav l khong m

trn tn s thay i 2 ln.

Gio trnh L thuyt iu khin t ng 1 Khi xy dng TTS logarit th cn lu l trc tung v trc honh khng ct

nhau ti tn s = 0. Hai trc s ct nhau ti mt tn s thch hp no , bi v khi 0 th lg() - : ngha l = 0 th s tng ng vi im xa v cng. Trn thc t, ngi ta thng s dng TTS bin logarit tim cn.

nghing ca c tnh tim cn thng dng l db/dc hay db/oc. Khi tc ng ln

u vo h thng T l mt tn hiu iu ho th lng ra ca h thng cng

thay i theo quy lut iu ho, nhng bin v pha ca lng ra v lng vo

s khc nhau. Ni chung, bin v pha lng ra s ph thuc vo tn s ca

lng vo. Ngi ta gi t s gia bin lng ra vi bin lng vo l m

un, cn gc lch pha gia lng ra vi lng vo l argument ca hm s

truyn tn s. V th, c tnh tn s bin din t s thay i ca t s cc

bin lng ra vi lng vo, cn TTS pha din t gc lch pha gia lng

ra vi lng vo. l ngha vt l i TTS.

3.2 Cc khu ng hc in hnh 3.2.1 nh ngha cc khu ng hc in hnh

Cc khu ng hc m phng trnh vi phn m t qu trnh ng hc ca

chng c bc nh hn hoc bng 2, c gi l khu ng hc in hnh.

c im ca cc khu ng hc in hnh l ch c mt u vo v mt u

ra, tn hiu u ra khng nh hng n tn hiu u vo.

51

W(s) X(s) Y(s)

Hnh3.2 Biu din khu ng hc in hnh. Cc khu ng hc in hnh bao gm: Khu nguyn hm, khu tch phn,

khu vi phn, khu tr. Khu nguyn hm gm cc khu: Khu khuch i

(khu khng qun tnh), khu qun tnh bc mt, khu qun tnh bc hai (Khu

giao ng). Sau y ta kho st cc khu ng hc in hnh trn.

3.2.2.Cc khu nguyn hm.

a)Khu khuch i

- Khu khng qun tnh l khu m phng trnh ng hc c dng:

Gio trnh L thuyt iu khin t ng 1 y = K.x - Hm truyn t ca khu. W(s) = K - Cc c tnh thi gian. Hm qu :h(t) = K.1(t).

Hm trng lng : g(t) = K.(t).

K

0 t

h(t) K.(t)

0

g(t)

t

Hnh 3.3. c tnh thi gian ca khu khng qun tnh

- Cc c tnh tn s.

TBP

K 0 P()

jQ() 20lgK

0 lg

L() TBL

A()

()BT PT

00

K

x

Hnh 3.4: c tnh tn s ca khu khnng qun tnh

Hm truyn tn s : W(j) = K. c tnh BT : A() = K. c tnh PT : () = 0. c tnh BTL : L() = 20.lgK

b) Khu qun tnh bc nht: - Phng trnh vi phn:

x.Kydt

dyT =+ (3.21)

52

Gio trnh L thuyt iu khin t ng 1 Trong K l h s truyn ca khu.

T l hng s thi gian ca khu. - Hm truyn t : W(s) = K/ (Ts+1)

- Cc c tnh thi gian :

Hm qu : h(t) = K( 1-e-t /T) .1(t)

Hm trng lng : g(t) = dh(t)/dt = K.e-t / T.1(t)/T

K

0 t

h(t) .K

0 t

g(t)

T

T

Hnh 3.5: c tnh thi gian ca khu qun tnh bc nht - Cc c tnh tn s:

BTK

0

TBP jQ()

0K

P()lg

-20db/dec

0

L()

x

20lgK

lgc

TBL

PT ()0

-/2

A()

Hnh 3.6: c tnh tn s ca khu qun tnh bc nht TTS bin pha : W(j) = Y(j)/X(j) =K/ (Tj + 1) TTS bin : A() = [ P2 () + Q2()] 1/2 =k/ (T22 + )1/2 TTS pha () = arctgT TTS bin logarit (TBL L() = 20lg A()

c) Khu qun tnh bc hai:

- Phng trnh:

53


Kxydt

dyT2

dt

ydT

2

22 =++ (3.22)

Trong : T l hng s thi gian.

K l h s truyn. l h s tt dn tng i ( 0).

- Hm truyn t: 1Ts2sT

K

)s(X

)s(Y)s(W

22 ++==

- Cc c tnh thi gian:

Trng hp th nht: 0 < < 1. Hm qu : )]tsint(cose1[K)t(h t

=

Hm trng lng: tT

Kethtgt

sin.)(')(

==

Trong : T

= v T

1 = Trng hp ny c tnh qu c dng dao ng tt dn v c gi l

khu dao ng. Trng hp th hai: = 0.

Hm qu : )tcos1(K)t(h 1= Hm trng lng : g(t) = K1sin1t Vi 1=1/T : Tn s ring dao ng. c tnh qu dao ng iu ho

v c gi l khu dao ng iu ho. Trng hp th ba : 1.

Hm qu : Khi = 1 vi l nghim kp ca phng trnh c trng.

)e)t1(1(K)t(h t= Khi > 1 vi 1, 2 l hai nghim thc ca phng trnh c trng.

+

= t12

1t

12

2 21 ee1K)t(h

Hm trng lng:

Khi = 1 vi l nghim kp ca phng trnh c trng. t2 teK)t(k =

54

Gio trnh L thuyt iu khin t ng 1 Khi >1 vi 1, 2 l hai nghim thc ca phng trnh c trng.

)ee(K)t(k tt

12

21 21 =

Hnh 3.7: c tnh thi gian ca khu bc hai

g(t

0 t =1

0 < < 1

t

K

h(t)

0 < 1

0

- Cc c tnh tn s:

-40db/d

20lgK

0lg

=0L(

0 < <

=0

=1 >1

-

(0-/2

0 < < =1 >1

0

>1 0 < < 1

A() K

0 K

P(

jQ()

=1 0 < < 1

>1

Hnh 3.8: c tnh tn s ca khu bc hai TTS bin pha : W(j) = Y(j)/X(j) =K/( 1-T22 +j2T) TTS bin :

A() = [ P2 () + Q2()] 1/2 =K/[(1-T22 )2 +(2T)2]1/2 TTS pha () = -arctg [2T/(1-T22 )]. TTS bin logarit (TBL L() = 20lg A().

3.2.3 Khu tch phn.

- Phng trnh vi phn ca khu tch phn.

= xdtKy hoc Tx

Kxdt

dy == (3.23) Trong T = 1/K c gi l hng s thi gian tch phn.

- Hm truyn t ca khu.

55

Gio trnh L thuyt iu khin t ng 1 Bin i phng trnh vi phn sang ton t Laplace ta c :

Ts

1

)s(X

)s(Y)s(W ==

- Cc c tnh thi gian. Hm qu : h(t) = Kt.

Hm trng lng : k(t) = h(t) = K.

Hnh 3.9: c tnh thi gian ca tch phn - Cc c tnh tn s.

Hm truyn tn s : Kj

Tj

TjjW === 11)(

Nh vy hm truyn tn s ca khu tch phn ch c pn o m khi thay i t 0 n m khng c phn thc.

c tnh BT: = T1

)(A

c tnh PT : 2

)( =

c tnh BTL : L() = lgA() = -20lgT - 20lg y l phng trnh ca mt ng thng ct trc tung ti im c

tung bng - 20lgT v c nghing bng - 20db/dec.

56

Hnh 3.10: c tnh tn s ca khu tch phn

0 t

h(t)

K

0 t

g(t)

tg=K

BT

-/2

PT( 0

A()

0 TBP

P(0 = =0 lg

-20db/d

0

L(-20lgT

TBL

Gio trnh L thuyt iu khin t ng 1 3.2.4.Khu vi phn.

- Phng trnh khu vi phn l tng: y(t) = Kdx(t)/dt.

- Phng trnh khu vi phn bc mt: y(t) = KTdx(t)/dt + Kx(t).

- Hm truyn t:

Khu vi phn l tng: G(s) = Ks

Khu vi phn bc mt: G(s) = C(s)/R(s) =K(Ts + 1)

- Cc c tnh thi gian:

Khu vi phn l tng:

Hm qu : h(t) =K(t) Hm trng lng: g(t) = dh(t) /dt = Kd(t)/dt Khu vi phn bc mt:

Hm qu : h(t) =K.1(t) + KT(t) Hm trng lng: g(t) = dh(t) /dt = Kd(t)/dt + K(t)

57

Hnh 3.11: c tnh thi gian ca khu vi phn l tng 0 t

h(t)

t

T.(t)

0

g(t)

T.(t)

- Cc c tnh tn s:

Khu vi phn l tng :

TTS bin pha : G(j) = C(j)/R(j) =-j K TTS bin : A() = [ P2 () + Q2()] 1/2 =K TTS pha () = /2 TTS bin logarit (TBL) L() = 20lg A()=20lgK


Hnh 3.12: c tnh tn s ca khu vi phn l tng

0

A()

TBP

0

()

0 P()

jQ()

20db/dec

0 lg

L()

BT

=0

=

/2

20lgT

TBL

PT

Khu vi phn bc mt :

TTS bin pha : G(j) = C(j)/R(j) =-j KT + K TTS bin : A() = [ P2 () + Q2()] 1/2 =K(1 + 2T2 )1/2 TTS pha () = arctg T. TTS bin logarit (TBL) L() = 20lg A()=20lgK

3.2.5 Khu tr

Khu chm sau l khu ng hc m sau mt khong thi gian xc nh th

lng ra lp li lng vo v tn hiu khng b mo.

Phng trnh ng hc ca khu tr c dng :

y(t) = x(t-) Cc phn t thuc khu tr nh bng ti , ng ng dn nhit , ng ng

dn cht lng

- Hm truyn t ca khu tr:

W(s) = Y(s) /X(s) = e-s

- c tnh thi gian:

Hm qu : h(t) = 1(t-) Hm trng lng: g(t) = dh(t)/dt = (t- ) - Cc c tnh tn s:

TTS bin pha : G(j) = e-j TTS bin : A () = 1 TTS pha : () = -

58

Gio trnh L thuyt iu khin t ng 1 TTS bin lgarit : L() = 20lgA() = 0

59

()

A() A(),()jQ()

P()t

h(t)

Hnh 3.13. c tnh qu v cc c tnh tn s ca khu tr 3.3 M hnh ZPK (Zero, Pole and Gain) Ta xt hm truyn sau:

mnpspspszszszsK

asasasabsbsbsbsG

n

m

nnnn

mmmm

=

++++++++=

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

...

...)(

21

21

22

110

22

110

(3.24)

t : (3.25) m

mmmn

nnn

bsbsbsbsBasasasasA

++++=++++=

...)(

...)(2

21

10

22

110

A(s) l mu s ca hm truyn, B(s) l t s ca hm truyn.

- im khng (Zeros) l l cc gi tr lm cho hm truyn G(s) bng 0 hay l

nghim ca phng trnh B(s) = 0. Cc im khng c k hiu l zi (i: 1m).

- im cc (Poles) l cc gi tr lm cho hm truyn khng xc nh hay l

nghim ca phng trnh A(s) = 0. Cc im cc c k hiu l pi (i: 1m).

- H s khuch i tnh (Gain) k hiu l K.

V d 1: Tm cc im cc, im khng v h s khuch ica h thng ca

hm truyn sau:

)3)(5()2)(1(5)( ++

++=sssssG (3.26)

H s khuch i K = 5.

- im cc: A(s) = (s+5)(s+3) = 0 suy ra p1 = -5 v p2 = -3

- im khng: B(s) = (s+1)(s+2) = 0 suy ra z1 =-1 v z2 = -2

V d 2:

Ngoc ThanhHighlight

Gio trnh L thuyt iu khin t ng 1 Ta c hm truyn sau

)5(2)( +

+=ssssC (3.27)

Phn tch thnh tng cc phn s bc nht

5)5(2

++=++

sB

sA

sss (3.28)

Quy ng mu s v ng nht hai v ta c

(A+B)s + 5A = s + 2 (3.29)

Gii h phng trnh:

A + B = 1

5A = 2

Suy ra: A = 2/5, B = 3/5

55

35

2

)5(2)( ++=+

+=ssss

ssC (3.30)

Hnh 3.14 : S b tr cc im cc v im khng

p ng u ra:

tetc 553

52)( += (3.31)

trong : 52 l thnh phn cng bc

te 553 l thnh phn t do.

60

Gio trnh L thuyt iu khin t ng 1 Mt s kt lun:

1. im cc ca hm truyn u vo ca h thng quyt nh dng ca p ng

cng bc.

2. im cc ca hm truyn h thng quyt nh dng ca p ng t do.

3. p ng u ra c dng hm m nu c im cc nm trn trc thc. te

4. im cc v im khng quyt nh bin ca c p ng cng bc v

p ng t do.

V d 3: Cho h thng c hm truyn nh sau:

61

)5)(4)(2()3(

++++

ssss

)(sCssR 1)( =

Hnh 3.15 :H thng i tng lm v d 3

Tm hm p ng u ra c(t) bao gm hai thnh phn p ng t do v p ng

cng bc.

Gii:

- Kim tra xem cc im cc ca h thng to ra thnh phn p ng t do tun

theo quy lut hm m.

- im cc u vo to ra thnh phn p ng cng bc.

Ta c: 542

)( 4321 ++++++= sK

sK

sK

sKsC

(3.32 )

p ng

t do p ng

cng bc

Thc hin bin i Laplace ngc ta c:

c(t) = K1 + K2e-2t + K3e-4t + K4e-5t (3.33)

p ng

t do p ng

cng bc

3.4 H thng bc nht

H thng bc 1 khng c im khng c biu din nh sau:

Ngoc ThanhHighlight

Ngoc ThanhHighlight


62

asa+

j

Hnh 3.16: H tthng bc nht v phn b im cc nu tn hiu u vo l bc thang n v R(s) = 1/s th p ng u ra C(s) l:

)()()()(

assasGsRsC +== (3.34)

Thc hin bin i Laplace ngc ta c p ng u ra biu din trn min thi

gian l

c(t) = cf(t) + cn(t) = 1 e-at (3.35)

- im cc u vo ti thi im ban u to ra p ng cng bc cf(t) = 1

- im cc h thng ti a tao ra p ng t do cn(t) = - e-at.

Hnh 3.17: p ng u ra ca h thng bc 1 vi tn hiu bc thang n v

G(s) R(s) C(s)

- a

a) b)

0.3

0.2

0.1

0.6

0.5

0.4

0.9

0.8

0.7

1

a1

a2

a3

a4

a5

t

dc ban u = 1/hng s thi gian = a

Tr

x(t)

Ts

0

Gio trnh L thuyt iu khin t ng 1 ti thi im t = 1/a ta c

63.037.011)(

37.0

11

11

===

==

=

=

=

atat

at

atat

etx

ee (3.37)

T vic kho st c tnh ca i tng bc ta c cc khi nim sau:

- Hng s thi gian (constant time): gi 1/a l hng s thi gian ca p ng.

Hng s thi gian c th c hiu nh l khong thi gian m e-at gim 37%

gi tr ban u hay l khong thi gian p ng tn hiu bc thang dn v tng

ti 63% gi tr xc lp.

Nghch o ca hng s thi gian gi l tn s (1/s). V vy ta c th gi

hng s a l tn s hm m. Hng s thi gian c xem nh l c tnh p

ng thi gian ca h thng bc 1 v vy n c quan h vi tc ca h thng

tng ng vi tn hiu bc thang n v u vo.

- Thi gian tng Tr (rise time): thi gian tng c nh ngha l thi gian m

c c tnh mp m i t 01. n 0.9 gi tr xc lp.

Thi gian tng c tnh bng s sai lch gia hai thi im c(t) = 0.9 v

c(t) = 0.1.

aaaTr

2.211.031.2 == (3.38)

- Thi gian xc lp hay thi gian n nh Ts (settling time): thi gian xc lp l

khong thi gian m p ng t n v sai s trong khong 2%. Vi c(t) =

0.98 thay vo cng thc v rt ra c

a

Ts4= (3.39)

Hm truyn ca h thng bc 1 qua thc nghim:

Trn thc t khng d dng tm c hm truyn ca h thng bi v cc thit

b trong h thng kh c th xc nh c. V vy hm truyn ca h thng c

th xc nh c bng cch xc nh quan h gia u vo v u ra thng qua

phn tch ng c tnh ca i tng khi cho p ng u vo l tn hiu bc

thang n v. Hm truyn c th xc nh ngay car khi ta khng bit c cu

trc bn trong ca i tng. 63

Gio trnh L thuyt iu khin t ng 1 vi tn hiu vo l hm bc thang n v ta c th tnh c hng s thi gian

v cc gi tr xc lp.

Xt v d sau:

asKsG +=)( (3.40)

Cc p ng u ra:

asa

K

sa

K

assKsC ++=+= )()( (3.41)

Nu ta xc nh c h s khuch i K v a t phng th nghim ta s xc

nh c hm truyn ca i tng.

Gi s ta c p ng sau:

Bin 0.8

0.7

0.6

0.5

0.1 0.30.2 0.5 0.70.4 0.6 0.8 Thi gian (s)

0.4

0.3

0.2

0.1

Hnh 3.18 : ng c tnh p ng ca h thng bc nht p ng ca h thng bc nht khng c qu iu chnh v sai lch im

khng. T ng p ng ta xc nh hng s thi gian

- Gi tr xc lp l gi tr m ng p ng t n bng 0.72.

- Hng s thi gian l thi gian m ln bng 63% gi tr xc lp v

bng 0.63 x 0.72 = 0.45 hay bng 0.13 (s) suy ra a = 1/0.13 = 7.7

- p ng cng bc t n gi tr xc lp K/a = 0.72 suy ra K = 5.54

- Lc ta c hm truyn ca h thng l

64


7.754.5)( += ssG (3.42)

Hm truyn ny cng rt gn vi hm truyn ca p ng trn

75)( += ssG (3.43)

3.5 H thng bc 2 )(sG

65

ssR 1)( =

c(t)

ssR 1)( =

bassb++2

tt eetc 146.1854.7 171.1171.01)( +=

0.5

0

1j C(s)

-7.854 99

92 ++ ss

)(sGa)

- 1.146

C(s) t

b)1 2 3 4 5

ssR 1)( =

0.8

0.6

0.4

0.2

1.4

1.21

j

- 1

8j

8jC(s)

929

2 ++ ss

c) 0 543 1 2

)47.98cos(06.11

)8sin888(cos1)(

0=+=

te

ttetc

t

tc(t)

)(sG

t

Ngoc ThanhHighlight


66

Hnh 3.19 : Cc h thng bc hai v p ng vi tn hiu bc thang n v Ta c hm truyn tng qut ca h thng bc hai :

12)( 22 ++= TssT

KsG (3.44) trong : K l h s khuch i.

T l hng s thi gian.

l suy gim.

3.5.1 H thng p ng xung tt dn (Overdamped)

y l p ng khng c dao ng trong khong gi tr n nh nhng t

ti dao ng gii hn tt dn lu hn.

j

- j3

99

2 +ss

sR 1)( =

)(sG

C(s)

j3

c(t) = 1 cos3tc(t)

2

1

t0

1 2 3 4 5 d)

c(t)c(t) = 1 3e-3t e-6t

0.20.40.6

1

- 3

j 0.8)(sG

969

2 ++ ss

ssR 1)( = C(s) t

0 1 2 3 4 5e)

Ngoc ThanhHighlight

Gio trnh L thuyt iu khin t ng 1 Khi khu qun tnh bc hai c hai im cc thc th bao gm 2 khu qun tnh

bc mt ni tip nhau. Vi iu kin > 1ta c

11)(

2

2

1

1

++= sTKx

sTKsG (3.45)

Hai im cc l: p1 = -1/T1 v p2 = -1/T2Xt p ng u ra sau:

)46.1.1)(854.7(9

)99(9)( 2 =+=++= sssssssC (3.46)

- p ng u ra c mt im cc t gc to (do c p ng tn hiu bc

thang n v).

- Hai im cc thc ca h thng.

- im cc u vo s to ra thnh phn p ng cng c. Mi im cc

ca h thng s to ra p ng t do c dng hm m trong tn s hm

m chnh bng v tr cc im cc.

p ng u ra s c dng:

c(t) = K1 + K2e - 7.854t + K3e - 1.146t (3.47)

ng c tnh ca h thng bc hai tt dn th hin hnh 3.19b

3.5.2 H thng p ng di tt dn (Underdamped)

y l p ng c dao ng trong khong ng bao suy gim. H thng

cng c nhiu ng bao th p ng t ti trng thi n nh cng lu.(Xem

hnh 3.19c)

Ta xt phng trnh c tnh:

01222 =++ TssT (3.48) Khi < 1 thi phng trnh (3.48) s c hai nghim phc lin hp hai

nghim ny l hai im cc ca hm truyn.

Tj

T

jT

jT

p

jT

jT

p

2

2,1

2

2

2

1

2

1

1;

1

1

==

==

+=+=

(3.49)

67


Tj

2

2,11 =Qu trnh qa xy ra trong khu bc hai l qu trnh dao ng

l khu dao ng bc 2. p ng thi gian bao gm bin hm m gim to

bi phn thc ca im cc h thng v dng sng hnh sin to bi phn o ca

im cc h thng.

c(t) ng c tnh hm m gimto bi phn thc ca im cc

ng c tnh hnh sin to biphn o ca im cc

t

Hnh 3.20: p ng bc hai to bi cc nghim phc Hng s thi gian ca hm m bng phn thc ca im cc h thng. Gi tr

ca phn o l tn s thc ca dao ng hnh sin. Tn s dao ng hnh sin

c gi l tn s suy gim ca dao ng wd. p ng n nh c quyt nh

bi im cc u vo t gc to . Chng ta gi p ng ny l p ng

di tt dn m tin ti gi tr n nh qua p ng thi gian gi l dao ng

suy gim.

3.5.3 H thng p ng khng b nht (Undamped)

Nu im cc tin gn v khng cng b, ng bao gim cng lu, lc ta

c dao ng khng tt.

H thng bc hai ny s c: im cc nm gc to do p ng tn hiu

bc thang u vo v hai im cc ca h thng ch c phn o ( = 0).

T hnh 3.19d

)9(9)( 2 += sssC (3.50)

68

Gio trnh L thuyt iu khin t ng 1 Hai im cc p1,2 = j3 to ra p ng dao ng hnh sin m tn s ca n

bng v tr ca cc im cc nm trn trc o.

p ng u ra l:

))]tan(cos(11[))sin(cos1()( 12

+=+=

tKtteKtc t (3.51)

Thay vo ta c

c(t) = 1- cost3t (3.52)

3.5.4 H thng p ng tt dn ti hn (Critically Damped Response)

y l p ng t ti gi tr n nh nhanh nht. Gi tr gii hn lun lun

bng 1.

Ta c p ng sau:

22 )3(9

)96(9)( +=++= ssssssC (3.53)

p ng ny c mt im cc nm ti gc to v hai im cc thc.

c(t) = 1 3te - 3t e 3t (3.54)

Xem dng p ng hnh 3.19d

3.5.5 Tm p ng t do

p ng tt dn: Cc im cc: hai im thc 1, 2p ng: (3.55) tt eKeKtc 21 21)( +=

p ng di tt dn Cc im cc: 2 nghim phc d jd

p ng:

==

d

dd

t tAetc d 1tan);cos()( (3.56)

p ng khng b nht Cc im cc: 2 im cc o j

p ng:

== 1tan);cos()( tAtc (3.57)

p ng tt dn ti hn Cc im cc: 2 im cc thc (kp) 1

69

Gio trnh L thuyt iu khin t ng 1 p ng: (3.58) tt eKeKtc 11 21)( +=

3.6 Mt s vn chung v h thng bc hai Trong phn ny ta s xem xt hai khi nim ca hai thng s h thng bc 2

c dng miu t ng c tnh p ng thi gian. l tn s t do

(natural frequency) v h s tt dn (damping ratio).

- Tn s t do (Natural Frequency, n): l tn s ca dao ng trong h thng

m khng c s tt dn.

- H s tt dn (Damping ratio ):

Tn s suy gim hm m Chu k t do (seconds)

Hng s m 21= (3.59)

Tn s t do (rad/second) =

Biu din h thng bc hai theo hai thng s n v

bassbsG ++= 2)( (3.60)

i vi h thng khng b nht ta c cc im cc nm trn trc o

bsbsG += 2)( (3.61)

Theo inh ngha tn s dao ng t do n l tn s ca dao ng trong h thng.

V vy cc im cc nm trn trc o l bj . Suy ra 2nn bhaybj == (3.62)

Vi gi thit h thng di tt dn im cc phc c phn thc l a/2. ln

ca gi tr ny chnh l tn s gim hm m

70

suy ra na 2= (3.64) Vy hm truyn l

22

2

2)(

nn

n

sssG

++= (3.65)

Tn s suy gim hm m

nn

a

2== (3.63)

Tn s t do (rad/second) =


V d:

Cho hm truyn sau:

362.436)( 2 ++= sssG (3.66)

So snh hai cng thc (3.66) v (3.65) ta c:

35.02.426362

====

n

nn

Ta tm cc im cc ca h thng:

Phng trnh c trng l: s2 + 4.2s + 36 = 0

C hai nghim phc:

122,1 = nns (3.67) ng c tnh p ng t gi tr ca

T a = 2n v bn = suy ra

ba

2= (3.68)

Ta c cc p ng tng ng vi gi tr ca nh sau:

71

c(t)

j 21 nj

n 21 njj

n

10


H thng tt dn

21 + nn j

21 nn

c(t)

>1 t

Hnh 3.21 : p ng bc hai theo h s tt dn

3.7 H thng bc hai di tt dn (Underdamped) H thng di tt dn, m hnh vt l ph bin, c cc p ng n nht nn

c xem xt c th hn. nh ngha cc thng s p ng ca h thng di tt

dn theo thi gian v xem xt mi quan h vi v tr cc im cc.

Trc tin ta tm p ng ca h thng bc hai vi p ng tn hiu bc

thang n v

22321

22

2

2)2()(

nnnn

n

ssKsK

sK

ssssC

++

++=++= (3.69)

gi thit < 1 v thc hin bin i ta c

)1()(

11

)(1)( 222

2

2

++

+++=nn

nn

s

s

ssC (3.70)

Thc hin php bin i Laplace ngc

)1cos(1

11

1sin1

1cos1)(

2

2

2

2

2

=

+=

te

ttetc

nt

nt

n

n

(3.71)

trong :

=

2

1

1tan

Khi cng nh th p ng dao ng cng nhiu.

72


c(t)

c0.98c

1.02c

cmax

0.9c

0.1c

%

tTp

Tr

Ts

Hnh 3.22: p ng bc hai ca h thng di tt dn

Ngoi hai khi nim h s suy gim v tn s p ng t do n ta c thm cc

khi nim sau:

- Thi gian nh Tp (Peak Time): l thi gian m c(t) t max u tin.

- Phn trm qu iu chnh: %OS (Percent Overshoot): l khong m dng

sng vt qu gi tr n nh c.

- Thi gian tng Tr (rise time): thi gian tng c nh ngha l thi gian

m c c tnh mp m i t 0.1 n 0.9 gi tr xc lp.

- Thi gian xc lp hay thi gian n nh Ts (settling time): thi gian xc lp

l khong thi gian m p ng t n v sai s trong khong 2%.

a) Tnh Tp

{ }

)1()(

11

)1()(

2)()(

222

2

2

222

2

22

2

++=++=

++==

nn

n

nn

n

nn

n

ss

ssssCtcL &

(3.72)

Bin i Laplace ngc ta c:

tetc tn n 22

1sin1

)( =

& (3.73)

Cho 0)( =tc& suy ra 73


ntn = 21 (3.74) hay

21 = n

nt (3.75)

Khi n = 1 ng c tnh t gi tr max

21

= np

T (3.76)

b) Tnh phn trm qu iu chnh

T hnh v 3.22 ta c

100S% max xc

ccO

= (3.77)

cmax l gi tr khi ng c tnh t gi tr max ti thi im Tp

2

2

1

2

1max

1

sin1

cos1)(

+=

+==n

n

e

eTcc p (3.78)

i vi tn hiu bc thang n v

c = 1 (3.79)

Thay vo cng thc (3.77) ta tm c phn trm qu iu chnh

100%21 xeOS

n

= (3.80)

Suy ra

( )( )100%ln

100%ln

22 OS

OS

+=

(3.81)

c) Tnh Ts tm c Ts ta phi tm c thi gian m c(t) t n v gi n nh

trong khong 2%

T cng thc (3.71) ta tnh bin ca c(t) t n 0.02

02.01

12

= tne (3.82)

74


vi gi thit 1)1cos( 2 = tn ti Ts.

Suy ra ( )

nsT

2102.0ln = (3.83)

Ly xp x cng thc (3.83)

aT

ns

24 == (3.84)

d) Tnh TrTm nt bng cch cho c(t) = 0.9 v c(t) = 0.1. Ly gn ng ta c thi gian

tng nTr.

V d: Cho hm truyn sau

10015100)( 2 ++= sssG (3.85)

Tnh Tp, %OS, Ts v Tr.

Gii:

T hm truyn ta tnh c 75.0,10 == n Thay vo cng thc tnh Tp

475.075.0110

14.31 22

===

n

pT

838.2100100% 22 75.01

1075.01 === xexeOS x

n

533.01075.0

44 ===x

Tn

s

Ta c bng sau:

75


H s suy gim Thi gian tng thng thng

0.1 1.104

0.2 1.203

0.3 1.321

0.4 1.463

0.5 1.638

0.6 1.854

0.7 2.126

0.8 2.467

0.9 2.883

Da vo bng trn ta tnh c thi gian tng thng thng xp x 2.3 suy ra

Tr = 0.23 v n = 10.

76

Gio trnh L thuyt iu khin t ng 1 Bi tp chng 3 1. Hy xc nh hm trng lng g(t) v hm qu h(t) ca nhng h tuyn tnh c hm truyn t sau a)

4321)( 2 ++

+=ss

ssG b))51)(31(

12)(ss

ssG +++=

2. Tm v tr cc im cc, im khng v v trn mt phng phc a)

22)( += ssG b) )4)(3(

1)( ++= sssG

c))14)(7(

)2(5)( +++=ss

ssG d) 92)( 2 +

+=sssG

3. Tm hm truyn v im cc ca h thng sau

[ ]

==

+

=

000

)0(001

)(100

420100012

xxy

tuxx&

4. Tm cc thng s ca h thng bc 2 OS%,, prsn TTT a)

12012120)( 2 ++= sssG b) 01.0002.0

01.0)( 2 ++= sssG 5. Tm p ng u ra c(t) khi bit tn hiu tc ng l tn hiu bc thang n v a)

)5(5)( += sssC b) )4(

4)( += sssC

c))16(

16)( 2 += sssC d) )168(16)( 2 ++= sssC

77


CHNG 4: CC PHNG PHP GIM THIU

H THNG A CP Mc ch: trn thc t cc h thng k thut c biu din bng cc s khi rt phc tp, tm c quan h gia tn hiu u vo v u ra ca h thng tc l phi tm c hm truyn t ca h thng. Do ta phi tm cch rt gn h thng tm c hm truyn chung ca ton b h thng. 4.1 S khi ca mt h thng

C(s)

Hm truyn ca

G(s)

R(s) u vo

(Input) u ra

(Output)

Hnh 4.1: S khi ca h thng Quy nh:

- K hiu tn hiu u vo: R(s). - K hiu tn hiu u ra: C(s). - K hiu cc hm truyn con: Gi(s) - K hiu hm truyn h thng: G(s).

Quan h ca tn hiu u vo v u ra c biu din di dng hm truyn (transferfunction):

)()()(

sRsCsG = (4.1)

Hai dng biu din: - S khi. - hnh tn hiu Graph

4.1.1 H thng dng ni tip

H thng c gi l mc ni tip nu tn hiu ra ca phn t trc l tn hiu vo ca phn t sau. Tn hiu vo ca h thng l tn hiu vo ca phn t u tin. Tn hiu ra ca h thng l tn hiu ra ca phn t cui cng.

G1(s) G2(s) G3(s) G4(s)C(s) R(s)

G1(s)xG2(s)xG3(s)xG4(s)C(s)R(s)

Hnh 4.2: S khi ca h thng ni tip

78

Gio trnh L thuyt iu khin t ng 1 V d: Ta c m hnh nh sau:

Hnh 4.3: H thng ghp ni tip.

Hnh 4.3a) hm truyn c tnh:

11

11

1

1'

1 1

1

)()()(

CRs

CRsVsVsG

+== (4.2)

Hnh 4.3 b) hm truyn c tnh:

22

22

1

22 1

1

)()()(

CRs

CRsVsVsG

+== (4.3)

Hnh 4.3 c) ta tnh c hm truyn ca h thng bng mch vng hoc theo nt:

2211122211

2

2211

1

2

1111

1

)()()(

CRCRs

CRCRCRs

CRCRsVsVsGT

+

+++

== (4.4)

Nhng nu tnh theo cng thc ca s ni mc ni tip

22112211

2

221112

1

2

111

1

)()()()()(

CRCRs

CRCRs

CRCRsGsGsVsVsGT

+

++

=== (4.5)

Ta thy s khc nhau l do gia hai h thng tn ti mt h s t l. khc phc gia hai h thng ta mc thm mt khu khuch i nh hnh 4.3 d).

79

Gio trnh L thuyt iu khin t ng 1 4.1.2 H thng dng song song(Parallel Form)

H thng mc song song l h thng c tn hiu vo ca h thng l tn hiu vo ca cc phn t thnh phn, cn tn hiu ra ca h thng bng tng i s ca cc tn hiu thnh phn.

G1(s)

G2(s)C(s) R(s)

G3(s)

80

]Hnh4.4: S khi ca h thng mc song song

[ )()()()()( 321 sGsGsGsRsC ++= (4.6) 4.1.3. H thng dng phn hi (Feedback Form)

H thng c mch mc phn hi gm hai mch: mch thun v mch phn hi. Tn hiu ra ca mch thun l tn hiu ra ca h thng v l tn hiu vo ca mch phn hi. H thng c hai dng phn hi:

- Phn hi m: E(s) = R(s) C(s) . - Phn hi dng: E(s) = R(s) + C(s).

G4(s)

G1(s) G2(s) G3(s)C(s) R(s)

Hnh 4.5: S khi ca h thng c phn hi

Hnh 4.6: a) H thng phn hi m b) H thng phn hi dng

c) Hm truyn ca h thng c phn hi

Gio trnh L thuyt iu khin t ng 1 - Mch phn hi n v:

G(s)

R(s) C(s) E(s)

Hnh 4.7: S khi h thng phn hi n v

E(s) = R(s) C(s) (4.7) Mt khc:

)()()(

sGsCsE = (4.8)

Hm truyn ca h thng c tnh l:

)(1)(

)(sG

sGsGe m= (4.9)

Cc k nng bin i s c bn: - Chuyn tn hiu u vo: T trc ra sau mt khi:

X(s)

X(s)

G(s) G(s

G(s

C(s) R(s)C(s)R(s)

T sau mt khi ra trc mt khi:

- Chuyn i tn hiu ra:

T trc mt khi ra sau khi :

81

G(sR(s) C(s)

X(s)

G(sC(s) R(s)

)(1sG

X(s)

)(1sG

G(sR(s)

C2(s

C1(s

G(sR(s)

C2(s

C1(s

Gio trnh L thuyt iu khin t ng 1 C1(s) = R(s) C2 (s) = R(s).G(s)

T sau mt khi ra trc khi :

G(sR(s)

C2(s

C1(sG(s

R(s)

C2(s

C1(sG(s

C1(s) = C2(s) = R(s).G(s) - Cc b cng lin nhau c th i ch cho nhau hoc cng xp chng li:

82

Hoc l

X1 C

X3X2

X1 C

X3 X2

X1 C

X3X2

C = X1 - X2 + X3 V d 1: Rt gn h thng nh hnh sau

H1(s)

G1(s G2(s) G3(s)+

-

H2(s)

H3(s)

+

+-

+C(s)R(s)


H1(s)

G1(s G2(s) G3(s)R(s) C(s)

H3(s)(a)

G1(s G2(s)+G3(s

H1(s)-H2(s)+H3(s)

+

-

(b)

C(s)R(s)

+

-

H2(s)

-+

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

3213121

3121

sHsHsHsGsGsGsGsGsG

++

(c)

C(s)R(s)

Hnh 4.8 : Hnh bin i cc s khi c bn. V d 2: Rt gn s khi p dng cc quy tc di chuyn tn hiu

H3(s)

G1(s G2(s) G3(s) R(s)

H2(s)

H1(s)

+

-

+

-

V2(s

V6(s

V1(s +

-

+

V8(sV7(s

V5(sV4(s

V3(s

C(s)

R(s) G1(s G2(s) )()(1)(

33

3

sHsGsG

+H2(s)

H1(s)

+

-

+

V7(s

V3(s

V1(s

V6(s

V2(s +V4(s

)(1

2 sG

(a)

+

-

C(s)

83


G1(s)G2(s) )()(1)(

33

3

sHsGsG

+

)()(

1

2

sGsH

H1(s)

+

-

+ 1)(

1

2

+sG

V4(sV1(s

-

C(s)R(s)

(b)

G1(s)G2(s)

+

+

)()(1)(

1)(

1

33

3

2 sHsGsG

sG

)()()(

11

2 sHsGsH +

+ V4(s

-

C(s)R(s)

)()()()()(1

)()(

12122

21

sHsGsGsHsGsGsG

++

+

+

)()(1)(

1)(

1

33

3

2 sHsGsG

sG

V4(s(c)

C(s)R(s)

(d)

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

3312122

231

sHsGsHsGsGsHsGsGsGsG

++++ C(s)R(s)

(e) Hnh 4.9: Rt gn s p dng cc quy tc bin i

4.2 Phn tch v thit k h thng phn hi Mc ch : ng dng cc quy tc trn phn tch v thit k h thng bc 2.

Phn trm qu iu chnh, thi gian nh, thi gian tng c th c tnh ton t hm truyn ca h thng.

Xt h thng:

)( ass

K+

R(s) C(s)

-

+

Hnh 4.10: H thng c phn hi m

i tng c hm truyn l: )( ass

K+ (4.10)

Hm truyn ca h thng c tnh l:

84


KassKsT ++= 2)( (4.11)

Trong : K: H s khuch i (t l gia in p u vo v u ra)

Khi h s K thay i, cc im cc thay i qua 3 ch hot ng ca h thng bc hai: dao ng tt dn, tt tt dn ti hn v di tt dn. V d K bin i trong ri gia 0 v a2/4, cc im cc ca h thng l thc

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