Gii thiu
Cc h thng iu khin k thut hin i c nghin cu v ng dng vo cc lnh vc
k thut, sn xut cng nghip v i sng mt cch mnh m.
Rt nhiu mc tiu quan trng ca k thut khng th thc hin c vi cht lng
cao nu khng da vo iu khin h thng.
Mn hc : H thng iu khin k thut hin i dng cho i hc k thut khng
chuyn ngnh iu khin, thi lng 30 tit. V vy Chng ti bin son theo s tng
hp c chn lc rt gn nhng vn c gng bao trm c cc ni dung thit yu nht
khi cn phi gii quyt c bn cc vn iu khin mt h thng k thut. Cui mi
chng c ng dng phn mm Matlab gii quyt cc ni dung ca iu khin trn my
tnh.
Chng 1.
M u1.1 Cc h thng iu khin lin tc v iu khin ri rc.
1.2 Cc h thng iu khin lin h v iu khin kn.
1.3 Khng gian trng thi v hm truyn t.
1.4 M hnh ho ton hc ca cc h thng vt l.
1.5 Cc m hnh ca h thng iu khin.
1.6 Tuyn tnh ho cc h thng phi tuyn.
1.7 S dng MATLAB phn tch v thit k h thng iu khin.
Chng 2.
Hm truyn t 2.1 Cc biu din trong min tn s.
2.1.1. Hm truyn t ca h thng iu khin lin tc.
2.1.2. Hm truyn t ca h thng iu khin lin tc.
2.1 S khi.
2.2 i s s khi.
2.3 Graph tn hiu v qui tc Mason.
2.4 Cc h thng ly mu d liu.
2.5.1 Hm truyn h thng mch h.
2.5.2 Hm truyn h thng mch kn.
2.5.3 Hm truyn ca h thng iu khin kn.
2.6 Xc nh hm truyn trong Matlab.
Chng 3.
Khng gian trng thi.
3.1 Cc m hnh khng gian trng thi.
3.1.1 Cc m hnh khng gian trng thi v cc phng trnh vi phn.
3.1.2 Xc nh cc bin trng thi t hm truyn.
3.2 p ng thi gian t phng trnh trng thi.
3.2.1 Nghim trong min thi gian.
3.2.2 Tm nghim bng phng php bin i Laplace.
3.2.3 M hnh khng gian trng thi v hm truyn.
3.3 Cc M hnh thi gian ri rc.
3.3.1 Phng trnh sai phn v khng gian trng thi.
3.3.2 Hm truyn ri rc v m hnh khng gian trng thi.
3.3.3 Ri rc ho cc h thng lin tc.
3.3.4 Nghim ca cc phng trnh trng thi.
3.3.5 Tm nghim ca cc phng trnh trng thi bng bin i Z.
3.4 H phng trnh c trng v tr ring.
3.4.1 Tr ring.
3.4.2 Dng phn ly.
3.5 Xc nh khng gian trng thi bng MATLAB .
Chng 4.
n nh ca h thng iu khin tuyn tnh.
4.1 Khi nim n nh.
4.2 Tr ring ca h thng v n nh.
4.3 Tiu chun Lyapunov.
4.4 Tiu chun Routh- Hurwitz.
4.5 Kim tra n nh i s ca h thng ri rc.
4.6 Kho st n nh trong min tn s.
4.7 ng dng MATLAB.
Chng 5.
Kh nng n nh v quan st c ca h thng
iu khin.
5.1 Kh nng quan st c ca cc h thng ri rc.
5.2 Kh nng quan st c ca cc h thng lin tc.
5.3 Kh nng iu khin c ca cc h thng ri rc.
5.4 Kh nng iu khin c ca cc h thng lin tc.
5.5 ng dng MATLAB.
Ti liu tham kho
[1]. Z. Gajic and M. Lelic, Modern Control Systems
Engineering,
Prentice Hall, Englewood Cliffs, New Jersey, 1996.
[2]. Kuo, B., Digital Control Systems, Saunder college
Publishing,
New York, 1992.
[3]. Kuo, B., Automatic Control Systems, Prentice Hall,
Englewood
Cliffs, New Jersey, 1991.
Chng I:
Cc vn c bn v
H thng Iu khin tuyn tnh v phi tuyn
1.1- H thng iu khin.
* iu khin (control): L tc ng ln i tng i tng lm vic theo mt mc ch
no .
* H thng iu khin (controller System): L mt tp hp cc thnh phn vt
l c lin h tc ng qua li vi nhau ch huy hoc hiu chnh bn thn i tng hay
mt h thng khc.
* Xung quanh ta c rt nhiu h thng iu khin nhng c th phn chia thnh
3 dng h thng iu khin c bn.
- H thng iu khin nhn to.
- H thng iu khin t nhin (bao gm iu khin sinh vt).
- H thng iu khin t nhin v nhn to.
Trong cc h thng i tng iu khin c th l h thng vt l, thit b k thut,
c ch sinh vt, h thng kinh t, qu trnh v.v... i tng nghin cu ca ta l
cc thit b k thut gi l iu khin hc k thut.
Mi h thng (hoc phn t ca h thng) k thut, u chu tc ng ca bn ngoi v
cho ta cc p ng ra. Gi tc ng vo l u vo, tc ng ra l u ra( hoc tn hiu
vo, tn hiu ra).
Hnh 1-1
1.2- Phn loi h thng iu khin.
* Vic phn loi h thng iu khin (Controller System) c rt nhiu hnh
thc tu theo gc nhn nhn nh gi: phn loi theo tn hiu vo, theo cc lp
phng trnh vi phn m t qu trnh ng lc hc ca h thng. Theo s vng kn
trong h, v.v... Tuy nhin y ch l tng i.
* Theo tnh cht lm vic v ni dung c bn ca iu khin, h thng iu khin
c hai loi c bn trong vic phn tch tnh nng ca h. H thng iu khin kn v
h thng h, c phn bit bng quan h ca tc ng iu khin i vi h (tn hiu vo)
v p ng (tn hiu ra).
+ H thng iu khin h: Tc ng iu khin (tn hiu vo) c lp vi u ra (tn
hiu ra).
+ H thng iu khin kn: L h thng c tc ng iu khin ph thuc u ra (tn
hiu ra).
a)
b)
Hnh 2.1: a) H thng iu khin h.
b) H thng iu khin kn..
* c tnh ca h thng h:
- chnh xc c quyt nh bi iu chnh (cn) tn hiu v c duy tr chnh xc iu
chnh c khng.
- Nhy cm vi cc tc ng bin i xung quanh: nhit, dao ng, rung, xung
lc, in p, v.v...
- p ng chm khi tn hiu vo thay i.
+ u im:
. n gin, gi thnh thp (dng khi chnh xc va phi).
. Vn mt n nh khng nghim trng lm .
* c tnh ca h thng kn (phn hi).
(Phn hi l tnh cht c trng ca h thng kn, n cho php u ra hoc cc bin
s iu khin khc trong mt h c so snh vi u vo ca h (hoc phn t trong h)
hoc u vo ca mt phn t nm ngoi h nhm to tc ng iu khin thch hp theo
quan h hm s no gia u ra v u vo). H phn hi iu chnh li mt cch t ng p
ng ra (u ra) nhm gim thiu sai s.
- Nng cao chnh xc iu khin, c kh nng to li u ra.
- Tc p ng nhanh.
- chnh xc ph thuc vo cc iu kin lm vic.
- iu khin mm.
- Gim tnh phi tuyn v nhiu.
- Gim nhy cm tn hiu vo, ra i vi cc thay i tnh cht ca h.
- Tng b rng di tn (dy tn s ca u vo h c p ng tt).
- C khuynh hng dao ng hoc khng n nh .
1.3- Cc m hnh din t h thng iu khin.
tin vic nghin cu v cc vn iu khin cn s dng cc s (m hnh) din t cc
thnh phn ca h thng sao cho r rng mi mi quan h bn trong v ngoi h
thng d dng phn tch, thit k v nh gi h thng.
Thc t s dng cc m hnh sau l ph bin v thun tin:
1) H thng cc phng trnh vi phn v cc quan h ton hc khc m t h thng
(thng khng trc quan khi m t h thng).
2) S khi.
3) Graph tn hiu.
(S khi v Graph tn hiu l cch biu din bng ho din t mt h thng vt l
hoc mt h phng trnh ton c trng cho cc phn t ca h thng - Din t mt cch
trc quan hn).
* V mt l thuyt mi h thng iu khin u c th din t bng cc phng trnh
ton. Gii cc phng trnh ny v nghim ca chng s din t trng thi ca h
thng. Tuy nhin vic gii phng trnh thng kh tm nghim (c trng hp khng
tm c) lc cn t cc gi thit n gin ho nhm dn ti cc phng trnh vi phn
tuyn tnh thng H iu khin tuyn tnh lin tc.
* Phn ln k thut iu khin hin i, l s pht trin ca cc m hnh ton hc
cho cc hin tng vt l. Sau da vo cc m hnh ton hc nghin cu cc tnh cht
ca h thng iu khin.
1.4- S khi.
* S khi c biu th bng cc khi lin kt vi nhau din t mi quan h u vo
v u ra ca mt h thng vt l.
* S khi thun tin din t mi quan h gia cc phn t ca h thng iu
khin.
V d:
a)
b)
Hnh 4-1. c)
* Cc khi c th l mt thit b hoc dng c v c th l mt hm (chc nng -
function) xy ra trong h thng.
Khi: K hiu thut ton phi thc hin u vo to u ra.
ng ni: ng ni gia cc khi biu th i lng hoc bin s
trong h thng.
Mi tn: Ch tiu ca dng thng tin hoc tn hiu Cc khi ni tip
nhau th u ra ca khi trc l u vo ca khi sau.
* Hm truyn t:
T l gia u vo v u ra ch phng thc m tn hiu truyn qua khi.
G1 = hm truyn ca khi.
Gh = hm truyn ca h.
* im t: Biu hin thut ton cng hoc tr k hiu bng mt vng
trn u ra ca im t l tng i s ca cc u vo.
Hnh 4.2a
* im tn: Cng mt tn hiu hoc mt bin s phn ra nhiu nhnh ti im gi l
im tn, tc l ti u ra p ln nhiu khi khc k hiu l mt nt trn en.
Hnh 4.2b
* Cu trc s khi ca h thng iu khin kn.
Hnh 4.3
Hnh (4.3) din t mt h thng iu khin kn bng s khi. Cc khi m t cc
phn t trong h c ni vi nhau theo quan h bn trong ca h thng.
* Cc bin s ca h:
(1) Gi tr vo V: tn hiu ngoi p vo h.
(2) Tn hiu vo chun R: rt t gi tr vo V l tn hiu ngoi h p ln h iu
khin nh mt lnh xc nh cp cho i tng. R biu th cho mt u vo l tng dng
lm chun so snh vi tn hiu phn hi B.
(3) Bin s iu khin M (tn hiu iu chnh): l i lng hoc trng thi m phn
t iu khin G1 p ln phn t (i tng) iu khin G2 (qu trnh c iu khin).
(4) Bin s ra C (tn hiu ra): l i lng hoc trng thi ca i tng (hoc
qu trnh) c iu khin.
(5) Tn hiu phn hi B: l mt hm ca tn hiu ra C c cng i s vi vo chun
R c tn hiu tc ng E.
(6) Tn hiu tc ng E (cng gi l sai lch hoc tc ng iu khin) l tng i
s (thng l tr) gia u vo l R vi phn t B l tn hiu p ln phn t iu
khin.
(7) Nhiu u: l tn hiu vo khng mong mun nh hng ti tn hiu ra C. C
th vo i tng theo M hoc mt im trung gian no (mong mun p ng ca h i vi
nhiu l nh nht).
* Cc phn t ca h:
(1) Phn t vo chun GV: chuyn i gi tr vo V thnh tn hiu vo chun R
(thng l mt thit b chuyn i).
(2) Phn t iu khin G1: l thnh phn tc ng i vi tn hiu E to ra tn
hiu iu khin M p ln i tng iu khin G2 (hoc qu trnh).
(3) i tng iu khin G2 l vt th, thit b, qu trnh m b phn hoc trng
thi ca n c iu khin.
(4) Phn t phn hi H: l thnh phn xc nh quan h (hm) gia tn hiu phn
hi B v tn hiu ra C c iu khin (o hoc cm th tr s ra C chuyn thnh tn
hiu ra B (phn hi).
(5) Kch thch: l cc tn hiu vo t bn ngoi nh hng ti tn hiu ra C. V
d tn hiu vo chun R v nhiu u l cc kch thch.
(6) Phn hi m: im t l mt php tr E = R - B
(7) Phn hi dng: im t l php cng: E = R + B
(iu khin kn gm hai tuyn: Tuyn thun truyn tn hiu t tc ng E n tn
hiu ra C. Cc phn t trn tuyn thun k hiu G (G1 , G2, ...) tuyn phn hi
truyn t tn hiu ra C n phn hi B cc phn t k hiu l H (H1 , H2 ,
...).
1.5- Cc h thng iu khin lin tc v gin on.
Cc h thng thc c m t trng thi tnh hoc ng lc hc. Cc h thng tnh
thng c din t bi h thng cc phng trnh i s. Trong iu khin k thut cc h
thng tnh khng din t y trng thi ca h thng. V vy ngi ta dng cc phng
trnh vi phn/sai phn m t trng thi ng lc hc ca h thng (c bit nh l cc
h thng vi cc tham s cc b hoc tp trung) hoc cc phng trnh vi phn o hm
ring (nh l cc h thng c cc tham s phn tn).
Trong ni dung gio trnh ta nghin cu cc h thng c m t bi h cc phng
trnh vi phn/sai phn tuyn tnh, ngha l cc tham s ca h thng c lp tuyn
tnh.
V d h thng ng lc hc c m t di dng cc phng trnh vi phn/sai phn v
hng:
(t) = fc(x(t)) , x(to) = xo
(1-1)
x(k +1) = fd (x(k)) , x (ko) = xo
(1-2)
y: t : bin thi gian lin tc.
k : bin thi gian gin on.
Ch s e: (continuous- Time) - thi gian lin tc.
d: (discrete - Time) - thi gian gin on.
Nu h thng chu tc ng ca ngoi lc, hay cc tc ng vt l khc. Ta ni n
chu ti ng iu khin v phng trnh vi phn/sai phn m t trng thi ng lc ca
h thng.
(t) = fc (x(t), u(t)) ; x(to) = xo
(1-3)
x(k+1) = fd (x(k), u(k)) ; x(ko) = xo
(1-4)
y: u(t) ; u(k) ng vai tr bin iu khin. Vi mc ch ca iu khin ta
thay i bin iu khin nhn c cc p ng ca h thng k thut theo yu cu nh vy,
nhn chung vn chnh ca iu khin c th m hnh ho theo dng sau: tm bin iu
khin bng cch gii h thng phng trnh vi phn c trng ca h.
Nu cc h phng trnh vi phn (1-1) ( (1-4) l tuyn tnh ta gi h thng l
tuyn tnh. Nu l phi tuyn ta gi l h thng phi tuyn. Vic nghin cu h
thng phi tuyn tng i kh. Trong thc t, ngi ta tm cch tuyn tnh ho.
Trong phm vi gio trnh ny, chng ta ch nghin cu h thng iu khin tuyn
tnh.
1.5.1- M hnh ton:Cc h thng ng lc hc tuyn tnh c bin thi gian lin
tc hay gin on c m t bi cc phng trnh vi phn v sai phn vi h s hng. Cc
m hnh ton hc ca cc h thng c mt u vo v mt u ra nh sau:
+ an-1 + ... + a1 + ao (y(t) = u(t) (1-5)
v y(k+n) + an-1 (y(k+n-1) + ... + a1y(k+1) + aoy(k) = u(k)
(1-6)
y: n - bc ca h thng.
y - p ng (u ra ca h thng) (System output).
u - tc ng t bn ngoi vo h thng (System input).
- Cc tc ng t bn ngoi ca h thng c iu khin bi cc tc ng bn trong t
trng thi ban u ca n (iu kin u).
+ i vi h thng thi gian lin tc cc iu kin u l cc gi tr bit trc ca
cc bin u ra n bc (n-1) i vi thi gian u to.
y(to) ; , ...,
(1-7)
Trong lnh vc thi gian gin on cc iu kin u l:
y(ko) , y(ko + 1), ..., y (ko + n-1)
(1-8)
y ta ch nghin cu cc h thng lin tc vi cc h s hng ai , i = 0, 1,
..., n-1, .
Trong thc t cc u vo c th c cho di dng hm vi phn.
+ an-1 + ... + a1 + ao (y(t) =
= bm + bm-1 + ... + b1 + bou(t) (1-9)
v y(k+n) + an-1.y(k+n-1) + ...+ a1.y(k+1) + aoy(k) =
= bm u(k+m) + bm-1 . u(k+m-1) + ...+ b1 u(k+1) + bo u(k)
(1-10)
y: ai , i = 0, 1, ..., n-1 v bj, j = 0, 1, 2, ..., m l cc h s
hng.
* Cc phng trnh t (1-5) n (1-10) c gi l cc m hnh ton hc ca h thng
iu khin.
1.5.2- Cc phng trnh trng thi v hm truyn t.Khi phn tch v thit k h
thng iu khin tuyn tnh thng s dng mt trong hai hnh thc sau:
+ i vi lnh vc thi gian s dng hm trng thi.
+ Trong lnh vc tn s dng hm truyn t.
Nh trn, ta xt h phng trnh vi phn, sai phn o hm n bc n (h thng bc
n) ; n thc cht l trng thi ca cc bin. Cc trng thi ca bin c m t nh l
vect x. Cc phng trnh trng thi c m t di dng sau (h thng tuyn
tnh).
(t) = A.x(t) + B.u(t) ; x(o) = xo
y(t) = C.x(t) + D. u(t)
(1-11)
v x(k+1) = A. x(k) + B.u(k) ; x(o) = xo
y(k) = C.x(k) + D. u(k)
(1-12)
y: A, B, C, D l cc ma trn h s hng c kch thc.
An(n , Bn(r , CP(n , DP(r
Cc h phng trnh vit dng (1-11); (1-12) cc phng trnh trng thi ca h
thng iu khin.
+ Hm truyn t ca h thng.
* Hm truyn t ca h thng i vi h thng iu khin lin tc mt u vo v mt u
ra c nh ngha:
- L t s ca bin i Laplace ca u ra vi bin i Laplace ca u vo vi gi
thit ton b cc iu kin u ng nht bng khng (iu kin dng).
G(s) = (1-13)
i vi h thng vt l thc cc ch s trong hm truyn n ( m.
* Trong lnh vc thi gian gin on (iu khin ri rc) vic bin i Z ng
vai tr ca bin i Laplace:
Hm truyn c dng sau:
G(z) = (1-14)
* i vi h thng nhiu u vo nhiu u ra vi r u vo, p u ra, cc hm truyn
l cc phn t ca ma trn cp p(r phn t , vi ch s i ca phn t th i ca u
vo, ch s th j ca phn t th j u ra.
G11(s)G12(s).....G1r(s)
G21(s)G22(s).....G2r(s)
G(s) =..........Gji(s).....(1-15)
....................
GP1(s)..........GPr(s)
y: Gji(s) = ; cc u vo khc ui(s) u coi l bng khng.
(Nguyn l c lp tc dng).
* Mt cch tng t vi h thng iu khin gin on ta c hm truyn ca h thng
nhiu u vo nhiu u ra.
G11(z)G12(z).....G1r(z)p(r
G21(z)G22(z).....G2r(z)
G(z) =..........Gji(z).....(1-16)
....................
GP1(z)..........GPr(z)
y: s - s phc - bin Laplace.
z = eS.T - bin ca php bin i z.
1.6- M hnh ton hc ca h thng thc.
1.6.1. H thng c kh dch chuyn thng.
a)
S m hnh ho h thng ng lc hc c dao ng l dch chuyn thng. Ta c th
vit hai phng trnh ca chuyn ng nh sau:
F1 = m1 + B1 + k1(y1 - y2)
(1-17)
v F2 = m2 + B2+ k2y2- B1- k1 (y1-y2) (1-18)
y ta c th coi F1 , F2 l hai u vo.
y1, y2 l hai u ra, ta c th vit li hai phng trnh trn.
m1 + B1 + k1y1 - B1 - k1y2 = F1
(1-19)
- B1 - k1y1 + m2 + (B2 + B1) + (k2 + k1) y2 = F2 (1-20)
T cc phng trnh (1-19), (1-20) cc dng bin ca phng trnh trng thi c
t nh sau:
x1 = y1 ; x2 = ; x3 = y2 ; x4 =
u1 = F1 ; u2 = F2
(1-21)
Nh vy phng trnh trng thi ca h thng trn c vit nh sau:
1 = x2
2 = - x1 - x2 + x3 + x4 + u1
3 = x4
4 = x1 + x2 - x3 - x4 + u2v y1 = x1
y2 = x3
Vit di dng ma trn.
10100x100
2=-
-
(x2+
0u1
30001x300u2(1-22)
4
-
-
x40
v
x1
Y1=1000x2 (1-23)
Y200 10x3
x4
Nh vy y l mt h thng tuyn tnh.
1.6.2- M hnh ca h thng iu khin con lc ngc.
H thng con lc ngc l mt nhm h thng ng lc hc uc s dng ph bin.
a)
Gi s cho con lc l tng c di l mang khi lng m1 (tp trung) xe c khi
lng tp trung m2. C lc ngoi F tc ng, xe ch c dch chuyn l x, dch
chuyn ca con lc l gc (.
- p dng nh lut 2 Newton.
m2. + m1 = F
(1-24)
- Phng trnh cn bng mmen quay quanh im A.
m1. l . cos(() . d2(x + lsin() - m1lsin( = m1glsin( (1-25)
y: g = 9,8 m/s2 gia tc trng trng.
= cos( ; = - sin( .
= cos( - sin( . (1-26)
= -sin( - cos( .
Thay (1-26) vo (1-24) v (1-25) ta c m hnh ton hc ca h iu khin bc
2.
(m1 + m2) - m1lsin( + m1l .cos( = F
Cos( . + l . = g.sin( (1-27)
(1-27) ch ra h thng iu khin phi tuyn.
1.7- Tuyn tnh ho h thng phi tuyn.
Trong thc t khng c mt h thng vt l no c th m t tuyt i chnh xc bng
phng trnh vi phn h s hng tuy nhin nhiu h phi tuyn c th xp x hoc coi
nh tuyn tnh trong tng on lm vic. C nhiu phng php c p dng cho vic
tuyn tnh ho h thng phi tuyn. Phng php trung bnh gn im lm vic. Phng
php tuyn tnh ho iu ho v phng php sai lch nh.
1.7.1- Phng php trung bnh gn im lm vic.
y l phng php n gin c dng trong thit k cc h thng khi c tnh trn
khng th xp x ho c bng cc hm gii tch.
Phng php ny p dng cho cc h c nhng phn t ch phi tuyn trng thi
tnh, quan h gia u ra y vi u vo u l trng thi xc lp (n nh).
Gi thit trong on: - uM < u < um c tnh phi tuyn c th xp x
ho bng ng thng.
Trong : y = K . u ; k = = tg( ; ( l dc.
1.7.2- Phng php tuyn tnh ho iu ho.
Phng php ny c dng khi h c mt phn t tuyn tnh ni sau mt phn t phi
tuyn lm vic ch t dao ng. Cc tn hiu trong h l lm tun hon theo thi
gian.
Phng php ny da trn c s khai trin hm sng thnh chui hm dng sin
(chui Fonricr) iu ho c tn s l (, 2(, 3(, ... c bin v gc pha khc
nhau. Gi thit cc hm iu ho bc cao khc (2(, 3(, ...) c bin nh b qua
ch gi li thnh phn iu ho bc nht (() (gi thit lc) ngha l:
a)
Trong : u(t) = Um . sin ((t + ()
y(t) = Ym1 . sin ((t + ()
Trong Um = Ym1 v ( - ( = ( c gi l iu kin cn bng iu ho.
1.7.3- Phng php sai lch nh.Theo phng php ny vic tuyn tnh ho c
thc hin bng cch khai trin hm phi tuyn thnh chui Taylor ti vng ln cn
im n nh (tng ng vi ch xc lp). Ch kho st cc sai lch bc nht trong
chui . Sai lch so vi trng thi nnh cng nh th vic nh gi cc qu trnh ca
phn t phi tuyn c sai s cng b sau khi bin i tuyn tnh.
a) H thng (bc nht) phi tuyn.
(t) = f(x(t) , u(t) )
(1-28)
Gi thit rng h thng lm vic trng thi xc lp vi qu o xn(t) khi n c
iu khin bi tn hiu vo un(t). Chng ta gi xn(t) v un(t) l qu o danh
ngha v u vo danh ngha theo phng trnh (1-28) ta c:
n(t) = f(xn(t) , un(t) )
(1-29)
By gi ta gi thit rng thay i ca h phi tuyn (1-28) ln cn qu o danh
nh mt lng nh (v cng b).
x(t) = xn(t) + (x(t)
(1-30)
Lng bin i v cng b ny l do thay i u vo:
u(t) = un(t) + (u(t)
(1-31)
T cc phng trnh (1-28), (1-30), (1-31) ta c:
n(t) + ((t) = f(xn(t) + (x(t), un(t) + (u(t)) (1-32)
S dng khai trin Taylor vi cc i lng (x(t), (u(t) ta s c:
n(t) + ((t) = f(xn(t), un(t)) + (xn , un) (x(t) +
+ (xn , un) (u(t) + cc thnh phn bc cao. (1-33)
(Cc thnh phn bc cao l cc i lng v cng b (x2 , (u2, (x.(u, (x3...)
c b qua, t y ta c:
((t) = (xn , un) (x(t) + (xn , un) (u(t) (1-34)
Nh vy bng vic trnh by xp x vi (x(t) ta tin hnh tuyn tnh ho theo
sai lch bc nht c phng trnh xp x bc nht (1-34).
t: ao = - (xn , un); bo = (xn , un)
(1-35)
Ta c phng trnh m t h thng tuyn tnh:
((t) + ao(t)(x(t) = bo(t). (u(t)
(1.36)
iu kin u ca h thng c tuyn tnh ho c xc nh.
(x(to) = x(to) - xn(to) (1.37)
b) H phi tuyn bc 2:
= f( x, , u, )
(1.38)
Vi gi thit rng:
x(t) = xn(t) + (x(t); (t) = n(t) + ((t)
u(t) = un(t) + (u(t); (t) = n(t) + ((t) (1.39)
Tng t ta c:
n + (= f (xn + (x, n + ( , un + (u, n + () (1.40)
p dng khai trin Taylor ln cn cc im danh ngha: xn , n , un , n v
ta c:
((t) + a1((t) + ao(x(t) = b1((t) + bo(u(t) (1-41)
Cc h s xc nh theo:
a1 = - (xn , n , un , n ), ao = - (xn , n , un , n )
b1 = (xn , n , un , n ), bo = (xn , n , un , n ) (1-42)
Cc iu kin u c xc nh.
(x(to) = x(to) - xn(to) ; ((to) = (to) - n(to)
V d: Cho h thng phi tuyn.
= Sin( - u.cos( = f((, u)
Trong : ( = ((t) ; u = u(t)
y l m hnh ton ca thanh thng ng cn bng, u: lc ngang; ( l gc lch
khi phng thng ng.
y l h thng ng lc hc bc 2. Trng thi danh nh ca n:
n(t) = (n(t) = 0 ; un(t) = 0 ; s dng (1-42) ta c:
a1 = - = 0, ao = - = - (Cos( + Usin() = -1
b1 = = 0 ; bo = = - Cos( = -1
Vy phng trnh tuyn tnh ho:
dy: (((t) = ((t) , (u(t) = u(t)
ng thi (n(t) = 0, un(t) = 0
Bi tp: Tuyn tnh ho h thng con lc ngc.1.8- ng dng php bin i
Laplace gii cc phng trnh
vi phn h s hng.Phng php 1:
ng dng i vi cc phng trnh n gin cp thp (dng bng bin i L v cc tnh
cht ca php bin i gii).
1- M hnh ho ton hc h cho.
2- Bin i Laplace cho tng s hng.
3- T phng trnh bin i rt ra nghim di dng Laplace.
4- Bin i ngc nhn c nghim l hm thi gian.
V d: Gii phng trnh ca h m t.
Rx + l = E ; x(o) = 0
L(Rx + l) = L(E)
RX(s) + l(X(s).S - x(o)) =
X(s) (R + lS) =
X(s) = =
Bin i ngc: L( (1 - e-at)) =
x(t) =
Phng php 2:
Phng php 2.1: ng dng cho lp phng trnh.
= x trong an = 1
Vi cc iu kin u = yok
k = 0, 1, 2, ..., (n-1) ; yo : cc hng.
1) Bin i Laplace ca phng trnh vi phn ny l:
{ai [SiY(s) - S(i-1-k) . y ok ]} = x(s) (1)
2) T y rt ra u ra:
Y(s) = +
Trong : aisi = ansn + ...+ a1s1 + ao ; a thc c trng.
3) Bin i Laplace ngc phng trnh trn c nghim.
V d: Gii phng trnh vi phn.
+ 4 + 3y = u(t)
iu kin u: u(t) hm bc thang n v
y(o+) = -1 ; = 2
Ly laplace c hai v ca PT ta c:
3Y(s) + 4(S.Y(s) + 1) + (s2.Y(s) + S - 2) =
Y(s) (S2 + 4S + 3) = - S - 2 =
Phng php 2.1: ng dng gii cc phng trnh c dng.
= (an = 1 ; m ( n)
Vi iu kin u: x =
1) Tng t ta c bin i Laplace:
[ai (SiY(s) - S(i-1-K) . y)] = [bi(Si.X - S(i-1-K).x)
2) Rt ra u ra Laplace:
Y = Y(s) = X(s) - +
3) Bin i Laplace ngc ta c kt qu.
9- Khai trin phn thc (ng dng bin i Laplace ngc).
Xt phn thc: P(s) =
Trong : an = 1 ; m ( n
Gi P l nghim ca phng trnh c trng mu s.
+ Nu khng c nghim lp th phn thc trn c khai trin theo:
P(s) = bn +
Trong : Ci1 = (S + Pi). P(s)
V d: P(s) = =
P(s) = b2 + +
b2 = 1;
C11 = (S + 1) = 1
C21 = (S + 2) = -2
P(s) = 1 + +
L-1 [P(s)] = ((t) + e-t + (-2) e-2t
+ Nu c n1 nghim lp -P1 ; n2 nghim lp -P2 , ..., nr nghim lp -Pr
v ni = n.
Khai trin phn thc ca hm hu t l:
P(s) = bn +
EMBED Equation.3
EMBED Equation.3
K = (1, 2, ..., n-1)
bn = 0 nu m < n
H s CiK = .
V d: Khai trin phn thc:
P(s) =
P = b3 + + +
b3 = 0
C11 = . = = - 1
C12 = (S+1)2 . P(s) = = 1
C21 = (S+2) . P(s) = = 1
P = - + +
L-1[P] = -e-t + + e-2t = e-t (e-t + t - 1)
V d: Y(s) = = +
S2 + 2S + 1 = (a + c) S2 + (2a + b) S + 2b + c
ng nht thc:
a + c = 1
a = c =
2a + b = 2 ( b =
2b + c = 1
( Y(s) = . + .
Y(s) =
. + . +
y(t) = 4/5 u(t) cos(t) + u(t) . sin(t) + e-2t
H thng (hoc phn t ca h thng)
R
Cc tc ng vo
Cc p ng
C
G1
G2
M
C
M
G2
G1
R
H
Ra
B
G1
Phn t G
Vo
A
A
G2
B
C
y = EMBED Equation.3
d
dt
x
x
+
+
y
(x+y)
(x-y)
y
-
+
x
(x+y-u)
y
+
+
x
x
x
x
u
-
x
C
C
C
C
M
G2
G1
E
R +
V
GV
B
H
-
System
u(t)
input
y(t)
output
m2
m1
F1
F2
k1
B1
k2
B2
y2(t)
y1t)
m2
u
F
(
l
lcos (
y
x
lsin (
m1g
m1g sin(
x
Nonlinear
System
u(t)
Element
Linearization
y(t)
EMBED Equation.3 (t) - ((t) = - u(t)
Y(s) = EMBED Equation.3
11
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