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Chng 4
IU KHIN M
Khi nim v logic m c gio s L.A Zadeh a ra ln u tin nm 1965, ti
trng i hc Berkeley, bang California - M. T l thuyt m c pht trin v
ng dng rng ri. Nm 1970 ti trng Mary Queen, London Anh, Ebrahim
Mamdani dng logic m iu khin mt my hi nc m ng khng th iu khin c bng
k thut c in. Ti c Hann Zimmermann dng logic m cho cc h ra quyt nh.
Ti Nht logic m c ng dng vo nh my x l nc ca Fuji Electronic vo 1983,
h thng xe in ngm ca Hitachi vo 1987. L thuyt m ra i M, ng dng u tin
Anh nhng pht trin mnh m nht l Nht. Trong lnh vc T ng ho logic m ngy
cng c ng dng rng ri. N thc s hu dng vi cc i tng phc tp m ta cha bit
r hm truyn, logic m c th gii quyt cc vn m iu khin kinh in khng lm
c.
4.1. Khi nim c bn hiu r khi nim M l g ta hy thc hin php so snh
sau : Trong ton hc ph thng ta hc kh nhiu v tp hp, v d nh tp cc s
thc R, tp cc s nguyn t P={2,3,5,...} Nhng tp hp nh vy c gi l tp hp
kinh in hay tp r, tnh R y c hiu l vi mt tp xc nh S cha n phn t th
ng vi phn t x ta xc nh c mt gi tr y=S(x). Gi ta xt pht biu thng
thng v tc mt chic xe mt : chm, trung bnh, hi nhanh, rt nhanh. Pht
biu CHM y khng c ch r l bao nhiu km/h, nh vy t CHM c min gi tr l mt
khong no , v d 5km/h 20km/h chng hn. Tp hp L={chm, trung bnh, hi
nhanh, rt nhanh} nh vy c gi l mt tp cc bin ngn ng. Vi mi thnh phn
ngn ng xk ca pht biu trn nu n nhn c mt kh nng (xk) th tp hp F gm cc
cp (x, (xk)) c gi l tp m.
4.1.1. nh ngha tp m Tp m F xc nh trn tp kinh in B l mt tp m mi
phn t ca n l mt cp gi tr (x,F(x)), vi x X v F(x) l mt nh x :
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F(x) : B [0 1] trong : F gi l hm thuc , B gi l tp nn.
4.1.2. Cc thut ng trong logic m
cao tp m F l gi tr h = SupF(x), trong supF(x) ch gi tr nh nht
trong tt c cc chn trn ca hm F(x). Min xc nh ca tp m F, k hiu l S l
tp con tho mn : S = SuppF(x) = { xB | F(x) > 0 } Min tin cy ca
tp m F, k hiu l T l tp con tho mn : T = { xB | F(x) = 1 } Cc dng hm
thuc (membership function) trong logic m C rt nhiu dng hm thuc nh :
Gaussian, PI-shape, S-shape, Sigmoidal, Z-shape
0
0.2
0.4
0.6
0.8
1trapmf gbellmf trimf gaussmf gauss2mf smf
0
0.2
0.4
0.6
0.8
1zmf psigmf dsigmf pimf sigmf
Hnh 4.1:
1
min tin cy
MX
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4.1.3. Bin ngn ng
Bin ngn ng l phn t ch o trong cc h thng dng logic m. y cc thnh
phn ngn ng ca cng mt ng cnh c kt hp li vi nhau. minh ho v hm thuc v
bin ngn ng ta xt v d sau : Xt tc ca mt chic xe mt, ta c th pht biu
xe ang chy: - Rt chm (VS) - Chm (S) - Trung bnh (M) - Nhanh (F) -
Rt nhanh (VF) Nhng pht biu nh vy gi l bin ngn ng ca tp m. Gi x l gi
tr ca bin tc , v d x =10km/h, x = 60km/h Hm thuc tng ng ca cc bin
ngn ng trn c k hiu l : VS(x), S(x), M(x), F(x), VF(x)
Nh vy bin tc c hai min gi tr : - Min cc gi tr ngn ng : N = { rt
chm, chm, trung bnh, nhanh, rt nhanh } - Min cc gi tr vt l : V = {
xB | x 0 } Bin tc c xc nh trn min ngn ng N c gi l bin ngn ng. Vi mi
xB ta c hm thuc: x X = { VS(x), S(x), M(x), F(x), VF(x) } V d hm
thuc ti gi tr r x=65km/h l: X (65) = { 0;0;0.75;0.25;0 }
VS S M F VF
0 20 40 60 65 80 100 tc
1
0.75 0.25
Hnh 4.2:
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4.1.4. Cc php ton trn tp m Cho X, Y l hai tp m trn khng gian nn
B, c cc hm thuc tng ng l X, Y, khi : - Php hp hai tp m: XY + Theo
lut Max XY(b) = Max{ X(b) , Y(b) } + Theo lut Sum XY(b) = Min{ 1,
X(b) + Y(b) } + Tng trc tip XY(b) = X(b) + Y(b) - X(b).Y (b) - Php
giao hai tp m: XY + Theo lut Min XY(b) = Min{ X(b) , Y(b) } + Theo
lut Lukasiewicz XY(b) = Max{0, X(b)+Y(b)-1} + Theo lut Prod XY(b) =
X(b).Y(b) - Php b tp m: cX (b) = 1- X(b)
4.1.5. Lut hp thnh
1. Mnh hp thnh V d iu khin mc nc trong bn cha, ta quan tm n 2 yu
t: + Mc nc trong bn L = {rt thp, thp, va} + Gc m van ng dn G = {ng,
nh, ln} Ta c th suy din cch thc iu khin nh th ny: Nu mc nc = rt thp
Th gc m van = ln Nu mc nc = thp Th gc m van = nh Nu mc nc = va Th
gc m van = ng Trong v d trn ta thy c cu trc chung l Nu A th B. Cu
trc ny gi l mnh hp thnh, A l mnh iu kin, C = AB l mnh kt lun. nh l
Mamdani: ph thuc ca kt lun khng c ln hn ph thuc iu kin Nu h thng c
nhiu u vo v nhiu u ra th mnh suy din c dng tng qut nh sau: If N =
ni and M = mi and Then R = ri and K = ki and .
2. Lut hp thnh m Lut hp thnh l tn gi chung ca m hnh biu din mt
hay nhiu hm thuc cho mt hay nhiu mnh hp thnh.
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Cc lut hp thnh c bn + Lut Max Min + Lut Max Prod + Lut Sum Min +
Lut Sum Prod
a. Thut ton xy dng mnh hp thnh cho h SISO Lut m cho h SISO c dng
If A Then B Chia hm thuc A(x) thnh n im xi , i = 1,2,,n Chia hm
thuc B(y) thnh m im yj , j = 1,2,,m Xy dng ma trn quan h m R
R=
),(......)1,(............
),2(......)1,2(),1(......)1,1(
ymxnyxn
ymxyxymxyx
RR
RR
RR
=
rnmrn
mrr
mrr
......1............
2......211......11
Hm thuc B(y) u ra ng vi gi tr r u vo xk c gi tr B(y) = aT.R , vi
aT = { 0,0,0,,0,1,0.,0,0 }. S 1 ng vi v tr th k. Trong trng hp u vo
l gi tr m A th B(y) l: B(y) = { l1,l2,l3,,lm } vi lk=maxmin{ai,rik
}.
b. Thut ton xy dng mnh hp thnh cho h MISO Lut m cho h MISO c
dng: If cd1 = A1 and cd2 = A2 and Then rs = B Cc bc xy dng lut hp
thnh R: Ri rc cc hm thuc A1(x1), A2(x2),, An(xn), B(y) Xc nh tho mn
H cho tng vct gi tr r u vo x={c1,c2,,cn} trong ci l mt trong cc im
mu ca Ai(xi). T suy ra H = Min {A1 (c1), A2(c2), , An(cn) } Lp ma
trn R gm cc hm thuc gi tr m u ra cho tng vct gi tr m u vo: B(y) =
Min {H, B(y)} hoc B(y) = H. B(y)
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4.1.6. Gii m Gii m l qu trnh xc nh gi tr r u ra t hm thuc B(y)
ca tp m B. C 2 phng php gii m : 1. Phng php cc i Cc bc thc hin : -
Xc nh min cha gi tr y, y l gi tr m ti B(y) t Max G = { yY | B(y) =
H } - Xc nh y theo mt trong 3 cch sau : + Nguyn l trung bnh + Nguyn
l cn tri + Nguyn l cn phi
Nguyn l trung bnh: y = 2
21 yy +
Nguyn l cn tri : chn y = y1 Nguyn l cn phi : chn y = y2 2. Phng
php trng tm im y c xc nh l honh ca im trng tm min c bao bi trc honh
v ng B(y). Cng thc xc nh :
y =
S
S
(y)dy
)(
dyyy trong S l min xc nh ca tp m B
y1 y2
y
H
G
Hnh 4.3:
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Phng php trng tm cho lut Sum-Min Gi s c m lut iu khin c trin
khai, k hiu cc gi tr m u ra ca lut iu khin th k l Bk(y) th vi quy
tc Sum-Min hm thuc s l B(y) =
=
m
kkB y
1'
)( , v y c xc nh :
y = ( )
=
=
=
=
=
=
=
=
m
kk
m
kk
m
kyB
m
kkB
S
m
kkB
S
m
kkB
A
M
dyy
dyyy
dyy
dyyy
1
1
1 S'
1'
1'
1'
)(
)(
)(
)(
(4.1)
trong Mi = S
')( dyyy kB v Ai =
S'
)( dyykB i=1,2,,m
Xt ring cho trng hp cc hm thuc dng hnh thang nh hnh trn :
Mk = )3333(6 12222
122 ambmabmm
H+++
Ak = 2H (2m2 2m1 + a + b)
Ch hai cng thc trn c th p dng c cho lut Max-Min Phng php cao T
cng thc (4.1), nu cc hm thuc c dng Singleton th ta c:
y =
=
=
m
kk
m
kkk
H
Hy
1
1 vi Hk = Bk(yk)
y l cng thc gii m theo phng php cao.
y m1 m2
a b
H
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4.1.7. M hnh m Tagaki-Sugeno
M hnh m m ta ni n trong cc phn trc l m hnh Mamdani. u im ca m
hnh Mamdani l n gin, d thc hin nhng kh nng m t h thng khng tt.
Trong k thut iu khin ngi ta thng s dng m hnh m Tagaki-Sugeno (TS).
Tagaki-Sugeno a ra m hnh m s dng c khng gian trng thi m ln m t linh
hot h thng. Theo Tagaki/Sugeno th mt vng m LXk c m t bi lut : Rsk :
If x = LXk Then uxBxxAx kk )()( += (4.2) Lut ny c ngha l: nu vct
trng thi x nm trong vng LXk th h thng c m t bi phng trnh vi phn cc
b uxBxxAx kk )()( += . Nu ton b cc lut ca h thng c xy dng th c th m
t ton b trng thi ca h trong ton cc. Trong (4.2) ma trn A(xk) v
B(xk) l nhng ma trn hng ca h thng trng tm ca min LXk c xc nh t cc
chng trnh nhn dng. T rt ra c : += ))()(( uxBxxAwx kkk (4.3) vi
wk(x) [0 , 1] l tho mn chun ho ca x* i vi vng m LXk Lut iu khin tng
ng vi (4.2) s l : Rck : If x = LXk Then u = K(xk)x V lut iu khin
cho ton b khng gian trng thi c dng:
=
=
N
k
kk xxKwu
1)( (4.4)
T (4.2) v (4.3) ta c phng trnh ng hc cho h kn: xxKxBxAxwxwx
lkklk ))()()()(()( += V d : Mt h TS gm hai lut iu khin vi hai u vo
x1,x2 v u ra y. R1 : If x1 = BIG and x2 = MEDIUM Then y1 = x1-3x2
R2 : If x1 = SMALL and x2 = BIG Then y2 = 4+2x1 u vo r o c l x1* =
4 v x2* = 60. T hnh bn di ta xc nh c :
LXBIG(x1*) = 0.3 v LXBIG(x2*) = 0.35 LXSMALL(x1*) = 0.7 v
LXMEDIUM(x2*) = 0.75
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T xc nh c : Min(0.3 ; 0.75)=0.3 v Min(0.35 ; 0.7)=0.35 y1 =
4-360 = -176 v y2 = 4+24 = 12 Nh vy hai thnh phn R1 v R2 l (0.3 ;
-176) v (0.35 ; 12). Theo phng php tng trng s trung bnh ta c:
77.7435.03.0
1235.0)176(3.0=
+
+=y
4.2. B iu khin m
4.2.1. Cu trc mt b iu khin m Mt b iu khin m gm 3 khu c bn: + Khu
m ho + Thc hin lut hp thnh + Khu gii m
Xt b iu khin m MISO sau, vi vct u vo X = [ ]Tnuuu ...21
0.7 1
0.3
1 0.75
0 60 100 0 4 10
0.35
X y
R1 If Then
Rn If Then
H1
Hn
Hnh 4.4:
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4.2.2. Nguyn l iu khin m
Cc bc thit k h thng iu khin m. + Giao din u vo gm cc khu: m ha v
cc khu hiu chnh nh
t l, tch phn, vi phn + Thip b hp thnh : s trin khai lut hp thnh
R
+ Giao din u ra gm : khu gii m v cc khu giao din trc tip vi i
tng.
4.2.3. Thit k b iu khin m
Cc bc thit k: B1: nh ngha tt c cc bin ngn ng vo/ra. B2: Xc nh cc
tp m cho tng bin vo/ra (m ho). + Min gi tr vt l ca cc bin ngn ng. +
S lng tp m. + Xc nh hm thuc. + Ri rc ho tp m. B3: Xy dng lut hp
thnh. B4: Chn thit b hp thnh.
Hnh 4.5:
e B y
lut iu khin
Giao din u vo
Giao din u ra
Thit b hp thnh
X e u y BK M I TNG
THIT B O
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B5: Gii m v ti u ho. Nhng lu khi thit k BK m
- Khng bao gi dng iu khin m gii quyt bi ton m c th d dng thc hin
bng b iu khin kinh in.
- Khng nn dng BK m cho cc h thng cn an ton cao. - Thit k BK m
phi c thc hin qua thc nghim.
Phn loi cc BK m i. iu khin Mamdani (MCFC) ii. iu khin m trt
(SMFC) iii. iu khin tra bng (CMFC) iv. iu khin Tagaki/Sugeno
(TSFC)
4.2.4. V d ng dng
Dng iu khin m iu khin h thng bm x nc t ng. H thng s duy tr cao
bn nc mt gi tr t trc nh m hnh bn di.
M hnh :
Ba b iu khin m (control) s iu khin : bm, van1, van2 sao cho mc
nc 2 bn t gi tr t trc (set).
S simulink:
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S khi iu khin:
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Thit lp h thng iu khin m : Xc nh cc ng vo/ra : + C 4 ng vo gm :
sai lch e1, e2; o hm sai lch de1, de2 + C 3 ng ra gm : control1,
control2, control3 Xc nh bin ngn ng : Sai lch E = {m ln, m nh, bng
khng, dng nh, dng ln} E = {NB, NM, ZR, PM, PB} o hm D = {gim nhanh,
gim va, khng i, tng va, tng nhanh} D = {DF, DM, ZR, IM, IP} iu khin
C = {ng nhanh,ng chm,khng i,m chm,m nhanh} C = {CF, CS, NC, OS, OF}
Lut iu khin : + Khi controller1 v controller2 : (Hai khi ny ch khc
nhau lut hp thnh)
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Lut hp thnh m Max Min
DE Khi controller1 ERROR DB DM ZR IM IB NB OF OF NC NM OS ZR OF
OS NC CS CF PM CS PB NC CF CF DE Khi controller2 ERROR DB DM ZR IM
IB NB CF CF NC NM CS ZR CF CS NC OS OF PM OS PB NC OF OF
+ Khi control3 y l khi iu tit lu lng cho bn 2, ta a ra mc u tin
nh sau : Khi sai lch bn 1 ln th van2 s iu tit sai lch ny nh ri mi n
bn 2. If error1=NB and de1=DB Then control=CF If error1=NB and
de1=DM Then control=CS If error1=NB and de1=ZR Then control=CS If
error1=NM and de1=DB Then control=CS
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If error1=PB and de1=IB Then control=OF If error1=PB and de1=IM
Then control=OF If error1=PB and de1=ZR Then control=OF If
error1=PM and de1=IB Then control= OF If error1NB and error2=NB and
de1DB and de2=DB Then control=OF If error1NB and error2=NB and
de1DB and de2=DM Then control=OF If error1NB and error2=NB and
de1DB and de2=ZR Then control=OF If error1NB and error2=NM and
de1DB and de2=DB Then control=OS If error1NB and error2=NM and
de1DB and de2=DM Then control=OS If error1PB and error2=PB and
de1IB and de2=IB Then control=CF If error1PB and error2=PB and
de1IM and de2=IB Then control=CS
Kt qu p ng vi cc thng s h thng : - Chiu cap bn height=1m - Din
tch y area = 0.125m2 - Lu lng max pump maxflow = 1lit/s - Din tch
ng dn pipe area = 0.001m2
mc nc t Zdat=[0.5 0.3] mc nc ban u Zinit=[0 0]
z (m)
thi gian (s)
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mc nc t Zdat=[0.5 0.4] mc nc ban u Zinit=[0.8 0]
4.3. Thit k PID m
C th ni trong lnh vc iu khin, b PID c xem nh mt gii php a nng
cho cc ng dng iu khin Analog cng nh Digital. Vic thit k b PID kinh
in thng da trn phng php Zeigler-Nichols, Offerein, Reinish Ngy nay
ngi ta thng dng k thut hiu chnh PID mm (da trn phm mm), y chnh l c
s ca thit k PID m hay PID thch nghi.
4.3.1. S iu khin s dng PID m :
Hnh 4.6:
thi gian (s)
z (m)
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M hnh ton ca b PID:
u(t) = Kpe(t) + dt
tdeKdxxeK Dt
I)()(
0
+
GPID(s) = sKs
KK DIP ++
Cc tham s KP, KI, KD c chnh nh theo tng b iu khin m ring bit da
trn sai lch e(t) v o hm de(t). C nhiu phng php khc nhau chnh nh b
PID ( xem cc phn sau) nh l da trn phim hm mc tiu, chnh nh trc tip,
chnh nh theo Zhao, Tomizuka v Isaka Nguyn tc chung l bt u vi cc tr
KP, KI, KD theo Zeigler-Nichols, sau da vo p ng v thay i dn tm ra
hng chnh nh thch hp. 4.3.2. Lut chnh nh PID:
+ Ln cn a1 ta cn lut K mnh rt ngn thi gian ln, do vy chn: KP ln,
KD nh v KI nh.
thi gian
Tn hiu ra
b1
c1
d1
a2 b2
a1
t
Hnh 4.7
u e x y BK PID
B CHNH NH M
THIT B CHNH NH
I TNG
dtde
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+ Ln cn b1 ta trnh vt l ln nn chn: KP nh, KD ln, KI nh. + Ln cn
c1 v d1 ging nh ln cn a1 v b1.
4.3.3. V d ng dng Matlab Xy dng b PID m iu khin l nhit. Hm truyn
l nhit theo
Zeigler-Nichols : G(s) = 1+
LsKe Ts
, tuyn tnh ho G(s)= )1)(1( ++ LsTsK
.
Cc bc thit k : 1. Xc nh bin ngn ng: u vo : 2 bin + Sai lch ET =
o - t
+ Tc tng DET = T
iEiET )()1( +, vi T l chu k ly mu.
u ra : 3 bin + KP h s t l + KI h s tch phn + KD h s vi phn S lng
bin ngn ng ET = {m nhiu, m va, m t, zero, dng t, dng va, dng nhiu}
ET = { N3, N2, N1, ZE, P1, P2, P3 } DET = { m nhiu, m va, m t,
zero, dng t, dng va, dng nhiu} DET = { N31, N21, N11, ZE1, P11,
P21, P31 } KP/KD = { zero, nh, trung bnh, ln, rt ln } = {Z, S, M,
L,U} KI = {mc 1,mc 2,mc 3,mc 4, mc 5} = {L1,L2,L3,L4,L5}
N3 N2 N1 ZE P1 P2 P3
-12 -8 -4 0 4 8 12 0C
ET
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2. Lut hp thnh:
C tng cng l 7x7x3=147 lut IF THEN Lut chnh nh KP
N31 N21 N11 ZE1 P11 P21 P31
-0.6 -0.4 -0.2 0 0.2 0.4 0.6 0C/s
DET
Z S M L U
0 0.25 0.5 0.75 1 KP KD
L1 L2 L3 L4 L5
1 1.2 1.4 1.6 1.8 KI
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KP DET N31 N21 N11 ZE1 P11 P21 P31
ET
N3 U U U U U U U N2 L L L L L L L N1 M M M M M M M ZE Z Z Z Z Z
Z Z P1 M M M M M M M P2 L L L L L L L P3 U U U U U U U
Lut chnh nh KD: KD DET
N31 N21 N11 ZE1 P11 P21 P31
ET
N3 U U U U U U U N2 L L M M M L L N1 M M M M M M M ZE Z Z Z Z Z
Z Z P1 M M M M M M M P2 L L M M M L L P3 U U U U U U U
Lut chnh nh KI: KI DET
N31 N21 N11 ZE1 P11 P21 P31
ET
N3 L1 L1 L1 L1 L1 L1 L1 N2 L3 L2 L2 L1 L2 L2 L3 N1 L4 L3 L2 L1
L2 L3 L4 ZE L5 L4 L3 L2 L3 L4 L5 P1 L4 L3 L2 L1 L2 L3 L4 P2 L3 L2
L2 L1 L2 L2 L3 P3 L1 L1 L1 L1 L1 L1 L1
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Biu din lut chnh nh KP trong khng gian
3. Chn lut v gii m + Chn lut hp thnh theo quy tc Max-Min + Gii m
theo phng php trng tm. 4. Kt qu m phng Vi cc thng s : K=1; T=60;
L=720 T y theo Zeigler-Nichols ta tm ra c b ba thng s {KP, KI, KD }
th di y s cho ta thy s khc bit ca iu khin m so vi iu khin kinh
in.
Tham s theo Zeigler-Nichols
Tham s PID m
t (s)
T (0C)
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4.4. H m lai H m lai (Fuzzy Hybrid) l mt h thng iu khin t ng
trong thit b iu khin bao gm: phn iu khin kinh in v phn h m
4.4.1. Cc dng h m lai ph bin: 1. H m lai khng thch nghi
2. H m lai cascade
3. Cng tc m iu khin h thng theo kiu chuyn i khu iu khin c tham s
i hi thit b iu khin phi cha ng tt c cc cu trc v tham s khc nhau cho
tng trng hp. H thng s t chn khu iu khin c tham s ph hp vi i
tng.
Hnh 4.8
B K I TNG B tin X l m
Hnh 4.9
x
u
u + y
BK M
BK KINH IN I TNG
Hnh 4.10
x u
y
B iu khin n
B iu khin 1
BK M
i tng
-
Chng 4 : iu khin m
Trang 271
4.4.2. V d minh ho Hy xt s khc bit khi s dng b tin x l m iu khin
i tng gm khu cht ni tip vi khu )2.01()( ss
KsG
+= . Chn BK PI vi
tham s KP = 10, TI = 0.3sec.
S dng Simulink kt hp vi toolbox FIS Editor ca Matlab m phng h
thng trn. p ng h thng khi khng c b m:
x + y x
DE
E sT
KI
R1
+
i tng
u -u G(s) B m
-
Chng 4 : iu khin m
Trang 272
Th vi cc gi tr u v K khc nhau cho thy c tnh ng ca h s xu i khi
vng cht rng hoc h s khuch i ln. hiu chnh c tnh ng ca h thng ta a vo
b lc m nh hnh v trn. Xy dng lut iu khin vi 2 u vo v mt u ra nh
sau:
DE x NB NS ZE PS PB
E
NB NB NS NS NS NS NS ZE ZE NB NS ZE PS PB PS PS PS PS PS PB PB
PS PB
Tt c 18 lut c khun dng nh sau: Nu E = x1 v DE = x2 Th x = x3
Trong x1, x2, x3 {NB, NS, ZE, PS, PB}
-1 0 1 E -20 0 20 DE
NB NS ZE PS PB
-1 0 1 x
NB NS ZE PS PB
-
Chng 4 : iu khin m
Trang 273
T hai th trn ta thy c b m ci thin rt tt c tnh ng ca h thng. Th
vi nhiu u khc nhau ta s thy p ng hu nh khng ph thuc vo u .
4.5. H m mng nron v ng dng 4.5.1. Mng nron nhn to Mng nron l s
ti to bng k thut nhng chc nng ca h thn kinh con ngi. Mng nron gm v
s cc nron lin kt vi nhau nh hnh sau
Hai c tnh c bn ca mng nron l: + Qu trnh tnh ton c tin hnh song
song v phn tn trn nhiu
nron gn nh ng thi. + Tnh ton thc cht l qu trnh hc, ch khng phi
theo s nh
sn t trc.
M hnh ton ca mng nron nhn to : (Artifical Neural Networks)
y l m hnh iu khin dng MISO, vi u vo l n tnh hiu X={x1,x2,xn }T,
u ra l tn hiu y c xc nh:
y(t) = =
n
kkk txwf
1))(( ,
trong l ngng kch hot nron, wk l cc trng s, f l hm kch hot.
X y B tng
Hm phi tuyn
Nhn Axon
Hnh 4.11 Khp ni
-
Chng 4 : iu khin m
Trang 274
4.5.2. Cu trc mng nron Nguyn l cu to ca mt mng nron l bao gm
nhiu lp, mi lp bao gm nhiu nron c cng mt chc nng. Sau y l cc dng
lin kt mng c bn: a) Mng truyn thng (Feedforward Neural
Networks)
b) Mng c hi tip:
4.5.3. Mt s mng nron c bn 1. Mng MLP (Multilayer perceptron) C
rt nhiu cng trnh nghin cu v mng MLP v cho thy nhiu u im ca mng ny.
Mng MLP l c s cho thut ton lan truyn ngc v kh nng xp x lin tc. Thut
ton lan truyn ngc: Tp d liu cho c n mu (xn,dn), vi mi n, xn l tn
hiu u vo, dn l u ra mong mun. Qu trnh hc l vic thc hin cc tiu ho hm
G sau:
G = =
N
n
nG1
, vi Gn =
N
q nnxdqxyq
N 12))()((1
Q l s nt ti lp ra ca mng. Cn trng s lin kt mng c iu chnh theo
php lp sau :
x1
x2
y
Lp vo Lp b che Lp ra
-
Chng 4 : iu khin m
Trang 275
=+Gkwkw )()1( , trong >0 l hng s t l hc.
Mng MLP l mt gii php hu hiu cho vic m hnh ho, c bit vi qu trnh
phc tp hoc c ch cha r rng. N khng i hi phi bit trc dng hoc tham s.
2. Mng RBF (Radial basis functions) Biu din ton hc ca RBF
=
+=1
00 )()(
N
kkk RxCCxF
trong C : vct cha trng s RBF R : vct cha cc tm RBF : hm c s hoc
hm kch hot ca mng F(x) : hm nhn c t u ra ca mng C0 : h s chch || ||
: chun Euclide Nh kh nng xp x cc hm phi tuyn bt k vi chnh xc tu ,
mng nron, c bit l mng RBF l cng c quan trng cho m hnh ho h thng v
cho iu khin thch nghi cc h thng phi tuyn.
4.5.4. Nhn dng m hnh v iu khin s dng mng nron 1. Nhn dng thng s
m hnh
Nhn dng thng s chnh l qu trnh luyn mng. Tn hiu sai s yye ~= l c
s cho luyn mng, l thi gian tr.
2. iu khin s dng mng nron Ta c nhiu cu trc iu khin s dng mng
nron nh:
e(k) u(k)
y(k)
)(~ ky Mng nron
i tng K
Hnh 4.12
-
Chng 4 : iu khin m
Trang 276
+ iu khin theo vng h + iu khin theo vng kn + iu khin vi m hnh
tham chiu + iu khin theo thi gian vt qu (over time) + B iu khin vi
quyt nh h tr ca mng nron
3. ng dng mng RBF nhn dng h ng lc hc phi tuyn Xt h ng hc phi
tuyn ca TK uxgxfx )()( += (4.5) Gi s TK l n nh vng h, vct trng thi
x l quan st c. Cn tm m hnh xp x (4.5). Chn ARn x m l ma trn n nh,
ta vit li (4.5) dng :
uxgAxxfAxx )())(( ++= Theo tnh cht xp x ca mng RBF cho hm phi
tuyn: Nu s lng cc nt trong lp n l ln th f(x) - Ax v g(x) c th xp x
bng cc mng RBF sau: f(x)- Ax = W*S(x) v g(x) = V*S(x) trong W* Rn x
N v V* Rn x N l cc ma trn trng s ca cc t hp tuyn tnh trn. N xc nh s
lng nt trong mt lp RBF ca mng. S(x) = [ S1, S2, , SN ]T, vct cc hm
c s sau:
( ) 2122 += kkk CxS , vi k = 1, 2, 3,N Tm Ck Rn v rng k Rn c bit
trc. Ta vit li (4.5) nh sau: uxSVxSWAxx )(*)(* ++= Vy m hnh ca i
tng c th c m t bng phng trnh:
e r u
yd
y
e
M hnh tham chiu
TK BK bng mng nron
Hnh 4.13: iu khin vi m hnh tham chiu v sai s lan truyn qua
TK
-
Chng 4 : iu khin m
Trang 277
uxVSxWSxAx )()(~~ ++= trong W Rn x N, V Rn x N l cc ma trn c lng
ca W*, V*, x~ Rn l c lng trng thi ca x. Gi xe = xx
~
, We = W*-W, Ve = V*-V Phng trnh sai s c lng s l : uxSVxSAxx
eeee )()(W ++= (4.6) Thut ton nhn dng s dng hm Lyapunov:
)(21)(
21
21),,( eTeeTeeTeeee VVTrWWTrPxxVWxL ++= (4.7)
vi P l ma trn i xng xc nh dng. C th xc nh ma trn Q i xng xc nh
dng tho phng trnh Lyapunov sau: PA+ATP = - Q. Thay (4.6) vo (4.7) v
ly o hm ta c:
)(21)(
21)()()(
21
e
Tee
Tee
Te
Te
Te
Te
TTe VVTrWWTruPxVxSPxWxSxPAPAxL +++++=
Chn : eT
e
Te
Te PxWxSWWTr )()( = (4.8)
uPxVxSVVTr eT
e
Te
Te )()( = (4.9)
th : eTeeee QxxVWxL 2
1),,( = (4.10)
Do cc ma trn W* v V* l ma trn hng nn t (4.8), (4.9) ta suy ra
thut nhn dng m hnh nh sau:
=
=
n
kekikjij xPSW
1
=
=
n
kekikjij uxPSV
1
vi i = 1,2,,N v j = 1,2,,N, Pij l phn t ca ma trn Lyapunov P. T
(4.7) ta thy rng L(xe,We,Ve) 0 T (4.10) nhn c 0),,( eee VWxL V vy
xe(t) 0, We 0, Ve 0, hoc xx , W W*, V V* khi t . tnh ton n gin c th
chn : A = aI, Q = qI, P = pI, vi a > 0, q > 0 v I l ma trn n
v Khi thut ton nhn dng m hnh n gin nh sau:
-
Chng 4 : iu khin m
Trang 278
eijij xpS=W uxpSV eijij =
T phng trnh Lyapunov rt ra : a
qp2
= >0.
hi t n trng s thc, h ng lc phi c giu thng tin u vo. V th a s u
vo c chn ngu nhin.
4.5.5. Kt hp mng nron v h m Qua phn tch trn ta c th thy c nhng u
nhc im ca mng nron v iu khin m nh sau: Tnh cht Mng Nron B iu khin
m
Th hin tri thc Thng qua trng s c th hin n trong mng
c th hin ngay ti lut hp thnh
Ngun ca tri thc T cc mu hc T kinh nghim chuyn gia
X l thng tin khng chc chn
nh lng nh lng v nh tnh
Lu gi tri thc Trong nron v trng s ca tng ng ghp ni nron
Trong lut hp thnh v hm thuc
Kh nng cp nht v nng cao kin thc
Thng qua qu trnh hc Khng c
Tnh nhy cm vi nhng thay i ca m hnh
Thp Cao
T ngi ta i n vic kt hp mng nron v iu khin m hnh thnh b iu khin m
- nron c u im vt tri.
Vo Ra
Mng nron X l tn hiu nron vo c lng trng thi D bo trng thi Nhn dng
h thng
B iu khin m
iu khin Ra quyt nh
Kin trc kiu mu ca mt h m-nron
-
Chng 4 : iu khin m
Trang 279
4.5.6. Thut ton di truyn (GA)
Gii thiu Thut ton di truyn l thut ton ti u ngu nhin da trn c ch
chn lc t nhin v tin ha di truyn. Nguyn l c bn ca thut ton di truyn
c Holland gii thiu vo nm 1962. C s ton hc c pht trin t cui nhng nm
1960 v c gii thiu trong quyn sch u tin ca Holland, Adaptive in
Natural and Artificial Systems. Thut ton di truyn c ng dng u tin
trong hai lnh vc chnh: ti u ha v hc tp ca my. Trong lnh vc ti u ha
thut ton di truyn c pht trin nhanh chng v ng dng trong nhiu lnh vc
khc nhau nh ti u hm, x l nh, bi ton hnh trnh ngi bn hng, nhn dng h
thng v iu khin.
Thut ton di truyn cng nh cc thut ton tin ha ni chung, hnh thnh
da trn quan nim cho rng, qu trnh tin ha t nhin l qu trnh hon ho
nht, hp l nht v t n mang tnh ti u. Quan nim ny c th xem nh mt tin
ng, khng chng minh c, nhng ph hp vi thc t khch quan. Qu trnh tin ha
th hin tnh ti u ch, th h sau bao gi cng tt hn (pht trin hn, hon
thin hn) th h trc bi tnh k tha v u tranh sinh tn.
Cc php ton ca thut ton di truyn
1. Ti sinh (Reproduction) Ti sinh l qu trnh chn qun th mi tha
phn b xc sut da trn thch nghi. thch nghi l mt hm gn mt gi tr thc
cho c th trong qun th. Cc c th c thch nghi ln s c nhiu bn sao trong
th h mi. Hm thch nghi c th khng tuyn tnh,khng o hm, khng lin tc bi
v thut ton di truyn ch cn lin kt hm thch nghi vi cc chui s.
Qu trnh ny c thc hin da trn bnh xe quay roulette (bnh xe s x) vi
cc rnh c nh kch thc theo thch nghi. K thut ny c gi l la chn cha m
theo bnh xe roulette. Bnh xe roulette c xy dng nh sau (gi nh rng,
cc thch nghi u dng, trong trng hp ngc li th ta c th dng mt vi php
bin i tng ng nh li t l sao cho cc thch nghi u dng).
-
Chng 4 : iu khin m
Trang 280
- Tnh thch nghi fi, i=1 n ca mi nhim sc th trong qun th hin
hnh,vi n l kch thc ca qun th (s nhim sc th trong qun th).
- Tm tng gi tr thch nghi ton qun th: =
=
n
iifF
1
- Tnh xc sut chn pi cho mi nhim sc th: Ff
p ii =
- Tnh v tr xc sut qi ca mi nhim sc th: =
=
i
jji pq
1
Tin trnh chn lc c thc hin bng cch quay bnh xe roulette n ln, mi
ln chn mt nhim sc th t qun th hin hnh vo qun th mi theo cch
sau:
- Pht sinh ngu nhin mt s r (quay bnh xe roulette) trong khong
[01]
- Nu r < q1 th chn nhim sc th u tin; ngc li th chn nhim sc th
th i sao cho qi-1 < r qi
V d 4.5.6: Xem xt dn s c 6 nhim sc th vi gi tr tng thch nghi ton
qun th l 50 (bng 1), bnh xe roulette trong hnh 4.14. By gi ta quay
bnh xe roulette 6 ln, mi ln chn mt nhim sc th cho qun th mi. Gi tr
ngu nhin ca 6 s trong khong [01] v cc nhim sc th tng ng c chn c cho
trong bng 2.
Nhim sc th
Chui m ha
Tr thch nghi f(i)
Xc sut chn pI
V tr xc sut qi
1 01110 8 0.16 0.16 2 11000 15 0.3 0.46 3 00100 2 0.04 0.5 4
10010 5 0.1 0.6 5 01100 12 0.24 0.84 6 00011 8 0.16 1
Bng 1: Cc nhim sc th v cc gi tr thch nghi
-
Chng 4 : iu khin m
Trang 281
Hnh 4.14: Bnh xe roulette
S ngu nhin 0.55 0.1 0.95 0.4 0.8 0.7 Nhim sc th 4 1 6 2 5 5
Bng 2: Qun th mi Qua v d trn ta thy rng, c th s c mt s nhim sc
th c chn nhiu ln, cc nhim sc th c thch nghi cao hn s c nhiu bn sao
hn, cc nhim sc th c thch nghi km nht thi dn dn cht i.
Sau khi la chn c qun th mi, bc tip theo trong thut ton di truyn
l thc hin cc php ton lai ghp v t bin.
2. Lai ghp (Crossover) Php lai l qu trnh hnh thnh nhim sc th mi
trn c s cc nhim sc th cha - m, bng cch ghp mt hay nhiu on gen ca
hai (hay nhiu) nhim sc th cha - m vi nhau. Php lai xy ra vi xc sut
pc, c thc hin nh sau:
- i vi mi nhim sc th trong qun th mi, pht sinh ngu nhin mt s r
trong khong [01], nu r < pc th nhim sc th c chn lai ghp.
1
2
6
5
4 3
-
Chng 4 : iu khin m
Trang 282
- Ghp i cc nhim sc th chn c mt cch ngu nhin, i vi mi cp nhim sc
th c ghp i, ta pht sinh ngu nhin mt s nguyn pos trong khong [0m-1]
(m l tng chiu di ca mt nhim sc th - tng s gen). S pos cho bit v tr
ca im lai. iu ny c minh ha nh sau:
b1b2bposbpos+1bm c1c2cposcpos+1cm
- Chuyn i cc gen nm sau v tr lai.
b1b2bposcpos+1cm c1c2cposbpos+1bm
Nh vy php lai ny to ra hai chui mi, mi chui u c tha hng nhng c
tnh ly t cha v m ca chng. Mc d php lai ghp s dng la chn ngu nhin,
nhng n khng c xem nh l mt li i ngu nhin qua khng gian tm kim. S kt
hp gia ti sinh v lai ghp lm cho thut ton di truyn hng vic tm kim n
nhng vng tt hn.
3. t bin (Mutation) t bin l hin tng c th con mang mt (s) tnh
trng khng c trong m di truyn ca cha m. Php t bin xy ra vi xc sut
pm, nh hn rt nhiu so vi xc sut lai pc. Mi gen trong tt c cc nhim sc
th c c hi b t bin nh nhau, ngha l i vi mi nhim sc th trong qun th
hin hnh (sau khi lai) v i vi mi gen trong nhim sc th, qu trnh t bin
c thc hin nh sau:
- Pht sinh ngu nhin mt s r trong khong [01] - Nu r < pm, th t
bin gen .
t bin lm tng kh nng tm c li gii gn ti u ca thut ton di truyn. t
bin khng c s dng thng xuyn v n l php ton tm
V tr lai
-
Chng 4 : iu khin m
Trang 283
kim ngu nhin, vi t l t bin cao, thut ton di truyn s cn xu hn
phng php tm kim ngu nhin.
Sau qu trnh ti sinh, lai v t bin, qun th mi tip tc c tnh ton cc
gi tr thch nghi, s tnh ton ny c dng xy dng phn b xc sut (cho tin
trnh ti sinh tip theo), ngha l, xy dng li bnh xe roulette vi cc rnh
c nh kch thc theo cc gi tr thch nghi hin hnh. Phn cn li ca thut ton
di truyn ch l s lp li chu trnh ca nhng bc trn.
Cu trc ca thut ton di truyn tng qut
Thut ton di truyn bao gm cc bc sau:
- Bc 1: Khi to qun th cc nhim sc th.
- Bc 2: Xc nh gi tr thch nghi ca tng nhim sc th.
- Bc 3: Sao chp li cc nhim sc th da vo gi tr thch nghi ca chng v
to ra nhng nhim sc th mi bng cc php ton di truyn.
- Bc 4: Loi b nhng thnh vin khng thch nghi trong qun th.
- Bc 5: Chn nhng nhim sc th mi vo qun th hnh thnh mt qun th
mi.
- Bc 6: Nu mc tiu tm kim t c th dng li, nu khng tr li bc 3.
-
Chng 4 : iu khin m
Trang 284
4.6. ng dng iu khin m trong thit k h thng
4.6.1 iu khin m khng thch nghi (Nonadaptive Fuzzy Control) 1. B
iu khin m tuyn tnh n nh SISO Phng trnh bin trng thi ca h SISO
)]([)()()(
)()()(
tyftutcxty
tbutAxtx
=
=
+=
Thay phng trnh cui vo hai phng trnh trn ta c h m vng kn nh
sau:
Thit k BK m n nh SISO Bc 1: Gi s y(t) c min gi tr l khong U=[ ],
chia U ra 2N+1 khong Ak nh hnh v bn di:
x1 x2 xN+1 x2N+1 y
A1 A2 AN AN+1 AN+2 A2N A2N+1
Hnh 4.16: Hm thuc ca BK
i tng K
x u y
A
b c
BK m
f(y)
Hnh 4.15: Cu trc h SISO
-
Chng 4 : iu khin m
Trang 285
Bc 2: Thnh lp 2N+1 lut m IF THEN c khun dng IF y = Ak THEN u =
Bk trong k = 1,2,.,2N+1 v trng tm y ca khong m Bk l:
++=+==
=
12,...,2010
,...,10
NNkNk
Nky (4.11)
Bc 3: Chn lut hp thnh tch, gii m theo phng php trung bnh trng s,
ta c lut iu khin nh sau:
+
=
+
=
== 12
1
12
1
)()()( N
k A
N
k A
y
yyyfu
k
k
vi y tho (4.11) v )(ykA
c nu trong Hnh 4.16. 2. B K m tuyn tnh n nh MIMO Phng trnh bin
trng thi ca h MIMO:
)()()()()(
tCxtytButAxtx
=
+= (4.12)
Gi s h c m u vo v m u ra th u(t) = (u1(t),,um(t))T c dng : uk(t)
= - fk[y(t)] (4.13) vi k=1,2,,m v fk[y(t)] l h m m u vo 1 u ra. M
hnh h thng c cu trc nh Hnh 4.15, nhng thay cho cc s b,c bi cc ma
trn B,C, hm v hng f bi vct f = (f1,f2,,fm)T. Thit k BK m n nh MIMO
Bc 1: Gi s u ra yk(t) c min gi tr l Uk = [k k], vi k=1,,m. Chia Uk
ta 2N+1 khong ilkA v thit lp hm thuc nh Hnh F.2
Bc 2: Thnh Lp m nhm lut m IF THEN, nhm th k cha
=
+m
i kN
1)12( lut dng:
IF y1= 11lA And . And ym= mlmA , THEN u= m
llkB
...1
Trong li=1,2,,2Nk+1; k=1,2,,m v trng s mllky...1 ca tp m
mllkB
...1
c chn nh sau:
-
Chng 4 : iu khin m
Trang 286
++=+==
=
12,...,2010
,...,2,10...1
kkk
kk
kkll
k
NNlNl
Nly m (4.14)
Bc 3: Chn lut hp thnh tch, gii m theo phng php trung bnh trng s,
ta c lut iu khin:
+
= = =
+
= =
+
=
== 12
1 1 1
12
1 1...
12
1
1
1
11
1
))((...))((...
)( Nl
m
l
m
i iA
iN
l
m
i All
kN
lkk
mil
i
m
mil
i
m
y
yyyfu
(4.15)
vi k=1,2,,m.
3. B iu khin m ti u Phng trnh trng thi
0)0()()()(
xx
tButAxtx=
+= (4.16)
vi x Rn v u Rm, v ch tiu cht lng dng ton phng:
[ ]dttRututQxtxTMxTxJ T TTT ++=0
)()()()()()( (4.17)
vi M Rn n, Q Rn n, R Rm m l cc ma trn xc nh dng. Ta xc nh u(t)
dng nh (4.15), vi u(t) = (u1,u2,,um)T
+
=
+
= =
+
= =
+
=
== 12
1
12
1 1
12
1 1...
12
1
1
1
11
1
))((...))((...
)( Nl
N
l
n
i iA
iN
l
n
i All
kN
lkk
n
nil
i
n
nil
i
m
x
xyxfu
(4.18)
Chng ta cn xc nh thng s nllky...1 cc tiu J.
Ta nh ngha hm m c s b(x) = (b1(x), , bN(x))T vi:
+
=
+
= =
=
= 12
1
12
1 1
1
1
1))((...
)()( N
l
N
l
n
i iA
n
i iAl
n
nil
i
ili
x
xxb
(4.19)
vi li = 1,2,,2Ni+1; l = 1,2,,N v =
+=n
i iNN
1)12( . Ta nh ngha
ma trn thng s Rm N nh sau :
-
Chng 4 : iu khin m
Trang 287
[ ]TTmTT = ,...,, 21 (4.20) vi NTk R
1 cha N thng s nllky...1
, c bc ging nh bl(x). Ta vit li tn hiu iu khin m dng u =
(u1,u2,..,um)T = (-f1(x),,-fn(x))T nh sau: u = b(x) (4.21) Gi ta gi
s = (t). Thay (4.21) vo (4.16) v (4.17) ta c : [ ])()()()(
txbtBtAxtx += (4.22) v hm ch tiu cht lng l :
[ ]dttxbtRttxbtQxtxTMxTxJ T TTTT ++=0
))(()()())(()()()()( (4.23)
V vy vn cn gii quyt by gi l xc nh (t) ti u c tiu ho J. Xt hm
Hamilton:
)]([)()(),,( xbBAxpxbRxbQxxpxH TTTT +++= (4.24)
Ta c: 0)()()(2 =+=
xpbBxbxbRH TTT
Suy ra : 11 )]()()[(21
= xbxbxpbBR TTT (4.25)
Thay (4.25) vo (4.24) ta c: pBBRpxxAxpQxxpxH TTTT 12 )]()([),(
++= (4.26)
trong : )()]()()[(21)( 1 xbxbxbxbx TT = (4.27)
p dng nguyn l cc tiu Pontryagin ta c:
pBBRxxAxp
Hx T12 )]()([2
+=
= (4.28)
pBBRpx
xxpAQx
x
Hp TTT 1)(]1)(2[2
=
=
(4.29)
Gii hai phng trnh vi phn (4.27) v (4.28) ta s c x*(t) v p*(t), t
ta xc nh c:
11 ))](())(())[(()(21)( = txbtxbtxbtpBRt TTT (4.30)
-
Chng 4 : iu khin m
Trang 288
V b m ti u s l: )()( xbtu = (4.31) Cc bc thit k BK m ti u: Bc 1:
Xc nh hm thuc )( iA xili , vi li = 1,2,,2Ni+1 v I = 1,,n. Chn dng
hm thuc l Gaussian. Bc 2: Tnh hm m c s bl(x) theo (4.19) v tnh (x)
theo (4.27), xc nh tr o hm :
x
x
)(
.
Bc 3: Gii (4.28) v (4.29) c x*(t) v p*(t), tnh *(t) theo (4.30)
vi t[0 T]. Bc 4: Xc nh BK m ti u t (4.31) V d ng dng: Hy thit k v m
phng h thng Qu bng v n by nh hnh v sau:
Thit k BK m iu khin qu bng di chuyn t im gc O n mc tiu (v tr t)
cch O khong r. Chn bin trng thi nh sau:
TT xxxxrrx ),,,(),,,( 4321== v y = r = x1 Phng trnh bin trng thi
c chn l:
ux
xxx
x
x
x
x
x
+
=
1000
0
)sin(4
3241
2
4
3
2
1
Chn M=0, Q=I, R=I, Ni=2 vi i=1,2,3,4. Chn hm thuc dng:
])(2exp[)( 2iil
ip
liiiA
xxx =
O
r u
Hnh 4.17
-
Chng 4 : iu khin m
Trang 289
Trong i=1, 2, 3, 4; li=1, 2, 3, 4, 5 v )1( += iiili lbax i vi a1
= a2= - 2, a3=a4=-1, b1=b2=1, b3=b4=0.5. Chn = 0.7143, = 9.81. Kt
qu m phng vi 3 mc tiu khc nhau:
4. iu khin m c h thng gim st
Thit k b gim st Xt h thng phi tuyn c cho bi phng trnh vi phn:
uxxxgxxxfx nnn ),...,,(),...,,( )1()1()( += (4.32) trong Tnxxxx
),...,,( )1( = l vct trng thi ra, u R l tn hiu iu khin, f v g l cc
hm cha bit, gi thit g > 0.Gi s ta c BK m: u = ufuzz(x)
mc tiu
iu khin
i tng
B K m
B K gim st
Hnh 4.18
-
Chng 4 : iu khin m
Trang 290
Gi s |x(t)| Mx, x vi Mx = const. Khi thm b gim st th tn hiu iu
khin h thng s l: u = ufuzz(x) + I*us(x) (4.33) trong I* = 1 nu
|x(t)| Mx, I* = 0 nu |x(t)| < Mx. Ta cn thit k b gim st us(t).
Thay (4.33) vo (4.32) ta c: x
(n) = f(x) + g(x)ufuzz(x) + g(x)I*us(x) (4.34)
Gi s ta lun xc nh c hai hm fU(x) v gL(x) sao cho |f(x)| fU(x) v
0 < gL(x) g(x).
t : [ ]xkxfxg
u T= )()(1
(4.35)
Trong k = (kn,kn-1,..,k1)T R. Ta vit li (4.34) nh sau:
[ ]sfuzzTn uIuugxkx ++=)( (4.36)
t
=
121 .........
10...00000.....................
00...010000...0010
kkkk
A
nn
=
g
b0...
0
Vit (4.36) dng vct : ][ sfuzz uIuubAxx ++= (4.37)
Xt hm Lyapunov : PxxV T21
= (4.38)
Trong P l ma trn i xng xc nh dng tho phng trnh Lyapunov : QPAPAT
=+ (4.39) T (4.37), (4.39) v xt trng hp |x| Mx , ta c:
-
Chng 4 : iu khin m
Trang 291
s
Tfuzz
Tsfuzz
TT PbuxuuPbxuuuPbxQxxV ++++= )(][21
(4.40)
Ta cn tm us 0V , kt hp phng trnh trn vi (4.6.25) ta c:
++= fuzz
TU
L
Ts uxkfgPbxsignu )(
1)( (4.41)
Thay (4.41) vo (4.40) ta s c 0V . V d (4.6.1.4) Thit k h thng c
b gim st gi cn bng cho con lc ngc. M hnh:
Phng trnh trng thi: 21 xx = (4.42)
u
mm
xml
mm
x
mm
xml
mm
xxmlxxg
x
c
c
c
c
)cos34(
cos
)cos34(
sincossin
12
1
12
1122
1
2
+
++
+
+
= (4.43)
Thit k b gim st u tin ta tm fU v gL, ta c
22
22
12
1122
1
21 0366.078.15
1.105.0
32
1.125.08.9
)cos34(
sincossin
),( xx
mm
xml
mm
xxmlxxg
xxf
c
c +=
+
+
+
=
Hnh 4.19
2x=
mgsin
=x1
l
mc u
-
Chng 4 : iu khin m
Trang 292
chn 2221 0366.078.15),( xxxf U += con lc n nh th gc x1 = 200.
Suy ra Mx = 200.
1.1)20cos
1.105.0
32(1.1
20cos),(02
0
21 =
+xxg
chn gL(x1,x2) = 1.1 Chn cc thng s thit k nh sau:
a = pi/18, k1 = 2, k2 = 1 , Q =
100010
Gii phng trnh Lyapunov (4.39) ta c : P =
155515
Thit k BK m c ufuzz(x). T (4.41) ta s c BK c gim st h con lc
ngc. Dng simulink ca matlab chy m phng ta s thy c tnh u vit khi c v
khng s dng b gim st.
5. iu khin m trt
1. Nguyn l iu khin trt Xt h thng phi tuyn uXgXfx n )()()( +=
(4.44) y(t) = x(t) trong u l tn hiu iu khin, x l tn hiu ra, TnxxxX
),...,,( )1( = l vct trng thi. Trong (G.1) f(X) l hm cha bit v b
chn bi mt hm bit:
)()()( XfXfXf += (4.45) v )()( XFXf (4.46) 0 < g0 <
g(X)
-
Chng 4 : iu khin m
Trang 293
lm c iu ny ta a ra hm trt sau:
xadtdx
adt
xda
dtxdS
n
n
nn
n
012
2
21
1
... ++++=
(4.48)
trong n l bc ca i tng. Cc h s a0, a1, , an-2 phi c chn sao cho a
thc c trng ca phng trnh vi phn S=0 l a thc Hurwitz. Phng trnh S=0 m
t mt mt trong khng gian trng thi n chiu gi l mt trt ( Sliding
surface). Ta cn xc nh lut iu khin u sao cho S 0 c x 0. i vi iu khin
bm mc tiu, ta cn xc nh lut iu khin u = u(X) sao cho trng thi ca h
thng vng kn s bm theo trng thi mong mun
( )Tndddd xxxX )1(,...,, = Gi e l sai lch gia tn hiu ra v tn hiu
t:
( )Tnd eeeXXe )1(,...,, == Mc tiu iu khin l trit tiu e khi t .
nh ngha hm trt :
eadtde
adt
eda
dted
eSn
n
nn
n
012
2
21
1
...)( ++++=
(4.49)
trong n l bc ca i tng iu khin, cc h s a0, a1, an-2 c chn sao cho
a thc c trng ca S(e)=0 l a thc Hurwitz. S dng phng php Lyapunov,
chn hm V xc nh dng nh sau:
2
21 SV = (4.50)
SSV = (4.51) V xc nh m ta chn lut iu khin u sao cho: Khi S>0
th S
-
Chng 4 : iu khin m
Trang 294
Vi lut iu khin nh vy, h thng s n nh theo tiu chun Lyapunov, lc
ny mi qu o trng thi ca h thng bn ngoi mt trt s c a v mt trt v duy
tr mt cch bn vng.
2. H thng iu khin trt m Xt h thng (4.44), ta cn xc nh lut iu
khin u a ng ra ca h thng bm theo theo gi tr mong mun cho trc y(t)
yd(t) hay ni cch khc l ( ) 0)()()( = idii yye , i = 0,1,,n-1 Da vo
t tnh ca b iu khin trt ta cn thc hin hai bc sau: Bc 1: Chn mt trt S
Bc 2: Thit k lut iu khin cho h thng ri vo mt trt S = 0 v duy tr ch
ny mi mi.
Gi ( ) ( )TnTn tetetetetetete )(),...,(),()(),...,(),()( )1(21
== Chn hm trt:
ebdtdeb
dtedb
dted
eSn
n
nn
n
012
2
21
1
...)( ++++=
(4.53)
Trong b0, b1,,bn-2 c chn sao cho nghim ca a thc c trng 0...
01
22
1=++++
bpbpbp nnn
u nm bn tri mt phng phc.
Mt trt S c cho b phng trnh S(e) = 0, lut iu khin u c chn sao cho
0)(
-
Chng 4 : iu khin m
Trang 295
3.Thit k b iu khin m trt bc hai Xt h thng phi tuyn bc hai sau:
uXgXfx )()( += (4.54) y = x (4.55) trong ( )Txx,X = l vct trng thi,
u l ng vo iu khin y(t) l ng ra ca h thng. Mc tiu ca iu khin l xc nh
lut iu khin u ng ra ca h thng bm theo qu o mong mun yd(t) vi sai s
nh nht. Lut iu khin u gm 2 thnh phn: u = ueq + us (4.56) Thnh phn
ueq c thit k nh sau:
[ ]etytXfg
tu deq += )(),(
1)( , (>0) (4.57)
Thnh phn us c chn l:
[ ])()1()),((
1)( tutXFg
tu eqs ++ (4.58)
Trong ),( tXf l gi tr c lng ca f(X,t) F(X,t) l cn trn ca sai s c
lng 0 < g0 < g(X) < g1 10 ggg =
0
1
gg
=
Lut iu khin m c thit k nh sau:
>
-
Chng 4 : iu khin m
Trang 296
H qui tc m c khun dng nh sau: R1 : Nu S0 Th )()(2 tutu +=
(4.61)
Chn lut hp thnh tch, gii m theo phng php trng tm, lut iu khin u
c xc nh nh sau:
=
=
=r
ii
r
i
ii
S
tuStu
1
1
)(
)()()(
(4.62)
Vi r : s lut m
=
=
n
jAi SS ij
1)()(
)(SijA
l hm thuc c dng Gaussian nh sau:
4. Thit k BK m trt cho h thng nng vt trong t trng
M hnh: Hnh 4.22 minh ho mt h thng nng vt bng t trng, t trng c to
ra t cun dy qun quanh li thp, cun dy nhn p iu khin u.
Hnh 4.21 : Dng hm thuc m ha S
-
Chng 4 : iu khin m
Trang 297
Phng trnh ton m t h thng
=
+=
=
2
))((
hiCmg
dtdv
m
dtihLdRiu
vdtdh
(4.63)
Trong : h : v tr hn bi (m) v : vn tc hn bi (m/s) i : dng in qua
cun dy (A) u : in p cung cp cho cun dy (V) R, L : in tr v in cm cun
dy (, H) C : hng s lc t (Nm2/A2) m : khi lng hn bi (Kg) g : gia tc
trng trng. (m/s2) in cm ca cun dy l mt hm phi tuyn ph thuc vo v tr
ca hn bi
hCLhL 2)( 1 += (4.64)
L1 l in cm ca cun dy khi hn bi rt xa. Chn bin trng thi nh
sau:
Hnh 4.22 : H thng nng vt trong t trng
-
Chng 4 : iu khin m
Trang 298
x1 = h, x2 = v, x3 = i (4.65) Vct trng thi ca h thng X = (x1,
x2, x3)T T (4.63), (4.64) v (4.65) ta c phng trnh trng thi:
+
+=
=
=
uLx
xx
LC
xLR
x
x
x
m
Cgx
xx
1221
3233
2
1
32
21
(4.66)
im cn bng ca h thng l nghim ca h ( )0,0,0 321 === xxx Gii ra c
Xb = [x1b, 0, x3b ]T , vi C
gmxx bb 13 =
Gi Xd = [ x1d, x2d, x3d ]T l vct trng thi mong mun. Mc tiu ca h
thng l a X tin v Xd vi sai s nh nht.
Thit k BK trt Thc hin php i trc nh sau:
=
=
=
2
1
33
22
111
x
x
m
Cgz
xz
xxz d (4.67)
Lc ny ta cn xc nh lut iu khin u sao cho Z = (z1, z2, z3)T tin v
(0,0,0)T khi t , khi y X Xd. Kt hp (4.66), (4.67) v mt s php bin i
ta c:
( )
+
+
+
+=
=
=
uzgm
CxzLL
RxzL
Cxz
zzgz
zz
zz
ddd
)()(2
)(212 3
111111
233
32
21
(4.68)
-
Chng 4 : iu khin m
Trang 299
t
+
+
+=
+=
LR
xzLC
xz
zzgzf
uzgm
CxzL
zg
dd
d
)(21)(2)(
)()(2)(
1111
23
311
(4.69)
T (4.68) v (4.69) ta c m hnh ng hc ca h thng trong h to mi nh
sau:
+=
=
=
uzgzfzzz
zz
)()(332
21
(4.70)
Ng ra ca h thng trong h ta mi l: dxxze 111 == (4.71)
Mi quan h ng vo v ng ra: uzgzfe )()()3( += (4.72) Hai hm f(z),
g(z) tng ng trong h to ban u l f1(x), g1(x):
=
+
=
21
31
2
1
3
131
232
1
2)(
212)(
LmxCx
xg
x
x
LR
LxC
x
xx
m
Cxf
(4.73)
Ta vit li (4.72) trong h to ban u: uxgxfe )()( 11)3( += (4.74)
Chn mt trt nh sau: eaeaeS 01 ++= (4.75) Vi a1, a0 c chn sao cho a
thc t trng ca phng trnh S = 0 l Hurwitz. T (4.75) v (4.70) ta c:
10213 zazazS ++= (4.76) Ly o hm ca S theo thi gian ta c: 203110213
)()( zazauzgzfzazazS +++=++= (4.77) Chn lut iu khin u nh sau:
-
Chng 4 : iu khin m
Trang 300
[ ])()()(1
102132031 zazazWsignzazazfzg
u ++= (4.78) Thay (4.78) vo (4.77) ta c: )()( 10213
SWsignzazazWsignS =++= (4.79) Nu chn W l hng s dng th ta s c 0
-
Chng 4 : iu khin m
Trang 301
Thit k BK trt m cho h thng nng vt trong t trng Trong phn thit k
BK trt ta bit lut iu khin u nh sau:
= )(1 20
2
1
311
1
SWsignxax
x
m
Cgafg
u
vi S c xc nh t (4.80), f1 v g1 c xc nh t (4.73). Do trong lut iu
khin c hm sign nn gy ra hin tng dao ng, khc phc nhc im ny ta thm
khu x l m trong b iu khin thay th cho hm sign. Chn lut iu khin u =
ueq + us , vi:
= 20
2
1
311
1
1xa
x
x
m
Cgafg
ueq (4.82)
Hnh 4.24: V tr v p iu khin khi tn hiu t l hng
-
Chng 4 : iu khin m
Trang 302
Cc bc xy dng b m: Bc 1: M ho mt trt S
Bc 2: Xy dng h qui tc m: R1: If S is zero Then u1 = ueq R2: If S
is pos Then u2 = ueq + C0 R3: If S is lpos Then u3 = ueq + C1 R4:
If S is neg Then u4 = ueq C0 R5: If S is lneg Then u5 = ueq C1 C0,
C1 l cc hng s dng C0 > C1 Bc 3: Gii m Bng phng php gii m trng
tm, lut iu khin u c xc nh:
=
=
= 5
1
5
1
ii
i
iiu
u
(4.83)
Trong i l ng ca qui tc th i :
)()()()()(
ln5
4
3
2
1
SSSSS
eg
neg
lpos
pos
zero
=
=
=
=
=
(4.84)
Hnh 4.25:Hm thuc vi 5 tp m
-
Chng 4 : iu khin m
Trang 303
Kt qu m phng
S dng 3 tp m, chn C0 = 350.
Hnh 4.27 V tr v p K khi tn hiu t l hng s
Hnh 4.26: V tr v p K khi tn hiu t l xung vung
-
Chng 4 : iu khin m
Trang 304
S dng 5 tp m, chn C0 =100 v C1 = 350.
Hnh 4.28 V tr v p K khi tn hiu t l xung vung
Hnh 4.29: V tr v p K khi tn hiu t l hng s
-
Chng 4 : iu khin m
Trang 305
S dng 7 tp m, chn C0 = 100, C1 = 200 v C2 = 350.
Kt lun - Vic thm BK m trit tiu hin tng dao ng. - p ng h thng tt
hn. - Chn 5 tp m l thch hp nht khi xy dng BK m.
Hnh 4.30: V tr v p K khi tn hiu t l xung vung
Hnh 4.31: V tr v p K khi tn hiu t l hng s
-
Chng 4 : iu khin m
Trang 306
4.6.2. iu khin m thch nghi (Adaptive Fuzzy Control) M hnh c bn
ca BK m thch nghi:
Phn loi cc BK m thch nghi: +BK m thch nghi gin tip +BK m thch
nghi trc tip +BK m thch nghi hn hp 1. Thit k BK m thch nghi gin
tip
ym
f, g
i tng x
(n) =f(x)+g(x)u, y=x
BK m )|(/])|([ )( gTnmfI xgekyxfu ++=
Lut thch nghi
IT
g
Tf
uPbe
Pbe
2
1
=
=
iu kin ban u f(0), g(0)
Hnh 4.33: H thng K m thch nghi gin tip
e
r y
ym
u
M hnh tham chiu
i tng
B iu khin m
Lut thch nghi ),( eh = Hnh 4.32
-
Chng 4 : iu khin m
Trang 307
Phng trnh trng thi uxxxgxxxfx nnn ),...,,(,...,,( )1()1()( +=
(4.85)
y = x (4.86) trong u R l u vo, y R l u ra, x = (x1,x2,,xn)T l
vct trng thi; f(x) v g(x) l hai hm m t cha bit c din t qua lut m:
Nu x1 = rF1 v v xn =
r
nF Th f(x) = Cr (4.87) Nu x1 = rG1 v v xn =
r
nG Th f(x) = Ds (4.88) Thit k BK m Nu f(x) v g(x) c bit trc th
vic thit k kh n gin nh ni cc phn trc, ta s c lut iu khin nh
sau:
[ ]ekyxfxg
u Tnm ++= )()()(
1 (4.89)
vi Tnmm eeexyyye ),...,,( )1( === v Tnn kkkk ),...,,( 11= Thay
(4.89) vo (4.85) ta c : 0...)1(1
)(=+++ ekeke n
nn
Chn k sao cho e(t) 0 khi t , khi y y ym. Khi f(x) v g(x) cha bit
r th ta thay bi h m )( xf v )( xg . nng cao chnh xc th ta phi mt s
thng s ca )( xf v )( xg t do. Gi s ta chn hai thng s fMf R v g
Mg R l t do, ta k hiu nh
sau : )|()( fxfxf = v )|()( gxgxg = , thay vo (4.89) ta c:
[ ]ekyxfxg
uu Tnmfg
I ++==)()|()|(
1
(4.90)
xy dng BK (4.90) ta phi xc nh )|( fxf v )|( gxg , iu ny c thc
hin qua 2 bc sau: Bc 1: Vi mi bin xi (i=1,2,,n), nh ngha pi tp m
iliA (li=1,,pi) v qi tp m iliB (li=1,,qi).
Bc 2: Xc nh )|( fxf t =
n
i ip
1 lut m dng:
Nu x1 = 11lA v . v xn = nlnA , Th n
llEf ...1 =
-
Chng 4 : iu khin m
Trang 308
Xc nh )|( gxg t =
n
i iq
1lut dng:
Nu x1 = 11lB v . v xn = nlnB , Th n
llHg ...1 =
Chn thit b hp thnh tch, hm m dng singleton, gii m theo phng php
trung bnh trng s, ta c:
= = =
= ==
=1
1
11
1
1 1 1
1 1...
1
))((...))((...
)|( pl
p
l
n
i iA
p
l
n
i iAll
fp
lf
n
nil
i
n
nil
i
n
x
xyxf
(4.91)
= = =
= ==
=1
1
11
1
1 1 1
1 1...
1
))((...))((...
)|( ql
q
l
n
i iB
q
l
n
i iBll
gq
lg
n
nil
i
n
nil
i
n
x
xyxg
(4.92)
Cho thng s nllfy...1
v nllgy...1 t do, v th ta c th dn vo f v g , ta vit
li (4.91) v (4.92) nh sau: )()|( xxf Tff = (4.93)
)()|( xxg Tgg = (4.94) trong (x) lf vct
=
n
i ip
1chiu v (x) l vct
=
n
i iq
1chiu, vi
thnh phn l1ln c cho bi:
= = =
=
=1
1
1
1 1 1
1... ))((...
)()( p
l
p
l
n
i iA
n
i iAll
n
nil
i
ili
n
x
xx
(4.95)
= = =
=
=1
1
1
1 1 1
1... ))((...
)()( q
l
q
l
n
i iB
n
i iBll
n
nil
i
ili
n
x
xx
(4.96)
Ta thy f v g c chn da theo (4.87) v (4.88), do f v g thay i lin
tc, ta cn tm f v g cc tiu ha sai s e. Thit k lut thch nghi Thay
(4.90) vo (4.85) v sau mt vi bin i ta c:
[ ] [ ] IgfTn uxgxgxfxfeke )()|()()|()( ++= (4.97) t :
-
Chng 4 : iu khin m
Trang 309
=
=
10...
0
,,
............
10...0000.....................
00...010000...0010
11
b
kkk
A
nn
(4.98)
Ta vit li (4.97) dng vct:
[ ] [ ]{ }Igf uxgxgxfxfbAee )()|()()|( ++= (4.99) nh ngha cc
thng s ti u nh sau:
=
=
)()|(arg supmin1
xfxf fRXpR
fnn
i if
(4.100)
=
=
)()|(arg supmin1
xgxg gRXqR
gnn
i ig
(4.101)
t :
[ ] [ ] Igf uxgxgxfxfw )()|()()|( += (4.102) Ta vit li (4.99) nh
sau:
[ ] [ ]{ }wuxgxgxfxfbAee Iggff +++= )|()|()|()|( (4.103) Thay
(4.93) v (4.94) vo (4.102) ta c phng trnh ng hc vng kn din t mi lin
h gia sai s e v thng s f v g.
[ ]wuxxbAee ITggTff +++= )()()()( (4.104) Ta cn tm lut thch nghi
chnh nh f v g sao cho cc tiu ho e,
ff , gg . Xt phng trnh lyapunov:
)()(21)()(
21
21
21
++= gg
Tggff
Tff
T PeeV
(4.105)
vi 1 v 2 l cc hng s dng, P tho phng trnh: ATP + PA = - Q vi Q l
ma trn n n , xc nh dng. Ly o hm V dc theo qu o h thng ta c:
-
Chng 4 : iu khin m
Trang 310
[ ])()(121
11
xPbePbwePeeV TfT
ffTT
+++=
[ ]ITgTgg uxPbe )()(1 22
++ (4.106)
cc tiu ho e, ff , gg , tng ng cc tiu V, ta chn lut thch nghi sao
cho 0
-
Chng 4 : iu khin m
Trang 311
f(x1,x2) g(x1,x2) x1 F11 F12 F13 F14 F15 x1 F11 F12 F13 F14 F15
x2 x2 F11 - 8 - 4 0 4 8 F11 1.26 1.36 1.46 1.36 1.26 F12 - 8 - 4 0
4 8 F12 1.26 1.36 1.46 1.36 1.26 F13 - 8 - 4 0 4 8 F13 1.26 1.36
1.46 1.36 1.26 F14 - 8 - 4 0 4 8 F14 1.26 1.36 1.46 1.36 1.26 F15 -
8 - 4 0 4 8 F15 1.26 1.36 1.46 1.36 1.26
Vit chng trnh M-file hay dng simulink ca Matlab m phng kt qu v d
trn.
2. Tht k BK m thch nghi trc tip 1.M hnh
u=uD
ym
I TNG x(n) = f(x) + bu, y = x
BK M uD = T(x)
LUT THCH NGHI )(xpe nT =
iu kin u (0)
Hnh 4.34
-pi/6 -pi/12 0 pi/12 pi/6 x1
x2
F11 F12 F13 F14 F15
-
Chng 4 : iu khin m
Trang 312
Phng trnh trng thi m t i tng buxxxfx nn += ),...,,( )1()(
(4.109) y = x (4.110) Trong f l hm bit b l hng s dng cha bit. Ta cn
thit k BK u = uD(x|) da trn h m v lut thch nghi chnh nh thng s .
Lut m c dng nh sau: NU x1 = rP1 v v xn =
r
nP , TH u = Qr (4.111) Trong riP v Qr l cc tp m, r = 1,2,,Lu.
2.Thit k BK m
+ Bc 1: Vi mi bin xi (i=1,2,,n) ta nh ngha mi tp m iliA
(li=1,2,,mi) .
+ Bc 2: Xy dng h m uD(x|) t =
n
i im
1 lut dng:
IF x1= 11lA and and xn= nlA1 , THEN uD = n
llS ...1 (4.112) Trong li = 1,2,,mi, i = 1,2,..,n. S dng lut hp
thnh tch, m ho singleton, gii m theo phng php trung bnh trng s, ta
c:
= ==
= ==
=n
nil
i
n
nil
i
n
m
l
n
i iA
m
l
m
l
n
i iAll
u
m
lD
x
xyxu
1 11
1 1...
1
])([...])([...
)|(1
1
11
1
(4.113)
Chn nlluy...1
nh thng s c th chnh nh v ta a nlluy...1
vo thnh phn ca vct thng s , t lut iu khin c xc nh: )()|( xxu TD
= (4.114) 3.Thit k lut thch nghi Xem u* nh l BK l tng (4.89) trong
phn (4.6.2.1), vi g(x) = b, ta c:
)]|([)( xuubeke DTn += (4.115) Ma trn A c nh ngha nh (4.98), b =
(0,,0,b)T, ta vit li (4.115) dng vct nh sau: )]|([ xuubAee D+=
(4.116) nh ngha thng s ti u * :
-
Chng 4 : iu khin m
Trang 313
=
uxuDRxR n
)|(supminarg
(4.117)
vi =
=
n
i im
1
t : w = uD(x|*) u* (4.118) T (4.114) v (4.118) ta vit li (4.116)
nh sau: bwxbAee T += )()( (4.119) Xt phng trnh Lyapunov
)()(22
1
+= TTbPeeV (4.120)
trong P l ma trn xc nh dng tho: QPAPAT =+ (4.121) o hm (4.120) v
s dng cc biu thc (4.119) v (4.121) ta c:
TTTT bwxPbeQeeV )(])()[(21
+= (4.122)
Xem pn l ct cui ca ma trn P, t b = (0,,0,b)T, ta c eTPb = eTpnb.
Ta vit li (4.122) nh sau:
bwpexpebQeeV nTnTTT += ])([)(21
(4.123)
T (4.123) tho mn 00
nn h (*) l khng n nh. Chn = 1 v nh ngha cc tp m nh sau:
-
Chng 4 : iu khin m
Trang 314
Xy dng 2 lut m nh sau: NU x=N2, TH u(x) = PB NU x=P2, TH u(x) =
NB Trong PB(u)= exp(-(u-2)2) v NB(u)= exp(-(u+2)2). Vit chng trnh
Matlab hoc dng simulink thy c p ng trong hai trng hp c v khng c lut
m.
ng dng : Xy dng BK tc ng c DC
M HNH BK TC NG C DC
Tc mong mun ym
Tc thc y
COM
PWM
My tnh (B iu khin m thch nghi trc tip)
Vi x l (AT89C52) ng c DC Encoder
Hnh 4.35
-
Chng 4 : iu khin m
Trang 315
M hnh gm c : 1. ng c DC 14V, tc Max 2100vng/pht, lm vic khng ti.
2. Cm bin tc Incremental 200xung/vng.
3. Vi x l AT89C52, tn s xung clock 11.059MHz, chu k my )(
000.059.1112
sTVXL = , c nhim v o tc ng c gi v my tnh
iu khin p cp cho ng c bng phng php PWM. 4. Chu k PWM = 1024 TVXL
(1.1ms), chu k ly mu 46.080TVXL (50ms), tc port ni tip
19200Kbps.
5. H s thch nghi thay i tu thuc vo sai lch m
m
yyy
= . Khi
sai lch 1% th = 0, khi
-
Chng 4 : iu khin m
Trang 316
Bng lut hp thnh:
BIN NGN NG GIA TC ( )212 x ( )222 x ( )232 x ( )2m2 x2
BIN NGN NG TC
( )111 x 1,1 2,1 3,1 2m,1 ( )121 x 1,2 2,2 3,2 2m,2 ( )131 x 1,3
2,3 3,3 2m,3
( )1m1 x1 j,m1 2,m1 3,m1 j,m1
2. Lut hp thnh: Xt lut hp thnh th (i,j), vi i = 1m1, j = 1m2 IF
)( 111 xx i AND )( 222 xx i THEN jiu ,= 3. Gii m:
Chn thit b hp thnh Max Product , phng php gii m cao. Gi tr r u
ra PWM iu khin ng c:
Hnh 4.36: Cc tp m )(2 xj ca bin ngn ng gia tc.
-
Chng 4 : iu khin m
Trang 317
( ) ( )
( ) ( )
= =
= =
=
1m
1i
2m
1j
2j21
i1
1m
1i
2m
1j
2j21
i1j,i
x.x
x.x.
u (4.125)
4. Lut cp nht thng s:
( ) ( )
+=+=
ij,ij,i
21j,i2T
j,i
k1k
)x,x(.p.E. (4.126)
Trong : i,j : Thng s cn cp nht lut hp thnh th (i,j). ),( eeE = :
Vct sai s, vi sai s e = ym y , vi ym l vn tc t. p2 : l ct th 2 ca
ma trn P c c t phng trnh Ricatti (4.121).
Vi
=
21
10kk
A , k1, k2 c chn sao cho phng trnh
s2+k1s+k2=0 c nghim nm bn tri mt phng phc. Cc th nghim trong bi
c chn vi 201.0 js = .
>0 l h s cp nht
( ) ( ) ( )( ) ( )
= =
=
1 2m
1k
m
1l2
l21
k1
2i21
i1
21j,i
x.x
x.xx,x : h s xc nh t v IF ca lut
hp thnh th (i,j).
Kt qu m phng v nhn xt:
Ghi ch : Trong cc th bn di, ng lin nt l tc mong mun ym ng cn li
l tc thc.
Trng hp 1: Chn 0 = 0.5; )1202
sin(5001400 tympi
+= , i,j = 20, vi
i = 1m1, j = 1m2. Cc tp m cho bi Hnh 4.36 v Hnh 4.37
-
Chng 4 : iu khin m
Trang 318
b. Sai s ng ra
a. Gi tr PWM
Hnh 4.37: Cc tp m )(1 xi ca bin ngn ng tc .
Hnh 4.38: Kt qu iu khin ca Trng hp 1.
c. p ng ng ra ca m
d. p ng c phng to
-
Chng 4 : iu khin m
Trang 319
Nhn xt: T cc th Hnh 4.38 ta thy rng: tc thp, gi tr PWM thay i t
nhng tc thay i nhiu; tc cao gi tr PWM thay i nhiu nhng tc thay i t.
B iu khin m ban u c thit k m khng da trn nhiu thng tin v i tng,
nhng cht lng iu khin l kh tt d i tng l phi tuyn.
a. p ng trng hp 2a
b. p ng trng hp 2b
Hnh 4.39: Kt qu iu khin ca Trng hp 2
c. Sai s trng hp 2a
d. Sai s trng hp 2b
-
Chng 4 : iu khin m
Trang 320
Trng hp 2: 0 = 0.5 ; )302
sin(5001400 tympi
+= ( trng hp 2a) v
)602
sin(5001400 tympi
+= (trng hp 2b); i,j = 20, vi i = 1..5, j = 1..5. Cc tp m vn nh
trng hp 1. (Xem kt qu Hnh 4.39)
Nhn xt: Vi cng h s cp nht v cc gi tr ban u I,j, khi tc mong mun
ym bin thin nhanh hn th tc thc y khng bm theo kp dn n sai s ln. Do
lut cp nht ph thuc vo ym nn ta cn hiu chnh li thng s 0 cho ph
hp.
a. p ng khi 0 = 0.2
b. p ng khi 0 = 0.5
c. p ng khi 0 = 0.8
d. p ng khi 0 = 1.2
Hnh 4.40: Kt qu iu khin Trng hp 3
-
Chng 4 : iu khin m
Trang 321
Trng hp 3: i,j = 20; vi i =1..5, j = 1..5; )602
sin(5001400 tympi
+= , cc
tp m nh Trng hp 1, 0 ln lt l 0.2, 0.5, 0.8, 1.2. Nhn xt:
Vic tng 0 s lm cho lut cp nht nhy hn vi sai s, do vy p ng h thng
s tt hn. Tuy vy tc thp, khi 0 tng s lm cho tc ng c b dao ng ln hn.
S dao ng t l thun vi vic tng 0 .
Bng kinh nghim qua cc trng hp xt ta thy rng p ng tc ph thuc vo
nhiu yu t nh: s lng tp m, h s 0, i,j, tc bin thin ca tc mong munT
ta a ra vic la chn cc thng s cho ph hp ti u p ng ca h thng.
Trng hp 4: 0 = 2.5; )602
sin(5001400 tympi
+= ; i,j c chn nh bng
bn di, s dng 7 tp m cho bin tc v 5 tp m cho bin gia tc.
BIN NGN NG GIA TC ( )212 x ( )222 x ( )232 x ( )141 x ( )252
x
BIN NGN NG TC
( )111 x 10 10 10 10 10 ( )121 x 10 10 10 10 10 ( )131 x 20 20
20 20 20
( )141 x 30 30 30 30 30 ( )151 x 45 45 45 45 45 ( )161 x 60 60
60 60 60 ( )171 x 80 80 80 80 80
-
Chng 4 : iu khin m
Trang 322
Kt lun chung i tng ng c DC c iu khin bng phng php PWM l i tng
phi tuyn. Mt BK m thch nghi c thit k hp l s iu khin tc ca ng c bm
theo nhiu dng tc mong mun khc nhau. Nhng kinh nghim, thng tin bit v
i tng s rt hu ch trong vic tm ra BK thch nghi ti u. Cc thng s quyt
nh cht lng h thng l : h s 0, gi tr ban u i,j, tn hiu mong mun ym Vi
mi thng s c mt tc dng ring, vic tm ra b thng s ti u cn da vo kinh
nghim v kin thc v h thng iu khin.
c. p iu khin (%PWM)
d. Sai s ng ra
a. p ng ng ra (0..1200s)
b. p ng c phng to
Hnh 4.41: Kt qu iu khin trng hp 4
-
Chng 4 : iu khin m
Trang 323
4.7. H thng iu khin tch hp
Ngnh iu khin hc ra i v pht trin t rt sm, c bit l trong 2 thp nin
gn y vic ng dng L thuyt m v Mng nron to ra nhiu phng php iu khin mi
vi c tnh linh hot v thng minh hn. Cng ngh m v cng ngh mng nron l
hai tr ct chnh to nn cng ngh tch hp mi, cng ngh tnh ton mm (Soft
computing). 4.7.1. Khi nim Mt s phng php c s dng trong ngnh iu khin
hc:
K Kinh in & Hin i K Thng minh PID GA Ti u Nron Thch nghi M
Bn vng
Mi phng php u c nhng im mnh v hn ch nht nh, v vy ngi ta thng c
xu hng kt hp chng li vi nhau to ra mt m hnh iu khin c kh nng p ng
cao vi cc i hi thc t. Vic kt hp ny cho ra mt phng php iu khin mi l
iu khin tch hp. iu khin tch hp : iu khin kt hp phng php kinh in hoc
hin i vi phng php iu khin thng minh.
4.7.2. Mt s h thng tch hp iu khin s dng PID m iu khin m - thch
nghi, m - ti u. S dng h m - nron nhn dng & ti u h thng. ng dng
thut ton GA trong thit k h thng iu khin. ..
phn 4.3 ta trnh by v cch thit k b PID m, phn 4.6 ni v vic tch hp
cng ngh m trong iu khin. Sau y ta trnh by v ng dng gii thut GA
trong iu khin thng qua mt v d.
-
Chng 4 : iu khin m
Trang 324
4.7.3. ng dng thut ton GA thit k b iu khin PID ti u H2/H 1. M t
bi ton Da vo hai bi ton cc i ha d tr n nh v cc tiu ha hm nhy ca iu
khin ti u H, bi ton thit k b iu khin PID ti u H2/H c m t nh
sau.
Cho h thng iu khin PID nh trong hnh 4.42. M hnh P(s) ca i tng
trong bi ton ny c gi thit l c mt sai lch 0(s) c biu din theo m hnh
sai s nhn u ra.
Hnh 4.42: H thng iu khin PID vi sai s nhn u ra
B iu khin PID c dng nh sau:
skskksC 321 /)( ++= (4.127)
Sai s m hnh 0(s) c xem nh n nh nhng khng bit r rng.Gi s 0(s) b
chn nh sau: ,)()( 00 jj
-
Chng 4 : iu khin m
Trang 325
Trong ch tiu cht lng thng c s dng kt hp vi iu khin ti u H l ch
tiu tch phn ca bnh phng sai lch (ISE) hay cn gi l phim hm H2:
0
2 )(min dtteC
(4.130)
vi e(t) l sai s iu chnh trong h thng hnh 4.42. Nh vy, mc tiu ca
bi ton tng hp b iu khin PID kt hp vi iu khin ti u H2/H l tm b iu
khin PID sao cho cc tiu (4.130), ng thi tha mn iu kin rng buc n nh
bn vng (4.129). Tng t, i vi bi ton cc tiu ha hm nhy gim nh hng ca
nhiu n cht lng ca h
+
)()(1
1sCsP
(4.131)
vi l mt gi tr v hng nh hn 1. c trng cho mc nh hng ca nhiu tc ng
n tn hiu ra ca h.
gim nh hng ca nhiu trong dy tn s m nhiu tp trung th iu kin
(4.131) tr thnh:
+
)()(1
)(sCsP
sW (4.132)
vi W(s) l hm trng gim nh hng ca nhiu trong dy tn m nhiu tp
trung.
Nh vy, i vi bi ton cc tiu ha hm nhy, vic xc nh BK PID kt hp vi
iu khin ti u H2/H l tm b iu khin PID sao cho cc tiu (4.130), ng thi
tha mn rng buc n nh bn vng (4.132). 2. C s thit k
Trong trng hp bnh thng i vi h danh nh (khng xt n sai s m hnh ca
i tng cng nh nhiu d), tn hiu sai s iu chnh E(s), trong h thng hnh
4.42 c dng nh sau:
-
Chng 4 : iu khin m
Trang 326
)()(1)()(
0 sCsPsR
sE+
= (4.133)
Theo nh l Parseval, chun bc hai ca mt tn hiu x(t) v nh Fourier
X(j) ca n c quan h nh sau:
pi
djXdttxtx2
22
2(
21)()(
== (4.134)
V vy ta c:
=
0
2 )(min dtteJC
=
pi
djE 2)(21
thay s = j ta c:
=
j
jkkkdssEsEjJ )()(2
1min
321 ,, pi
++
=
j
jkkkds
sCsPsCsPsRsR
j )]()(1)][(()(1[)()(
21
min321 ,, pi
=
j
jkkkds
sAsAsBsB
j )()()()(
21
min321 ,, pi
(4.135)
A(s) v B(s) c th c biu din nh sau: =
=
m
k
kk sasA
0)( ,
=
=
1
0)(
m
k
kk sbsB ; phng trnh (4.20) c vit li nh sau:
==
=
=
=
j
jm
k
kk
m
k
kk
m
k
kk
m
k
kk
m dssasa
sbsb
jkkkJ
00
1
0
1
0321
)(
)(
21),,(pi
(4.136)
-
Chng 4 : iu khin m
Trang 327
Vic xc nh J trong (4.136) c th c tnh mt cch n gin bng l thuyt
thng d nh sau:
- Xc nh tt c cc im cc pk ca E(s). - =
k psEsEsJ
k
)()(Re
Gi tr ca Jm(k1, k2, k3) c th tm c trong Newton, 1957.
10
20
3211 2),,(
aa
bkkkJ =
210
2200
21
3212 2),,(
aaa
ababkkkJ +=
)(2)2(),,(
213030
32203020
2110
22
3213aaaaaa
aabaabbbaabkkkJ+
++=
)(2)()2()2()(
321421
23040
432241
2043020
2141031
222103
20
23
4aaaaaaaaa
aaaaabaaabbbaaabbbaaaaabJ
+
+++++=
Phim hm H2 c dng nh sau:
),,(min 321,, 321
kkkJJ mkkkm = (4.137)
vi Jm(k1, k2, k3) l hm s ca cc thng s PID (k1, k2, k3), v m l bc
ca i tng.
T nh ngha chun H:
)(sup)(),0(
jAsA
(4.138)
iu kin n nh bn vng (4.129) c vit li nh sau:
))()(1))(()(1()()()()()()(
sup)()(1)()()(
00
0000
),0[0
00
jCjPjCjP
jjjCjCjPjPsCsPssCsP
++
=
+
-
Chng 4 : iu khin m
Trang 328
),0[)()(
sup
=
)()(
sup),0[
= 1< (4.139)
Vi () v () l nhng a thc thch hp ca . ngha ca (4.139) l nu gi tr
ln nht ca ()/() nh hn 1, th h thng trong hnh 4.42 n nh vi mi )()(
00 jj
-
Chng 4 : iu khin m
Trang 329
c gi tr cc tiu ca phim hm H2, ng thi tha mn rng buc n nh bn vng
H.
3.1. Biu din nhim sc th
Thut ton di truyn lm vic trn cc nhim sc th (nhng chui s), ch
khng phi chnh bn thn thng s . Mi tp thng s (k1, k2, k3) ca b iu
khin PID s c m ha v ghp li thnh mt nhim sc th. Vic m ha c th c thc
hin bng nhng chui s nh phn hoc thp phn. Trong lun vn ny, s dng phng
php m ha thp phn.
3.2. Hm thch nghi v hm nh gi
Hm nh gi c nh ngha nh sau:
),,,(),,( 321321 kkkJkkkE m= Dkkk 321 ,, (4.144)
V phi ca (4.144) l phim hm H2 m chng ta mun cc tiu, hm nh gi ch
c xc nh trong min n nh (D) ca h thng. Mc tiu ca chng ta l tm (k1,
k2, k3) trong D cc tiu (4.144). Tng ng vi mi nhim sc th ta s c c mt
gi tr ca hm nh gi E(k1,k2, k3). Sau gi tr nh gi c nh x thnh gi tr
thch nghi F(k1,k2, k3) cho ph hp vi thut ton di truyn (tm kim gi tr
cc i). Qu trnh tm kim gi tr nh nht ca Jm(k1, k2, k3) tng ng vi qu
trnh tm kim gi tr ln nht ca F(k1, k2, k3). Nhim sc th c Jm(k1,
k2,k3) nh hn s c gi tr thch nghi ln hn. Sau mi th h, thut ton di
truyn s to ra nhng con chu tt hn, ci thin gi tr thch nghi, do nu
thch nghi ca thut ton di truyn tt hn th s tm c mt b iu khin PID tt
hn. V vy ta c:
),,(1),,(
32132
