B IU KHIN M CHO I TNG L NHIT VI THNG S CA L L:
iu Khin PID M Cho i Tng L Nhit
B IU KHIN PID M CHO I TNG L NHIT VI THNG S CA L L:
K=110,T=80,L=60:
Trong K: li tnh
T: hng s thi gian
L: thi gian tr
A.B iu khin PID:
1.CC KHI NIM V IU KHIN PID:
B iu khin PID c s dng rt rng ri trong thc t iu khin nhiu loi i
tng khc nhau nh nhit l nhit, tc ng c, mc cht lng trong bn cha l do
b iu khin ny c s dng rng ri l v n c kh nng trit tiu sai s xc lp,
tng p ng qu , gim vt l nu cc tham s b iu khin c la chn thch hp. Do
s thng dng ca n nn nhiu hng sn xut thit b iu khin cho ra i cc b iu
khin thng mi rt thng dng. Thc t cc phng php gii tch rt t c s dng do
vic kh khn trong xy dng hm truyn i tng.
B iu khin PID c hm truyn lin tc nh sau:
Kp = h s khuch i t lTi = thi gian tch phn
Td = thi gian vi phn
Ki = h s khuch i tch phn
Kd = h s khuch i vi phn
2.Quy tc Ziegler-Nichol cho cc b iu khin PID:
Ziegler v Nichols ngh cc qui tc xc nh cc gi tr h s khuch i t l
Kp, thi gian tch phn Ti, v thi gian vi phn Td da trn cc c tnh p ng
qu ca mt i tng cho trc. Vic xc nh cc thng s ca b iu khin PID hay
chnh nh cc b iu khin PID c th c chnh nh bi cc k s v i tng.C hai
phng php chnh nh ca Ziegler-Nichol. Trong tt c hai phng php ny chng
ta t mc ch vt cc i bng 25% trong p ng bc.
Hnh 1: ng cong p ng bc n v, vt cc i 25%:
Phng php th nht.
Trong phng php ny chng ta t c bng thc nghim p ng ca i tng vi tn
hiu vo bc n v, nh c hnh v 2. Nu i tng khng c khu tch phn v cng khng
c cc cc lin hp phc tri, th ng cong p ng bc n v trng ging ng cong ch
S, nh hnh v 3( Nu p ng khng c dng ch S th phng php ny khng p dng
oc). Cc ng cong p ng bc ny c th c to ra bng thc nghim hoc t m phng
ng hc ca i tng.ng cong ch S c th c c tnh ha bi hai hng s, thi gian
tr L v hng s thi gian T. Thi gian tr v hng s thi gian c xc nh bng
cch v tip tuyn ti im un ca ng cng ch S. Xc nh giao im ca tip tuyn
vi trc thi gian v ng C(t)=K. Hm truyn C(s)/U(s) khi c th c xp x bng
mt h thng bc nht c tr truyn t.
Hnh 2: p ng bc n v ca i tng.
Hnh 3: p ng dng S.
Phng php th nht ca Ziegler-Nichol c:
V vy b iu khin PID c mt cc ti gc ta v im khng kp ti s=-1/L
Cc tham s ca b iu khin PID c tnh theo phng php nc ca
Ziegler-Nichol nh bng 1: Bng 1B iu khinKpTiTd
P
0
PI
0
PID
2L0.5L
Phng php th hai: Trong phng php ny chng ta t Ti= v Td=0. s dng
tc ng iu khin t l (nh hnh 4), tng Kp t 0 dn gi tr gii hn Kct, khi
tn hiu ra dao ng iu ha. (Nu tn hiu ra khng dao ng iu ha vi bt c gi
tr Kp no , th phng php ny khng s dng c.) V vy, h s khuch i gi hn
Kct v chu k tng ng Pct c xc nh bng thc nghim(hnh 5). Ziegler v
Nichols ngh t cc gi tr ca cc thng s Kp,Ti v Td theo cc cng thc
trong bng 2.
Hnh 4: H thng vng kn vi b iu khin t l
Hnh 5: Dao ng iu ha vi chu k Tgh
bng 2:B iu khinKpTiTd
P0.5Kgh 0
PI0.45Kgh
0
PID0.6Kgh0.5Tgh0.125 Tgh
p dng cc qui tc ca Ziegler-Nichol va iu khin l nhit
B iu khin PID c hm truyn dng lin tc nh sau:
3.t trng ca b iu khin PID:
iu khin t l (Kp) c nh hng lm gim thi gian ln v s lm gim nhng
khng loi b sai s xc lp. iu khin tch phn (Ki) s loi b sai s xc lp
nhng c th lm p ng qu xu i. iu khin vi phn (Kd) c tc dng lm tng s n
nh ca h thng, gim vt l v ci thin p ng qu .
nh hng ca mi b iu khin Kp, Ki, Kd ln h thng vng kn c tm tt bng
bn di (bng 3).Bng 3
p ng
vng knThi gian
lnVt lThi gian
xc lpSai s
xc lp
KpGim Tng Thay i nhGim
KiGimTngTngLoi b
KdThay i nhGimGimThay i nh
4.Xc nh hm truyn l nhit
Ta xc nh hm truyn gn ng ca l nhit theo nh ngha:
Tn hiu vo hm nc n v (cng sut=100%)
Tn hiu gn ng chnh l hm C(t)= f( t L )
Trong
Tm laplace ca hm C(s): Ta xt hm dch chuyn f(1- )1(t- ), vi 0. Hm
ny bng 0 khi t< .. Hm f(t)1(t) v hm f(t- )1(t- ) c v trn hnh 6:
Hnh 6:f(t) v hm f(t- )1(t- )
Theo nh ngha bin i laplace ca f(t- )1(t- ) l:
Dng phng php bin i T1=t- ta c
V vy vi 0
Trong ta ly Laplace ca hm
VY
SUY RA :
HM TRUYN L NHIT L:
Thng s l nhit l K=110,T=80,L=60i ra n v s l T=4800s,L=3600s,
Hm truyn B iu khinKpTiTd
P
0
PI
0
PID
2L0.5L
G(s)=
Da vo bng ta tnh :
(Kp=1.2=0.001455, Ti=2L=7200sec, Td=0.5L=1800sec
Kp=0.01455, Ki=0.0000000202, Kd= 26.19
B.B iu khin m:
1.Gii thiu m:
Trong nhng nm gn y cc h m c nhng bc tin nhanh chng. H thng m c p
dng vo nhiu h thng khc nhau: iu khin. x l tn hiu truyn thng , ch to
vi mch, cc h chuyn giaTrong , c nhng ng dng c gi tr nht thuc v lnh
vc iu khin. Tuy nhin, c rt nhiu ngi s dng b iu khin m nhng li khng
hiu r v nhng vn c bn ca h m. iu ny dn ti vic s dng khng hiu qu b iu
khin m.
2. Cc khi nim v iu khin m:
2.1 nh ngha tp m:
Tp m F xc nh trn nn B l mt tp cc phn t ca n l mt cp gi tr Vi x
thuc X v l mt nh x:
:B([0,1]
l mt min lin thuc ca bin x nhn gi tr trong on [0,1]C rt nhiu dng
hm lin thuc: Gausian, dng PI, dng S, dng tam gic u.Hnh 6: gi tr hm
lin thuc tam gic u theo ng vo x
2.2 Bin ngn ng:
L thnh phn ch o trong cc h thng s dng logic m. y cc thnh phn ngn
ng ca cng mt ng cnh c kt hp vi nhau. V d: ta c cc m t v vn tc xe nh
sau
Vn tc xe l nhanh; hay vn tc xe l trung bnh; hay l chm.
Vn tc chnh l bin ngn ng. Nhanh, trung bnh, chm l cc gi tr ngn ng
ca bin vn tc.
2.3 Cc php ton trn tp m:Cho X v Y l hai tp trn khng gian nn B, c
cc hm thuc tng ng l khi : php hp hai tp m:X
-Theo lut Max:
( 2.3.1)-Theo lut Sum: (2.3.2)-Tng trc tip:
(2.3.3)Php giao hai tp m: XY Theo lut Min: (2.3.4) Theo lut
Lukasiewics: (2.3.5) Theo lut Prod: (2.3.6)Php b tp m: (2.3.7) 2.4
Mnh hp thnh m:Quay li mnh kinh in, gia mnh hp thnh pq v cc mnh iu
kin p, kt lun q c mi quan h sau:
PQp(q
001
011
100
111
Bng 4: Bng gi tr ca mnh logic ko theoNh vy, mnh hp thnh kinh in
pq l mt biu thc logic c gi tr tha mn p=0
EMBED Equation.3 =1
q=1
EMBED Equation.3 =1
p=1& q=0
EMBED Equation.3 =0
t ba tnh cht trn ta rt ra c:
p1p2
EMBED Equation.3
EMBED Equation.3
q1q2
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3 Mnh hp thnh m c cu trc nh sau:
Nu x=A th y=B
Hay
Trong , [0.1] l hm thuc ca tp m u vo A trn tp nn X,
[0.1] l hm thuc ca B trn tp nn Y.Suy din n thun: Gi tr ca mnh hp
thnh m l mt tp m nh ngha trn nn Y v c hm thuc :Y[0.1] tha mn:
ch ph thuc vo v
=o
EMBED Equation.3 =1
=1
EMBED Equation.3 =1
=1 & =0
EMBED Equation.3 =1
EMBED Equation.3 =1
EMBED Equation.3
EMBED Equation.3 Do mnh kinh in lun ng khi p sai nn s chuyn i
thnh mnh m s dn ti nhng nghch l, v d nh mnh sau:
Nu nh sng= ti th n bt
Trong trng hp nh sng = nng ta c mnh :
Nu nh sng = nng th n bt mnh ny vn c gi tr ng
Nh vy, d c nng hay ti tri bt n l u bt hp l. gii quyt vn ny
Mamdani a ra nh l:
ph thuc ca kt lun khng c ln hn ph thuc ca iu kin.T y hnh thnh
quy tc hp thnh Mamdani l:
==
Hay =
(2.4.1)Tng ng l quy tc MIN v PROD.Thot nhn hai quy tc hp thnh
trn c dng gn ging nh cng thc(2.3.1) v(2.3.3) xc nh hm thuc ca tp
giao hai tp m. Tuy nhin chng li khc nhau bn cht l trong khi tp m kt
qu ca quy tc hp thnh c nh ngha trn tp nn ca B, cn ca li c nh ngha
trn hai tp nn tch ca hai tp nn A v B. Ngoi ra, gi tr ca phc thuc gi
tr r u ra cn th khng.
2.5 Lut hp thnh m:
Lut hp thnh c hiu theo ngha l tp hp cc mnh hp thnh.
ng vi mt gi tr r u vo xo th thng qua php suy din m ta c n tp m u
ra tng ng vi n mnh ca lut hp thnh. Ta k hiu:
Lut hp thnh l R.
Cc mnh hp thnh l: R1,R2,....,Rn
Tp m u ra tng ng l:
u ra ca lut hp thnh l c gi tr l:
Nu nh php ton hp thnh ny c tnh theo quy tc Max v cc c tnh theo
quy tc Min th ta c lut hp thnh Max-Min. tng t vy ta cn c cc lut hp
thnh khc l: Lut hp thnh Max-Prod
Lut hp thnh Sum-Min Lut hp thnh Sum-Min
Lut hp thnh Sum-Proda. thut ton xy dng mnh hp thnh cho h
SiSO:
Lut m cho h SISO c dng If A then B
Chia hm thuc thnh n im xi, i=1,2,..,n
Chia hm thuc thnh m im yj, j=1,2,m
Xy dng ma trn quan h m R.
Hm thuc u ra ng vi gi tr r u vo xk c gi tr , vi
aT={0,0,0,,0,1,0,..,0,0}. S 1 tng ng vi v tr th k.Trong trong hp u
vo l gi tr m A, th l ={l1,l2,l3,,lm} vi lk =maxmin{ai,rik}.
b. Thut ton xy dng mnh hp thnh cho h MISO:
Lut mcho 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 , ,, ,
Xc nh tha mn H cho tng vct gi tr r u vo x={c1,c2,,cn} trong ci l
mt trong cc im mucua t suy ra
Lp ra ma trn R gm cc hm thuc gi tr m u ra cho tng vct gi tr m u
vo :
hoc
2.6 Gii m:Gii ml qu trnh xc nh gi tr r no c th chp nhn c t hm
thuc ca gi tr m (hay xc nh gi tr r u ra ng vi gi tr u vo). C hai
phng php gii m chnh l phng php cc i v phng php trng tm.
2.6.1 Phng php cc ai:
Theo t tng cho rng gi tr r u ra i din cho tp m phi l gi tr c xc
sut thuc tp m ln nht, phng php cc i giii m gm c hai bc:
B1: Xc nh min cha gi tr r , gi tr r l gi tr m ti hm thuc t gi tr
cc i( cao H ca tp m), tc l min:
Vi Y l tp nn ca tp m .
B2: Xc nh c th chp nhn c t G.
thc hin bc hai ta c ba nguyn l:
Nguyn l trung bnh.
Nguyn l cn tri.
Nguyn l cn phi.
Ta k hiu:
v
a. Nguyn l trung bnh:
Nguyn l trung bnh c s dng trong trng hp G l mim lin thng v nh vy
y s l gi tr c ph thuc ln nht.
b. Nguyn l cn tri:
Gi tr r s ly gi tr cn tri ca G. Gi tr r ly theo nguyn l cn tri
ny s ph thuc tuyn tnh vo tha mn ca lut iu khin quyt nh.
c. Nguyn l cn phi:
Gi tr r s ly bng gi tr cn phi ca G. Gi tr r ly theo nguyn l cn
phi ny cng s ph thuc tuyn tnh vo p ng vo ca lut iu khin quyt
nh.
-Sai lch gi tr r gia ba phng php trn s cng ln khi tha mn H ca
lut iu khin quyt nh cng nh.
2.6.2 Phng php trng tm:
Phng php trng tm s cho ra kt qu l honh ca im trng tm ca min c
bao bi trc honh v ng .
Cng thc xc nh theo phng php im trng tm nh sau:
(2.6.1)
Cng thc ny cho php ta xc nh gi tr vi s tham gia ca tt c cc tp m
u ra ca mi lut iu khin mt cch bnh ng v chnh xc, tuy nhin li khng n
tha mn ca lut iu khin quyt nh v thi gian tnh chm.
Phng php im trng tm cho lut hp thnh Sum-Min:
Gi s c m lut iu khin c trin khai, k hiu gi tr m cho ng ra ca lut
iu khin th k l th vi quy tc Sum-Min gi tr r s c xc nh l: trong
v
(2.6.2)Xt trng hp cc hm thuc hnh thang:
2.6.3 L
Hnn 7: Hm thuc hnh thang.
Phng php cao:
T cng thc (2.6.1), nu cc hm lin thuc c dng singleton th ta
c:
cng thc c th s dng cho tt c cc lut hp thnh: Max-Min, Sum-Min,
Max-Prod v Sum_Prod.
p dng l thuyt thit i tng l nhit:
3. B iu khin m cho l nhit:
Hnh 8: S khi b iu khin m cho l nhit.
Vi s khi ca b iu khin m c bn nh trn, ta rut ra 5 boc cn tin hnh
tng hp mt b iu khin m:
nh ngha tt c cc bin ngn ng vo/ra.
nh ngha tp m cho cc bin vo /ra.
Xy dng cc lut iu khin.
Chn thit b hp thnh.
Gii m.
Trong , bc 1 v 2 lm cng vic ca khu m ha, bc 3 thit lp b my suy
din m da trn lut hp thnh m,bc 4 v 5 lm cng vic gii m nhn c gi tr r
ng ra.
3.1 nh ngha cc bin vo/ra:
a. Sai s: ET
ET(m nhiu, m, m t, khng, dng t, dng, dng nhiu)
Vi cc k hiu tng ng sau:
NB: m nhiu, NM:m,NS: m t,ZE: khng
Ps: dng t,
PM: dng,
PB: dng nhiu,
ET{NB,NM,NS,ZE,PS,PM,PB}
b. o hm sai s: DETTng t DET{ NB,NM,NS,ZE,PS,PM,PB}
c. Sai phn cng sut: DP
DP{ NB,NM,NS,ZE,PS,PM,PB}
3.2 Tp m cho bin ngn ng:DET
Et:C= gi tr t gi tr o
DP:%
Hnh 9: Biu din tp m vi hm thuc dng tam gic u.
3.3 Lut iu khin:
Th hin trong bng m:
Bng mDPDET
ETNBNMNSZEPSPMPB
NBNBNBNBNBNMNSZE
NMNBNBNMNMNSZEPS
NSNBNMNMNSZEPSPM
ZENBNMNSZEPSPMPB
PSNMNSZEPSPSPMPB
PMZEZEPSPMPMPBPB
PBZEPSPMPBPBPBPB
Nhng in nghing gch di l nhng lut c bn nht cn phi xc nh. Ta c th
pht biu mt mnh hp thnh nh sau:
Lut hp thnh bng m:
IF ET is NB and DET is NB then DP is NB tng t ta lp c bn m:3.4
Chn lut hp thnh: Lut hp thnh c chn l lut Max-Min.3.5 Gii m:
Bng phng php cao theo cng thc:
3.1
tha mn Hijc theo cng thc (2.3.4), yij l gi tr vt l tng ng vi gi
tr ngn ng.C. Thit k b iu khin m cho l nhit:xy dng b PID m iu khin l
nhit. Hm truyn l nhit theo Zeigler-Nichols:
ng dng matlab cho l nhit: tuyn tnh ha
Cc bc thit k:
1. xc nh bin ngn ng:
u vo 2 bin
-Sai lch ET= o-t -Tc tng DET=, vi T l chu k ly mu. -u ra : 3 bin
KP h s t l
KI h s tch phn
c tnh phn PID
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 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}
2. Lut hp thnh:
C tng cng l 7x7x3=147 lut IFTHEN
Lut chnh nh KP
Lut chnh nh KD
Lut chnh nh KI
NG DNG MATLAB: BIU DIN KP:
S KHI:
SVTHHUNH VN HNG TD05
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T
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Text
L Nhit
B iu khin PID
