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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_1243447183.vsdText

L

T

K

a

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Text

L Nhit

B iu khin PID