Khai ni~m nhi phan cho bi~t cac d?i luQ'Ilg v~t ly ho~c tin hi?u s6 (bi~n nhi phan) co th~ t6n t?i & m<}t trong hai tr?ng thai 1 ho~c 0. Lam th~ nao d~ k~t hgp cac hai ho~c nhi~u bi~n nhi phan d~ dua ra k~t qua TRUE ho~c FALSE, tucmg ung 1 va 0. Be} di~u khi~n l~p trinh dua ra quy~t dinh dga tren k~t qua cua cac tr?ng thai logic.
Cac ho?t d<}ng dugc thl,Ic hi?n b&i cac thi~t bi s6 va dugc dga tren ba ham chuc nang logic ca ban AND, OR, va NOT. Cac chuc nang nay k~t hgp cac bi~n nhi phan d~ t?o cac cau l~nh. M6i chuc nang co m<}t quy tile xac dinh k~t qua cua cau l?nh (TRUE ho~c FALSE) va m<}t bi~u tugng
d?i di?n cho no. K~t qua cua cau l?nh g9i la dfru ra (Y) va cac di~u ki~n cua cau l?nh dugc g9i la dfiu vao (Ava B). Ca hai dfru vao va dfru ra d?i di?n cho hai bi~n tr?ng thai.
3.2.1 Cltfrc niing AND
Hinh 3.1 la ky hi?u cua c6ng AND. Ngo ra c6ng AND la TRUE (1) chi khi t~t ca cac ngo vao la TRUE (1).
Dfiu ra Hinh 3 .1. Ham chuc nang AND
C6ng AND co m<}t s6 lugng ngo vao khong gi&i h?n, nhung no chi co m9t ngo ra. Hinh 3.2 cho th~y m9t c6ng AND co hai ngo vao A, B va m<)t ngo ra Y. L~p bang k~t qua ngo ra dàa tren cac ngo vao dugc xac djnh theo m<)t quy lu~t g9i la bang tr?ng thai. Vi d\l 3. I la m<)t ung dl,lng cua chuc nang AND.
Ban2 tnm ~ thai AND Ne:o vao
A B
y 0 0
0 I
1 0
I l
Hinh 3.2. Ham AND 2 ngo vao va bang tn,mg thai Vi di! 3./:
Ne:o ra y
0 0 0 1
Bi~u di~n c6ng logic, bang tr?ng thai va cac SCY d6 ID?Ch cho m<)t h? th6ng bao d<)ng: coi se bao n~u hai ngo vao, cac nut nh~n PBl va PB2 cung muc 1 (ON).
Giiii:
_ L _ PBI
---0 0----1
_ L _ PBI
---0 )----t.. _ _,
M?ch di~n d?ng c6ng logic
PBt M&
M&
Dong Dong
PB2 Coi bao dom!
M& coi tat
Dong coi tat
Ma Coi tat
Dong Coi keu
34
LI
3.2.2 Ham chuc niing OR
L2 Coi bao d(mg
Chu-ong 3: Khai ni?m vi logic
PBI PB2
_ j _ _ j _
~ o----0
Coi bao d(mg
Hinh 3.3 la ky hi?u cua c6ng OR. Ngo ra c6ng OR la TRUE (1) n~u m(>t ho~c nhi~u d§u vao . la TRUE(!).
D§u ra
Hinh 3.3. Ham chuc nang OR
Nhu chuc nang AND, chuc nang OR cfing co s6 lugng ngo vao khong gioi h~n nhung chi co m(>t ngo ra. Hinh 3.4 cho thiy m(>t c6ng OR co hai ngo vao A, B, m(>t ngo ra Y va bang tr~g thai. Vi d1,1 3.2 la m(>t ung d1,1ng cua chuc nang OR.
Bang trang thai AND N ovito N20 ra
A B y
y 0 0 0
0 I I
I 0 1
l I 1
Hinh 3.4. Ham OR va bang tr~mg thai Vi dlJ 3.2
Bi~u di~n c6ng logic, bang tr~ng thai va cac sa d6 m1;1ch cho m(>t h? th6ng bao d(>ng: cai se bao n~u m(>t trong hai ngo vao cua no len muc I, nut nhin PB I ho~c PB2 la muc I (ON).
_ J _ PB2
- - 0 0---\
PB
- - 0
Cai bao d(>ng
M1;1ch di~n d1;1ng c6ng logic LI
PBI _ j _
PB2 _ j _
L2 Cai bao d9ng
M1;1ch di?n d1;1ng Ladder.
PBl PB2 Coi• bao d{>n2
Mo Mo Cai tat
Mo Dong Cai keu
Dong Mo Cai keu
Dong Dong Cai keu
Cai bao d(>ng
ChU011g 3: Khai nifm vi Logic 35
3.2.3 Ham chirc niing NOT
Hinh 3.5 la ky hi~u cua c6ng NOT. Ngo ra c6ng NOT la TRUE (1) n~u ngo vao la FALSE (0) va nguqc !.;ti. K~t qua cua ngo ra luon la nghich dao cua tr.;tng thai ngo vao, vi v~y no con duqc g9i la c6ng dao.
Ham chuc nang NOT, khong gi6ng nhu cac ham AND va OR, chi c6 m<)t ngo vao. No it khi duqc su di,mg m9t minh ma duqc k~t hqp v&i c6ng AND ho~c OR. Hinh 3.6 cho thfry ho.;tt d9ng cua c6ng NOT va bang tr~mg thai cua no. Luu y rfing A co g.;tch tren diu la NOT A.
D~u vao ~ 0~u ra
Hinh 3.5. Ham NOT
Ham Not
A--01--A NOT Ngo vao A 0 1 Hinh 3.6. Ham NOT va bang tr.;tng thai
Ngo ra
A 0 0
Tho.;tt nhin, khong d~ dang hinh dung ung d\lllg cua cac chuc nang NOT. Tuy nhien, c6ng NOT don gian va kha hfru ich:
1. NOT duqc su di,mg khi yeu du muc O ( di~u ki~n OFF, FALSE, LOW) phai kich ho.;tt m<)t s6 thi~t bi.
2. NOT duqc su di,mg khi yeu du muc 1 (di~u ki~n ON, TRUE, HIGHT) phai dung m<)t s6 thi~t bi.
Hai vi d1,1 sau day cho thfry cac ung d\lllg cua chuc nang NOT su di,mg k~t hqp v&i chuc nang AND va OR, vi dt,I dfiu tien cho thfry chuc nang NOT duqc su dt,Ing m9t minh.
Vi dlf 3.3
Bi~u di~n c6ng logic, bang tr.;tng thai va cac so d6 m.;tch cho m<)t h~ th6ng van solenoid (Vl ):
van Vl se ma (ON) n~u cong tic SI B.~T va n~u cong tAc muc L1 KHONG DONG (ch~t long chua d.;tt d~n muc).
vff-~1--. l - - - J 7 ' 0- ~ 1
,- . - - - ' " '
I ã. I
I I
I , I
I Chuyen m.;tch muc I l nu&c LI ______ I
---<JC1---...r-... S1 LI
M.;tch di~n d.;tng c6ng logic Vl
Luu y: Trong vi dt,I nay, cong tAc muc L 1 thuong ma, nhung no dong !.;ti khi ffi\JC ch~t long d.;tt d~n LI. M.;tch di~n yeu du m9t relay pht,I (CR 1) d~ th\Ic hi~n tin hi~u L 1 thuong dong. Khi LI
36 Chu-ong 3: Khai ni?m vd logic dong (ON), CRI duqc kich ho?t, do do mo ti~p di~m CRI-I va t~t VI. SI la ON khi mu6n kich ho?t h~ th6ng.
LI
LI
CRI-I
L2 CRI
VJ
L2 0 0 I 1
L2 (LI) -
0 I I 0 0 I 1 0
M?ch di~n d?ng Ladder Bang tr~g thai
Vi d1:13.4
VI 0 0 I 0
Hi~n thi c6ng logic, bang tr~g thai va cac so d6 m?ch cho M th6ng bao d(mg: Cai se keu n~u nut nh§n PB 1 la I (ON) va PB2 la O (OFF).
.2E.
- - 0 J - - - ~
- - 0 ~
M?ch di~n d?ng c6ng logic
LI
PBI
___L PB2 Coi bao d9ng
L2
PBl PB2 Coi bao dqng
Mo Ma Cai tat
Mo Dong Cai tat
Dong Mo Cai keu
Dong Dong Cai keu
PBI PB2 ___L
~ J - - - - \ J ~ - - ,
-=-v Coi bao d9ng
Ml;lch di~n dl;lng Ladder M~ch di~n
Liru y: Trong vf dà nay, nut nhĐn PB2 su dàng ti~p di~m thuong dong co nghfa nhu chuc nang NOT.
Hai vf dl) tren cho th§y c6ng NOT dugc d(lt o m9t ngo vao cua c6ng AND. C6ng NOT cfing co th~ duqc d~t angora cua mc;,t c6ng AND. Khi do, ngo ra cua nose phu nh~n, ho~c dao nguqc k~t qua ngo ra c6ng AND. Mc;,t c6ng bj phu nh~ k~t qua nhu v~y duqc g9i la c6ng NAND. Hinh 3.7 cho th§y ky hi~u logic va bang tr?ng thai cua no.
Ham NAND Ngo vao Ngo ra
y A B y
0 0 I
0 1 1
I 0 0
I I 0
Hinh 3. 7: Ham N AND va bang tr?ng thai
Chircrng 3: Khai ni¢m vJ Logic 37 Nguyen tiic tucmg ttJ ap d\mg niu m()t ky hi?u NOT dugc d~t 6 d!u ra cua m()t cling OR.
Ngo ra binh thucmg bi phu nh~n va ham chuc nang m&i dugc gc;>i la cling NOR. Hinh 3.8 cho th~y
ky hi?u va bang tr?ng thai cua no.
Ham NOR Ngo vao Ngo ra
A B y
0 0 l
0 l 0
l 0 0
l l 0
Hinh 3.8: Ham NOR va bang tr~g thai
Chung ta cfing dn tim hi~u v~ cac ky thu~t Boolean d~ viit cac bi~u thuc rut gc;>n cho cac cau l?nh logic phuc t?P khi t?o m()t chucmg trinh di~u khi~n tr?ng thai Boolean ho~c chucmg trinh Ladder.
Nam 1849, m(>t nguoi Anh ten George Boole da phat tri~n d:;ii s6 Boolean. Mvc dich cua d?i
s6 nay la d~ h6 trq trong ly lu~n logic, hinh thuc tri~t hc;>c. No cung dp each dcm gian d~ viit chucmg trinh phuc t:;ip cua "cac tr~g thai logic" duqc dinh nghia nhu cac cau l?nh TRUE ho~c FALSE.
Khi logic s6 duqc phat tri~n vao nhfrng nam 1960, d?i s6 Boolean da chung minh duqc SlJ dan gian d~ phan tich va th~ hi?n cac cau l?nh logic, vi tit ca cac h? th6ng ky thu~t s6 sir dvng m()t khai ni?m logic TRUE/FALSE. Do m6i quan h? gifra logic va Boolean nen doi khi cling logic gc;>i la c6ng Boolean. mQt s6 c6ng k~t n6i duqc v&i nhau gc;>i la m(>t m?ng Boolean ho~c th~m chi m(>t ngon ngfr PLC duqc gc;>i la m(>t ng6n ngfr Boolean.
Hinh 3.9 tom tilt cac toan tu Boolean co ban lien quan d~n cac ham logic co ban AND, OR va NOT. Cac toan tu nay sfr dvng chfr in hoa d~ bi~u di~n cho tin hi?u ngo vao, ngo ra, m(>t d.1u nhan (•) d~ d:;ii di?n cho c6ng AND va m()t d.1u c(>ng (+) d~ d:;ii di~n cho c6ng OR. M(>t chfr cai co g,;ich ngang tren dftu d:;ii di?n cho c6ng NOT.
Ki hi~u Logic Cau l~nh D~i s6 Boolean
A Y =A* B
y Y la 1 Niu A AND B la 1 OR
B Y =AB
A l>- Y Y la 1 Niu A OR B la 1 Y=A+B
B -
NOT Y la 1 N~u A la 0
A 0 A Y la O N~u A la 1 y = Al7l
Hinh 3.9. D:;ii s6 Boolean cua cac cling AND, OR, NOT
Trong hinh 3.9, c6ng AND co hai tin hi?u ngo vao (Ava B) va m(>t ngo ra tin hi?u (Y). Dfru ra co th~ duqc th~ hi~n b~ng cau l?nh logic:
Y la 1 n~u A va B la I .
38 Chu-m,g 3: Khai ni¢m v~ logic
Bi~u thuc Boolean tucmg ung la:
Y=A•B
Duqc d9c la: Y bi1ng A AND v6i B. Bi~u ttrQ'Ilg Boolean ( •) co th~ duqc go bo va bi€u thuc duqc vik Y = AB. Tucmg t\l', n~u Y la k~t qua cua OR A va B, bi~u thuc Boolean la:
Y= A+B
Duqc d9c la: Y bi1ng A OR v6i B. Trong chuc nang NOT, Y la nghich dao cua A, bi~u thuc Boolean la:
Y = Alzl
Duqc d9c la: Y bing NOT A. Bang 3.3 minh h9a Boolean co ban hol;lt d(mg cua AND, OR, va NOT. Bang cling minh h9a each cac chuc nang nay c6 th€ duqc k~t hqp d~ c6 th€ t?o duqc b§t ky k~t hqp logic ma nguoi dung mong mu6n.
1. Cac c6ng CO' ban: Cac c6ng logic CO' ban thlJC hi~n cac chuc nang logic CO' ban. M6i ham logic dtrqc bi€u di~n du6i dl;lng m(>t bang trl;lilg thai va bi~u thuc Boolean cua n6.
D D A B AB A B ABl11 A =~ ãA , B A+B A ..IL -A ,[y B A+ B111 A+B A B111
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 1 0 0 1 0 0 1 0 0 1
1 0 0 I 0 0 1 0 0 1 0 0 1 0
1 1 1 1 1 1 1 1 1 1 1 1 1 1
2. Cac c6ng kSt hqp: B§t ky SIJ k~t hqp cac chuc nang di~u khi€n co th~ duqc th€ hi~n trong cac thu~t ngu Boolean su di,mg ba toan tu dcm gian: ( • ), ( + ), va (-).
DC AB D Y • AB+C
D AB
B D v• Ali+ c
D C A+B D Y • (A+ B)C
;r>A+B
D Y=(A+B)C
3 Cac quy tic dl;li s6 Boolean: Cac ham logic di~u khi~n c6 th~ k~t hqp khac nhau tu dcm gian d~n phuc tl;lp cua cac bi~n d§u vao. Tuy nhien, SIJ k~t hqp d6 d~u phai tuiin thu cac quy tic sau day. C~c quy tic nay la k~t qua cua SIJ k~t hqp dcm gian cua bang trl;lng thai CO' ban va c6 th€ SU dt,mg de dcm gian h6a cac ml;lch logic.
Chirang 3: Khai ni¢m vd Logic
Commutative Laws A+B = B+A
AB=BA
De Morgan's Laws A + B171171 = AB171
AB171rn = Arn + Brn
Arn = A, 1171 = 0, 0111 = 1
A + Al1JB = A + B
AB + AC + B Cl1l = AC + B Cl1l
Associative Laws A+(B+C) = (A+B)+C A(BC) = (AB)C
ãDistributive Laws A(B+C) = AB+AC A+BC = (A+B)(A+C)
Law of Absorption A(A+B) = A+AB = A
39
4. Trinh tv ho?t d{mg va nhom cac d!u hi~u: Trong ho?t d(mg Boolean cua cac c6ng logic AND, OR, NOT, trinh tà thàc hi~n la quan tr9ng. Trinh t\I nay se anh hm'mg dSn gia tri logic kSt qua cua bieu thuc. Xem xet ba tin hi~u d~u vao A, B va C. KSt hgp chung trong bi~u thuc Y = A + B • C co the d~n dSn sai tin hi~u ngo Y tuy thu(>c vao thu tl,T ho?t d(>ng duqc th1,Tc hi~n. Thvc hi~n chuc nang OR tm6c chuc nang AND duqc viSt (A + B) • C, va thvc hi~n chuc nang AND tru6c chuc nang OR duqc viSt A + (B • C). KSt qua cua hai ho?t d(>ng nay khong gif>ng nhau.
Thu tv uu tien trong bieu thuc Boolean la chuc nang NOT (dao nguqc) dAu tien, thu hai la AND va cubi cung la OR, trir khi ch(mg duqc chi ra b~ng each nhom cac d!u hi~u nhu dfiu ngo?c don, ngo?c vuong, ngo?c moc. Theo cac quy tiic nay, n~u sv k~t hgp A+ B • C, khong co dtlu nhom, se luon duqc thvc hi~n la A + (B • C). V 6i dfiu ngo?c, ro rang la B AND v6i C tru6c khi OR k~t qua v6i A. BiSt duqc thu tv thvc hi~n, sau do, lam cho no co the viSt don
gian nhu A + BC, ma khong SQ' sai sot. .
Khi 1~ vi~c v6i cac bieu thuc logic .Boolean, sir dvng cac dfru hi~u nhom la each dung ph6 biSn. Tuy nhien, n~u cac dfru hi~u xay ra theo C?P, chung thuong khong gay ra vfrn d@ nJu ch(mg da duqc d~t dung theo logic mong mubn. Hai biSn AND se duqc trong ngo?c la khong dn thiSt b&i vi chuc nang AND se duqc thvc hi~n tru6c. N~u hai tin hi~u ngo vao duqc OR tru6c AND, chung phai duqc d?t trong dfru ngo?c dcm.
De dam bao thu tl,T thich hgp cua vi~c thvc hi~n m(>t bi~u tht'.rc, SU' d\lfig dJu ngO?C dcm () lam cac dtlu hi~u nhom. NSu cac d!u hi~u b6 sung duqc yeu du la dfru ngO?C vuong [], va sau do la dfiu ngo?c moc {} duqc su dvng. M(>t minh ho? v@ vi?c su dvng cac dJu hi~u nhom duqc th~ hi~n du6i day:
Yl = Y2 + Y5 [Xl(X2 + X3)] + {Y3[Y4(X5 + X6)]}
AB=A+ B
A
AB Y=A+B
B
andMB= AB
A
Y=A+B Y = AB