TRNG I HC KT-KT HI DNG KHOA IN T - TRUYN THNG Hc phn: iu khin logic Ging vin: ThS. L Th N Gii thiu mn hc 5/20/20122 -S VHT: 3 (2:1) -i tng: SV ngnh in t truyn thng, chuyn ngnh in Cng Nghip -Ti liu hc tp:Bi ging iu khin logic - Ti liu tham kho:1. Gio trnh iu khin lgc v ng dng Nh xutbnkhoahcvKthut-PGS.TS Nguyn Trng Thun;2. Cc loi cm bin trong k thut v o lng MC TIU HC PHN 5/20/20123 Sinh vin c kh nng: + Phn tch, tng hp, thit k cc mch iu khin tun t trong thc t nh mch cu trc, bng ti, vv + c hiu cc bn v iu khin cc thit b in, cc my cng c trong cng nghip Ni dung mn hc 5/20/20124 Gm 3 chng: Chng 1: C s ton hc Chng 2: Tng hp mch nChng 3: Tng hp mch kpCHNG 1: C S TON HC 5/20/20125 1.1. L thuyt i s boole 1.1.1. t vn -Trongcucsng:ccsvt,hintngthng biu hin hai mt i lp nhau. VD: Mt vt p - xu; Nc sch hay bn,

-TrongiukinKT-XH:thnggpbitonm d liu vo ch c th nm 1 trong 2 trng thi i khng nhau. VD: ng sai;Tt - xu;t - r - Trongkthut(cbitlkthutinviu khin) cc phn t iu khin lun mt trong hai trng thi tc ng hoc khng tc ng, ng hoc ct, VD:Rle, cng tc t, vv -Trongtonhc,lnghahaitrngthii lp ca mt s vt hin tng ngi ta dng hai gi tr 0 v 1; ON OFF; TRUE FALSE; Ct ng1.1. L thuyt i s boole 1.1.1. t vn -GiathkXIX,GeorgeBoole- nhtonhc ngiAnhxydngcstonhctnh ton cc hm v bin ch ly hai gi tr 0 v 1.i s lgc = i s Boole 5/20/20126 1.1.2.Miquanh giaisboole vccphnttc ng gin on - isBoolecngdngvthchin rngrithngquahnhviiukhincacc thit b Rle . -Rlechcthmttronghai trngthiquan st c l tip im ng hoc m v v nguyn tc khng c hin tng chp chn gia ng v m. CHNG 1: C S TON HC CHNG 1: C S TON HC 5/20/20127 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim a. Bin Lgc TrongisBoole,ccbincgilbin Logc nu chng ch c hai gi tr, c trng cho haitrngthiikhngcamthintngv c k hiu bng hai ch s 0, 1. b. Mch logic (Hm lgc)- nhngha:Mchlogicbaogmsghpni caccphntvtl,nhmthchinccquan h logic xc nh trc. M ch LgcABCQ1Q2CHNG 1: C S TON HC 5/20/20128 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim -NuQ1,Q2chphthucvogitrccbin vo th mch gi l mch logic t hp: Q1 = Q1(A, B, C); Q2 = Q2 (A, B, C) - NuQ1,Q2cnphthucvotrngthibn trong tthiimxtthmchgilmch logic dy: Q1 = Q1(A, B, C,t ); Q2 = Q2 (A, B, C, t) a. Bin Lgc b. Mch logic (Hm lgc)c. Thit b Lgc +Thitblogiclccthitbchaitrngthiv thc hin nhim v bin i tn hiu.VD:Rle,Cngtctctipimvccloi rle khng tip im l cc phn t gin on. CHNG 1: C S TON HC 5/20/20129 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic (hi, v, giao) +nhngha:thchinphptnhhi(gil phpnhnlogic)giaccbinA,B,Cu vo. Bin ra l: Q = A.B.C + K hiu phn t v & BAQBAQ( Q = A.B )( Q = A.B )+ Bng gi trABQ 000 010 100 111 CHNG 1: C S TON HC 5/20/201210 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic (hi, v, giao) + K hiu phn t hoc + Bng gi trb. Php cng logic (tuyn, hp, hoc)+ nh ngha: Thc hin php tnh tuyn (cn gi l php cng lgc) gia cc bin vo A, B, C, Bin ra Q = A + B+ C + BAQ>1BAQ ( Q = A + B )( Q = A + B )ABQ 000 011 101 111 CHNG 1: C S TON HC5/20/201211 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic (hi, v, giao) + K hiu phn t hoc o+ Bng gi trb. Php cng logic (tuyn, hp, hoc+ nh ngha: Thc hin php tnh ph nh ca bin vo A. c. Php nghch oA1 A AAA 01 10 CHNG 1: C S TON HC5/20/201212 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic (hi, v, giao) + K hiu phn t v o+ Bng gi trb. Php cng logic (tuyn, hp, hoc+nhngha:lmchthchinhaiphptnh logiclintipnhau:phpnhn,knlphp ph nh. c. Php nghch od. Php V o BA& BAQ ( Q = A.B )Q ( Q = A.B )ABQ 001 011 101 110 CHNG 1: C S TON HC5/20/201213 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic (hi, v, giao) + K hiu phn t hoc o+ Bng gi trb. Php cng logic (tuyn, hp, hoc+nhngha:lmchthchinhaiphptnh logiclintipnhau:phpcnglogic,tiptheo l php ph nh c. Php nghch od. Php V o e. Php hoc o ( Q = A + B )( Q = A + B )BA>1BAQQABQ 001 010 100 110 CHNG 1: C S TON HC5/20/201214 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic (hi, v, giao) + K hiu + Bng gi trb. Php cng logic (tuyn, hp, hoc+ nh ngha: L mch thc hin php tnh XOR. u ra Q s bng 1 khi A v B khng bng nhauc. Php nghch od. Php V o e. Php hoc o f. Cng khng ng t BA( Q = A B ) QABQ 000 011 101 110 CHNG 1: C S TON HC5/20/2012 15 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic a. Php nhn logic + K hiu + Bng gi trb. Php cng logic (tuyn, hp, hoc+nhngha:lmchthchinhaiphptnh lin tip: php tnh XOR, k n l ph nh. S c ga tr l 1 khiA v B l tng ngc. Php nghch od. Php V o e. Php hoc o f. Cng khng ng t g. Cng ng t( Q = A B ) QBAABQ 001 010 100 111 CHNG 1: C S TON HC5/20/201216 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh cht ca php ton lgc a. Tnh giao honGi s x1, x2, x3 l cc bin lgc ta c: x1 + x2 + x3 = x2 + x3 + x1 x1x2= x2x1

b. Tnh kt hp (x1 + x2)+ x3 = x1 + (x2+ x3) (x1 . x2). x3 = x1 . (x2. x3) c. Tnh phn phi (x1 + x2). x3 = x1 . x3 + x3 . x2 x1. x2 + x3= (x1 +x3).(x2 + x3) (*) Chng minh (*) bng bng sauCHNG 1: C S TON HC5/20/201217 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh cht ca php ton lgc c. Tnh phn phi (x1 + x2). x3 = x1.x3 + x1.x2 x1.x2 + x3= (x1 +x3).(x2 + x3) (*) x1x2x3x1 . x2 x1 . x2 + x3 x1 +x3 x2 + x3 (x1 +x3).(x2 + x3) 00000000 00101111 01000010 01101111 10000100 10101111 11011111 11111111 CHNG 1: C S TON HC5/20/201218 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh cht ca php ton lgc d. nh lut De Morgan *) Dng n gin Nghch o ca mt tng bng tch cc nghch o Nghch o ca mt tch bng tng cc nghch o V d: B A B A . = +B A B A + = .Ta chng minh tnh ng n ca biu thc trn bng cch thnh lp bng di yCHNG 1: C S TON HC5/20/201219 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh cht ca php ton lgc d. nh lut De Morgan *) Dng n gin x1x2 10011111 10110001 11001001 01100000 +21 x x1x2x2 1x x+ 2 1x x2 1.x xCHNG 1: C S TON HC5/20/201220 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh cht ca php ton lgc d. nh lut De Morgan *) Dng tng qut Nghch o ca mt hm bt k s cho mt hm khctngngnu:Thayccbintrong hmbngnghchoccbinn,cco binnthnhccbinnvittcdu cng thnh du nhn v cc du nhn sang du cng v tr ca n. V d:

) ).( (2 1 2 1 2 1 2 1x x x x x x x x + + = + += + +2 1 2 1x x x xCHNG 1: C S TON HC5/20/201221 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh cht ca php ton lgc e. Mt s biu thc thng dng trong i s logic1A + 0 = A10A.B = B.A 2A.1 = A11A+B = A + B 3A.0 = 012A(A + B) = A 4A + 1 = 113 .B + A.B = B 5A + A = A14 (A+ B)(B+ ) = B 6A.A = A15(A + B + C) = ( A + B ) + C 7 + A = 116A.B.C = ( A.B) .C 8 . A = 017 9A + B = B + A18 B A B A + = .B B A . A = +AAAAACHNG 1: C S TON HC5/20/201222 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh chtca php ton lgc 1.2.4. S nguyn l *) Biu thc cu trc (hm cu trc) nh ngha: Biu thc cu trc l biu thc cho bit cu trc bn trong ca h ang xt. V d: *) S cu trc - L mt dng biu din ca biu thc cu trc. Nhnvocththyngaysnitiphay song song ca cc bin lgc. T biu thc cu trc s cu trcCh : Nhn l ni tip Cng l song song ab b a ab b a f + + = ) , (CHNG 1: C S TON HC5/20/201223 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh chtca php ton lgc 1.2.4. S nguyn l *) S cu trc T biu thc cu trc ta c s cu trc nh saua b ya ba bV d: ab b a ab b a f + + = ) , (CHNG 1: C S TON HC5/20/201224 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.2.1. Khi nim 1.2.2. Cc php ton i vi bin Logic 1.2.3. Cc tnh chtca php ton lgc 1.2.4. S nguyn l *) S nguyn l s dng cc phn t lgc C C B A B A y ) )( ( + + + =V d:S nguyn l sau: A 1ABA +BA+B+CC1Cy = (A+B)(A+B+C)C CHNG 1: C S TON HC5/20/201225 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.3. Mt s khi nim v l thuyt tmt hu hn 1.3.1. t vn ivingithitk,hthngiukhin (HTK) c coi nh hp en. Trong iu khin hc,hpenccoinhlitngnghin cu: Cn phi xc nh cu trc ca hp en khi bit c cc tn hiu vo/ra. HTKABCQ1Q2 Thit b iu khin lm vic theo nguyn tc gin on th hp en vi u vo/ra xc nh s c gi l mt tmt hu hn. CHNG 1: C S TON HC5/20/2012 26 1.1. L thuyt i s boole 1.2. Cc hm c bnCa i s Logic 1.3. Mt s khi nim v l thuyt tmt hu hn 1.3.1. t vn 1.3.2. Khi nim *) Mch n (t hp): Mch n l mt tmt hu hn m tn hiu ra chphthucvotnhiuvo(haynicchkhc ng vi mt t tn hiu vo ch c mt trng thi ra xc nh). *) Mch kp (hay h dy): Mchkplmttmthuhnmtnhiu rakhngchphthucvotnhiuvomcn ph thuc vo trng thi trc ca h thng .CHNG 2. TNG HP MCH N 5/20/201227 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l nh ngha:Bng chn l cho bit quan h u vo v u ra ca mch n. Numch c n bin vo v1 bin ra th bng biu din c:S ct = n +1 S hng = 2n + 1 c im ca cch biu din ny: - R rng, d nhn, t nhm ln. - Di dng, cng knh khi bin s ln. Vd1: Mtmchn c 3 bin vo l a, b, c mt binralQ.Quanhgiauvovuranh sau: CHNG 2. TNG HP MCH N 5/20/201228 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l Bng chn l: abcQ 0001 0011 0100 0110 100x 1011 110x 111x Ch:Nhngnhduxlgatrhm khng xc nh (c th l 0 hoc 1) CHNG 2. TNG HP MCH N 5/20/2012 Mchc1binra v4binvo,vy bngchnlc5 ctv17hngnh sau: 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l Bi tp p dng: V d 2: T biu thc lgc:D ABC D C B A D C B A D C B A Y + + + =Hy lp bng chn l cho mch Lgc trn?Bi gii ABCDY 00001 00010 00100 00110 01000 01011 01100 01111 10000 10011 10100 10110 11000 11010 11101 11110 CHNG 2. TNG HP MCH N 5/20/2012 30 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l Vd3:Mtnthngcyucusau:Mt qut in ch quay khi c du bi trn v lng bo him. Hy vit bng chn l? Bi gii Nhn xt: C 3 bin vo: a, b, c v mt bin ra: Q Tn hiu voTn hiu ra K hiu nghaK hiu ngha a = 0 a = 1 Qut khng c in Qut c in Q = 0 Q = 1 Qut khng chy Qut chy b =0 b = 1 Qut khng c du Qut c du c = 0 c = 1 Qut cha c lng bo him Qut c lng bo him CHNG 2. TNG HP MCH N 5/20/201231 2.1. Biu din mch n Lp bng chn l ABCQ 0000 0010 0100 0110 1000 1010 1100 1111 T yu cu cng ngh rt ra nhn xt Q = 1 khi tt c cc tn hiu vo a, b, c u c tn hiu l 12.1.1. Biu din bng bng chn l CHNG 2. TNG HP MCH N 5/20/201232 2.1. Biu din mch n V d 4: C3loaavobkhuchichaiura,1 u l S4, 1 u l S8. - Nu 2 trong 3 loa cng hot ng th a vo S4. - Nu c 1 loa th a vo S8. - C 3 loa cng hot ng th khng a vo. Hy phn tch tn hiu vo ra v lp bng chn l ? 2.1.1. Biu din bng bng chn l CHNG 2. TNG HP MCH N 5/20/201233 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l 2.1.2. Biu din mch n bng hm tuyn chun ton phn v hm hi chun ton phn*) Cch vit hm dng Tuyn chun ton phn: - Chquantmnthpbinmhmcgitr bng1.Slnhmbng1cngchnhlstch ca cc t hp bin (hay cn gi l hi c bn). - Trongmihicbn,ccbincgitrbng1 c gi nguyn, cn cc bin c gi tr bng 0 th clygitro:nghalx=1thtrongbiu thc hi c bn s c vit l x v ngc li. - Hm tuyn chun ton phn s l tng cc hi c bn(Tonphnvtrongcchicbnsc mt ca tt c cc bin vo). CHNG 2. TNG HP MCH N 5/20/201234 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l 2.1.2.Biudin mchnbng hmtuynchun tonphnvhm hichunton phnV d 5: Cho mch n c biu din di dng bng chn l. Hy xc nh hm tuyn chun ton phn? ABCQ 0001 0010 0100 0111 1001 1010 1101 1110 c ab c b a bc a c b a c b a f + + + = ) , , (CHNG 2. TNG HP MCH N 5/20/201235 2.1. Biu din mch n 2.1.1. Biu din bng bng chn l 2.1.2.Biudin mchnbng hmtuynchun tonphnvhm hichunton phn ,,,*) Cch vit hm di dng Hi chun ton phn- Ch quan tm n t hp bin m hm c gi tr bng0.Slnhmbng0sltngcacct hp bin (hay cn gi l Tuyn c bn). - Trong mi tuyn c bn, cc bin c gi tr bng 0thcginguyn,cnccbincgitr bng 1 c ly o.-Hmhichuntonphnsltchcacc Tuyn c bn p dng cho VD5: Xc nh hm hi chun ton phn? ) ).( ).( ).( ( ) , , ( c b a c b a c b a c b a c b a f + + + + + + + + =CHNG 2. TNG HP MCH N 5/20/201236 2.1. Biu din mch n ,,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng phng php gii tch *) Vic rt gn hm thng p dng mt s nh l sauV d 6: Ti gin hm t hp sau: Bi gii C B A D C B A BD A C A Y + + = .) () ( ) () (B C D A B C YC B C D A C B A B C A YC B A D C B A D C A B C A CA A YC B A D C B A D B A C A Y+ + =+ + + =+ + + + =+ + + + =CHNG 2. TNG HP MCH N 5/20/201237 2.1. Biu din mch n ,,,2.2. Tng hp mch n 2.2.1.Tiginhm thpbngphng php gii tch V d 7: Rt gn biu thc sau Bi gii C B YC B A B A YC C B A B A Y=+ + =+ + + =) )( () )( (A 1ABA +BA+B+CC1Cy = (A+B)(A+B+C)C CHNG 2. TNG HP MCH N 5/20/201238 2.1. Biu din mch n ,,,2.2. Tng hp mch n Vd8:Mtcngtycntuynnhnvinphi tho mn mt trong cc iu kin sau: 1. Di 30 tui, trnh vn ho i hc tr ln, sc kho tt. 2.Trn30tui,chacnttnghipihc,sc kho tt. 3. Di 30 tui, sc kho tt, bit mt ngoi ng 4. Tt nghip i hc, bit mt ngoi ng 5. C sc kho tt 2.2.1.Tiginhm thpbngphng php gii tch CHNG 2. TNG HP MCH N 5/20/201239 2.1. Biu din mch n ,,,2.2. Tng hp mch n Vd9:mtngcbmnchotng c cn phi tha mn mt trong s cc iu kin nh sau: 1/ C in, n cng tc Start, Rle nhit khng b tc ng. 2/ C in, in p khng vt qu 220V 3/ n cng tc Start, Rle nhit khng b tc ng, in p khng vt qu 220V. 4/ in p khng vt qu 220V. 2.2.1.Tiginhm thpbngphng php gii tch CHNG 2. TNG HP MCH N 5/20/201240 2.1. Biu din mch n ,,,2.2. Tng hp mch n 2.2.2.Phngphptithiuhohmlgctheo thut ton a.Tnghpmchnbngphngphpdng bng Karnaugh *) Quy lut gp (dn) cc - Cc trong mt vng gp nhn cng mt gi tr. - Strongmtvnggpphil2k(vik= 1,2,3,... cng ln cng tt v k chnh l bin i tr trong vng s mt i).- Vng gp ny phi khc vng gp kia t nht mt . 2.2.1.Tiginhm t hp bng phng php gii tch CHNG 2. TNG HP MCH N 5/20/201241 2.1. Biu din mch n ,,,2.2. Tng hp mch n 2.2.2.Phngphp tithiuhohm lgctheothut ton *) Cch thc hin ti gin Mun dng ti gin l dng tng ca cc tch: - Lp vng lin kt cha 2k lin k nhau c cng gi tr lgc 1. -VitbiuthcLgicchomivnglinktva thnh lp, biu thc l tch ca ch cc bin vo c gitrkhngthayitrongvng,ccbintrong biuthccthlchnhn,nugitrcabin vngbng1,hocphidngphnh,nubin c gi tr 0 - Cui cng hm ti gin l tng cc cc biu thc Lgc vng . 2.2.1.Tiginhm thpbngphng php gii tch CHNG 2. TNG HP MCH N 5/20/201242 2.1. Biu din mch n ,,2.2. Tng hp mch n V d 10: Hy dng bng Karnaugh tng hp mch sau: abc bc a c ab c b a c b a c b a f + + + + = ) , , (2.2.2.Phngphp tithiuhohm lgctheothut ton 2.2.1.Tiginhm t hp bng phng php gii tch 1 c=10 011 101 b=1 a=1 A B Bi gii CHNG 2. TNG HP MCH N 5/20/201243 2.1. Biu din mch n ,,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2.Phngphp tithiuhohm lgctheothut ton a. Tng hp mch n bng phng php dng bng Karnaugh *) Ch : PP ny dng rt thun tin ti gin ho cc hmLgic c s bin t 5 tr xung Bng gn t hp bin vo cho cc ca bng Karnaugh

DE ABC 000010110100101111011001 00 01 11 10 AB CD00011110 00 01 11 10 Hm 4 bin Hm 5 bin CHNG 2. TNG HP MCH N 5/20/201244 2.1. Biu din mch n ,,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2.Phng phptithiuho hmlgctheo thut ton b)Tithiuhahmlogicbngphngphp Quine Cluskey *Mt s nh ngha: - Hiscpcbn(HSCCB)ltchccbinu vo v o bin n.- Nguyn t ct yu (NTCY) l tch c s bin l t nht hm c gi tr bng 1 hoc c gi tr khng xc nh. - Hm tuyn chun thu gn l tuyn cc NTCY - Hm tuyn chun ti thiu l tuyn ca cc NTCY mnckhnngbaophhtccHSCCB.Nl hm c di ngn nht v phc tp b nht.+ di D th hin bng s nguyn t ct yu.+ phc tp F l s k hiu ca bin. CHNG 2. TNG HP MCH N 5/20/2012 2.1. Biu din mch n ,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2.Phng phptithiuho hmlgctheo thut ton b)TithiuhahmlogicbngphngphpQuine Cluskey * Cc bc tin hnh: Bc 1: M ha cc hi s cp c bn Quy c: - Bin no nghch o th thay bng 0 - Bin no khng nghch o th thay bng 1 Bc 2: Lp hm tuyn chun thu gn: - Sp xp cc t hp theo m nh phn theo th t cc ch s 1 trong t hp tng dn t 0,1,2,... (Bng B) - Lpbngbiudinccgitrhmbng1vcc gi tr khng xc nh(Bng A) - So snh mi t hp th i vi t hp th i+1, nu 2 t hp ch khc nhau mt ct, th kt hp hai t hp thnh mt t hp mi, ng thi thay ct s khc nhau ca hai t hp c bng mt (-) (Bng C) CHNG 2. TNG HP MCH N 2.1. Biu din mch n ,,2.2. Tng hp mch n 5/20/2012 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2.Phng phptithiuho hmlgctheo thut ton Bc 2: Lp hm tuyn chun thu gn(t): - T bng C chn ra cc t hp ch khc nhau 1 ch sv c cng gch ngang (-) trong 1 ct. em cc t hp ny kt hp vi nhau s c 1 t hp mi Bng D. Cc t hp cui cng ny chnh l cc nguyn t ct yu ca hm cho. Bc 3: Lp hm tuyn chun ti thiu: Bng cch Lp bng - Mi hng s tng ng vi 1 NTCY - MicttngngvimtHSCCB(bquacchicgi tr khng xc nh). - nhdu*hocxvocctrongbngngvicc NTCY bng 1 - Xt tng ct, ct no ch c mt du * th NTCY ng vi n l NTCY quan trng - Hm tuyn chun ti thiu l tuyn cc NTCY quan trngCHNG 2. TNG HP MCH N 2.1. Biu din mch n ,,2.2. Tng hp mch n 5/20/2012 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2.Phng phptithiuho hmlgctheo thut ton V d 1: Tng hp mch lgic sau: abc c ab c b a c b a c b a c b a c b a F + + + + + = ) , , (Bi gii Bc 1: M ha cc hi s cp c bn abc c ab c b a c b a c b a c b a c b a F + + + + + = ) , , (M ha HSCCB 000 001100010 110 111 Bc 2: Lp hm tuyn chun thu gn: Bng ABng BBng CBng D S thp phn S nh phn (abc) S ch s 1 S thp phn S nh phn Lin kt T hp Lin ktT hp 0000000000,100-0,2,4,6- -0 1001110010,20-00,4,2,6- - 0 201020100,4-000,100- 410041002,6-106,711- 6110261104,61-0 7111371116,711- CHNG 2. TNG HP MCH N 5/20/201248 2.1. Biu din mch n ,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2. PP ti thiu ho hm lgc theo thut ton Bc 3: Lp hm tuyn chun ti thiu**** ** ** c abc b ac b ac b ac b aabccb aabab c b a c b a Ftt + + = ) , , (C D=3; F=5 T bng D, ta c biu thc ca hm tuyn chun thu gn l: ab c b a c b a Ftg + + = ) , , (CHNG 2. TNG HP MCH N 5/20/201249 2.1. Biu din mch n ,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2. PP ti thiu ho hm lgc theo thut ton * V d 2: Ti thiu ho hm f(a,b,c,d) vi cc nh (hi c bn) bng 1l: L = 2, 3, 7, 12, 14, 15; v cc nh gi tr hm khng xc nh l N = 6, 13 Bi gii Bc 1 v 2: Lp bng v xc nh hm tuyn chun thu gnnh sau: CHNG 2. TNG HP MCH N 5/20/2012 50 2.1. Biu din mch n ,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2. PP ti thiu ho hm lgc theo thut ton b) Ti thiu ha hm logic bng PP Quine Cluskey Bng ABng BBng CBng D S thp phn S nh phn (abcd) S ch s 1 S thp phn S nh phn (abcd) Lin kt T hp Lin kt T hp 200101200102,3001-2,3,6,70-1- 300112300112,60-102,6,3,70-1- 60110601103,70-116,7,14,15-11- 1211001211006,7011-6,14,7,15-11- 701113701116,14-11012,13,14,1511-- 13110113110112,13110- 14111014111012,1411-0 15111141511117,15-111 13,1511-1 14,15111- ab bc c a d c b a f + + = ) , , , (Hm tuyn chun thu gn l: CHNG 2. TNG HP MCH N 5/20/201251 2.1. Biu din mch n ,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2. PP ti thiu ho hm lgc theo thut ton b) Ti thiu ha hm logic bng PP Quine Cluskey - Bc 3: Lp hm tuyn chun ti thiuTa c bng sau 237121415 (x)(x)x bc xxx ab (x)xx c ad c b a cd b abcd a d c ab d abc abcdNhn vo bng ta c hm tuyn chun ti thiu l: c a ab d c b a Ftt + = ) . . . (CHNG 2. TNG HP MCH N 5/20/201252 2.1. Biu din mch n ,,2.2. Tng hp mch n 2.2.1. Ti gin hm t hp bng pp gii tch 2.2.2. PP ti thiu ho hm lgc theo thut ton *Vd3:Tnghpmchnsautheophng php Quine - Cluskey:y z x z y x xyz z y x z y x f ) ( ) , , ( + + + + =z xy z y x xyz z y x z y x f + + + + = ) ( ) , , (* V d 4: Hy tng hp mch sau bng 3 PP: gii tch, ccn, v Quine Cluskey. yz x z y x z y x z y x xyz z y x f + + + + = ) , , (KIM TRA 5/20/201253 ,,Cu 1: Tng hp mch n sau theo phng php Quine - Cluskey:z xy z y x xyz z y x z y x f + + + + = ) ( ) , , (Cu 2: Hy tng hp mch sau bng 3 PP: gii tch, ccn, v Quine yz x z y x z y x z y x xyz z y x f + + + + = ) , , (CHNG 3. TNG HP MCH KP ,,3.1. Phng php gii tch 3.1.1. Khi nim 5/20/201254 -Cc thit b hot ng theo nguyn tc gin on,gi tr logic khng ch ph thuc vo tng hp bin m cn ph thuc vo min thi gian m nhot ng.(Ngha l,nhngthiimkhcnhauvicngmtthp bin vo, hm logic c th cho nhng gi tr khc nhau)Do , thi gian cng l mt bin tc ng vo h iu khinvthpbinvoXcthcoiltpcctn hiu vo xi v thi gian t: X = { x1, x2, , t}Khi : y = f(x1, x2, , t) c gi hm Boole thi giankhostntachcnkhosttrongkhongthi gian m h khng thay i trng thi (t=1) y = f(x1, x2, , 1)CHNG 3. TNG HP MCH KP ,,3.1. Phng php gii tch 3.1.1. Khi nim -Mchkplmchmtnhiurakhngchph thuctnhiuvomcnphthucvotrngthi trc ca chnh h thng . 5/20/201255 - Nu s khong thi gian gy ra chuyn trng thi ca h l K th t hp bin ca h ti K.2n (n s bin vo khng ph thuc vo thi gian).- Nu gi khong thi gian l t vi (K 1) t 0 th khi : y = f(x1, x2, , t)KL: Nhng hm logic biu din di quan h thi gian v quan h th t c gi mch kp.CHNG 3. TNG HP MCH KP ,,3.1. Phng php gii tch 3.1.1. Khi nim 5/20/201256 - Nu s khong thi gian l k vi 0 ti t (k 1) th hm logic tng qut xt trong ton min thi gian s l: f(x1, x2, , t) = f0. t0+ f1. t1+ + fk-1. tk-1 (3.1) 3.1.2. Cc bc thc hin Biuthc(3.1)trnfilbiuthclogiccahmf(x1,x2, , xn) trong khong thi gian tivi 0 i (k -1)V d 3.1: Tng hp hm logic cho bng sau: x100110011 x201010101 tt0t0t0t0t1t1t1t1 f10100001 CHNG 3. TNG HP MCH KP ,,3.1. Phng php gii tch 3.1.1. Khi nim 5/20/201257 3.1.2.Ccbc thc hin V d 3.1: Tng hp hm logic cho bng sau: x100110011 x201010101 tt0t0t0t0t1t1t1t1 f10100001 T biu thc (3.1) ta c: f = f0. t0 + f1. t1 Trong :f0 = +2 1 2 1. . x x x x f1 = x1.x2

Bi gii Vy hm logic cho trong bng c biu din chung cho ton b min thi gian ang xt l:

f = ( +2 1 2 1. . x x x x ).t0 + x1.x2.t1 CHNG 3. TNG HP MCH KP 5/20/201258 ,,3.2. Phng php bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi -Bngchuyntrngthilbngmtqutrnh chuyn i trng thi, bao gm : Gi s h iu khin c: + n : bin vo (l cc tn hiu iu khin t ngi vn hnh, ca thit b chng trnh hoc cc tn hiu pht ra ca ccthit b cng ngh) +m:binra(ltnhiuktqucaqutrnhiu khin v c ghi ct u ra) + k : s trng thi trong cn c ca h Th bng chuyn trng thi c: (k+1) s hng v (2n + m + 1) s ct3.1. Phng php gii tch CHNG 3. TNG HP MCH KP ,,3.2. Phng php bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 5/20/201259 3.1. Phng php gii tch Trng thi Tn hiu voTn hiu ra x1x2...Y1Y2 k1 k2 ... Ch:-Ccgiaonhaucamtbinvovcc hng trng thi s ghi trng thi ca mch.-Numttrngthictntrngthimch trng vi tn hng th l trng thi n nh (trng thibnvng).Nutrngthikhngtrngvitn hng th l trng thi khng n nh (khng bn vng). CHNG 3. TNG HP MCH KP ,,3.2. Phng php bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 5/20/201260 3.1. Phng php gii tch 3.2.2.Tnghpmchkptheophngphp bng chuyn trng thi Cc bc thc hin: Bc1:Phntchtnhiuvo/ravlpGraph chuyn trng thi ca h Bc2: Thnh lp bng chuyn trng thi (Din t cc yu cu cng ngh thnh k hiu kiu bng) Bc 3: Thnh lp bng trng thi rt gn Bc 4: Xc nh bin trung gian v tm hm lgc Bc5:Tmhmlgccaccbinrakhicmt ca cc bin trung gian Bc 6: Lp s iu khin v s ng lc Bc 7: Thuyt minh h s iu khin cng ngh cho CHNG 3. TNG HP MCH KP ,,3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 5/20/201261 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bng chuyn trng thi V d p dng V d 3.2: Hy thit k h thng iu khin cu trc theo yu cu cng ngh sau bng phng php bng trng thi. AXBCLPTma1aobob1CHNG 3. TNG HP MCH KP ,,3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 62 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bng chuyn trng thi AXBCLPTma1aobob1Bi gii Bc 1: Phn tch tn hiu vo/ra a0 : l tn hiu bo trng thi chuyn ng i xunga1 : L tn hiu bo trng thi chuyn ng i ln b0 : l tn hiu bo trng thi chuyn ng sang phi b1 : L tn hiu bo trng thi chuyn ng sang tri Ch:XvL;PvTlcccpqutrnhingc nhau. Tc l nu c X th khng c L, nu c P th khng c T v ngc li. n gin chng ta c th coi: a = 0 ng vi a0 a = 1 ng vi a1 b = 0 ng vi b0b = 1 ng vi b1 Vy tn hiu vo ch cn l a v b. - C 4 tn hiu vo l : CHNG 3. TNG HP MCH KP 5/20/201263 ,,- C 4 tn hiu ra : X : c cu i xung P : c cu sang phi L : c cu i ln T : c cu sang tri - Graph chuyn trng thi c vit nh sau :Tn hiu voTn hiu ra XLPTab;001000100100; ;000010;0100014 1 2 3AXBCLPTma1aobob13.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bng chuyn trng thi CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bng chuyn trng thi Bc 2: Thnh lp bng chuyn trng thi- S ct = s t hp bin vo+ s bin ra +1 = 22 + 4 + 1 = 9 (ct) - S hng =s trng thi + 1= 4 +1 = 5 (hng) Vy ta c bng chuyn trng thi nh sau Trng th iBin vo Bin raab123413422341X L P T10 1110 0 00 00 00 0 00CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi Bc 3 : Thnh lp bng trng thi rt gn Nguyn tc rt gn : Hai hng tng ng nhau th rt gn li thnh 1 hng. Hai hng c gi l tng ng khi c s trng thi v kt qu u ra nh nhau hoc c th suy ra c nhau.b4431 122aBin voTrng th i+ 43 + 2Vy t bng trng thi trn ta rt gn thnh bng nh sau:CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi Bc4:Xcnhbintrunggianvtmhm lgc ca n- Bin trung gian c nhim v phn bit trng thi ra khi t hp bin l nh nhau.- X bin trung gian da vo cng thc: 2Smin(n-1) Trong :+ Smin l s bin trung gian ti thiu cn;+ n l s hng ca bng trng thi khi rt gn +(n-1) l s trng thi ra cn phn bit- Trong v d trn ta c: n = 3 Smin = 1.Vy s lng bin trung gian cn dng l 1 (k hiu ly).Strngthicnphnbitl(n-1)=2(l trng thi i xung X v trng thi sang phi P). CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi Bc4:Xcnhbintrunggianvtmhm lgc ca n- T bng trng thi rt gn, ta c thmhobintrunggiannh sau: 1y243*Tmhmlgccabintrunggian:lpbng Karnaugh cho bin trung gian:+Ghigitrcabin trunggianvovtrcc trngthitngng(c haitrngthibnvngv khng bn vng). a1by101 00y b a y f + = ) (CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi Bc5:Tmhmlgcca ccbinrakhi cmt ca cc bin trung gian.Ghi gi tr ca bin ra vo vo v tr ca tng trng thitngngvchquantmtitrngthibn vng.- Hm xung (X)y0 00 1ba(X)y b X f = ) (- Hm ln (L)ya1 00 0(L)b f(L) = a (khng c bin y v y l trng thi khng cn phn bit) CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi Bc5:Tmhmlgcca ccbinrakhi cmt ca cc bin trung gian.- Hm sang phi (P)- Hm sang tri (T)ya0 10 0(P)by a P f = ) (ya0 00 1(T)bf(T) = b CHNG 3. TNG HP MCH KP ,,AXBCLPTma1aobob15/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi Bc 6: Lp s iu khin v s ng lc aba yLPTbbaStartDgStopDgyyyX*u, nhc im: - D dng - Di dng, d nhm ln khi tn hiu u vo tng. - Qu trnh hiu chnh phc tp. CHNG 3. TNG HP MCH KP ,,5/20/2012 3.2. PP bng trng thi 3.2.1. Biu din mch kp bng bng chuyn trng thi 3.1. PP gii tch 3.2.2.Tnghp mchkptheoPP bngchuyntrng thi *Vd2:Tnghpmchkpsaubngphng php bng trng thi: mACPTBLXBc 1: Phn tch tn hiu vo ra, lp Graph 0001Tn hiu voTn hiu ra 0100111000 001000 ab 101 3 2XLPT; ; ; ;104CHNG 3. TNG HP MCH KP 5/20/201272 ,,3.1. Phng php bng trng thi 3.1.1. Biu din mch kp bng bng chuyn trng thi 3.1.2. Tng hp mch kp theo phng php bng chuyntrng thi *V d 2 - Gii -B1: Phn tch tn hiuvo ra, lp Graph 00mA101011CPTBLX0001Tn hiu voTn hiu ra 0100111000 001000 ab 101 3 2XLPT; ; ; ;104CHNG 3. TNG HP MCH KP 5/20/201273 ,,3.1. Phng php bng trng thi 3.1.1. Biu din mch kp bng bng chuyn trng thi 3.1.2. Tng hp mch kp theo phng php bng chuyntrng thi *V d 2 - Gii -B2,3: Trng th i1234443312211 00 11 0 0 00 0100 0 00Bin voabBin raX L P TThu gn 2Trng th i3 41 ++ 423 13Bin vo1abCHNG 3. TNG HP MCH KP 5/20/201274 ,,3.1. Phng php bng trng thi 3.1.1. Biu din mch kp bng bng chuyn trng thi 3.1.2. Tng hp mch kp theo phng php bng chuyntrng thi *V d 2 - Gii -B4: M ho bin trung gian 31y42y0110a1000bay b y f + = ) (00mA101011CPTBLXCHNG 3. TNG HP MCH KP 5/20/201275 ,,3.1. Phng php bng trng thi 3.1.1. Biu din mch kp bng bng chuyn trng thi 3.1.2. Tng hp mch kp theo phng php bng chuyntrng thi *V d 2 - Gii -B5: y b T fb L fy a X fa P f====) () () () (00mA101011CPTBLXCHNG 3. TNG HP MCH KP 5/20/201276 ,,3.1. Phng php bng trng thi 3.1.1. Biu din mch kp bng bng chuyn trng thi 3.1.2. Tng hp mch kp theo phng php bng chuyntrng thi *V d 3 Tng hp mch kp sau bng phng php bng trng thi P PLXTmCHNG 3. TNG HP MCH KP 5/20/201277 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng phng php hm tc ng (Dy cc bin c) 3.2.1. Khi nimVi cc hm lgc xy ra tun t, cc bin c trong s mchkp xy ra theo dng thi gian, (tc l theo cc khong thi gianni tip nhau). Do vy dycc s kin c th c m t di dng mt k hiu hm di yf = +A(+X, +Y) +B(Y,+Z) B ( - Z, X, + Y) + C ( -Y+T) + *) ngha ca dy bin c s kin - S xut hin ca tn hiu A lm cho X hoc Y hot ng - B xut hin lm cho Y ngng hot ng *) Ch : -A, C ch c du (+), ltn hiu dng xung, tn hiu ang c mun mt ch cn khng cung cp in -B c c (+), (-), gi l tn hiu dng th c ngha khi ang xut hin mun mt i th ta phi tc ng CHNG 3. TNG HP MCH KP 5/20/201278 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng phng php hm tc ng. (Dy cc bin c) 3.2.1. Khi nim*) Ch 1-Ngi ta thng dng nhng k hiu u bng ch ci A, B, C ... k hiu tn hiu vo -Ngi ta thng dng nhng k hiu cui bng ch ci T, U, Z, X, Y ... k hiu tn hiu ra -Ngi ta thng dng nhng k hiu gia bng ch ci P, Q, R... k hiu tn hiu trung gian *) Ch 2i vi nhng bin vo c du + hoc - ng trc k hiuA, B, ... ch r tn hiu xut hin hay mt i do cc yu tiu khin t ngoi (c th do cng ngh)CHNG 3. TNG HP MCH KP 5/20/201279 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpBc 1: Xc nh tn hiu vo ra ca cng nghBc 2: Xc nh im n, im nc i ca h thng - im n: l im ti ch c mt tn hiu ra - im nc i: L im ti c 2 tn hiu ra tr lnBc 3: Xc nh bin trung gian 2Smin n (n l s trng thi cn phn bit ti im nc i) Xc nh hm lgc ca bin trung gianBc 4: Vit hm tc ng cho bin ra v bin trung gian Xc nh chu k hot ng ca bin ra: Chu k hotng ca bin ra l t khi xut hin n lc mt i. Mi chu k hot ng gm 2 giai on: ng, ct. Hm tc ng c k hiu:F(x) = fng (x) ) (x fcatBc 5: V s h thng iu khinCHNG 3. TNG HP MCH KP 5/20/201280 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpV d p dngTng hp cng ngh sau theo phng php hm tc ngAXBCLPTmCHNG 3. TNG HP MCH KP 5/20/201281 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpV d p dng Bi giiAXBCLPTmBc 1: Phn tch tn hiu vo ra -Cc tn hiu vo l: A, B, C thit b in tng ng l cngtc gi hn hnh trnh. Tn hiu khi ng (m), thit b intng ng l nt n -Cc tn hiu ra l X, L, P, T. Thit b in tng ng l cccng tc t X, L, P, T.Bc 2: Xc nh im nc i, im n- Ti im A: Gp A ln th nht tn hiu ra l X Gp A ln th hai tn hiu ra l P Vy im A l im nc i - Ti im B, C c mt tn hiu ra nn l im n CHNG 3. TNG HP MCH KP 5/20/201282 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpV d p dng Bi giiAXBCLPTmBc 3: Xc nh bin trung gian2Smin 2 Smin 1 Vy cn t nht 1 bin ph (y) phn bit 2 trng thi *) Hm y xc nh nh sau-Gi s khi gp A ln th 2, y xut hin c ngha l (y) xut hin khi bt u gp B v tn hiu mt khi gp C - Hm tc ng ca bin trung gian (y) c xc nh nh sau: F (y) = B+y(y: l bin duy tr) Fc (y) = C C y B y Fc y Fd y f ). ( ) ( ). ( ) ( + = =CHNG 3. TNG HP MCH KP 5/20/201283 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpV d p dng Bi giiAXBCLPTmBc 4: Vit hm tc ng ca cc bin ra khi c bin ph yB x y A X FB X Fcx y A X Fd). ( ) () () ( ) (+ = =+ =Vi bin ra XVi bin ra L ) ).( ( ) () () ( ) (y A l B L FAy L Fcl B L Fd+ + = =+ =Vi bin ra P C p Ay P FC P Fcp Ay P Fd). ( ) () () ( ) (+ = =+ =Vi bin ra T ) ).( ( ) () () ( ) (y A t C T Fy A T Fct C T Fd+ + = =+ =CHNG 3. TNG HP MCH KP 5/20/201284 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpV d p dng Bi giiAXBCLPTmBc 5: Xc nh s iu khin yy tcapyxyayablacDgStopDgStartbtcbplxyCHNG 3. TNG HP MCH KP 5/20/201285 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.2.1. Khi nim3.2.2. Cc bc tng hpV d 2: Tng hp cng ngh sau bng phng php hm tc ngmC BALXT TPCHNG 3. TNG HP MCH KP 5/20/201286 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) Phn bi tp Hy tng hp cc cng ngh sau theo phng php hm tc ng (m l nt nhn khi ng) Hnh cHnh aXLPTLXmmC B ALXT TPF E DHnh dD B FCATPmXL LHnh bXEPXLXLTLXAXBLPTmXLCHCHNG 3. TNG HP MCH KP 5/20/201287 ,,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phng php Grafcet 3.3.1. Khi nim Grafcet l t vit tt ca ting Php: Graphe fonctionnel decommande estape transition c ngha l: hnh chc nng,cho php m t cc trng thi lm vic ca h thng v biudin qu trnh iu khin vi cc trng thi chyn bin t trng thi ny sang trng thi khc, l mt Grafcet nh hng v c xc nh bi cc phn t sauG = { E, T, A, M }Trong : + E = { E1, E2, Em} l tp hp m trng thi (giai on) ca h thng v mi trng thi c k hiu bng mt hnh vung c nh s+ T = { t1, t2, , ti} l tp hp i chuyn trng thi c biudin bng .Gia hai trng thi lun tn ti mtchuyn trng thiCHNG 3. TNG HP MCH KP 5/20/201288 V d ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phng php Grafcet 3.3.1. Khi nim G = { E, T, A, M }+ A = {a1, a2, , an} l tp hp n cung nh hng ni gia trng thi+ M = {m1, m2, , mm} l tp hp cc gi tr (0,1) nu trng thi no c m=1 th trng thi l ang hot ng(k hiu m+) v ngc li (k hiu m-).12XCHNG 3. TNG HP MCH KP 5/20/201289 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet Chng ta c th m t qu trnh vn ng ca mt tmat bng mt biu tp hp bi cc thnh phn hp thnh sau y: *) Trng thi (A) - Trng thi ban u: Vi h dy lun c trng thi ban u + L trng thi h bt u chu trnh hot ng + K hiu nh sau: - Trng thi bt k + K hiu: bng vng trn n hoc hnh vung n + Gn vi trng thi bt k c th c hnh ng ra CHNG 3. TNG HP MCH KP 5/20/201290 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet *) Trng thi (A) *) Hnh ng-Khi nim: Tn hiu hot ng a ra t h thng s lm hot ng c cu chp hnh, n c th l cc cng tc t, nam chm in, cc van in t, ng c chp hnh, tn hiu bo ng, - K hiu: Y:- V d Trng thi 1, c hnh ng l YCHNG 3. TNG HP MCH KP 5/20/201291 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet *) Trng thi (A) *) Hnh ng*) Cc chuyn tip (iu kin) - Khi nim: Cc chuyn tip l cc iu kin lgic, cc iu kin ny c trng bng nt gch ngang trn s 21X- c im: khi chuyn tip tch cc (tc ng) h thng s chuyn i trng thi- ngha:X l iu kin chuyn tip t trng thi 1n trng thi 2 CHNG 3. TNG HP MCH KP 5/20/201292 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet *) Trng thi (A) *) Hnh ng*) Cc chuyn tip (iu kin) *) Cc lin h c hng 12X-Khi nim: L mi tn ch s lin h c hng gia trng thi vi chuyn tip hoc chuyn tip vi trng thi.- th hin s tch cc ca trng thi ngita dng du chm t bn trong trng thi 12X.- ngha: Cho bit c hng vn ng ca h thng mngCHNG 3. TNG HP MCH KP 5/20/201293 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet V d: Cch xy dng cu trc biu Xt bi ton v d: Bi ton cnh tay my, lm nhim v chuyni tng t v tr A n v tr B. *) Yu cu cng ngh: Ti A xut hin i tng, tay myh xung kp vt sau nng ln v c chuyn n B, ti B h xung v nh vt. M hnh hot ng ca h thng nh sau: A Bt c t dt at bm ( n tSt ar t )CHNG 3. TNG HP MCH KP 5/20/201294 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet V d: Cch xy dng cu trc biu *) B tr thit b chuyn ng v cm bin - Tay my nng ln h xung do ng c khng ng b ko - Tay my sang tri sang phi do ng c khng ng b ko - Kp v nh vt th do ng m van thu lc - Hn ch ln xung ca tay my do 2 tip im tA, tB - Hn ch tri phi ca tay my do 2 tip im tC, tD - Nhn bit i tng bng cm bin quang *) M t chu k hot ngL: lnT: triK: kp vt X: xungP: phiN: Nh vt CHNG 3. TNG HP MCH KP 5/20/201295 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet V d: Cch xy dng cu trc biu *) M t chu k hot ng(tA, tC,PDA, M)XtB(2)(1)Kp vtt1LtAPXtDtBLT(3) (4)(5)Nh vtt2t1 t2A Bt c t dt at bm ( n tSt ar t )CHNG 3. TNG HP MCH KP 5/20/201296 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet V d: Cch xy dng cu trc biu *) M t chu k hot ngA Bt c t dt at bm ( n tSt ar t )*) La chn trang b in + CTH: Cng tc t h tay my + CTN: Cng tc t nng tay my + CTP: Cng tc t chy phi + CTT: Cng tc t chy tri + M: Nt nhn khi ng + PDA: Cm bin quang, nhn bit i tng v tr A + VTL: Van thu lc kp v nh vt + tA, tB: Hn ch hnh trnh nng, hn ch hnh trnh h + tC, tD: Hn ch hnh trnh chy tri, hn ch hnh trnh chy phi CHNG 3. TNG HP MCH KP 5/20/201297 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.1. Khi nim 3.3.2. Cc thnh phn cu trc ca mt Grafcet V d: Cch xy dng cu trc biu *) Xy dng mng Grafcet,3.3. Tng hp mch kp theo phngphp Grafcet Hthng ngh012X345678KLPxNLT iu kin utB.tCtA.tCtD.tAtB.tDtA.tDtC.tAa) Grafcet cho phn thao t c268b) Grafcet cho phn iu khintC.tA7tA.tD45tB.tDtD.tA3tA.tCt1 / 2 /LtB.tC1CTH0Hthng ngh(tA, tC,PDA, M)VTLL = CTNCTPCTHLVTLL = t2 / 6 / CTNCTTt1t2t2t2t1t15/20/201298 CHNG 3. TNG HP MCH KP 5/20/201299 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.3. Cc quy tc vn ng ca Grafcet -Chuyn trng thi: H thng chuyn trng thi ny sang trngthi khc phi tho mn ng thi hai yu t: + Trng thi trc ang tch cc + Chuyn tip (iu kin) phitch cc. Thiu mt trong hai trng thi trn, h thng s khng chuyn tip c -Khi xy ra chuyn trng thi th trng thi mi c xc lp v trng thi c b xo b hot ng. Khi trng thi mt i th hnh ng gn vi trng thi cng mt i. a(b+c)=3210V d -Khng chuyn tip c dotrng thi 2 khng tch ccCHNG 3. TNG HP MCH KP 5/20/2012100 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.3. Cc quy tc vn ng ca Grafcet V d a(b+c)=032.Khng chuyn tip c do iu kin a(b+c) = 0 (iu kin khng tch cc).a(b+c)=1.23Chuyn tip c do c trng thi v iu kin u tch cc32.a(b+c)=01Chuyn tip vt qua, iu kin a(b +c) khng cn ngha naCHNG 3. TNG HP MCH KP 5/20/2012101 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.4. Mt vi dng cu trc mch Grafcet thng gp*) Mch phn k OR 213t12 t13Khi trng thi 1 hot ng,nu chuyn tip t12 tho mn thtrng thi 2 hot ng, cn nuchuyn tip t13 tho mn thtrng thi 3 hot ng, ta c:m2+ = t12.m1 m3+ = t13.m1 m1- = m2 + m3 CHNG 3. TNG HP MCH KP 5/20/2012102 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.4. Mt vi dng cu trc mch Grafcet thng gp*) Mch phn k OR *) Mch hi t OR 5 46t46 t56 Nu trng thi 4 hot ng vchuyn tip t46 tho mn thtrng thi 6 hot ng, cng nh vynu trng thi 5 hot ngv t56 tho mn th trng thi 6hot ng. Ta c:m6+ = t46.m4 +t56.m5

m4- = m-5 = m6 CHNG 3. TNG HP MCH KP 5/20/2012103 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.4. Mt vi dng cu trc mch Grafcet thng gp*) Mch phn k OR*) Mch hi t OR *) Mch phn k AND t778 9Nu trng thi 7 hot ng v t7 tho mn th c hai trng thi8 v 9 cng hot ng, ta c: m8+ = m9+= t7.m7

m7- = m8 m9 CHNG 3. TNG HP MCH KP 5/20/2012104 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.4. Mt vi dng cu trc mch Grafcet thng gp*) Mch phn k OR*) Mch hi t OR *) Mch phn k AND *) Mch hi t AND Nu trng thi 10 v 11 cng hotng v iu kin chuyn tiptrng thi t12 tho mn thtrng thi 12 hot ng, ta c:m12+ = t12.m10 .m11

m10- = m11- = m12 CHNG 3. TNG HP MCH KP 5/20/2012105 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.5. Cc bc thc hin- Bc 1: Phn tch tn hiu vo/ra, Lp Grafcet- Bc 2: Dng phn t trung gian c hai u vo l Trig Smicht thit k mch. Mi trng thi s dng 1 trig smicht. Trig Smicht l phn t c hai tn hiu vo l S+ v S- vc mt u ra l S. K hiu nh hnh vTrigo SmichtS-S+S+-u ra S c th nhn mt trong hai mc lgc l 1 v 0 S = 0 khi u vo (-) nhn gi tr 1 S = 1 khi u vo (+) nhn gi tr 1 C ngha l: S+ = 1 S = 1 tng ng vi trng thi Eiang hot ng, nu S- = 1 S = 0 tng ng vi trngthi Ei khng hot ng.CHNG 3. TNG HP MCH KP 5/20/2012106 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.5. Cc bc thc hinV d p dngDng phng php tng hp mch kp bng Grafcet tnghp cng ngh sau: LXPmTS cng nghtay m yABCCHNG 3. TNG HP MCH KP 5/20/2012107 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.5. Cc bc thc hinV d p dngLXPmTS cng nghtay m yABCBc 1: Phn tch tn hiu vo/ra + C 5 tn hiu vo a: Tn hiu ra lnh sang phi b: Tn hiu ra lnh i xung c: Tn hiu ra lnh i ln d: Tn hiu ra lnh sang tri g: Tn hiu bo liu + C 4 tn hiu ra P: Tn hiu tay my i sang phi X: Tn hiu tay my i xung L: Tn hiu tay my i ln T: Tn hiu tay my i sang tri CHNG 3. TNG HP MCH KP 5/20/2012108 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.5. Cc bc thc hinV d p dngLXPmTS cng nghtay m yABCBc 1: Lp Grafcet S1: Trng th i khi um gaS2: Trng th i tay m y i sang phi PbcbS3: Trng th i tay m y i xung XS4: Trng th i tay m y i ln LS5: Trng th i tay m y i sang tr i TT mch Grafcet ta cS+1 = S5.g + m(Trng thi khi u hotng khi c tn hiu khing m hoc l c trng thi tay my i sang tri v c tn hiu liuS-1 = S2

Trng thi khi u ngng hot ng khi tay my sangphi hot ng CHNG 3. TNG HP MCH KP 5/20/2012109 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.5. Cc bc thc hinV d p dngLXPmTS cng nghtay m yABCBc 1: Lp Grafcet S1: Trng th i khi um gaS2: Trng th i tay m y i sang phi PbcbS3: Trng th i tay m y i xung XS4: Trng th i tay m y i ln LS5: Trng th i tay m y i sang tr i TTng t ta c: S+1 = S5.g + m S-1 = S2 S+2 = S1.aS-2 = S3 S+3 = S2.bS-3 = S4 S+4 = S3.cS-4 = S5 S+5 = S4.bS-5 = S1 CHNG 3. TNG HP MCH KP 5/20/2012110 ,3.3. Tng hp mch kp theo phngphp Grafcet 3.4.5. Cc bcthc hinV d p dngBc 2: Dng Trig Smicht biu din mch nh sauS+1 = S5.g + m S-1 = S2 S+2 = S1.aS-2 = S3 S+3 = S2.bS-3 = S4 S+4 = S3.cS-4 = S5 S+5 = S4.bS-5 = S1 S1-+mgS5+S2-a+-S3b cS4-+bS5+-RPD1+12V +12VD2RX+12VD3RL+12VD4RTRPRXRLRTPXLTCHNG 3. TNG HP MCH KP 5/20/2012111 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.5. Cc bc thc hinNhn xt-u im: Khng cn phi xc nh bin trung gian,mch nh gn -Nhc im: Do mch in tr nn tnh n nh khng cao,trong thc t t c s dngCHNG 3. TNG HP MCH KP 5/20/2012112 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.3.6. Tng hp mch kp bng phng php kt hp Hm tc ng v Grafceta. Cc bc thc hin Bc 1: Xc nh mng Grafcet Bc 2: Xc nh cc trng thi ging nhau v thu gn mngGrafcet Bc 3: Xy dng cc trng thi Si: Khi no xut hin,khi no mt i Bc 4: Tm ra cc hm tc ng f(Si) tng ng theo cngthc sau) ( ) ( ) (i c i d iS f S f S f =Bc 5: V s iu khinCHNG 3. TNG HP MCH KP 5/20/2012113 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.6. Tng hp mch kp bng phng php kt hp Hm tc ng v Grafcetb. V d p dngPT TXLBmA CDCHNG 3. TNG HP MCH KP 5/20/2012114 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.6. Tng hp mch kp bng phng php kt hp Hm tc ng v Grafcetb. V d p dngBc 1: Xc nh mngGrafcet PT TXLBmA CDAS2: Trng th i tay m y i sang phi PCS3: Trng th i tay m y i sang tr i TBS4: Trng th i tay m y i xung XDS5: Trng th i tay m y i ln LmS1: Trng th i khi uBS6: Trng th i tay m y i sang tr i TABc 2: Xc nh cctrng thi ging nhau v thu gn mngGrafcetDo v tr 3 v 6ging nhau nn tac mng thu gnnh sau: CHNG 3. TNG HP MCH KP 5/20/2012115 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.6. Tng hp mch kp bng phng php kt hp Hm tc ng v Grafcetb. V d p dngPT TXLBmA CDDo v tr 3 v 6ging nhau nn tac mng thu gn S2: Trng th i tay m y i sang phi PS3: Trng th i tay m y i sang tr i TS4: Trng th i tay m y i xung XACBDS5: Trng th i tay m y i ln LmS1: Trng th i khi uBAMng Grafcet thu gnCHNG 3. TNG HP MCH KP 5/20/2012116 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.6. Tng hp mch kp bng phng php kt hp Hm tc ng v Grafcetb. V d p dngPT TXLBmA CD*) Bc 3 v bc 4: Lp hmS2: Trng th i tay m y i sang phi PS3: Trng th i tay m y i sang tr i TS4: Trng th i tay m y i xung XACBDS5: Trng th i tay m y i ln LmS1: Trng th i khi uBAMng Grafcet thu gn2 1 3) . ( ) 1 ( S S A S m S f + + =3 2 1). . ( ) 2 ( S S A S S f + =1 4 3 5 2. ). . . ( ) 3 ( S S S B S C S S f + + =5 4 3). . ( ) 4 ( S S B S S f + =3 5 4). . ( ) 5 ( S S D S S f + =S+1 = m + S3.A S-1 = S2 S+2 =S1.A S-2 = S3 S+3 = S2.C + S5.BS-3 = S4 + S1 S+4 = S3.B S-4 = S5 S+5=S4.D S-5 = S3 Vy f(P) = f(S2); f(T) = f(S3); f(X) = f(S4); f(L) = f(S5) CHNG 3. TNG HP MCH KP 5/20/2012 ,3.1. Phng php bng trng thi 3.2. Tng hp mch kp bng PP hm tc ng. (Dy cc bin c) 3.3. Tng hp mch kp theo phngphp Grafcet 3.4.6. Tng hpmch kp bngPP kt hp Hm tc ng v Grafcetb. V d p dng*) Bc 5: V s iu khinS3mStopS3S5S2S3S4S1AS1S2S2AS1CS2S3S5BS4 S1BS3S4S5S5S4DS3S2PTS3XS4LS5CHNG 4. MT S S IU KHIN NG C C BN 5/20/2012118 ,4.1. S iu khinng c khng ngb 3 pha 4.1.1. S m my trc tip ng c khng ng b 3 pha apkr nk bdmkkr n3 4 5 1 2220V / 24V c ca b cCHNG 4. MT S S IU KHIN NG C C BN 5/20/2012119 ,4.1. S iu khinng c khng ngb 3 pha 4.1.1. S m my trc tip ng c khng ng b 3 pha * Gii thch cc k hiu:+ AP: p t mt 3 pha dng bo v ngn mch cho ng c + K : Khi ng t, dng ng ct mch in cho ng c + RN: L rle nhit, dng bo v qu ti cho ng c + D, M: L nt n t phc hi, ngha l khi thi khng nhn ntna th cc nt n ny tr v trng thi ban u *) Nguyn l hot ng ca s -Chy ng c: Sau khi ng AP, ta nhn nt n M, cun ht K c in, ng cp tip ng lc cp in cho ng c KB v ng cp tip im thng h 3 - 4 duy tr in cho cun ht K khi thi nhn nt M -Mun dng ng c: n nt D ct in cun ht k, ng c s dng liCHNG 4. MT S S IU KHIN NG C C BN 5/20/2012120 ,4.1. S iu khinng c khng ngb 3 pha 4.1.1. S m my trc tip ng ckhng ng b 3 pha 4.1.2. S m my ng c khng ng b 3 pha bngphng php i ni Y/A t ng k Y22 20 24k b16 18k A9k12 10r n4 614k8ap4 3 dm5k ARthk 7r n6k Ak yk yRthRth1115 13220V / 24V c c0CHNG 4. MT S S IU KHIN NG C C BN 5/20/2012121 ,4.1. S iu khinng c khng ngb 3 pha 4.1.1. S m my trc tip ng ckhng ng b 3 pha 4.1.2. S m my ng c khng ng b 3 pha bngphng php i ni Y/A t ng - Nguyn l hot ng ca s : ng AP, sau nhn nt n M, cp in cho cun ht K, Rthv cun ht KY c in, khi :+ ng c s khi ng vi cch mc hnh Y + Sau khi ht thi gian t, cp tip im thng ng mchm (5 - 9) m ra ngt in cho cun ht KY, ng thi cptip im thng m ng chm (3 - 11) ng li cp in cho cun ht ca KA, cp tip im ng lc ca KA ng li, ng c chuyn sang ch lm vic vi cch u hnh A. -Tt ng c: Nhn nt n D, cun ht khi ng t K mt in v ngt in li cp cho ng cCHNG 4. MT S S IU KHIN NG C C BN 5/20/2012122 ,4.1. S iu khinng c khng ngb 3 pha 4.1.1. S m my trc tip ng ckhng ng b 3 pha 4.1.2. S m my ng c khng ng b 3 pha bng PP i ni Y/A t ng 4.1.3 S iu khin ng c khng ng b 3 pha o chiu k bapr nk tdk tmtr nk tk nk nmnk nk tk n1 3 2 5 7 91113c c220V / 24V CHNG 4. MT S S IU KHIN NG C C BN 5/20/2012123 ,-Khi mun ng c chy thun: ng AP, sau nhn nt MT xy ra qu trnh in t ng nh sau: 4.1. S iu khinng c khng ngb 3 pha 4.1.1. S m my trc tip ng ckhng ng b 3 pha 4.1.2. S m my ng c khng ng b 3 pha bng PP i ni Y/A t ng 4.1.3 S iu khin ng c khng ng b 3 pha o chiu *) Nguyn l hot ng ca s + Cun ht Khi KT c in, v ng tip im ng lc thng h KT li v ng c chy theo chiu thun + Tip im thng h KT 3 - 5 ng li duy tr in cho cun ht KT lun c in khi thi nhn nt MT + Tip im thng ng KT 11 - 13 m ra ngt in cun ht KN m bo trong qu trnh ng c ang chy thun thkhng th chy ngc c.+ Mun dng ng c, hoc cho ng c chy ngc ta nhn ntD, ng c s dng li ngay. -Khi mun ng c chy ngc, n nt MN, qu trnh in xy ra tng t, v ng c s chy ngcCHNG 4. MT S S IU KHIN NG C C BN 5/20/2012124 ,4.1. S iu khinng c khng ngb 3 pha 4.2. S iu khin ng c mt chiu kch t c lp *) S m my ng c in mt chiu kch t c lpqua 3 cp in tr ph theo nguyn tc thi gian m c h ngl cc cc d10 11c ckm c h iuk hins mm y n gc 1 c h iuq ua 3 c p in t r phs ? d?ng ro le th?i gian16 15c ckc dk 3r 3k 2r 2k 1r 112 13c ck12dm14Rtgc ck t 8k 3r t g 3r t g 3k 2r t g 2r t g 17 r t g 2c ck 15k6r t g 143 r tgCHNG 4. MT S S IU KHIN NG C C BN 5/20/2012125 ,4.1. S iu khinng c khng ngb 3 pha 4.2. S iu khin ng c mt chiu kch t c lp *) S o chiu quay ng c c cm c h ngl c8c c2k n13mnk n15k tmtm c h iu k hin17k n12k tr tg9 11 k t220V / 24V 10c c3 d 1 5mtk tk n7 mnc cc c6k t 4Rtg14k t10k nc c8CHNG 4. MT S S IU KHIN NG C C BN 5/20/2012126 ,4.1. S iu khinng c khng ngb 3 pha 4.2. S iu khinng c mt chiukch t c lp Phn bi tp Bi 1: Hy thit k s m my ng c khng ng b bng phng php ct dn 3 cp in tr ph mch rtotheo nguyn tc thi gian Bi 2: Hy thit k s hn ch dng khi ng ng cmt chiu qua 3 cp in tr ph Bi 3: Thit k s hm ng nng ng c mt chiu kcht c lp Bi 4: Thit k s hm ng nng ng c khng ng b 3 pha. Bi 5: Thit k s o chiu ng c mt chiu kch tc lp