vi mach phan 2

Bi ging Vi mch s Bin son Ng Vn Bnh

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Phn 2: H logic tun t

2.1. Khi nim H logic tun t l h logic c u ra khng ch ph thuc vo trng thi hin ti ca

u vo m cn ph thuc vo cc trng thi lch s ca u vo, n gin cc trng thi ny c th hin bng trng thi trong ca h.

H tun t s thc hin hm chuyn cc trng thi vo v trng thi trong hin ti thnh cc trng thi trong v trng thi u ra mi, sau mt thi gian tr trng thi trong mi ny tr thnh trng thi trong hin ti v li lp li qu trnh tnh trng thi trong v trng thi u ra mi.

Nu s thay i trng thi ch xy ra khi c mt tn hiu tham kho gi l xung nhp (clock) th h c gi l h ng b v nhng h c trng thi thay i khng cn xung nhp c gi l h khng ng b. S khi ca c 2 loi ny c v hnh sau:

Trong hnh v cho thy mch t hp vo dng tnh trng thi trong mi t trng thi vo hin ti v trng thi trong hin ti. Mch nh trng thi trong cho php lu tr trng thi trong ca h, mch ny c th c xung nhp hoc khng. Mch t hp ra l mch tnh hm ra t trng thi trong v trng thi vo hin ti, nu trng thi ra ca h ch c tnh t trng thi trong th h thuc loi Moore v ty thuc c vo trng thi vo th l h loi Mealy. 2.2 Cc phng php biu din h tun t 2.2.1 M t chc nng


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Mt h logic bt k c th c xem l mt hp en c mt cng vo v mt cng ra. Quy lut bin i ca tn hiu vo v ra c m t bng mt s mnh c th. V d m t hot ng ca mch iu khin n giao l ng t. Mch gm c 3 u ra iu khin 3 n: (), xanh (X) v vng (V) v 2 u vo: Mt ng vo thi gian v mt nt nhn chuyn trng thi. Nguyn l lm vic c m t nh sau: - Nt nhn s khng c tc dng khi n xanh ang sng (X = 1) - Nu nhn nt trong khi n vng hoc ang sng (V = 1) hoc = 1) th sau mt thi gian T1 s chuyn sang xanh. - Nu n xanh ang sng th h thng s t ng chuyn sang vng sau khong thi gian T2 v sau sang sau khong thi gian T3. - Tng t, nu n ang sng th h thng s t ng chuyn sang vng sau khong thi gian T2 v sang xanh sau khong thi gian T3. 2.2.2 M t bng hm

Theo s khi, h logic tun t c th c m t bi mt b gm - Tp trng thi trong ca h k hiu l S = {Si} nu s lng trng thi l hu

hn. - Tp cc dy vo gy ra s thay i trng thi ca h k hiu l X = {Xi}. - Tp cc trng thi ra Y = {Yj}. - Hm ra theo h Moore Y = Fy (S) hoc hm ra theo Mealy Y = Fy (S, X). - Hm vo cn gi l hm chuyn S = Fs (S, X).

2.2.3 Bng th thi gian L mt h th biu din tng tn hiu vo, ra v xung nhp vi cng gc thi

gian. V d th thi gian sau y l ca mt b m ng b mod 2

2.2.4 Bng gin trng thi

Gin trng thi cho thy s chuyn t trng thi ny sang trng thi khc ca h. Gin ny gm cc nt v cc nhnh c hng, mi nt biu din mt trng thi, ni gia hai nt l mt nhnh c hng i din cho hng chuyn trng thi ca h, nt ti gc ca nhnh l trng thi trong hin ti, nt ti ngn ca nhnh l trng thi trong mi,


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tn trng thi trong c ghi ti nt tng ng, trn mi nhnh ghi tn hiu vo gy ra s chuyn trng thi ca h.

Trong h Moore tn hiu ra ch ph thuc vo trng thi trong nn c th ghi chui ra tng ng vi tng trng thi trong trn nt i din cho trng thi trong. Vi h Mealy tn hiu ra ph thuc c vo trng thi trong v tn hiu vo hin ti nn tn hiu ra c ghi bn cnh dy vo gy ra s chuyn trng thi tng ng.

V d xt my bn hng t ng c yu cu nh sau: - My cho php mi ln b vo mt ng 5 xu hoc 2 ng 5 xu. Nu s tin b

vo bng hoc ln hn 15 xu th my s m ca a hng ra:

- Trong s khi k hiu N l tn hiu b tng ng 5 xu, D l tn hiu b mi ln 2 ng 5 xu v tn hiu reset h thng v trng thi ban u. My s pht tn hiu m ca khi: - 3 N (N, N, N) - 2 N v 1 D (N, N, D) - 1 N v 1 D (N, D) - 1 D v 1 N (D, N) - 2 D (D, D) H ny c th c biu din bng gin Moore hoc Mealy


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2.2.5 Bng bng chuyn trng thi

L phng php thng c x dng khi thit k h tun t. Bng chuyn trng thi l mt bng m t s chuyn sang trng thi mi v to hm ra mi ca mt h c N trng thi trong khi xut pht t mt trng thi trong v mt chui vo xc nh.

Cu to bng chuyn trng thi ca h Moore v Mealy khng ging nhau: Vi h c N trng thi trong th bng chuyn trng thi c N + 1 hng, ct u ca mi hng tnh t hng th hai l tn cc trng thi trong.

S t hp vo khc nhau ph thuc vo di ca chui vo gy ra s chuyn trng thi trong. Gi s chui vo c n k hiu s t hp bin vo l M = 2n.

Vi h Moore bng chuyn trng thi s c M+2 ct, ct u l tn trng thi trong, t ct th hai n ct M+1 th hng u l t hp gi tr ca dy vo, cc hng khc ti mi v tr l trng thi trong mi c chuyn t trng thi trong ct u cng hng vi n di tc ng ca t hp vo hng u cng ct. Ct cui cng l gi tr ca hm ra ng vi cc trng thi trong ct u ca bng

V d vi h Moore c gin trng thi v bng chuyn trng thi tng ng hnh sau:


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hnh 2.5 Bng chuyn trng thi h Moore

Vi h Mealy s ct ca bng l 2M+1 trong M+1 ct u ging nh h Moore, M ct cui cng ti mi v tr s l gi tr ca hm ra ng vi chui tn hiu vo ghi hng u cng ct kt hp vi trng thi trong ghi ct u cng hng vi v tr ang xt.

Hnh 2.6 Bng chuyn trng thi h Mealy

2.2.6 Bng lu thut ton - ASM ( algorith state machine) Gm cc khi, mi khi bao gm mt nh trng thi, cc nh iu kin v cc nh ra, hnh 2.7 trnh by mt khi in hnh, khi ny i din cho mt trng thi trong v ng chuyn n cc trng thi trong khc t n cng nh ng chuyn t trng thi khc n n.


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Hnh 2.7 Mt khi trong lu thut ton

nh trng thi: c xem nh nh php ton to ra cc tn hiu ra khi h ang trng thi trong i din bi khi ang xt. Pha trn bn tri nh ny l tn trng thi trong v pha trn bn phi l m dng m ha tn ny, bn trong nh l cc tn hiu ra c to ra khi ang trng thi ny. V h c th lm vic vi logic m hoc dng nn phi thm vo phn u ca tn hiu ra ch L hoc H cho bit tn hiu ra mc thp hoc cao. Tn hiu ra c th c to ra ngay lp tc hoc sau mt thi gian tr cho n khi c xung nhp mi, hai trng hp ny c phn bit bng cch thm tip u ng I vo tn tn hiu ra. nh iu kin: Thc hin vic th mt u vo quyt nh vic r nhnh t khi hin ti sang cc khi khc cc khi iu kin c th ni tip nhau nu cn th nhiu u vo v d hnh 2.8a trng thi A ch chuyn sang B khi hai iu kin I0 v I1 u bng 1 cn mi trng hp khc s chuyn n C. nh ra: L nh to tn hiu ra khi h trng thi trong hin ti v phi tha mn iu kin th u vo xc nh, nh ra phi sau nh iu kin v cha danh sch cc tn hiu ra.


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B kim tra chn - l

Hnh 2.8 V d v lu thut ton Hnh 2.8b l l thut ton mt b kim tra chn - l gm mt u vo X, hai trng thi Even v Odd c m ha l 0 v 1 v u ra Z s bng 1 khi h trng thi Odd. 2.3 Lu m hnh Moore v Mealy


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Xt gin trng thi hnh 2.9 Hai h ny s cho cng mt dy ra nu xut pht t S0 v tc ng cng mt dy vo.

Hnh 2.9 Gin trng thi Moore v Mealy

Vi h Mealy do gi tr ra c tnh t trng thi trong kt hp vi gi tr vo hin

ti nn thng c t trng thi hn. V d xt trng hp tc ng u vo l 2 s 1 lin tip th Moore phi cn 2 trng thi mi c th phn bit c v sau o cho kt qu ra l 1 cn Mealy ch cn mt trng thi v thm mt tn hiu vo 1 na mi cho ra 1. Vic gim s trng thi lm cho h c cu trc n gin hn nn h Mealy thng c x dng mc d vic tnh hm ra phc tp hn h Moore. i vi h Mealy cn phi quan tm n tnh phc tp thi gian cho cc tn hiu ca n v thi gian to hm ra lm tr thi gian to trng thi trong mi. V khi c tn hiu vo l trng thi mi c to ra ngay nn c khi gi tr ra cha c tnh xong t trng thi trong hin ti th trng thi trong thay i v pht sinh nhiu trong tn hiu ra. 2.4 Chuyn gia 2 m hnh Moore v Mealy Moore v Mealy l 2 m hnh ton hc ca cng mt h tun t, do lun tn ti mt thut ton chuyn gia hai m hnh ny vi nhau. Chuyn t Mealy sang Moore Qu trnh chuyn i gm cc bc sau: 2.4.1. ng vi mi cp :trng thi trong - tn hiu ra ca Mealy ta thay bng mt trng thi trong tng ng Q ca Moore. 2.4.2 Thnh lp bng chuyn trng thi Moore km theo tn hiu ra tng ng vi mi trng thi Q . V d: Chuyn t m hnh Mealy sau y sang Moore


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Hnh 2.10 a. Gin Mealz

b. Bng trng thi c. Bng trng thi rt gn

a. Xc nh trng thi Moore S0/0 = Q0 S1/0 = Q1 S1/1 = Q2 S2/0 = Q3 b. Thnh lp bng chuyn trng thi Trc tin, xc nh tn hiu ra tng ng vi cc trng thi trong, chnh l tn hiu ra trong cp trng thi / tn hiu ra ca Mealy

Hnh 2.11 Bng tn hiu ra


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Hnh 2.12 Bng trng thi Moore

Chuyn t Moore sang Mealy

Qu trnh chuyn t Moore sang Mealy n gin hn v ch cn ghi thm bn cnh mi trong bng trng thi cc tn hiu ra tng ng, sau tin hnh ti thiu ha trng thi ca h 2.5 Thit k h logic tun t Qu trnh thit k thng bao gm cc bc nh saui 2.5.1 M t yu cu thit k Trong phn ny nhim v thit k c m t bng ngn ng hoc bng lu thut ton, ni chung l cha c hnh thc ha. 2.5.2 Hnh thc ha Chuyn cc yu cu trn thnh mt hnh thc m t hot ng ca h thnh bng trng thi, gin trng thi. Rt gn cc trng thi ca mch c c s trng thi t nht, mt im cn lu l lc ny h cha c m ha nh phn. 2.5.3 M ha nh phn M ha cc tn hiu vo, cc trng thi trong v tn hiu ra. 2.5.4 Xc nh h phng trnh ca mch Xc nh h phng trnh v ti thiu ha cc phng trnh ny. Nu h thit k c dng F-F th ty theo loi F-F xc nh phng trnh kch tng ng. 2.5.5 V s mch T h phng trnh trn v s mch thc hin. Thit k h tun t t gin trng thi Cc bc thit k 1) M ha tn hiu vo, trng thi trong v tn hiu ra to cc tp tn hiu vo X, tp trng thi trong Q v tp tn hiu ra Y. 2) Xc nh h phng trnh tn hiu ra Yi = Fi (X, Q), i vi h Mealz phng trnh ny c xc nh trn cc cung v i vi Moore c xc nh trn cc nh, sau tin hnh ti thiu ha. 3) Xc nh h phng trnh kch cc F-F v ti thiu ha cc phng trnh ny. i vi mt FF Qi bt k, s thay i trng thi t Qi n Qi+ ch c th c 4 kh nng nh sau:


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Hnh 2.13 K hiu cc cung biu din s chuyn trng thi t Qi n Qi+ V d 1: Thit k mch m ng b mod 5 gin trng thi v m ha trng thi cho bn di y, thc hin mch bng: a. D - FF b. T - FF c. JK - FF d. RS - FF

Hnh 2.14 B m mod 5 a) Gin trng thi

b) Bng m ha Mch c 5 trng thi nn c m ha bng 3 bin nh phn tng ng vi 3 FF

nh trong bng m ha hnh 2.14, bc tip theo l xc nh phng trnh kch cho cc FF trong trng hp:

1. Dng D - FF nh du cc nh c : Q1 = 1: nh (4) Q2 = 1: nh (2); (3) Q3 = 1: nh (1); (3)


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S o mach logic


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S mch b m ng b mulo 5 dng D - FF


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4. Dng RS - FF

Phng trnh u vo S S = Ton + Cc cung loi 1 Phng trnhy u vo R R = Toff + Cc cung loi 0 Suy ra: S1 = Ton1 + [Cc cung loi 1] = (3) + [0] R1 = Toff1 + [Cc cung loi 0] = (4) + [ (0), (1), (2), (3) ] S2 = Ton2 + [Cc cung loi 1] = (1) + [2] R2 = Toff2 + [ Cc cung loi 0] = (3) + [(0), (4)] S3 = Ton3 + [Cc cung loi 1] = (0) + (2) + [0] R3 = Toff2 + [ Cc cung loi 0] = (1) + (3) + [(4)]


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V d 2: Thit k h tun t ng b c ng vo X, ng ra Y v gin trng thi km theo vi cc trng thi oc m ha nh sau: S0 = 00; S1 = 01; S2 = 10, m 11 khng dng

a. Dng D - FF b. Dng JK - FF


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c. Dng RS - FF Bi gii a/ Dng D - FF

b/ Dng JK - FF

c/ Dng RS - FF


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2.6 Ti thiu ha trng thi L gim s kng trng thi trong ca h n mc t nht m vn khng lm thay

i chc nng ca h. Vic ti thiu ha s lm gim trng thi trong dn n gim phc tp, gim gi

thnh v tng tin cy ca h. C s ca phng php da trn khi nim trng thi tng ng. Hai trng thi

c gi l tng ng khi vi cng mt t hp bin vo u c tn hiu ra v cc trng thi chuyn n ging nhau, c hai phng php ti thiu ha l phng php kim tra hng v phng php bng ko theo.

2.6.1 Phng php kim tra hng (Row matching methode) Phng php ny cho php thc hin th cng trn giy, nhng ch thch hp vi cc

h c t trng thi, cc bc tin hnh nh sau: T bng chuyn trng thi ca h nhm cc trng thi c cng gi tr tn hiu ra vi

nhau v cc trng thi chuyn n ging nhau thnh mt trng thi chung, sau thnh lp bng chuyn trng thi mi.

Lp li cng vic trn cho n khi khng cn c th nhm c na, h c ti thiu ha ln nhm cui cng.

V d mt h nhn dng m c mt u vo X v mt u ra Y, tn hiu ra Y ca h ch bng 1 khi xut hin dy tn hiu vo X = 110 hoc X = 010. Bng chuyn trng thi c trnh by nh sau:

Hnh 2.15 Gin trng thi

Hnh 2.16 Bng chuyn trng thi


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T bng trng thi cho thy S3 v S5 l 2 trng thi tng ng v S4 v S6 cng l cp trng thi tng ng v c tn hiu ra v cc trng thi chuyn n ging nhau, gp chng thnh trng thi chung S3 v S4 v v bng trng thi mi

S/X 0 1 0 1 S0 S1 S2 0 0 S1 S3 S4 0 0 S2 S3 S4 0 0 S3 S0 S0 0 0 S4 S0 S0 1 0

Hnh 2.17 Bng trng thi mi

T bng trng thi cho thy S1 v S2 l 2 trng thi tng ng, tip tc gp thnh mt trng thi chung

S/X 0 1 0 1 S0 S1 S1 0 0 S1 S3 S4 0 0 S3 S0 S0 0 0 S4 S0 S0 1 0

Hnh 2.18 Bng trng thi ti gin Trong bng trng thi mi cho thy khng th tm c cc trng thi tng ng

v s trng thi trong bng l ti gin. T nh ngha cp trng thi tng ng nh trnh by trn dn n mt gii hn

ca phng php kim tra hng. l trong trng hp hai trng thi xut pht chuyn i qua li ln nhau vi cng mt t hp bin vo.

Xt mch kim tra tnh l c mt u vo X v mt u ra Y, tn hiu ra bng 1 khi chui tn hiu vo l s l.

S/X 0 1 u ra S0 S0 S1 0 S1 S1 S2 1 S2 S2 S1 0


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Hnh 2.19 Gin v bng trng thi ca h kim tra l

T bng trng thi cho thy nu nhm S0 v S2 li vi nhau th hot ng ca mch cng khng thay i nhng li gim bt c mt trng thi nh hnh di y:

S/X 0 1 u ra S0 S0 S1 0 S1 S1 S2 1

T kt qu ny nh ngha v cp trng thi tng ng c sa li nh sau: Mt cp trng thi c gi l tng ng khi xut pht t chng h nhn c cc

tn hiu ra ging nhau v cng chuyn n cc trng thi ging nhau hoc tng ng hoc chuyn qua li ln nhau di tc ng ca cng mt t hp bin vo.

2.6.2 Phng php bng ko theo (Implication chart methode) Bng ko theo l mt bng cho php tm s tng ng ca tng cp trng thi

trong. Bng ko theo ban u ca h c n trng thi l mt bng c (n-1) hng v (n-1) ct, mi hng ghi tn mt trng thi trong, mi ct cng i din mt trng thi trong, giao im ca hng Si v ct Sj s cha cc cp trng thi chuyn n t cp trng thi


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ban u SiSj ti cng mt t hp bin vo, ly cng v d trn, ta c bng ko theo nh sau:

Hnh 2.21 Bng ko theo th nht

Gch b cc tng ng vi cc cp SiSj c tn hiu ra khng ging nhau (nhng cp ny khng tng ng)

Hnh 2.22 Bng ko theo th hai

Da trn cc cp khng tng ng loi b, kim tra li cc , xa cc no c cha cc cp khng tng ng


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Hnh 2.23 Bng ko theo th ba

Tip tc kim tra cho n khi khng cn no c th xa c na

Hnh 2.24 Bng ko theo cui cng

Da vo bng ko theo cui cng, kim tra tnh tng ng ca cp trng thi trong tng , hai trng thi cng tng ng vi trng thi th ba th tng ng nhau, t kt qu trn cho thy: S3 tng ng S5 nn c th thay bng S3 v S4 tng ng S6 nn thay bng S4. Dn n S1 cng tng ng S2 nn thay bng S1, bng trng thi ti gin nhn c nh sau:

S/X 0 1 0 1 S0 S1 S2 0 0 S1 S3 S4 0 0 S3 S0 S0 0 0 S4 S0 S0 1 0


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Hnh 2.25 Bng trng thi ti gin Bi tp 1/ Dng phng php bng ko theo n gin bng trng thi sau

2/ Thit k h tun t ng b dng JK - FF c 1 ng vo X v 1 ng ra Z v bng trng thi km theo sau y

3/ n gin trng thi ca mt h tun t c 2 ng vo, 1 ng ra v bng chuyn trng thi - ra nh sau


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2.7 M ha trng thi

M ha trng thi l gn mt ga tr nh phn cho mt trng thi trong ca h, gi tr ny c lu tr trong cc FF, khi h chuyn t trng thi ny sang trng thi khc cng tng ng vi vic chuyn cc m t gi tr nh phn ny sang gi tr nh phn khc.

cho mch kch FF c n gin th khi chuyn trng thi s FF thay i cng t cng tt, c bit l trong h tun t khng ng b nu c nhiu FF cng thay i nhng tc khng ging nhau s c th dn n sai trng thi rt nguy him.

t c yu cn ny th m ca cc trng thi k tip nhau phi tht gn nhau, tt nht l ch khc nhau mt bt, nguyn tc ny c gi l nguyn tc m ha k cn.

Vi h c n trng thi trong s c n! cch m ha bng n m nh phn. Vic chn cch m ha tt nht rt kh khn v thc ra cng cha c l thuyt hon chnh cho vn ny, di y l mt vi phng php thng dng

2.7.1 M ha th cng Phng php ny x dng mt bng lit k trng thi c cu trc ging bng

Karnaugh, cc FF c ghi trn 2 cnh, trong mi l trng thi trong c m tng ng trn 2 cnh V d: Mt h tun t c lu thut ton nh sau


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Hnh 2.26 Lu thut ton

V h c 5 trng thi nn phi dng m nh phn di 3 bt v cc gi tr ca m s c lu tr trong 3 FF, sau y l bng m ha trng thi ca h

Hnh 2.27 Bng m ha trng thi

Trc tin, chn s liu 00 m ha S0 v S1 v S2 l cc trng thi k S0 nn cng chn cc k cn l 010 v 100 m ha chng, trng thi S3 k S1 v S2 nn chn 110 l thch hp, ring i vi S4 va k S3 va k S1 nhng trong bng kghng c v tr no p ng cng lc 2 yu cn ny nn ch chn c k mt trong hai trng thi trn l 011.

Phn di y trnh by mt s lut tin v k cn, cc lut ny da trn c s v gi tr cc u vo v ra ca h, quy tc nh sau:

- u tin cao nht l cc trng thi cng chuyn n mt trng thi chung di tc ng ca cng mt t hp bin vo.

- u tin trung bnh l cc trng thi mi xut pht t mt trng thi chung di tc ng ca cc t hp vo khc nhau.

- u tin thp nht l cc trng thi cho cng t hp ra di tc ng ca cng t hp bin vo, hnh sau y minh ha cc lut u tin va nu trn.


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Hnh 2.28 Lut u tin

2.7.2 M ha nng bo m tnh k cn khi m ha c th tng s FF, mt trong cc phng php

nyl phng php m ha nng (one hot encoding). Vi phng php ny mt h c n trng thi s c m ha bng n FF, trong

trng hp ny m nh phn di n bt v ti mt thi im bt k ch c mt FF bng 1. Trong phng php ny mch kch cc FF tr nn phc tp v kh thc hin bng vi

mch ri nhng li thch hp vi cng ngh vi mch s lp trnh. V d mt h tun t c gin trng thi nh sau:

Hnh 2.29 Gin trng thi v bng lit k trng thi

Nguyn tc thnh lp hm chuyn trng thi trong mi


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Hm ra ca h = S0 M1R = S0 + S2 M1L = S1 S mch thc hin


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Hnh 2.30 S mch dng D - FF

2.7.3 Thc hin h tun t dng vi mch mt tch hp cao Nh trnh by cc phn trn, mt h tun t c th c thc hin bng cc FF

v cc vi mch ri chnh l cc cng logic NAND, NOR, EXOR.... Ngy nay, do cng ngh pht trin xut hin rt nhiu vi mch c mt tch hp rt cao, do cn thit phi x dng cc loi ny trong vic thit k mch logic nhm gim kch thc, gi thnh v tng tin cy. Trong phn ny s m t cch x dng ROM, PLA, PAL v b m cng nh cc vi mch s lp trnh khc tng hp h logic tun t. Mt h tun t c th c thc hin bng cc FF lu gi trng thi kt hp vi mt mch ROM hoc PLA thc hin mch t hp vo v ra.

2.8 Thc hin bng ROM

Khi dng ROM thc hin h t hp th a ch ca ROM l cc tn hiu vo v cc trng thi trong t ng ra ca cc FF.

Nu tn hiu vo c n thnh phn v h c m FF biu din cc trng thi trong th s bt a ch ca ROM phi l n + m, gim s lng tn hiu kch FF, trong thc t thng chn FF D.


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Tn hiu ra ca h v tn hiu kch FF l ng ra ca ROM, nu mch c b tn hiu ra th s u ra ca ROM phi l b + m, v cc FF u ra l loi D nn c th dng thanh ghi lm FF.

Hnh 2.31 S thc hin h tun t dng ROM

Mi mt nh ca ROM cha thng tin v cc trng thi trong mi v tn hiu ra

ca h tnh t trng thi trong hin ti v gi tr ca tn hiu vo. VD mt h tun t c bng chuyn trng thi nh sau

Hnh 2.32 Bng chuyn trng thi

Bng trng thi c chuyn thnh bng s tht ca ROM c trnh by hnh sau,

trong trng thi trong hin ti v tn hiu vo l a ch ca ROM cn tn hiu ra v trng thi trong mi l ng ra ca ROM.

T v d trn cho thy thc hin cn phi dng mt ROM c 16 a ch v 4 ng ra. i vi nhng h c nhiu trng thi trong, mt ROM c th khng s chn cn thit, trong trng hp ny c th gii quyt bng cch phn gii h thnh ra nhiu h nh c th tn dng cc vi mch ROM thng dng c sn trn th trng hin nay.


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a ch ROM X Q2 Q1 Q0

u ra ca ROM Z D2 D1 D0

0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

1 1 0 0 1 0 1 X 0 0 1 1 0 1 X X

0 0 1 1 1 0 0 X 0 1 1 1 1 0 X X

0 1 0 0 0 0 0 X 1 0 0 0 1 0 X X

1 1 0 1 1 0 0 X 0 0 0 1 0 0 X X

Hnh 2.33 Bng s tht ca ROM

2.9 Thc hin bng PLA

Vi PLA c th thc hin cc hm logic dng AND - OR bt k c s u vo v ra ph hp vi PLA. Trc tin cn phi ti thiu ha s trng thi v m ha cc trng thi ca h, tip theo l ti thiu ha cc hm ra v hm kch FF v lp trnh PLA

u vo ca PLA l cc tn hiu vo v cc ng ra ca FF, u ra ca PLA l cc tn hiu ra v tn hiu kch FF (dng FF - D tit kim u ra)


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Hnh 2.34 S thc hin dng PLA

Trong nhiu trng hp hm ra hoc hm chuyn trng thi c s bin vo nhiu hn s u vo ca PLA th cn phi x dng mt s u ra ca PLA a tr li cc ng vo v hm ra l cc u ra cn li. 2.10 Thc hin bng PAL

Trong PAL c th c cc FF ghi nh trng thi cc u ra, ng ra ca cc FF ny c a tr v lm u vo ni b bn trong PAL, u ra ca PAL cn c thm mch logic 3 trng thi, vi cc PAL loi ny c th x dng u ra cc FF a ngc tr li lm u vo ca trng thi hin ti gim s chn u vo ca PAL.

Hnh 2.35 Mt cu trc ca PAL16V8 c FF - D


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Hnh 2.36 Mt cu trc ca PAL20V8 c FF - D


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2.11 Thc hin bng b m

V b m l mt h tun t c th thay i trng thi bi xung vo v nh kh nng np li gi tr ban u cho cc FF nn c th x dng b m thc hin cc h tun t c s trng thi khng ln hn s trng thi ca b m.

Gi s trng thi hin ti ca b m l n, trng thi ny c gi nguyn nu khng c tn hiu vo, nu c tn hiu vo m trng thi s tng ln 1 thnh n+1, nu c tn hiu reset b m s tr v 0, nu c tn hiu np b m s chuyn v trng thi c m cn np.

Hnh 2.36 Gin chuyn trng thi ca b m

tng hp h logic tun t bng b m, trc tin cng phi ti thiu ha v m ha trng thi trong ca h, sau t bng chuyn trng thi va tm c thnh lp bng chuyn trng thi cho b m, cc hm ra ca bng ny l tn hiu cho php m, reset, np v m cn np. V d mt h tun t c bng chuyn trng thi ti thiu ha nh sau:

Trng thi

hin ti Trng thi k tip

X = 0 X = 1 Hm ra

X = 0 X = 1 S0(000) S1 S2 1 0 S1(001) S3 S4 1 0 S2(010) S4 S4 0 1 S3(011) S5 S5 0 1 S4(100) S5 S0 1 0 S5(101) S0 S0 0 1 S6(110) S0 1

Hnh 2.37 Bng chuyn trng thi ca h Trong bng chuyn trng thi ca b m Z l hm ra ca h, CLR l tn hiu xa b m, LD l tn hiu np b m, A, B, C l m cn np, X l tn hiu vo ca h, Q0, Q1, Q2 l trng thi trong hin ti v Q0+, Q1+, Q2+ l trng thi trong mi


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Hnh 2.38 Bng chuyn trng thi ca b m

T bng chuyn trng thi ca b m tin hnh ti thiu ha hm ra v cc tn hiu iu khin b m sau thc hin bng cc mch logic ri, ROM hoc PLA nh trnh by phn trn Khi dng b m thit k h logic tun t th s tn hiu iu khin b m thng nhiu hn s tn hiu kch FF - D, phng php ny ch c li khi qu trnh chuyn trng thi ca h ging nh qu trnh tng gim ca b m v s lm gim tn hiu iu khin do khng cn np trng thi mi. iu ny c th t c do s kho lo trong khi m ha 2.12 H tun t khng ng b L cc h c iu khin bi cc s kin ngu nhin, linh kin thc hin l cc cng v cc FF khng ng b nh RS - FF hoc D - FF, cch thit k h cng ging nh h ng b. Tuy nhin, trong h khng ng b cn phi lu trnh 2 hin tng l hin tng chu k v hin tng chy ua.

2.12.1 Hin tng chu k L hin tng h thay i trng thi lin tc khng dng ti mt t hp bin vo no c ngha l h hon ton mt n nh, hin tng ny d dng nhn thy nu trong bng trng thi c t nht mt ct khng tn ti mt trng thi n nh no c (trng thi c gi l n nh khi ng vi mt t hp bin trng thi mi bng trng thi c)


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V d mt h tun t c bng trng thi sau dy, h c m ha bng 2 bin A v B Hnh 2.39 H khng ng b c hin tng chu k

Gi s u tin h ang trng thi S3 (AB = 10) v X = 0, sau X chuyn ln 1, h s chuyn trng thi theo chu k sau: S3 -> S0 -> S1 -> S2 -> S3 -> S0...

2.12.2 Hin tng chy ua

Hin tng ny xy ra khi trong h c nhiu FF thay i cng lc khi h chuyn t trng thi ny sang trng thi khc v nu tc cc FF khng ging nhau s to thnh cc ng chuyn n trng thi n nh khng ging nhau thm ch s chuyn n trng thi khng mong mun. Do , hin tng ny c chia lm 2 loi: Chy ua khng nguy him v chy ua nguy him. a. Chy ua khng nguy him V d mt mch khng ng b c bng trng thi nh sau

Hin tng chy ua khng nguy him Gi s mch ang trng thi S0 v X = 0, mch s chuyn trng thi theo cc hng sau y: - Nu A v B tc bng nhau: S0 -> S2 -> S3 v n nh ti S3.


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- Nu A nhanh hn B: S0 -> S3 v n nh ti S3. - Nu A chm hn B: S0 -> S1 -> S2 -> S3 v n nh ti S3. Mc d qu trnh chuyn trng thi khc nhau nhng cui cng h cng n nh ti S3, hin tng chy ua ny l khjng nguy him b. Chy ua nguy him V d xem bng chuyn, trng thi ca h sau:

Hnh 2.41 Chy ua nguy him

Gi s h ang trng thi S0 v X chuyn t 0 ln 1, cc hng chuyn trng thi ca h nh sau: - Nu tc A bng B: S0 -> S1 y l trng thi ng. - Nu A chm hn B: S0 -> S2, y l trng thi sai. - Nu A nhanh hn B: S0 -> S3 trng thi S3 l trng thi kha v h khng th thot ra c d c thay i tn hiu vo. Trng hp ny l chy ua nguy him. 2.13 n gin v m ha trng thi i vi h ng b mc ch ca qu trnh n gin v m ha trng thi l gim s trngthi n s l65ng t nht c th c, ring trong h khng ng b cn phi lu n hin tng chu k v chy ua trnh hin tng chu k th phi lu sao cho vi mi t hp bin vo phi lun lun c mt trng thi n nh trnh hin tng chy ua phi m ha sao cho vi mi chuyn bin trng thi ch c mt bin trong t m thay i m thi V d xt h khng ng b c mt ng vo X v gin trng thi c m ho nh sau


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Hnh 2.42 Gin trng thi c chy ua T hnh v cho thy khi h thay i t trng thi S2 sang S0, c 2 bt ca t m u thay i v s xy ra hin tng chy ua, khc phc cn phi a thm vo mt trng thi trung gian sao cho qu trnh chuyn ny ch c 1 bt thay i, kt qu nh sau

Hnh 2.43 Gin c thm trng thi trung gian

T hnh v cho thy khi thm trng thi trung gian t S2 sang S0 phi 1ua S3 v t m ch thay i mt bin v t trng thi trung gian sang trng thi khc khng cn iu kin, trong trng hp h khng cn s trng thi chn lm trung gian th phi tng thm s lng bt trong t m


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2.14 Bi tp 1. V s khi cho bit cch ni ROM v cch D flipflop ci t bng 1.1. xc nh bng chn tr cho ROM dng php gn trng thi nh phn trc tip. Bng 1.1 bng trng thi vi nhiu ng ra v ng vo

Trng thi hin ti Trng thi k Gi tr ra hin ti

S0 S1 S2 S3

X1X2= 00 01 10 11 S3 S2 S1 S0

S0 S1 S2 S3 S3 S0 S1 S1 S2 S2 S1 S0

X1X2= 00 01 10 11 00 10 11 01 10 10 11 11 00 10 11 11 00 00 01 01

2. Bng trng thi 1.2 s c ci t dng PLA v cc D flipflop Bng 1.2 Z ABC X= 0 1 0 1 000 S0 S1 S2 0 0 110 S1 S3 S2 0 0 001 S2 S1 S4 0 0 111 S3 S5 S2 0 0 011 S4 S1 S6 0 0 101 S5 S5 S2 1 0 010 S6 S1 S6 0 1

a) V s khi. b) Xc nh ni dung ca PLA theo dng bng dng gn trng thi bng 1.2. c) Nu thay PLA bng ROM th cn ROM kch thc bao nhiu?

3. Ci t bng trng thi 1.2 dng PAL 16R4. nh du cc X trn bng sau ca s 16R4 ch mu cu ch. 4. Lm tng t 1.2 vi bng trnh thi sau: PS NS Z X= 0 1 X= 0 1 A A E 0 0 B C B 0 1 C A F 0 0 D C B 0 1 E F E 0 0


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F A F 0 0

Suy ra cc chng trnh D dng php gn trng thi A= 000, B= 111, C=110, D= 101, E=100 v F= 010.

5. Thit k 1 h tun t nhn 1 s BCD 8421 vi 3 cho mt s nh phn 5 bit. V d nu vo l 0111 th ra s l 10101. Nhp v xut ca h l ni tip vi LSB i trc. Gi s nhp l 0 thi im clock th 5, v reset h sau bit th 5.

a) Suy ra bng trng thi ca s trng thi ti thiu. b) Thit k h dng JK flipflop v cc cng NAND v NOR. c) Thit k h dng PLA v cc D flipfop. Lp bng PLA. d) Thit k h dng PAL. Cho bit PAL v khun mu cu ch.

6. Ci t bng trnh thi sau dng PAL 16R4

a. Thc hin gn trng thi nh phn v suy ra cc phng trnh trng thi thit k cho D flipflop v cc bin ra.

b. C th ci t cc phng trnh a) bng 1 PAL 16R4 c khng? Nu khng, tm mt gn trng thi mi m c cc phng trnh trng thi k c t s hng hn. Ch ra cc phng trnh cn ci t bng trng thi bng PAL 16R4. Bng 1.6

PS NS

XY= 00 01 10 11 Z1Z2

00 01 10 11

Trng thi hin ti

Trng thi k X = 0 1

Output Z

S0 S1 S3 0 S1 S2 S5 0 S2 S1 S6 1 S3 S1 S4 0 S4 S4 S4 0 S5 S2 S4 0 S6 S1 S4 0


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A B C D E F G H i

A b c d B c g e f i g h d d d a g c a b i f h g b g c a i h i h h i a a

0 0 1 0 1 0 0 1 1 1 0 0 0 1 0 1 1 0 1 1 1 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0

7. Mt bnh xe hi c 3 n ui bn tri v 3 n ui bn pi m nhp nhy theo mu duy nht ch r tri v r phi.

Mu r tri (LEET) Mu r phi (RIGHT) LC LB LA RA RB RC LC LB LA RA RB RC 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Thit k h tun t Mooe iu khin cc n ny. H c 3 ng vo: LEEt, RIGHT

v HAZ. LEET v RIGHT c t cng tc tin hiu r cc ti x v khng th ng thi bng 1. Nh ch trn, khi LEET =1 n nhp nhy theo mu LA sng; LA v LB sng; LB v LC sng; tt c iu tt v ri qu trnh ny lp li. Nu cng tc t LEET sang RIGHT ( hoc ngc li) xy ra gia chui nhp nhy, h tc thi i v trng thi ngh IDLE (tt c cc n tt) v ri bt u chui mi. HAZ c cng tc hazard, v khi HAZ=1, tt c 6 n nhp nhy tt v m ng b. HAZ ly u tin nu LEET hoc RIGHT cng ang ON. Gi s tn hiu clock kh dng bng tc nhp nhy mong mun.

a. V gin trng thi (8 trng thi). b. Ci t h dng 6 D flipflop, v thc hin php gn trng thi sao cho mi

ng ra flipflop li trc tip 1 trong 6 n. c. Ci t h thng 3 D flipflop, dng cch rt gn trng thi. d. Ch n kinh t ca nhiu flipflop hn v nhiu cng hn trong (b) v (c).

ngh 1 PLD thch hp ch mi trng hp. e. Ci t b m ln/xung BCD 4 bt (0,1,.....,9) dng XORPAL thch hp.

B m c cc ng vo iu khin U (=1 m ln), v D (=1 m xung), nhng khng c cc ng vo np. Suy ra cc phng trnh PAL v ch khun mu cu ch PAL.


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f. Ci t b m ln nh phn modulo 11 dng X PAL. Chui m l 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,0 ... Vit cc phng trnh trng thi k vi dng thch hp s dng vi X PAL.

g. Mt thanh ghi dch N- bit tng t vi 74178 s c ci t bng 22V10 Gi tr ti a ca N l bao nhiu? (cc) cu ch no nn cho chy mi

output macrocell? 7. Mt h tun t c 1 ng vo (X) v 2 ng ra (Z1 v Z2). Ng ra Z1= 1 khi nhn c chui vo 101, ng ra Z2= 1 khi nhn c chui v 011. H c c im l mt khi xy ra Z2= 1 th Z1= 1c th khng bao gi xy ra v ngc li. Hy tm gin trng thi Mealy v bng trng thi (c s trng thi ti thiu l 8). 8. Tng t bi 7 nhng Z1= 1 vi chui vo 010, Z2= 1 vi chui vo l 100. (S trng thi ti thiu l 8). 9. Mt h tun t c 2 ng vo (X1,X2) v 1 ng ra (Z). Ng ra gi gi tr khng i tr khi c 1 trong cc chui vo sau xy ra:

a) Chui vo X1 X2=00, 01 lm cho Z =0. b) Chui vo X1 X2= 01, 11 lm cho Z= 1. c) Chui vo X1 X2= 10, 11 lm cho Z o gi tr c ( ngha l trcs l 0 th

by gi l 1 v ngc li). Suy ra gin trng thi Moore v bng trng thi. Ci t h bng: a) ROM, b) PLA. 10. Tng t bi 7 nhng vi:

a) Chui vo X1 X2= 01, 00 lm cho Z= 0. b) Chui vo X1 X2= 11, 00 lm cho Z= 1. c) Chui vo X1 X2=10,00 lm cho Z o gi tr c.

11. Mt h tun t c ng vo (X) v mt ng ra (Z). Vez gin trng thi Mealy cho cc trng hp sau:

a) Ng ra Z= 1 nu tng s bit 1 nhn c chia ht cho 3 (ta xem 0, 3, 6, 9... chia ht cho 3).

b) Ng ra Z=1 nu tng s bit 1 nhn c chia ht cho 3 v tng s bit 0 nhn c mt s chn >0 (9 trng thi).

12. Mt h tun t c 1 ng vo (X) v mt ng ra (Z). V gin trng thi Mealy cho cc trng hp sau:

a) Ng ra Z= 1 nu tng s bit 1 nhn c chia ht cho 4 (ta xem 0, 4, 8, 12... chia ht cho 4).

b) Ng ra Z=1 nu tng s bit 1 nhn c chia ht cho 4 v tng s bit 0 nhn c l 1 s l (c 9 trng thi).

13. Mt h tun t c 2 ng ra. Cc ng vo (X1,X2) biu din s nh phn 2 bit N. Nu gi tr hin ti ca N ln hn gi tr trc th Z1 = 1. Nu gi tr hin ti ca N nh hn gi tr trc th Z2 = 1. Cc trng hp khc th Z1 = Z2 =0. Khi cp gi tr vo u tin nhn c th khng c gi tr trc ca N th ta xem nh trng hp vi Z1 =Z2 = 0.

a) Tm bng trng thi Mealy ca h (s trng thi ti thiu bao gm trng thi bt u l 5). Ci t h bng PLA.


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b) Tm bng trng thi Moore ca h ( s trng thi ti thiu l 11). Ci t h bng PLA.

14. Mt h tun t c 2 ng vo v 2 ng ra. Cc ng vo (X1,X2) biu din s nh phn 2 bit N. Nu gi tr hin ti ca N cng vi gi tr N trc ln hn 2 th Z1= 1. Nu gi tr hin ti ca N nhn vi gi tr trc ca N m ln hn 2 th Z2= 1. Cc trng hp khc th Z1 =Z2 = 0. Khi nhn c cp gi tr vo u tin th xem nh gi tr trc ca N= 0.

a) Tm gin trng thi Mealy v bng trng thi ca h ( s trng thi ti thiu l 4). Ci t bng PLA.

b) Tm bng trng thi Moore ca h ( s trng thi ti thiu l 11 nhng vi p s c s trng thi 16 chp nhn c).

15. Mt h tun t Moore c 1 ng vo v 1 ng ra. Khi chui vo l 011 th ng ra Z= 1 v gi gi tr 1 cho n khi chui vo 011 xy ra mt ln na th ng ra Z= 0. Ng ra Z gi gi tr 0 cho n khi 011 xy ra ln th 3. V d chui vo:

X = 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 1 1

Cho chui ra Z = 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1

Suy ra bng trng thi ( c s trng thi ti thiu 6). Ci t h bng PLA.

16. Tm gi tr vo ca 1 h tun t gm cc nhm 5 bit. Mi nhm 5 bit biu din BCD loi m 2 trong 5 ( c 2 bit 1 trong nhm 5 bit). Sau khi nhn 5 bit, h cho tr ra l 1 v reset nu nhm 5 bit l m 2 trong 5 hp l, cc trng hp khc th ng ra bng 0 v reset. H c mt ng vo v mt ng ra. Suy ra gin trng thi Mealy ( c s trng thi ti thiu l 13).

17. Mt h tun t Mealy c 2 ng vo v 1 ng ra. Nu tng s bit 0 nhn c 4 v ti thiu 3 cp tr vo xy ra th ng ra bng 1 cp vo cui trong chui tr vo. Khi ng ra Z =1 xy ra th h reset. Suy ra gin trng thi v bng trng thi. Ch r ngha ca mi trng thi. V d S0 ngha l reset, S1 ngha l cp tr vo l 11...

V d:

Chui vo: X1= 1 1 1 0 0 0 1 1 1 0 0 0 1 1 0 0 0 1 0

X2= 1 0 0 0 0 0 1 1 1 1 1 1 0 1 0 0 0 1 0

Chui ra: Z = 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1


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18. H tun t Moore c 1 ng vo v 1 ng ra. Ng ra Z =1 nu tng s bit 1 nhn c l l v tng s bit 0 nhn c l chn > 0. Suy ra gin trng thi v bng trng thi (s trng thi ti thiu l 6). Ci t h bng PLA.

19. Mt h tun t c 1 ng vo X v 1 ng ra Z. Ng ra Z hin ti bng gi tr vo X trc 2 chu k clock. V d: X = 0 1 0 1 1 0 1 0 1 1 0 1 0 0 0 1 Z = 0 0 0 1 0 1 1 0 1 0 1 1 0 1 0 0

20. Mt h tun t c 1 ng vo X v 1 ng ra Z. Ng ra hin ti bng gi tr vo trc 3 chu k clock. V d: X = 0 1 0 1 1 0 1 0 1 1 0 1 0 0 0 1 Z = 0 0 0 0 1 0 1 1 0 1 0 1 1 0 1 0 ba gi tr u ca Z =0. tm gin trng thi Mealy v bng trng thi ca h. Ci t h bng PLA.

21. Mt h tun t c mt ng vo l X v 2 ng ra S v V. X biu din s nh phn 4 bit N m LSB c nhp trc, S biu din s nh phn 4 bit bng N + 2 m ng ra s cho LSB ra trc. thi im tr vo th 4 xy ra thh V = 1 nu N + 2 ln hn biu din 4 bit ( b trn [overflow]), ngc li th V = 0. H lun lun reset sau khi nhn c bit th 4 ca X. Tm gin trng thi Mealy v bng trng thi ca h. Ci t h bng PLA. V d: X = 0111 ( y l 1410vi LSB i trc) S = 0000 (v 14 + 2 = 1610 =10000 V = 0001

22. Mt h tun t c 1 ng vo l X v 2 ng ra D v B. X biu din s nh phn 4 bit N m nhp vo vi LSB i trc, D LSB i trc. thi im tr vo th 4 xy ra th B = 1 nu N -2 nh hn 0 (mn = Borrow ), ngc li th B = 0. H lun lun reset sau khi nhn c bit th 4 ca X. Tm gin trng thi Mealy v bng trng thi. Ci t h PLA. V d: X = 0001 1000 1100 D = 0110 1111 1000 B = 0000 0001 0000

23. Mt h tun t c 1 ng vo X v cc ng ra Y v Z. Cp YZ biu din 1 s nh phn 2 bit bng s bit 1 nhn c l 3 hoc khi tng s bit 0 nhn c l 3. Tm gin trng thi Moore v bng trng thi ca h. Ci t h bng PLA. 1.32 Mt h tun t c mt ng vo X v cc ng ra Yv Z.Cp YZ biu din mt s nh phn 2 bt bng s cp bt 1 lin tip nhn c ng vo, v d chui 0110 cha mt cp bt 1, chui 011110 cha 2 cp bt 1 , v chui 0110111 cha 3 cp bt 1 lin tip. H reset khi tng s cp bt mt lin tip l 4. Tm gin trng thi moore v bng trng thi ca h. Ci t h bng PLA. Reset Chui vo: X=010110111001010101110110110010


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Chui ra: Y= 000000011111111111000000011111 Z=000011101111111111000011100000 Chui vo: X=11111111 Chui ra: Y=00110001 Z=01010010 Reset Hng dn: bo m h reset cc v d trn 24. Mt h tun t dng iu khin hot ng ca mt my bn hng $0,25 (25 xu). H 3 ng vo N, D v Q v 2 ng ra R v C. B pht hin tin ng trong my bn hng ng b vi clock ca h tun t ta thit k. B pht hin tin ng s cho ra 1 cho N,D hoc Q (N=nickel = 5 xu, D= Dime = 10 xu v Q = quarter = 25 xu), khi ta cho vo 5 xu, 10xu hay 25 xu. Mi ln ch c ti a ng ra l 1 b pht hin tin ng. Khi khch hng a tin vo th my bn hng kim tra thy nu ti thiu 25 xu th giao hng cho khch tr tin d theo 5 xu. Vi mi gi tr ra l mt C th my pht ra ng 5 xu cho khch hng. Mn hng ch c xut ra khi h cho ng ra R = 1. (C = change = thi tin v R = return = giao hng). H s reset sau khi giao hng. V d: Khch hng nht mt ng 5 xu, 1 ng 10 xu v mt ng 25 xu. Cc ng vo v ra ca h nh sau: N = 0001000000000000 Ng vo D = 0000000100000000 Q = 0000000000100000 Ng ra R = 0000000000000010 C = 0000000000011100 Ch l c th c cc gi tr khng gia cc gi tr vo. Suy ra bng trng thi moore ca h, v mi trang thi ch ra khch hng a vo bao nhiu tin hoc thi li bao nhiu. Ci t h bng PLA. 25 a) Suy ra bng trng tht ca h tun t Mealy i mt chui cc bt ni tip t m NRZ sang m NRZI. Gi s rng chu k xung nhp bng thi gian bt nh hnh E.1.35.

b) Lp li a) vi h tun t moore c) V gin nh th cho p s a) dng dng sng vo NRZ nh hnh trn.

Nu ng vo hi thay i sau cnh clock, hay ch cc ch dng sng ra m cc glitch (gai/xung nhiu) [cc gi tr ra sai] c th xy ra.

d) V gin nh th cho p s b), dng cng cc dng sng vo nh c). C nhn xt g?


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26. Thit k mt h tun t ng b dng PLA v D flipflop c kch cnh ln m cho ng ra Z t l vi tn s xung nhp CK ( Z = CK/n) nh c minh ha gin nh th hnh sau:

a) Tn s tn hiu ra Z t l nh th no vi tn s CK? b) Tm bng trng thi thit k. c) S dng gng trng thi a = 00, b = 01, c = 10. Cho tt c cc trng thi khng s

dng v a (00). V s mch thit k. 27. Thit k h tun t ng b Moore c 2 ng vo X1 v X2 v mt ng ra Z. Kki X1 = 0 v X2 = 1 th ng ra Z = 1. Nu tip theo ng vo X2 = 1 th ng ra Z vn tr l 1, cn cc trng hp khc th Z = 0. Ci t h bng PLA v D flipflop kch cnh xung. 28. Xy dng mt khi SM c 3 bin vo (D,E,F), 4 bin ra (P,Q,R,S) v 2 ng ra. Vi khi ny, ng ra P lun lun l 1 v Q = 1 nu D = 1. Nu D v F l 1 hoc D v E l 0 th R = 1 v ly ng ra 2. Nu(D = 0 v E = 1) hoc (D = 1 v F = 0) th S = 1 v ly ng ra 1. 29. Xy dng mt khi SM c 3 bin vo (A, B, C), 4 bin ra (W, X, Y, Z) v 2 ng ra. Vi khi ny, ng ra Z lun lun bng 1 v W = 1 nu c 2 A v B bng 1, nu C = 1 v A = 0 th Y = 1 chn ng ra 1. Nu C = 1 hoc A = 1, th X = 1 v chn ng ra 2. 30. Chuyn cc gin trng thi hnh 2.1 v 2.2 sang cc lu SM.


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31. Chuyn cc gin trng thi (b pht hin tun t) hnh 2.3 v 2.4 sang lu SM. S dng cc ng ra theo iu kin cho hnh 2.3.

32. chuyn gin trng thi hnh E.2.5 sang lu SM. Kim tra mt bin duy nht trong mi hp quyt nh. Hy th ti thiu ha s hp quyt nh. 33. Chuyn gin trng thi hnh 2.23 sang lu SM. 34. Hon tt gin nh th sau cho lu SM hnh 2.25( b nhn nh phn), gi s St = 1

Clock

State

M

K

Ad

Sh

Hnh 2.6 Gin nh th ca hnh 2.25

S0


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35. Hon tt gin nh th sau cho lu SM hnh 2.20( b chia nh phn)

36. Thit k li b nh phn bt c lc no php cng xy ra bit nhn(M) s c t thnh 0. Nh vy. Nu M = 1 thi im clock cho trc v php cng xy ra, M s bng 0 thi im clock k. Nh vy ta c th lun lun cng khi M = 1 v lun lun dch khi M = 0. iu ny c ngha l mch iu khin khng phi i trng thi khi M = 1, v s trng thi c th c gim t 8 xung 5. V lu SM cho b iu khin nhn. 37. cho lu SM nh hnh sau.

a) V gin nh th cho clock, trng thi (S0,S1,S2), cc gi tr vo X1 v X2 v cc gi tr ra. Gi s l X3 = 0 v chui tun t vo cho X1X2 l 01, 00, 10, 11, 01, 10. Gi s l tt c cc thay i trang thi xy ra cnh ln ca clock, v cc gi tr vo thay i gia cc xung clock.

b) Nu dng ROM thay v PLA, th cn ROM kch thc bao nhiu? Vit 5 dng u tin ca bng ROM.

Clock

State

St

C

Sh

Stt

Hnh E.2.7 Gin nh th ca hnh 2.20

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38. Cho trc cc bin X1 v 2, cc bin trng thi Y2, Y2 v Y3 v cc bin ra Z1 n Z8, hy tm cc bin Moore v Mealy nu bit :

a) Z1 = Y1.Y2.X1 b) Z2 = Y1.Y31 c) Z3 = Y1.Y2.Y3 d) Z4 = Y3.X2.X1 e) Z5 = Y2 f) Z6 = Y1.Y3.X1 f)Z7 = Y1+Y2 h) Z8 = Y2+X2

vi mach phan 2

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