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7/24/2019 H tun t khng ng b

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Chng 3

HTUN TKHNG NG B

Gii thiu

Phn loi cc mch logic M hnh hthp


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M hnh htun t Chlm vic

cho htun tkhng ng bhot ng ng, ngita phi gii hn n lm vic chcbn (fundamentalmode) hoc chxung (pulse mode).

chcbn, mi mt trong cc tn hiu vo tngoichc ththay i ti 1 thi im t0 sang 1 hoc t1sang 0, v mch phi trong iu kin n nh (tt ccctn hiu trong mch phi n nh; ngha l chng phi tc gi trxc lp) khi xy ra mt sthay i tn hiuvo.

chxung Th d: xung dng th mi xung dng,sthay i t0 sang 1 v trv0 c to ra bng 1 tnhiu vo tbn ngoi, biu th1 skin nh th. Mchphi trong iu kin n nh khi mi xung xy ra

Chcbnca htun tkhng ng b

Ch : Mch phi n nh trc mi skin nhth v mi ln chc mt tn hiu vo tngoi cththay i t0 sang 1 hoc t1 sang 0

Chxungca htun tkhng ng b

Ch : Mch phi n nh trc mi skinnh th v mi ln chc mt tn hiu vo t

ngoi c thto 1 xung dng


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Mc tiu

Thit khtun tkhng ng bth kh hn hngbdo cc vn nh th.

Vi htun tng bth xung nhp gii quyt vnnh th.

Vi htun tkhng ng bth sphi cn cc kthut thit kc bit khcc vn nh th c t

cc tr hon khng bng nhau qua cc ng dn khcnhau trong h.

n gin ha phn tch v thit k, ta sgisrngcc htun tkhng ng bca ta lm vic ch

cbn.

3.1 PHN TCH

HTUN TKHNG NG B

3.1 PHN TCH HTUN TKHNGNG B

3.1.1 Phn tch hkhng ng bvi ccflipflop SR

3.1.2 Phn tch hcng khng ng b

3.1.3 Cc iu kin chy ua v lp vng

3.1.1 Phn tch hkhng ngbvi cc flipflop SR


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Th d: ta sphn tch hhnh sau bng cch theo di cc tnhiu dng chui vo

X1X2=00, 10, 11, 01, 11, 10, 00.

Giscc ng ra Q ban u ca flipflop l 0 (Q1Q2=00).

Hnh 3.5 Htr hon khng ng b

Phn tch bng bng chuyn trng thi vichui trvo cth(1)

Ct thnht ca bng 3.1 lit k cc gi trhin ti cacc bin vo v cc gi trra flipflop nhng thi imktip trong phn tch hbt u vi gi trban u(X1X2=00) v trng thi bt u (Q1Q2=00 ).

Mi thp cc gi trca cc ng vo mch v cc ngra bnh(cc ng ra flipflop trong trng hp ny) sc gi ltrng thi ton phn (total state) ca h

trnh nhm vi trng thi ton phn, ta sgi trngthi ca cc flipflop ltrng thi ni (internal state)

Phn tch bng bng chuyn trng thi vichui trvo cth(2)

Vi mi trng thi ton phn trong phn tch hnh 3.5,cc gi trra v cc gi trvo flipflop c tnh tondng cc phng trnh:

i vi trng thi ton phn cho trc, trng thi nikging nhtrng thi ni hin ti, skhng xy rathay i trng thi v ta ni rng htrng thi ton

phn n nh (stable total state)

Bng 3.1 Phn tch ca hnh 3.5

R1=X1X2S1=X1X2

R2=X1X2

S2=X1X2

Z1=Q1Q2 + X1(Q1 + Q2)Z2=Q1Q2 + X2 (Q1 + Q2)

Chui gi trvo v ra (tbng 3.1) l:X

1X

2=00, 10, 11, 01, 11, 10, 00

Q1Q2=01, 00, 10, 11, 01, 11, 10 tr hon khng ng b


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Phn tch bng bng chuyn trng thi Trc ht ta xy dngbng chuyn trng thi (transitiontable) m n cho ta thy cc trng thi kca flipflop l hm

ca trng thi hin ti v cc gi trvo. Ta c thto bngny nhcch lm cho hng bbng cch nh xccphng trnh trng thi kcho h. Vi hnh 3.5, cc phngtrnh trng thi kl:

. Mi ct trong bng ny tng ng vi thp cthcc gitrca cc bin vo, hoc tng ng vi trng thi vo (inputstate). Mi hng tng ng vi php gn cthcc gi trvocc bin bnh, hoc tng ng vi trng thi ni hin ti.Mi trong bng tng ng vi 1 trng thi ton phn, v gitrtrong l trng thi ni kcho trng thi

Hnh 3.6Bng chuyn trng thi

c

a hnh 3.5

Hnh 3.7 Cc bng dng (flow table)ca hnh 3.5

(a) nh nhn theo trng thi ni (b) nh nhn theo trng thi ton phn

Hnh 3.8 Bng dng v bng gi trracho hnh 3.5

Chui gi trvo v ra l:X1X2=00, 10, 11, 01, 11, 10, 00

Q1Q2=01, 00, 10, 11, 01, 11, 10


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Hnh 3.9 Gin nh th ca hnh 3.5

3.1.2 Phn tch hcngkhng ng b

Hnh 3.10 Hkhng ng bc xy dng tcng

dphn tch, ngi ta tp trung tt ccc tr hon lin hvimi ng hi tip vo mt hp c nh nhn delay. Ri talin hmt bin trng thi vi mi ng ra delay. Nu mt thiim cho trc, ng vo delay l 0 (hoc 1), ng ra sl 0 (hoc1) sau mt thi gian tr hon. T ng vo delay biu dintrng thi kca ng ra delay v tr hon vi ng ra Q1 c ngvo l Q1+.

Sau khi t cc bin trng thi, ta lp c bng dng. Hcm tbng cc phng trnh sau:

Cc phng trnh ny c in trong cc hnh 3.10(b) v (d), vbng dng ca hnh 3.10(c) c to nn bng cch thay thmithp cc bin trng thi bng k hiu trng thi. Cc ng vokhoanh trn biu din cc trng thi ton phn n nh. Vi mtchui vo bt kcho trc, cc chui trng thi v ng ra c thc xc nh thnh 3.10(b) hoc (c) v (d)

/ /


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Cch chn bin trng thiM hnh tng qut ca htun t

khng ng b

3.1.3 Cc iu kin chy ua v lp vng

Hnh 3.14 Mch v bng chuyn trng thi vichy ua khng ti hn v lp vng

Chy ua khng ti hn(noncritical race)

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Chy ua ti hn (critical race)

3.2 THNH LP V RT GN CC BNGDNG CBN

3.2.1 Thnh lp cc bng dng cbn

3.2.2 Rt gn cc bng dng cbn

3.2 THNH LP V RT GNCC BNG DNG CBN

Cch thit kcho hkhng ng bth tng tvi cch thit kcho cc hng b:

Xy dng mt bng trng thi (hay bng dng)tpht biu vn /bi ton

Rt gn bng c shng ti thiu, thchin gn trng thi (c thtng thm hng)

Ci t bng dng cc phn tlogic thch hp.Tuy nhin chi tit ca mi bc hi khc i vicc hkhng ng b.

3.2.1 Thnh lp

cc bng dng cbn

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Bng dng cbn(primitive flow table)

Thng thng im bt u tng hp hkhng ngbl lpbng dng cbn (primitive flow table) haycn gi lbng dng skhai.

Bng dng cbn c nh ngha nhbng dng m cchnh xc mt trng thi ton phn n nh trn mi hng.

Ta c thrt gn bng dng cbn thnh 1 bng c thng hn.

Tuy nhin, bo m c c bng c shng ti thiuth phn ln ngi ta bt u bng bng dng cbn hnl cgng xy dng trc tip mt bng c rt gn tu.

Qui c lp bng dng cbn

trnh cc vn nh th m c thpht sinh khi 2bin vo thay i ng thi, ta sgisrng mi lnchc 1 bin vo thay i v cc thay i gi trvocch nhau h slun lun t n trng thi tonphn n nh gia nhng thay i (hot ng chcbn).

Khi thnh lp bng dng cbn, mi thay i gitrvo phi lm cho 1 thay i trng thi v cho phpch1 trng thi n nh trn 1 hng ca bngdng cbn m thi.

Th d3.1Mt hkhng ng bc 2 ng vo v 1 ng ra. Chuivo X1X2=00, 01, 11 lm cho ng ra trthnh 1. Ri sthay i gi trvo ktip lm cho ng ra quay v0.Khng c chui vo no khc sto ra 1 ng ra.

Th d3.1Bng dng cbn hon tt

Ghi ch:

*: cc trng thi ny khng thdn n gi trra 1 nuban u khng reset

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Gin trng thi ca bng dng cbnTD 3.1

Skhc bit vi gin trng thi ca htun tng bl g?

Th d3.2Th dny minh ha thit kT flipflop vi xungnhp kch cnh hnh 3.18(a). Mc d flipflopny c sdng nh1 bphn trong cc hng b, thit kbn trong ca flipflop l vn khng ng b. Flipflop c 2 ng vo T v P.Flipflop si trng thi nu T=1 khi xung nhpP thay i t1 sang 0. Di tt ccc iu kinvo khc, Q gikhng i. Ta sgisrng Tv P khng thay i ng thi.

Th d3.2Bng dng cbn ca T FF

Th d3.3Mt tn hiu xung nhp C c chn cho qua bng mt tn hiu Skhc. Mng cho qua sao cho phi c nhng xung y xut hin ng ra Z ngay ckhi S c ththay i gia xung nhp.

Hnh sau chgin nh th ca mng sc thit k. Ngay ckhi c xung nhp, y l vn khng ng bbi v ng vo S

c cho php thay i bt clc no so vi xung nhp. Ta sgisrng S slun lun ON (dn=1) hay OFF (tt=0) ti thiu 2xung nhp y .

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Th d3.3

3.2.2 Rt gn ccbng dng cbn

Phng php rt gn bng dng cbnC 2 phng php rt gn bng dng cbn:

1. Phng php thnht cn 2 bctrc ht tm ra bngdng cbn c shng ti thiu v ri bng ny crt gn hn bng cch trn (hay gom) cc hng li.

2. Phng php thhai thc hin srt gn ton phntrong 1 bc bng cch p dng phng php rt gntng qut cho cc bng trng thi khng hon ton (cthm Introduction to Switching Theory ni dung LogicDesign, 3rd Edition ca Hill v Peterson, chng 13 v14).

Phng php thnht thng thc hin t tn cng hn v

sc bn y.

Rt gn sbbng dng cbnbng cch khcc trng thi ton phn n nh tha

Ta c thtm c bng dng cbn c shng ti thiubng cch khi cc trng thi ton phn n nh tha. lm iu ny, ta phi tm ra c cc trng thi ton phn nnh tng ng.

Hai trng thi ton phn n nh c gi l tng ng nuchng c cng sng vo v cc trng thi ni lin hltng ng. Nhvy 2 trng thi ton phn n nh cgi l tng ng nu

1. cc ng vo ca chng ging nhau,

2. c cc ng ra ging nhau, v

3. cc trng thi kca chng tng ng vi mi gitrvo c thc.

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Bng 3.7 Bng dng cbn sc rt gn

X1X2 Cc trng thi c khnng tng ng

00 (2, 6, 8 )

01 (5, 12)

11 (3, 10)

10 (4, 11)

Sau khi xt tip cc trng thicng 1 dng ta tm c cctrng thi tng ng cn li l:(2, 8) (5, 12) (3, 10) (4, 11)

Bng 3.8 Bng dng cbn c rtgn tbng 3.7

Rt gn bng dng cbnbng sbtrn (merger diagram)

Mi mt bng dng cbn c shng tithiu c xt trn c thc rt gnthm na bng cch trn cc hng (hay gi lgom cc hng)

Cc bc tin hnh:

1. i bng Moore thnh bng Mealy

2. Vsbtrn (merger diagram) (haycn c gi l bng sgom hng)

Bng 3.9Bng dng cbn Mealycho bng 3.4

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Strn v bng dng rt gn tbng 3.9

Bng dng rt gn

tbng 3.8

Bng dng rt gn thnh 3.18 Tm tt: Rt gn bng dng cbn

Ta c thrt gn bng dng cbn thnh bngMealy c shng ti thiu dng thtc sau:

1. Tm c bng dng cbn rt gn bngcch khi cc trng thi ton phn nnh tha.

2. Xy dng bng dng Mealy tbng dngcbn rt gn tm c bc 1.

3. Vsbtrn v dng n trnnhng hng tng ng v c c bngdng rt gn c shng ti thiu.

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3.3 GN TRNG THI V CIT CC BNG DNG

Trong vic chn gn trng thi cho cc hng b, itng chnh l n gin ha mch logic.

Tuy nhin, i vi cc hkhng ng bth cc itng chnh trong chn gn trng thi l ngn nga ccchy ua ti hn, v sn gin ha mch logic trthnh i tng thhai.

Sau khi thc hin gn trng thi khng c cc

chy ua ti hn, th ta c thci t hkhng ng bdng cc cng v cc S-R flipflop hoc chsdng cccng.

Th d Ta shon tt thit kcho th d3.1. Bng 3.10 cho

bng dng rt gn ca bng 3.9 vi cc trng thi kc t tn theo cc trng thi ni a, b, v c. Cc gitrra cho cc trng thi ton phn khng n nh sc in vo sau khi thc hin gn trng thi.

Gn trng thi

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Hon tt bng gi trra

Bng 3.12

Hnh 3.24(a) Ci t bng cng cho bng 3.12

Phng php tt suy ra ccphng trnh ng vo ca SR FF

Tbng sau, ta thy l khi Qi=0, Si=Qi+;

tng tkhi Qi=1, Ri=(Qi+).

Suy ra phng trnh SR bngphng php tt cho th d3.1

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Hnh 3.24(b) Ci t bng SR FF cho bng 3.12

3.3.2 Gn trng thi cho ccbng 3 hng v 4 hng

Hnh 3.27 Bng 4 hng tng qut vgin chuyn trng thi.

Hnh 3.28. (a) Gn trng thi vn nng cho bng 4 hng(b) Bng 4 hng c mrng

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Hnh 3.30. Bng dng vi cc dont care

3.3.3 Cc php gn trng thic hng chung

(Shared-Row assignments)

Hnh 3.31

Cc chuyn trng thi c lit k trong bng hnh 3.31(b) hm lcc trng thi trong mi tp kcn sau phi c t trong 1 chui cc kcn:

Ct 00: (a, c, e) (b, d)Ct 01: (a, b) (c, d, f)Ct 11: (b, c, f) (a, d, e)Ct 10: (a, c, d) (e, f)

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Hnh 3.33 Bng mrng ca bng hnh 3.31(a) dng php gn hnh 3.32(b)

3.3.4 Hon tt bng ra

Hon tt bng ra

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Hon tt bng ra (tt) Tm tt thtc thit k htun tkhng ng b

1. Cho trc vn , xc nh quan hcn c gia cc bin vo v ra.Suy ra bng dng cbn m c mt trng thi ton phn n nhtrn mi hng, v xc nh gi trra c lin quan vi mi trng thiton phn n nh.

2. Rt gn bng dng cbn c c shng ti thiu. Vic nythng c thc thc hin trong 2 bc. Trc ht xc nh cctrng thi ton phn tng ng v tm ra bng dng cbn c shng ti thiu. Ri hp nht (gom, trn) cc hng trong bng nytm ra bng rt gn cui cng. Strn c thhu ch trongvic chn cc hng sc hp nht.

3. Tm ra php gn trng thi m khi tt ccc chy ua ti hngia cc bin trng thi. Trong qu trnh ny c thcn mrngbng dng bng cch thm cc hng.

4. Lp bng chuyn trng thi bng cch thay thcc gi trc gn cc bin trng thi cho mi trng thi trongbng dng mrng. Nu sdng bng gi trra Mealy,in vo cc chkhng xc nh th cn trnh cc qu nht thi trong bng gi trra. Vcc bng trng thi kv cc bng gi trra tbng chuyn trng thi.

5. Nu khng sdng cc SR flipflop, tm mt ci tkhng c hazard cho mi hm trng thi kdng cccng logic kh dng.

6. Nu khng sdng SR flipflop, vcc bng gi trvoflipflop v tm mt ci t khng c cc hazard 0 chomi S v R. (nu sdng cc SR flipflop cng NAND,cc ng vo flipflop l cc b ca S v R v phi khng

c cc hazard 1).

7. Tm ci t khng c hazard ca cc hm ra.

8. Nu c cc hazard tt yu (essential hazard) trongbng dng, th thm cc tr hon (delay) trong ccng hi tip hoc sa i cu trc cng khicc hazard tt yu.

9. Kim tra li thit kca bn bng cch kim tra thcnghim hoc m phng trn my tnh.

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