Excitation Tables

SectionS.8 Design Procedure 229

,

~

_~

J , 0

IS

l£4Z ^I

J )§

I

I

Ciock

T

A

8

B'

)

FIGURE5.29

logic diagr am ofseq uence detector

230 Chapte r5 Synch ronous Sequential Logle

Table 5.12

Flip-flopExcitationTobles

Q( t ) Q(l

=

¹⁾

J «

Q(l) Q(t

=

¹⁾

~

0 0 0 X 0 0 0

0 I I X 0 I I

I 0 X I I 0 I

I I X 0 I I 0

(a )JK (b)T

how therequiredtransitionis achieved. There arefourpossibletransition sfrom the presentslate tothenextstare .Therequiredinputconditionsforeachofthe four transitionsarederivedfrom the informationavailable inthecharacteristictable.The symbolX in the tables represents a don'H arecondition, which meansthatitdoes not matterwhethertheinputis 1or O.

The excitationtablefor theJK flip-tlop isshownin part (a).Whenbothpresent"lateandnext state areO.theJinput mustremainat0andtheKinput canbeeither0or I.Similarly...henboth presentstateand next state arc 1.theKinputmust remain at O.whiletheJinputcanbe0 or 1.

Iftheflip-flop istohaveatransitionfromthe O-statetothe l-state,Jmustbeequalto I.since theJinputsetsthe flip-flop.However.inputKmaybeeither0orI.IfK

=

O. the J

=

Icon- ditionsets the flip-flopas req uired;ifK = Iand J = I.theflip-flop iscom plemented and goesfrom theOcstatctothe I-state asrequired.Therefore.the Kinputismarkedwith a don't- careconditionfor theO-to-Itransition.Foratransitionfromthe I-stateto the O-state. wemust haveK = I. since theK inputclears the llip-flop.However.theJ input maybeeither a or 1.

sinceJ

= a

hasno effectandJ

=

1 together withK

=

1complementstheflip-flopwith a re- sultant transitionfromthe l-statetothea-state .

The excitat iontable fortheTflip-Ilop is shown inpart(b).Fromthecharacteri...tic table.we findthatwheninputT = I.the stateoftheflip-flop iscomplemented. and whenT = O.the stateof the flip-flop remainsunchanged.Therefore.whenthe state of the flip-flopmustre- main thesame.the requireme nt isthatT = O.Whenthestateof theflip-flopha..to becom- plemented.Tmust equal I.

Synthesis Using JK Flip-F lops

ThemanualsynthesisprocedureforsequentialcircuitswithJKflip-flopsisthe same aswith Dflip-flops.exceptthat the inputequationsmustbeevaluatedfrom the present-statetothe next- state transition derivedfromtheexcitationtable.To illustratetheprocedure.we willsynthe- size thesequentialcircuitspecifiedbyTable 5.13.Inadditiontohaving columnsforthepresent state,input.and nextstate.asinaconventionalstatetable.the tableshowsthe nip-flopinput conditionsfromwhich theinput equationsare deri ved.Theflip-flopinputs are derived from thestate tableinconjunctionwith the excitationtablefortheJKflip-nap.Forexample.in the firstrow ofTable 5.13.wehave a transitionfor llip-flop A from 0 inthe present stateto 0 in the nextstate .In Table 5.12.fortheJKflip-flop.we find that atransition ofstates frompres- entstate atonextstate0 requiresthatinputJbe0andinputKbe a don't-care.So0andX are

Section S.B Design Proced ure 231

Table 5.1J

Srat~TabkandJK flip-FlopInputs

Present Nellt

State Input State Fllp.FlopInputs

A

• ^•

^A

•

^J•

'.

^J.

'.

0 0 0 0 0 0 X 0 X

0 0 I 0 I 0 X I X

0 I 0 I 0 I X X I

0 I I 0 I 0 X X 0

I 0 0 I 0 X 0 0 X

I 0 I I I X 0 I X

I I 0 I I X 0 X 0

I I I 0 0 X I X I

enteredin the firstrow underJ,4andK_A,respectively.Sincethefirstrowalsoshows a transi- tion forflip-flopBfrom0 inthepresent stateto 0in thenext state, 0 and X arcinserted into thefirstrowunderIn^{and KB,respectively,The second row ofthetableshowsatransition for} flip-flopBfrom0 in the presentstateto Iinthenextslate. From the excitat ion table. wefind thatatransition from0to IrequiresthatJbeIandKbea don't -care.so Iand Xarecopied into the second rowunderJ_IlandKn.respecti vely.The proce....iscontinued foreachrow in thetable and for each nip-flop.with theinputcondi tionsfromthe excitationtablecopied into theproper rowoftheparticular flip-flop being considered.

Thenip-flop inputsinTable 5.13 specifythe truth tableforthe input equationsasa func- tion of present stateA,present stateB.and input.r.The inputequationsare simplified inthe maps ofFig.5.30.The next -starevaluesare nOI used duringthesimplification,sincetheinput equation sare a function ofthc presentstate andtheinputonly.Notethe advantageof usingJ K- type nip-flops whensequential circu itsare designedmanually.Thefactthat there aresoman)' don 's-careentriesindicatesthat the combinationalcircuit for theinputequationsislikelytobe simpler,becausedon' t-carenunrermsusuallyhelp inobtaining simplerexpressions.Ifthereare unused statesinthestate table. there willbeadditionaldon't-care conditionsin the map.

The fourinput equationsforthepairofJKflip-flopsarclisted underthemapsofFig.5.30 . The logicdiagram (schematic)of the sequentialcircuitisdrawnin Fig.5.3 1.

Synthesis Using T Flip-Flops

Theprocedure for synthesizingcircuits usingTflip, flopswillbedem onstrated b),designin g abinary counter.An n-bitbinarycount er consistsof n flip-flops thatcancount inbinary from 0102" - I.Thestalediagramofathree-bit counterisshownin Fig .5.32.As seenfrom

e e

binary statesindicated in..idethecircles.the nip-flopOUIPUISrepeat the binarycountsequence witha returnto 000 after 111.Thedirected linesbetweencircles arenotmarked with input andoutput valuesas in otherstatediagram s.Rememberthatstate transitions inclocked se- quentialcircunsoccurduring a clock edge;the flip-flop..remainintheir present slate sif no clockisapplied.Forthat reason.the clock does nOIappear explici tly a!'>an input variablein

232 Chapte r S Synchronou sSequentia l lo gic

B B

B,

A

..

00

^-,

0'

_{' .}

II

^.,

10

0 X X X X

I ,

^"

^.. .. _, ..

, A

B, ^,

A ₀

..

_~, ^"01

..

II 1

7

10

'j'"

I , ^{- ,}

_X

^..

_X

^..

_X

..

I ~

,

A

,

I"- Bx'

B B

B"

A 00 OJ II 10

"

-, _II"',•

u

X X

I I '

I ,

^" _X

_{I l: ,} ^{-. , "} ..

, A

B"

A

..

_~I

_{I r}

0

_t

'

_'

^-.II^«

..

10

0 X X

A

I ,

^m,

I

^"

' ~ I

^"

~ ^..

^X

FIGURE5.30

MapsforJandKinputequations

, ^{J '}

v

J

^J

CIt

J ~ ; ,

J

CIk

11 .9 ,

A

A'

B

B'

Clod FI(j,URE5.:31

Logic diagram for sequenti al circuit withJI(flip-flops

Sectio nS.8 Design Procedu re 233

WI III

0 '.

@l _e---' e

FIGURE 5.32

State diagramof three-bit binarycounter

aslate diagram orslate table. From thispoint of view.thestate diagram of a counter doesnot havetoshow inputandoutput valuesalong the directedlines.The only input to the circuitis the clock.and the outputs arespecified bythepresentstate oftheFlip-fl ops. Thenext stateof a counterdependsentirelyon itspresent state,and thestatetransitionoccursevery time the clock goesthrough atransition.

Table5.14isthestatetableforthethree-bit binary counter.The three flip-flopsaresym- bolized byA2•A].andAo.Binary counters are constructed mostefficiently withTflip-flops because oftheir complement property.The flip-flopexcitation fortheTinputsisderivedfrom theexcitation table oftheTnip-flop and by inspection of thestate transition of thepresentstate tothe next state.As anillustration.consider the flip-flopinputentries forrow 001.The pres- entstate hereis 001andthe next stateis010.which isthenext countin thesequence. Com- paringthesetwo counts,wenotethatA2goes from 0 to O. soTA2ismarkedwith 0 because flip-napA2must not changewhenaclock occurs. Also,AI goesfrom 010I, soTAJismarked witha I becausethisnip-flop must becomplemented in the nextclock edge.Similarly,Ao goesfrom Ito 0,indicatingthaiit mustbecomplemented.soTAOismarkedwitha I.The last row. with present state II I.is compared with thefirstcount000,which isitsnext state. Going from all tsto allO'srequiresthat allthreeflip-napsbecomplemented.

Table5.14

5tate Table for Three-BitCounter

Present State Next State Flip-FlopInputs

A, AI A, A, A,

_Ao

T " T Al T"

0 0 0 0 0 I 0 0

0 0 I 0 I 0 0 I

0 I 0 0 1 I 0 0

0 I I I 0 0 I I

I 0 0 I 0 I 0 0

I 0 I I I 0 0 I

I I 0 I I I 0 V

I 1 1 0 0 0 I I

234 Chapter5 Synchronous Sequential Logic

~I

11 10

01

A1Au A

. ,

,

_{00 01}

1 1

¹⁰

" .

^.',

_{I'~ J~f:}

^m,

0 . ,;1//

,1 1 .

^{, m,} _~A"

_{~}~ ;~,}

"'

.

A

A

FIGURE 5.33

Mapsfor three-bit binarycounter

A,

FIGURE 5.34

logic diagramof three-bit binarycounter

Theflip-flopinputequationsaresimplifiedinthe maps ofFig.5.33.Note thaiTAOhas t's in all eight mintermsbecausethe leastsignificant bit ofthecounteris complementedwith each count.A Booleanfunctionthatincludesallmlntermsdefinesa constantvalue of J.The input equationslisted under each mapspeci fy the combinational part ofthecounter.in- cludingthesefunct ions with thethreeflip-flops,weobtain the logic diagramofthecount- er.as shownin Fig. 5.34.Forsimplicity,the resetsignalisnot shown, butbeawarethat every design sho uldinclude aresetsignal.