fyse420 digital electronics lecture 3 - jyväskylän...

FYSE420 DIGITAL ELECTRONICS

Lecture 3Lecture 3

1

DIGITAL LOGICCIRCUIT ANALYSIS& DESIGNNelson, Nagle, Irvin, CarrolISBN 0-13-463894-8

DIGITAL DESIGNMorris Mano

[1]

[2]Morris ManoFourth editionISBN 0-13-198924-3

Digital DesignPrinciples and PracticesFourth editionWakerly John F.ISBN 0-13-186389-4

[3]

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Definition :

Määritelmä :

Kaksi tilataulukkoa ovat identtiset, jos edellisen taulukonjokaista tilaa vastaa jokin jälkimmäisen taulukon tila siten,että kun lähdetään liikkeelle vastintiloja osoittavilta riveiltä,niin jokainen äärellinen sisäänmenosekvenssi aiheuttaa sa-mat ulostulosekvenssit kummassakin tapauksessa.

Reduction of Flow Tables

mat ulostulosekvenssit kummassakin tapauksessa.

3©Loberg

2,0 -,- 3,- 2,0

3,0 5,1 2,0 -,-

3,0 4,1 -,- 5,03

2

1

00 01 11 10s

Vaatimukset rivien yhdistämiselle

Ulostulot määrätylle syöttökombinaatiolle eivät saaolla ristiriidassa keskenään.

Syötöt eivät saa johtaa tilojen osalta uusiin rivienyhdistämisvaatimuksiin siten, että jälleen synnyt-


-,- 1,1 2,- -,-

-,- -,- 1,1 -,-5

4

Syötöllä 11 on ristiriita riveillä 2 (Z=0) ja 5 (Z=1).

S,Z

Ulostuloissa ei ristiriitaa millään syötöillä riveillä 1 ja 3, mutta syötöllä 10 saadaanyhdistämisvaatimus riveille 2 ja 5, joka todettiin edellä ristiriitaan johtavaksi.

yhdistämisvaatimuksiin siten, että jälleen synnyt-täisivät ristiriitaa edellisen kohdan mukaisesti.

4©Loberg

Tilataulukosta tulee etsiä systemaattisesti kaikki yhdistyvät riviparit.

Tämän jälkeen voidaan etsiä laajempia yhdistämismahdollisuuksia : 3 riviä, 4 riviä, …

Note !

Yhdistyvyys ei ole transitiivinen ominaisuus.


(a,b) ja (b,c) ovat yhdistyviä pareja, mutta siitä EI SEURAA, että a ja c olisivatmyös yhdistyviä.

Esimerkiksi edellissä taulukossa (1,2) ja (2,3) ovat yhdistyviä, mutta (1,3) EI OLE.

Jos c on yhdistyvä yhdistyvän parin (a,b) kummankin alkion kanssa erikseen( (a,c) ja (a,c) ovat yhdistyviä), on myös ((a,b),c) yhdistyvä joukko (a,b,c).

Yhdistyvä joukko on maksimaalinen ellei se ole minkään yhdistyvän joukon alijoukko

5©Loberg

Example 6

d,0 b,0

e,0 a,0

g,0 f,1c

b

a

0 1x

state s

b

c

d

States are equivalent √

States are not equivalent X

√

X

Completely specified sequential circuit.

Reduction of Flow TablesSystematic Reduction

a,1 d,0

a,1 d,0e

d

c,0 b,0f

a,1 e,0g

S,output

Flow table

d

e

f

g

a b c d e f

Empty Implication table

6©Loberg

b d,e States are equivalent √


First StepPlace a cross in any square corresponding to pair of states whose outputs are not equalfor every inputs

Second Step

Starting from the top square in the left

column and going down and then pro-

ceeding with next column.


Example 6

Systematic Reduction Completely specified sequential circuit.

c

d

e

f

g

a b c d e f

X

X

X

X

d,c

X

X

X

X

X

X

X

X

e,ca,b

d,e d,e

√

X X

X

Implication table

States are not equivalent XFrom the flow table we see that (a,b)

implies (d,e), so write (d,e) in the empty

square defined by column a and row b.

States (d,e) are equivalent because of

same next state and they have same

output.

7©Loberg

b

c

d

e

d,e

X

X

X

X X

√ States are equivalent √

States are not equivalent XSquares filled by implaied pair must be

checked.

It must be crossed out if it contains at

least one implaied pair that is not equivalent.


Example 6

Third Step


e

f

g

a b c d e f

X

X

d,c

X

X

X

X

X

e,ca,b

d,e d,e√ √

√

XXX X

X

Implication table

This procedure is repeated until no

additional squares can be crossed out.

The squares without cross definepairs of equivalent states.

(a,b) (d,e) (d,g) (e,g)

Try combine pairs of states into larger groups of equivalent states.

8©Loberg

b

c

d

e

d,e

X

X

X

X

X

X

X

X

√

√

States are equivalent √


Start from the right colum and findfirst square without cross.

(e,g)

Find next column without cross


Example 6


f

g

a b c d e f

X

d,c

X

X

X

e,ca,b

d,e d,e√ √

XXX X

X

Implication table

Column d

If (d,e) and (d,g) are equivalent pairs then we cancombine three pairs into a set of three equivalent states (d,e,g)

(a,b) (c) (d,e,g) (f)Try next column

9©Loberg

(a,b) (c) (d,e,g) (f)

d,0 b,0

b

a

0 1x

state s

d,0 b,0a,b

0 1x

A C D F

0 1x


Example 7


e,0 a,0

g,0 f,1

a,1 d,0

a,1 d,0e

d

c

b

c,0 b,0f

a,1 e,0g

S,output

Flow table

e,0 a,0a,b

g,0 f,1c

d,e,g

a,1

a,1

a,1

d,0

d,0

e,0

f c,0 b,0

A

C

D

0 1

F

D,0 A,0

D,0 F,1

A,1 D,0

C,0 A,0

Reduced flow table

S,output

s

10©Loberg

Incompletely specified sequential circuit.

In asynchronous sequential circuits, the primitive flow table is alwaysincompletely specified.

In synchronous sequential circuit is incompletely specified if certaincombinations of inputs or input sequences never occur because ofinternal or external constrains.

Incompletely specified states can be combined to

Reduction of Flow TablesSystematic Reduction

Incompletely specified states can be combined toreduce the number of states in the flow table.

a,1

b,-Compatible states

c,-

-,-or

Incompletely specified states that can be combined are said to be compatible.

11©Loberg

Determine all compatible pairs by using the implication table.

Find maximal compatibles with the use of a merger diagram.

Find a minimal collection of compatibles that covers all the states and is closed.

Reduction of Flow TablesSystematic Reduction Incompletely specified sequential circuit.

Merging rows

The set of chosen compatibles must cover all the states.

The set must be closed.

The closure conditon is satisfied if :

There are no implied states

The implied states are included within the set

12©Loberg

Example 8

x1x2

a

00 01 11 10

b

c,- a,0 b,-

-,-

-,-

a,- e,-b,1b √


(a,b) (a,c) (a,d) (b,e) (b,f) (c,d) (e,f)

Compatible pairs

c

d

e

f

b,-

b,-

-,-

-,-

a,- -,-

-,-

a,-

c,-

c,0

d,0

e,1

f,1

f,-

d,-

e,-

Primitive flow table

c

d

e

f

a b c d e

√

√

√

√

√

√

X

X

c,f

c,f

d,e

d,e

d,ec,f

d,ec,f

X

X

X

X

X

XX

X

Implication table

13©Loberg

b √

Compatible pairs

(a,b) (a,c) (a,d) (b,e) (b,f) (c,d) (e,f)

a

bf

(a,b) (a,c,d) (b,e,f)

Maximal compatible


Example 8

Systematic Reduction Incompletely specified sequential circuit.

b

c

d

e

f

a b c d e

√

√

√

√

√

√

√

X

X

c,f

c,f

d,e

d,e

d,ec,f

d,ec,f

X

X

X

X

X

XX

X

Implication table

c

d

e

Merger diagram

There are noimplied states.

(a,c,d) (b,e,f)

All six states from theflow table are included

Both conditions are satisfied

14©Loberg

Example of merger diagram

a

bh

A rectangle with its two diagonals

The four-state compatible ( a,b,e,f )

A triangle

The three-state compatible ( b,c,h )


Merger diagram

c

d

e

f

g

The three-state compatible ( b,c,h )

A line

The two-state compatible ( c,d )

A dot

A single state ( g ) which is not com-patible with any other state.

15©Loberg

b

c

d

√

√

√X d,e

a,d

b,c

√b,c X

a

be

Six compatible pairs (a,b) (a,d) (b,c) (c,d) (c,e) (d,e)

(a,b) (a,d) (b,c) (c,d,e)

Maximal compatibles


Example 9


d

e

a b c d

√

√ √

a,d

b,cX

Implication table

√b,c X

X

Merger diagram

cd

There are several implaid pair of states

16©Loberg

b √b,c

Because of implied states we have to construct the closure table.

(a,b) (a,d) (b,c) (c,d,e)

(b,c) (b,c) (d,e) (a,d)

Compatibles

Implied states


Example 9


c

d

e

a b c d

√

√

√ √

X d,e

a,d

b,cX

Implication table

√b,c X

X

(b,c)

Closure table

17©Loberg

(a,b) (a,d) (b,c) (c,d,e)

(b,c) (b,c) (d,e) (a,d)

(b,c)

Compatibles

Implied states

We choose the two compatibles (a,b) (c,d,e)

This set of compatibles will cover all fivestates.

but


Example 9


Closure table The set is not closed

The implaid pair of states for (a,b) is (b,c) but (b,c)is not included in the chosen set of (a,b) (c,d,e).

If we choose compatibles (a,d) (b,c) (c,d,e) the set of compatibles will satisfy theclosed-covering condition.

18©Loberg

Originalflow table

a

b

c

d

e

Reducedflow table


(a,d) (b,c) (c,d,e)


flow table flow table

Alternative set of compatibles

(a,b) (b,c) (d,e)

There may be more than one possible way of merging rows when reducing a primitiveflow table.

19©Loberg

Race-free state assignment

Critical races can be avoided by making a binary state assignment in such a way thatonly one variable changes at any given time when a state transition occurs in the flowtable.

x1x2

s

a

00 01 11 10

a b c a

a = 00 b = 01


Three-Row Flow-table example

a

b

c

b

a

a

a

b

b

c

ccc

c a

Flow table

c = 11

Transition diagram

20©Loberg

Critical race during transition from a to c ( a = 00 -> c = 11) (b=01)

ac

0011

b01

Non critical race during transition from c to a

Reduction of Flow TablesRace-free state assignment

x1x2

s

a

00 01 11 10

b

c

b

a

a

a

b

b

c

ccc

c a

Flow table

21©Loberg

A race-free assignment can be obtained if we add an extra row to the flow table.

x1x2

s

a

00 01 11 10

a b d a

a=00 b=01

Reduction of Flow TablesRace-free state assignment

x1x2

s 00 01 11 10

Modified Flow table

b

c

b

d

a b c

ccc

d a -c-

Transition diagram

c=11d=10

s

a

00 01 11 10

b

c

b

a

a

a

b

b

c

ccc

c a

Flow table

22©Loberg

A proper operation of sequential circuits

Fundamental mode

One input changing at any time

No critical race

Circuit is on stable state during transition of input variable.

Hazard free logic

Asynchronous Sequential Logic

Hazard free logic

Hazards are unwanted switching transients that may appearat the output of a circuit because different paths exhibit differentpropagation delays.

Hazards may result in a transition to a wrong stable state.

23©Loberg

Types of hazards

Static 1-hazard Static 0-hazard Dynamic hazard

1 1 1


0 0 0

24©Loberg

Hazard and its removal

BA

DC00

00 01 11 10

01 1


11

10

1 1 1

Minterms DCBA = 1101 and 1111 are covered by product term DCA

25©Loberg

The EndThe End

26©Loberg

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