DigDesignCh18L04.pdf

7/25/2019 DigDesignCh18L04.pdf

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Chapter 18

FUNDAMENTALMODE SEQUENTIAL

CIRCUITS


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Ch18L4- "Digital Principles and Design", Raj Kamal, Pearson Education,2006

2

Lesson 4

Race Free Assignments


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3

Outline

Race Free Assignment

Circuit Implementation


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4

xQ and

xQ

differing in more than one

latch during a transition

Memory (or delay) section next-state

variables when have the bits differingby more than one during a transition,the critical (indeterminate time forstable state) and non-critical(deterministic) races occur.

Critical type of race must not occur inasynchronous sequential circuit.


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Race free assignment Race free assignment is an assignment

of state variables such that next-statevariables when have the bits differing

by one during a transition.

Recall the bit changes in adjacent cells

in a Karnaugh map.

Only a single bit changes between twoadjacent cells


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Finding Non-Adjacencies in Transitions

for Different Input Conditions

Step 1: Build a Karnaugh map type

table for each set of X in a State table

Draw adjacency map for the

assignments of present and next statevariables for X = 0 and X =1,respectively. It shows in a cell the

present state by 0 (no change) and nextstate (change) by 1


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Map for X = 0 Using a flow Table

S (xQ)

S (xQ)

S0

00

S1

01

S0 00 0

-S1 01 0

S3 11 1

S2 10

S3

11

S2

10

0

1 0


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Map for X = 1 using the flow table

S (xQ)

S (xQ)

S0

00

S1

01

S0 00 0

S1 01

S3 11 1

S2 10

S3

11

S2

10

1

1

1

0

0


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Observation

It is observed that for the present state S2,the transition to S1 is not to the adjacent

cell. S2 and S1 have to be the neighbors

It is observed that for the present state S3,

the transition to S0 is not to the adjacent

cell. S3 and S0 have to be the neighbors


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Number of Races

Number of critical + non-critical races

= 3


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Map for X = 00 Two inputs Using for another

circuit flow table

S (xQ)

S (xQ)

S0

00

S1

01

S0 00 0

01S1 01

S3 11

S2 10

1

S3

11

S2

10

01

1 0


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12

Map for X = 01 using another circuit flow

table

S (xQ)

S (xQ)

S0

00

S1

01

S0 00 01

01S1 01

S3 11 1

S2 10

S3

11

S2

10

0

1 0


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Map for X = 11 Using for another circuit

flow table

S (xQ)

S (xQ)

S0

00

S1

01

S0 00 0

0S1 01

S3 11

S2 10

S3

11

S2

10

1

1

0 1

1 0


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14

Map for X = 10 using another circuit flow table

S (xQ)

S (xQ)

S0

00

S1

01

S0 00 0

0S1 01 1

S3 11

S2 10

S3

11

S2

10

-

01

01


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Observation

It is observed for X =01that for a presentstate S2, the transition to S1 is not to the

adjacent cell. S2 and S1 have to be the

neighbors

It is observed for X =11 that for the present

state S1, the transition to S2 is not to the

adjacent cell. S1 and S2 have to be the

neighbors


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Observation

It is observed for X =11 that for apresent state S0, the transition is non

existent due to two input changes and

is leading to unstable intermediate

states


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Number of races

It is observed for X =11 that for apresent state S0, the transition is non

existent due to two input changes and

is leading to unstable intermediate

states


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Finding Race free State assignments for

Different Input Conditions

Step 2:

Method 1 Replacement of an

unstable non-adjacent state by another

adjacent state in flow table so that inthe next cycle the stable state is

obtained.


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Modification of flow table is permitted

if it does not change the result finallyachieved


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Example 1: Flow Table for Y = X. xq2 +

xq1; xq1 =D and xq2 = xqn+1 = J. xqn +K. xqn

State Next State-Transition Outputs YO

(yq) [xq]X=0 [xq]X=1 X=0 X =1

S0 S0 S3 0 -

S3 S1 S0 1 1S

2S

1S

3 0 1

YO is present output state after the xq outputs

but before transition yq next state


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Map for X = 0 Using a modified flow Table

with new variable assignments .

S (xQ)

S (xQ)

S0

011

S1

101

S0 011 0

-S1 101 0

S3 111 1

S2 110

S3

111

S2

110

0

1 0

Neighbors

101 and

110


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Map for X = 1 Using a modified flow Table

with new variable assignments

S (xQ)

S (xQ)

S0

011

S1

101

S0 011 0

S1 101

S3 111 1

S2 110

S3

111

S2

110

1

1

1

0

0

Neighbors

111 and

011


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Example 1: S1 present state is

unresponsive to any input condition.So S1 can be assigned three state

variables, (xq0, xq1, xq2) = 101, in placeof 01. S3 to S0 transition in flow table,

means 111 to 110, transition, and that

means only one variable change. S0 =011, S3 = 111 are also neighbors.


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Race Free Assignment

Race condition will not now exist and

after intermediate cycles the state willbecome same as when race sets in due

to two or ore variable changes withmismatched delays


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Step 2:

Method 2A systematic method,

called one-hot method of race free

state assignments


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Action 1

Let number of rows in flow table = n. Use nstate variables. Assign each row (xq0, xq1

, , x

qn--1) in sequence as 0001, 00. 10,

, 10 00. In k-th row the k-th statevariable is 1. For example, if there are fourrows in the flow table, S0, S1, S2 and S3,

assign 1000, 0100, 0010 and 0001,respectively to (xq0, xq1

, , xqn--1

) with m= 4.


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Action 2

Fill present state column 1 contains thestates as per new state variable

assignments


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Action 3

Assign the stable state variables to thestate variable as per corresponding

state variables used in present state

column and leave presently unstable

state as such


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Example 2: Flow Table Actions 1 to 3State Transition Outputs (xq0, xq1) Y

(xq0, xq1) X=00 01 10 11 X=00 01 10 11

S0 1000 S1 1000 - S1 0 0 - 1

S1 0100 0100 S1 S0 S2 1 - 1 0

S2 0010 S3 S1 0100 0100 0 - 0 1

S3 0001 0001 S1 0001 S2 1 1 0 1

Row 2 column 5 0100 S1 is final state

after intermediate feedback cycles]


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Action 4

:For unstable state, now write theassignment after ORing with the

assignment for its next cycle transition.


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Ch18L4- "Digital Principles and Design", Raj Kamal, Pearson Education,2006 31

Action 5 Add extra row for those next state,

which have two variables as 1s. Theassignment for the same vertical

column of the state with two variable

1s can now be done so that next state

transition occurs to the same next state

occurs which was ORed with theunstable state assignment before. .


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Example 2: Flow Table Actions 4 and 5State Transition Outputs (xq0, xq1) Y

(xq0, xq1) X=00 01 10 11 X=00 01 10 11

S0 1000 0100 1000 - 0100b 0 0 - 1

S1 0100 0100 0100 1010a 0110c 1 - 1 0

S2 0010 0001 0100 0010 0100b 0 - 0 1

S3 0001 0001 0100 0001 0110 1 1 0 1

S4 1010

a

- - 0010 -S5 0110

c - - - 0100b


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New States

a

New assignment to unstable state afterORing with its next cycle state assignmentin column 1 for present states

b

New assignment to unstable state afterORing with its next cycle state assignmentin column 5 for present states

cNew row addition for an unstable stateassignment in row-2


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Action 6 Change the output column entry for the

state assigned two variables as 1s. Putthe next cycle stable state. This is to

prevent two times changes in an output

when adding another extra row in the

flow table.


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New Outputs

dNew output entry eNow it has next cycle stable state

output entry


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Summary


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Using Karnaugh map concept, we findthe number of races

By reassigning the state variables, weget race free assignments


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End of Lesson 4Race Free Assignments


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THANK YOU

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