Digital Design: FUNDAMENTAL MODE SEQUENTIAL CIRCUITS Part - III

Chapter 18

FUNDAMENTAL MODE SEQUENTIAL

CIRCUITS

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

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Lesson 4

Race Free Assignments


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Outline

• Race Free Assignment• Circuit Implementation


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x’Q and x’Q differing in more than one latch during a transition

• Memory (or delay) section next-state variables when have the bits differing by more than one during a transition, the critical (indeterminate time for stable state) and non-critical (deterministic) races occur.

• Critical type of race must not occur in asynchronous sequential circuit.


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Race free assignment • Race free assignment is an assignment

of state variables such that next-state variables 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 two adjacent 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 state variables for X = 0 and X =1, respectively. It shows in a cell the present state by 0 (no change) and next state (change) by 1


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Map for X = 0 Using a flow Table

S (x’Q)S (xQ)

S000

S101

S0 00 0

-S1 01 0

S3 11 1S2 10

S311

S210

01 0


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Map for X = 1 using the flow table

S (x’Q)S (xQ)

S000

S101

S0 00 0

S1 01

S3 11 1

S2 10

S311

S210

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 (x’Q)S (xQ)

S000

S101

S0 00 0

01S1 01

S3 11 S2 10

1

S311

S210

011 0


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Map for X = 01 using another circuit flow table

S (x’Q)S (xQ)

S000

S101

S0 00 01

01S1 01

S3 11 1 S2 10

S311

S210

0 1 0


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Map for X = 11 Using for another circuit flow table

S (x’Q)S (xQ)

S000

S101

S0 00 0

0S1 01

S3 11 S2 10

S311

S210

1

1

0 1

1 0


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Map for X = 10 using another circuit flow table

S (x’Q)S (xQ)

S000

S101

S0 00 0

0S1 01 1

S3 11 S2 10

S311

S210

-

01 01


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Observation

• It is observed for X =01that for a present state 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 a present 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 a present 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 in the next cycle the stable state is obtained.


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• Modification of flow table is permitted if it does not change the result finally achieved

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Example 1: Flow Table for Y = X. xq2 + xq1; xq1’ =D and xq2’ = xqn+1 = J. xq n +

K. xq n

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 1S2 S1 S3 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 (x’Q)S (xQ)

S0011

S1101

S0 011 0

-S1 101 0

S3 111 1S2 110

S3111

S2110

01 0

Neighbors101 and 110


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Map for X = 1 Using a modified flow Table with new variable assignments

S (x’Q)S (xQ)

S0011

S1101

S0 011 0

S1 101

S3 111 1

S2 110

S3111

S2110

1

1

1

0 0

Neighbors111 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 place of 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 will become same as when race sets in due to two or ore variable changes with mismatched delays


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• Step 2:Method 2—A 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 n state variables. Assign each row (xq0, xq1, … , xqn--1) in sequence as 00…01, 00. 10, …, 10… 00. In k-th row the k-th state variable is 1. For example, if there are four rows 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 the states as per new state variable assignments


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Action 3

• Assign the stable state variables to the state 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 (x’q0, x’q1) 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 the assignment after ORing with the assignment for its next cycle transition.


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Action 5• Add extra row for those next state,

which have two variables as 1s. The assignment 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 the unstable state assignment before. .


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Example 2: Flow Table Actions 4 and 5State Transition Outputs (x’q0, x’q1) 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 1010a - - 0010 -

S5 0110c - - - 0100b


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New States

• a New assignment to unstable state after ORing with its next cycle state assignment in column 1 for present states

• b New assignment to unstable state after ORing with its next cycle state assignment in column 5 for present states

• c New row addition for an unstable state assignment in row-2


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Action 6• Change the output column entry for the

state assigned two variables as 1s. Put the 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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Example 2: Flow Table Action 6State Transition Outputs (x’q0, x’q1) Y

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

S0 1000 0100 1000 - 0100 0 0 - 1

S1 0100 0100 0100 1010e 0110 1 - 0d 1d

S2 0010 0001 0100 0010 0100b 0 - 0 1 S3 0001 0001 0100 0001 0110 1 1 0 1

S4 1010a - - 0010 -

S5 0110c - - - 0100b


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New Outputs

• d New output entry• e Now it has next cycle stable state

output entry

Summary

• Using Karnaugh map concept, we find the number of races

• By reassigning the state variables, we get race free assignments

End of Lesson 4

Race Free Assignments


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THANK YOU