Post on 28-Mar-2018
ECE 301 – Digital Electronics
Derivation of Flip-Flop Input Equationsand
State Assignment
(Lecture #24)
The slides included herein were taken from the materials accompanying
Fundamentals of Logic Design, 6th Edition, by Roth and Kinney,
and were used with permission from Cengage Learning.
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Sequential Circuit Design1. Understand specifications
2. Draw state graph (to describe state machine behavior)
3. Construct state table (from state graph)
4. Perform state reduction (if necessary)
5. Assign a binary value to each state (state assignment)
6. Create state transition table
7. Select type of Flip-Flop to use
8. Derive Flip-Flop input equations and FSM output equation(s)
9. Draw circuit diagram
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Derivation of Flip-Flop Input Equations
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Derivation of FF Input Equations
Example #2:
Derive the Flip-Flop input equations for the FSM described by the following state table.
Assume that JK Flip-Flops are used in the design.
Excitation Table:
Q Q+ J K
0 0 0 x
0 1 1 x
1 0 x 1
1 1 x 0
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Example #2: FF Input Equations
State Table
Present Next State Output
State X = 0 X = 1 X = 0 X = 1
S0 S1 S2 0 1
S1 S2 S3 0 0
S2 S3 S0 1 0
S3 S0 S1 0 1
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Example #2: FF Input Equations
1. Assign a binary value to each state.2. Construct the state transition table.
A+B+ JAKA JBKB
AB X = 0 X = 1 X = 0 X = 1 X = 0 X = 1
00
01
10
11
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Example #2: FF Input Equations
3. Construct K-maps for Flip-Flop inputs.4. Derive the minimized FF input equation.
JA = KA =
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Example #2: FF Input Equations
3. Construct K-maps for Flip-Flop inputs.4. Derive the minimized FF input equation.
JB = KB =
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Derivation of FF Input Equations
Example #3:
Derive the Flip-Flop input equations for the FSM described by the following state table.
Assume that SR Flip-Flops are used in the design.
Excitation Table:
Q Q+ S R
0 0 0 x
0 1 1 0
1 0 0 1
1 1 x 0
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Example #3: FF Input Equations
State Table
Present Next State Output
State X = 0 X = 1 X = 0 X = 1
S0 S1 S2 0 1
S1 S2 S3 0 0
S2 S3 S0 1 0
S3 S0 S1 0 1
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Example #3: FF Input Equations
1. Assign a binary value to each state.2. Construct the state transition table.
A+B+ SARA SBRB
AB X = 0 X = 1 X = 0 X = 1 X = 0 X = 1
00
01
10
11
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Example #3: FF Input Equations
3. Construct K-maps for Flip-Flop inputs.4. Derive the minimized FF input equation.
SA = RA =
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Example #3: FF Input Equations
3. Construct K-maps for Flip-Flop inputs.4. Derive the minimized FF input equation.
SB = RB =
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State Assignment
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State Assignment
● After the number of states in the state table has been reduced …
● A binary value must be assigned to each of the states.
– State assignment (or state encoding)
– Binary value = state of Flip-Flops
● The cost of the logic required to realize the FSM is dependent on the state assignment.
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State Assignment
● Given: A FSM with three states.
● Requires: Two Flip-Flops (A and B)
– Can implement a maximum of four states.
● There are 4 x 3 x 2 = 24 possible state assignments.
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State Assignment
● For a FSM realized using symmetrical Flip-Flops (i.e. JK and SR)
– 3 unique state assignments for 3-state FSM
– 3 unique state assignments for 4-state FSM
Binary GrayCode
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State Assignment
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Guidelines for State Assignment
The author provides a set of guidelines by which the optimal state assignment can be selected.
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One-Hot State Assignment
● Sometimes, reducing the next-state logic is more important than reducing the number of Flip-Flops.
● One-hot state assignment may result in minimal next-state logic.
– Uses one Flip-Flop per state.
– Exactly one Flip-Flop is set to 1 for each state.
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Example: State Assignments
For a 4-state FSM, three possible state assignments are:
State Binary Gray-code One-hot
S0 00 00 0001
S1 01 01 0010
S2 10 11 0100
S3 11 10 1000
# of FF 2 2 4
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Example: State Assignments
Binary state-assignment:
Present Next State
State X = 0 X = 1
S0 S1 S2
S1 S2 S3
S2 S3 S0
S3 S0 S1
A+B+
AB X = 0 X = 1
00 01 10
01 10 11
10 11 00
11 00 01
State Transition TableState Table
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Example: State Assignments
Gray-code state-assignment:
Present Next State
State X = 0 X = 1
S0 S1 S2
S1 S2 S3
S2 S3 S0
S3 S0 S1
A+B+
AB X = 0 X = 1
00 01 11
01 11 10
11 10 00
10 00 01
State Transition TableState Table
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Example: State Assignments
One-hot state-assignment:
Present Next State
State X = 0 X = 1
S0 S1 S2
S1 S2 S3
S2 S3 S0
S3 S0 S1
A+B+C+D+
ABCD X = 0 X = 1
0001 0010 0100
0010 0100 1000
0100 1000 0001
1000 0001 0010
State Transition TableState Table
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Questions?