Derivation of Flip-Flop Input Equations and State...
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ECE 331 – Digital System Design Derivation of Flip-Flop Input Equations and State Assignment (Lecture #24) The slides included herein were taken from the materials accompanying Fundamentals of Logic Design, 6 th Edition, by Roth and Kinney, and were used with permission from Cengage Learning.
Transcript of Derivation of Flip-Flop Input Equations and State...
ECE 331 – Digital System Design
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.
Spring 2011 ECE 331 - Digital System Design 2
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?
