Synchronous design process

Synchronous Design Process1. Construct a state diagram and/or state/output table

corresponding to the word description or specification

2. Minimize the number of states

3. Choose a set of state variables and assign state variable combinations to the named states

4. Obtain the transition/output table

5. Determine the number of flip-flops and select the type of flip-flop to be used (D is often the default)

6. Construct the excitation table

7. Derive excitation equations

8. Derive output equations

9. Draw the logic diagram that realizes the equations

1

Design a clocked synchronous state machine

which detects a sequence of three or more

consecutive 1’s in a string of bits coming through

an input line.

Let the input be X and output be Z

Assuming Mealy machine design

Mealy machine with D flip flops

Moore machine with D flip flops

Moore machine with JK flip flops

Mealy machine with D flip flops (change of encoding)

Obtaining the state Diagram

Assume initial condition to be Z = 0Let the initial state be represented by state A If X = 0, then output Z = 0, same state A

= 1, then output Z = 0, goes to state B

A B0/01/0

X/Z

01

0

A

0

B

Mealy machine

Moore machine


Assume machine has moved to state B If X = 0, then output Z = 0, goes back to state A

= 1, then output Z = 0, goes to state C

A B0/01/0

X/Z

01

0

A

0

B

C1/0

0/0

1

0

C

0

Mealy machine

Moore machine


Assume machine has moved to state C If X = 0, then output Z = 0, goes back to state A

= 1, then output Z = 1, same state C

A B0/01/0

X/Z

01

0

A

0

B

C1/0

0/0

1

0

0C

0

1/1

0/0

0

1

1

1C1

0

Mealy machine

Moore machine

Obtaining the state/output table

State/output table

Next State S*, Z

State Input X

S 0 1

A A,0 B,0

B A,0 C,0

C A,0 C,1

State/output table

Next State S*

State Input X

S 0 1

A A B

B A C0

C0 A C1

C1 A C1

Output

Z

0

0

0

1

Mealy machine Moore machine

Assigning state variable to obtain transition/output

tableTransition/output table

Encoding A = 00, B = 01 C0 = 10,C1 = 11

Choosing D type flip flop

Next State Q1*Q0*, Z

State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 10,0

10 00,0 10,1

Next State Q1*Q0*

State Input X

Q1Q0 0 1

00 00 01

01 00 10

10 00 11

11 00 11

Output

Z

0

0

0

1

Transition/output tableMealy machine Moore machine

Constructing the excitation table

Excitation/output table

D1D0, Z

State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 10,0

10 00,0 10,1

D1D0

State Input X

Q1Q0 0 1

00 00 01

01 00 10

10 00 11

11 00 11

Output

Z

0

0

0

1

Excitation/output tableMealy machine Moore machine

Transferring onto K-maps to derive excitation

equations (Mealy Machine)State Input X

Q1Q0 0 1

00 0 0

01 0 1

11 X X

10 0 1

D1

State Input X

Q1Q0 0 1

00 0 1

01 0 0

11 X X

10 0 0

D0XQXQD 101


State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 10,0

10 00,0 10,1

D1D0, Z XQQD 010

Transferring onto K-maps to derive output equation

(Mealy Machine)Excitation/output table

State Input X

Q1Q0 0 1

00 0 0

01 0 0

11 X X

10 0 1

ZXQZ 1

State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 10,0

10 00,0 10,1

D1D0, Z

Circuit (logic) diagram

excitation equations

output equation

D Q

Q

Q0

Z

Q0' Q1

Clk

D Q

Q

Q0D0

D1 Q1

X

XQXQD 101XQQD 010

XQZ 1

Q1'

Mealy machine


equation (Moore Machine)State Input X

Q1Q0 0 1

00 0 0

01 0 1

11 0 1

10 0 1

D1

State Input X

Q1Q0 0 1

00 0 1

01 0 0

11 0 1

10 0 1

D0XQXQD 101

Excitationtable

XQXQD 010

D1D0

State Input X

Q1Q0 0 1

00 00 01

01 00 10

10 00 11

11 00 11


(Moore Machine)Excitation/output table

01 QQZ

D1D0

State Input X

Q1Q0 0 1

00 00 01

01 00 10

10 00 11

11 00 11

Output

Z

0

0

0

1



output equation

D Q

Q

Q0Z

Q0' Q1

Clk

D Q

Q

Q0D0

D1 Q1

X

XQXQD 010 XQXQD 101

01 QQZ

Moore machine


tableTransition/output table

Next State Q1*Q0*

State Input X

Q1Q0 0 1

00 00 01

01 00 10

10 00 11

11 00 11

Output

Z

0

0

0

1

Q Q* J

0 0 0

0 1 1

1 0 X

1 1 X

K

X

X

1

0

Excitation table for JK flip flop

Moore machine with JK flip flops


J1K1,J0K0

State Input X

Q1Q0 0 1

00 0X, 0X 0X, 1X

01 0X, X1 1X, X1

10 X1, 0X X0, 1X

11 X1, X1 X0, X0

Output

Z

0

0

0

1

Excitation/output tableTransition/output table

Next State Q1*Q0*

State Input X

Q1Q0 0 1

00 00 01

01 00 10

10 00 11

11 00 11

Output

Z

0

0

0

1


equationsState Input X

Q1Q0 0 1

00 0 0

01 0 1

11 X X

10 X X

J1

State Input X

Q1Q0 0 1

00 X X

01 X X

11 1 0

10 1 0

K1XQJ 01

Excitationtable

XK 1

J1K1,J0K0

State Input X

Q1Q0 0 1

00 0X, 0X 0X, 1X

01 0X, X1 1X, X1

10 X1, 0X X0, 1X

11 X1, X1 X0, X0


equationsState Input X

Q1Q0 0 1

00 0 1

01 X X

11 X X

10 0 1

J0

State Input X

Q1Q0 0 1

00 X X

01 1 1

11 1 0

10 X X

K0XJ 0

Excitationtable

XQK 10

J1K1,J0K0

State Input X

Q1Q0 0 1

00 0X, 0X 0X, 1X

01 0X, X1 1X, X1

10 X1, 0X X0, 1X

11 X1, X1 X0, X0


01 QQZ

J1K1,J0K0

State Input X

Q1Q0 0 1

00 0X, 0X 0X, 1X

01 0X, X1 1X, X1

10 X1, 0X X0, 1X

11 X1, X1 X0, X0

Output

Z

0

0

0

1



excitation equations output equation

J Q

Q

Q0Z

Q1'

Clk

J Q

Q

Q0J0

J1 Q1

X

01 QQZ XQJ 01 XK 1XJ 0 XQK 10

K0

K1

X'

K

K


table

Change of Encoding A = 00 B = 01 C = 11

Choosing D type flip flop

Next State Q1*Q0*, Z

State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 11,0

11 00,0 11,1

Transition/output tableMealy machine



D1D0, Z

State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 11,0

11 00,0 11,1

Mealy machine


equations (Mealy Machine)State Input X

Q1Q0 0 1

00 0 0

01 0 1

11 0 1

10 X X

D1

State Input X

Q1Q0 0 1

00 0 1

01 0 1

11 0 1

10 X X

D0XQD 01


State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 11,0

11 00,0 11,1

D1D0, Z XD 0


(Mealy Machine)Excitation/output table

State Input X

Q1Q0 0 1

00 0 0

01 0 0

11 0 1

10 X X

ZXQZ 1

State Input X

Q1Q0 0 1

00 00,0 01,0

01 00,0 11,0

11 00,0 11,1

D1D0, Z



output equation

D Q

Q

Z

Clk

D Q

Q

Q0D0

D1 Q1

X

XQD 01XD 0

XQZ 1

Mealy machine

26

Example 1: State Diagram

• Design the FSM for the given state diagram– Graphical version of states, inputs, transitions and

outputs

Assigned state diagramAssigned state diagram

W

X Y

Z

State State assignmentassignment

Given state diagramGiven state diagram

27

Example 1: State Table

• For each current-state, specify next-state(s) as a function of the present inputs

• For each current-state, specify the output(s) as a function of the present inputs

We often use Q0, Q1, We often use Q0, Q1, Q2, etc. to acknowledge Q2, etc. to acknowledge that state variables are that state variables are F-F outputsF-F outputs

State VariablesState Variables

28

Example 1: Alternate State Table

• State table, alternate format emphasizing state transitions

• Notation: A → Q(t) and A+ → Q(t+1), used to indicate the change in state variable needed for the desired transition

Q1(t) Q0(t) Q1(t) Q0(t) →→ Q1(t+1) Q0(t+1)Q1(t+1) Q0(t+1)

a more a more descriptive descriptive

notationnotation

( )

Example 1: Flip-Flop Excitation Tables

• Use these tables to “move” Q to next state• Used to design the memory control CL circuit

29We will use the D F-FWe will use the D F-F

Example 1: Flip-Flop Excitation Tables

• Why are we are interested in the D Flip-Flop for this design?

30

Q(t) Q(t+1) D Operation 0 0 0 Reset 0 1 1 Set 1 0 0 Reset 1 1 1 Set

We select D Flip-Flops because the D input is We select D Flip-Flops because the D input is simply the value of Q(t+1) that we desiresimply the value of Q(t+1) that we desire

Example 1: Block Diagram

• We need to use two D flip-flops

• The sequential circuit would look as shown

• We will design the CL for a 3-input, 3-output circuit– CL for Output (y)– CL for Next State

controls (D1 and D2)

31

x y

inputs = x, Q1, Q2inputs = x, Q1, Q2outputs = y, D1, D2outputs = y, D1, D2

Q1

Q2

D1

D2

(output)

(next state)

Example 1: Truth Table

• The truth table for this 3-input, 3-output circuit can be generated from the original state table

• We will need to minimize each output function using Karnaugh maps, mindful of overlap

32

01234567

mintermsminterms

y

B+A+

Example 1: Karnaugh Maps• Minimizing each of the three outputs:

33

1 11 1 1

1 1 1

A+ = x ∙ Q1 + x ∙ Q2 B+ = x ∙ Q1

y = x ∙ Q1 + x ∙ Q2

A = Q1 and B = Q2

Example 1: Circuit Diagram

• The circuit that results from these equations is shown below:

34

A+ = x ∙ Q1 + x ∙ Q2

B+ = x ∙ Q1

y = x ∙ Q1 + x ∙ Q2

y = x ∙ (Q1 + Q2)

A+ = x ∙ (Q1 + Q2)

35

Design: Example 2• Given the state diagram as follows, design the

sequential circuit using JK Flip-Flops

A

B

C

D

A = 01B = 10C = 11D = 00

state assignment

36

Design: Example 2• Note that this is a state diagram for a Mealy machine

inputsinputs outputsoutputs

37

Design: Example 2• Block diagram, signal identification

We need to determine what

goes in here?

Input, x

1-bit

SequentialLogicCircuit

CP

KB = Ax + A’x’

38

Design: Example 2• From state table, get input flip-flop function

KB = Ax + A’x’

A xBxx B

x

39

Design: Example 2• Input flip-flop function

• Logic Diagram

Input, x

1-bit

CP

x

AB

JAKA

JBKB

01

0

A

0

B 0

0

0C

1

0

1

1

1C

1

0

101 overlap Moore machine

CLK

QA

QB

QC

QD

0 1 2 3 4 8 9 10 11 12 0

CLK

QA

QB

QC

QD

0 1 2 3 4 8 9 10 11 12 0

CLKCLK

QD

QA

QB

QC

5 6 7 8 9 10 11 12 13 14 5

LD

Initial Loading

Counting starts

Reloading

1110 Moore machine

0

10

A

0

B 10

C

0

1

0

D1

01

E0

01

0

44

Next State S*

State Input X

S 0 1

A A B

B A C0

C A C1

D A C1

Output

Z

0

0

0

1

D Q

Q

Q0

Z

Q0'

Q1

Clk

D Q

Q

Q0D0

D1 Q1

D Q

Q

D2 Q2

X

X'

46

Next State S*

State Input X

S 0 1

A A B

B A C0

C0 A C1

C1 A C1

Output

Z

0

0

0

1

Synchronous design process

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