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Chapter 5 Sequential Logic Sequential Logic Combinational circuit outputs depend on present inputs. Sequential circuit outputs depend on present inputs and the state of the memory elements. Outputs Inputs Combinational Circuit Memory Sequential Circuit

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slide1

Chapter 5

Sequential Logic

slide2

Sequential Logic

Combinational circuit outputs depend on present inputs.

Sequential circuit outputs depend on present inputs and the

state of the memory elements.

Outputs

Inputs

Combinational

Circuit

Memory

Sequential Circuit

slide3

Flip-Flop - a storage (memory) element. A flip-flop circuit has

two outputs, one normal form and one complemented. It may have

one or two inputs, depending on type. Clock, preset and clear

functions may also be present. One flip-flop stores one binary bit.

R

Q

S R Q Q

1 0 1 0

0 0 1 0

0 1 0 1

0 0 0 1

1 1 0 0

Q

S

NOR

X Y Z

0 0 1

0 1 0

1 0 0

1 1 0

1) Logic 1 on S sets Q to 1.

2) Logic 1 on R resets Q to 0.

3) Logic 1 on both S, R produce an

indeterminate result.

slide4

Clocked RS Flip-Flop

Q S R Q ( T+1)

R

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 Indeterm

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 Indeterm

Q

CP

Q

S

Characteristic Table

S Q

CP

R Q

slide5

S

SR

Q(t + 1) = S + RQ

SR= 0

Q

0 0 d 1

1 0 d 1

Q

SR must equal 0 so that S and R

cannot both be 1 simultaneously

R

Q = present state

Q(t + 1) = next state

Indeterminate states can be considered as don’t cares since

either a 1 or a 0 may result after S and R both equal 1.

slide6

D Flip-Flop

D

Q

CP

Q

S

Q D Q(t+1)

0 0 0

0 1 1

1 0 0

1 1 1

D Q

CP

Q

D

1

Characteristic

Table

Q

1

Q( t + 1) = D

slide7

Q

Q

JK Flip-Flop

R

K

CP

S

J

slide8

Q J K Q ( t+1)

J Q

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

CP

K Q

J

Characteristic Table

0 0 1 1

Q(t+1) = J Q + KQ

Q

1 0 0 1

K

slide9

Q

Q

T Flip-Flop

R

T

CP

S

slide10

Q T Q(t + 1)

0 0 0

0 1 1

1 0 1

1 1 0

T Q

CP

Q

T

1

Q ( t + 1) = T Q + T Q

Q

1

slide11

Triggering of Flip-Flop

* Synchronous: behavior can be defined from the knowledge of

signals at discrete instants of time. A master clock is used, only

change during a clock pulse.

* Asynchronous: behavior depends on the order in which input

signals change. Can be changed at any time.

Level Triggering: Flip-Flop sensitive to pulse duration

Edge Triggering: Flip-Flop sensitive to pulse transition

(solves feedback timing problems)

1

0

Positive

edge

Negative

edge

slide13

Problem: Instability occurs if output of memory elements

(FF’s) are changing while output of combinational circuit

that go to the Flip-Flop inputs are being sampled by the

clock pulse.

Inputs

FF

CP

Output

Comb.

Logic

Propagation delay from Flip-Flop input to output must be

greater then clock pulse duration.

slide14

CP

S

Input

Y

Output

Q

Output state change occurs on the negative clock

transition.

slide15

Direct Inputs

Preset

J Q

PR

CP

K Q

CL

Clear

slide16

State Diagrams

State: The condition (values) stored in the flip-flops of a

sequential circuit.

Present State: The condition of the flip-flops prior to a clock

pulse.

Next State: Condition of the flip-flops after a clock pulse

 state

a

1/0

1/1

 transition between

states which occurs

with a clock pulse.

0/0

b

d

1/0

0/0

0/1

X/Y  x= input which causes

transition.

y = output during present

state.

0/0

c

1/0

slide17

State Table

Contains the same information as a state diagram

Next State

Output

X = 0 X = 1

X = 0 X = 1

Present state

a 0 0 c b 0 0

b 0 1 c b 1 0

c 1 0 d c 0 0

d 1 1 c a 0 1

In general, a state table for m flip-flops will have 2m

rows, one for each state. The next state and output section

will have 2n columns each for n inputs.

External outputs can be taken from flip-flop output

or from logic gates. If there are no logic gates for output, the

output columns can be deleted. The output is then read

directly from the present state of the flip-flops.

slide18

State Reduction

Goal: reduce the number of flip-flops in a sequential circuit.

Next State

Output

X = 0 X = 1

X = 0 X = 1

Present state

a a b 0 0

b c d 0 0

c a d 0 0

d e f 0 1

e a f 0 1

f g f 0 1

g a f 0 1

Equivalent states = same input gives the same output for each

state and the transitions go to the same state.

In example above state g = state e . . . . .

slide19

State Reduction

. . . . . and state f = state d.

Next State

Output

X = 0 X = 1

X = 0 X = 1

Present state

a a b 0 0

b c d 0 0

c a d 0 0

d e f 0 1

e a f 0 1

f g f 0 1

g a f 0 1

e

slide20

State Reduction

. . . . . and state f = state d.

Next State

Output

X = 0 X = 1

X = 0 X = 1

Present state

a a b 0 0

b c d 0 0

c a d 0 0

d e f 0 1

e a f 0 1

f g f 0 1

g a f 0 1

d

d

e

slide21

State Assignment

Some state assignment schemes can reduce the

combinational portion of a sequential circuit. However, there

are no state assignment techniques which guarantee minimization.

Suggestion:

1) For counters use the sequence of binary numbers

for state assignments.

2) Otherwise, make arbitrary assignments.

slide22

Flip-Flop Excitation Tables

Characteristic tables define next state - when the input

and present state are known .

Excitation tables define the input conditions required

to make a desired transition from a known present state to a

a desired next state.

Q S R Q ( T+1)

Q Q(t + 1) S R

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 Indeterm

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 Indeterm

0 0 0 d

0 1 1 0

1 0 0 1

1 1 d 0

SR Excitation Table

SR Characteristic Table

slide23

Q J K Q ( t+1)

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

Q Q(t + 1) J K

0 0 0 d

0 1 1 d

1 0 d 1

1 1 d 0

JK Excitation Table

JK Characteristic Table

Q Q(t + 1) D

Q D Q(t + 1)

0 0 0

0 1 1

1 0 0

1 1 1

0 0 0

0 1 1

1 0 0

1 1 1

D Excitation Table

D Characteristic Table

slide24

Sequential Design Process

Goal: obtain a logic diagram or list of Boolean functions

which describe a sequential circuit meeting the design

specification.

1) Write a word description. Draw a state diagram.

2) Construct a state table.

3) Perform state reduction.

4) Make state assignments.

5) Determine number of ff required and assign a letter to each.

6) Select type of ff to use.

7) Construct circuit excitation and output tables.

8) Reduce Boolean functions.

9) Draw the logic diagram.

slide25

Example

Design a circuit to implement the following state diagram.

0/1

a

1/1

1/0

0/0

1/1

b

d

1/1

0/0

c

0/1

slide26

2) Conduct a state table.

Next State

Output

X = 0 X = 1

X = 0 X = 1

Present state

a a b 1 1

b c b 0 1

c c d 1 1

d d a 0 0

3) Perform state reduction. None possible.

slide27

4) Make state assignments.

a = 0 0

b = 0 1

c = 1 0

d = 1 1

5) Determine number of flip-flops required

4 states required 2 flip-flops (2n = 4)

6) Select flip-flop type. Choose T. (TA, TB)

slide28

7) Construct excitation and output tables.

Inputs of combinational

circuit

Output of combinational

circuit

Present State Input Next State FF inputs Output

A B X A B TA TB Z

0 0 0 0 0 0 0 1

0 0 1 0 1 0 1 1

0 1 0 1 0 1 1 0

0 1 1 0 1 0 0 1

1 0 0 1 0 0 0 1

1 0 1 1 1 0 1 1

1 1 0 1 1 0 0 0

1 1 1 0 0 1 1 0

a

a

b

b

c

c

d

d

slide29

Q Q(t + 1) T

Using:

0 0 0

0 1 1

1 0 1

1 1 0

8) Reduce Boolean Functions.

B

0 0 0 1

TA = A B X + A B X

A

0 0 1 0

X

slide30

B

0 1 0 1

TB = B X + AX + A B X

A

0 1 1 0

X

B

1 1 1 0

Z = B + A X

A

1 1 0 0

X

slide31

9) Draw the logic diagram.

T Q

T Q

A

B

Q

Q

TA A A TB B B

X

Input

Z

Output

Conbinational Circuit

slide32

A

B

X

TA

A

B

X

B

X

TB

A

X

A

X

Z

B

slide33

Example

Design a sequential circuit using the following state table:

Next State

Output

X = 0 X = 1

X = 0 X = 1

Present state

0 0 0

0 0 1 0 0 1 0 1 0 0 0

0 1 0 0 1 1 1 0 0 0 0

0 1 1 0 0 1 1 0 0 0 0

1 0 0 1 0 1 1 0 0 0 1

1 0 1 0 0 1 1 0 0 0 1

1 1 0

1 1 1

3 flip-flop are requires for 5 states. There will be 3 numbered

states (000, 110, 111). Letter the ff A,b, C. Construct an

excitation table for RS ff.

slide34

Present State Input Next State Flip-Flop Inputs Output

A B C X A B C S A R A S B R B S C R C Y

0 0 0 0 d d d d d d d d d d

0 0 0 1 d d d d d d d d d d

0 0 1 0 0 0 1 0 d 0 d d 0 0

0 0 1 1 0 1 0 0 d 1 0 0 1 0

0 1 0 0 0 1 1 0 d d 0 1 0 0

0 1 0 1 1 0 0 1 0 0 1 0 d 0

0 1 1 0 0 0 1 0 d 0 1 d 0 0

0 1 1 1 1 0 0 1 0 0 1 0 1 0

1 0 0 0 1 0 1 d 0 0 d 1 0 0

1 0 0 1 1 0 0 d 0 0 d 0 d 1

1 0 1 0 0 0 1 0 1 0 d d 0 0

1 0 1 1 1 0 0 d 0 0 d 0 1 1

1 1 0 0 d d d d d d d d d d

1 1 0 1 d d d d d d d d d d

1 1 1 0 d d d d d d d d d d

1 1 1 1 d d d d d d d d d d

slide35

Use the SR excitation table to complete the circuit excitation

table:

Q Q(t +1) S R

0 0 0 d

0 1 1 0

1 0 0 1

1 1 d 0

Reduce Boolean function:

C

CX

AB

d d 0 0

0 1 1 0

d d d d

d d d 0

SA = BX

B

A

X

slide36

C

CX

AB

Others

SB = A B X

RB = B C + B X

S C = X

R C = X

Y = A X

d d d d

d 0 0 d

d d d d

0 0 0 1

RA = CX

B

A

X

slide37

Draw the logic diagram:

Y Output

SA

A

S Q

X

Input

A

CP

RA

A

R Q

SB

B

S Q

B

RB

CP

R Q

B

SC

C

S Q

B

CP

RC

R Q

CP

slide38

Don’t Care States

0/0

0 0 1

0/0

??

0/0

0 0 0

1/0

1 0 1

0 1 1

0 1 0

0/0

1/0

1/0

??

0/0

1 1 0

1 0 0

1/1

1 1 1

??

slide39

Don’t Care States

  • What about don’t care states?
  • Self-Correcting: all unused states eventually lead to valid
  • states. (some designs may specify self-correcting in one clock
  • pulse.)
  • Self-starting: initial state on power up is specified. Usually a
  • master reset input is used. It is customary to provide a master-
  • reset input whose purpose is to initialize the states of all
  • flip-flops in the system. Typically, the master reset is a signal
  • applied to all flip-flops asynchronously before master-reset
  • signal, but some may be set to 1.
  • Q: What if, because of a noise signal the circuit finds itself in
  • an invalid state? In that case it is necessary to ensure that the
  • circuit eventually goes into one of the valid states so it can resume
  • normal operation.
slide40

Don’t Care States

  • It was stated previously that unused states in a sequential
  • circuit can be treated as don’t-care conditions.
  • Once the circuit is designed, the m flip-flops in the system can
  • be in any of 2 possible states.
  • Some of these states were taken as don’t cares conditions.
  • The circuit muse be investigated to determine the effect of these
  • unused states. The next state from invalid sates can be determined.
slide41

0/0

0/0

0 0 1

0/0

0/0

0/0

0 0 0

1/0

1 0 1

1/0

0 1 1

0 1 0

0/0

1/0

1/0

0/0

1 1 0

0/0

1 0 0

1/1

1/1

1 1 1

slide42

Design Problem

0/0

0 0 1

0/0

1/1

1/0

1 0 0

0 1 1

0/0

0/0

1/1

0 1 0

1/1

0/0

1/1

0 0 0

slide43

Example Problem: Design with T Flip-Flops

Present State Input Next State Flip-Flop Inputs Output

A B C X A B C TA TB TC

0 0 0 0 0 1 1 0 1 1 0

0 0 0 1 1 0 0 1 0 0 1

0 0 1 0 0 0 1 0 0 0 0

0 0 1 1 1 0 0 1 0 1 1

0 1 0 0 0 1 0 0 0 0 0

0 1 0 1 0 0 0 0 1 0 1

0 1 1 0 0 0 1 0 1 0 0

0 1 1 1 0 1 0 0 0 1 1

1 0 0 0 0 1 0 1 1 0 0

1 0 0 1 0 1 1 1 1 1 0

1 0 1

1 0 1

1 1 0

1 1 0

1 1 1

1 1 1

Need

3 Flip-Flops

to Get 5 States

Unused

States

slide44

Don’t Care States ??

0/0

0 0 1

0/0

1/1

1 1 1

1/0

1 0 0

0 1 1

0/0

0/0

1/1

1 1 0

0 1 0

1/1

0/0

1 0 1

1/1

0 0 0

slide45

C

C

CX

CX

AB

AB

0 1 1 0

0 0 0 0

d d d d

1 1 d d

1 0 0 0

0 1 0 1

d d d d

1 1 d d

B

B

A

A

TA = A + BX

X

X

TB = A + B C X + B C X + B C X

C

CX

C

CX

AB

AB

1 0 1 0

0 0 1 0

d d d d

0 1 d d

0 1 1 0

0 1 1 0

d d d d

0 0 d d

A

A

Z= AX

TC = AX + CX + A B C X

X

X

slide46

To Gates

Prob 6-20 CONTD

A

TA

T

B

X

Q

Z

A

Q

A

X

A

To Gates

B

B

C

X

To Gates

TB

T

B

C

Q

X

Q

B

C

B

X

To Gates

To

Gates

A

A

C

TC

T

Q

X

Q

C

X

*Note Clock Pulse

is assumed

To

Gates

C

slide47

Prob CONTINUED

Now, we must analyze circuit to confirm that unused states are in

fact “Don’t Cares”

Analyzes Procedure is:

1) Start w/Logic Diagram

2) Formulate FF input Equations

3) Derive next state equations using FF characteristics EQ.

4) Formulate transition table (state table w/binary #’s)

5) Formulate state diagram.

1) Logic Diagram on previous Page

2) TA = A + BX QA = A

TB = A + BC X + B C X + B C X QB = B

TC = AX + CX + A B C X QC = C

slide48

3) T F-F Char. EQ: Q(t +1) = TQ + TQ

A = (A + B X) A + (A + BX) A

A B X + [ A (B + X) ] A

A B X + (A B + A X) A

A B X

B = (A + B C X + B C X + B C X) B + (A + B C X + B C X + B C X) B

AB + B C X + [ A (B + C + X) (B + C + X) ( B + C + X) ] B

AB + B C X + [ BA (B + BC + BX) (BC + BX) (BC + BX) ]

AB + B C X + [B A + A B C + A B X) (BC + BX) (BC + BX)

AB + B C X + [A B C + A B X + A B C + A B X + A B C X + A B X) (B C + B X)

AB + B C X + A B C X + A B C X

CX

C

AB

1

B = A B + A C X + A B C X

1

1

B

A

1 1 1 1

X

slide49

1

C = (AX + CX + A B C X) C + ( A X + C X + A B C X) C

A C X + A B C X +[ (A + X) (C + X) (A + B + C + X) ] C

A C X + A B C X + (C A + C X) (C X) (C A + C B + C + C X)

A C X + A B C X + ( C A X + CX) ( CA + CB + C + CX)

A C X + A B C X + C A X B + C A X + C X A + C X B + CX

A C X + A B C X + C A X ( B + 1) + C X A + C X B + C X

A C X + A B C X + A C X + A CX + B C X + C X

CX

C

AB

1

C = CX + A C X + A B X

1

1

B

1

A

1

1

X

slide50

4)

Present State Input Next State Output

A B C X A B C

0 0 0 0 0 1 1 0

0 0 0 1 1 0 0 1

0 0 1 0 0 0 1 0

0 0 1 1 1 0 0 1

0 1 0 0 0 1 0 0

0 1 0 1 0 0 0 1

0 1 1 0 0 0 1 0

0 1 1 1 0 1 0 1

1 0 0 0 0 1 0 0

1 0 0 1 0 1 1 0

1 0 1 0 0 1 1 0

1 0 1 1 0 1 0 0

1 1 0 0 0 0 0 0

1 1 0 1 0 0 1 0

1 1 1 0 0 0 1 0

1 1 1 1 0 0 0 0

slide51

0/0

5)

0 0 1

1/0

0/0

1/1

1/0

1 0 0

0 1 1

0/0

0/0

1/1

0/0

1 1 1

1/1

1/0

0 1 0

1 0 1

0/0

1/1

1/0

0/0

1 1 0

0 0 0

0/0

Unused state are “Don’t cares” since they

all go to valid states. Other 5 states produce

same state diagram as original.

slide52

Alternate way to get next states for “Don’t Cares”

0 1 1 0

1 0 0 0

0 0 0 0

0 1 0 1

d 12 d 13 d 15 d 14

d 12 d 13 d 15 d 14

1 1 d11 d10

1 1 d11 d10

TA

TB

1 0 1 0

0 1 1 0

0 0 1 0

0 1 1 0

d 12 d 13 d 15 d 14

d 12 d 13 d 15 d 14

0 1 d11 d10

0 0 d11 d10

TC

Z

slide53

PS Input NS Output

ABC X A B C Z

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

slide54

Example: Design a circuit to recognize this bit pattern:

… 0 1 1 0 1 … When the position is recognized, return to a

starting point (or points). Circuit should be self- starting and

self - correcting.

Y = 1 when pattern

is recognized

Sequential

circuit

X

Input

pattern

slide55

1/0

1/0

1/0

0/0

g

0/0

0/0

b

a

0/0

h

1/0

a = 1’st digit = 0, correct

b = 1’st digit = 1, incorrect

c = 2’nd digit correct

d = 3’rd digit correct

e = 4’th digit correct

f = all digits correct

pattern recognized

g = unused state.

h = unused state.

1/0

c

1/0

0/1

0/0

d

0/0

1/1

e

1/0

b

slide56

Output

Next State

Present State X = 0 X = 1 X = 1 X = 0

A B C A B C A B C Y Y

a 0 0 0 0 0 0 0 1 0 0 0

b 0 0 0 0 0 0 0 0 1 0 0

c 0 1 0 0 0 0 0 1 1 0 0

d 0 1 1 1 0 0 0 0 1 0 0

e 1 0 0 0 0 0 1 0 0 0 0

f 1 0 1 0 0 0 0 0 1 1 1

g 1 1 0 0 0 0 0 0 1 0 0

k 1 1 1 0 0 0 0 0 1 0 0

Date reduction shows b, g, h as equivalent states; however, 3

flip-flops are still required.

Implement the circuit with JK flip-flops.

slide57

Q Q(t+1) J K

0 0 0 d

0 1 1 d

1 0 d 1

1 1 d 0

slide58

Present State Input Next State Flip-Flop Inputs Output

A B C X A B C J A K A J B K B J C K C Y

0 0 0 0 0 0 0 0 d 0 d 0 d 0

0 0 0 1 0 1 0 0 d 1 d 0 d 0

0 0 1 0 0 0 0 0 d 0 d d 1 0

0 0 1 1 0 0 1 0 d 0 d d 0 0

0 1 0 0 0 0 0 0 d d 1 0 d 0

0 1 0 1 0 1 1 0 d d 0 1 d 0

0 1 1 0 1 0 0 1 d d 1 d 1 0

0 1 1 1 0 0 1 0 d d 1 d 0 0

1 0 0 0 0 0 0 d 1 0 d 0 d 0

1 0 0 1 1 0 1 d 0 0 d 1 d 0

1 0 1 0 0 0 0 d 1 0 d d 1 1

1 0 1 1 0 0 1 d 1 0 d d 0 1

1 1 0 0 0 0 0 d 1 d 1 0 d 0

1 1 0 1 0 0 1 d 1 d 1 1 d 0

1 1 1 0 0 0 0 d 1 d 1 d 1 0

1 1 1 1 0 0 1 d 1 d 1 d 0 0

slide59

C

C

d d d d

1

d d d d

B

B

1 1 1 1

d d d d

A

A

1 1 1

d d d d

X

X

KA= B + C + X

C

C

d d d d

1

1 1 1

B

d d d d

1 1 1 1

B

A

d d d d

d d d d

A

X

JB = A C X

X

KB= B + C + X

slide60

C

C

d d

d d 1

1 d d

d d

1

B

B

1 d d

d d

1

A

A

1 d d

1

d d

JC = BX + AX

KC = X

X

X

C

Y = ABC

B

A

1 1

X

slide61

A

J Q

A

CP

A

K Q

cp

X

B

J Q

A

B

Y

CP

K Q

cp

C

J Q

A

CP

C

K Q

cp

CP

slide62

A Clocked sequential circuit has 2 inputs, X1, and X2, and one

output, Z. The output should equal 1 if and only if the inputs

agree and have agreed for an even non-zero (2,4,6, …) number

of clock pulses since the last pulse when they disagreed.

Convert the two inputs into a single input, D, that equals 1

when X1 & X2 disagree.

X1

d

Z

X2

CLK

slide63

1/0

1/0

A

NS

0/0

A Input A Output FF Inputs

B

0 0 1 0 1 d

0 1 0 0 0 d

1 0 0 1 d 1

1 1 0 0 d 1

0/1

2 states: How many

F-F? (1)

D

1 0

d d

X1

X2

Z

J Q

A

J = D

A

CP

D

K Q

cp

1

D

1 0

d d

1 0

d d

A

A

K = 1

K = AD

slide64

Prepare a state diagram for a sequential circuit whose output,

Z, equals 1 if and only if input sequence has either 1001 or 11

as its most recent subsequence.

0/0

1/1

One’s

1/0

A

B

1/1

0/0

0/0

1/0

0/0

1&0

D

C

1 & 0 & 0

slide65

A single-input, single-output clocked sequential circuit is to

produce an output of 1 coincident with a Ø - input, provided

that the Ø - input is immediately preceded by at least two

consecutive 1-inputs

1/0

0/0

PS Input NS Outputs

A

C

A B X A B JA KA JB KB Z

0/1

0 0 0 0 0 0 d 0 d Ø

0 0 1 0 1 0 d 1 d Ø

0 1 0 0 0 0 d d 1 0

0 1 1 1 0 1 d d 1 0

1 0 0 0 0 d 1 0 d 1

1 0 1 1 0 d 0 0 d 0

1 1 0 d d d d d d d

1 1 1 d d d d d d d

1/0

0/0

B

1/0

A = 00

B = 01

C = 10

slide66

B

B

d d d d

0 0 1 0

KA = X

JA= BX

1 0 d d

A

d d d d

A

B

B

0 1 d d

0 0 0 0

KB = 1

Z = AX

0 0 d d

A

1 0 d d

A

X

X

slide67

J Q

A

J Q

B

K B

1

K B

slide68

Try different state assignment:

PS Input NS Outputs

A B X A B JA KA JB KB Z

1 1 0 1 1 d 0 d 0 0

1 1 1 0 1 d 1 d 0 0

0 1 0 1 1 1 d d 0 0

0 1 1 0 0 0 d d 1 0

0 0 0 1 1 1 d 1 d 1

0 0 1 0 0 0 d 0 d 0

1 0 0 d d d d d d d

1 0 1 d d d d d d d

B

B

0 1 d d

d d 0 1

KA = X

JA= X

d d d d

1 0 d d

A

A

X

X

slide69

B

B

d d d d

0 0 1 0

JB = X

KB =BX

1 0 d d

d d d d

A

A

X

X

Z=AX

X

J Q

A

J Q

B

K B

K B

slide70

State Equations

Sometimes called “Next-State Equations”

- A(t + 1) = (AB + AB + AB)X + Abx

Next State

of a F-F Boolean Functions specifying present state

condition that make the next state of the

Flip-Flop be 1.

(that is, if this function = 1, the next clock

pulse will cause the output, 1, of the flip-flop

to be 1]

- Can be derived from a State Table or Logic Diagram

- State equations plus output function(s) fully specify a sequential

circuit.

slide71

State Equation Derivation

- From a state table:

Next State Output

Present State x = 0 x = 1 x = 0 x = 1

AB AB AB y y

0 0 0 0 0 1 0 0

0 1 1 1 0 1 0 0

1 0 1 0 0 0 0 1

1 1 1 0 1 1 0 0

slide72

Read off next-state columns for A, Alternating from input

Ø to input 1 and put directly into K-Map

0 0 0 1 1 1 1 0

0 0 0 1 1 1 1 0

0

1

0

1

1

1

A

1

1 1 1

A

1

1

a)

a)

A( t + 1) = Bx’ + (B + x’) A

= Bx’ + (Bx’)’ A

B( t + 1) = A’x + (A’ + x) B

= Ax + (Ax’)’ B

Form of Rs flip-flop: q(t + 1) = S + R Q

slide73

State Equation Derivation

- Form a logic Diagram:

CP

x’

A

R Q’

B’

1

B

x

A’

S Q

2

x

B’

R Q’

A’

3

x’

B

A

4

S Q

slide74

- For flip - flop B: R = AX

S = A X

-Substitute into RS F-F characteristic equation:

Q(t + 1) = S + RQ

Gives

B(t + 1) = Ax + (Ax) B

(which is same as got with state table derivation)

slide75

Use of state Equations

- Have to use when doing analyses, such as checking validity of

design that used “Don’t Cares”

- Convenient for design when circuit is alrady specified in this

form or when the equations are easily derived from the state

table.

-- Easy eith D flip-flops

-- Sometimes convenient with JK flip-flops

-- Possible with RS and T flip-flops but requires

considerable algebraic manipulation.

slide76

State Reduction: Another Method

PS NS Output

X=0 X=1

A d c 0

b f h 0

c e d 1

d a e 0

e c a 1

f f b 0

g b h 1

h c g 1

slide77

- Go thru & set up a table

- If intersection of too

states is at an “x”, they

cannot be equal; Put an x

in that block.

- Continue passes thru

table until nothing left to

cross out

- Blocks with no X reflect

equal States.

d- f

c -h

b

c

d

e

f

g

h

a b c d e f g

Example:

a  b iff d  f and c  h

a  c because outputs differ