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Analyze the sequential-state-machine shown in figure. Obtain state diagram for the same.

Mumbai University > Electronics Engineering > Sem 3 > Digital Circuits and Design

Marks: 10M

Year: Dec 2015

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The output Y depends only on present state of memory elements.

Therefore, it is a Moore circuit.

To draw state diagram:-

Consider four states of output:-

00,01,10,11

1.Assume AB=0.0

When X=0,

$J_{A}$=B=0 $J_{B}$=A=0
$K_{A}$=B.X’=0.1=0 $K_{B}$=A+X=0+0=0
.·. $A_{n+1}$­=0 .·. B$_{N+1}$=0

.•. Next State=00

When X=1,

$J_{A}$=B=1 $J_{B}$=A=0
$K_{A}$=B.X’=0.0=0 $K_{B}$=A+X=0+1=1
.·. $A_{n+1}$­=0 .·. B$_{N+1}$=1

.•. Next State=01

.•. Output Y=A’.B’=1.1=1

2.Assume AB=01

When X=0,

$J_{A}$=B=1 $J_{B}$=A=0
$K_{A}$=B.X’=1.1=1 $K_{B}$=A+X=0+1=1
.·. $A_{n+1}$­=0 .·. B$_{N+1}$=1

.•. Next State=01

When X=1,

$J_{A}$=B=1 $J_{B}$=A=0
$K_{A}$=B.X’=1.0=0 $K_{B}$=A+X=0+1=1
.·. $A_{n+1}$­=0 .·. B$_{N+1}$=1

.•. Next State=01

.•. Output Y=A’.B’=1.0=0

3.Assume AB=10

When X=0,

$J_{A}$=B=0 $J_{B}$=A=0
$K_{A}$=B.X’=0.1=0 $K_{B}$=A+X=0+0=0
.·. $A_{n+1}$­=1 .·. B$_{N+1}$=1

.•. Next State=11

When X=1,

$J_{A}$=B=0 $J_{B}$=A=1
$K_{A}$=B.X’=0.0=0 $K_{B}$=A+X=1+1=1
.·. $A_{n+1}$­=1 .·. B$_{N+1}$=1

.•. Next State=11

.•. Output Y=A’.B’=0.1=0

4.Assume AB=11

When X=0,

$J_{A}$=B=1 $J_{B}$=A=1
$K_{A}$=B.X’=1.1=1 $K_{B}$=A+X=1+0=1
.·. $A_{n+1}$­=0 .·. B$_{N+1}$=0

.•. Next State=00

When X=1,

$J_{A}$=B=1 $J_{B}$=A=1
$K_{A}$=B.X’=1.0=0 $K_{B}$=A+X=1+1=1
.·. $A_{n+1}$­=0 .·. B$_{N+1}$=0

.•. Next State=00

.•. Output Y=A’.B’=0.0=0

Now,

State Table:-

Present Next State Output
State X=0 X=1 Y=A'.B'
00 00 01 1
01 11 01 0
10 11 11 0
11 00 00 0

State Diagram:-