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

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written 8.9 years ago by

aksnitald • 20

• modified 8.9 years ago

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:-

enter image description here

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