written 7.2 years ago by
aksnitald
• 20
|
•
modified 7.2 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:-