written 7.3 years ago by
aksnitald
• 20
|
•
modified 7.3 years ago
|
--> State Diagram will be,
States are as follows,
a=00, b=01, c=10
State table:-
Present |
Next |
State |
Output |
Output |
State |
X=0 |
X=1 |
X=0 |
X=1 |
a |
a |
b |
0 |
0 |
b |
c |
b |
0 |
0 |
c |
a |
b |
0 |
1 |
To design a circuit:-
Input |
Present |
State |
Next |
State |
|
F/F |
I/P |
|
O/P |
X |
B |
A |
$B_{n+1}$ |
$A_{n+1}$ |
$J_{A}$ |
$K_{B}$ |
$J_{A}$ |
$K_{A}$ |
0 |
0 |
0 |
0 |
0 |
0 |
X |
0 |
X |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
X |
X |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
X |
1 |
0 |
X |
0 |
0 |
1 |
1 |
X |
X |
X |
X |
X |
X |
X |
1 |
0 |
0 |
0 |
1 |
0 |
X |
1 |
X |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
X |
X |
X |
0 |
1 |
1 |
0 |
0 |
1 |
X |
1 |
1 |
X |
1 |
1 |
1 |
1 |
X |
X |
X |
X |
X |
X |
X |
Input equation for $J_{A}$ :-
X\BA |
00 |
01 |
11 |
10 |
0 |
$0_{0}$ |
$0_{2}$ |
$1_{6}$ |
$1_{4}$ |
.•. $J_{A}$ = BX |
1 |
$X_{1}$ |
$X_{3}$ |
$X_{7}$ |
$X_{5}$ |
Input equation for $K_{A }$:-
X\BA |
00 |
01 |
11 |
10 |
0 |
$X_{0}$ |
$X_{2}$ |
$X_{6}$ |
$X_{4}$ |
.·. $K_{A}$ =B’X |
1 |
$1_{1}$ |
$X_{3}$ |
$X_{7}$ |
$X_{5}$ |
Input equation for $K_{B}$ :-
X\BA |
00 |
01 |
11 |
10 |
0 |
$X_{0}$ |
$1_{2}$ |
$1_{6}$ |
$1_{4}$ |
.·. $K_{B}$ =AX' |
1 |
$X_{1}$ |
$X_{3}$ |
$X_{7}$ |
$X_{3}$ |
Circuit Diagram :-