Example of Circuit Synthesis -- Part I
Finite state machine
State table
| A(t) |
B(t) |
x |
A(t+1) |
B(t+1) |
z |
JA |
KA |
JB |
KB |
| 0 |
0 |
0 |
0 |
0 |
1 |
0 |
? |
0 |
? |
| 0 |
0 |
1 |
0 |
1 |
0 |
0 |
? |
1 |
? |
| 0 |
1 |
0 |
1 |
0 |
0 |
1 |
? |
? |
1 |
| 0 |
1 |
1 |
0 |
0 |
1 |
0 |
? |
? |
1 |
| 1 |
0 |
0 |
0 |
1 |
0 |
? |
1 |
1 |
? |
| 1 |
0 |
1 |
1 |
0 |
0 |
? |
0 |
0 |
? |
Boolean functions to implement sequential circuit
| z = Sigma(0, 3) |
z = d(6, 7) |
z = A' B' x' + B x |
| DA = Sigma(2, 5) |
DA = d(6, 7) |
DA = A x + B x' |
| DB = Sigma(1, 4) |
DB = d(6, 7) |
DB = A' B' x + A x |
| JA = Sigma(2) |
JA = d(4, 5, 6, 7) |
JA = A' B |
| KA = Sigma(4) |
KA = d(0, 1, 2, 3, 6, 7) |
KA = x' |
| JB = Sigma(1, 4) |
JB = d(2, 3, 6, 7) |
JA = A' x + A x' |
| KB = Sigma(2, 3) |
KB = d(0, 1, 4, 5, 6, 7) |
KB = 1 |