ANS :
1) K-MAPS and simplification
For F1 consider it as F1(A,B,C,D)
F1=$\bar{\mathbb{A}}~\overline{\mathbb{B}}~\overline{\mathbb{D}}~+~\bar{\mathbb{B}}~\overline{\mathbb{C}}~\overline{\mathbb{D}}~+~A\overline{\mathbb{B}}~\overline{\mathbb{D}}~+~\bar{\mathbb{A}}~B~\overline{\mathbb{C}}~D$
For F2,
F2=$\bar{\mathbb{A}}~\overline{\mathbb{B}}~D~+~A~B~D~+~A~\overline{\mathbb{B}}~\overline{\mathbb{D}}$
For F3,
2) PLA Programming table:-
PRODUCT TERM NO. |
PRODUCT TERM |
INPUTS |
|
|
|
OUTPUTS |
A |
B |
C |
D |
F1 |
F2 |
F3 |
1 |
$\bar{\mathbb{A}}~\overline{\mathbb{B}}~\overline{\mathbb{D}}$ |
0 |
0 |
- |
0 |
1 |
- |
- |
2 |
$\bar{\mathbb{B}}~\overline{\mathbb{C}}~\overline{\mathbb{D}}$ |
- |
0 |
0 |
0 |
1 |
- |
- |
3 |
$A~\overline{\mathbb{B}}~D$ |
1 |
0 |
- |
0 |
1 |
- |
- |
4 |
$\bar{\mathbb{A}}~B~\overline{\mathbb{C}}~D$ |
0 |
1 |
0 |
1 |
1 |
- |
- |
5 |
$\bar{\mathbb{A}}~\overline{\mathbb{B}}~D$ |
0 |
0 |
- |
1 |
- |
1 |
- |
6 |
$A~B~D$ |
1 |
1 |
- |
1 |
- |
1 |
- |
7 |
$A~\overline{\mathbb{B}}~\overline{\mathbb{D}}$ |
1 |
0 |
- |
0 |
- |
1 |
- |
8 |
$\bar{\mathbb{A}}~\overline{\mathbb{B}}~\overline{\mathbb{C}}$ |
0 |
0 |
0 |
- |
- |
- |
1 |
9 |
$\bar{\mathbb{B}}~\overline{\mathbb{C}}~D$ |
- |
0 |
0 |
1 |
- |
- |
1 |
10 |
$\bar{\mathbb{A}}~B~D$ |
0 |
1 |
- |
1 |
- |
- |
1 |
11 |
$A~B~\overline{\mathbb{D}}$ |
1 |
1 |
- |
0 |
- |
- |
1 |
3) Implementation of logic circuit :