written 9.5 years ago by
teamques10
★ 70k
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modified 9.5 years ago
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Step 1: List all minterm in the binary form
| Minterm |
Binary Representation |
| $M_1$ |
0 0 0 1 |
| $M_3$ |
0 0 1 1 |
| $M_7$ |
0 1 1 1 |
| $M_9$ |
1 0 0 1 |
| $M_{10}$ |
1 0 1 0 |
| $M_{11}$ |
1 0 1 1 |
| $M_{13}$ |
1 1 0 1 |
| $M_{15}$ |
1 1 1 1 |
| Minterms |
Binary Representation |
| $M_1$ |
$0 0 0 1$ |
| $M_3 \\ M_9 \\ M_{10}$ |
$0 0 1 1 \\ 1 0 0 1 \\ 1 0 1 0$ |
| $M_7 \\ M_{11} \\ M_{13}$ |
$0 1 1 1 \\ 1 0 1 1 \\ 1 1 0 1$ |
| $M_{15}$ |
$1 1 1 1$ |
Step 2 : Arrange Minterms according to categories of 1’s.
$\\ \;$
Step 3: Comparing each binary number with every term in higher category and if they differ by one position put a check mark and copy the term in the next column with ‘_’ in that position.
| Minterm |
Binary Representation |
| 1,3 |
0 0 0_1 √ |
| 1,3 |
0 0 0_1√ |
| 3,3 |
_ 0 0 1 √ |
| 3,7 |
0 _ 1 1 √ |
| 3,11 |
_ 0 1 1 √ |
| 9,11 |
1 0 _ 1 √ |
| 9,13 |
1 _ 0 1 √ |
| 10,11 |
1 0 1 _ |
| 7,15 |
_ 1 1 1 √ |
| 11,15 |
1 _ 1 1 √ |
| 13,15 |
1 1 _ 1 √ |
Step 4 : Report procedure until a single pass through cycle yields no further elimination.
| Minterm |
Binary Representation |
| 1,3,9,11 |
_ 0 _ 1 |
| 3,7,11,15 |
_ _ 1 1 |
| 9,11,13,15 |
1 _ _ 1 |
Step 5 : List Prime Implication.
| Prime Implication |
Binary Representation |
| $A\overline{B}C$ |
1 0 1 _ |
| $\overline{B}D$ |
_ 0 _ 1 |
| C D |
_ _ 1 1 |
| A D |
1 _ _ 1 |
Step 6 : Select the Minimum number of prime Implication which must cover all the minterms.

$Y=A\overline{B}C+\overline{B}D+CD+AD$
Final Expression is Y=(1 0 1 _ )+(_ 0 _ 1)+(_ _ 1 1)+(1 _ _ 1)