Logic gateMedium
Question
The correct truth table for the given input data of the following logic gate is :
Options
A.
| Inputs | Output | |||
|---|---|---|---|---|
| A | B | C | D | Y |
| 1 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 |
B.
| Inputs | Output | |||
|---|---|---|---|---|
| A | B | C | D | Y |
| 1 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 |
C.
| Inputs | Output | |||
|---|---|---|---|---|
| A | B | C | D | Y |
| 1 | 1 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
D.
| Inputs | Output | |||
|---|---|---|---|---|
| A | B | C | D | Y |
| 1 | 1 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 |
Solution
$${Y = \overline{(\overline{A \cdot B}) \cdot (C + D)} = \overline{\overline{A \cdot B}} + (\overline{C + D}) }{Y = (A \cdot B) + (\overline{C + D})}$$
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